Neutralising sulfuric acid but not boric acid I have another question here which refers to me asking how to separate boric acid from other compounds, and this is partially a follow-up.

Suppose I was to go down the route of making boric acid by the reaction of:
$$\ce{Na2B4O7·10H2O + H2SO4 → Na2SO4 + 4 H3BO3 + 5 H2O}$$

Which in actual fact, I just did 10 minutes ago and failed miserably.
The issue I'm having is that no matter what happens in a home-experiment, you're going to have contaminants, and in this case, they are the sodium sulfate and left-over sulfuric acid.

I added sodium bicarbonate in to remove the sulfuric acid which was stupid because I required so much but then realised that it will also neutralise the boric acid. So next I am going to attempt to neutralise the acid with sodium hydroxide.

But sodium hydroxide will also target the boric acid and just change it back into borax. Is there a way I can neutralise the sulfuric acid without touching the boric acid?

You've got to know that sulfuric acid is a strong acid and boric acid is a very weak acid. So, sulfuric acid can neutralize most of the bases where boric acid cannot. So, you have to take a base that can be neutralized by only sulfuric acid and not boric acid. So, you should probably take group 2 hydroxides and try to react with the unreacted sulfuric acid because sulfuric acid reacts with group 2 hydroxides whereas boric acid can't.

$$\ce{M(OH)2 +  H2SO4 -> M2SO4 + 2H2O}$$

$$\ce{M(OH)2 +  H3BO3 ->nothing}$$

where $\ce{M=Be,Mg,Ca,Sr,Ba}$.

So, which group 2 hydroxide to use? Group 2 hydroxide react with the sulfuric acid to form group 2 sulfates. Now, solubility of group 2 sulfates decreases down the group. So, you should use calcium hydroxide so the calcium sulfate that forms will not dissolve in the solution due to its low solubility and form a precipitate. You can separate that precipitate by filtration.

Formation of calcium borate is highly unlikely and so only calcium sulfate will only form.

Also, barium hydroxide is the best chemical for this process because barium sulfate is very insoluble. But barium hydroxide is costly if you buy it or making it is difficult due to the cost of other reagents. Instead, calcium hydroxide is your best bet.


Addition of acid to 3-methyl-2,3-dihydropyran When a general acid, H-X reacts with the alkene in an addition reaction, which regiochemistry is expected?

In addition reactions X attaches to the most substituted carbon whilst H bonds to the carbon with fewest alkyl groups. In this case I would believe the third carbon to be the most substituted carbon due to it being bonded to three other carbons. And the second carbon to be where the hydrogen will bond. However, apparently it should be the other way around. And this would be the compound formed:

Why is the second carbon the most substituted in this case? Does it have to do with oxygen being more electronegative and thus it is seen as a higher prioritized substituent?

Apparently, you're using the Markovnikoff's rule here. Recall that it is only a basic rule intended for simple unsaturated hydrocarbons, and fails at more advanced examples involving heteroatoms or rings.

In this case, the first step is the addition of $\ce{H+}$ ion from the acid to the pi bond, which leads to the formation of a carbocation. The carbocation's positive charge is most stable next to the oxygen atom. Can you tell why?

In the second step, the $\ce{X-}$ ion will simply attach at the position where the positive charge was the most stable, leading to a product as you've shown in the second image.


What is the relation between partial pressure and concentration? So, one of my books said that partial pressure plays more significance in determining the gradient than concentration. So, it said it was possible for a gas to go from high partial pressure at point A to low partial pressure at point B, even though there is low concentration at point A and high concentration at point B.

How is this even possibly? I thought partial pressure and concentration were always related.

MATHEMATICALLY

At point A

The pressure according to ideal gas equation is $P_A = C_ART_A$. Here $C_A (= n/V)$ is concentration at A.

At point B

Similarly, the pressure according to ideal gas equation is $P_B = C_BRT_B$. Here again $C_B$ is concentration at B.

Now, $P_A$ > $P_B$

So, $C_ART_A$ > $C_BRT_B$

i.e. $C_A/C_B$  > $T_B/T_A$  .....equation 1

Now as the condition is that concentration at A is lower than the the concentration at B. Therefore we can say that

$C_A/C_B$ < $1$ .....equation 2

Now from equation 1 and 2 we have

$T_B/T_A$ < $1$

which gives us $T_A$ > $T_B$

So in the end, yes, your book was right. It is possible if the temperature at A is more than that at B.

"HOW TO FIND IT IN SECONDS" or "I HATE TO DERIVE BY USING EQUATIONS"

Think of two containers A and B which are connected by a pipe. Both are connected with the help of a pipe which is closed (i.e. it doesn't let the gases exchange). Now we know that concentration of gas in A is lower than the the concentration of gas in B. If we want more pressure of gas in A then the velocity of its particles should be more. Now it is possible if it has more thermal energy. So the temperature should be more in A.


How can AuCl3 exist? The reduction potential for $\ce{Au^3+ + 3e- -> Au}$ is $1.503 \text{ V}$. The reduction potential of $\ce{Cl2 + 2e- -> 2Cl-}$ is $1.36 \text{ V}$. So, the $E^o$ for $\ce{2Au^3+ + 6Cl- -> 2Au + 3Cl2}$ is $0.143 \text{ V}$.

If the reaction $\ce{2Au^3+ + 6Cl- -> 2Au + 3Cl2}$ has a positive E standard ($E^o$ = $0.143 \text{ V}$), then how is it possible for $\ce{AuCl3}$ to exist in aqueous solution without $\ce{Au^3+}$ oxidizing $\ce{Cl-}$ to $\ce{Cl2}$?

Gold(III) chloride does not exist as $\ce{AuCl3}$. According to Wikipedia, the name gold trichloride is a simplification of the name, which is referring to the empirical formula, $\ce{AuCl3}$. The X-ray defraction studies revealed that the chemical exists as a chloride-bridged dimer, $\ce{Au2Cl6}$ (Ref.1):

In $\ce{Au2Cl6}$, each gold center is square planar, and the bonding in $\ce{Au2Cl6}$ is considered somewhat covalent (Wikipedia). Yet, your suggestion is somewhat true, since a recent research has shown that $\ce{Au2Cl6}$ is susceptible to decompose to metalic $\ce{Au}$ and $\ce{Cl2}$ gas at high temperatures or expose to $\ce{e-}$-beam irradiation (Ref.2).

Also, another research group (Ref.3) has shown that:

  Gold chloride, supplied as $\ce{HAuCl4}$ solution, is rapidly adsorbed on the silicate surfaces, the Au(III) is reduced to metallic gold, and gold particles grow on the surface. SEM images show agglomerates of gold unevenly distributed on the surface of the silicates, including in some areas forming agglomerates, especially on quartz and feldspar.

References:

E. S. Clark, D. H. Templeton, C. H. MacGillavry, "The crystal structure of gold(III) chloride," Acta Cryst. 1958, 11, 284-288 (https://doi.org/10.1107/S0365110X58000694).
Keita Kobayashi, Ryo Kitaura, Katsuhiro Sasaki, Kotaro Kuroda, Takeshi Saito, Hisanori Shinohara, "In Situ Observation of Gold Chloride Decomposition in a Confined Nanospace by Transmission Electron Microscopy," Materials Transactions 2014, 55(3), 461-465 (https://doi.org/10.2320/matertrans.MD201308).
Sima Mohammadnejad, John L. Provis, Jannie S. J. van Deventer, "Reduction of gold(III) chloride to gold(0) on silicate surfaces," Journal of Colloid and Interface Science 2012, 389(1), 252-259 (DOI: 10.1016/j.jcis.2012.08.053).


Why does Fluorine have a high affinity for Calcium? I keep hearing from several websites and videos that Fluorine seems to like bonding to Calcium specifically; can someone please tell me why? I can't seem to find a clear response on the internet. Fluorine has a high affinity to Calcium, more than other elements. Why is that?

Actually it's not the question of affinity but insolubility of CaF2. As this salt has solubility product of ~4×10−11 which is mostly caused by nice packing and strong ionic interaction (see enter link description here)


How to figure out the most stable conformation of the diastereomer? Question 1: Which of the following Newman projections represent the most stable confirmation of the diastereomer of the molecule given below?

I guess my main question is looking only at the Newman projection, how does one determine the stereochemistry?
In the molecule to the right, I get for the first carbon, ($\ce{Et, H, Me}$) R (I flipped it after moving the $\ce{H}$ back) and for $\ce{C}$2 ($\ce{Et, Br, Me}$) I get S.

Question 2: Which of the following Newman projections represents a confirmation of the enantiomer of the molecule given below?

In the same vein, for the second question, I thought I should flip all the stereocenters to get the enantiomer since it is a mirror image. But I am not getting the answers the professor provided to us.

To answer your professor's questions, there is no need to determine the R/S configuration of the two carbon atoms. We have to check the clockwise/anticlockwise arrangement of the substituents around the carbon atom, but checking specifically whether it is R or S is unnecessary.
Before moving further, let's have a look at the following image so that we are clear on how to visualise Newman Projections:

Since in the options provided in the question the carbon atom having bromine is at the back, we'll have to look at the molecule in the following manner:

Hence the Newman Projection obtained would be this:

Now, in question 1, option A can be rejected as it's an entirely different molecule. Option 4 has the atoms in the same order as above, so it is the exactly same molecule. In option 2 and 3, the arrangement of atoms is same for the back carbon but different for the front carbon. Therefore, without finding R or S configuration, we can say that options 2 and 3 are the diastereomers!
Out of these two structures, option 3 is more stable as the bulky ethyl groups are at the farthest position from each other.
Moving on to question 2:
You are right in flipping the arrangement about both the carbon atoms and you should get the answer that way. Maybe you are doing some mistake in getting the right Newman projection. Look again at the Newman projection I've provided above and then compare the options with it to see which option has different arrangement of substituents on both the carbons.


What emits the least radiation when hot? What element with a high vapor point the emits to least light or radiation when heated to its vapor pressure point in a vacuum?

Ideally I would like to know in 0 gravity and 0 pressure which elements can be get the hottest around the sun before the vapor pressure point is reached and/or emit the least black body radiation when hot.

https://space.stackexchange.com/questions/35486/shielding-capability-of-liquid-metals

Technically, a smooth or molten, unoxidized metal surface (regardless of chemistry) makes a good candidate.  This thesis from Auburn University reports the emissivities of both oxidized and unoxidized metals at temperatures typically encountered in industrial processes; note the low values for all the unoxidized metals studied in the IR range where most radiation is given off.


Dehydrobromination of 3-bromo-3-cyclopentylhexane and number of possible alkene products I faced a question like this:

  The total number of alkene products possible from the dehydrobromination of 3-bromo-3-cyclopentylhexane is:

The answer is given 5. But I only found 3.

3-cyclopentylhex-2-ene
3-cyclopentylhex-3-ene
(hexan-3-ylidene)cyclopentane

I can't find the other 2. Any hint?

The structures (i) and (ii) have two isomers each,namely E and Z isomers.

For the structure (iii) there is a plane of symmetry (marked in the diagram) so it is not a geometrical isomer.

Thus Number of isomers=2 (of structure i) +2 (of structure        ii) + 1(of structure iii)= 5 isomers


how do electromeric / inductive effect theories compare with resonance Back in high school, long time and many seas away, I learned about electrons moving within organic molecules based on electromeric and inductive effect. Googling them today, I found that they are deemed obsolete by IUPAC, but still used, mainly in Russia and India.

True, today, for my limited chemistry needs, I can make do with resonance. But how did +E, -E +I and -I become obsolete? Is a mere statement of the various resonance structures enough to fully convey the information stated by describing those effects, or were they rejected for some inaccuracy?

Also, were these effects part of Chelintsev's "Marxist chemistry"? That would explain their historical and geographical extent.

The concepts of electromeric and inductive effects are not obsolete.

Only the term "electromeric" is obsolete and has been subsumed in the terms "mesomeric effect" (+M and -M) and "resonance effect" (see iupac source). The term "isomeric effect" is still in use and I learned it not a decade ago along with "mesomeric" and "resonance".

Amusingly, your SE question is my first result in google for "Marxist chemistry". From the little I have just read in this book, Marxist chemistry had apparently more to do with refusal of the concept of resonance than anything else. So... maybe? Maybe not.


Differences between alkyl halides and alkyl tosylates What are the functional differences between alkyl halides alkyl tosylates? Both halides and tosylates are good leaving groups, yes. But is there chemistry that one will do and the other won't? Etc.? I'm not worried about their relative leaving group abilities. My prof treats the two as equivalent but I can't help but wonder if this is an oversimplification. Wikipedia doesn't point out any differences. The only difference I can think of is that an alkyl tosylate is an ester ... and this opens up the world of ester chemistry - i.e. hydrolysis with hot concentrated acid. Where as I can't see alkyl halides reacting in hot concentrated acid.

Both can be leaving groups. In general, TsO- are better leaving groups, thats why they are often specifically used as so. Halides can be used in many more reactions, like preparation of organometallic compunds.
TsO- more common function is to transform an alcohol group in a good leaving group. Formation of an halide, usually is not the way to go in this case.

But their basic reactivity is very similar.
If you dive into more complex reactivity, you can complicate things to a very very large extent...


Technique to measure the volume of a crystal I am doing a project on crystal growth. I am making potash alum in two different conditions, one in the fridge and the other one in room temperature. I wanted to how can I measure their volume? Techniques?

Potash alum is soluble in water (~14g/100mL at 20C) but not in acetone. I would submerge the sample in a known volume of acetone and measure displacement. If you want to be super thorough, cool the acetone first.

I wouldn't use reported density because you don't necessarily know if you made a pure product etc.


IChO problem involving the condensation point of NF3, NHF2, NH2F This is a problem from the 2016 International Chemistry Olympiad (IChO):

  Which of $\ce{NF3}$, $\ce{NHF2}$, and $\ce{NH2F}$ condenses at the lowest temperature?

The answer book says that it is $\ce{NF3}$. But I don't understand why. In $\ce{NF3}$, $\ce{N}$ should have the greatest positive charge of the three molecules; therefore, it should have the highest b.p., no?

Maybe the answer lies in somewhere in the geometry of ions. Can't figure out what. Any hint?

The interactions do not limit to only dipole-dipole, which means positive charge is not everything.

What's special about $\ce{NF3}$? It has no hydrogens. As a result, it cannot form hydrogen bond, therefore its boiling point (or condense point) is far lower than the others'.


Structure of compound The synthetic scheme below makes reference to a reagent $\ce{(CH_3)_3COCl}$  that is added at some points in the synthesis.

What kind of structure does $\ce{(CH_3)_3COCl}$  have? The closest structure I can imagine is that of $\ce{(CH_3)_3CCOCl}$ with an extra carbon for the 3 methyl groups to be attached to. $\ce{(CH_3)_3COCl}$ seems to imply that the 3 methyl groups are attached to the carbonyl carbon (which is impossible).

Image source: IChO 2015, Problem 6

Tert-butyl hypochlorite, cas no 507-40-4

Reagent for N-chlorination of amines and amides


Does dissociation of molecules into atoms change the volume of mixture? For example, if $\ce{H2}$ is converted to two $\ce{H},$ does the volume of the mixture change?

If it doesn't, does it imply that the work done is $0$ because work is $P\,\mathrm dV?$

Short answer is "not necessarily". If we isolate V in the ideal gas law, we have $$V=\frac{nRT}{P}.$$

If $n$ doubles and nothing else changes, then $V$ does also double. But the doubling of $n$ could also be offset by a doubling of $P$ or a halving of $T$, either of which would result in no change in the volume.

And yes, if the volume increases, pressure-volume work has been done, and if the volume remains constant, no pressure-volume work has been done.  That isn't the same as "no work has been done", as there are other types of work besides pressure-volume work.


Will a crystal of iodine lower the yield of an Grignard Reaction I discussed the following today:

If adding an iodine crystal to a Grignard to make the reaction go. will this tiny crystal force some of the metal or reagent to be used
up. ergo, therefore, there is a lower yield (I know it is microscopic
change, this is not the question)

It is more about if it takes some of the metal and uses it up, maybe forming Mg-Iodide, so it cannot be used in the Grignard reaction, and I am not sure if Mg-Iodide cannot be used or if it actually catalyzes the reaction.
It is in the first step of forming phenyl-Magnesium bromide (or other reagent of the same type) reagent for use in step 2 of Grignard. I write bromide on purpose, so you can see that it is another halide than iodide.
Could it be the iodine catalyzes the reaction, or actually, some of the phenylmagnesium bromide is turned to phenyl magnesium iodide, which can also be used in the reaction? Ergo there is no loss in yield at all?
Imagine for step 1, forming only the Grignard reagent, I used iodine equivalents to 10% of, for example, of bromobenzene, and used Mg (not in excess), but the exact amount of Mg equal to bromobenzene, as 1:1 ratio, and the reaction normally went to 100% yield (it does not) but imagine it. Would I lose 10% from the iodine crystals?

can you tell me what the iodine crystals forms in the reaction and will it able to be used in the reaction or not?
will it catalyze the reaction instead?

Will you show me, with drawings, the intermediates or products of the iodine crystal?
Will it interfere with the yield, as it takes some of the magnesium and forms most likely $\ce{Mg-I}$ that is not used in the reaction, or is it I am asking you? It doesn't give any hints regarding lowering of yield. Will it help catalyze or form Grignard reagent with iodide? Why?
A pointer to what I am actually asking is. what happens to the iodine that is not used to remove hydroxide or etch the surface then what comes out of that will it continually as catalytic etch give $\ce{Mg-I}$ then Grignard reagent with bromide and back or what happens with it? will it use op the Mg turnings lowering yield.

No. Iodine is used to help magnesium react with the bromoderivate. If pure magnesium is added to a solution containing a bromoderivate, it should react quickly. But sometimes it does not react at all, because magnesium is covered by a thin and protective layer of magnesium oxide. Added iodine will etch this layer and the underlying metal. This reaction breaks the oxide layer, so that the metal can get in contact with the bromoderivate, and quickly react.


Number of octahedral and tetrahedral voids present in a HCP structure The given image depicts the tetrahedral and octahedral voids in HCP structure.

For the tetrahedral part

I figured out that there would be 6 tetrahedral voids in total.

  My visualisation: In the image, 3 inverted and 3 upright tetrahedron can be formed. Is this correct?

For the octahedral part

It is hard to visualise the octahedrons for an atom. And so, I can't figure out the number of octahedral voids in HCP.

  Also, I want to calculate how much part of the octahedron (see figure) is in each HCP structure.

Finally, the image shows only one octahedron. Can you help me in visualising the other octahedrons?

This picture is from How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?:

Let's say the red circles represent a layer of atoms above, and the pink circles a layer of atoms below. The T's and O's represent voids with tetrahedral and octahedral coordination located between layers.

  My visualisation: In the image, 3 inverted and 3 upright tetrahedron can be formed. Is this correct?

It is correct that there are inverted and upright tetrahedra. In the picture above, the T's squeezed between red circles are inverted, and the T's between pink circles are upright.

I can see eight tetrahedra with their center in the unit cell reposted below. Six have one of the orange atoms (in the image posted by the OP) as their tip, and two have all orange atoms as their base.

Complicating things, there are also tetrahedra with their center at the edge of the unit cell, making counting them difficult (see below).

  It is hard to visualise the octahedrons for an atom. And so, I can't figure out the number of octahedral voids in HCP.

The atoms forming the octahedral coordination sphers are hard to visualize in the picture of the question because the unit cell does not fully contain a full set of six. Here is the view from the top again, with the base of the unit cell shaded (red circles correspond to orange atoms in the unit cell shown form the side):

Because we are only showing two layers of atoms, this reflects only half of the unit cell. In this half of the unit cell, there are three voids with octahedral coordination. To visualize them, you need three "pink" atoms at the base of the unit cell (the center one and two vertices of the hexagon) and three "red" atoms (located at c = 1/2, i.e. the second hexagonal layer). Of these three, two are in the unit cell and one is in a neighboring one. If you complete the unit cell by adding another layer, you get a total of six octahedrally coordinated voids in the cell. Because none are on the edge of the cell, this is the total number.

  Also, I want to calculate how much part of the octahedron (see figure) is in each HCP structure.

I'm not sure how to do that. I can say that two of the coordinating atoms are in our unit cell, one is in the neighboring cell, and three are on the border of the unit cell. So the larger part of the volume is probably within the unit cell.

  Number of octahedral and tetrahedral voids present in a HCP structure

We already know that there are 6 octahedrally coordinated voids per unit cell. Let's get the ratio of the two types of voids. If you look at the "red/pink" pictures, you notice that T's and O's come in horizontal rows, with one row of O's next to two rows of T's. So there are twice as many T's than O's, or 12 tetrahedrally coordinated voids per unit cell.

You can also figure this out in a more complicated way. There are 8 T-voids with center inside the unit cell, and 12 T-voids with center on the edge. These are shared among three unit cells touching at these centers (3-fold axis). So the math is 8 plus (12 divided by 3) is a total of 12, as we already figured out.

Or you can make an argument how many tetrahedral coordination spheres an atom partakes in. Each atom forms the tip of two (up and down) and the base of six (three directions, up and down), that is eight. A tetrahedron has four vertices, so two tetrahedra per atom. The unit cell contains six atoms, so twelve tetrahedra per unit cell.

Guide to related questions

Same question for cubic-close packing: How are the number of tetrahedral voids twice the number of octahedral voids in a CCP structure?
Counting atoms and voids is much easier in that case, but the answer is the same.

Shape of the voids and visualization: Distinguishing between octahedral and tetrahedral holes


Absorption and emission at same wavelength? Is it possible for a molecule to absorb and emit at the same wavelength? What is the reason behind it?
I’m working on charged tin porphyrins and got the excitation and emission (fluorescence) wavelength around 400 nm.

Yes it is very common particularly in the more rigid type of molecule. The best example is chlorophyll and this overlap of absorption and emission leads to energy transfer in photosynthesis.  The reason for the overlap is that the excited state potential energy profile is very similar in shape but slightly shifted (as displacement) to that of the ground state and this results in a 'mirror image' spectrum. The spectra below are from anthracene.

You can see even in this rather crude figure that there can be absorption and emission at the same wavelength.


Stability of cyclobutyl methyl carbocation I know that cyclopropyl methyl carbocation is exceptionally stable compared to say, benzyl carbocation.

But I want to know how stable is cyclobutyl methyl carbocation compared to say, tertiary/secondary carbocations? Or, will it go rearrangement breaking the ring?

George Olah and coworkers have examined the cyclobutyldicyclopropylmethyl and the cyclobutylmethyl cation.

(Yes, this is an awfully short answer and I can't really provide more information on the topic.)


Bridging Hydride Covalent Model Electron Count - Hartwig How can a bridging hydride be considered a 3 electron donor in the covalent model for electron counting?
On page 3 of John Hartwig's "Organotransition metal chemistry" textbook [1], he lists the number of donated electrons for common ligands according to the two standard models of electron counting in organometallic chemistry:
The ionic model: ligands are characterized according to their formal charges after cleaving the M-L bond under the assumption that both electrons reside on the ligand.
The covalent model: all ligands are treated as neutral, and the number of electrons they donate (1 $\ce{e^-}$ "X-ligands", 2 $\ce{e^-}$ "L-ligands", 3 $\ce{e^-}$ "LX-ligands" and so on) are determined by distributing the electrons from the M-L bond such that you generate a neutral ligand (he phrases this as a neutral organic group).
In Table 1.1 [1, pp. 3–4] a bridging hydride is listed as donating 4 electrons in the ionic model (in which it would be described as an "anionic, 4-e ligand"), and 3 in the covalent model.

The latter makes no sense to me. When the bonds connecting the bridging hydride are severed, the only way to produce a neutral ligand would be to generate atomic hydrogen, meaning 3 electrons from the M-H-M system would remain with the metals. How then, do we arrive at the hydride donating 3 electrons in this scheme? Is this a typo? Or is it written to somehow indicate that the 3 "donated" electrons are actually from the metals? Or the H-M composite ligand which we can view as then binding to the third metal?
Reference

Hartwig, J. F. Organotransition Metal Chemistry: From Bonding to Catalysis; University Science Books: Sausalito, California, 2010. ISBN 978-1-891389-53-5

A bridging hydride is both an X- and an L-type ligand. Since hydrogen has a valency of 1, the hydride itself can only be an X-type ligand for one of the metal centers. Then we treat the M-H σ bond as an L-type ligand to the other metal center. In the ionic method, this is 4 electrons. In the covalent method, this is 3 electrons.
EDIT:
It was noticed that in the table you provided, the example is actually for a bridging halide, not a hydride. This case is similar, but not exactly like the case of hydride. The main difference is that the halide has lone pairs, so those act as the L-type ligand for the second metal center instead of the M-X σ bond.


Is the atom the smallest particle, which takes part in chemical reactions? According to modern atomic theory, the atom is the smallest particle which can take part in a chemical reaction. But during the formation of hydronium ion, $\ce{H+}$ ion reacts with $\ce{H2O}$ to form $\ce{H3O+}$ ion. Here the $\ce{H+}$ ion, which is smaller than hydrogen atom, takes part in a chemical reaction. So is that postulate invalid?

Protons definitely participate in chemical reactions.  Free protons are generally not present in liquid water because a free proton is extremely reactive, but in the upper atmosphere or in other situations where the density of matter is low there can be free protons which participate in chemical reactions.
However, I completely agree with Jori that an electron, which is much smaller than a proton, takes part in chemical reactions.  Gain or loss of one tiny electron can completely change the properties of a large metalloenzyme.
How small is an electron vs. a proton? In terms of mass, the electron is 1836 times lighter, but what about in terms of radius. According to Hans Dehelt's Nobel Lecture:

Proton radius [is] $0.86 \times 10^{-15}$m...
Today everybody “knows” the electron is an
indivisible atomon, a Dirac point particle with
radius R = 0....
But is it?

And goes on to explain that it is still possible the electron could have a radius on the order of $10^{-20}$ m.
See also Limit on Electron Substructure
Also, photons, which have no mass, participate in chemical reactions.


What is the enol form of 4-methylcyclohex-2-en-1-one? What is the enol form of this compound?

I have made an attempt, but it doesn't look quite right.

Which hydrogen should be taken off (left or right hand side of the carbonyl)? And I don't think the ring is aromatic so I don't know how it is stabilized.

There are two possible enol forms of this ketone. The first one is:

This is the one you drew, which shows the α hydrogen being abstracted. It is perfectly valid.

The other one is:

Note how the alkene next to the carbonyl group "extends" the range of the carbonyl group. This allows the γ hydrogen (i.e. the one three carbons away) to be removed, and then you can "push" electrons into the carbonyl group through the alkene group. In general, this is a property of α,β-unsaturated ketones (i.e. ketones which have a double bond between the α and β carbons).

Note that in this enol form, the methyl substituent is now in the same plane as the C=C next to it. This means that, if you were to reverse the tautomerisation to get back the keto form, there's an equal chance of the hydrogen coming in from above the plane and from below the plane, and racemisation would occur. In practice, there is a rapid equilibrium between keto and enol forms, and this means that if you were to dissolve this compound in any kind of acid (or base), the racemisation would occur pretty quickly.

Here, I have drawn the mechanisms for acid-catalysed tautomerisation. I assume you are now able to draw the mechanisms for the reaction in base (which in general forms an enolate instead of an enol).

At this stage, you might wonder why we cannot abstract the other α hydrogen. There's a couple of reasons. Firstly, it's attached to a $\mathrm{sp^2}$ carbon, and the C-H bond is therefore somewhat stronger. Secondly, even if you could take off that proton and form the "enol", you would have two double bonds next to each other. That's a huge problem because those double bonds want to be linear, but don't quite have the space to do so since there's a ring:

Regarding the stability of the two legitimate enol forms, I would not say that they are any more stable than the keto form. In general, the keto form is more stable. There are cases in which the enol form is relatively stabilised, most often when there is aromaticity or extra conjugation in the enol form (other factors such as solvent will also influence the position of equilibrium):

However, that is not the case here. The starting keto form is as conjugated as any enol you can form.


Does the catalyst alter the Potential energy Surface? Catalyst provides alternative pathway to the reactants. Does it also mean that the potential energy surface is altered?

As you said, catalysts do not alter the PES, but extend it to other dimensions, which usually provide new reaction pathways with lower reaction barriers. If you consider a surface catalyst and an impinged molecule. Considering the full systems PES, the pure molecule PES is a part of it (for large distances of the surface to the molecule). If we bring the molecule closer to the surface, we get new partial PES which have lower reaction barriers.


What is the molecular mechanism behind oncotic pressure? I can't wrap my head around the idea of oncotic pressure and osmolarity, logically.

I imagine a blood vessel. It is filled with proteins, like a ton of proteins and solutes. So this means, according to oncotic pressure and osmolarity, that water will diffuse into this blood vessel.

I can't understand that. If I have a glass full of water, and I drop a bunch of rocks into the water (my proteins) the water will flow out of the glass. It will exert a force, a very basic force of buoyancy, that will lead water out of the glass.

How exactly is oncotic pressure generated?

The molecular basis of osmotic pressure (oncotic pressure is a special type of osmotic pressure) is similar to filtration but on a very small scale.

Some membranes are partially permeable: they let small molecules pass but block larger ones. This asymmetry means that a flow of small molecules is created. Imagine things at the molecular level where a lot of individual motions are essentially random. The big molecules can't pass through the membrane but small molecules like water can pass through. If one side of the membrane consists of small molecules only and the other side has small and large molecules, then random motion will tend to lead to a net flow from of small molecules into the side containing large ones just because the concentration of small molecules is lower on the large molecule side. There are fewer opportunities for random small molecules to move from that side to the small molecule only side because there are fewer small molecules per unit volume on that side.

Obviously I'm simplifying a little, but this is the essential reason behind osmotic pressure.


Penny turns green with vinegar What is the chemical explanation for pennies in vinegar turning green? What is the explanation at a macroscopic and atom-molecular level?

The procedure is:

Fold a paper towel so that it fits inside a dish.
Place a penny on the paper towel.
Pour vinegar over it so that the paper towel is fully saturated.

Experiment

This question may explain about what happens with the pennies at a microscopic level: What is the product of a copper and vinegar reaction?

Also, in his paper, Stephen DeMeo presents the electrochemical perspective:

  The mechanism that has been proposed involves the formation of a free radical and an electrochemical corrosion process (1). [...], suffice it to say that the mechanism does involve formation of a strong oxidizer, peroxoacetic acid ($\ce{CH3COOOH}$), from dissolved oxygen, as well as the production of a copper oxide and a copper ion.

He also explains why the reaction just happens in the presence of oxygen:

  Since copper is below hydrogen in the electrochemical series, copper is reluctant to displace hydrogen ions from electrolytes.  In terms of standard electrode potential, copper has a potential of 0.34 volts ($\ce{Cu -> Cu^{2+} + 2e-}$), while hydrogen has a greater potential of 0.00 volts ($\ce{2 H+ + 2e- -> H2}$). The electromotive force of the cell would be negative and therefore the reaction would be nonspontaneous under standard conditions. When oxygen is involved in the reaction between copper and acetic acid, the electrochemical series and the standard electrode potentials cannot be used as a guide to determine if the reaction is spontaneous.  A generalized relationship between copper metal and acids is nicely described by Massey: "Although copper metal will dissolve readily in an oxidizing acid such as nitric acid it will not dissolve in a nonoxidizing acid unless either an oxidizing agent or a suitable complexing agent is present also" (6).

Reference: DeMeo, S. Does Copper Metal React with Acetic Acid? J. Chem. Educ. 1997, 74, 844-846. DOI: 10.1021/ed074p844


Can nucleophilic attack be faster than rearrangement? Let's say in a reaction, unimolecular (SN1), we have $\ce{CH3-CH(CH3)-CH2+}$ carbocation, and a nucleophile $\ce{Nu-}$. Generally the carbocation is first rearranged and then there is attack of $\ce{Nu-}$, but can't it be that a very strong nucleophile, strongly attracted towards the cation, attacks the cation befor it rearrange to give kinetically controlled product (KCP) (at least at lower temperatures) ?

You are asking what would happen if Something like 1-bromo-2-methylpropane were to undergo an $\mathrm{S_N1}$ reaction. The simple answer is: Your premise won’t happen.

The first step in an $\mathrm{S_N1}$ reaction is the formation of a carbocation by the leaving group leaving. A general equation for this would be:

$$\ce{R3C-Br <=> R3C+ + Br-}$$

Where this reaction has its equilibrium and thus what concentration of the carbocation can be expected, depends on the relative stabilities of reactant and carbocation (the bromide is always the same so it doesn’t matter). Now primary carbocations are so unstable (in a thermodynamic sense) that thermodynamics predicts an equilibrium far to the left for primary haloalkanes:

$$\ce{RCH2-Br <<=> RCH2+ + Br-}$$

(This reaction arrow doesn’t do it justice. It should be a humungous left-pointing arrow and a minute right-pointing one.)

For tertiary or otherwise stabilised carbocations, we can expect a significant concentration of the right-hand side of the equation at equilibrium. Thus, we have a potential carbocation concentration that can potentially react with another nucleophile in an $\mathrm{S_N1}$ reaction. But primary haloalkanes will not even allow the carbocation to form, or in a different view, will pull back the bromide to regenerate the haloalkane before any nucleophile has a chance of attacking the carbocation.

That’s why primary haloalkanes predominantly react via the $\mathrm{S_N2}$ mechanism: There simply isn’t any significant amount of carbocation present for $\mathrm{S_N1}$ or any type of Wagner-Meerwein rearrangement.


What is the purpose of calculating other orbitals than HOMO and LUMO? In this video of benzene orbitals, a lot of different orbitals are presented. Even orbitals that, I assume, could never be occupied. What is the purpose of calculating all these orbitals? Isn't all the chemical information needed (bonding, reactions) coming from the HOMO and LUMO (FMO theory)?

Here is an image of a few of the orbitals shown in the video:

  Energy diagram of the molecular orbitals explored for the D6h
  structure (left) and the D3h distorted geometry (right)
  (from David Casanova  and  Pere Alemany, Phys. Chem. Chem. Phys. 2010, 12, 15523-15529. DOI:10.1039/C0CP01326A)

A practical answer: in most of molecular quantum chemistry, the cost of obtaining unoccupied ("virtual") orbitals is next to nothing because of the algorithms involved. In short, one diagonalizes some matrix (i.e. finding eigenvalues and eigenfunctions, which are the orbital energies and orbitals) and the virtual orbitals are usually delivered anyway.

A more theoretical answer: While a lot of chemistry involves only the ground state, there is UV/vis spectroscopy and photochemistry. For those fields, virtual orbitals are immensely useful, because they will be used for the description of the electronic transitions that happen in those fields. These transitions are not necessarily between HOMO and LUMO because of symmetry, energy scales involved etc.


Melting point of ocean salt: eutectic? I'm interested in the melting point of ocean salt. While sodium chloride has a melting point of 801 °C, ocean salt contains 15 % other ions, mainly sulphate, magnesium, calcium and potassium (the exact composition seems to differ depending on the water source).

Does ocean salt form a eutectic mixture with a lower melting / boiling point? Are there perhaps publications actually testing ocean salt melting points? Thanks!

Yes, the melting point of ocean salt is going to be lower than pure NaCl at 801 °C.

In sea water you don't have compounds but rather dissolved ions. Other than the most abundant Cl- and Na+ (in this order), sea water contains Mg2+, K+, Ca2+ and SO42- (not in this order). Thus, we can assume that the solid residue would be NaCl, MgCl2, CaCl2 and KCl.

Let's have a look at several phase diagrams, taken from http://www.crct.polymtl.ca/fact/pdweb.php:

So yes, adding additional cations other than Na+ will reduce the melting point. Note however, that you will get a lower melting point (solidus) and a higher melting point (liquidus) that in between you will have an equilibrium mixture of solid crystals and melt.

Other than Cl- there's also SO42- and CO32- (as both carbonate and bicarbonate). This will also lower the melting point:

The melting point of the sea salt mixture will be lower than the decomposition of the sulfates and carbonates, so you will dissolve them into the liquid salt rather than degas them as SO2 and CO2 (at least initially).

It's also interesting to see that as you heat the mixture, you will end up with intermediate compounds, such as (Na,K)Cl and (Na,Mg)Cl. As to the exact melting point of the mixture, I couldn't find any experimental data in my quick search. It is possible that thermodynamic modeling would be able to give a prediction. But remember, as the mixture is eventually dominated by NaCl it will not be too far off from the melting point of the pure compound. My uneducated guess is that you will only start seeing any significant melting at around 700 °C and above.

Note of caution - if you do decide to actually melt sea salt, be sure to take the necessary precautions. Even though SO2 will mostly dissolve in the melt, some of it will degas. Molten halides are extremely corrosive and will react with atmospheric moisture, sometimes explosively. They will react with your crucible, possibly causing leakage of really nasty stuff. Unless you know what and how you're doing it, don't do it.


Deriving the entropy of mixing of a non-ideal solution In the earlier posted question named Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics the entropy of mixing was found by assuming an ideal solution. The user Porphyrin mentioned that in case the molecules in the solution do interact with each other, the free energy would have an additional term (n1+n2)x1x2w, with

  w≈(2E12−E11−E22) and E12E12 is the interaction energy between molecules of type 11 and 22 when they are adjacent and similarly for
  E11E11 and E22E22 all of which are negative. From a consideration of
  the energies involved ww is expected to be positive.

My question is: how do we go about finding an answer for the entropy of mixing when we do not assume that the solution is ideal?

The calculation of the mixing energy is extremely complicated and only approximate solutions have been obtained as far as I am aware. I shall attempt to outline one of these methods*.

All molecules are assumed to be spherical and of equal size and situated on a lattice with co-ordination number z . Each pair of neighbours AA of type A molecules contributes energy $w_{AA}$, similarly $w_{BB}$ for B molecules and energy $w_{AB}$ for AB pairs. The total energy due to these  interactions
is
$$E_T=N_{AA}w_{AA} + N_{BB}w_{BB}+N_{AB}w_{AB}$$
where $N_{AA}$ is the number of AA pairs and so on. Counting the number of  $A$ interactions and number of B ’s gives
$$zN_A=2N_{AA} + N_{AB}  ~~~\text{and }~~~ zN_B=2N_{BB} + N_{AB} $$

and thus  combining these equations gives
$$E_T=\frac{zN_Aw_{AA}}{2}+\frac{zN_Bw_{BB}}{2}+wN_{AB}=E^T_{AA}+E^T_{BB}+wN_{AB}$$

where $w=w_{AB}-w_{AA}/2-w_{BB}/2$ is the energy gain on mixing per pair of AB neighbours.

In statistical mechanics the partition function is the key to all other thermodynamic functions. In this case partition function is $$
Z= \sum_{N_{AB}}Z_AZ_Bg(N_A,N_B,N_{AB})e^{(-E^T_{AA}-E^T_{BB}-wN_{AB}/KT)}$$

and where $g(..) $ is the number of arrangements of $N_A$  A’s , $N_B$ B ’s and $N_{AB}$  pairs of molecules. Separating out the terms due to pure A and B molecules gives

$$Z=Z’_AZ’_B \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/KT}$$

and as the Helmholtz energy is
$F=-kT\ln(Z)=F_A+F_B+\Delta F$
and for mixing only the $\Delta F$ term is important the others being the energy of each pure component.
Thus
$$\Delta F = -kT\ln\left(\sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT}  \right) $$

The change in energy can be obtained from the Gibbs-Helmholtz equation $(E = F-T(dF/dT))_V$ and is
$$\Delta E= w\frac{\sum_{N_{AB}} N_{AB} g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT} } { \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT} } $$

and this is an equation that seems to have no exact solution and so this is where approximations have to be made.

The difficulty is in the exponential terms, as they contain $N_{AB}$ the aim is therefore to replace these with an average value, and thus the exponential terms become constant wrt $N_{AB}$. It can be shown that
$$\sum_{N_{AB} }g(N_A,N_B,N_{AB}) =\frac{(N_A + N_B)!}{N_A!N_B!}$$
and
$$\bar N_{AB}=w\frac{\sum_{N_{AB}}N_{AB}g(N_A,N_B,N_{AB})}{\sum_{N_{AB}}g(N_A,N_B,N_{AB})} = z\frac{N_AN_B}{N_A + N_B} $$

By replacing $N_{AB}$ by $\bar N_{AB}$ in the partition function produces

$$Z=Z’_AZ’_B \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-w\bar N_{AB}/KT)}
 = \frac{(N_A+N_B)!}{N_A!N_B!}\exp\left(-\frac{zw}{kT}\frac{N_AN_B}{N_A+N_B}\right)$$

the change in energy is then

$$\Delta E = zw\frac{N_AN_B}{N_A+N_B}= w\bar N_{AB}$$

and this can be re-written as molec fractions to produce
$$ \Delta E= z(N_A+N_B)x_1x_2w$$

based on text in Rushbrooke ‘Introduction to Statistical Mechanics’


Why are correlation consistent basis sets used with DFT calculations? Reading some papers, there is a bit of a curious observation that does not exactly make a lot of sense to me.

In Quantum Chemistry, different methods exist to carry out geometry optimisation and energy calculations. Quantum chemists can have a huge field day arguing which method and basis set to use.

A common type of calculation used for larger (though large is relative to one's computational resources) is density functional theory (DFT) - because it provides good results on a reasonable time scale.

Now here is where it gets weird:
Every method has its basis set - basis sets geared towards DFT and basis sets geared towards couple cluster methods. It is generally recommended not to use a correlation consistent (cc) basis set with a DFT method (and I guess conversely, a basis set aimed at DFT should not be used with a coupled cluster method).

Why then would people benchmark methods using seemingly inappropriate basis sets. E.g. a cc basis set with an M06 functional.

(See for example: Edward G. Hohenstein, Samuel T. Chill and C. David Sherrill; J. Chem. Theory Comput. 2008, 4 (12), 1996–2000)

It is generally recommended not to use a cc basis set with a DFT method (and I guess conversely, a basis set aimed at DFT should not be used with a coupled cluster method).

This statement glosses over some specifics that might be important.

There is nothing technically wrong with using correlation-consistent or ANO basis sets with DFT, unless the basis is deficient to begin with (cc-pVDZ). That is, using cc-pVQZ to optimize the geometry of a small organic molecule will not lead to more "incorrect" results than 6-31G* unless linear dependencies arise in the basis due to over-completeness.
However, it is most certainly inefficient to use these basis sets for general usage (SCF-type methods) due to using general contraction rather than segmented contraction. Most modern integral algorithms still have a difficult time dealing with highly-contracted basis sets. This is one reason to use the TURBOMOLE/Ahlrichs/Karlsruhe Def2 and Frank Jensen's segmented polarization-consistent (pcseg-n) basis sets.
"Using a basis set aimed at DFT should not be used with a coupled cluster method" is more tricky. The TURBOMOLE basis sets are certainly used for MP2 and coupled cluster calculations, however they are more appropriate for "single" calculations (not CBS extrapolation), and only with the largest basis one can afford. They were designed to capture the total mean-field energy, which decays exponentially, not the correlation energy, which decays like ~$(L+1)^{-3}$, where $L$ is the maximum angular momentum being considered. The correlation-consistent and atomic natural orbital basis sets are better for coupled cluster because they were designed from the beginning to recover the correlation energy.
The use of Pople-style (STO-nG, 3-21G, 6-31G*, 6-311++G(d,p), etc.) basis sets would be incorrect with wavefunction-based methods, perhaps even for qualitative understanding depending on how "difficult" your system is. Like the TURBOMOLE basis sets, they are designed to capture the mean-field atomic energies, but many have constraints (identical s- and p-type exponents, the "SP" functions that are often seen) due to the computational resources available at the time of their creation. Many of the older basis sets (notably 6-31G and not 6-311G) were optimized to use Cartesian rather than pure d-type functions, leading to potential inconsistencies in results. It is also not as obvious how to systematically improve the quality of the basis sets, and none are larger than triple-$\zeta$. Ultimately, the virtual MO space created by these basis sets is not sufficient to recover the correlation energy, and the benefits of using coupled cluster are completely lost.


Is the triplet state of the cyclopentadienyl cation really aromatic? While checking the ASAP articles of the Journal of the American Chemical Society I came across a very interesting contribution by Costa et al. who synthesised a $\ce{BF3}$ stabilised cyclopentadienyl cation 1 and confirmed that its singlet state was indeed antiaromatic as predicted by various definitions.[1] The authors, however, also included the following sentence:

  According to Baird’s rule, “the lowest triplet state for $4n$ rings is aromatic since the bonding energy is significantly greater than for the diradical reference structure”.[1]

This statement confuses me. In my introductory organic chemistry classes, I was taught the $4n+2$ rule along with simplified Hückel analyses to predict the relative energy levels. It was always pointed out that $4n$ (if planar and a single π system) represents an unstable, antiaromatic state which the system will avoid by distortion. This was explained by the triplet, biradicallic structure of the $4n$ systems. Yet here by Costa et al. I read that $4n$ triplets are actually aromatic.

How can I reconcile this new information with what I was taught previously? Reading the article and the references has not been able to help me.

Reference:

[1]: Costa, P.; Trosien, I.; Mieres-Perez, J.; Sander, W. Isolation of an antiaromatic singlet cyclopentadienyl zwitterion. J. Am. Chem. Soc. 2017, 139 (37), 13024–13030. DOI: 10.1021/jacs.7b05807.

TL;DR Yes. For cyclic conjugated systems in the $T_1$ state (i.e. lowest energy triplet excited state), $4n$ electrons means aromatic and $4n+2$ means anti-aromatic. The $T_1$ state of the cyclopentadienyl cation, which has 4 pi-electrons, is therefore aromatic.

Aromaticity

The (quite understandable) confusion here arises over the interpretation of the word "aromatic" or "anti-aromatic". These terms can be used differently depending on context, and there is no unified physical description of (anti)aromaticity (although progress has been made in recent years, e.g. nucleus-independent chemical shifts, etc.)

When anybody claims that something is aromatic, it is therefore imperative to check the context, and to figure out what their basis for assigning aromaticity is. In this case, the assignment of the triplet excited state of the cyclopentadienyl cation as aromatic by Costa et al.1 is based on Baird's rules. In turn, looking up Baird's article2 reveals that his assignment of (anti)aromaticity is based on the concept of Dewar resonance energies (DREs).

The concept of DREs has been explained by Baird elsewhere.3 Essentially, it is based on the comparison of idealised cyclic structures with a reference acyclic structure; so, benzene is compared with hexatriene, and cyclobutadiene ($D_\mathrm{4h}$) is compared with butadiene. For the cyclopentadienyl cation, the relevant comparison is with a penta-2,4-dien-1-yl cation.

The question thus arises as to how this comparison can be made. The way that is commonly taught, at introductory level, is to use Hückel molecular orbital (HMO) theory. So, let us use cyclobutadiene as an illustration. Note that within HMO theory the singlet and triplet configurations are of the same energy, as electron-electron repulsions are not taken into account.

The resonance integral $\beta$ is negative. So, singlet cyclobutadiene is anti-aromatic because it has less π-bonding energy than the singlet state of its acyclic counterpart butadiene. However, triplet cyclobutadiene is aromatic because it has more π-bonding energy than triplet butadiene does.

There are two things of note here: firstly, the Jahn–Teller distortion in cyclobutadiene isn't relevant for the purposes of finding whether it is aromatic or not; the aromaticity is assigned based on a hypothetical planar structure with no bond length alternation. Secondly, the aromaticity is determined with respect to the corresponding electronic state of the acyclic reference. That is to say, the triplet state of cyclobutadiene is compared with the triplet state of butadiene and not the singlet state of butadiene.

Perturbational MO theory

Qualitatively, we have reached the correct conclusion. However, this was not the approach that Baird used. We will restrict our discussion now to that of triplet states, since it is relevant to the question. Baird used a more sophisticated method involving perturbation theory. That is, Baird used MO theory to calculate the stabilisation/destabilisation that would occur upon bringing together two radicals to form either the cyclic compound or the acyclic compound.

This is confusing, so a bit more explanation is in order. The triplet state of butadiene can be described by the promotion of a π electron to a π* electron. Upon this promotion, the C=C bond twists until it has a dihedral angle of 90°, which corresponds to the breakage of the π bond.

In the triplet state of butadiene (1 ($T_1$)), the π systems of the allyl and methyl radical fragments are orthogonal to each other; they are simply joined by a C–C σ bond. Hence the total π-bonding energy of 1 ($T_1$) is simply equal to the sum of the individual π-bonding energies of 2 and 3.

On the other hand, the triplet state of cyclobutadiene (4 ($T_1$)) can still be considered to be an allyl fragment plus a methyl fragment.4 The difference is that, due to the planarity of the ring, the two radical systems are forced to overlap in a π manner (depicted by the red lines).

Now, if this overlap is stabilising in nature, then the π-system of 4 ($T_1$) is more stable than the π-system of 1 ($T_1$), and therefore we can say that 4 ($T_1$) is aromatic. Conversely, if the overlap is destabilising, then it is anti-aromatic.

Finally, we get to the crux of the answer. (I make no apology for taking so long to get to it - I thought the buildup was pertinent.) Baird first constructed the MOs of the radical fragments 2 and 3, and labelled each MO with their symmetry under a reflection. The mirror plane is drawn in light blue; antisymmetric MOs are in red and symmetric MOs in green.

Then, Baird identified two types of orbital interactions in the cyclic structures that could lead to either stabilisation or destabilisation. The first type is called a Type I interaction, and it involves the overlap of the two SOMOs on the two fragments. In this case, since the two SOMOs are of different symmetries, they cannot overlap and therefore the Type I interaction leads to neither stabilisation nor destabilisation.

The second type of interaction (I leave the reader to guess what it was called) involves the interaction of the SOMO on one fragment with doubly occupied or empty orbitals on the other fragment. In this case, the SOMO of the allyl radical does not find any symmetry match on the methyl radical, and so there is again neither stabilisation nor destabilisation. However, the SOMO of the methyl radical can overlap with both symmetric MOs of the allyl radical; this is a stabilising interaction.

Overall, therefore, the π-type overlap between the two radical fragments leads to a favourable interaction. As such, triplet cyclobutadiene is considered to be aromatic.

For the cyclopentadienyl cation, one can construct an analogous diagram, starting from an ethene radical cation and an allyl radical.

Again the SOMOs do not match in symmetry, so the Type I interaction is necessarily zero. And again, the Type II interaction is stabilising; this time, both SOMOs find symmetry matches on the partner fragment. Based on this analysis, the triplet state of the cyclopentadienyl cation is indeed aromatic.

Triplet benzene

At this stage one may wonder: when two orbitals interact, the only possibility of destabilisation is if both orbitals are fully filled. Since both interactions involve orbital overlap between not-fully-filled orbitals, how can any of them be destabilising, and how can anything be anti-aromatic? To address this issue, and for the sake of completeness, let us analyse the case of benzene. We expect the triplet state of this 6π system to be anti-aromatic. The interaction diagram may be constructed from two allyl radicals:

First, we see that by symmetry, the Type II interaction is identically zero. Now, we turn to the Type I interaction. Surely the overlap of two SOMOs must be a bonding interaction? The astute reader will notice, though, that we are talking about a triplet state of benzene. The overlap of the two SOMOs still forms a bonding and an antibonding orbital, but if we were to put both electrons into the bonding orbital, we would obtain a singlet state.

As such, the Type I interaction leads to both the bonding and antibonding MOs having one electron each. Since the antibonding MO is more antibonding than the bonding MO is bonding, this is a net destabilisation. Hence, triplet benzene may be assigned anti-aromaticity.

Further reading

The concept of excited state aromaticity has been reviewed by Rosenberg et al.5, which is a good read (I found the explanation more accessible than that in Baird's original article2).

Notes and references

Costa, P.; Trosien, I.; Mieres-Perez, J.; Sander, W. Isolation of an Antiaromatic Singlet Cyclopentadienyl Zwitterion. J. Am. Chem. Soc. [Online early access]. DOI: 10.1021/jacs.7b05807.
Baird, N. C. Quantum organic photochemistry. II. Resonance and aromaticity in the lowest 3ππ* state of cyclic hydrocarbons. J. Am. Chem. Soc. 1972, 94 (14), 4941–4948. DOI: 10.1021/ja00769a025.
Baird, N. C. Dewar resonance energy. J. Chem. Educ. 1971, 48 (8), 509–513. DOI: 10.1021/ed048p509.
Technically, there is one hydrogen less on both fragments, but the C–H bonds do not contribute to the π framework so they can be ignored - after all, we never bothered taking the C–H bonds into account, not even in the Hückel approach. The point is that whether it is a methyl fragment or a methylene fragment, the π system remains the same (likewise for the allyl fragment).
Rosenberg, M.; Dahlstrand, C.; Kilså, K.; Ottosson, H. Excited State Aromaticity and Antiaromaticity: Opportunities for Photophysical and Photochemical Rationalizations. Chem. Rev. 2014, 114 (10), 5379–5425. DOI: 10.1021/cr300471v.


Explain the enolisation process of 5,8-dihydroxynaphthalene-1,4-dione I have no idea really how to get started...

Where does the hydrogen for the first deprotonation come from?

My working thus far:

There is no need for any catalyst apart from acid or base. Also, the $\ce{^{13}C}$ is just there to tell you that it is not the same molecule. Also, this is not a zwitterion, it is tautomerism since there is pi bond movement involved.
Don't deprotonate any carbons. They're all $\ce{sp^2}$ and won't take kindly to deprotonation. Just push electrons from one oxygen to the other through the conjugated system.
Try doing this one first. In acid, your general tautomerism mechanism would look like this:

Protonate the carbonyl group.
Push electrons from phenol oxygen.
Deprotonate the carbonyl formed on the other side.

In base, it would be:

Deprotonate the phenol.
Push electrons from phenoxide oxygen.
Reprotonate the phenoxide formed on the other side (strictly this shouldn't happen, but whatever).

If you can do this one, the mechanism for your original question just consists of the same thing repeated. You can "fill in" the parts of the molecule that I omitted to see what I mean.


Can an ethyl ester of L-proline be synthesised from the amino acid and ethyl alcohol in toluene with azeotropic distillation of water? I was told that the ethyl ester of L-proline may not be synthesized from L-proline and ethyl alcohol in refluxing toluene, with an acid catalyst such as p-toluensulfonic acid for example, because toluene and ethanol and in this case water as well form an azeotrope. Is this true?

I do not have a supply of pure thionyl chloride, and that's why I am trying to use this route. I appreciate if anyone may answer this question.

I think this is unlikely to work under the conditions described as the catalytic PTSA is tied up as its proline salt. If you reflux under D-S conditions you will just remove the EtOH due to the azeotrope with Toluene.


Halogenation of alkenyl boranes When a terminal alkenyl borane is succumbed to halogenation where the halogen and the base are added in two separate steps, the geometry of the double bond is changed. When they are added simultaneously, the double bond geometry is retained. Why is it so?

I remember learning that these two reactions proceed via different mechanisms. I'm going to use an alkenylborane to illustrate them.

Organoboranes are electron-deficient and electrophilic; when hydroxide is added together with iodide, the first step is the formation of an anionic ate complex. This is now nucleophilic via the $\ce{B-C}$ $\sigma$ bond, and can therefore react with the electrophilic iodine:

This mechanism is somewhat analogous to the hydroboration-oxidation $\ce{H2O2}$ step, where the carbon "nucleophile" is intramolecularly transferred from boron to an electrophilic oxygen atom. This step is an intermolecular transfer of carbon from boron to iodine instead.

Incidentally, I remember that with alkylboranes the migration occurs with inversion of stereochemistry, but I cannot seem to find a reference. Given the premise of the question, I would hypothesise that alkenylboranes react with retention of stereochemistry (which would not be surprising since stereochemistry at $\mathrm{sp^2}$ and $\mathrm{sp^3}$ centres, and the stereoelectronic requirements for inversion/retention, are entirely different).

As for the other reaction, hydroxide isn't added first, so iodine (electrophilic) has to react with the double bond (nucleophilic). The subsequent steps are best described by a diagram instead.

Again I am a little frustrated at not being able to find a reference, but I am fairly sure this was how I learnt it. I literally looked up all reactions of $\ce{BR3}$ with iodine on Reaxys but couldn't find anything on it, so...


Why the pH of xanthan gum in aqueous solution is 7 (neutral) when it has glucuronic and pyruvic acids in its chemical structure? Since xanthan gum has both glucuronic and pyruvic acids in its chemical structure, it should release $\ce{H+}$ ions upon solvation in an aqueous solution and make it acidic. However, the meausured $\ce{pH}$ is $7$.

Pyruvic and glucuronic acids are both pretty strong acids (pK's of respectively 2.5 and 3.28) but that does not mean that the polymer of xanthan gum will be. Those protons are still there but they may be more tightly bound in the polymer than in the free acids. To find out try titrating a solution with strong base. Eventually you should start prying some of those protons off and you should see an inflection point. OTOH as there are so many carboxylic protons in the polymer the inflection points would probably smear and not be visible but you should see  some buffering above some pH if my hypothesis is correct.

I should make clear that this is purely speculation on my part but it sounds like an interesting experiment.


Fundamentals of Li-ion Functionality I would like to have an in-depth understanding of Li-ion batteries functionality, at atomic level. Can anyone help me to understand it? I have included some known background to support my questions.

For a fully charged battery:

The theory says that all (let's assume an ideal case) the Li-ions are in the anode, right? But are they really ions at this point or simply neutral atoms where the numbers of electrons is equal to the number of protons? In other words, are there ions or atoms in the anode? See figure below for a single Li-"ion":

Do we say that the electrical energy is stored in the electrons or is this a false statement, considering that batteries store chemical energy by definition? If we say that it is only chemical energy, how does one make the conversion between electrical energy and chemical energy? Is it stored in bonds?
Is the anode "more energetic" than the cathode in this case, considering that there are more electrons in the anode?

When discharging:

The theory says that the load "steals" the electrons which travel through the load and then "land" back into the cathode. Is the electrical energy from the electrons used by the load and then they go back to the cathode "empty"? (According to the definition of internal energy, the work is done by kinetic energy due to movement of electrons or molecules).

Fully discharged battery

Now we are in the situation where "all" the ions are back to the cathode. At this point, are there less electrons in the cathode or why is the battery discharged? Popularly, the battery is "out of fuel", but how do we actually translate this into energetic phenomena?

I appreciate any type of help, recommendation or mathematical explanation.

And please don't tell me that I need to go and read the basics of thermodynamics and stuff, because I did that, but none of the books/pages that I found describe these things at an atomic level. However, any recommendation of a book/page that does that will be highly appreciated!

Thank you very much in advance!

Paul

Based on the Wikipedia article, I am inclined towards thinking that the lithium on the electrodes is neutral. The lithium ions move through the electrolyte to facilitate the two half-reactions. The electrons in the half reaction participate in closing the external circuit, so the lithium ions and electrons are moving in the same direction, except one does so inside the battery and the other outside the battery).

Onto your questions:

The lithium in the anode is neutral, not an ion.
There's no traditional bond per se, but it's stored in the energy difference between $\mathrm{LiC}_{6} + \mathrm{CoO}_{2}$ versus $\mathrm{C}_{6} + \mathrm{LiCoO}_{2}$. There is an interaction here, but I wouldn't call it a "bond."
Well, the anode or cathode isn't more energetic, that's not quite the right term. The $\mathrm{LiC}_{6}$ is more energetic than other species. It just happens that this happens to be the anode of a charged battery. Once the battery discharged, the anode is still that side, but it doesn't have more energy.
Electrical energy doesn't work that way per se. We generally say we build up a potential difference $V = \frac{\Delta E}{q}$. The extraction of energy from this difference is dependent on the circuit. If you have an incandescent light bulb, the kinetic energy of the traveling electron is converted to heat and light, for example.
When the battery is discharged, the lithium atoms have moved to the metal oxide. That's the thermodynamically more stable of the two substrates for the lithium to be adsorbed onto.

I've never looked this up before, but this is actually reall cool. We'll basically using the thermodynamics of lithium preferring to be adsorbed onto one electrode over the other. In order for the lithium atom to get there, it has to move as an ion through the electrolyte and the missing electron moves through the external circuit. I think your confusion is here: the system wants to move the lithium atom, so we'll extract something useful out of it by forcing it to move the lithium ion and the electron separately.


What does HTO decay into? I'm aware that tritium undergoes beta decay forming Helium-3. But in the case of HTO, what does it decay into? I don't suppose it decays into HeHO.

Very shortly after a radioactive element decays in a molecule, you could consider that for a brief moment, it does indeed form a molecule with the new element in the same place as its originating radioisotope. So yes, after the decay of $HTO$ you could get $HeOH^+$ via the unusual reaction $TOH → HeOH^+ + e^- + \bar\nu_e$. Of course, though, this is a very unstable compound and it would immediately participate in further chemical reactions, very likely releasing the helium atom at some point. Even if the molecule were relatively stable after having one of its atoms transmuted, the molecule might still decompose as the nuclear decay releases a lot of energy, which can tear the comparatively fragile molecule to pieces.

A curious tidbit is that, in fact, the decay of tritiated compounds has been considered as a means to produce legitimate helium compounds, given that directly reacting helium with other substances has been fruitless other than for very trivial species, such as $HeH^+$. The problem is that tritium has a rather long half life of about 11 years, so it would be quite hard to produce reasonable quantities of helium compounds in a reasonable timespan, along with the likelihood of the helium compound being very poorly bound and the emission of vast quantities of nuclear decay energy which could easily decompose the fleeting substance.


Change in water vapor with change in pressure and volume Given a closed (air tight) container (Volume $V_0$) held at the room temperature ($T_0$) and pressure ($P_0$). The container contains 100% saturated water vapor with no air (at $T_0$ and $P_0$).
Now assuming the container is compressible (say a balloon whose volume can be changed by applying uniform external force/pressure how does the water vapor inside the container change when we apply external pressure on its surface ($P_1$) and its volume changes ($V_1$).
Assuming the Ideal Gas Law I assume none of the water vapor would condensate with any such change in external pressure and volume or i am missing something?
Edit 1: the container contains no air and no liquid water either. Just water vapor at 100% saturation (whatever its value is at initial room temperature and pressure i.e. $T_0$ and $P_0$).

If water is introduced in an empty container, the water will slowly evaporate. The pressure $p$ will slowly increase from $0\ \mathrm{Pa}$ to a maximum which is $3167\ \mathrm{Pa}$ at $25\ \mathrm{^\circ C}$. This corresponds to a molar concentration of water vapor equal to: $${c = p/RT = \frac{3167\ \mathrm{J\ m^{-3}}}{8.314\ \mathrm{J\ mol^{-1}\ K^{-1}}\times298\ \mathrm K} = 1.278\ \mathrm{mol\ m^{-3}}}$$
Expressed in gram per cubic meter, the concentration of water is:  $ c = 23.0\ \mathrm{g\ m^{-3}}$.
This means that between $0$ to $23\ \mathrm{mg}$ water can be evaporated in a $1$ liter container at $25\ \mathrm{^\circ C}$, and that a maximum amount of $2.3\ \mathrm{mg}$ water can be evaporated in a $0.1$ liter container. As a consequence, if the gas volume of your water vapor is compressed from $1$ liter to $0.1$ liter without changing the temperature, $23\ \mathrm{mg} - 2.3\ \mathrm{mg} = 20.7\ \mathrm{mg}$ water will be condensed as a liquid on the inner wall of the container.


Is a sequence of two Wagner-Meerwein rearrangements possible? We've just studied about the Wagner-Meerwein rearrangement in carbocations.
Our teacher told us that it occurs in order to stabilize the carbocation intermediate.

Knowing the carbocation stability order,

  $\mathrm{Tertiary > Secondary > Primary} > \ce{CH3+}$

I decided to test the "limits" of this particular rearrangement reaction.

I drew a primary carbocation (A):

Upon subjecting it to a Wagner-Meerwein hydrogen-shift (in my head), I obtained the secondary carbocation (B):

Which was all fine and dandy; well within what I've been taught.

But seeing the potential for additional stability another Wagner-Meerwein rearrangement could provide, I conducted a Wagner-Meerwein alky(methyl)-shift and ended up with the following tertiary carbocation (C):

I started with a primary carbocation, changed it to a secondary carbocation and finally ended up with a tertiary carbocation.

Now I showed this to my teacher, but he said that in this particular case the carbocation (B) won't convert to carbocation (C); i.e- carbocation (A) isn't going to undergo two Wagner-Meerwein rearrangements (1,2-anionotropic shifts), because in effect it'd be a 1,3-shift and not the 1,2-shift as the Wagner-Meerwein rearrangement calls for.

A few days later, I took a different sheet with carbocations (B) drawn on it and carried out a Wagner Meerwein alkyl(methyl)-shift to obtain carbocation (C) just as before [except that this time around, my starting molecule was B not A]. I showed this to my teacher again, but the rearrangement I carried out in this case doesn't seem to perturb him at all. His argument is that, since my starting intermediate is carbocation (B), to reach carbocation (C), it undergoes only one Wagner-Meerwein rearrangement which is perfectly fine, unlike my first case which involved two rearrangements.

Now my question is, if he's right, then how would carbocation (B) "know" whether it already underwent a Wagner-Meerwein rearrangement [from carbocation (A)] before considering rearranging itself to become the more stable, carbocation (C)? How does it "know" if it's the starting intermediate or not.

There is no reason why a sequence of two Wagner-Meerwein rearrangements wouldn’t happen — assuming low concentrations and thus the absence of directly capturing nucleophiles. In fact, a professor of mine once said:

  The only secondary carbenium ion is the prop-2-yl ion; all others will rearrange to form tertiary carbenium ions.

(He did explicitly exclude carbenium ions stabilised by other means from that discussion; of course an oxocarbenium ion is well-known as is an allyl cation.)

A very well-known example of this is the biosynthesis of lanosterol, in which the enzyme lanosterol synthase takes squalene-epoxide, orders it in a boat-chair-boat conformation, catalyses a domino sequence of carbocations being captured by double bonds, followed by a sequence of four Wagner-Meerwein rearrangements and finally deprotonation to generate the backbone ring-structure of chloresterols. See the following scheme.

If nature can do it, it will happen in the lab at least to a minute extent.


Glass electrode pH measurement How is the potential difference between the outer and inner surface of the glass bubble in a glass electrode measured if an Ag/AgCl wire is used as the indicator electrode and Ag/AgCl/KCl is used as the reference electrode? In the Nernst equation for the reaction, hydronium ions don't participate.

You measure an electric potential difference. It can be considered sum of three different effects. If you could isolate them, you will have the electromotive force of both reference electrodes. If they can be isolated, each of them will generate a potential difference at open circuit (when the stationary state is reached) equal in magnitude to the emf. You always will have this difference of potential plus the potential difference at both sides of the membrane. Just add potential differences due to the stationary condition reached because of the emf.


How many electric vehicles can our current supply of lithium power? This Reddit comment contends

There was 870 tons of lithium mined in 2019 in the US. That enough for under 20,000 cars, assuming you're using ~50 kg per car. So that would be 40,000 in 2023, assuming all of it is used in cars. In reality only a fraction is; most is used in consumer electronics etc.

The global car production is about 100,000,000 per year, meaning you'd need like 5,000,000 tons of lithium just for cars. Current WORLD production is 77,000 tons, although there's a large reserve.

Proliferation of electric cars is, as far as I can see, completely dependent on a large step forward in battery tech.

What exactly does there's a large reserve entail? How much Li does the "large reserve" contain?

Are these calculations correct?

You assume there is X lithium and all of it cost Y. And then there would be no lithium to mine. Thats not quite right. As you are ready to pay more, harder to get sources become profitable and become part of available reserves. Sea has almost unlimited amount of lithium, but it is about x5 as expensive to mine.
The total lithium content of seawater is very large and is estimated as 230 billion tonnes
(wiki)


Abundant substances which I may use like coupling agents What is the most abundant substances which I may use like coupling agent for Wood plastic composite with HDPE polymer? I don't want maleated polyolefin because they are not so abundant in my country.

If you don'y want to use maleated polyolefin (the best coupling agent for this composite), you can use oxidized polyethylene, ethylene propylene rubber, ethylene-co-glycidyl methacrylate or polyolefin elastomer.


Excluded Volume in van Der Waals equation I was just reading over the EOS topic when where I stumbled upon excluded volume.
Although the pictorial representation of the hard sphere model clearly shows that the excluded volume is 4 times the volume occupied by molecules, but what i am not unable to understand is that why do we take the sphere with diameter equal to one of the molecules in the first place.

So, my question is that why do we have to take the whole spehere..

I guess this has something to do with viewing the molecules in 3-dimensions, but i am unable to visualize.

We assume that the molecules are identical hard spheres with diameter $d$. The excluded volume is emphatically not the volume of the sphere, but rather the volume excluded to the other particles by the presence of that sphere.

When two particles collide, they touch each other at some point on their surface, and the distance between the center of one molecule to that of the other is $d$. Indeed, two particles cannot get closer than a distance $d$ to each other, and hence the excluded volume of each particle is the volume of a sphere of diameter $d$—-we must take into account the radii of both colliding spheres.

Edit: This suggests that the excluded volume should be eight times as large as the original volume of a sphere, not four times. However, a collision event involves two particles, and this excluded volume is shared, in a sense, by both particles. We must therefore divide by a factor of two, and that gives the resulting multiplier of four.


Uses of oxidizing and reducing flames Why do we use oxidizing and reducing flame in different analytical tests?

For example, in the charcoal block reduction test, we use reducing flame. Why? How does this affect the salt to be reduced into metal?

As the name suggests the oxidizing flame is used to oxidize probes, while the reducing flame is used to reduce probes. Hereby oxidation is done by oxygen, while reduction is done by reducing agents in the flame, such as carbon monoxide or hydrogen.

In the charcoal block reduction test with a blowpipe one often will oxidize the probe in the first step to get rid of nonmetal elements in the probe, and then will use the reduction flame to reduce oxidic metal compounds in the melt to the elemental metal. Volatile metals will hereby be precipitated at the colder regions of the charcoal as oxides. Metal and oxide precipitate can then be further analyzed.


Fractional Distillation Why does the long length of the fractionating column in fractional distillation enable more effective separation of two liquids in a mixture?

(basically what's the purpose of the fractionating column being long in relation to the effectiveness of the fractional distillation process)

It simply spreads the temperature change over a longer distance, enabling separation of distillates with boiling points close to each other. "Generally the component parts boil at less than 25 °C from each other under a pressure of one atmosphere."

Imaging reading a mercury thermometer one cm tall as compared with one 100 cm tall: it would be difficult to distinguish temperatures on the smaller one. In the fractionation column below, it helps keep the outputs separate.


Are there any compound lists searchable by elemental content? As I say in the question, I am looking for lists of compounds that I can search based on elemental content. Essentially, I want to find candidate compounds for boron doping liquid scintillator, we have a few likely candidates already but we want to see the available chemical space and see if there are better/cheaper candidates. I want to avoid high electron densities, so essentially I want to find compounds that only have boron, carbon, and hydrogen. I will filter later for hazard/reactivity/flammability.

Does such a list / service exist? If not, is there a general list of compounds with their formulas that I could then parse with a python script or something?

After some more searching, followed by asking reddit, I got a few good suggestions.
ChemSpider's advanced search method allows you to, among other things, search for elements containing specific elements and lacking other elements.

Also, for large lists of compounds I can use python tools to extract the tables from the CRC Handbook pdfs and then, search that data.


Differences between formulae for dipole–dipole interaction energy I came across two formulae for dipole-dipole interaction energies on Chemistry LibreTexts — Dipole-Dipole Interactions:

$$V=-\frac{2\mu_1 \mu_2}{4\pi\epsilon_0r^3}\tag{3}$$

and

$$V=-\frac{2\mu^2_\ce{A}\mu^2_\ce{B}}{3{(4\pi\epsilon_0)}^2r^6}\frac{1}{k_\mathrm{B}T}\tag{7}$$

Which of the above two is the formula for the interaction energy of dipole-dipole interactions? Also, what is the difference between the two formulae?

They are both dipole-dipole energies as from the link but their contexts are different. Eq. 3 (as numbered in the LibreTexts link),
$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}} \tag{3}$$
is specifically for the case when the dipoles are aligned. The more general Eq. 4
$$V(r) = - \frac{\mu_{1}\mu_{2}}{4\pi\epsilon_{0}r^{3}_{12}} (\cos\theta_{12} - 3\cos\theta_{1}\cos\theta_{2}) \tag{4}$$
is preferable because it includes the dependence on the orientation of the dipoles in space. And I believe Eq. 7
$$V = - \frac{2\mu_{\mathrm{A}}^{2}\mu_{\mathrm{B}}^{2}}{3(4\pi\epsilon_{0})^{2}r^{6}} \frac{1}{k_{\mathrm{B}}T} \tag{7}$$
is the energy derived by thermally averaging the dipole–dipole energies when the dipoles are fixed in space but are free to rotate.
So Eq. 3 is the dipole–dipole energy when the dipoles are both in a specific orientation, Eq. 4 reveals the orientational dependence of the dipole-dipole energy, and Eq. 7 is the thermally averaged energy for two dipoles that have a fixed separation $r$ but are free to rotate.


Most energy-dense commercial battery available What's the most energy-dense and currently commercially available chemical battery technology? Would it be the Lithium-Ion Polymer or has it been surpassed already?

There's is a chemical non-rechargeable battery which is more energy-dense than LiPo, even if it's less power-dense?

I'm trying to have a general sense of the current state-of-the-art technology, as the Wikipedia articles and tables seems to be very dated.

Edit: I meant "dense" as in respect to mass. I believe that "specific energy" would be the right terminology.

This has a list of common battery types with energy densities.

Lithium ion primary batteries appear to have the highest specific energy at around 300 Wh/kg. Lead-acid batteries have a specific energy around 10% that of lithium-ion batteries, but are extremely reliable (depth of discharge, self-discharge, etc.), and do not have nearly the same heating/fire issues (though they can explode if you mess with them), which is one of the reasons they have endured so long in automotive applications.

More exotic chemistries have higher specific energies, but I am unsure as to whether they are available to the average consumer.


Why does chocolate melt so easily? (Yes, I know the question sounds super-trivial... but bear with me here)

Most chocolates (especially milk-chocolate) tend to melt into this sticky (albeit delicious) mess at temperatures slightly above room temperature. Why does this happen?

From my experience (as a high-school student), most organic solids (at room temperature) tend to either sublime or decompose at moderate temperatures; by "moderate", I mean a temperature range between 25 degrees Celsius (typical room temp) and 50 degrees Celsius (a very hot day in Arizona).

Now why do I find this particular property (ease of melting) of chocolate intriguing all of a sudden? Because chocolate is an organic cocktail of all sorts of compounds such as sugars, lipids, proteins, alkaloids, blah blah blah; yet in the moderate temperature range, it doesn't sublime nor does it decompose (chill the molten chocolate, and you'll still get chocolate)... it first softens and then melts, unlike most other organic substances in that temperature range.

So, what makes chocolate so special? And why does milk chocolate (tend to) melt faster than dark chocolate (higher-cocoa content).

tl;dr The main structural component of chocolate is cocoa butter, which is a blend of fatty acids (primarily oleic, palmitic, and stearic acids). Cocoa butter has multiple crystal structures, and manufacturers target a specific form which melts at around 33 °C. The fact that chocolate is about 90% sugar and cocoa butter is why chocolate is (chemically) insensitive to temperature changes.

While chocolate itself is a giant mess of hundreds of different compounds, the primary structural material of chocolate is fat. Wikipedia lists a typical composition (by mass) of about 60% sugar, 30% fats, and 10% proteins/other. So it's not terribly surprising that chocolate doesn't sublime or decompose. You probably wouldn't expect a ball of butter + sugar to suddenly decompose at 50 °C, and for the most part, chocolate is chemically pretty similar.

Now, what's interesting about cocoa butter is that it can form multiple crystal structures, and it's actually a bit tricky to get the right crystal structure out. You can read about the process in this RSC article if you'd like. The one that most chocolatiers are after is polymorph V, which is glossy and melts just below body temperature.

However, a different polymorph, polymorph, VI is the most stable form of cocoa butter, and it has a melting point that's a bit higher. It also tends to have a faint white crust on it (known as "fat bloom"). As a result, it's not as pleasant to look at, not as pleasant to eat (since it takes some chewing on to get it to melt), and generally undesirable.

Since form VI is the most stable, all chocolate eventually goes "bad" and becomes not as delicious as the maker originally intended. If you've ever bitten into an old chocolate bar and thought it was a bit waxy and unmelty, you've experienced this interconversion. I've been told by someone who did research on crystal structure kinetics that chocolate manufacturers spend lots and lots of money on research in hopes of being able to slow down or stop this process.

Now, I'd like to take stab at addressing two points in the question:

  chill the molten chocolate, and you'll still get chocolate

Yes, but is it the exact same chocolate as you got before? What if you take the molten chocolate and put it in the freezer, or, heaven forbid, drop it in liquid nitrogen? Does the resulting chocolate still melt at the same temperature? My suspicion (based on this chart) is that you'll have generated some of the lower-melting cocoa forms, and thus the chocolate will get soft and melt at a much lower temperature than previously.

  And why does milk chocolate (tend to) melt faster than dark chocolate (higher-cocoa content)?

I don't know. If I were to hazard a guess, I would say it's because milk chocolate tends to contain more sugar than dark chocolate, and so you're seeing some form of freezing-point depression. However, if it turns out that, at similar concentrations of sugar, milk chocolate still melts more quickly, I'd have no idea.


What happens in an exothermic reaction at the atomic level? Exothermic Reactions are those chemical reactions in which heat is released.

How does this happen?

What I mean is where does  the heat energy come from? Which form of energy is converted to heat energy? What makes some reactions release heat? What affects the amount of heat energy released?

Thanks!

What happens in an exothermic reaction at the atomic level?

In a chemical reaction, bonds are broken and bonds are formed. For example, when elemental oxygen and hydrogen form water, we are breaking bonds in the elements and making heteroatomic bonds connecting hydrogen and oxygen atoms:
$$\ce{O2 + 2H2-> 2H2O}$$

To emphasize the bonds, we could also write:

$$\ce{O=O + H-H + H-H -> H-O-H + H-O-H}$$

In a thought experiment, we could first separate the elements into "isolated" atoms, and then recombine them to form water. Bond dissociation energies estimate how much energy it takes to separate bound atoms into "isolated" atoms. In the case of the example reaction, it takes less energy to break all the bonds of the reactants (one double bond connecting oxygen atoms and two single bonds connecting hydrogen atoms) than to break all the bonds of the products (four single bonds between oxygen and hydrogen atoms). As a result, there is an excess of energy that becomes available, resulting in an exothermic reaction.

  What affects the amount of heat energy released?

The reaction follows a more subtle path than completely ripping molecules apart into atoms and then making new molecules. However, because enthalpy is a state function, the result in terms of reaction enthalpy is the same. So the only thing that affects the reaction enthalpy is the set of reactants and products.

The amount of heat depends on how you set up the reaction. If the temperature before and after the reaction is the same, if the pressure is constant and there is no non-PV work, it is equal to the reaction enthalpy. If the reaction is used to do work (electrochemical cell, steam engine etc), if it is a photochemical reaction, if it gives off light, or if any other of the conditions given in the previous sentence are not met, that will affect the amount of heat transferred.

  Which form of energy is converted to heat energy?

The energy "comes from" the different energy of electronic states in the reactants and products. You can call this chemical energy, or a combination of kinetic and potential energy of the electrons. We see that electrons going from a higher energy state to a lower energy state give off energy when considering atomic spectra (sodium giving off a yellow light in the Bunsen burner, for example). In the case of reactions, electronic states are different between reactants and products not because they are in excited states, but because of the different configuration of nuclei giving rise to different electronic states.

  How does this happen?

Unless the reaction gives off energy directly (by emitting electromagnetic waves - rather rare), the available energy gets transformed into kinetic energy of the nuclei (e.g. molecular vibration), and then is transferred to the surrounding by the common ways of heat transfer until the system is at thermal equilibrium again. If the reaction is endothermic, some of the activation energy (provided by molecular collisions) is not "returned to the pool" but instead used to get the electrons into their higher energy states in the products (compared to the reactants), leading to a net heat transfer from surroundings to the chemical reaction.


What's the difference between pharmacotherapy and chemotherapy Is it fair to say that pharmacotherapy (the use of pharmaceuticals to treat disease) is a subset of chemotherapy (the use of chemicals to treat disease)?

If not, why? Aside from the fact the former involves a subset of available chemicals and the latter all chemicals, what is the difference between the two?

I would like to believe that the major difference between these terms is based on their backgrounds and time however nowadays there are used interchangibly, without paying much attention to actual definitions.
Background

The term chemotherapy was coined by Ehrlich at the beginning of the
20th century to describe the use of synthetic chemicals to destroy
infective agents.  In recent years, the definition of the term has
been broadened to include antibiotics-substances produced by some
microorganisms (or by pharmaceutical chemists) that kill or inhibit
the growth of other microorganisms. It can also be  broadened to
include agents that kill or inhibit the growth of cancer cells.

Chemotherapy is the term originally used to describe the use of drugs that are 'selectively toxic' to invading microorganisms while having minimal effects on the host.
The term also embraces the use of drugs that target tumours and, in fact, has now come to be associated specifically with that branch of pharmacology (Oncology). Although, in the public mind at least, chemotherapy is usually associated with cytotoxic anticancer drugs that cause unwanted effects such as loss of hair, nausea and vomiting.
Pharmacotherapy is a more recent term which generally means the treatment of diseases through the administation of drugs. However there is more to it, the term also describes a more patient-focussed approach in which pharmacists treat patients through drug management with one or more specific outcomes.
The specific goals of pharmacotherapy (primary therapeutic outcomes include):
• Cure of disease (e.g., bacterial infection).
• Reduction or elimination of symptoms (e.g., pain from cancer).
• Arresting or slowing of the progression of disease (e.g., rheumatoid
arthritis, HIV infection).
• Preventing a disease or symptom (e.g., coronary heart disease).
Other important outcomes of pharmacotherapy include:
• Not complicating or aggravating other existing disease states.
• Avoiding or minimizing adverse effects of treatment.
• Providing cost-effective therapy.
• Maintaining the patient’s quality of life.
A systematic approach is consistently applied to each case. The steps involved in this approach include:

Identifying real or potential drug therapy problems

Determining the desired therapeutic outcome(s)

Evaluating therapeutic alternatives

Hope this helps
References

RANG AND DALE’S Pharmacology 8th edition: Basic principles of antimicrobial chemotherapy

Pharmacotherapy Casebook


What workflow can I use on to extract cartesian coordinates of many molecules from both .spartan and .out (Gaussian) files? I'm searching for a workflow for extracting cartesian coordinates for multiple moelcules from both a single file (e.g., several conformers from one .spartan file or one multiple molecules, each from multiple Gaussian output files). Any insight would be appreciated. I'm using Ubuntu (14- & 16.04), on a machine on which I do not have either Gaussian or Spartan installed.

Thank you kindly

The only tool that I know for reading Spartan files (besides Spartan) is Jmol. With some programming or scripting, you will probably be able to save conformers to other files in a workflow - I don't think there is a pre-existing script.

There are multiple open source tools that read Gaussian output files. I would suggest either:

Open Babel - GUI, command-line and Python
cclib - using Python

For example to extract a bunch of Gaussian files to XYZ Cartesian:

obabel *.log -oxyz -m

The -m flag indicates "multiple" and the -oxyz indicates that you want XYZ-format files. More is in the manual.