Determining structure factors for crystals from lattice and motif I understand that the structure factor for a crystal is given by:

$$F_{hkl} = \sum_r f_{r} e^{2{\pi}i(hx_{r}+ky_{r}+lz_{r})} $$

My textbook (''Structures of crystals'', Glazier) suggests that you can calculate structure factors if you know the lattice type and motif of the crystal, simply by finding the product of the lattice and the motif structure factors.

Can this method be used generally, and if so, do you need to define the motif relative to a centre of symmetry in order to get real structure factors for centrosymmetric crystals ?

Q1) A crystal is a convolution of a lattice and motif, so by the convolution theorem, the fourier transform of the crystal (i.e. your structure factor) is the product of the fourier transform of the lattice and the fourier transform of the motif. That is, this is general.
Q2) If the crystal is centrosymmetric* then the structure factor is real - this is a property of the fourier transform (the imaginary sine terms have opposite signs and equal other and are therefore zero). If your lattice and motif generate a centrosymmetric crystal then your fourier transform will be real. The motif doesn't need to be centrosymmetric to generate a centrosymmetric crystal.
*and the atomic form factors are real, but this isn't relevant.


Non-computational theoretical chemistry Theoretical chemistry is nowadays almost synonymous with computational (and usually quantum) chemistry.
Are there any examples of recent (let's say post 2000) research in chemistry that is theoretical (i.e. non-experimental) but does not use extensive computations (i.e. HPC), does not aim to produce better computational methods, or does not investigate mathematical foundations of computational chemistry methods?
Something I imagine can exist are models or model Hamiltonians in quantum chemistry that are analytically solvable (or with minimal help from computers), or abstract mathematics like topology applied to chemistry.

I think I can answer this in the affirmative. There are articles which find connections between abstract mathematics and chemistry, sometimes even bypassing physics altogether. Of course, these kinds of articles are considerably rarer, but they're sprinkled out there.
I can think of two articles which I'd love to discuss, but I literally do not have the necessary background, it just goes way over my head. The first one is Quantum Interference, Graphs, Walks, and Polynomials, Chem. Rev. 2018, 118, 10, 4887–4911. This is some rather pure graph theory which is not related to cheminformatics. In particular, I find it interesting that the connectivity in azulene is comparatively unusual, and this probably is deeply connected to its unusual photophysical properties. And then there's The Rouse Dynamic Properties of Dendritic Chains: A Graph Theoretical Method, Macromolecules 2017, 50, 10, 4007–4021, more graph theory with a little bit of physics, and whose content eludes me entirely.
Surely some amount of computer assistance was employed in deriving the results for these papers, because there's no reason to give up that tool. However, I don't think this counts as "extensive computations" in the sense you may have had in mind.


What does 'a' stand for in 'a-C' according to research paper lingo? I came across this paper but I cannot figure out what is a-C as it is not defined. The C here is obviously Carbon. Regarding 'a' prefix My best guess is that it stands for amorphous. Is this correct?

a for amorphous is totally correct. Also used in other publications like here.


How are there only 8 structural isomers of C5H11Br? Everywhere I see mentioned, that there are only 8 structural isomers of $\ce{C5H11Br}$. That is, the simple $\ce{C5}$ straight chain, $\ce{C4}$ with substituted methyl and $\ce{C3}$ with two substituted methyl groups, giving us a total of 8 isomers. But what about substituted ethyl groups? Or substituted bromo ethyl group like $\ce{C2H4Br}$? Am I missing something? Also is there a general formula to know the number of such structural isomers of haloalkanes?

You'll probably need to draw all the possible isomers to see it, but with only 5 carbons to work with the ethyl isomers are redundant with the methyl isomers: i.e. 1-bromo-2-ethyl-propane is the same thing as 1-bromo-2-methyl-butane.

Remember we are only talking about structural isomers, and ignoring stereoisomers.


Why does only one reaction occur when so many are possible? Why does only one reaction take place, although so many would be possible? I'm more of a physics and math guy, so simply "accepting" it without a proper explanation is kind of hard for me. For example:
$$\ce{Na2CO3 + HCl -> NaCl + H2O + CO2}.$$

Why can't $\ce{H+}$ combine with $\ce{CO3^2-}$ to form a compound?

I'm going to highschool, and I am great at physics and mathematics, and I dislike chemistry since it's mostly memorizing. Please describe it in terms I can understand. Also, this is something that I genuinely want to understand rather than memorize, it is not related to school.

Your example is not quite the case when multiple reactions are possible. Also, $\ce{H+}$ not only can, but actually does combine with $\ce{CO3^2-}$ to form a new particle: $\ce{HCO3-}$, the hydrocarbonate anion. This is where it would end if you'd use just one tiny droplet of $\ce{HCl}$. Otherwise, that anion would attract another $\ce{H+}$ and form carbonic acid $\ce{H2CO3}$, which is unstable and decomposes into $\ce{H2O}$ and $\ce{CO2}$, to produce the summary equation which you already know.

As for your question in general, in most cases when so many reactions are possible, they indeed occur all at once. It just so happens that they go at different rates, and one of them outruns the others by a wide margin, so you may consider it "the only one" and safely ignore the rest. Or it does not, and you end up with a bucketful of brown goo. Such examples rarely make it to the textbooks, though.

All this is especially big deal in the field of organic chemistry, where all reactions produce more or less "side products". Read the details on just about any reaction, and you'll see these yield figures to the right: 96%, 92%... or maybe less if the authors were less lucky. That's how much of our reactants went down the desired pathway. The rest went someplace else.


Relative reactivity of alkyl and aryl halides in nucleophilic-substitution reactions In case of alkyl halides, the order for relative rates of reactions is $\ce{RI > RBr > RCl}$.
The reason is clear that $\ce{I-}$ is a better lg because of negative charge stabilisation and also $\ce{C-I}$ bond is weaker than $\ce{C-Br}$ bond which in turn is weaker than $\ce{C-Cl}$ because of overlap of orbitals.
But why is it the reverse order in case of aryl halides, $\ce{ArF > ArCl > ArBr > ArI}$?
Why is it so even though $\ce{C-F}$ bond strength is higher and $\ce{F}$ is also not able to stabilise negative charge?

There are three mechanisms to discuss here: $S_N2$, $S_N1$ for $RX$ and nucleophilic aromatic substitution by addition-elimination for $ArX$:
$S_N2$ and $S_N1$:
(Image from https://www.quora.com/Why-steric-hindrance-doesnt-affect-Sn1-reaction)

$S_N2$ is concerted (one step), so this one step is rate determining, or the slowest of the mechanism. In $S_N1$, formation of the carbocation is rate determining. In both rate determining steps, the leaving group is being kicked off. This means the easier it is for the $C-X$ bond to break, the faster the reaction goes. The bond enthalpy decreases as size of halogen increases, giving the order you stated above.
Aromatic addition-elimination:
(Image from https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/#seven)

In this case, step 1 is rate determining, which does not involve the leaving group departing. The halogen, however, can speed up the reaction by stabilising the negative charge formed during this step. The stronger the inductive effect of the halogen - which outweighs the +M mesomeric effect - the more it can draw the negative charge from the ring, stabilising it. The bond strength is irrelevant here, as the step during which it is cleaved is faster than the previous step regardless, and is always 'waiting around' for the first step. This gives the reverse order to the previous one.


Can drinking a lot of water be fatal? I vaguely recall having heard that drinking too much water can, over time, prove fatal to the human body.

Nothing special about the water; not distilled or de-ionized or anything … just plain ol' water.

Now the reason that accompanied this "fact", was that drinking too much water serves to dilute, and thereby, disrupt the electrolyte balance in all (living) body tissue. This in turn, messes with the sodium-potassium pump present in cells leading to loss of functionality of cells.

Now I can't seem to find the source of that information, but I do recall it wasn't particularly trustworthy.

So what I want to know is:

Is it true, that drinking a lot of water can be fatal?
If so, is it because the electrolyte concentration inside and outside cells are "diluted"?
Does this affect all (living) tissue equally?**

Edit:

I fail to see why this was put on hold as a personal medical question. I vote to reopen this on the grounds that it does not:

a) Ask for advice on treating water intoxication

b) Suggest, promote or request, in any way, alternative forms of medicine.

Based on what I gathered from this Wikipedia article, Yes.

Drinking copious amounts of water can prove fatal. The proper term is "Water intoxication".

When you start taking in a lot of water (by "a lot" I mean more water than your body can excrete via sweat or urine), the interstitial fluid that bathe the cells that form your (living) tissue end up getting "diluted", i.e- the concentration of ions like $\mathrm{Na^{+}}$ is greatly reduced. The result? The concentration of ions in the interstitial fluid is far lower than the concentration of ions inside the cells it surrounds. Thinking of this in another way; the concentration of water in the interstitial fluid is far higher than the concentration of water inside the cells.

Ever soaked dry raisins in water? Over time they begin to absorb water and swell up as a consequence of the concentration gradient set up between the tissue in the raisin (Low concentration of water) and the water in the bowl. This phenomenon, where water flows from a region of higher concentration (of water) to a region of lower concentration (of water) across a semi-permeable membrane is called osmosis.

It's pretty much the same thing in the case of water intoxication. A concentration gradient has been set up between cells and the interstitial fluid that surrounds it. So water begins to flow into the cell, and the cell swells up quite a bit, resulting in a build-up of turgor pressure

This increase in turgor pressure will be seen to varying degrees in all sorts of (living) tissue, especially tissue that is highly vascular. An important example would be neural tissue, particularly that in the brain. This swelling up of tissue results in an increase in intracranial pressure which can lead to loss of functionality over time. This is what makes water intoxication so lethal.

There isn't really much about the chemistry of water that's responsible for this, apart from osmosis that is.

Addendum:

Drinking a ton of water isn't the only way for water to build up in potentially lethal amounts in the body. Hyponatremia, which refers to a condition of low (below-normal) sodium levels in body fluids, can also lead to water accumulating in the body in excess.


What allows all 3 states of H₂O to be present at the same time? How is it possible for there to be a glacier floating on water but under clouds? Depending on the temperature, shouldn't there only be one state of water?

At the triple-point of a substance, all three phases can be present. Since it takes additional energy to go from ice at 0 C to liquid water at 0 C, there can be a considerable overlap for the two substances, e.g. 1 g ice with 99 g water or 99 g ice with 1 g water, but for thermodynamic equilibrium, the triple point is very narrow.

BTW, the first well-know American chemist, Josiah Willard Gibbs, pioneered in defining the relationship of temperature, pressure and state. See  Gibbs' phase rule for more information.


Ethyl alcohols use on polycarbonate and acrylics I have an interesting question about the use of ethyl alcohol on polycarbonate windshields.  I have had friends asking me about the use of Pledge Multi-Surface cleaner on their windshields. Some insist it will degrade, yellow, or even crack and soften their windshields. Others have been using it for years with no apparent effect.

Pledge does have a small amount of ethyl alcohol, but does it have enough to degrade and ruin motorcycle windshields or does ethanol do any damage anyway?

Pledges' MSDS states ethyl alcohol weight at 0.01-1.00  (http://www.scjohnson.ca/pdfViewer.aspx?pkMSDSId=367&language=en).

So what are the effects, if any, on using this cleaner on polycarbonate windshields?

Thanks

Not able to comment....

I think the main problem is propyl alcohol (not present in this cleaner). Propyl alcohol (propanol) is present in a lot of cleaning solutions and known to damage polycarbonate and acrylics.

Check this chemical resistance sheet:

Sheet

30% solution containing ethanol (ethyl alcohol) lightly attacks the plastic. However, a 10% (propanol) propyl alcohol already does the same damage. That is a factor 3!


How to write a wave function for infinite potential well with different width than from 0 to a? I am working on the perturbation theory problem with the infinite potential well. However, I am not sure how to write a wavefunction for different than the infinite potential well from $0$ to $a$ and in my problem the well is from $0$ to $2a$.

Is it necessary to solve the potential well all over again? Maybe it is possible just to apply the transformation as $a\rightarrow2a$, for the initial wavefunction

$$\psi(x)=\sqrt\frac{2}{a}\sin\left(\frac{n \pi x}{a}\right)$$ ?

Well, yes; the original length $a$ is just a symbol and you can replace it with whatever other quantity you wish. The relevant wavefunctions are thus just

$$\psi_n = \sqrt{\frac{1}{a}}\sin\left(\frac{n\pi x}{2a}\right)$$

You can verify that these wavefunctions are still normalised correctly by explicit integration.


What is the mechanism for the creation of P(CH2OH)4+ cation? I know that the initial reactants are phosphine and menthanal in an acid ($\ce{H+ A-}$) and that the first step in the reaction is the HOMO of phosphine (the lone pairs) and the LUMO of methanal (the π* orbital).

Then, the electrons that just moved to the oxygen atom react with $\ce{H3O+}$, so the oxygen is protonated. I'm not sure what follows, though.

The acid serves as a catalyst to protonate the formaldehyde oxygen.  This makes the carbonyl carbon more electrophilic, a more reactive target for the weakly nucleophilic lone pair of electrons on phosphorous.  Here is a diagram of the basic reaction mechanism.

The proton transfer is thought to be intramolecular according to this link.


pH Curves- Anomalies In class, the other day, we were discussing pH curves. Assume that I have HCl and NaOH of concentration 1 mol/mL. Now, consider that I have 50mL of HCl, and begin adding small quantities of NaOH. I observe that the pH increases, and once the volume added of NaOH reaches 50mL, the pH of the resulting solution becomes 7.

However, when mathematically calculating, if I added 49.99999 mL of NaOH, I find that the pH is way over 7 (the concentration of H+ becomes 50-49.99999 divided by 99.99999), which seems to be an anomaly. In essence, when finding the limit of - log (H+) as [H+] approaches 0, it needs to equal 7, but it doesn't. Does this mean that there should be a horizontal asymptote at pH = 7? Can somebody please explain this?

You are neglecting the autoionization of water.

The autoionization of water has a $K_\text{eq}$ of $10^{-14}$ at room temperature and hence contributes $10^{-7}$M $\ce{H+}$ to solution, which is small (and thus negligible) when we deal with "regular" quantities of acid and base. Since we are not dealing with regular quantities of acid and base here, we must take into account the protons generated from autoionization; the addition of $10^{-7}$M $\ce{H+}$ will resolve any anomalies in pH calculations.


Hygroscopic Substances and Drying Agents My book says:

  Most of the hygroscopic substances are drying agents

Based on what I read on several websites, I think dehydrating agents also act as drying agents. If so, all dehydrating agents would also be drying agents and that would contradict my book's statement. So, is there another type of hygroscopic substances or is the book wrong, or is my information or logic wrong?

P.S. I don't think the tags I've used are suitable. I just put the following tags as Wikipedia said that hygroscopy is caused due to either absorption or adsorption. I would appreciate if anyone could help me out.

Read closer:

  Most of the hygroscopic substances are drying agents

not

  all drying agents are hygroscopic substances.

First  Terminology:
Hygroscopic: Absorbs water from air
Drying agent: Removes water from organic substance

Your book did not exclude other materials from being considered drying agents. It just said many of the substances that are hygroscopic are also useful as drying agents.

Take for example anhydrous calcium chloride calcium sulfate hemihydrate (also called plaster or DrieriteTM) and phosphorous pentoxide. All of these are hygroscopic (sequester water from air)

$$\begin{align}
\ce{CaCl2 (s) + H2O (fl) & -> CaCl2 (aq)}\\
\ce{CaSO4.\tfrac12 H2O (s) + H2O (fl)&  -> CaSO4.2H2O (s)}\\
\ce{P2O5 (s) + H2O (fl) &-> HPO3 (cd)}\
\end{align}$$

Calcium chloride is deliquescent (hygriscopic to the point that it forms a liquid) and calcium sulfate hemihydrate is hygroscopic (stays solid but still absorbs water) and both can be used as drying agents for most solvents. Phosphorous pentoxide on the other hand cannot be used as a drying agent for some materials such as alcohols and styrene.

$$\ce{P2O5 (s) + EtOH -> EtH2PO4 (cd) + HEt2PO4 (cd)}$$
$$\ce{PhCHCH2 ->[P2O5] [PhCHCH2]_n \tag{explosive}}$$


What is three dimensional aromaticity? I have recently come across a statement which states:

  The relative stability of the 1,3-dehydro-5,7-adamantanediyl dication is ascribed to its three-dimensional aromaticity. [1]

My understanding of aromatic nature is, that the molecule must be cyclic and planar, every atom in the ring must be $\ce{sp^2}$-hybridized and the molecule must have [4n+2] $\pi$ electrons, where $n\in\mathbb{N}$ (Hückel’s rule). The molecular orbitals also have to have the correct orientation and phase for aromaticity (2).

In my understanding 1,3-dehydro-5,7-adamantanediyl dication does not follow these rules for Y-aromaticity (3).

So my question is: What is three-dimensional aromaticity and how does it stabilise the 1,3-dehydro-5,7-adamantanediyl dication?

References

Firme, C. L.; Antunes, O. A. C.; Esteves, P. M. Electronic nature of the aromatic adamantanediyl ions and its analogues. J. Braz. Chem. Soc. 2008, 19 (1), 140–149. DOI: 10.1590/S0103-50532008000100020.
LibreTexts
What is Y-aromaticity? Is the trinitromethanide anion aromatic?

Aromaticity in Organic chemistry textbooks covers the Huckel version, which abide by the rules you have correctly listed above. However, these rules are mainly a tool to identify an organic molecule which has a high delocalization of electrons across a ring structure (a valuable tool in identifying a molecule's stability/reactivity/etc...).

So, for 3-dimensional aromaticity, the molecule must reach a certain threshold of delocalization in the ring structure (which, in 3D, becomes more of a sphere). Rules such as trigonal planar hybridization in 3D are no longer useful, and thus different approaches are taken to measure the resonance state of a system, and thus its level of aromaticity.

Nozawa et al. (2016, cited below) discusses use of "[t]he bond length alternation (BLA) in cyclic compounds" as "a good indicator of aromaticity, as it allows an evaluation of the degree of effective $\pi$-electron delocalization." These values, using a "harmonic oscillator model of aromaticity (HOMA)," are close to 1 for aromatic molecules by definition, and Nozawa obtains these from a variety of experimental and computational techniques to compare aromaticity in normally anti-aromatic structures.

For your above structure, as some comments pointed out as well, they are mesomeric, or resonance structures of the molecule illustrating the ability for the electrons to delocalize to an aromatic state. Firme et al's article mentions "delocalization indexes among bridged atoms," what I assume to be yet another similar measure of aromaticity related to the BLA using HOMA quantification.

Hope this helps! Please let me know if there is any confusion, as this response is a bit packed with summary detail.

References:
1. Nozawa, R. et al. Stacked antiaromatic porphyrins. Nature Communications 7, 13620 (2016).


An automated approach to visualize several xyz structures I have several .xyz files (each composed of several molecules) and I'd like to find an automated way to visualize and save them. I'd appreciate any guidance.

You can use VMD to do this.
Note that the following is only tested on Linux with VMD version 1.9.2, i don't know how it works out on Windows or Apple OS.
The following is from a script that i used to render a trajectory. The first thing to do is to load your xyz file as trajectory. I.e. in a format so that vmd allows you to "slide" through the structures in its normal view. Then execute the script in the tlc console. The script iterates over the steps in incremenets of 2 and assumes that you have 2000 structures. The iteration variable is i. The actual command that changes the frames is the animate goto $i command. You would need to adjust that to fit your requirement. I used the tachyion renderer and saved the output figures in tga format. You will most likely have to adjust the path to the renderer binary which is the path /software/vmd-1.9.2/lib/vmd/tachyon_LINUXAMD64 in my case. VMD saves first a .dat file that contains the scene and then applies the renderer to this dat file. This is why i am setting two path variables, first fin for the .dat file and then fout for the actual tga file.
for {set i 0} {$i < 2001} {incr i 2} {
animate goto $i
set fin [format "pic_%04d.dat" $i]
set fout [format "pic_%04d.tga" $i]
render Tachyon $fin "/software/vmd-1.9.2/lib/vmd/tachyon_LINUXAMD64" -aasamples 12 $fin -format TARGA -o $fout -res 1920 1080
}

The nice part about this is that you can implement other commands during your loop, like a zoom or a rotation. All you have to do, is find out what the tcl command in vmd is.


Synthesis of benzophenone Would his be a valid synthesis? I'm just unsure about the oxidative clevage step. I know that there is a benzylic hydrogen. So should oxidative clevage work here? What happens to the fragment that is lopped off? Is it also oxidized?

Your basic strategy to involve a Friedel-Crafts acylation is correct, but your route is a bit circuitous.  You could acylate benzene directly to produce benzaldehyde in one step using carbon monoxide and hydrochloric acid to generate formyl chloride in situ.

While it is "just another" Friedel-Crafts acylation, it has it's own name, the Gatterman-Koch reaction, after its discoverers.  Once in hand, you could oxidize benzaldehyde to benzoic acid.

An alternate approach would be to 1) brominate benzene to form bromobenzene, 2) form the Grignard reagent with magnesium and 3) react it with carbon dioxide to produce benzoic acid.

In either case, with benzoic acid in hand, you can procede as you've indicated, on to benzophenone.


Attack of electrophile on toluene Does hyperconjugation in toluene influence the attack of an electrophile in any way?

If not, where would the electrophile prefer to attack on toluene?

Looking at related answers, it seems like hyperconjugation leads to delocalisation of electrons over the ring, like in resonance. But hyperconjugation is the interaction between $\sigma$ and $π$ orbitals, how does it lead to delocalisation over the ring? I cannot understand it in terms of the molecular orbital theory.

Lastly, would the electrophile prefer to attack at Ortho position due to greater $+I$ effect and (if possible) hyperconjugation effect? Please correct me if I'm missing something.

sorry that was my mistake. the correct mechanism is this


Strange "far out" orbitals in HF and DFT calculations. Why do they occur? I calculated some orbitals and I get quite strange results with orbital geometries I wouldn't expect from MO theory. A more experienced colleague just said: "That's always the case with big basis sets, just ignore those and search for the ones that make sense" which is a very unsatisfying answer and he couldn't explain what's the cause or what they mean. I also noticed that these occur more often/with lower energy when using HF but they also occur using DFT.

Can anyone explain what happens here and why?

Here is one example with the first two looking like I would expec.

The first 5 unoccupied orbitals using HF (6-311+G(d,p)) including energies in eV:

As already mentioned in the comments you have to be careful when interpreting the meaning of the virtual orbitals in a ab-initio or DFT calculation. In the calculation the only orbitals that are optimized are the occupied orbitals while the virtual (unoccupied) orbitals give you a guess of the shape you expect from them if they were occupied. To look at it from a simple perspective, somehow you are using the LCAO method, you obtain as many MO as AO you are using in your basis set, but you only optimize the MO that are occupied.

There is a second point that may be relevant to your question. In many programs the default orbitals you get in your calculation are the canonical orbitals (delocalized orbitals) that even if occupied may not have an evident chemical interpretation on its own. For those porpouses it may be better to use localized orbitals (e.g. NBO) that may be easier to interpret.

Two book recommendations:

C. J. Cramer, Essentials of Computational Chemistry. Theories and
Models, Wiley (2004).
Frank Weinhold, Clark R. LandisDiscovering chemistry with natural
bond orbitals, Wiley, 2012


Bromine with 10 electrons On a test question, my organic chemistry professor gave us a really big molecule and told us to give list the formal charges on each atom. There was a $\ce{Br}$ with 3 lone pairs and a double bond to carbon (which then had a single bond to a $
\ce{H}$ and a $\ce{N}$). On the key, he labeled it with a + charge and I was wondering how is that possible (or if it is a mistake)? Shouldn't a $\ce{Br}$ with that many electrons be highly negative or does it go onto the next orbital allowing it to accommodate for this?

I suspect that the issue here is with counting the number of lone pairs.
I think the structure you are referring to looks something like the right hand one shown below.

As you can see, this structure can be described by the two resonance forms shown. The double bond arises because the bromine 'donates' one of its lone pairs to form another bond to the carbon, shifting the positive charge onto the more electronegative bromine and hence stabilising the ione. Since this is the 'more stable' resonance contributor it is usually shown as the structure of the ion in drawings but the actual structure is more accurately represented as a resonance hybrid of the two.


Inorganic chemistry handbook with reaction lists There is a handy handbook on common inorganic substances' properties called "Reactivity of Inorganic Substances" (eds. R. A. Lidin, V. A. Molochko, and L. L. Andreeva). It was originally written in Russian and then translated into English (see the review here). The entries in this handbook go like this (sorry for the Russian, couldn't find the English version):

You see that the compound's name is followed by the brief summary of its physical properties (color, solubility in water), general notes on its stability under ambient atmosphere, and then the list of the chemical reactions which were considered by the authors to be the most important or illustrative for this particular substance.

The question is - are there any reputable handbooks that are similar to this one, with not only the summary of the physical properties, but the summary of possible chemical reactions as well, or is this book one-of-a-kind?

An Intro in Russian edition of Reactivity of Inorganic Substances by Lidin et al. claims its uniqueness:

Настоящее пособие носит информационно-справочный характер, не имеет аналогов в отечественной и зарубежной химической литературе, охватывает все разделы неорганической химии, изучаемые в химических вузах.
(English) This reference handbook has no analogues among domestic and foreign chemical literature, covers all inorganic chemistry sections studied in university chemistry course.

However, Turova's Inorganic Chemistry in Tables comes pretty close.
It doesn't only list reactions per element groups, but also arranges them in visual blocks; available both in Russian and English [1, 2]:

References

Турова, Н. Я. Неорганическая Химия в Таблицах; Высший химический колледж РАН: Москва, 1997.
Turova, N. Inorganic Chemistry in Tables; Springer: Heidelberg ; New York, 2011. ISBN 978-3-642-20486-9.


Are these two space groups in a group–subgroup relationship? Given two crystal space groups, how can one determine if they are in a group-subgroup relationship? The specific case at hand is P 21/m 2/m 2/a (aka P m m a, #51) and P m m 2 (#25), but knowing how to solve this question in a general manner will sure prove helpful in the future as well…

The Pmma is a minimal non-isomorphic supergroup of Pmm2; or you can say that Pmm2 is one of maximal non-isomorphic sub-groups of Pmma.

How to solve the question:

(1) fast and dirty: take International Tables for Crystallography vol. A go to your group (in mine edition Pmma starts on p.274). On second page describing space group you can find Maximal non-isomorphic subgroups. The third listed is [2] Pmm2 (25) 1; 2; 7; 8.

You can see that the operation in  #50 and #25 have a lot of similarities. No.1 you can ignore as it's [x,y,z] but 2, 7 and 8 for #50 are [-x+.5,-y,z], [x,-y,z] and [-x+.5,y,z]. As one can see the #25 has following general operators: (2) [-x,-y,z] (3) [x,-y,z] and (4) [-x,y,z].

(2) The hard way: Mathematics is explain in the same volume in chapter 13.1. Isomorphic subgroups. Of course the knowledge of group theory and matrix calculus is to your advantage ;-)


How is dissolving a physical change and not a chemical change? When salt dissolves in water it dissociates and it is no longer in a lattice structure. The electrostatic forces are no longer there and are replaced by water-ion interactions. To me that sounds like old bonds breaking and new bonds forming. Shouldn't the reaction look something like this? $$\ce{NaCl + H2O -> Na+ + Cl- + H2O}$$

This is indeed a bit of a grey area. Consider the following: chemical or physical?

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

There is a tendency in more advanced chemistry to view the formation / breaking of ionic lattices as a physical change, because there are no differences in the electron shells of the atoms. Ions are considered as independent species. As such we could consider the sodium and nitrate irrelevant and reduce the above to simply

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

Note that I have shown the charges all the way through, to indicate that there are no covalent, dative, or metallic bonds made or broken.

Reactions such as these are definitely "chemical":

$$\ce{Na + \frac{1}{2}Cl2 -> Na+ Cl-}$$
(metallic bonds in $\ce{Na}$ and covalent bonds in $\ce{Cl2}$ broken, electron transferred from $\ce{Na}$ to $\ce{Cl}$.)

$$\ce{C2H4 + H2 -> C2H6}$$
(double bond of $\ce{C2H4}$ opened up, bond of $\ce{H2}$ broken, new covalent bonds formed.)

$$\ce{SOCl2 + 2H2O -> H2SO3 + 2HCl}$$ (covalent bonds broken, new bonds made.)

This last reaction sets up another point. It is usually done with excess water, so the products (which are themselves acids) will then dissolve in water, thereby protonating it:

$$\ce{H2SO3 + H2O -> H3O+ + SO3H-}$$
$$\ce{HCl + H2O -> H3O+ + Cl-}$$

These particular dissolutions are more likely to be seen as a chemical change than the dissolution of $\ce{Na+Cl-}$, because there is a rearrangement of covalent bonds (even though it's only a $\ce{H+}$ that moves from one molecule to the other.)

Finally, $\ce{H2SO3}$ has another equilibrium, again involving covalent bonds being broken and made, and therefore a chemical change:
$$\ce{H2SO3 -> H2O + SO2 (g)}$$

Conclusion

Making or breaking of metallic, covalent or dative bonds, or the formation/destruction of new ions or molecules, is always considered a chemical change.

The simple dissolution of an ionic salt, where none of the above mentioned bonds are made or broken, and no new ions or molecules are formed/destroyed, is frequently considered a physical change.

Note the care taken to always write $\ce{Na+Cl-}$ (and never $\ce{NaCl}$) in this answer.


Are Grignards Chiral? Can Grignards be chiral? I've seen them drawn as ionic structures in class. I've also seen little asterisks on certain positions of Grignards (a common designation of chirality). So this begs the question - are Grignards chiral?

From what I understand Grignards are best described as polar covalent rather than ionic, but that still leaves one more question: can ionic compounds be chiral?

I read the question as asking whether a stereocenter bearing a C-Mg bond will keep its configuration. The answer is yes, but not indefinitely. Unfunctionalized (alkane) Grignards of this type racemize/epimerize with a half life on the order of hours. Specific compounds will have more or less stability. By analogy to organolithium compounds, I would expect having Lewis basic atoms in the proximity to stabilize the configuration. I'm not sure if this is still an active field of research, but it certainly was for a long time. Try searching "configurational stability of Grignard reagents" or similar search terms.

There's an additional complication with these chiral Grignard reagents: they can't be prepared by the usual method of oxidative addition of Mg(0) to an alkyl halide. The mechanism of that reaction goes by a single electron transfer, through a radical intermediate, destroying any stereocenter. Chiral Grignards have to be synthesized using less direct methods.

Ref: Side Reactions in Organic Synthesis: A Guide to Successful Synthesis Design, pp. 200-202.


Are there any descent symmetry tables for character tables of linear structures? I know about tables containing information about descending in symmetry, i.e. from larger point  groups to their sub-groups.

We can see such tables in this answer, or in this PDF.

My question is - does any similar table exist even for $C_{\infty v}$ and $D_{\infty h}$ point groups?

Here they are, finally I found them in some random presentation for a lecture:

$$\begin{array}{c|c} \hline
D_{\infty \mathrm h} & D_\mathrm{2h} \\ \hline
\mathrm{\Sigma_g^+} & \mathrm{A_g} \\
\mathrm{\Sigma_g^-} & \mathrm{B_{1g}} \\
\mathrm{\Pi_g} & \mathrm{B_{2g} + B_{3g}} \\
\mathrm{\Delta_g} & \mathrm{A_{g} + B_{1g}} \\
\mathrm{\Sigma_u^+} & \mathrm{B_{1u}} \\
\mathrm{\Sigma_u^-} & \mathrm{A_u} \\
\mathrm{\Pi_u} & \mathrm{B_{2u} + B_{3u}} \\
\mathrm{\Delta_u} & \mathrm{A_{u} + B_{1u}} \\ \hline
\end{array}$$

$$\begin{array}{c|c} \hline
C_{\infty \mathrm v} & C_\mathrm{2v} \\ \hline
\mathrm{A_1 = \Sigma^+} & \mathrm{A_1} \\
\mathrm{A_2 = \Sigma^-} & \mathrm{A_2} \\
\mathrm{E_1 = \Pi} & \mathrm{B_1 + B_2}\\
\mathrm{E_2 = \Delta} & \mathrm{A_1 + A_2} \\ \hline
\end{array}$$


What does it mean for an equillibrium to move? Le Chatelier's principle was described to me as:

  If a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

I do not understand the phrase, "...the position of the equilibrium moves to counteract the change."

How can the equilibrium "move?" I know that the equilibrium constant cannot change so that cannot be it.

Let's keep this simple. Think about adding vinegar (acetic acid) to water.

$$\ce{CH3COOH + H2O <=> H3O^+ + CH3COO^-}$$

The equilibrium equation is

$$\ce{K_a} = \dfrac{\ce{[H^+][CH3COO^-]}}{\ce{[CH3COOH]}}$$

Le Chatelier's principle simply says that if I add more acetic acid to the water then the solution gets more acidic. However if I add more water, then the solution gets less acidic. Hence the equilibrium moves.(In this case it works to a point. Adding a drop of water to glacial acetic acid isn't an aqueous solution...)


Least likely product of a hydrocarbon cracking reaction? The hydrocarbon $\ce{C17H36}$ can be cracked. Which compound is the least likely to be produced in this reaction?
A: $\ce{C3H8}$ B: $\ce{C4H8}$ C: $\ce{C8H16}$ D: $\ce{C16H34}$

Is it D because cracking's purpose is the breaking of long hydrocarbon chains into smaller ones?

There are two ways to look at this question.  One is simple carbon atom counts.  Another is the saturation of the products.

Carbon atom counts.  The question doesn't specify which of the 24894 isomers of heptadecane is to be cracked.   Suppose it is "normal" or n-heptadecane.  Cracking usually begins with the homolytic cleavage of a carbon-carbon bond.  To make a $\ce{C16}$ product from n-heptadecane by cleaving carbon-carbon bonds, there are only two potential bonds to cleave, one at one end of the molecule, another at the other end.  In either case, to obtain a $\ce{C16}$ product, only a single cleavage would be allowed, because further reactions would break more C-C bonds and change the product.  Cracking is a high temperature reaction that is entropically driven.  The system is increasing the disorder.  Imagine rolling four 6-sided dice, and the total comes up 17.  Then you take away either one or two of the dice (whether its one or two is random).  What is the probability that the remaining dice sum to 16, compared to say 8 or 4?

Saturation.  Choices B and C are unsaturated hydrocarbons.  Choices A and D are saturated hydrocarbons.  Both saturated and unsaturated compounds can be formed by cracking, but most modern cracking processes tend to focus on making unsaturated hydrocarbons.

Thus, primarily because of simple statistical unlikelihood, but also because of saturation considerations, D is the correct answer.

Wikipedia's article on cracking is very helpful.  See here.$%edit$


Changing the order of the reaction with change in concentration The reaction

$$\ce{2 A + B + C -> D + 2 E}$$

is found to follow the rate law as

$$r = k[\ce{A}][\ce{B}]^2[\ce{C}]^0$$

If the concentration of $\ce{A}$, $\ce{B}$ and $\ce{C}$ increases two times then the rate of reaction becomes eight times higher:

$$r' = k\cdot 2[\ce{A}]\left(2[\ce{B}]\right)^2\left(2[\ce{C}]\right)^0 = 8k[\ce{A}][\ce{B}]^2$$

Note: there is no change in the order of reaction. I want to change the order of this reaction by changing the concentration of reactants. Can I do this?

Because if we double the concentration of $\ce{A}$, $\ce{B}$ or $\ce{C}$, it does not change the order. But if we increase  the concentration of $\ce{A}$ by 4 times, then we can change order by

$$\left(4[\ce{A}]\right)^2 = \left(2[\ce{A}]\right)^{2+2} = \left(2[\ce{A}]\right)^4$$

You would change the reaction speed, but not the order.

The order of the reaction can be changed by reagent concentration, if it is result of multiple reaction steps with different orders. By variation of concentrations you can manipulate the relative reaction speed,
as the order is often ruled by the slowest reaction.

Let suppose the 2 step reaction
$$\begin{align}
\ce{2A + B &-> C} \\
\ce{A + A &->[slow] D} \\
\ce{B + D &->[fast] C} \\
\end{align}$$

For small $c_A$, the reaction speed
$$\frac{dc_C}{dt}=k_{low A}.{c_A}^2$$.

For high $c_A$, the first reaction becomes faster then the second.

$$\frac{dc_C}{dt}=k_{high A}.{c_A}^2.c_B$$.

You can also change the effective order of reaction, if some reactant is in about constant abundance and is not explicitly considered.

E.g. The speed of the reversed esterification
$$\ce{R1COOR2 + H2O -> R1COOH + R2OH}$$ is the reaction kinetics of the 2nd order. But in abundance of water, it is effectively a reaction of the 1st order:
$$\frac{dc_{\ce{R1COOR2}}}{dt}=-k_1.c_{\ce{R1COOR2}}$$.


Would the End Point and Equivalence Point be Labelled the Same? I created two graphs of two different titrations : One graph was the titration curve of acetic acid with sodium hydroxide (weak acid-strong base) and the other was that of nitric acid with sodium hydroxide (strong acid-strong base). I've marked the equivalence point of the acetic acid curve to be at a pH of ~8.5, while the nitric acid curve has a pH of 7.

Knowing this, where would I label the endpoint of the titration curves? In the same general location? On most of the photos I've researched, people don't tend to label the end point but it's required for this assignment. For the titration, I didn't use an color indicator like methyl red or anything but rather a pH meter. If anyone can give an idea, that would be greatly appreciated.

The endpoint and equivalence point are different.

According to this Wikipedia article , "The endpoint (related to, but not the same as the equivalence point) refers to the point at which the indicator changes colour in a colourimetric titration."

Essentially, the endpoint is when the titration visually ends in an experimental setting, but the equivalence point is when the titration actually ends, or when the moles of acid are equilalent to the moles of base.

Since you didn't use an indicator, you didn't have an endpoint.

Also, if you did use an indicator, you wouldn't really know what pH the titration ended at because you don't have a pH meter to measure the pH with.

Unless you had both a pH meter AND indicator (which seems kind of useless), you shouldn't have to label the endpoint on any titration curve.


The oxidation of aldehydes and alpha-diketones with peroxy compounds Recently, I have been reading up on the Baeyer-Villiger (BV) oxidation. The oxidation is most commonly discussed for ketones. However, the oxidation also works for aldehydes and $\alpha$-diketones (Rojas, 2015).

The above image is Scheme 2.2 from Rojas (2015).

I have mainly two queries:

Does the oxidation of aldehydes to carboxylic acids by peroxy compounds, such as peracids, proceed via a BV oxidation where the hydride is the migrating group? This would be seemingly different from the oxidation mechanism for the oxidation by $\ce {MnO4^-}$ or $\ce {CrO3}$ where the aldehyde is oxidised when it is in the gem-diol form.
Why does the oxidation of $\alpha$-diketones yield the corresponding anhydride and not an $\alpha$-ketoester? This is extremely puzzling because the acyl group is terribly bad at stabilising any partial positive charges on the acyl carbon. How is it possible that such a BV oxidation can occur?

Reference

Rojas, C. M. (2015). Molecular Rearrangements in Organic Synthesis (1st ed.). Hoboken, New Jersey: John Wiley & Sons, Inc.

The Baeyer-Villiger oxidation is conducted under acidic conditions such as an alkyl peroxide in the presence of a mineral acid (e.g.; H2SO4) or a peracid (e.g.; peracetic acid or m-chloroperbenzoic acid). In the reactions described in the diagram, the reactions of aldehyde 1 or α-diketone 7 are initiated by proton of the carbonyl oxygen followed by addition of peroxide (an alkylperoxide is illustrated). Ultimately, the oxygen of the RO-moiety is protonated and is the effective leaving group (ROH) in the rearrangement. In both sequences, the red route is favored; the blue route disfavored. Cation 3 is a more stable species than cation 5 because of hyperconjugation with the alkyl group or resonance stabilization if R'=aryl. Cation 3 affords a carboxylic acid while cation 4 would lead to formate ester 6.
In the case of peroxyhemiketal 8, acyl migration leads to cation 9 which is more stable than species 11, which has a positive charge adjacent to a carbonyl group, which is destabilizing. Thus, anhydride 10 is formed in preference to α-ketoester 12.


Solubility of silver acetate in the presence of nitric acid What will be the effect on the solubility of $0.1\ \mathrm{mol}$ of $\ce{CH3COOAg}$ in presence of $1\ \mathrm{l}$ of $0.1~\mathrm{M}\ \ce{HNO3}$ solution?

  $$K_\mathrm{sp} (\ce{CH3COOAg}) = 10^{-8}\\
K_\mathrm{a} (\ce{CH3COOH}) = 10^{-5}$$

I know that silver acetate will react with nitric acid to form silver nitrate and acetic acid. So the solubility of $\ce{CH3COOAg}$ will increase. But in the end, I am not able to find out the final solubility (as concentration of $\ce{H+}$ and $\ce{NO3-}$ is coming different).

Maybe I have an error somewhere?

The process you are trying to investigate is:

$$\ce{AgOOCCH3(s) + HNO3(aq) <=> AgNO3 (aq) + CH3COOH (Aq)}$$

with the net ionic equation:

$$\ce{AgOOCCH3(s) + H+(aq) <=> Ag+(aq) + CH3COOH(aq)}$$

The equilibrium constant is

$$K_c = \dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]}$$

Note that this is law of mass action can be derived from the $K_a$ and the K_{sp}$ you have:

$$\begin{aligned}
K_{a}&=\dfrac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]}\\
K_{sp}&=[\ce{Ag+}][\ce{CH3COO-}]\\
K_c &= \dfrac{K_{sp}}{K_a}=\dfrac{[\ce{Ag+}][\ce{CH3COOH}]}{[\ce{H+}]}
\end{aligned}$$

The problem is now an ICE (Initial-change-equilibrium) problem.

$$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{H+}] & [\ce{Ag+}] & [\ce{CH3COOH}]\\ \hline
I & 0.1\ \mathrm{M} & 0\ \mathrm{M} & 0\ \mathrm{M}\\
C & -x & +x & +x \\
E & 0.1 -x \ \mathrm{M} & x \ \mathrm{M} & x \ \mathrm{M}\\ \hline
\end{array}$$

Plug values into the law of mass action, and solve for $x$, the concentration of $\ce{Ag+}$, and then you can determine the solubility.


What are some general tests that would determine an unknown compound? I know a little bit about unknown compounds but I am still new to the topic here. I know that you could use filtration, chromatography and distillation. You can also combine the unknown with others to form a compound. And you can measure properties and such and such to figure out what the compound is.

If I were to do this, what should I watch out for, and what would be the best way to proceed? I would appreciate it if you don't go "PhD" on me.

This question will be ambiguous for most people, but I am looking for a general idea and what you would normally do, instead of "what I would do in this situation".

As usual, it depends on what is available.

Mass spectrometry, $^{1}\ce{H}$ and $^{13}\ce{C}$ NMR, IR, UV/VIS spectroscopy, would be obvious choices in organic chemistry nowadays. XRD of a suitable crystal would be nice to have!

Back in the days, (undergrad) laboratories didn't have sophisticated analytical instruments and you were happy to find a melting point apparatus, a refractometer and an IR spectrometer.

You would have used IR spectroscopy to identify funcional groups, possibly supported your hypothesis by using staining reagents for TLC, like Dragendorff's reagent and then started to synthesize derivatives from which you knew that they show sharp, well-defined melting points.

You would have converted

amines to sulfonamides
carbohydrates to osazones
ketones and aldehydes to 2,4-dinitrophenylhydrazones or semicarbazones

and compared the melting points of the derivatives with tabulated values. Those were the days…


Crystal field splitting in complexes other than octahedral or tetrahedral Is there a methodology to developing the crystal field splitting diagram for various crystal fields?  I know the octahedral and tetrahedral diagrams, but if I'm asked to draw something like square pyramidal, trigonal prismatic, etc. that I haven't memorized, then is there a process I can go through?  I searched Google, but am having little luck finding anything that shows the process.

Thank you!

Yes, the methodology is the same that is used to derive the crystal field theory a priori. Your course should have taught you that when it introduced crystal field theory, otherwise it is not worth its money.

Crystal field theory assumes that the ligands will approach the central metal in a certain manner and that these ligands will be point-shaped negative charges. Before the ligands approach, all orbitals of the metal’s same subshell will be degenerate, i.e. have the same energy. Because these orbitals have an orientation in space (e.g. the $\mathrm d_{z^2}$ orbital can be seen as ‘pointing’ its lobes in the direction of all the coordinate axes), the approach of the ligands will cause interactions between them and the negatively charged electrons that make up the orbitals. Therefore, any orbital that is pointed towards the approaching ligand will be destabilised (and all others stabilised to a certain degree).

In the octahedral case, the ligands approach from $\pm x,\pm y$ and $\pm z$. I have already pointed out that the $\mathrm d_{z^2}$ orbital would be pointing towards them; the $\mathrm d_{x^2-y^2}$ orbital does, too. Therefore, these two are destabilised while the other three orbitals $\mathrm d_{xy}, \mathrm d_{xz}$ and $\mathrm d_{yz}$ are stabilised.

Applying this general principle to other geometries is left as an exercise to the reader.


What makes ice slippery? I've heard and read (multiple times) two theories have have been around for a while, both of them obvious nonsense.

One states that pressure and friction make the ice melt (turn to water) where the skater's blade comes in contact with it. This is patent nonsense since a puck that weighs very little, or a pencil that weighs next to nothing, slip along the ice as well as any skater. I also find it hard to believe that the temperature of the ice (several degrees below freezing) would suddenly jump several degrees in the split second that it takes for the skate to come in and out of contact with it.

The other theory is that the ice surface itself has the same qualities as water due to some hydrogen molecules still in search of home. This is also nonsense: pour water over a concrete floor and try to skate on it; see how far you'll get.

So my question is: why is ice slippery?

This is a question has been one of interest since 1850, and I found a paper that nicely summarizes some of the hypotheses that you yourself listed, as well as what is considered to be true. I've paraphrased it below.

  Michael Faraday was the first to propose that there exists a film of water on ice that freezes when in contact with other pieces of ice, but remains liquid by itself. Though he performed several experiments that supported his claims, most of which involved sticking two pieces of ice together, his ideas were largely dismissed until C. Gurney also suggested that this film of water on play a role in ice's slipperiness.

  Gurney hypothesized that molecules, inherently unstable at the surface due to the lack of other molecules above them, migrate into the bulk of the solid until the surface becomes unstable, which prompts the formation of the liquid phase. This came to be accepted, and Charles Hosler compiled very compelling evidence on the the forces required to pull ice spheres apart. He discovered that at vapor pressures below ice saturation, no adhesion occurred below $-4\:^{\circ} \mathrm{C}$. From this they inferred that the expected roughness of the surfaces was removed by the presence of a liquid film whose thickness was sufficient to provide a smooth surface of contact.

  More experiments were performed trying to determine the exact thickness of the water film at different temperatures, and the phenomenon was found to occur at temperatures even below $-22\:^{\circ} \mathrm{C}$.

  Here is a depiction of what the surface of ice looks like at a molecular level.

Rosenberg, R. Why Is Ice Slippery? Phys. Today Physics Today. 2005, 58, 50–54.


Stereochemistry + isomer check We start with compound M:

[Please imagine green bonds as dotted bonds going into the plane]
From compound M, we are told that the 3 membered ring is opened u using sodium azide yielding a mix of isomers N and O.
I drew the following:

and

[Again, please imagine the green bonds as dotted bonds and the blue
bonds as wedge bonds]
Given are the following two molecules:

Are my solutions the same as the ones given? If they aren't, where did I go wrong?
Many thanks!

First of all, for both products, you have not given the O atom in the carbonyl position. In organic chemistry, we assume that there is a C atom if no atom is mentioned specifically. Now for this reaction, the carbonyl group will be there as it is.
Now we can consider the reaction. The epoxide is away from the plane, i.e. below the plane. For the steric reason, it is more feasible for azide to attack from the upper side of the plane. Now again, azide can attack both C centre of the 3-membered epoxide ring. In that case, we'll be having two different products. But in both cases, we will have the azide group on the above of the plane and O atom in the below of the page.
Now if I assume that you have mistakenly skipped the O atom in your both products, then your first molecule matches with the first answer given below. But  the second answer given by you is not correct and doesn't match with the second answer. You can have the green and blue coloured bond in the same place and then you have to alternate the position of OH group and azide group. Then that will match the answer given below and the answer will be correct.


What is the difference between radical initiator and radical inhibitor? I am confused between these two concepts. My book gives $\ce{O2}$ as an example of a radical inhibitor and peroxide as an example of an initiator. But I don’t know why this is so? Both have $\ce{O-O}$ bonds so what is the difference?

Also aren’t all radical initiators in some sense? Aren’t radicals in general reactive and that they always create more radicals?

The confusing thing about initiators and inhibitors of radical reactions is that both are often radicals.

But it isn't being a radical that is important: it is what happens when that radical reacts in the reaction conditions. If the radical reacts with another radical to give something that is no longer capable of propagating the radical reaction, it is an inhibitor. If it reacts to give more radicals it is an initiator.

Which compounds can do the job of either depends a lot on the specific reaction.


Why NaBH4 does not reduce carboxylic acid while LiAlH4 does? My teacher told - "To reduce carboxylic acid  to alcohol, $\ce{LiAlH4}$ is used. But since $\ce{LiAlH4}$ is an expensive reagent, it is commercially reduced by forming an ester and then to alcohol". I asked that we can also use $\ce{NaBH4}$ as it also contains 4 hydrogen bonds. He said that $\ce{LiAlH4}$ is a strong reducing agent that $\ce{NaBH4}$. I was not satisfied by his answer. I want an answer in terms of a reaction mechanism.

The answer is polarity.

Consider the two ions $\ce{AlH4-}$ and $\ce{BH4-}$ (as they exist as ions in a solution).

Now the electro-negativity of $\ce{Al}$ is $1.5$, and that of $\ce{B}$ is $2$, and that of $\ce{H}$ is $2.1$.

So the electro-negativity difference ($\Delta\mathrm{EN}$) between $\ce{Al-H}$ bond is $0.6$ and that of $\ce{B-H}$ bond is $0.1$. It is clearly visible that the $\Delta\mathrm{EN}$ of the $\ce{Al-H}$ bond is greater than that of the $\ce{B-H}$ bond.

Now considering the mechanism, due to the small polarity induced by the carbonyl group of the carboxylic acid, the $\ce{H}$ atom in $\ce{Al-H}$ bond is readily broken due to its high polarity, while the $\ce{H}$ in $\ce{B-H}$ can't do the same because it has a low polarity.

This is the reason why $\ce{NaBH4}$ cannot be used instead of $\ce{LiAlH4}$ for reducing carboxylic acids to alcohols.


Explanation of terminology in equation used for energy calculation using Hartree–Fock method I found in the literature that the energy of the Hartree–Fock method is given by the following equation:
$$E_\mathrm{HF} = \int (\Psi^* \hat{H}\Psi)\,\mathrm d\tau$$
The term $\Psi$ is the wave function, but what would the term $\Psi^*$ be?
Does anyone have a simple of explanation of what this may be?

For every $x$ and $t,$ $\Psi(x,t)$ is a complex number. $\Psi^*$ is the conjugate of that number, no more, no less. The reason it seems like sometimes it's only the $t$ part that gets conjugated is simply that often it is the only part of the wavefunction that is complex. Let's use an example:
$$\Psi = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{-\mathrm iE_n t}.$$
We want to calculate $\Psi^*$. Well, since the conjugate of the product of two numbers is the product of their conjugates (that is, $(zw)^* = z^* w^*),$ let's do it step by step.
First we need to conjugate $\displaystyle\sqrt{\frac{2}{a}}$, but since it's a real number, it is equal to its conjugate. So we leave it alone and move on. Now we need to conjugate $\displaystyle\sin\left(\frac{n\pi x}{a}\right),$ but again, this is a real number, because $\sin x$ is real whenever $x$ is real. The last part is $\mathrm e^{-\mathrm iE_n t}$. This is actually complex, so we need to conjugate it, and its conjugate is $\mathrm e^{\mathrm iE_n t}.$ So putting it all together, we have
$$\Psi^* = \sqrt{\frac{2}{a}}\sin\left(\frac{n\pi x}{a}\right)\mathrm e^{\mathrm iE_n t}.$$


Stereoelectronic effects in carbamate carbonyl oxygen attack on halide 2-Bromoethyl (2-bromoethyl)carbamate can react through either Path A or Path B to form salt 1 or salt 2. In both instances the carbonyl oxygen functions as the
nucleophile in an intramolecular alkylation. What is the preferred reaction
path for the transformation in question?

I would go with A, but I don't know how to explain it.

Path A resembles a 5-enolexo-exo-tet reaction which is allowed by Baldwin rules while path B represents a 5-enolendo-exo-tet reaction which is disfavored. The orbital interactions of path B suffer severe geometrical constraints (even in the conformation where the C-Br bond is pointing towards the viewer when looking at the attached image). Path A is presented in the attached image.


Radial wave-functions and spherical harmonics of p-orbitals in a certain hypothesis I know that the s-orbitals of H atom are spherically symmetrical. But the p,d or f orbitals aren't. So, the H-atom needn't necessarily be spherically symmetrical, as the wave-function would be a linear combination of all the orbitals. But, in a certain hypothesis, if I assume the H-atom to be spherically symmetrical, what could I possibly comment on the specific orientations of $p_{x}$, $p_{y}$ and $p_{z}$ orbitals in space? The radial and angular wave-functions might/mightn't change for the p-orbitals, but how to comment on their spatial orientations? Can someone help me out in this?

In "real life", the hydrogen atom has its only electron in the s orbital. And unless it gets to an excitated state, that's the only thing that matters (whether you're studying the energy of the atom or the bounds it can make in terms of energy or symmetry).
Nevertheless, the H atom is likely to have a spherical symmetry, so any of its p-orbitals will be orthogonal to the others, so their linear combination is likely to keep the overall spherical(-ish) symmetry of the atom.
Here is a representation of such orbitals :

(Reference)
Hope this helps!


A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water. What is the total number of atoms in the compound A sample of copper(II) sulfate pentahydrate (CuSO4·5H2O) contains 0.360 g of water.
What is the total number of atoms in the compound

what's the idea of this question?

what i know is that atoms number is found by multiplying the Avogadro number by the mol's number

??

You have to calculate the amount of substance with the mass of water.

$n[H_2O]=\frac{m}{M}=\frac{0.360\,g}{18\,g} \cdot mol = 0.02\, mol$

$1~ n[CuSO_4\cdot 5 H_2O] = 5~n[H_2O]$

$1~ n[CuSO_4\cdot 5 H_2O] = 0.004\, mol$

In 0.004 mol copper sulfate pentahydrate you have 0.36 g water.

1 molecule of copper sulfate pentahydrate is composed of 21 atoms and the Avogadro constant is $N_A = 6.022 \cdot 10^{23}\, mol^{-1}$.

Now you can simply multiply all values.

$\text{#}Atoms = 0.004\, mol \cdot 6.022 \cdot 10^{23}\, mol^{-1} \cdot 21 = \underline{5.06 \cdot 10^{22}}$


Formaldehyde (37%) storage Should Formaldehyde (37%) be stored at 5 degrees C or lower?
Will it damage the Formaldehyde?

In the cold some paraformaldehyde (formaldehyde polymer) or trioxymethane (a cyclic condensation product of three formaldehydes) may precipitate.
It will depend on how much methanol has been added to avoid the polymerization products.
Normally it will contain 10 to 15% methanol.

In reality, it is not necessary to store formaldehyde solution in the cold.


Calculating amount of base needed to achieve a specific pH in a mixed ethanol-water solvent I'm looking for some way to calculate how much base I will need to add to an unbuffered solvent consisting of a mixture of ethanol and water in order to basify it to a specific reading (10) on an a pH probe calibrated with an aqueous standard.

I have several mixtures of ethanol and water, ranging from about 85% ethanol to nearly 100% ethanol, and I plan to basify these solutions.  Using the method defined by ASTM D6423, one of these solutions was measured to have a pHe of 5.2 due to the presence of some acidic species, but I don't expect these species to act as buffers.

Now, I'm aiming to get these solutions to a pHe of approximately 10, and in practice I'm just going to add the base and check the pHe as I go.  However, motivated by the fact that I'll potentially need to do this on quite a large scale at some point, I'm curious if it's possible to put a mathematical backing behind this.

If this were a pure water solvent, I would calculate $[\ce{OH^-}]=10^{-8.8}$, since the self ionization constant of water is $K_W = 10^{-14}$ and $K_W \approx [\ce{OH^-}][\ce{H^+}]$.  From there it's relatively simple to calculate the amount of base I should need to add to reach $[\ce{OH^-}] = 10^{-4}$, resulting in a pH of 10.

If this were a pure ethanol solvent, I would do something similar using the self ionization constant for ethanol, except nobody has published one as far as I can tell.  This answer also failed to find one.  The Wikipedia page on the leveling effect depicts ethanol having a wider acid-base discrimination window than water, so I would expect $K_{EtOH}$ to be somewhat higher than $K_W$, but no numbers are given.  I checked the source of the graphic as well, and they also give no numbers.

Finding $K_\ce{EtOH}$ wouldn't even fully solve the problem though, as there is a non-negligible amount of water in the solvent.  The only source I was able to find discussing pH in mixed-solvent system was this one, which discusses using a correction factor ($\delta$) to allow the measurement of something analogous to pH in mixed solvent systems using an electrode calibrated in an aqueous standard.  However they only determined this correction factor in solutions of up to about 72% ethanol, and the relationship is decidedly non-linear, as shown by this plot:

We know that when our probe reads 10 we'll get good results with the rest of the process, so presumably this means in the math I would want to target a pH of $10+\delta$, were $\delta$ known.

This is as far as I've been able to get with my own literature search.  If anyone here has any insight into this problem I would greatly appreciate it!

EDIT: Additional detail.  I just had some 200 proof ethanol measured by the ASTM D6423 method, and the pHe was measured as 6.5.  This is counter to what I expected.  I expected it to have a pHe greater than 7 since pure ethanol should be a somewhat weaker acid than water based on everything I've read.  I'm not certain if this is helpful information at all, but it's another data point.

I'm looking for some way to calculate how much base I will need to add to an unbuffered solvent consisting of a mixture of ethanol and water in order to basify it to a specific reading (10) on an a pH probe calibrated with an aqueous standard.

I think you have summarized the main issue yourself. The reading of a pH probe in a basic alcoholic solution is just a positive real number. It has no significance whatsoever. The value of 10 means nothing. The pH probe has no idea that it is being dipped in an aqueous organic solution. I am sure the probe will take a very long time to equilibrate to give a stable "real number" output. Since those numbers such as pHe =10 have no theoretical basis it is hard to calculate anything. This is confirmed by a quick look at the standard method.

The ASTM D6423 method says " Readings are taken at exactly 30 s because the meter reading will change throughout the analysis due to glass-electrode voltage drift caused by dehydration of the glass-electrode membrane. Because alcohol dehydrates the glass-electrode membrane, time spent soaking in alcohol solution should be minimized.
4.3 The electrode is soaked in water-based pH 7 buffer between readings to prepare it for the next sample. This rehydrates the glass electrode, a necessary step to preserve the electrode's response characteristics.
4.4 The electrodes are cleaned/rehydrated at least every ten samples by alternately soaking several times in 1 mol/L NaOH solution and 1 mol/L H2SO4 (or 1 mol/L HCl) for about 30 s each. New electrodes are treated by this procedure before first use."

The practical way to "calculate" perhaps is to do a acid base titration. Calculate the titer for each batch with a suitable indicator and add the amount of base accordingly.


7,7-dichlorobicyclo[4.1.0]heptane NMR solving i made a synthesis of this material in the lab and now i got the H-NMR specta , and by litreture i know that i got the right product.
im having trouble to match the hydrogens to the peaks,i tried to look at the molecule in it chair form, i know that the 2 hydrogens(A) attached to the cl cl group are in the highest ppm than the other 4 (B), but on the last 4 i got 2 different multiplets, which is a match to which hydrogens? i added the  spectrum

I would start by looking at the cis rather than trans isomer as the starting point for interpreting the NMR spectrum. It is much more likely to be the product you got from this reaction. Also, be sure to identify all the chemically distinct hydrogens in your structures.


Naming of asymmetric organic molecules with two nitroxides When writing my diploma thesis I had to name my molecules following IUPAC nomenclature rules. There were several molecules like the following one, that contained two TEMPO groups, but were not symmetrical. The following is one example:

While I was able to find IUPAC recommendations on how to name radical organic molecules I could not find sufficient information for this case and ACD/Labs was no help either. I went ahead and replaced the two TEMPO groups with COOH, as this is a high ranking moiety, got a name for it and replaced the corresponding parts of the name with what I found (carboxy to ylooxy; carboxylic acid to oxidanyl) in the recommendations.

This was the name I came up with:

  (4-(4'-((adamantan-1-yl)((1-ylooxy-2,2,6,6-tetramethylpiperidin-4-yl)carbamoyl)carbamoyl)-N-((adamantan-1-yl)carbamoyl)-[1,1'-biphenyl]-4-carboxamido)-2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl

In the meantime, the results have been published but I always wondered, if this was correct. I checked on SciFinder (CAS No. 1947339-77-6) and an index name has still not yet been assigned.

Is there a more elegant way of finding a name for a molecule like this?

The structure can be redrawn somewhat more clearly:

Let's call the basic structure, where R1 = R2 = H, ‘the skeleton’.

If your name was, after fixing some enclosing parentheses to square or curly brackets, even the ‘preferred IUPAC name’, it's not a nice one, because it does not reflect the symmetry of the skeleton, and repeats the TEMPO and adamantane building block names.

(From now on, I'm gonna use the blockquote for IUPAC organic nomenclature rules 2013 generated names, or the book citations, if section name included.)

"Almost preferred" IUPAC name, substituted carbamoyl biphenyl dicarboxamide

The skeleton is named

  N4,N4′-dicarbamoyl[1,1′-biphenyl]-4,4′-dicarboxamide.

This preferred name is obviously based on the biphenyl dicarboxylic acid diamide

  [1,1′-biphenyl]-4,4′-dicarboxamide

substituted with carbamoyl ($\ce{NH2-CO\bond{-}}$) groups at each nitrogen.

Simple monosubstituted derivatives of the skeleton (there are only two kinds of them initially) are named

  N4,N4′-dicarbamoyl-N4-methyl[1,1′-biphenyl]-4,4′-dicarboxamide

and

  N4-carbamoyl-N4′-(methylcarbamoyl)[1,1′-biphenyl]-4,4′-dicarboxamide.

R2 (“TEMPO-4-yl”) can be named,

  2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl (group)

according to the rules of making the radical to be prefix (e.g. when there is yet more prioritized radical):

  P-71.5 Prefixes denoting radicals

  (…)

  Examples:

  (…)

  $\ce{-O^.}$

  oxyl (preselected prefix)
  ylooxidanyl
    (not ylohydroxy)

  (…)

  P-71.7.d.1 (…)

  (3-oxylbenzoyl)oxyl (PIN)
  [3-(ylooxidanyl)benzoyl]oxidanyl

however the rules do not make a good distinction between residues (prefixes) and radicals.

So the resulting name would be
N4-(adamantan-1-yl)-N4′-[(adamantan-1-yl)carbamoyl]-N4′-(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)-N4-[(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)carbamoyl][1,1′-biphenyl]-4,4′-dicarboxamide. But it still redundantly repeats some building block names.

Unfinished IUPAC name based on substituted "diallophanylated biphenyl"

  P-66.1.6.1.1.4 Carboxylic acid derivatives of urea

  Two carboxylic acids are related to urea; have been known as ‘allophanic acid’ for $\ce{H2N-CO—NH-COOH}$ and ‘hydantoic acid’ for $\ce{H2N-CO-NH-CH2-COOH}$. These names are no longer recommended. Preferred IUPAC names for these two acids and their derivatives are formed systematically.

  Examples:

  $\ce{H2N-CO—NH-COOH}$
  carbamoylcarbamic acid (PIN)
  (aminocarbonyl)carbamic acid

  $\ce{H2N-CO—NH-CO\bond{-}}$
  carbamoylcarbamoyl (preferred prefix)
  [(aminocarbonyl)amino]carbonyl

The allophanoyl- prefix can be found in the older IUPAC 1979 nomenclature, rule C-971.

The skeleton could be named

4,4′-diallophanoyl[1,1′-biphenyl], or
4,4′-bis(carbamoylcarbamoyl)[1,1′-biphenyl],

if you want.

But now we would have to repeat the two differently substituted allophanoyls in the name. Or derive a name from the symmetrical one, but how to number it?

Urea based IUPAC name

Closer look at your structure reveals that it in fact is two N,N′ asymmetrically substituted urea

  P-66.1.6.1.1 Urea and its substitutive derivatives

  P-66.1.6.1.1.1 The compound $\ce{H2N-CO-NH2}$ has the retained named ‘urea’, Which is the preferred IUPAC name, With locants N and N′, as shown above the structure below. The systematic name is ‘carbonic diamide’. The locants 1, 2, and 3 have been used in the past and may be used in general nomenclature.

  $$\begin{array}{ccc}
\ce{H2\overset{$N$}{\underset{1}{N}}-\underset{2}{C}O-\overset{$N'$}{\underset{3}{N}}H2} & \hskip{3em} &
\ce{H2\overset{$N$}{N}-CO-\overset{$N'$}{N}H2} \\
\text{urea (PIN)} & &
\text{carbonic diamide}
\end{array}
$$

  $$\bbox[0.5em,border:1px solid black]{\text{Numerical locants for urea are no longer used in the IUPAC preferred names.}}$$

molecules, N,the-other-one's-N′-interconnected with

  [1,1′-biphenyl]-4,4′-diyldicarbonyl (group).

And that's a correct name, used e.g. in

  2,2′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)dipropanoic acid.

(If these propanoic acids were interconnected by different carbons, 2,3′ (there's no other option), such multiplicative name would be no longer a preferred one; see e.g. P-15.3.2.1.1. But never mind, we're making a nice one.)

The skeleton now can be named ([1,1′-biphenyl]-4,4′-diyldicarbonyl)diurea. (More precisely N,N′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)diurea. Or N,N′′-…? And what about such multiplicative names of asymmetrical derivatives? There's no definitive answer, but this would be for another question.)

So we'll use the other urea numbering, the numerical, non-preferred one.

Now we get a nice name:
1,3′-([1,1′-biphenyl]-4,4′-diyldicarbonyl)bis[1-(adamantan-1-yl)-3-(2,2,6,6-tetramethyl-1-oxylpiperidine-4-yl)urea]

(But I would like similarly short and non-redundant name that would end like “…bis[(2,2,6,6-tetramethylpiperidin-1-yl)oxidanyl]” yet more.)


Atomic mass ratios - Is the problem providing enough info? One gram of Hydrogen reacts with exactly (almost) 8 grams of oxygen to produce $\ce{H2O}$. Another single gram of Hydrogen reacts with 16 grams of Oxygen to produce $\ce{H2O2}$.

Can the ratio $\frac{m_O}{m_H}$ be determined from this information? If not, what else do chemists need to know to find this ratio?

Are you asking whether we can figure out the ratio of the mass of an oxygen atom to a hydrogen atom from this information? If so, then the answer is yes, assuming that you know that $\ce{H2O}$ and $\ce{H2O2}$ are the formulas for each compound, and that atoms exist and are indivisible in chemical reactions.

$$2n\space \rm{atoms \space H} = 1 \space g \space H$$
$$1n \space \rm{atoms \space O} = 8 \space g \space O$$

Divide the second equation by the first:

$$\frac{1n \space \rm{atoms \space O}}{2n\space \rm{atoms \space H}} = \frac{8 \space g \space O}{1 \space g \space H}$$

Cross multiply to get a 16:1 mass ratio. You can confirm this by looking at $\ce{H2O2}$, where the ratio of atoms is 1:1 and mass is again 16:1.

If you didn't know the formulae, then you could assume that the ratio was some integer multiple of 8:1, but without more data you wouldn't know which. This is because you would not know whether water had a 1:1 atom (or mole) ratio of $\ce{H}$ to $\ce{O}$, or if it was something else.

Incidentally, what you are describing is the law of multiple proportions, and it is one of the key findings that led to the development of atomic theory.


Is the density of solute same as the density of solution? This problem came up while solving some problems on mass percentage and mole concept.

Essentially, our teacher said that the $w/w\%$ and the $v/v\%$ for a given solution will be the same, because in order to obtain $v/v\%$ there will be a mutual division by density.

However, the density of solute is different from density of solution, right? For example, the problem we were solving involved $\ce{NaOH}$ dissolved in water. But $\ce{NaOH}$ as a compound has a characteristic density, different from the density of the solution. So $w/w$ and $v/v$ should have different values, right?

I'm not sure if I'm able to clarify what I'm thinking, but I need help with this nonetheless.

The density of the solution isn't needed for either wt/wt% nor vol/vol%.

wt% = (grams solute)/(grams solution)

vol% = (vol solute)/(vol solution)

For vol% it only really makes sense if mixing two liquids like ethanol and water.


Is it possible to melt a diamond? I am school student. In my school text book there is a fact about diamond. That is

  Each carbon atom is covalently bonded to four other carbon atoms. A lot of energy is needed to separate the atoms in diamond. This is because covalent bonds are strong, and diamond contains many covalent bonds. This makes diamond's melting point and boiling point very high.

If a diamond is a single molecule, how does a single molecule melt? Is that fact correct? Please tell me the answer!

The melting point of diamond is certainly an exotic topic. Diamond is slightly less stabile than its allotrope, graphite, and will in most circumstances turn to graphite way before its melting point. If you have enough pressure it will stay diamondlike and it will at some point melt.

A diamond is not a single molecule unless you stretch the definition of molecule. It is usually defined as the smallest fundamental unit, and for diamond I'd call the tetragonal carbon unit cell the smallest fundamental unit, or maybe even a carbon atom. Diamond, having one of the highest values for heat transfer among substances would certainly melt almost like a single molecule though. But in theory, one and one carbon decides to leave behind it's friends bound by the diamond structure boundaries behind to seek their fortune in the hippie camp that is called liquid.


What is Y-aromaticity? Is the trinitromethanide anion aromatic? The wikipedia article on trinitromethane claims that:

  There is some evidence that the anion (which obeys the $4n+2$ Hückel rule) is aromatic.

Unfortunately the citation is behind a paywall so I can't read it but my first obvious question is how can the trinitromethane anion obey Hückel's rule? Two of the requirements for the rule are that the molecule must be planar (or nearly planar) and must be cyclic and I fail to see how either of these conditions are satisfied since the anion is roughly trigonal pyramidal around the carbon and definitely not cyclic.

This appears to about Y-aromaticity. All the papers linked are behind paywalls for me so can someone please explain what Y-aromaticity is and how it applies to the example given. I found this previous answer which gives some discussion of Y-aromaticity in nitrate ions. If someone could provide a similar MO analysis of the trinitromethane anion this might be very interesting.

Introduction

It is fairly obvious that the statement given by Wikipedia cannot be entirely correct.

  There is some evidence that the anion (which obeys the 4n+2 Hückel rule) is aromatic.

The trinitromethane anion cannot obey the Hückel (4n+2) rule because it is neither monocyclic nor planar. And assigning a number of π electrons is fuzzy at best, so the (4n+2) part is also questionable. Hückel's rule is a very strict set, but an aromatic compound does not necessarily have to fulfil it. The easiest compound to understand this is naphtalene, which is clearly aromatic, but does not fulfil the criteria of Hückel's rule, i.e. it is not monocyclic. In extension to this, helicene is an aromatic compound that fails two of the given conditions. Another case is thiophene, which due to the included sulfur certainly does not entirely consist of trigonally hybridised (sp2) atoms (see Bent's rule).[1]

But can the anion still be regarded as aromatic? In order to explain that, we need to look at the concept of Y-aromaticity.

What is Y-aromaticity?

The term and the concept is not without controversial discussion. It is an attempt to give the "traditional" concept of resonance energy an new, fancy and more catchy name. It is often used to explain the extraordinary stability of the guanidinium cation, which has tree equivalent resonance structures, i.e. the electronic ground state is heavily delocalised.
Conventional Lewis structures employ two electron two centre bonds only, which therefore cannot represent the highly delocalised electronic structures, e.g. of aromatic compounds. Hence the concept of resonance had to be introduced to account for such stabilising effects. Obviously the same holds true for any compounds, that exhibit excessively delocalised electronic structures.
The term Y-aromaticity has been used for compounds, that are Y-shaped, with a trigonal planar coordinated (carbon) centre. In addition to that the π orbitals hold six electrons.[2] These compounds have a similar orbital picture and could therefore be grouped.

The electronic structure of Y-delocalised π systems

The π orbitals of a trigonal planar coordinated centre can be generalised by the following orbital picture, which was generated for the nitrate anion at the DF-BP86/def2-SVP level of theory.

What we can clearly see is, that there is one totally symmetric bonding orbital. Two degenerate (orthogonal) slightly bonding (or non-bonding) orbitals and another totally symmetric anti-bonding orbital.

The remarkable stabilisation comes from the lowest lying π orbital, which is delocalised over all atoms, just like in the more obvious cases like benzene. Additionally, such compounds often retain their planarity by undergoing substitution reactions, rather than addition reactions.

Pro and contra of Y-aromaticity

The concept itself allows the discussion of certain electronic properties in quite simple terms. It is related enough to the general concept of aromaticity so that it can easily be understood at the same level of teaching. Like aromaticity, it allows predictions - to a certain degree - about the reactivity of such compounds.

Peter Grund makes the following assessment, arguing in favour of the concept of acyclic aromaticity.[3]

  Despite recent trends, it would appear useful to designate systems as "aromatic" if they represent a delocalized system exhibiting peculiar chemical stability, in order to focus attention on such systems even when they are buried in complex structures. Since guanidine and its derivatives appear to possess really exceptional stability, and since their physical and chemical properties appear to be dominated by the tendency to retain the closed-shell, Y-delocalized 6 pi-electron configuration, we suggest that it is profitable to consider such substances as possessing a special type of "aromatic character," despite their acyclic nature.

For a general understanding of this feature it certainly does not hurt to keep it in the back of ones head.[4]

However, the applicability of this concept is limited to a quite small subset of compound (the same holds true for Hückel's rule, too). It therefore seems to be quite a superfluous concept, since it is in a more general sense already covered by resonance stabilisation.
After all, electronic structures are often not well enough described with simple models and in some cases these simple models can lead to wrong assumptions. This is one of these cases, where one can only use this concept as long as its restrictions are well known.

Some people might argue, that the naming itself is the problem and it may lead to confusion. I can partly agree with that. That does not change the fact, that the description in resonance, valence bond and molecular orbital theory terms will produce this result. In general I think the term aromaticity is unfortunate right in the beginnings. The concept has changed and extended quite a few times already. I often see problems when the term "Hückel rule" is used synonymously with aromaticity.

There is no true theoretical and at the same time all general explanation for aromaticity. It is often determined on a case by case basis experimentally. This limitation also has to apply to any derivations of it.

Electronic structure of the trinitromethane anion (TNM⊖)

Trinitromethane is a quite acidic substance, which can in part be attributed to the presence of the well delocalised π system. Cioslowski et. al. have investigated the anion and came to this conclusion. In their paper they also use the term Y-aromaticity.[5]

  Our calculations, however, show that at the HF/6-31G* level, the carbon atom is coplanar with the nitrogens while the oxygens of the nitro groups are rotated out of plane. The result is a structure with $D_3$ symmetry. The $\ce{C(CN)3^-}$  anion is likewise found to be planar, while the $\ce{CF3^-}$ ion exhibits a
  pyramidal geometry. In addition, we observe in $\ce{C(CN)3^-}$ a shortening of the carbon-carbon bond and a lengthening of the carbon-nitrogen bond relative to $\ce{CH(CN)3}$. We also find an analogous shortening of the carbon-nitrogen bond and lengthening of the nitrogen-oxygen bond in $\ce{C(NO2)3^-}$ relative to $\ce{CH(NO2)3}$. In contrast, the $\ce{C-F}$ bond length increases upon going from $\ce{CHF3}$ to $\ce{CF3^-}$. The planarity of the $\ce{C(NO2)3^-}$ and $\ce{C(CN)3^-}$ ions and the changes in bond lengths are obvious consequences of Y-aromaticity in these systems.

Obviously the bonding situation in the trinitromethane anion is a little more complicated than in guanidine, as the nitro group itself cannot be described by a single Lewis structure. The out of plane rotation of these groups further complicate things, but since the perpendicular $C_2$ axis is retained it is still somewhat possible to compare the resulting orbitals to a planar species. Therefore when you look at the orbitals you still find a remarkable resemblance to the picture above. For this purpose I have included the same schematic drawings next to the orbitals that resemble these best.

It is noteworthy, that because of the propeller structure there is no strict separation between σ and π orbitals, since the latter require a horizontal mirror plane, which is absent here.
But that is not all. Try counting the number of π electrons. This is far from trivial and it is in no case that makes sense six.[6] Don't let your brain explode about that fact, we are clearly dealing with a borderline case at this point.

And we have not even started to talk about the possibility of bonds between the oxygens of the nitro group.[7]

Take home message for TNM⊖

The trinitromethane anion has a strongly delocalised electronic system. It is extraordinarily stable due to this. If you accept the concept of Y-aromaticity as valid, then it is still a borderline case, where additional considerations have to be taken into account.

Notes, References and further reading

Bent's Rule

Bent, H. A. Chem. Rev. 1961, 61 (3), 275–311. DOI: 10.1021/cr60211a005.
What is Bent's rule?
Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?

The six π electrons are probably the reason for the statement in the Wikipedia article.
Gund, P. J. Chem. Educ. 1972, 49 (2), 100. DOI: 10.1021/ed049p100.
I have used this concept already various times to explain certain electronic structures. From a personal point of view I find it quite helpful.

Delocalization of pi electrons in nitrate ion
Why is the bond order in the SO₃ molecule 1.33 and not 2?
Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

Cioslowski, J.; Mixon, S. T.; Fleischmann, E. D. J. Am. Chem. Soc. 1991, 113 (13), 4751–4755. DOI: 10.1021/ja00013a007.
There are several possibilities, that can be considered to count electrons towards the number of π electrons.

The probably easiest way would be to flatten the molecule for this thought experiment, then count electrons perpendicular to the plane. This gives you two electrons for the central carbon (sp2), no electrons from the nitrogens (sp2) and two electrons each from the oxygens (sp). That gives you a total of 14 π electrons.
Maybe as simple as this, but somewhat the easy way out would be to disregard all contributions from the nitro groups, leaving only 2 π electrons.
Regard the nitro groups as π-backdonor. Since only one of the intrinsic π orbitals can effectively overlap with the central carbon, each of them would donate two electrons (maybe less, because backdonation). Plus two from the central carbon, makes up to eight electrons. This certainly is a very fuzzy approach.
Count four electrons per oxygen, which is kind of cheeky, but it could work. This would give 26 π electrons.
Write a Lewis structure and count only the double bonds (× 2), which gives a total of eight electrons.
I am almost certain, there are still more ways to count the electrons.

In [5] they analysed the anion with the quantum theory of atoms in molecules. They found bond critical points between the oxygen atoms. How these can be explained and what they mean is certainly a story for another time.


Why are thiolates more nucleophilic than alkoxides? Why is $\ce{RS-}$ (a thiolate) more nucleophilic than $\ce{RO-}$(an alkoxide)?

According to my logic, the larger size of the sulfur atom will allow the negative charge to be dispersed more as compared to the negative charge on the oxygen atom. Thus, shouldn't $\ce{RS-}$ be less nucleophilic as compared to $\ce{RO-}$ ion?

It's true that thiolate ion is more nucleophilic than alkoxide ion. You are going in a right way. The order of nucleophilities is according to polarisability of atom. As we know, the size of atoms goes on increasing down the group.

atomic size: $\ce{S>O}$

Due to increase in size of atoms, the electrons are more loosely bound to nucleus. Hence, the atoms are now easy polarise. As a result, electrons are easily polarisable down a group.

Thus, $\ce{RS-}$ of thiolate ion is more polarisable than $\ce{RO-}$ of alkoxide ion. Therefore, thiolate ion is more nucleophilic than alkoxide ion.

The polarisability of nucleophile is more important in case of nucleophilic substitution reaction where the nucleophile electrons are polarised by the positive charge of electrophile and bond is formed during the reaction.

As nucleophilicity is a kinetic factor. More stronger the nucleophile, greater is its rate constant. The following is relative rate constant of both nucleophile on reaction with iodomethane by using methanol as reference whose rate constant is taken as unity:

$\ce{CH3-O-};~k  =  2\cdot 10^6$

$\ce{CH3-S-};~k  =  1\cdot 10^9$


Nomenclature of organic compounds containing complex side chains How do we decide the sequence of substituents in a organic compound if one of the substituents is complex. Should the complex chain must always be written first while writing the IUPAC name or are there any other rules governing it?

The following IUPAC names got me confused:

7-(1,2-dimethylpentyl)-5-ethyltridecane

AND

5,5-dimethyl-6-(1,1-dimethylbutyl)-6-pentyltridecane

In the first compound the complex substituent is written first although p(pentyl) comes after e(ethyl).

In the second compound, the complex substituent is written later although m(methyl) comes after b(butyl).

Simple prefixes are arranged alphabetically disregarding any multiplicative prefixes. The multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’.
However, the name of a compound substituent is considered to begin with the first letter of its complete name.
For example, ‘1,2-dibromobutyl-’ is considered to begin with ‘d’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.
P-14.5.2 The name of a prefix for a substituent is considered to begin with the first letter of its complete name.

Therefore, the correct alphanumerical order corresponds to the name 7-(1,2-dimethylpentyl)-5-ethyltridecane (not 5-ethyl-7-(1,2-dimethylpentyl)tridecane) since the compound substituent name ‘dimethylpentyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘e’.
(Nevertheless, the preferred IUPAC name is 5-ethyl-7-(3-methylhexan-2-yl)tridecane.)

In your second example, the correct alphanumerical order corresponds to the name 6-(1,1-dimethylbutyl)-5,5-dimethyl-6-pentyltridecane (not 5,5-dimethyl-6-(1,1-dimethylbutyl)-6-pentyltridecane) since the compound substituent name ‘dimethylbutyl’ starts with ‘d’, and ‘d’ is earlier alphabetically than ‘m’.
(Nevertheless, the preferred IUPAC name is 5,5-dimethyl-6-(2-methylpentan-2-yl)-6-pentyltridecane.)


How does increasing temperature reduce the efficiency of solar cells? I've seen several answers and trying to confirm.

1) Does temperature increase the ground state energy of electrons once they reach the positively doped semiconductor, thus reducing the voltage between excited electrons from negatively doped semiconductor and ground state they will reach on positive side.

-This was the most compelling and easy to understand answer for me.

-However I'm not sure if it makes sense that "ground states" would rise with increasing temp.

2) Does temperature increase vibrations in wires and semiconductors, such that excited electrons following voltage difference lose energy in collisions along their path from N-doped to P-doped?

Combination of the two? Am I way off?

I am a molecular biology student with some basics in photochemistry. Somewhat familiar with solar cells and looking to dive deeper on the subject so feel free to reference.

While perhaps better suited to Physics SE, I will quote from Sze's Physics of Semiconductor Devices:

  As the temperature increases, the diffusion lengths in Si and GaAs will increase ... and the minority lifetime increases with temperature. The increase in minority-carrier diffusion length causes an increase in $J_{L}$. However, $V_{oc}$ will rapidly decrease because of the exponential dependence of the saturation current on temperature. The increase in the "softness" (roundness) in the knee of the $I-V$ curve as temperature increases will also degrade the fill factor. Therefore, the overall effect causes a reduction of efficiency as the temperature increases.

Here, $J_{L}$ is the current caused by light generation of electron-hole pairs. So, it goes up. But, the device is still a diode and must follow the standard diode equation, so the open circuit voltage ($V_{oc}$) drops. This then also impacts the power output.

In brief, neither of your proposed reasons is the basis for it. Instead, you have to probe deeper into the physics of semiconductor devices.


Why is activation enthalpy found and not just rate of reaction? What is the benefit of finding the activation enthalpy and not the rate of reaction?

One of the Factors Influencing reaction rate is given below (Source)

  Temperature: Usually conducting a reaction at a higher temperature delivers more energy into the system and increases the reaction rate by causing more collisions between particles, as explained by collision theory. However, the main reason that temperature increases the rate of reaction is that more of the colliding particles will have the necessary activation energy resulting in more successful collisions (when bonds are formed between reactants). The influence of temperature is described by the Arrhenius equation. As a rule of thumb, reaction rates for many reactions double for every 10 degrees Celsius increase in temperature, though the effect of temperature may be very much larger or smaller than this.

Reaction rate , Activation energy are connected to temperature through Arrhenius equation. We are interested in Activation Energy as it gives an idea what temperature the reaction starts. We want to know $E_{a}$ in order to manipulate reaction rate essentially. Catalysts or Inhibitors work essentially on this principle.

Just looking at rate wont help us in varying the rates to our interest.


Nomenclature of tetrahydropyran ring What's the logic behind the following nomenclature: 3,4,5,6-tetrahydro-2H-pyran?

Why not simply 2,3,4,5,6-pentahydropyran? Isn't the hydrogenation of the heterocycle obvious by including hydro- prefix? Specifying 2H-pyran seems to be redundant, unless it carries information about which position was hydrogenated first (which isn't the case).

There are two isomers of pyran, 2H-pyran and 4H-pyran.

In principle, it is possible to describe the corresponding saturated compounds using the prefix ‘hydro’, which yields the systematic names ‘3,4,5,6-tetrahydro-2H-pyran’ and ‘2,3,5,6-tetrahydro-4H-pyran’. However, these two names actually describe the same compound. In this case, the correct choice is ‘3,4,5,6-tetrahydro-2H-pyran’ because ‘2H’ is lower than ‘4H’ (note that indicated hydrogen atoms have priority over ‘hydro’ prefixes for low locants).

All locants are omitted in compounds or substituent groups in which all substitutable positions are completely substituted or modified, for example, by hydro, in the same way. Thus, in this case, the systematic name is simplified to ‘tetrahydro-2H-pyran’. Furthermore, omission of indicated hydrogen is permitted in general nomenclature if no ambiguity would result. Thus, in this case, a correct name is just ‘tetrahydropyran’.

Anyway, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), preferred IUPAC names of saturated heteromonocyclic compounds are formed by the extended Hantzsch–Widman system. Therefore, the preferred IUPAC name of the given compound is ‘oxane’.


Why does the structure of haemoglobin change when the ferrous ion changes to ferric ion? It is said that $\ce{Fe^2+}$ can bind oxygen while $\ce{Fe^3+}$ cannot. Why is that so? $\ce{Fe^3+}$ has an extra electron, it could bind more easily to the oxygen. And how and why does the structure of heme protein change with the ferric ion?

The $\ce{Fe^{II}}$ ion is held in the centre of the heme's porphyrin ring and is also attached to a nitrogen atom on a nearly histidine group which is part of the protein. When there is no oxygen present the Fe ion is in a high spin state. The Fe has two electrons in the $e_g$ orbital and is 5 coordinate. The other 4 bonds are to the porphyrin N atoms making the geometry square pyramidal. In this state the Fe is just a little too big to fit into the porphyrin ring which becomes a distorted, bending towards the histidine. The Fe is 0.07 - 0.08 nm out of the plane and the Fe-N distance 0.22 nm (to the imidazole N on histidine F8). When an oxygen molecule binds the Fe becomes low spin, (the $e_g$ orbitals are now empty) and 6 coordinate, and  has a slightly smaller radius and so fits into the porphyrin ring which becomes planar. As the Fe has moved on binding oxygen, the histidine is also moved and is pulled towards the porphyrin ring by $\approx$ 0.075 nm. This motion is transmitted and amplified through the protein to the other three binding sites which affects their ability to bind. This is called an 'allosteric' interaction as it is 'non-contact' between the sites.
The picture (b) is self explanatory. The structure (a) shows the protein backbone as cylinders and the porphyrins and Fe at atoms. The Fe is yellow.

(Picture from 'Protein Physics' A. Finkelstein & O. Ptitsyn, publ. Academic Press.)