Calculating the ionic polarizability from the Clausius-Mossoti relation The concentration of atoms in a crystal is $N_d = \pu{2.24e28 m-3}$, and the relative dielectric constant is $\varepsilon_{r} = 5.62$. The crystal has a refractive index of $n = 1.5$. Using the Clausius-Mossoti relation,

$$
\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} = \frac{N\alpha}{3\varepsilon_{0}},
$$

I got that

$$
\alpha_\mathrm{i} = \pu{8.26e-40 F m2}.
$$

However, the correct answer is

$$
\alpha_\mathrm{i} = \pu{3.7e-40 F m2}.
$$

Was the refractive index useless info here, or have I missed something?

Yep, you missed quite an important part: Clausius-Mossoti equation in the form you've written is limited  and holds true for the dielectric that consists of individual particles. However, the problem says you are dealing with a crystal, which, by the way, is most likely sodium chloride $\ce{NaCl}$, judging from its dielectric constant and refractive index.

In the first Clausius-Mossoti equation you've written $\alpha$ consists of two main components:

electronic polarizability $\alpha_\mathrm{e}$ caused by the displacement of the electron shell of the atom relative to the nucleus under the action of the field;
ionic polarizability $\alpha_\mathrm{i}$ caused by the displacement of ions with respect to other ions.

For the ionic crystal Clausius-Mossoti equation can be written as

$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-} + \alpha_\mathrm{i})\, ,\label{eq:1}\tag{1}$$

where $\alpha_\mathrm{e^+}$ and $\alpha_\mathrm{e^-}$ are electron polarizabilities of positive and negative ions in crystal lattice. Their sum can be determined using a Lorentz–Lorenz equation, which is derived from the fact that the effect of an external electromagnetic field with relatively high frequencies corresponding to the visible and ultraviolet range of the spectrum results in the displacement of only the electron shells relative to atomic nuclei, while the more massive particles (atoms and ions) do not have time to shift from the occupied positions. Correspondingly, only electronic polarization contributes to the polarization of the medium (e.g. $\alpha_\mathrm{i}\approx0$), and the refractive index is related to the electron polarizability of the particles by the Lorentz-Lorentz formula:

$$\frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}(\alpha_\mathrm{e^+} + \alpha_\mathrm{e^-})\label{eq:2}\tag{2}$$

Subtracting \eqref{eq:2} from \eqref{eq:1}, one eliminates the electronic constituent from the equation:

$$\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2} = \frac{N}{3\varepsilon_0}\alpha_\mathrm{i}\tag{3}$$

and ionic polarizability is actually determined as follows:

$$
\begin{align}
\alpha_\mathrm{i} &= \frac{3\varepsilon_0}{N}\left(\frac{\varepsilon_{r} - 1}{\varepsilon_{r} + 2} - \frac{n^2 - 1}{n^2 + 2}\right)\\
&= \frac{3\cdot\pu{8.85e-12 F m-1}}{\pu{2.24e28 m-3}}\left(\frac{5.62 - 1}{5.62 + 2} - \frac{1.5^2 - 1}{1.5^2 + 2}\right)\\
&= \pu{3.7e-40 F m2}\tag{4}
\end{align}
$$


Forming half equations I'm having trouble finding the ionic and half equations for the following equation:

$\ce{2Sr + O2 -> 2SrO}$

From this I would have to tell which elements are being oxidised and reduced.

This is IGCSE.

$\ce{SrO}$ is an ionic compound comprising $\ce{Sr^2+}$ ion and $\ce{O^2-}$ ion.

The oxidation state of $\ce{Sr}$ changed from $0$ to $+2$, and $\ce{O}$ from $0$ to $-2$.

Therefore, $\ce{Sr}$ is oxidized while $\ce{O2}$ is reduced.

The ionic equation is just that, since there is no individual ion in the equation ($\ce{SrO}$ is a solid!).

Are you really told to find a half equation for a reaction not performed underwater?


Is there a substance I can use to coat a projectile to eliminate friction heat? Let's say that I was going to shoot a can of beer into space from one of those "vacuum cannons." I would expect that the friction with the air around the projectile would cause the whole thing to just melt and burn (explode).

Is there such a substance that, if used to coat the projectile of my contraption, would eliminate the heat problem by basically "consuming the heat in a reaction" on the way out of the atmosphere?

Try using carbon fiber in a uniquely woven shape. Maybe apply a dual layer with the top layer having a unique shape for heat dissipation. There have been a few carbon fiber helix designs that maintained structure and supported weight with a 2000 F acetylene/butane flame directed on it.

Another option is go with layers of powdered graphite. See if you can convert thermochemical properties with blends. Maybe you can add a coolant system for your beer can with the excess energy stored.

If you are really trying to have your beer survive extreme heat,"Thermo-electric Reactions and Currents Between Metals in Fused Salts" by Thomas Andrews is probably a good start for prototype blends. I think google has a free .pdf available through Google books.

Here is a graph that might help with selecting your surface metals, if you ever take this project on seriously:

If nothing works, at least your efforts should land you a pretty hardcore beer-koozie.

Final deciding factors come down to the degree of effort you are wishing to exert and material expense.


Will PAEK polymers hydrolyze in concentrated sulfuric acid? One of my duties at the company where I work is to perform intrinsic viscosity (IV) experiments on PAEK (poly(aryl ether ketone)) polymers. These polymers don't dissolve well in most solvents, so the solvent for these experiments is concentrated sulfuric acid. I've found that if I leave the mixture to dissolve overnight and/or at too high of a temperature, the resulting IV is much lower (approximately half) than if I run the experiment as soon as all the material has dissolved at lower temperatures. The PAEK compounds I'm referencing are ones like PEK, PEEK, and PEKK (the results I've detailed above were seen primarily, maybe exclusively, with PEKK). The structures for these polymers is below.

PEK:

PEEK:

PEKK:

My thought on this was that with extra time and heat, the water in the sulfuric acid could attack the carbon alpha to the ether oxygen, hydrolyzing the polymer and creating shorter chain lengths. One of my colleagues disagrees, but to be honest, I haven't asked her to explain why.

Aromatic sulfonation can also occur when PAEK polymers are dissolved in concentrated sulfuric acid, as documented here, but I wouldn't expect that to have such a large impact on the viscosity (and I would also expect the effect of sulfonic acids to increase viscosity, not decrease it, though I'm less confident about that). My background is organic chemistry, not polymer chemistry/materials science, so I'm not super familiar with how different aspects affect things like viscosity. I do know shorter chains leads to a lower viscosity, though.

So here's my question: Is it likely that a significant amount of hydrolysis is occuring, or is it more likely some other factor causing this effect?

I resist concluding this but nothing else I can think of makes sense. The only way the polymer solution to become less viscous is through a cleavage of bonds reducing the MW of the polymer or dilution. Cleavage at the ether linkages would require reduction in an oxidizing environment or nucleophilic aromatic substitution which this polymer lacks the electron-withdrawing groups to effect, so those mechanisms should be eliminated (I concur with your colleague). Thus we can deduce that the cleavage must be occurring at the carbonyl groups.

To cleave the polymer at the carbonyl groups, the PAEK is likely undergoing an electrophilic aromatic substitution at the carbonyl groups. Though not commonly utilized, it is possible to substitute any electrophile on an aromatic ring in including acyl groups (especially with sulfuric acid which provides a high concentration of $\ce{H+}$ electrophiles) as shown in Figure 1 — essentially Friedel–Crafts acylation in reverse.

Figure 1. Electrophilic Aromatic Substitution Removing an Acyl Group.


Why is a rearrangement observed for the secondary alcohol but not the primary in these examples? I am thinking that maybe only hydrogen shifts are possible and methyl shifts are not allowed in such situation? Is that the reasoning? but then again in the second example the bromide could shift to a secondary position after doing a hydrogen shift.

You need to consider the possible and plausible reaction mechanisms for both substrates. They are both nucleophilic substitutions in first approximation, so we need to decide whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$ mechanisms are more plausible.

Primary electrophilic carbons are very likely to undergo $\mathrm{S_N2}$ reactions while $\mathrm{S_N1}$ is strongly disfavoured since primary carbocations are less stable than secondary or tertiary ones.

Secondary electrophilic carbons can react according to both mechanisms, technically. However, both mechanisms are not ideal: the steric strain is rather large so $\mathrm{S_N2}$ reactions are slower and secondary carbocations are still rather unstable compared to tertiary ones, so $\mathrm{S_N1}$ reactions are slower.

In this case, we have the possibility to stabilise the less stable second intermediate in the scheme shown below by a Wagner-Meerwein rearrangement. We can let both hydrogen or methyl migrate, but the hydrogen migration gives us the most stable carbocation (a tertiary one) so it will prevail. Multiple other migration paths will likely lead to minor products.

The primary carbon will not, as stated above, form a primary carbocation since it is too unstable. Hence only the bimolecular substitution will occur. If no primary carbocation is generated, no Wagner-Meerwein rearrangement can occur. And the mechanism of an $\mathrm{S_N2}$ reaction does not allow bromide’s (or anything else’s) migration.

The secondary carbon could undergo bimolecular substitution, too. The reaction rate will probably be slower than that of fragmentation and rearrangement, though, since the latter are unimolecular processes. And other shifts (e.g. methyl followed by hydride) are possible but again slower since you need to shift twice. All in all, a higher proportion of side-products may be expected.

In no case will the bromide shift after substitution; any shift would need to occur before.


Why is dichloromethane immiscible in water? We know that like dissolves like. And dichloromethane is a polar solvent and water is also a polar solvent. Also there ought to be a strong hydrogen bonding between the chlorine and hydrogen atoms. So why is it immiscible?

The physical causes of solubility of substances in each other can semiquantitatively be expressed by the Hansen solubility parameters.

$\begin{array}{}
\text{Water}:&&\delta_d=15.6&\delta_p=16.0&\delta_h=42.3&R_0=47.8\\
\text{Dichloromethane}:&&\delta_d=18.2&\delta_p=6.3 & \delta_h=6.1&R_0=20.2
\end{array}$

$$R_0=20.2,\ \ \ R_a=37.8$$

$$RED=\frac{R_a}{R_0}=1.87$$

RED < 1 would show solubility of both substances in each other. But RED > 1, so both substances are little soluble in each other.

The solubility parameters above show, the energies from polar and from hydrogen bridging bond intermolecular attractions are higher in water.

Generally, chlorinated hydrocarbons have lower energies from polar and hydrogen bridging bond intermolecular attractions.

The strongest hydrogen bridging bonds are formed between the atoms of the strongly electronegative elements $\ce{F}$, $\ce{O}$ and $\ce{N}$. But the atoms of other electronegative elements, e.g. $\ce{Cl}$-, $\ce{S}$- and $\ce{C}$-atoms, are capable of weak hydrogen bridging bonds at suitable molecular structure.

The atomic radii of the third period elements ($\ce{P}$, $\ce{S}$, $\ce{Cl}$) are much larger than those of the second period elements ($\ce{N}$, $\ce{O}$, $\ce{F}$). $\ce{Cl}$-atoms have lower charge density therefore. They are weaker proton acceptors therefore.

Hydrogen bridging bonds can be proved e.g. by molecular spectroscopy.

Li Bian: Proton Donor Is More Important Than Proton Acceptor in Hydrogen Bond Formation: A Universal Equation for Calculation of Hydrogen Bond Strength. J. Phys. Chem. A 107 (2003) (51) 11517–11524

Weinhold, F.; Klein, R. A.: What is a hydrogen bond? Mutually consistent theoretical and experimental criteria for characterizing H-bonding interactions. Mol. Phys. 110 (2012) (9-10) 565-579

Gilli, G.; Gilli, P.: The Nature of the Hydrogen Bond: Outline of a Comprehensive Hydrogen Bond Theory. Oxford University Press, Oxford, 2013


Are there more positive azeotropes than negative azeotropes? During a lecture on Flash and Distillation, my professor (which worked as a Chemical Plant Engineer for more than thirty years, specifically on separation processes) pointed out that the most frequent type of azeotropes he encountered in his work was positive-type. At first it seemed just common practice among azeotropic distillation et similia, however page 6-180 of CRC Handbook of Chemistry and Physics says:

  Nearly 90% of the binary azeotropic data show a pressure maximum (positive).

So, I'd like to know why it's true or not that nature could be somehow biased toward a specific or more frequent type of azeotrope.

Summary. Positive-deviation azeotropes mean that like interactions are much stronger than unlike interactions, whereas negative-deviation azeotropes mean the opposite. The former case is more common because most substances interact better with themselves than with other substances---like dissolves like.

(So I've only just rephrased the question at a lower level: one could equally well ask why like should dissolve like, but believing this statement on faith is perhaps more palatable than just accepting that there are more positive-deviation azeotropes.)

First, a few definitions.

A binary solution showing positive deviations (negative deviations) to Raoult's law is one in which the dew-point curve in a pressure vs. mole fraction graph lies above (below) the line connecting the pure vapor pressure of the two species.

A binary azeotrope is positive deviation (negative deviation) or minimum boiling (maximum boiling) if it shows positive deviations (negative deviations) to Raoult's law and exhibits a local maximum (minimum) in the dew-point curve in a pressure vs. mole fraction graph. From solution thermodynamics, one can prove that the vapor and liquid mole fraction of either species is equal at these extrema.

Physically speaking, positive deviations (negative deviations) from Raoult's law imply that like interactions are stronger than (weaker than) unlike interactions in the solution, for the solution's dew point pressure is greater than (less than) that of an ideal solution.

Azeotropes are extreme forms of these positive deviations (negative deviations), when the dew-point curve further exhibits a maximum (minimum), such that at some composition the solution's dew point pressure is actually greater (lesser) than the pure vapor pressure of either species.

We expect, therefore, that positive-deviation azeotropes are more common than negative-deviation ones because it is more likely for two species to have stronger like than unlike interactions.


In the mass spectrum of niacin (3-pyridine-carboxylic acid), what ion corresponds to the peak at m/z = 105? The mass spectrum can be found here.

I am interested to know what ion this is and why the peak at $m/z = 105$ is higher than the peak at $m/z = 106$ (which I believe corresponds to $\ce{C5H4NCO+}$).

From the general interpretation of electron ionization mass spectra, it would seem logical to report the $m/z$ 105 ion as a loss of $\ce{H_{2}O}$, thus $\ce{[C_{6}H_{3}NO]^{\bullet +}}$. As correctly suggested in the question, the peak at $m/z$ 106 is likely arising from a loss of $\ce{OH^{\bullet}}$ through an $\alpha$-cleavage of the $\ce{C-O}$ bond.

So the issue is: why would niacin (or nicotinic acid) undergo this $\ce{H_{2}O}$ loss? It is not observed for other similar molecules such as benzoic acid, which starts by a $\ce{OH^{\bullet}}$ loss. I did find two relevant publications in the literature on this point. Neeter and Nibbering[1] performed $\ce{D}$ labelling experiments, showing that the transferred hydrogen atom originates mostly from the $\alpha$ position (ortho from both the carboxylic acid and the nitrogen atom). Opitz[2] measured the appearance energy (AE) for $m/z$ 105 at 10.94 eV, more than 0.6 eV lower than the AE for $m/z$ 106 at 11.58 eV. He also suggests, based on metastable dissociation data showing loss of CO to $m/z$ 77, two possible structures for the $m/z$ 105 ion. So to answer the question: (a) formation of the $m/z$ 105 ion is thermodynamically favored compared to the formation of the $m/z$ 106 ion.

Based on this (scarce) literature evidence, I would suggest, based on labelling experiments, that some proton transfer occurs between the $\alpha$ cycle hydrogen and the hydroxyl group. This exchange can either move back and forth, or lead to a loss of water through formation of a cyclic structure.

References

Neeter, R.; Nibbering, N. M. M. Mass spectrometry of pyridine compounds. II: Hydrogen exchange in the molecular ions of isonicotinic and nicotinic acid as a function of internal energy. Org. Mass Spectrom. 1971, 5 (6), 735–742. DOI: 10.1002/oms.1210050612.
Opitz, J. Electron-impact ionization of benzoic acid, nicotinic acid and their n-butyl esters: an approach to regioselective proton affinities derived from ionization and appearance energy data. Int. J. Mass Spectrom. 2007, 265 (1), 1–14. DOI: 10.1016/j.ijms.2007.04.014.


link between acidity and [H3O+] Suppose I have an aqueous solution (1) of $\ce{CH3COOH}$. The chemical species in the solution will be $\ce{H2O, H3O+, CH3COO-}$ and, since it's a weak acid, some $\ce{CH3COOH}$ will remain.

Suppose I have another solution (2) containing initially water and $\ce{HCl}$. The chemical species in the solution will be $\ce{H2O, H3O+}$ and $\ce{Cl-}$ since it's a strong acid. Suppose the initial quantity of $\ce{HCl}$ was calculated so that the obtained concentration of $\ce{H3O+}$ in solution (2) is the same than in solution (1). But in solution (1), in addition to the $\ce{H3O+}$ ions, we have some molecules of another acid, the remaining $\ce{CH3COOH}$. Then the $\mathrm{pH} = - \log [\ce{H3O+}]$ is the same for (1) and (2), while (1) contains more "acidic" molecules.

How is pH a measure of acidity then ? It seems to me that if for practical purposes I had to use one of these solutions and wanted to avoid protons being exchanged, I would not consider these two solutions equally.

$\mathrm{pH}$ is a measure of acidity but defining acidity as the concentration of dissoluted protons in solution. This is quite a bit different from defining acidity as the amount of acid in solution.

Taking your example, the appropriate concentration of $\ce{HCl}$ would be something like $1~\mathrm{mM}$ while the corresponding acetic acid concentration is maybe $1~\mathrm{M}$. (These numbers are most certainly not correct, and maybe even the dimension is off by one or two orders of magnitute, but they can still serve well as an example.) So you could say ‘but there’s one thousand times as many acid molecules in the acetic acid solution!’

When considering the actual effects of an acid, the concentration of the original acidic species is irrelevant. Vinegar is, for example, usually $5~\%$ acetic acid and can be added to food (tough guys might even drink it) — I wouldn’t try that with $5~\%~\ce{HCl}$ since that contains a much higher active proton concentration.

What you can do, though, is consider the higher concentration of weak acids a type of buffer: Dilute acetic acid and the $\mathrm{pH}$ will hardly change while diluting $\ce{HCl}$ will have an immediate effect on $\mathrm{pH}$. So a weak acid can supply the same weak source of protons for a longer time.

And even though real-world analogies should be avoided in any case since they often give a wrong impression: If you jump down the staircase of a high building two steps at a time there will be likely no effect (small effect but buffered — acetic acid) just like if you only jump the two steps once (small effect unbuffered — strongly diluted $\ce{HCl}$), but if you try to jump the whole staircase (large effect unbuffered — $\ce{HCl}$ in the same concentration as acetic acid) you won’t remain unaffected.


Why do methane molecules have 15 degrees of freedom? I've just been learning a little thermal chemistry including the equipartition theorem. As part of this, my textbook discusses how to figure out the degrees of freedom for different chemicals.

The answer to one of the exercises says methane has 15 degrees of freedom: 3 for translation, 3 for rotations and 9 for vibrations. I am confused by two of these.

Firstly the text says that diatomic molecules like oxygen gas only have 2 rotational degrees of freedom, because they can't rotate about their axis of symmetry. However methane has an axis of symmetry, so why does it have 3 rotational degrees?

Secondly I don't understand where all the vibration degrees of freedom come from. The molecule has 4 bonds, so wouldn't each bond have one degree for kinetic energy and one for potential energy, for a total of 8 degrees, not 9?

15 is there because you have 5 atoms which can move independently, and we live in a 3D space, and $3\ \times\ 5\ =\ 15$.

3 of these are translational degrees of freedom, which you seem to agree with, so I'll just leave it at that.

3 are for rotations, because the methane molecule is not linear. Axes of symmetry are irrelevant: indeed, you may rotate methane around the axis of symmetry as well as around any other axis. Being or not being diatomic is hardly relevant either, except for the fact that all diatomic molecules are necessarily linear; however, the converse is not true.

Subtracting 3 and 3 from 15, we get 9, and that's how many vibrational degrees of freedom we have, because they can't be anything else. To visualize them is another story, and a great deal more difficult. Counting bonds doesn't help much. The separation of degrees into kinetic and potential is not a thing at all. You don't have to do that anyway. Write $15\ -\ 3\ -\ 3\ =\ 9$ and call it a day.


Designing synthetic route to dicarbonyl compound It's the last one, where 1,4-dimethylcyclohexane breaks apart into... something.

Here's the workup I have. I want to put out there that I'm not concerned about getting the optimal yield yet. Just making the product.

Free radical addition of $\ce{Br}$ to the 2 position.

Removal of that $\ce{Br}$ and the formation of a double bond via E2 using tBuO-

Breaking apart the structure via ozonolysis.

However, no matter where I place the double bond, I can't get the correct configuration of double bonded Oxygens (ketones and aldehydes, right?). So, I must be doing something wrong. Thoughts?

Everything you said is correct except for the first step.  Free radical bromination is reasonably selective, tertiary positions are about 20 time more reactive than secondary carbons, so the 1-bromo, 1,4-dimethyl compound is formed.  Everything's a go from there!


How does ring size affect SN2 reactions? Suppose we have to compare the $\mathrm{S_N2}$ reactivity order for cyclopropyl and cyclopentyl chloride. How can we do that?

I think that cyclopentyl should be more reactive towards $\mathrm{S_N2}$ because of larger bond angle and therefore lesser steric hindrance to the atttacking nucleophile but I am not sure about this.

Also, can the trend be generalised i.e. does increasing ring size lead to faster $\mathrm{S_N2}$?

TL;DR - Bond angle strain gets a lot worse in going from cyclpropyl chloride to the transition state for the 3-membered ring example, than for cyclopentyl chloride going to the corresponding transition state.

The rate for each reaction will be determined by the free energy barrier in each case. Cyclopropyl chloride is strained because the carbons "want" to be tetrahedral, with sp3-hybridized orbitals (109° bond angles), but are forced to have much more acute (60°) bond angles within the ring. However, at the transition state, the carbon undergoing nucleophilic attack "wants" to be trigonal planar, with sp2-hybridized orbitals (120°), but is still geometrically-constrained to 60° bond angles within the ring. This makes the free energy barrier for SN2 very large.

Cyclopentyl chloride has bond angles close to the ideal 109°. There will be some bond angle strain at the transition state, but the effect won't be nearly as bad as for the three-membered ring.

Now, can the effect be generalised? Mostly. As expected, cyclobutyl chloride reacts faster than cyclopropyl, but slower than 5 or 6-membered rings. Five vs six is a tricky one. Cyclopentyl chloride is actually slightly more reactive. If you look at cyclohexane, it is very stable in the chair conformation. All nice bond angles, no eclipsing interactions. At the transition state for SN2 with cyclohexyl chloride, bond angle strain is introduced and there is steric interaction between the incoming nucleophile and the adjacent axial hydrogens. Cyclopentyl chloride doesn't start off quite so perfect - there are some eclipsing interactions. Also, the transition state may actually even relieve some of these eclipsing interactions.

Beyond six-membered rings, things get very complicated quickly. You could apply the same approach, but the number of possible conformations quickly rises from 8-membered and above. It becomes difficult to take all the possible factors and conformations into account, so without high-level calculations, it is difficult to make predictions.

If you have access, the reactivity of 3-6 member cycloalkyl chlorides was included in a publication by Rablen et. al..[1] It includes calculated energy barriers for SN2 reactions of these cycloalkyl chlorides with cyanide ion as nucleophile.

Reference:

Rablen, P. R.; Mclarney, B. D.; Karlow, B. J.; Schneider, J. E. How Alkyl Halide Structure Affects E2 and SN2 Reaction Barriers: E2 Reactions Are as Sensitive as SN2 Reactions. J. Org. Chem. 2014, 79 (3), 867–879 DOI: 10.1021/jo4026644.


Determine the oxidation number of sulfur in SF6 I've just started learning about redox chemistry and oxidation numbers. I can work out most of the oxidation number questions, but I don't know how to determine the oxidation number of sulfur in $\ce{SF6}$. This is not an ionic compound, and doesn't contain $\ce{H}$ or $\ce{O}$, so it seems as if I have no starting point.

If you could, please answer the question and provide a simple explanation.

In the case of $\ce{SF6}$, sulfur would have the oxidation number of +6 because the charge being applied to the fluorine is +6. Similarly fluorine would consequently have an oxidation number of -1 since $$6x + 6 = 0 \Rightarrow x = -1$$

(the right-hand side is equal to zero since that happens to be the net charge on the overall chemical formula).


Why does the distance between atoms in a hydrogen bond impact the strength of the bond? Wouldn't the distance be irrelevant as after they are attracted they will just close the gap? E.g. once 2 magnets are close enough to be attracted, isn't the distance irrelevant to the strength of the bond between the magnets as they will just become flush against each other, making the distance between any 2 magnets that are attracted to each other equal.

Or is it perhaps talking about the other groups stopping the hydrogen and other atom becoming close?

Context:

Of the following, the factor that has the least effect on the strenght of a hydrogen bond in a molecule is the

type of atom to which the hydrogen atom is covalently bonded.
distance that the hydrogen atom is from the atom on the other part of the molecule
number of other hydrogen atoms atttracted to the atom on the other part of the molecule
type of atom in the other part of the molecule to which the hydrogen atom is attracted

(answer is 3, hence 2 must have some effect).

The main attractive force between the nuclei and the electrons is electromagnetic, positive and negative attracting. When two atoms are "pushed" together each one shares its own electrons with the other one's nuclei, and they bond. So why wouldn't they nuclei just get as close to each other as possible, so that they each would feel as many electrons as possible? At the same time, the electrons are repelling each other, and the nuclei repel (same charges repel). The bond length is a function of this "equilibrium" that is reached between the attraction of the nuclei to the electrons, and the repelling of the nuclei  and the repelling of the electrons. Therefore, if they begin to "close the gap" as you say, there will be more repelling than attracting, and they will move farther away from each other, and return to the lowest energy distance, the original bond length.


Why is HCl used in hydrolysis of sucrose? I am aware that $\ce{HCl}$ can be used to help speed up the hydrolysis of sucrose but I haven't really been able to find out why. I've read somewhere that the glycosidic link in sucrose is broken off and that the $\ce{H+}$ ions from the $\ce{HCl}$ is responsible for this.

I've also come across the ether link during my research and would like to know how this plays a part in the hydrolysis of sucrose and if the $\ce{HCl}$ interacts with it.

The mechanism really isn't much more complicated than what you describe.

The $\ce{H+}$ binds to the acetal linkage. Water then binds to the fructose ring, breaking its bond with the acetal linking $\ce{O}$. Finally, the water molecule on the fructose loses a proton, restoring the catalyst. As Jan mentioned in the comments, the specific acid used is irrelevant so long as it is strong enough to sufficiently acidify the solution.
(Credit to Khan Academy for the image)


How to find the concentration of barium ions in a mixture of barium sulfate and barium thiosulfate? I have a hard time with selective precipitation problem.

  A mixture of $\ce{BaSO4}$ and $\ce{BaS2O3}$ is shaken with pure water until a saturated solution is formed. Both solids remain in excess. What is $[\ce{Ba^{2+}}]$ in the saturated solution? $K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$ and $K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$.

My solution:

Because $K_\mathrm{sp}$ of $\ce{BaSO4} < \ce{BaS2O3}$, $\ce{BaS2O3}$ will dissolved first. When $\ce{BaS_2O_3}$ dissolved, the concentration of $[\ce{Ba^{2+}}] = [\ce{S2O3^2-}] = \sqrt{K_\mathrm{sp}(\ce{BaS2O3})} = 2 \times 10^{-5}$.

That means, in the solution, there are:

\begin{align}
[\ce{Ba^2+}] &= 2 \times 10^{-5}\\
[\ce{S2O3^2-}] &= 2 \times 10^{-5}\\
[\ce{SO4^2-}] &= \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} = \frac{9 \times 10^{-11}}{2 \times 10^{-5}} = 4.5 \times 10^{-6}
\end{align}

But the solution is $\ce{[Ba^2+]} = \sqrt{4.9 \times 10^{-10}}$.

What have I missed?

EDIT 1 :
\begin{align}
\ce{BaSO4 &<=> Ba^2+ + SO4^2- }\\
\ce{BaS2O3 &<=> Ba^2+ + S2O3^2- }
\end{align}

The Solubility Product Equation :

\begin{align}
K_\mathrm{sp}(\ce{BaSO4}) &= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
K_\mathrm{sp}(\ce{BaS2O3}) &= [\ce{Ba^2+}] \times [\ce{S2O3^2-}]
\end{align}

When the solids dissolves, it will produce $\ce{Ba^2+}$ ion, $\ce{SO4^2-}$ ion and $\ce{S2O3^2-}$ ion.

$\ce{BaSO4}$ will stop dissolving when $\ce{Ba^2+}$ ion concentration in solution when multiply with $\ce{SO4^2-}$ ion concentration is equal to solubility product of $\ce{BaSO4}$, which is $K_\mathrm{sp}(\ce{BaSO4}) = 9 \times 10^{-11}$

And at the same time,

$\ce{BaS2O3}$ will stop dissolving when $\ce{Ba^2+}$ ion concentration in solution when multiply with $\ce{S2O3^2-}$ ion concentration is equal to solubility product of $\ce{BaS2O3}$, which is $K_\mathrm{sp}(\ce{BaS2O3}) = 4 \times 10^{-10}$

Solubility of $\ce{BaSO4}$ in water is
\begin{align}
 s &= \sqrt{K_\mathrm{sp}(\ce{BaSO4})}\\
 s &= \sqrt{9 \times 10^{-11}} \\
 s &= 3 \times 10^{-5.5}
\end{align}

Solubility of $\ce{BaS2O3}$ in water is
\begin{align}
 s &= \sqrt{K_\mathrm{sp}(\ce{BaS2O3})}\\
 s &= \sqrt{4 \times 10^{-10}}\\
 s &= 2 \times 10^{-5}
\end{align}

Now, when the solids is mixed in water, There should be $[\ce{SO4^2-}] =3 \times 10^{-5.5} $ and $[\ce{S2O3^2-}] =2 \times 10^{-5} $.

Now I confuse what ion concentration should I use to determine $[\ce{Ba^2+}]$

EDIT 2:

\begin{align}
K_\mathrm{sp}(\ce{BaSO4}) &= [\ce{Ba^2+}] \times [\ce{SO4^2-}] \\
\frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]} &= [\ce{SO4^2-}] \\
\end{align}
and
\begin{align}
K_\mathrm{sp}(\ce{BaS2O3}) &= [\ce{Ba^2+}] \times [\ce{S2O3^2-}] \\
\frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} &= [\ce{S2O3^2-}] \\
\end{align}

From @santimirandarp answer, I got that :

\begin{align}
[\ce{Ba^2+}] &= [\ce{S2O3^2-}] + [\ce{SO4^2-}]\\
[\ce{Ba^2+}] &= \frac{K_\mathrm{sp}(\ce{BaS2O3})}{[\ce{Ba^2+}]} + \frac{K_\mathrm{sp}(\ce{BaSO4})}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &= \frac{4 \times 10^{-10}}{[\ce{Ba^2+}]} + \frac{9 \times 10^{-11}}{[\ce{Ba^2+}]}\\
[\ce{Ba^2+}] &= \frac{4.9 \times 10^{-10}}{[\ce{Ba^2+}]} \\
[\ce{Ba^2+}]^2 &= 4.9 \times 10^{-10} \\
[\ce{Ba^2+}] &= \sqrt{4.9 \times 10^{-10}} \\
\end{align}

My problem solved. Thank You


Why must polar molecules be asymetrical? As I understand, polarity in molecules comes from the difference in electronegativity or the ability of the atoms to attract eletrons so the electrons spend more time or are more probable to gravitate more around the atom with stronger electronegativity right? So I wonder why polar molecules must necessarily have an asymetrical structure? (as I've read on multiple sites). Shouldn't a symetrical atom with strong electronegativity with 2 atoms bonded linearly 180° on both sides which are significantly less electronegative, be polar since the electrons would gravitate more towards the center and the sides would then be more positively charged? What am I missing here?

Why is the first molecule (1) not possible? Is it becuase of the less polar atom on the left side repells electrons just as much as the one on the right? But then the electrons would still orbit the middle atom more because its more elctronegative, right?

If you start with the wikipedia definition of chemical polarity, which is standard (if not IUPAC) usage:

  In chemistry, polarity is a separation of electric charge leading to a molecule or its chemical groups having an electric dipole moment, with a negatively charged end and a positively charged end.

then it's not hard to understand why a molecule such as $\ce{CO2}$ containing a center of inversion lacks a dipole moment and is therefore "nonpolar", as are, by this definition, other molecules with high symmetry properties and lacking a net dipole moment:

  The only groups compatible with a dipole moment are Cn, Cnv and Cs. In molecules belonging to Cn or Cnv the dipole must lie along the axis of rotation.


Syn Dihydroxylation of Alkenes The answer makes sense to me. However, what if $\ce{OsO4}$ were to attack from the other side of the double bond? Why does the answer key not show that possibility? Would we not get an other molecule?

If you look carefully you'll see that your compound has stereocenters, yet is achiral.  Both stereocenters have the opposite configuration.  This is the meso-form, these are achiral, and the hydroxy groups are prochiral.  Look carefully for the rotoinversion center!


Is 4-chlorotoluene also an IUPAC name for 1-chloro-4-methylbenzene? I think that toluene is a common name for 1-methylbenzene. And in IUPAC nomenclature, we must provide formal prefixes and suffixes along with correct numbering so the only correct option must be 1-chloro-4 methylbenzene.

The use of 'toluene' as a parent hydrocarbon in nomenclature is described in Rule P-22.1.3 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013.

  These names [toluene, xylene, and mesitylene] are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not freely substitutable; toluene is substitutable under certain conditions, but only for general nomenclature. (emphasis mine)

"General nomenclature" refers to any name generated by the use of IUPAC principles, rules, and conventions, whereas the "preferred IUPAC name" (PIN) refers to a name that is preferred among two or more names generated from two or more IUPAC recommendations. The introduction P-10 reads:

  Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a 'general IUPAC name", in the context of a 'general IUPAC nomenclature'.

As stated in P-22.1.3, in the construction of a PIN for the molecule below, the use of 'toluene' as the parent hydrocarbon is not permitted; the PIN is 1-chloro-4-methylbenzene.

Regarding the substitution of toluene in general nomenclature, P-15.1.8.2.2 lists the characteristic groups that can be used in general nomenclature:

  [...] halides $\ce{-Br, -Cl, -F, -I}$, pseudohalides $\ce{-N3, -NCO}$ (and chalcogen analogues), $\ce{-NC}$, substituent groups derived from the halogen oxo acids $\ce{-ClO, -ClO2, -ClO3}$ (similarly for groups in which $\ce{Cl}$ is replaced by $\ce{Br}$ or $\ce{I}$), $\ce{-NO2}$ and $\ce{-NO}$, and $\ce{-OR}$ ($\ce{R}$ = alkyl groups), and chalcogen analogues, and $\ce{-SO-R}$ and $\ce{-SO2-R}$, and $\ce{Se}$ and $\ce{Te}$ analogues.

Since chloride falls under this list, the name 4-chlorotoluene is permissible in general nomenclature, but it is not the preferred IUPAC name.

Note that there is no space between the prefix 'chloro' and the parent hydride name 'toluene'.

Also note that there is no such thing as '1-toluene'; in the name '4-chlorotoluene', for example, the methyl group of toluene is automatically assigned the locant '1' (see figure below from the Recommendations). Therefore, the name '4-chloro-1-toluene' is incorrect.


How is traditional thermite powder stored such that it doesn’t corrode? I watched some guy talk about redox reactions. Then he was going to do the good old thermite demonstration. He just had a bottle of thermite mixture. I’m wondering, why didn’t it have a chance to corrode before the demonstration? I’ve always thought that aluminum metal gets a thin layer of corrosion in air.

Thermite powder is a heterogeneous mix and consists of fuel and oxidant, in you case aluminium $\ce{Al}$ and rust $\ce{Fe3O4}$. Separately, those can be corroded over time all right: aluminium can be oxidized and form aluminium(III) oxide, hydroxides and even some salts depending on what atmosphere thermite has been kept in. As you probably know, aluminium, being a reactive metal, doesn't corrode under ambient conditions as it's surface is protected by the thin layer of $\ce{Al2O3}$. Iron(II) is also capable of further oxidation, although rust is already the most stable form under ambient conditions.

However, it seems like you are using the term "to corrode" quite liberally to denote the act of burning of the thermite, e.g. the primary exothermic reaction

$$\ce{8 Al + 3 Fe3O4 → 4 Al2O3 + 9 Fe}$$

which only occurs when enough external energy (say, in a form of heat) is supplied. The thing is, not every reaction that is permitted by thermodynamics is also permitted by kinetics. Burning of a thermite is a heterogeneous process requiring to overcome activation barrier. Simply put, there are numerous solid particles or two phases being barely in contact, plus there is a protective layer on one phase ($\ce{Al2O3}$). Here, you need energy at least to increase thermal motion (promoting mass transfer between the phases) and compact the mix (ideally, melt to increase contact area).

That's why thermite won't easily go off, and it's a good thing making its storage safe and the working process predictable. There are only a few documented accidents [1, p. 62] resulting in ignition of thermite-alike mixtures, but none of them were spontaneous; instead, they were caused by the supply of substantial amount of external energy (heat, electricity, sparks, kinetic energy etc.), or third-party chemicals acting like catalysts:

  Aluminium-painted rusty surfaces are only hazardous if the coating has
  been heated to 150 °C: on cooling, the hazard remains (Eckart et al., 1971).
  Paints with cellulose nitrate as an additive are always hazardous. [...]

  Hawkins Colliery, 1950. The impact of a hand-held drilling machine cased
  in Elektron C alloy (94% Mg) upon rusty steel scaffolding ignited
  firedamp, burning 10 men, one fatally. This incident led to the initiation
  of a major research programme into the dangers of light alloys by the
  Safety in Mines Research Establishment. [...]

  SS Esso Durham, 1961. An explosion in one of the ship's tanks on 29th January was attributed to the impact of a brass object, of weight 361 b, onto a magnesium anode on the wall of the tank after a fall of 16 ft. From simulations of possible alternatives, it was considered likely that a grazing impact on the rough, dirty magnesium surface would have smeared the brass, which would then have reimpacted the anode a fraction of a second later, causing a thermite reaction and an explosion. The oxide necessary was probably provided by dirt in the area (ignitions were not obtained when impacting a clean magnesium surface). The possibility of ignition from impact onto hardened steel after a 40 ft fall was eliminated. [...]

  An aluminium anodizing works, Bletchley, Buckinghamshire, 1971. A ferrous metal object caused a spark on an aluminium grinding and polishing machine; the abrasive belts were lubricated by mineral oil, which caught light. The fire spread to the oil tank and extract ducting, causing £65000 worth of damage. [...]

  Horden Colliery, 1955. A steel weight was dropped on a friction prop with
  aluminium frictional surfaces, leading to ignition of firedamp.

  Glyncorrwg Colliery, 1955. A rusty steel wedge was struck from an aluminium roof bar, leading to ignition.

  United Kingdom, 1971. Propane was vented into a workshop during the filling of 140 lb propane cylinders when one was accidentally overfilled. The propane was ignited, and two people were injured in the subsequent explosion. The ignition source was believed to have been a spark from rusty steel cylinders being loaded outside, which, on examination, were
  shown to have aluminium smears on them. An aluminium roller conveyor was being used in the loading operation. A very light impact on such a smear is capable of generating a spark through a thermite reaction. [...]

  Ruhr Refinery, 1976. Two explosions within two hours of each other occurred on the 2nd December whilst switch-loading diesel into two tankers of carbon steel fabrication using an aluminium loading arm. The explosions were attributed either to a static electricity discharge or to a thermite reaction. [...]

  SS British Industry, 1963. The tanker had cathodic protection by incorporating a series of magnesium anodes in seven tanks. The cause of an explosion on the tanker was attributed to the impact of loose anode materials onto rusty deposits in the tank in which the explosion occurred. On examination after the explosion, many of the anodes were seen to be almost detached from their angle-iron bases, and bits had broken away. It
  was deemed that detachment could have been effected by a washing machine whipping violently in the tank, and that there would have been flammable concentrations of crude oil vapour in the tank.

References

Bond, J. Sources of Ignition: Flammability Characteristics of Chemicals and Products; Butterworth-Heinemann: Oxford; Boston, 1991. ISBN 978-0-7506-1180-0.


What does the neutralisation of acids and bases mean? From what I found in textbooks and google is that neutralisation is when acids and bases react to produce salt and water. Most examples I have learnt do so, however a reaction like:

$$\ce{NH3 + HCl -> NH4Cl }$$

It does not produce water.

So is neutralisation defined as only when salt and water produce a neutral salt and water or is it just produce any salt and water. And if it is any then how can does the example above work?

There are many theories put forward to explain acid-base reaction.
According to Arrhenius acid-base theory. All acid liberate protons (H+ ions) when dissolved in water. All bases liberate Hydroxyl ions (OH- ions) in water.
When the reaction takes place H+ and OH- combine to form water, and the remainder of the ions (what are left after liberating H+ and OH- ions) form what is called a salt.
In such reactions, Water is always formed.

What you have shown is an example of a Bronsted acid-base reaction.
Bronsted acids are proton donors (have ionisable hydrogen ion).
Bronsted bases have the capability to accept them (either via hydroxyl ion liberation or by their lone-pair as in NH3)

Clearly, the formation of water is not a necessity in a Bronsted acid-base reaction.


Confusion in calculating $\Delta U$ from a bomb calorimeter In the book, it is mentioned the formula for $\Delta U$ in a bomb calorimeter without any derivation:

$$\Delta U = q_v = \frac{Q\times M\times \Delta T}{m}$$
where
$$Q=\textrm{heat capacity of calorimeter,}$$
$$M=\textrm{molecular mass of sample,}$$
$$m=\textrm{mass of sample used, and}$$
$$\Delta T=\textrm{change in temperature of water in the bath}$$

I am confused regarding this formula. Can anyone give me the derivation of this formula (or a corrected formula)?
[I am 11-grader and am studying chemical thermodynamics. I can distinguish between $C$ as an extensive property and $c$ and $C_m$ as intensive properties.]
Any help would be appreciated :)
NOTE: I know that a formula is $q_v=cm\Delta T$, I want to know how the book got to the formula mentioned previously.

The formula in the book is correct.  They are trying to get the change in internal energy per mole of sample.  From the first law, for this constant volume system (no work), $$\Delta U_{\textrm{total}}=q=C\Delta T$$where C is the heat capacity of the calorimeter.  This equation assumes that the heat capacity of the water in the bath is lumped into C, and that the temperature change of other parts of the calorimeter is the same as that of the water.
The number of moles of sample is m/M.  So, $$\Delta U_{\textrm{per mole}}=\Delta U_{\textrm{total}}\frac{M}{m}=C\Delta T\frac{M}{m}$$In their notation, they use the symbol Q to represent the heat capacity of the calorimeter C.


Hypothetical: What happens to water as pressure increases to infinity? I've asked a similar question here but the answer given shows the behaviour of water under general conditions.

I'd like to know what the behaviour of water is like as pressures increase towards infinity without being able to escape it's confinement.. i.e. a ball of water at the core of a galactic mass.. maybe this question is more for theoretical physics since we can't really measure or experiment?

Ron's answer gives us a good idea of what might happen in terms of "normal" chemistry, but if you really mean "increases to infinity", some very strange stuff happens. Suffice it to say, it doesn't stay water after a certain point. The intense temperatures created by the compression will cause the water to break apart, eventually no longer even having oxygen atoms due to nuclear reactions. Because we're talking about an externally applied pressure, the Chandrasekhar limit doesn't apply, so there is a point at which electrons and protons combine (when the electron degeneracy pressure is overcome) and a mass of neutrons remains. Neutrons themselves also have a degeneracy pressure (though we don't have good models to predict the exact pressure that has to be overcome). From here, we don't know what happens with as much certainty, but the formation of quark matter has been predicted.

Eventually, we reach a singularity. We can think of this as all the matter we had before being compressed into an infinitesimal volume with infinite density and our applied pressure ceases to mean anything. If we started with enough water, this would behave much like any other black hole, though micro black holes are hypothesized to have some special properties.


IUPAC rules on omission of locants in carboxylic acid I have been lately in debate with my teacher as to why did the IUPAC name the compound with the current official IUPAC name of 2-methylpropanoic acid. (CN: isobutyric acid)

My argument is that:

You can't attach a methyl in carbon 1 (because the bonds in carbon 1 will exceed 4)
You can't attach a methyl in carbon 3 (because it will become butanoic acid)
Placing the locant in a substituent with no other constitutional isomers will be useless when to be interpreted, will be a few more characters longer, and will be less pleasant for the eyes end ears (I mean, methylpropanoic acid a bit more pleasant to see and hear than 2-methylpropanoic acid, right?)

To verify my initial assumption, is the name of such compound 2-methylpropanoic acid? If it is, then we may proceed to the real question in the next paragraph.

So, where in the 2013 Blue Book or any updated IUPAC nomenclature guidelines with less complicated wordings in organic chemistry books (this has seniority over the former) can I see that you must not omit the locant in these cases (and why).

Additionally, how can I request to the Union to alter this notion to make it methylpropanoic acid because of the stated reasons? (It's a bit like ethan-1-ol and ethanol.)

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name is 2-methylpropanoic acid. The trivial name ‘isobutyric acid’ is no longer recommended (see Subsection P-65.1.1.2.4).

Concerning your objection

  You can't attach a methyl in carbon 3 (because it will become butanoic acid)

IUPAC nomenclature doesn’t work like that. Generally, the complete information about a structure is explicitly given by the name and does not rely on any implied information or hidden logic involving other structures.

The given parent structure is propanoic acid. It includes two kinds of substitutable hydrogen. Therefore, the locant cannot be omitted in substituted propanoic acid, e.g. 2-chloropropanoic acid or 3-chloropropanoic acid.

This principle applies to any substituent. The special situation that replacing the chloro substituent in 3-chloropropanoic acid with a methyl substituent would actually lead to the name butanoic acid instead of 3-methylpropanoic acid does not change the preferred name for 2-methylpropanoic acid.

Nevertheless, there are a few exceptional cases where locants are omitted, but only when there is no ambiguity. For preferred IUPAC names, the nomenclature rules explicitly stipulate when locants are omitted. Most of these rules apply to the omission of the locant ‘1’. One case is the example given in the question:

  (It's a bit like ethan-1-ol and ethanol.)

Homogeneous chains consisting of only two identical atoms such as ethane only have one kind of substitutable hydrogen. Therefore, the current IUPAC recommendations include a rule (in Subsection P-14.3.4.2) that the locant ‘1’ is omitted in monosubstituted ethane; i.e. ethanol (not ethan-1-ol) is the preferred IUPAC name. However, if any locants are essential for defining the structure, then all locants must be cited for the structure, e.g. 2-chloroethan-1-ol (not 2-chloroethanol; see Subsection P-14.3.3).

Coming back to carboxylic acids, acetic acid also has only one kind of substitutable hydrogen, i.e. all substitutable hydrogen atoms have the same locant. Therefore, explicitly according to Subsection P-14.3.4.6, this locant is omitted, e.g. chloroacetic acid (not 2-chloroacetic acid).


Why is potassium phosphate KH2PO4 in this reaction? This is a continuation of this question because the first thing that came in my mind is that why potassium phosphate in this reaction is $\ce{KH2PO4}$ and not $\ce{K3PO4}$?

In the wikipedia article of phosphate ion ($\ce{PO4^3-}$), it is written that:

  Aqueous phosphate exists in four forms. In strongly basic conditions,
  the phosphate ion ($\ce{PO4^3-}$) predominates, whereas in weakly
  basic conditions, the hydrogen phosphate ion ($\ce{HPO4^2-}$) is
  prevalent. In weakly acidic conditions, the dihydrogen phosphate ion
  ($\ce{H2PO4−}$) is most common. In strongly acidic conditions,
  trihydrogen phosphate ($\ce{H3PO4}$) is the main form.

So, $\ce{FeCl2}$ must somehow be acidic in aqueous solution so as to protonate potassium phosphate $\ce{K3PO4}$ to $\ce{KH2PO4}$. Is this statement true? This question throws some light on acidity of chromium chloride in aqueous solution but is it applicable to iron chloride as well?

In the previous question, the homework question was write the equation of reaction of Iron(II) chloride and potassium phosphate. So, students can assume the reaction to be $\ce{FeCl2 + K3PO4}$ to simply yield iron(III) phosphate or iron(II) phosphate and potassium chloride but why make the reaction complex by assuming potassium phosphate $\ce{KH2PO4}$.

Iron hydrogen phosphate are new in this world and has been recently synthesised and as not as stable as iron phosphates. So, can we just assume that reaction of Iron(II) chloride and potassium phosphate to yield iron (II or III) phosphate and potassium chloride?

I think that’s just another mistake the OP of the other question made. Potassium phosphate is $\ce{K3PO4}$ and any student using any hydrogenphosphate instead would lose marks if I were marking the exam.

If $\ce{KH2PO4}$ is meant to be added, use the correct term: potassium dihydrogenphosphate.

This nomenclature can be found in the current version of IUPAC’s Red Book, sections IR-8.2 and IR-8.4 and the anions in question are explicitly named in table IR-8.1 therein.


Relative rates of two acyclic dienes in the Diels-Alder reaction As for factors that affect the rate of the Diels Alder reaction, having an s-cis conformation is primary.

But in a case as shown below, what makes the second diene react faster compared to the third?

The only reason I can think of is that the second diene has substituents pointing outward so they cause less steric hindrance as it reacts with a dienophile thus has a lower activation barrier. Are there any other underlying reasons apart from hindrance?

You are correct that the having the diene in an s-cis conformation is extremely important.  Only from this conformation can the Diels-Alder occur. If there is a high concentration of the s-cis conformer the Diels-Alder will likely occur rapidly; if there is a low concentration the reaction will occur more slowly.

In cyclopentadiene, the diene component has no choice, it is locked in an s-cis conformation by the ring.  However with acyclic 1,3-dienes an equilibrium exists between the s-cis and s-trans conformations.

In 1,3-butadiene the amount of the s-cis conformation is relatively small (~2%) because of the steric interaction between the "internal" hydrogens at the ends of the chain in this conformation. In accord with this expectation, 1,3-butadiene is not as reactive as cyclopentadiene in the Diels-Alder reaction.

If we replace one of these "internal" hydrogens with a methyl group (penta-1,cis-3-diene), the steric interaction between the "internal" methyl and hydrogen in the s-cis conformation is even greater, further destabilizing the s-cis conformation and reducing the amount present at equilibrium.  On the other hand, replacing one of the "exterior" hydrogens with a methyl group (penta-1,trans-3-diene) doesn't cause any increased steric interference in the s-cis conformation and the amount of the s-cis conformer is about the same as 1,3-butadiene. Again, in accord with expectations, penta-1,trans-3-diene reacts at a rate similar to 1,3-butadiene, whereas penta-1,cis-3-diene reacts slower.

(image source)

Using this information we can make predictions for your compounds. We would expect cyclopentadiene to react the fastest because it is held in an s-cis conformation. In the case of cis, cis-2,4-hexadiene we now have an internal methyl-methyl interaction, which severely destabilizes the s-cis conformation, so there is very little present at equilibrium and this compound reacts the slowest in the Diels-Alder reaction. trans, trans-2,4-hexadiene only has an "internal" hydrogen-hydrogen interaction when in the s-cis conformation, therefore it should react about as fast as 1,3-butadiene - in between the extreme rates of the other two compounds.


Vapor pressure vs. ambient pressure I've managed to get myself thoroughly confused in thinking about the vapor pressure of a substance at a particular temperature $T$, and how it relates to the pressure of the substance if it is in a phase equilibrium at that same temperature $T$.

Consider two sealed, rigid containers.  All three containers are in thermal equilibrium with a large temperature reservoir at (say) 298 K.

We evacuate Container A completely, and place into it a sample of liquid water.  Some of the water boils away until the pressure of the water vapor attains the appropriate pressure on the phase boundary in the P-T diagram.  (We assume that the sample of water is appropriately sized so that some water is left in each phase once it reaches equilibrium.)  Once equilibrium is reached, the water vapor has reached a pressure $P_A$.
Into Container B (which is of the same size) we also place an indentical sample of water.  However, Container B also contains dry air at $P_\text{air} = 1$ atm.  The water evaporates, mixing with the air, until it reaches a particular vapor pressure.  (Again, assume that the sample of water is appropriately sized so that some water remains unevaporated in equilibrium.)  Once equilibrium is reached, the water vapor has partial pressure $P_B$.

  Does $P_A = P_B$ exactly?  Or, to put it another way, does the value of $P_\text{air}$ affect the partial pressure of the water vapor in equilibrium?  If $P_A \neq P_B$ precisely, is there some limit or approximation under which $P_A = P_B$ exactly? If so, what is it?

I haven't been able to find a definitive answer in my searching online, and it seems plausible to me that the presence of an external source of pressure on the water (namely the air in container B) might affect the phase transition properties of the water.

The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$

The additional pressure compresses the liquid and thereby increases its chemical potential, resulting in an increased vapor pressure. However, since $V_{m,~l}$ is relatively small,  the effect on the vapor pressure is also small except at very high pressures. The effect of air at atmospheric pressure and $\pu{298 K}$ on the vapour pressure of water ($\pu{3.2 kPa}$ in the absence of inert gas) is negligible, an increase of $\pu{2.3 Pa}$.

So $P_A \neq P_B$, but they become nearly identical when $PV_{m,l}\ll RT$.


What functional groups are subsets of the Alkoxy group? In a quiz, I was asked to find all of the combinations of organic molecule of $\ce{C4H6O2}$. The combination excludes -COOH, -OH, double bonds and the alkoxy group.

The only problem, as far as I am concerned, is the definition of the term "alkoxy group". According to this Wikipedia article, a Peroxy or Methelenedioxy should be one of the possible combinations allowed. However, the names do have the "-oxy" suffix in them. So I am uncertain whether to include Peroxy's and Methelenedioxy's in the list.

P.S. Below is a picture of the Methelenedioxy group and the peroxy group, respectively,  from the already mentioned Wikipedia article.

There is another Wkipedia article entitled "Alkoxy group".  In it they show the following molecules as containing alkoxy groups.

They include methylenedioxy, but not peroxide in their list.


Calculating the pH of a highly dilute solution of HCl For relatively high concentrations of $\ce{HCl}$, I usually just assume that $[\ce{H+}] = [\ce{HCl}]$, because $\ce{HCl}$ is a strong acid and is completely ionized in solution. By taking the negative logarithm of the concentration of $\ce{[H+]}$, the pH comes out more or less correctly.

However, at very low concentrations, this method doesn't work out so well, because water has an amount of auto-ionization, which is described by $K_\mathrm{w} = 10^{-14}$. My old method was just to say that

$$[\ce{H+}] = [\ce{HCl}] + \pu{10^-7 mol dm-3},$$

to include the auto-ionization of water in my calculations. I can't really describe why, but I believe this method to be wrong, because the number $\ce{10^{-7}}$ comes from water just by itself, not with $\ce{HCl}$ added to it.

Then I came across this website, which implies that the above method is an OK method, but mentions that "However, that more complex technique will not be discussed here.". What is that 'more complex technique', and how could I use it?

At first you have to think of where the protons come from. That is as you mentioned it, hydrochloric acid (for every chloride ion in solution there has to be a proton) and water (for every hydroxide ion from the autoprotolysis there also has to be a proton). $$\ce{[H+] = [OH-] + [Cl-]}$$

The concentration for the hydroxide ions can be derived from the autoprotolysis of water, as I've said before.
$$k_\mathrm w = \ce{[H+]~[OH-]}
\Rightarrow \ce{[OH-]}=\frac{k_\mathrm w}{\ce{[H+]}}$$

Using some funny math, one can find out, that the concentration of the chloride ions in solution can be described the following eqation. Recognizing shortly after that the hard way was not necessary yields to the simplification that hydrochloric acid is completely dissociated.
$$\ce{[Cl-]} = \frac{\ce{[HCl]0}~k_\mathrm a}{\ce{[H+]} + k_\mathrm a} \xrightarrow{k_\mathrm a \gg \ce{[H+]~~}} \frac{\ce{[HCl]0}~k_\mathrm a}{k_\mathrm a} = \ce{[HCl]0}~(=c_0)$$

This changes the first equation to be like
$$\ce{[H+]} = \frac{k_\mathrm w}{\ce{[H+]}} + c_0$$

Rearranging everything gives you a second order equation in $\ce{[H+]}~(=x)$
$$x^2-c_0~x-k_\mathrm w = 0$$
Wherefrom you get two solutions (from which only the positive one makes sense):
$$x = \frac{c_0}{2} + \sqrt{\frac{c_0^2}{4}+k_\mathrm w}$$

Let's compare this solution to your mentioned version. The following graph shows you the difference between both assumptions as $$\operatorname{abs}(f_1(c)-f_2(c))$$

$\hskip3.5cm$

A maximum difference of 0.1 pH between the easier and the "more complex" version is acceptable.


Stir bar size and volume to stir I'm looking for a table showing which size of bar to use for a given volume to stir.
Or a formula if there is one.

No chart I'm aware of.  However, an informal and incomplete survey indicates most chemists to this by trial and error.

For pipe-type stir bars in flat bottom flasks/beakers, and you want a 'fast' stir, generally you want one roughly two thirds the size of your magnet arm to minimum the motor moving ahead of the magnet spin and 'flinging' it around.  Obviously this is not possible if your flask is much smaller than the motor.  If your stir bar is too close to the size of your vessel, you have 'catching' issues. with the sizes.

For egg-type stir bars in round bottom flasks, the concavity of the surface will help, but not eliminate 'flinging'.If you need a fast stir, you need a bigger stir bar.  Personally, I've never needed larger than 3/4" because of my use preference, (also do not typically use roundbottoms larger than 1L).

Ultimately, however, remember stir bars are for mild to moderate stirring.  If you need vigorous, you are better served with a mechanical stirrer.  Additionally, if you are using very large (5L+) flasks, mechanical stirring is best.


Reactivity of benzyl halides towards nucleophilic substitution I experienced some difficulty as I was doing question 22 of this quiz on nucleophilic substitution at saturated carbons:

I picked option B but the answer turned out to be C. My rationale for picking B was that I thought that IV was going to be more reactive than I because I has an additional electron-donating methyl substituent which would increase the energy of the transition state in the $S_{N}2$ pathway.

Evidently, II and III are in the 3rd and 4th places respectively because alkyl bromides are less reactive than alkyl iodides and that phenyl bromides are not as reactive towards nucleophilic substitution as the $ \pi$ electron cloud of the neighbouring benzene ring would repel the nucleophile away.

What went wrong with my thinking?

As you already noted, phenyl bromide is least reactive towards nucleophilic substitution, so any answer suggestion that does not place III last is wrong a priori. The reasoning you gave does not quite add up with me so here is my version: no $\mathrm{S_N2}$ back attack is possible through the phenyl ring since that pathway is blocked by the other ring atoms; and no $\mathrm{S_N1}$ pathway is possible since that would require a very unstable phenyl cation with the empty orbital being of the $\mathrm{sp^2}$ type.

Thankfully, this also immediately tells us that II is penultimate, since no other options exist and the question boils down to whether I or IV is more reactive.

If you were to assume an $\mathrm{S_N2}$ pathway, then indeed I would lose to IV due to increased steric inhibition: the electrophilic carbon is not easily accessible from behind.

However, both reactive carbons are benzylic and the neighbouring phenyl ring is very good at stabilising ionic charges. Considering that you also have a rather weak nucleophile — methanol — it is very likely that this reaction will proceed under $\mathrm{S_N1}$ conditions, so the rate determining step is the formation of the cation. From this point of view, I gives the better intermediate cation than IV because of the additional inductive stabilisation from the methyl group.

Therefore, the correct ordering is $\mathrm{I>IV>II>III}$ or answer C.


Balancing the redox reaction Question:

Balance the following redox reaction:

  $$\ce{FeS2 + O2 -> Fe2O3 + SO2}$$

My Efforts:

I tried balancing with oxidation number method.

First of all i determined the oxidation state as follows:
$$\ce{Fe^{(2)}S2^{(-1)} + O2^{(0)} -> Fe2^{(3)}O3^{(-2)} + S^{(4)}O2^{(-2)} }$$

Here, all the element are either oxidized or reduced, so how to move ahead?

I tried balancing with Half reaction method.

$$\ce{Fe^{(2)} -> Fe2^{(3)}} \text Oxidation$$
$$\ce{S2^{(-1)} -> 2S^{(4)} } \text Oxidation$$
$$\ce{O2^{(0)} -> 2O^{(-2)}} \text Reduction$$

Same here, how to move ahead?

P.S. Oxidation state are mentioned in the brackets. Correct me if i am wrong anywhere.

FeS2 has an odd structure; the iron atom has a +2 oxidation number and each of the sulfurs has a -1 oxidation number.

This can be balanced by inspection.

4 FeS2 + 11 O2 --> 2 Fe2O3 + 8 SO2

Just to check, using oxidation numbers we get:

Sulfur goes from -1 to +4.  (Total change from sulfur = +40)

Iron goes from +2 to +3.    (Total change from iron = +4)

(I'm not sure where you got 4 for the oxidation number of iron in FeS2.)

Oxygen goes from 0 to -2.   (Total change from oxygen = -44)

ETA:
Half reactions are messy here, since both iron and sulfur are oxidized.  Similarly, using oxidation numbers is problematic (except to check the solution) because there are three substances changing oxidation states.  You could write a system of equations to describe it, but that's a lot more trouble than it's worth.

You know from the structure of iron(II) disulfide that there are twice as many sulfur atoms as iron atoms.  That means that the number of sulfur dioxide molecules must be four times the number of iron(III) oxide molecules.  With that relationship in mind, the smallest ratio of molecules that fits the pattern is the one I wrote above.

If you really want to use oxidation numbers, here's what I've come up with:

Let a equal the number of iron atoms, b equal the number of sulfur atoms, and c equal the number of oxygen atoms.

Iron increases its oxidation state by 1, sulfur by 5, and oxygen decreases by 2. So:

1a + 5b - 2c = 0

We also know that iron and sulfur are in a 1:2 ratio because they come from pyrite.

2 a = b

Substitution yields:

1a + 5(2a) - 2c = 0

11a - 2c = 0      Iron and oxygen are in a 2:11 ratio.

Oxygen must be even since it comes as O2, so its smallest possible value is 22 atoms.  Using the ratios listed above, we get 22 O: 4 Fe: 8 S.

ETA #2: Half Reactions

Fe+2 --> Fe+3 + e-

S-1 --> S+4 + 5 e-

O20 + 4 e- --> 2 O-2

We know from FeS2 that there must be twice as many sulfur atoms as iron, so the second equation has to be multiplied by two.

2 S-1 --> 2 S+4 + 10 e-

If we add together all of the species that are being oxidized (Fe and S) then we get an oxidation half reaction of:

Fe+2 + 2 S-1 --> Fe+3 + 2 S+4 + 11 e-

This must be multiplied by 4 and the reduction equation by 11 to balance the number of electrons.

4 Fe+2 + 8 S-1 --> 4 Fe+3 + 8 S+4 + 44 e-

11 O20 + 44 e- --> 22 O-2

Adding this mess together gives:

4 Fe+2 + 8 S-1 + 11 O20 --> 4 Fe+3 + 8 S+4 + 22 O-2

Since we know the structure of the molecules, putting it all back together is fairly straightforward, and it gives the same result as listed above.


What is the point group of hexachloridotitanate(III)? I found the answer to this question to be $D_\mathrm{3h}$ but I am not convinced. Can someone explain me with a diagram if possible and explain why it doesn't belong to the octahedral point group $O_\mathrm{h}$?

Getting straight to the point, I believe whichever reference you have that claims $D_\mathrm{3h}$ is incorrect.

Titanium(III) complexes were the first example presented in my coordination chemistry class to examine the energy difference between the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ orbitals, explain why it exists and where the colour of coordination compounds derives from including carefully examining the UV spectrum of titanium(III). Our lecturer, Professor Klüfers of the LMU Munich, explained that the hexaaquacomplex is practically octahedral in shape. However, there is not just a clean absorption band but rather a band with a shoulder. This is because the excited state of $\ce{[Ti(H2O)6]^3+}$ is Jahn–Teller distorted, having an uneven $\mathrm{e_g}$ population, meaning that two slightly different excitation energies exist.

One might go a step further and argue that even in the ground state a certain distortion should exist. If one were to pull (not push) the ligands in $z$-direction slightly closer to the central metal, this would destabilise any orbitals with $z$ contribution meaning that $\mathrm d_{xy}$ suddenly becomes the single most stable d orbital. This slight distortion which one might call anti-Jahn–Teller (since it is opposite to the classic Jahn–Teller distortion) could explain a reduction of symmetry from $O_\mathrm{h}$ to $D_{\mathrm{4h}}$.

$D_\mathrm{3h}$ symmetry is not consistent with octahedraloid coordination spheres at all. It is very common for pentacoordinated centres such as $\ce{PF5}$. The only way to have six ligands — as in hexachloridotitanate(III) — around a central metal in this point group would be a trigonal prism. That would be a very uncommon coordination sphere and I am sure I would have heard of it in the context of titanium(III) in said course above should it exist for this species. Indeed, the source orthocresol found and linked in the comments makes no mention of any distorted geometry at all:

  The low-temperature structure results from rotations of the $\ce{[TiCl6]^3-}$ octahedra around a 4-fold axis. (Emphasis and square brackets mine)

Source: Amels, R.; Kremer, S.; Reinen, D. Jahn-Teller effect of titanium(3+) in octahedral coordination: a spectroscopic study of hexachlorotitanate (TiCl63-) complexes. Inorg. Chem. 1985, 24, 2751–2754. DOI: 10.1021/ic00212a009.


Can organic acids be concentrated and used like concentrated sulfuric acid and similar mineral acids? Is it possible to make organic acids concentrated like inorganic acids like concentrated sulfuric acid and similar mineral acids?

They can. Vinegar is typically 4-18% acetic acid by mass, but of course concentrated acetic acid also exists with a concentration >90% by mass. It is corrosive and flammable, but it surely exists.


Al2Cl6 - Different length of bonds How can we justify the longer length of the dative bond by bridge Cl with Al?
Is it because to prevent steric hindrance or repulsions the Cl atom is farther away from other Al as compared to normal Cl?

We know that the number of electrons in both the bonds are same. One of the bond involves a covalent bond $\ce{Al-Cl}$ where as the other is a 3-centred bond $\ce{Al-Cl\bond{->}Al}$. Thus while providing a dative bond, the effective atomic charge of bridged chlorine decreases in comparison with that at the terminal position. Due to this difference in atomic charge, the terminal ones attract more and closer that the bridges.


Coordination number of chromium(III) oxalate complex The coordination number refers to the number of binders that interact with the central metal ion. So, in the complex $\ce{[Cr(C2O4)3]^3-}$ chromium is bound to three oxalate ions.

Why its coordination number is considered 6? So, is it an incorrect definition of coordination number?

Can you help me?

This seems to be the case of "show, don't tell".

IUPAC's definition of coordination number (C.N.) which applies to inorganic complexes:

  In an inorganic coordination entity, the number of σ-bonds between ligands and the central atom. π-bonds are not considered in determining the coordination number.

Let's have a look at the one of the first determined structures of trioxalatochromate(III) inorganic salts [1]:

Figure 1. Crystal structure of $\ce{[Cr(C2O4)3]^3-}$ anion (ICSD-109607). Color scheme: $\color{#909090}{\Large\bullet}~\ce{C}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#8A99C7}{\Large\bullet}~\ce{Cr}$.

Here, there are three oxalate $\ce{C2O4^2-}$ ligands arranged in distorted octahedral environment (depicted with semi-transparent coordination polyhedron).
Each oxalate is a bidentate chelating ligand, resulting in C.N. 6 of chromium center.

References

van Niekerk, J. N.; Schoening, F. R. L. The Crystal Structure of Ammonium Trioxalatochromate(III), $\ce{(NH4)3[Cr(C2O4)3].2H2O}.$ Acta Cryst 1952, 5 (4), 499–505. DOI: 10.1107/S0365110X52001428.


Feasibility of Electrophilic attack on the Nitrogen over the Phenyl ring of Aniline I have learnt that the '-NH2' substituent attached to the phenyl ring, owing to the substitutent's highly positive mesomeric effect, activates the phenyl ring quite a lot for Electrophilic Aromatic Substitution, so much so that it easily forms a tri-substituted product when Br+ is used as an electrophile with Aniline, and that a protecting group such as Acetyl (introduced using Acetic Anhydride) has to be introduced on the Nitrogen to deactivate the ring for just monobromination to occur.

This made me wonder that if such is the influence of the '-NH2' on the benzene ring then why would the Nitrosonium ion generated in diazotization prefer attacking the lone pair of Nitrogen instead of the electronically enriched benzene ring on the ortho or para position?

Initially I lazily (and wrongly) attributed it to perhaps some stronger bonds forming between similar atoms, i.e Nitrogen-Nitrogen bonds being stronger than the Carbon-Nitrogen bonds.

But, more interestingly I realized that this line of reasoning wouldn't apply to my above stated acetylation of the Nitrogen in Aniline, for in that case too there isn't an Aromatic Substitution taking place and there are no Nitrogen-Nitrogen bonds.

In short, what is the reason for the feasibility of Diazotization and Acetylation in the way they occur with Aniline? Why aren't electrophilic aromatic substitution reactions taking place? I may have erred in my basics, please help me out.

Electrophilic Aromatic substitution in Aniline will occur when the $\pi$-electron cloud of Benzene ring attacks the electrophile generated (mostly in-situ)  instead of the lone pair on the Nitrogen atom. Thus for occurance of Electrophilic aromatic substitution, there always remains a competition between the $\ce{N}$ atom and the $\pi$-electron rich benzene ring. You may think that the lone pair of nitogen remains fully delocalised through the Benzene ring, but there always remains a high enough electronic charge density on the Nitrogen atom to behave as a nucleophile and in fact in greater extent the electronic density of the ring (not always for Oxygen though, being highly electronegative, its less nucleophilic, but Nitrogen is certainly more nucleophilic).

Thus,you will observe that it is difficult to perform electrophilic aromatic substitution reaction on  unprotected Aniline. Thus, most of the reactions on Aniline are initiated by attacking of the elctrophile by Nitrogen atom instead of the ring. It is its general trend. Thus to do substitution in the ring, we first need to decrease electronic charge density on the Nitrogen atom and that's why we first treat the Aniline with $\ce{CH3COCl}$ in Pyridine medium to form acetanilide, and then perform any other reaction on the ring like Bromination, Chlorination, Alkylation etc.

You will also observe that Aniline doesn't undergo Friedel Crafts alkylation/acylation to a great extent. As at the moment you add Lewis acid, Nitogen atom behave as Lewis Base and forms the adduct, which make the ring highly electron deficient due to induced positive charge on Nitrogen.

Thus, the Diazotization  and Acetylation are the normal/expected reaction of Aniline as a effect of higher nucleophilicity of $\ce{N}$(due to lesser elctronegativity and weaker basicity). Instead if you have to perform Electrophilic aromatic substiution in the ring, you have to think about how to perform it.


What's happening to carbon electrodes? Carbon electrodes are very popular, because unlike metal electrodes, carbon is quite inert during electrolysis (I'm sure that has some explanation based on electronegativity).

However they too are not eternally lasting. Especially the pencil graphite bars will slowly release black chunks into the solution.

My question is, what's happening them. The black chunks seem like carbon, so it's probably not reacting. What's damaging the structure?

Pencil leads aren't pure graphite, they're generally graphite powder with various polymers to hold them together and make sure they have the correct hardness (after all, you wouldn't be able to use them to write if they didn't leave graphite on the paper or if they crumbled under slight pressure. This is why the much more rigid glassy carbon is a popular material for carbon electrodes. My guess is that there's no reaction with the electrode happening but gas being produced is causing mechanical damage to the electrode. If you work at less extreme potentials, you might be able to reduce this effect. Otherwise, pure graphite rods should be more durable. Of course, if you're using pencil leads, they're obviously cheap to replace so it might be easier to just use them until they fall apart, then switch to a new one.


Why does the reaction between 2‐methylbut‐3‐en‐2‐ol and hydrogen bromide show allylic shift? In the following reaction

shouldn't the minor product be the major one as the carbocation formed before attachment of bromine is 3°? Also, can someone tell me why the product includes only nucleophilic substitution as there is no other condition mentioned to judge whether it should show elimination or substitution.

Please mention if what favours allylic shift in such type of reactions.

Even thought the immediate carbocation formed is tertiary, the stability of the double bond should also be taken into account. A more substituted double bond is also favored/stable. In this instance it seems that it's a competition between the double bond's stability and the carbocation's stability. There could be other situations where this is not the case, but you would need experimental data to back up stability (there's no universal ranking system that I know of to say when a double bond outweighs a positive charge).

In short, I would say that the relative stability of a tertiary double bond (compared to a tertiary carbocation) is what would justify the major product being labeled as it is.


Why do SN1 reactions prefer tertiary carbons? I can understand that $\mathrm{S_N2}$ reactions prefer primary carbons, because of the steric hindrance that would be an obstacle.

But what is the exact reason that $\mathrm{S_N1}$ reactions prefer tertiary carbons ?

The intermediate in an SN1 reaction is a carbocation. The stability of that intermediate determines how favorable that particular reaction is, so a more stable intermediate means a more favorable reaction.

Carbocations are quite unstable on their own, but the inductive effect of nearby C-C bonds provide some electron density to balance out the positive charge, making the carbocation intermediate more stable. Tertiary carbons have the largest number of adjacent C-C bonds, the largest inductive effect, the most stable carbocation intermediate, and are thus favored in SN1.


Magnetic couplings using DFT in a spin frustrated system I’m working with a trimeric copper system that have the classic triangular shape and experiences spin frustration. This is my first time dealing with such system and, so far, from what I’m reading, I know that the spins resonate between ↑↑↑ ↔ ↑↓↑ ↔ ↑↑↓ and that the triangle have couplings $J1_{ab}$, $J2_{bc}$ and $J3_{ca}$.

My question is: how it is possible, using DFT, to obtain the three magnetic couplings ($J1_{ab}$, $J2_{bc}$, $J3_{ca}$)? I’m supposing that the strategy of breaking the molecule in three dimers and calculating a broken-symmetry is not going to work, right?

Thanks in advance

P.S.: I’m doing my calculations using the ORCA package.

DFT is fundamentally the wrong tool for a system like this. When you say that the spin 'resonates' between several different states what you are explicitly saying is that the wavefunction is strongly statically correlated. DFT is a single reference method, so any experimental agreement you get from it in a case like this is at worst wrong and at best right for the wrong reasons (and the range of functionals will likely cover that whole spectrum).

If we knew the exact density functional DFT would, of course, be exact. Unfortunately, real functionals are parametrized for mostly organic compounds and the occasional single metal, but certainly nothing as correlated as a spin frustrated Cu$_3$ cluster. If you want a result which is believable, you need to use a real multireference method. (It's not going to be a quick calculation, but you only need a single point so it's not too bad)


How long burned tobacco/cigarette smoke remain toxic? It's hard to evaluate Environmental Tobacco Smoke lifecycle, it can certainly react, be absorbed by organisms, or be diluted in rain, and be emitted again from the ground.
But in the end, I'm interested to understand (as a non expert) if the toxicity (cytotoxicity, genetoxicity) of the compounds decrease over time (especially how are they eliminated) in a global outdoor environment where we suppose emissions would stop

Several resources list the components Chemistry and Toxicology of Cigarette Smoke, Comparison between Marijuana and Cigarette Smoke, CAEPA (page6):

Gas and particle phase contain roughly the same constituents as explained in the first paper, some compounds are 1,3-butadiene, acrolein, ammonia, aromatic amines, benzene, CO, isoprene, nicotine, nitrosamines, PAHs (polycyclic aromatic hydrocarbons), pyridine, toluene, hydrogen cyanide...

What could be their half-life in outdoor environment?

Toxicity of tobacco combustion products in different phases
For answering your question I think is better split it in two. First thing you need to understand that toxicity depends on the method of administation of the toxic agent. Tobacco smoke is a physical state of the tobacco combustion products. It is an aerosol originating from combustion where its particles are liquid, gas and solid (tar). Quoting Chemistry Central Journal 2012, 6:98

Cigarette smoke is a complex and dynamic aerosol consisting of at least 5,600 chemicals and toxicants found across two phases, the particulate (tar) and vapour phase.

(I would like to add the gas phase...) so when the product of the combustion of the tobacco are in this physical state you have to look at a different method of administration in this case vapour inhalation, Dust and Mist Inhalation and gas inhalation these are the most potentially dangerous method of administration. It seems that after about 40 minutes the liquid and solid part of the aerosol deposit (of course this depends on the condition, T, wind etc. etc.) so is probable that after this period of time the toxicity of tobacco combustion products decrease considerably. However this is not true for all the combustion products. For example: carbon monoxide is recognize tobacco combustion product and it is a gas so it wont deposit after this time and it will be "removed by reactions with OH radicals (85%), by soils (10%), and by diffusion into the stratosphere" very slowly.
Benzene is a good example of how the toxicity depends on the method of administration. LC50 inhalation value of benzene is 10000 ppm TFLo Dermal is 0.92 mL/kg so it is a big difference if you wait that the areosol deposit or if you inhale it.
If you are interested in rate of decomposition of benzene it seems that biodegradation of benzene can reach 0.95% at day according to Chiang.
Regarding 1-3 butadiene according to William A. McClenny and Donald Whitaker:

i.e., a 10-6 lifetime risk level for cancer due to inhalation exposure
of 0.03 $g/m^{3}$ (12.4 pptv at 20 °C and 1 atm pressure).1 This compound
is very reactive in the ambient atmosphere with a short
atmosphericlifetime, estimated to be 2–3 hr.

The half-life of acrolein is 14.4 hr.
You can also find a very complete EPA report EPA/600/P-98/001F
October 2002.
However determine the half-life of all the compounds is a very complex task because is related to how the ecosystem respond to these compounds so to the actual capability to decompose them through different mechanism is strictly related with thousands of factors.


Acid-Temperature coagulation of milk I was making coffee with a cup of boiling milk which was 2 days old. I used instant coffee powder and some sugar for the making of it. Tried to make a very hot coffee but usually prefer medium hotness. When I mixed the hot milk with coffee powder, I immediately observed coagulation.
Then, I decided to use an unopened milk packet and boiled it the same way. But this time I used a relatively more roasted coffee powder with same amount of common sugar. Again I observed coagulation but the particles were smaller.

I know that milk is a colloid and can be coagulated by using an acidic compound.
The casein starts coagulating at pH 4.6,
But I don't think that coffee is acidic enough to lower the pH. As in my second experiment, I've used a lesser acidic coffee powder (more roasted) but still observe coagulation.

In fresh dairy milk, is there any relation between titratable acidity and heat coagulation?
Does hydrogen ion concentration determine the coagulation?
Is it the coffee power or the milk or the temperature?
How to avoid this effect?

Coagulation of milk is a result of casein micelles clumping together. This is favoured by higher temperatures and also by favour a pH of around 4.6 as this is the isoelectric point. At this isoelectric point, the amino acids within the casein micelle are electrically neutral. This means there is less electrostatic repulsion between the casein micelles, facilitating coagulation Furthermore, the addition of acid means that the kappa casein, i.e. hairs on the outside of the casein molecule which stabilise it, become protonated. This protonisation causes them to destabilise, making it easier for the micelle to fall apart and clump together with other micelles that have fallen apart. Relatively High temperatures also increase reaction rates, which overall speeds up the aforementioned processes as well.

Even if the coffee does not reduce the pH by much, any change in pH which leads to the pH being closer to the isoelectric point (4.6) of the casein micelle will increase the rate of the coagulation process.

Using old milk also increases coagulation, as milk naturally coagulates due to rennet, an enzyme, cleaving the kappa-casein, causing the casein molecules to destabilise.

To avoid coagulation during heating, perhaps try heating the milk gently. The ideal temperature at which milk coagulates is about 43ºC, so perhaps staying below that could work. This source also provides tips on how to avoid coagulation while heating milkAlternatively, you have yourself some cheese to enjoy :)


Explanation of bond angles in the aluminium chloride dimer In order to attain stability $\ce{AlCl3}$ dimerises to $\ce{Al2Cl6}$ whose structure is depicted as:

I want to understand why one of the bond angles is $79^\circ$ and the other $118^\circ$. Is it possible to justify this using Bent's rule?

The bonding in aluminum chloride ought to be considered largely ionic. The electronegativity of $\ce{Al}$ is $1.47$ vs $2.83$ for $\ce{Cl}$ (for comparison: $\ce{C}$ $2.50$; $\ce{Mg}$ $1.23$; $\ce{Na}$ $1.01$). Molecular orbitals and covalent bonding can be added for a fuller picture of electron densities.

Aluminum chloride solid (Melting point $\pu{193^\circ C}$, under pressure) has a cubic close-packed array of chlorine atoms with only $2/3$ of the octahedral holes between every other pair of chlorine planes occupied by the aluminum atoms. At low temperatures in the vapor state, aluminum chloride is a tetrahedrally coordinated dimer, and at very high temperatures it is a trigonal monomer. So aluminum can easily have $3$, $4$ or $6$ coordination. Whatever directionality the bonds have from hybridization of the $\ce{3s}$ and $\ce{3p}$ orbitals is mixed with plain electrostatic (ionic) attraction.

Imagine putting just the 6 chlorine atoms together. There will be two tetrahedral holes for aluminum (see image below). The radius ratio
$$\frac{\ce{Al^3+}}{\ce{Cl-}} = \frac{\pu{50 pm}}{\pu{181 pm}} = 0.276$$
is well within the stability range ($0.225 - 0.414$) for tetrahedral coordination. So you might expect all the $\ce{Cl-Al-Cl}$ bond angles to be $109^\circ$.

But while the void for the aluminum is big enough, the distance between the internal chlorine atoms would be only $\pu{300 pm}$ in this configuration; the chlorine atoms are squeezed (ionic radius $\pu{181 pm}$). In addition, the $\ce{Al-Al}$ distance is only $\pu{250 pm}$ and the $\ce{Al^3+\bond{-}Al^3+}$ repulsion would be extreme. (In the solid, the octahedral holes are farther apart) So both aluminum atoms and the internal chlorine atoms are pushed apart, consistent with imagining more $\ce{p}$ character in the central bonds and therefore less $\ce{p}$ character in the external chlorine atoms, so $\ce{Al}$ goes to $\ce{sp^2}$ (or nearly so). The internal $\ce{Cl-Al-Cl}$ bond angle is even less than $90^\circ$ (less than you would expect for two $\ce{p}$ orbitals), so π bonding needs to be considered.

Aluminum chloride is a strong Lewis acid. Its solubility in organic solvents is due to formation of complexes. It is not soluble in the way that carbon tetrachloride or hexachloroethane are soluble. The salt-like (ionic) character of solid aluminium chloride is still present in $\ce{Al2Cl6}$; the aluminum atoms are surrounded by a cocoon of chlorides. Solid aluminum chloride can be ground into particles and nano particles; $\ce{Al2Cl6}$ is like a pico particle of salt.


Effect of impurities on melting point This is a 10th-grade question.

  The melting point of a substance decreases with increase in presence of impurities in it. The melting point of ice decreases from 0 °C to -22 °C on mixing salt in it in proper proportion. That is why salt is added to make freezing mixtures to keep ice creams frozen.

I want to ask that suppose ice melts at 0 °C and we are now adding salt to lower melting point but now when we have lowered the melting point the ice will melt more quickly, then what is the purpose of mixing salt?

In the old fashioned process, we used buttermilk sugar and a bit of vanilla. We put them into a churn (which is just a small bucket with a mixing blade in it so you could hand mix the contents) and then added ice around the churn (outside the churn) to cool it down. And then you mixed and mixed and mixed and mixed. But there was a problem. As the ice melted, the water it created was at $\pu{32°F}$ and even though the ice was colder (a commercial freezer is often about $\pu{15°F}$ (but a home freezer is warmer)) the water kept the churn at around (usually above) $\pu{32°F}$ ($\pu{0°C}$). And ice cream (being an "impure" water mixture) is a liquid, not solid, at $\pu{32°F}$. So, this wouldn't work. Enter salt. If you added salt to the ice, then the ice would melt well below $\pu{32°F}$ and be able to cool the churn to the freezing point of the ice cream mixture (I think its about $\pu{25 °F}$, but my memory shouldn't be trusted). So, the salt and ice form a salt water mixture which can be well below $\pu{32°F}$, and so can cool to below $\pu{32°F}$. If you were to add salt to the recipe, then you'd have to cool it even MORE to get it to freeze (although a little salt makes it taste better, imho). So, salt is added to the cooling mixture and (supposedly) not to the recipe's ingredients. I guess I should point out that you need to consider walking over hot rocks vs walking in say an inch of water at the same temperature. Which would burn you faster? There's no question that your feet would have more contact with the water than the solid rocks, so your feet would be more damaged by the water (water at $180$ or $\pu{200 °F}$ WILL burn you). My point is you actually WANT the ice to melt into a salt solution, because that will better cool the churn - as long as there's enough salt so that the liquid is cold enough.


How does aminoguanidine interfere with the formation of advanced glycation endproducts? This says that aminoguanidine interferes with the formation of advanced glycation endproducts (AGEs) through interacting with 3-deoxyglucosone.

But how does it interact with 3-deoxyglucosone that prevents AGE formation?

I would expect the N-amino group of aminoguanidine to react with 3-deoxyglucosone at either of the adjacent, reactive carbonyl groups to form a stable guanidinylimine derivative (analogous to a hydrazone). This derivative, unlike 3-deoxyglucosone itself, would not be able to react with free amino groups on proteins and continue on the pathway to AGEs, but rather would be excreted as is.