Bonding and coordination of oxygen in a Ga2O3 crystal structure I'm trying to find the crystal structure of gallium oxide ($\ce{Ga2O3}$). However, I find the images of the crystal structure in the peer-reviewed journals problematic.

First, I hop to the wikipedia article of Gallium oxide and find this image:

There was no labeling of the color, but I think Cyan is Gallium, and Red is Oxygen. The trouble is Ga typically has 3+ and Oxygen has 2- valence. Yet in this image we see Ga have 4 and 6 bonds! How is this possible?

The source is from a paper:  J. Åhman, G. Svensson and J. Albertsson: A Reinvestigation of β-Gallium Oxide. In: Acta Cryst. (1996). C52, 1336-1338. (The image was made by a wiki admin on chemistry structure, who made many structures for wikipedia.)

So I check out the article and it has this figure:

That's basically unreadable, so I go to Acta Crystallographica Section C and it has an interactive 3D chem model viewer and a model for Gallium Oxide:

Again I see Gallium with 6 bonds and oxygens with 3 bond! Is this correct? please tell me how?

In Galazka's β-Ga2O3 for wide-bandgap electronics and optoelectronics,I find this description:

  Crystal structure of $\ce{β-Ga2O3}$

  $\ce{β-Ga2O3}$ crystallizes in the base-centred monoclinic system in the
  space group C2/m. The unit cell (figure 1) contains 20 atoms
  consisting of crystallographically inequivalent $\ce{Ga^3+}$ and $\ce{O^2−}$ atoms. In this low symmetry structure Ga atoms are coordinated tetrahedrally and
  octahedrally (Ga1 and Ga2, respectively), while O atoms are
  coordinated three and fourfold (O1/O2 and O3, respectively). O1 shares
  two bonds with Ga2 and one bond with Ga1, O2 shares three bonds with
  Ga2 and one bond with Ga1, while O3 shares two bonds with Ga1 and one
  bond with Ga2. The lattice parameters of $\ce{β-Ga2O3}$ are listed in table
  1. There are two easy cleavage planes: the (100) plane formed by O3 atoms and (001) plane formed by O1 atoms.

and this figure:

Can someone please explain the crystal structure of $\ce{Ga2O3}$, beta-gallium oxide?

Does it help if you think of this as an ionic compound? In ionic compounds, there are no directed bonds. The coordination number depends on the size of the ions, coupled with the need for electroneutrality.

For example, in calcium oxide, both calcium and oxygen (in red) have octahedral coordination, with 6 neighbors each:

Granted, the electronegativity difference in calcium oxide is 2.44, while in gallium oxide it is only 1.65, and not different from that of silicon oxide, which has a network covalent structure.

Also, consider structure like that of hexagonal ice, where each oxygen is surrounded by four hydrogen atoms, two connected by covalent bonds, and two connected by hydrogen bonds. In this case, the atomic distances are clearly different, and you would be correct to say that oxygen makes two bonds, with a tetrahedral coordination if you consider the hydrogen atoms from neighboring atoms.

To distinguish these three extremes (ionic bonds in one case, covalent bonds plus contacts with neighboring atoms, network covalent structure) and to categorize what is going on with gallium oxide, you would have to look at the gallium oxygen bond lengths, which should be available in the paper you cite.


Why is bromobenzene more stable than benzene? Benzene is 45 kJ/mol more energetic than bromobenzene. I understand that due to resonance in bromine to benzene in bromobenzene increase the stability but such an enormous amount of energy can't only be dealt by resonance. There must be another reason. What is it?

The corresponding section in Wade, L. G. Organic Chemistry, 8th Edition; Pearson Education: Glenview, IL, 2013; p 758. reads

  Bromination of Benzene Bromination follows the general mechanism for electrophilic aromatic substitution. (…)

  Formation of the sigma complex is rate-limiting, and the transition state leading to it occupies the highest-energy point on the energy diagram (Figure 17-1). This step is strongly endothermic because it forms a nonaromatic carbocation. The second step is exothermic because aromaticity is regained and a molecule of HBr is evolved. The overall reaction is exothermic by 45 kJ/mol (10.8 kcal mol).

Thus, the given value of 45 kJ/mol refers to the reaction enthalpy of the total reaction
$$\ce{Br2 + C6H6 -> C6H5Br + HBr}$$

The reaction is mainly so exothermic because of the formation of HBr. This can be shown using values for the standard molar enthalpy of formation taken from “Standard Thermodynamic Properties of Chemical Substances”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

$$\begin{align}
&\Delta_\mathrm f H^\circ(\ce{Br2(l)})&&=0\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{Br2(g)})&&=30.9\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{C6H6(l)})&&=49.1\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{C6H5Br(l)})&&=60.9\ \mathrm{kJ/mol}\\
&\Delta_\mathrm f H^\circ(\ce{HBr(g)})&&=-36.3\ \mathrm{kJ/mol}\\
\end{align}$$

although the exact value of 45 kJ/mol cannot be reproduced using these values.


Why are amino acids named amino acids? When learning about amino acids, the primary things that were pointed out were that they had an amino group and a carboxyl group, and side chain that varies from one to the other.

The amino group of one can attach to the carboxyl group of another, making a peptide bond and water.

Both groups are clearly important, but what is it about the amino group that that's what goes in the name "amino acid"? Is the amino group in some way more important? Is it just something about the way/order in which they were discovered/identified that the name just came about?

Why amino acid? Not carboxyl acid, or some mix of the two, or something else?

When learning about amino acids, the primary things that were pointed out were that they had an amino group and a carboxyl group...
Why amino acid? Not carboxyl acid, or some mix of the two, or something else?

You stated why it is called an amino acid, the name comes from the amine (amino) and the carboxylic acid (carboxyl group).
Amines are weak bases, so the acid portion of the name must come from the carboxylic acid.


Determine oxidation in organic chemistry through oxidation numbers? I want to determine the carbons that are oxidatized in an organic reaction.

Is it an acceptable way to first figure out the oxidation number of every carbon in reactants and products and then compare the oxidation numbers (for example: $\ce{+I -> +II} =$ Oxidation; electron loss)?* Or is my attempt to work with oxidation numbers false (for example: an aldehyde and keton are on one oxidation level, but their carbon has different oxidation numbers?). Maybe I should go over the oxidation levels of functional groups given in many textbooks as tabulars (but my structures aren't that easy and have not the most common functional groups)?

*I have read in many posts here, that this is a legitimate way — but there is "my problem" with the keton and the aldehyd, as mentioned above.

Yes, determining oxidation numbers and thereby finding out what is oxidised and what is reduced is a valid method, especially for reactions that include oxidations or reductions.

Your ‘problem’ with an aldehyde and a ketone is actually not a problem at all. Ketones have an oxidation state of $\mathrm{+II}$, which means that the only thing they could be oxidised to in a two-electron (i.e. non-radical) process would be $\ce{CO2}$ — hence why they are stable towards oxidation. Aldehydes, on the other hand, have an oxidation state of $\mathrm{+I}$, which means that they can easily be oxidised (to carboxylic acids, oxidation state $\mathrm{+III}$) by simplet two-electron processes.

Likewise, the different oxidation states of primary ($\mathrm{-I}$), secondary ($\mathrm{\pm 0}$) and tertiary ($\mathrm{+I}$) alcohols can be used to help understand why it makes sense to distinguish.

However, do note that many organic reactions — while definitely involving a change of oxidation states — do not fit into the classical redox paradigm well. Throughout the mechanism of a Horner-Wadsworth-Emmons reaction, the (tertiary) phosphonate carbon changes its oxidation state from $\mathrm{-II}$ to $\pm 0$ in two distinct, one-electron steps, while the aldehyde carbon goes from $\mathrm{+I}$ to $\mathrm{-I}$ also in two distinct steps — but nobody would go as far as calling it a redox reaction.


Why does Calcium have a higher ionization energy than Aluminium? Given their places on the periodic table I'd assume Aluminium has a higher ionization energy, because it has fewer energy levels, and is on a "righter" row on the periodic table, but in reality it is the opposite. Does anyone here know why?

Ionization energy is the minimum amount of energy needed to remove most loosely bound electron (valence electron).
The atomic number of Calcium  is 20. So it's orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s$2 3p^6 4s^2}$. This is the arrangement of electrons in the shells. No free electron is present in the above configuration, all the electrons are bonded. So for removal of an electron from the outermost shell higher amount of energy is required.
In case of Aluminium, the atomic number is 13. It's orbital configuration is $\mathrm{1s^2 2s^2 2p^6 3s^2 3p^1}$. One single electron is present unbonded in outermost shell, so it can be easily removed as compared to Calcium.
So, ionization energy of Calcium is higher than the ionization energy of Aluminium.


What is Polymer Environmental Resin (PER)? I'm no chemist and I probably wouldn't understand any technical answer but...

I'm investigating the environmental impacts of various materials as compared to PVC and this is one that crops up regularly as a more environmental alternative.

Unfortunately there does not seem to be any authoritative information on the web concerning this material.

I'm beginning to think that this is a green-washing propaganda move by the phthalate/heavy metal free PVC industry as a move avoid mentioning PVC in product advertising.

Anyway I would welcome any information on what exactly PER is and what it is made from.

Thank you.

PER is essentially PVC which has been plasticized and stabilized with acetyl tributyl citrate, instead of the host of phthalate-based plasticizers such as bis-diethylhexyl phthalate (DEHP) whose endocrine-disrupting properties give PVC its well-deserved bad reputation.

Citrate salts tend to be readily water soluble, and therefore this material is touted to be more biodegradeable than phthalate-plasticized PVC. I certainly cannot find any information to refute that claim. ATBC will further degrade into other acetates and butyrates, most of which are relatively common and harmless in the environment (though butyl compounds are infamously stinky; butyric acid is a favorite less-lethal offensive-odor weapon of law enforcement and groups like the Sea Shepherds). All PVC will eventually degrade as the plasticizer leaches out over time, unlike other types of plastics that are so ridiculously inert they'll likely be the fossils of our modern era, but again, the phthalates just aren't a great idea to have in the environment.

Of course, neither is the vinyl chloride monomer. What many proponents of PER fail to mention is that the monomer is worse for your health than the plasticizers. It's carcinogenic in relatively small doses, for one thing. It's also toxic to the liver in even smaller doses, measured in the low parts per million of body mass for an acute exposure. Lastly, the monomer is extremely flammable, though this is mainly an industrial hazard; the polymer is actually among the more flame-retardant commercial plastics, due to its chlorine content (in other words it's already highly oxidized, as opposed to something like polyethylene). None of this changes when the plasticizer does.

In short, while PER is a step forward, there are still plenty of inherent hazards in the production, long-term use and disposal of PER products. PER is also a relative specialty product, whose primary use is a foam product marketed for children's toys and yoga mats. It's not a silver bullet to replace PVC wholesale.


How fast does MgSO4 absorb water from the air? I had to do a lab to figure out the water of hydration of magnesium sulfate. However, after the heatings, I had a loss 0.45 g of water resulting in 0.63 g of anhydrous magnesium sulfate. This gives 0.02497 mol of water and 0.005237 mol of anhydrous magnesium sulfate. Thus, this would imply that 0.02497/0.005237 = 4.77 = 5 is the water of hydration.

However, my chemistry teacher specifically said that the actual number for the MgSO4 he gave us was 7. I couldn't figure out exactly where I went wrong, but my only guess would be that the wait of 5 minutes for cooling the substance down. Some research shows that MgSO4 is a great drying agent, and the humidity during the time I carried out the experiment was 75% humidity.

Doing the calculations, during the five minutes of cooling, it must have absorbed 0.01169 moles of water in order for 7 to drop to 4.77. Is this plausible given 75% humidity? I don't know anything about Chemical kinetics yet.

Edit: Here are the procedures.
1. Measure the mass of the empty crucible with the lid.

Add some amount of the (powdered) hydrated salt in the crucible, and measure it's total mass with the lid.
Heat the crucible for 20 minutes over a bunsen burner, with lid slightly open.
Wait for 5 minutes, and then measure the amount of mass after heating.

The relevant data is given above.

This is a great question, and unfortunately, I don't have an answer for you.

My first instinct would be to question how well you heated the material and under what conditions. Insufficient heating would mean that you might still have water remaining in the material.

Anhydrous, powdered magnesium sulfate will absorb water relatively quickly at ambient conditions, though it will depend on the temperature of the solid, how anhydrous it is and the humidity of the air. Something more to consider is that the crystalline form will absorb water at a slower rate. Were you dealing with powder or crystals?

Unfortunately, it isn't possible to judge the kinetics without more information. Perhaps you can provide your procedure?


Why do gases coexist as liquid and vapour? I have having difficulty in understanding the liquefaction of gases.In the graph
of isotherms of $\ce{CO2}$ I cannot understand how can gas coexist as liquid and  vapour under area of $\mathrm{XCEBY}$.

I have read the following statements in my textbook:

At $\pu{21.5 ^\circ C}$, $\ce{CO2}$ remains a gas up to point $\mathrm B$. At point B a liquid of specific volume appears.In fact,the gas and liquid  $\ce{CO2}$ coexist along the horizontal line $\mathrm{BC}$. The decrease of volume represents condensation of more and more $\ce{CO2}$ gas till point $\mathrm C$ is reached.

A careful study of isotherms reveal that it is possible to change gas to liquid by following a path in which a single phase is present throughout the process.For example,move vertically from $\mathrm{A}$ to $\mathrm{F}$ by increasing the temperature of the gas.Now move from point $\mathrm{F}$ to point $\mathrm{G}$ by increasing pressure at constant temperature.Move from point $\mathrm{G}$ to point $\mathrm{H}$ by lowering the temperature.As soon as we cross point $\mathrm H$ we get a liquid.

(i)Why $\ce{CO2}$ coexist as liquid and vapour in the area of $\mathrm{XCEBY}$?(instead of converting directly into liquid by condensation of all the molecules simultaneously)
(ii)Why under some range of pressure and temperature $\ce{CO2}$ converts directly into liquid without going through the two-phase region?

Vapor-liquid coexistence.

This is possible because changing the system's molar volume $V_\mathrm m$ does not affect the Gibbs free energy of either phase, so equality of the Gibbs free energy (that is, phase coexistence,) can be maintained for a range of $V_\mathrm m$; this is the coexistence region $\mathrm{XCEBY}$.

On the other hand, changing $p$ and $T$ will affect the Gibbs free energy, and if instead we plot this phase diagram on the $p,T$-plane, the entire coexistence region $\mathrm{XCEBY}$ will reduce to a single curve, the liquid-vapor coexistence curve.

In other words: Fix $T$. Only one value of $p$ results in phase coexistence; phase coexistence is a curve on the $p,T$-plane. A range of $V_\mathrm m$ results in phase coexistence; phase coexistence is a region on the $p,V_\mathrm m$-plane.

Phase transitions without coexistence.

This is made possible because the liquid and vapor phases share continuous translational symmetry. In contrast, a phase transition from the liquid phase to a solid, crystalline phase requires breaking this continuous symmetry, resulting instead in the discrete translational symmetry possessed by a crystalline lattice. This shared symmetry permits the possibility of continuously going from one phase to the other without an explicit phase transition.

This possibility is realized past the critical temperature, the temperature represented by the critical point $\mathrm E$. Past the critical isotherm identified by line segment $\mathrm{HE}$ is the supercritical fluid "phase", which shares properties of both liquid and vapor. Isotherms like $\mathrm{GF}$ do not represent a liquid-vapor transition, because the species remains as a supercritical fluid throughout. A liquid-vapor transition that bypasses the coexistence region, like $\mathrm{AFGHD}$, is a continuous transformation from liquid to supercritical fluid to vapor, rather than a discrete phase transition.


Color of pheomelanin? Do we know what is the wavelength of the color of pheomelanin?

Pheomelanin has a very broad absorption spectrum in the visible, with monotonically increasing absorption as the wavelength decreases until a maximum is reached at 300 nm.

I think that this would correspond to a dark red/brown colour.

Source: Phys. Chem. Chem. Phys. 2011, 13 (20), 9119–9127


Why is a [Cu(SCN)2] complex black? I've been creating various Copper(II) complexes using different ligands and predicting their relative hues using Crystal Field Theory by using the spectrochemical series to predict the extent of d orbital splitting on the Copper(II) ion and the resultant colour of the copper(II) complex.

When I added potassium thiocyanate ($\ce{KSCN}$) to a solution containing hexaaquacopper(II) the entire solution turned deep black. Doing some research online and knowing that $\ce{SCN-}$ tends to form complexes with planar geometry I'm fairly certain that the resultant copper(II) complex was $\ce{[Cu(SCN)4]}$.

Because of the relative strength of the $\ce{SCN-}$ ligand compared to the pale blue complex formed with $\ce{H2O}$ I assumed the solution would appear redder in hue as higher frequency wavelengths would be absorbed, however, I fail to see why the entire visible spectrum would suddenly be absorbed. Can this phenomenon be explained with Crystal Field Theory?

Also perhaps my solution was not dilute enough?

As the user above states, if you mix concentrated solutions of $\ce{Cu(II)}$ salts and $\ce{NCS}$ you get $\ce{Cu(NCS)_2}$ which is a black solid. I'd disagree with them that it's not longer a complex because it 100% is - and you are correct to assume that the local coordination is $\ce{Cu(NCS)_4}$. To be more precise, each copper is coordinated by $\ce{Cu(NCS)_2(SCN)_2}$, and there's actually a large Jahn Teller distortion so it could also be described as $\ce{Cu(NCS)_2(SCN)_4}$. They're right that the colour can't be easily explained by $d-d$ transitions (i.e. the kind you're thinking of w.r.t. ligand field splitting) though.

$\ce{Cu(NCS)_2}$ is black probably because of ligand to metal charge transfer (the same reason $\ce{Fe(III)NCS}$ is blood red) - i.e. on absorption you transiently form $\ce{Cu(I)}$ and $\ce{NCS}$ from $\ce{Cu(II)}$ and $\ce{NCS–}$. It's more complex that that for sure, but I think it's fair to say no one knows yet, because we only worked out the structure two years ago.

If you want a lot more information about $\ce{Cu(NCS)_2}$ we published on it here https://journals.aps.org/prb/abstract/10.1103/PhysRevB.97.144421 (on the arxiv at https://arxiv.org/abs/1710.04889).


Is the effective nuclear charge only defined for valence electrons? I've just been taught about the effective nuclear charge and screening effect in school. I'd like to ask and confirm if the effective nuclear charge is only defined for valence electrons. If not then why don't we consider the repulsion from the electrons further away from the nucleus than the test electron, which by common sense should add to the attractive force by nucleus?

$Z=Z_{eff}+\sigma$, where $\sigma$ is shielding constant calculated according to Slater's rule. Since Slater's rule can be applied to any electron regardless of whether or not it's a valence electron, I think $Z_{eff}$ and $Z$ can be defined for any electron too, $Z$ in particular remaining constant.
The effect of outer electrons not being taken is also explained by Slater's rules empirically.


Cannizzaro reaction versus Nitroaldol reaction In Clayden et al. on p. 696 the following nitroaldol reaction is listed:

Later in the same chapter they explain the Cannizzaro reaction. Why is the yield of this nitroaldol reaction particularly good, while a Cannizzaro reaction is also competing for the benzaldehyde? Is the nitroaldol reaction that much faster?

I looked in some literature and they mostly seem to use nitrogen bases (cyclohexylamine, $\ce{Et3N}$, $\ce{n-BuNH2}$, methylamine, etc.) for these kind of reactions (nitroaldol reaction with aromatic electrophile). The reported yields are also typically lower. See for example the "Literature method" section of the synthesis of 1-phenyl-2-nitropropene on Erowid (which seems to be some kind of precursor to amphetamines, so there is plentiful information about it in comparison to the 1-phenyl-2-nitroethene synthesis listed in Clayden. Can the yields be so different due to the steric hindrance of the additional methyl group?). What is the benefit of these nitrogen bases over $\ce{NaOH}$? Clayden et al. only mentions them with the comment that $\ce{NaOH}$ is not really necessary since nitrogen bases are sufficiently strong for nitroaldol reaction, because the typical pKa value of short chain nitro alkyl compounds is ~10.

Is the nitroaldol reaction that much faster?

My guess would be that the nitroaldol is faster than the Cannizzaro.  The Cannizzaro reaction involves

attack by hydroxide on the carbonyl to reversibly form the tetrahedral intermediate
alignment of the tetrahedral intermediate with a second aldehyde molecule (entropically costly)
hydride transfer in a redox reaction (relatively high energy, even complexed hydride is not particularly stable) to form molecules (alcohol and carboxylic acid) that are not that different in stability than the starting material

The nitroaldol involves

generation of the nitroalkyl carbanion
attack by the carbanion on the carbonyl to form a tetrahedral intermediate
capture of a proton by the negatively charged oxygen to produce a neutral molecule
elimination of water to produce a more stable (more conjugated) molecule than we started with (e.g. energetically downhill)

The nitroaldol doesn't require any special molecular alignment after the tetrahedral intermediate is formed and is energetically downhill.

  Can the yields be so different due to the steric hindrance of the
  additional methyl group?

Further alkylation at the nitroalkyl carbanion center will not help the reaction in a steric sense just as you suggest, but also a secondary carbanion will be less stable (higher energy, more difficult to form) than a primary carbanion.

  What is the benefit of these nitrogen bases over $\ce{NaOH}$?

They are somewhat milder reagents than hydroxide and therefore less likely to produce side reactions.


Chemical nomenclature of ions I'm a very beginner in chemistry. The book I'm studying says that an ion is an atom, or a collection of atoms, that has an electric charge, then it writes about the difference between molecules and ions. What I'm wondering is, if I consider a unit of, say, sodium chloride (by unit I mean one atom of sodium plus one atom of chlorine), should I say "a molecule of s. chloride" or "an ion of s. chloride"?

It is said: one formula unit of sodium chloride.


Is there a temperature where salt doesn't dissolve in water? I know that salt dissolving in water is driven mostly by entropy. However, at lower temperatures, entropy is less of a factor in controlling reactions. Is there a temperature where salt doesn't dissolve in water?

Solubility is an equilibrium phenomenon, which means that its temperature dependence is a function only of the enthalpy change of the reaction, not the entropy. See the van't Hoff equation for the mathematical relationship.

If by salt you mean sodium chloride, the dissolution in that case is only very slightly endothermic, so the solubility decreases very slowly with temperature.

That means that if you go low enough in temperature, you will reach a point where the solubility theoretically approaches zero. However, the water will freeze long before the temperature gets that low, and solubility will no longer have meaning.


Software recommendations for overlaying molecular structures I am looking for a program to overlay the structure of small molecules as an aid for my students in comparing the results of optimization geometries using different optimization methods.

At this stage, we are thinking of atomic coodinates and a visual comparison. However, if it can be complemented with a quantitative comparison that would a welcomed extra.

We work with the Windows versions of Gaussian 09 and GaussView 5.

You can use the "compare" command in the Jmol program, either online or as free-standing program. A tutorial on how to use the program for this purpose is here, http://proteopedia.org/wiki/index.php/Superposition_with_jmol. It outputs the root mean square of the pairwise differences (RMSD) as well as the superposed coordinates.


Different Equivalent Weight for the same compound Let us say that we have the following two reactions:
$$\ce{A + O2 -> B}$$ where $\ce A \text{ and } \ce B $ are different oxides of metal $\ce M$.
Weights of $\ce A$ and $\ce B$ are $\pu{5.72 g}$ and $\pu{6.36 g}$.
$$\ce{B + H2->M + H2O}$$
The weights of $\ce B$ and $\ce M$ are $\pu{4.77 g}$ and $3.81$ g, respectively.

So I was trying to infer some information from the given data and this is what I did:

In the first reaction we can calculate the equivalent weight of $\ce A$ and $\ce B$ to be $\pu{71.5 g}$ and $\pu{79.5 g}$ respectively.
In the second reaction we can calculate the equivalent weight of $\ce B$ and $\ce M$ to be $\pu{39.75 g}$ and $\pu{31.7 g}$ respectively.

If we compare the equivalent weight of $\ce B$ in the two reactions, the equivalent weight of $\ce B$ in the first reaction is twice the equivalent weight in second reaction. What exactly does this mean?

Equivalent weights are considering a redox equation with one mole equivalent of electrons. But some redox reactions require more than one mole equivalent of electrons to be equilibrated.

Thus, if you have the second reaction that leads to a weight equivalent of $1/2$ of that for the first reaction, it means that the reaction 2 has consumed $1/2$ of  $B$ compared to reaction 1 for one mole equivalent of electrons.

You still cannot proceed to the molecular weights on these basis, but you can now try to consider that the most likely case is that the first one involved only one electron and the second one two electrons. Thus you can write that this is consistent with:

First equation:

$$\ce{A \rightarrow B + e-}$$

(equilibrated by $\ce{\frac 14 O_{2} + e- \rightarrow {1\over 2} O^2-}$).

And for the second equation:

$$\ce{B + 2 e- -> M}$$

(equilibrated with $\ce{H2 + O^2- -> H2O + 2 e-}$).

One this is said, one should consider the molar weight of $\ce M$ to be twice the equivalent weight of $\ce M$ calculated from the second equation, leading to a molar weight of 63.5 g/mol. Quite unsurprisingly, this is the molar weight of copper, so $M = Cu$. And going backwards, $\ce{B = CuO}$ with copper in its second oxidation state. We can check this, as the molecular weight of CuO is 76.5 g/mol, exactly the equivalent weight calculated for reaction 1. And thus $\ce{A = Cu2O}$ with copper in its first oxidation state.


How to calculate the amount of aluminium that has been oxidised? I have a sample of pure aluminium (100%), 1 kg, stored in "normal condition"
After some time the weight of sample has increased by 1%.
I assume that aluminium has oxidized to $\ce{Al2O3}$

I know the weight before and after, and my question is: how to calculate how much aluminium has been oxidised?

Initially, 1 kg of $\ce{Al}$ is present.
After some time, oxidation increases the weight by 1%, or 10 g (e.g. 10 g of oxygen has been added).
The atomic mass of oxygen is 16.0 g/mol.
The molar ratio of $\ce{Al}$ to $\ce{O}$, from the molecular formula $\ce{Al2O3}$, is $\mathrm{\frac{2~mol~\ce{Al}}{3~mol~\ce{O}}}$.

$$\mathrm{\frac{10~g~\ce{O}}{16.0~g~O/mol~O}=0.625~mol~O}$$

$$\mathrm{\frac{2~mol~\ce{Al}}{3~mol~O}*0.625~mol~O=0.417~mol~\ce{Al}}$$

$$\mathrm{\frac{27.0~g~\ce{Al}}{mol~\ce{Al}}*0.417~mol~\ce{Al}=11~g~\ce{Al}}$$

Note that even though only a single significant figure is given for each of steps 1 and 2 of the problem, I assumed that they should read "1.0 kg" and "1.0%" rather than just "1 kg" and "1%" for the purposes of illustrating the above calculation. Thereby, 0.42 mol or 11 g of aluminum was oxidized to $\ce{Al2O3}$.


Can rate constant depend upon the concentration of reactants (or other species involved in the reaction)? Transition state theory tells us that the rate constant of an elementary step is $$k_\mathrm{r} = \frac{\kappa k_\mathrm{B}T}{h}\exp\left\{\frac{- \Delta G^\ddagger}{RT}\right\},$$
where $\Delta G^\ddagger$ is the activation free energy of the step, i.e. the difference between the Gibbs free energy of the transition state and the reactant state.
Then the rate law would be $$r = k_\mathrm{r}[A]^a[B]^b[C]^c\dots,$$
where $A$, $B$, $C$, ... are all the reactants that come together in that elementary step. So far so good.
But I have come across reactions where changing the concentration of a reactant (by a large amount) in a multi-step reaction seems to change the rate determining step of the whole reaction.
The explanation offered was that changing concentration changed the Gibbs free energy of the reactant or intermediate states (as Gibbs free energy depends on the concentration in solution). The states which contain the species (whose concentration  is changed) are affected whereas others are not, so the Gibbs free energies of activation are changed, which means the rate constants are changed, which can change the rate determining step.
But if concentrations can affect $k_\mathrm{r}$ then it seems like the rate law cannot be written like that.
Does this sound right, or am I overthinking this?

It seems possible that you mix the terms. There is the reaction rate constant -- in your equation $k_r$, and there is the rate of reaction -- in your equation $r$ and sometimes expressed by $v$.
Depending on the order of a reaction, the overall rate of reaction may, but need not depend on the concentration of the reactants available.  There are reaction types known by experience to proceed typically one kinetic order (e.g., heterogeneous hydrogenation); however, reading the stoichiometric reaction equation often does not provide sufficient information and the rate law has to be determined experimentally.
Examples of kinetic reaction order are zero, one, two, three, but fractional orders are known, too.  It is possible to perform reactions which are of second order ($v = k_r[\ce{A}][\ce{B}]$) in a pseudo-first order regime if the concentration of reagent A is so much much higher, than the one of [B], that you detect after consumption of B still much of A.  This however assumes that the reaction still proceeds by the same pathway, thus requires experimental evidence to be supported.
This contrasts to $k_r$, which is independent from the concentration of the reagents, however depends on the temperature $T$ (keyword Boltzmann, and even more, Arrhenius equation).


Amorphous conductor transparent to XRD I want to conduct in-situ XRD of a cathode in my battery using this coin cell with a kapton viewing window. I assume this cell configuration is the best I can find for my purpose, and that I can't easily make a better cell. However, the viewing window isn't super helpful if I still need to use a stainless steel current collector, as I have previously.

Thus it seems I have two options:

Make a donut-shaped anode and separator, and XRD through the hole in the donut to hit the cathode. However, I worry that the cathode may not fully lithiate in the center, because it is not as close to in contact with my anode.
Find a current collector which is transparent to XRD. Does anyone have a suggestion for such a current collector?

You might have success with amorphous carbon sheets[1]. It's not cheap, but it's effective. It's the material used in the windows of the AMPIX cell[2] developed at Argonne. It does react at low potentials, though, so might not work for anode studies.

Have you considered aluminum? It's crystalline, but cubic so there aren't very many peaks to worry about.

This paper[3] might be helpful as you work on your experiments. You're right to think that not having a current collector for part of the electrode would affect the degree of lithiation; it's actually worse than that. Those kapton window cells are convenient, but always put an asterisk next to your results.

I'm doing a fair amount of in-situ characterization for my PhD. If you can share more about you goals, I'd be happy to chime in.

[1] http://www.htw-germany.com/products.php5?lang=en&nav0=3&nav1=6

[2] http://scripts.iucr.org/cgi-bin/paper?S0021889812042720

[3] https://pubs.acs.org/doi/abs/10.1021/acs.jpclett.5b00891


Identifying rearrangement reaction? I was asked to solve this problem, and the answer key shows this:

I'm not sure what's going on here, but it looks like some sort of a rearrangement and reduction. Is it an internal reaction between the ether and the alkene?

This is an example of the Claisen rearrangement.  It is basically the oxa-analogue of the Cope rearrangement.  The Claisen rearrangement can occur with aromatic or non-aromatic allyl vinyl ethers.  The reaction mechanism involves a concerted, intramolecular, [3,3]-sigmatropic rearrangement.  In the aromatic case, the rearrangement is followed by a proton shift (keto-enol tautomerization) to yield the final product with the aromaticity restored.

Here's a diagram of the aromatic Claisen rearrangement

Here's a diagram for the non-aromatic counterpart.


What is Sodium Chloride like in gas state? Since sodium chloride is sodium and chlorine atoms bonded as a lattice and there are no discrete molecules, doesn't that mean in gas state, sodium chloride is simply sodium and chlorine atoms separate from each other, resulting in a mixture of sodium gas and chlorine gas?

No, in the gas phase sodium chloride exists as a monomer (the sodium chloride molecule) along with its dimer $\ce{Na2Cl2}$.  The dimer has a roughly rectangular shape and is quite floppy with chlorines located diagonally across from each other.  The dimer makes up about 27% of the mix.  All of the bond lengths, etc. can be found in this thesis.  Go to the end and you'll see the full paper that appeared in JACS with all of the data.


How to explain covalent bonds with Bohr's atom model? I was reading about covalent bonding and got to know that in this type of bonding, the atoms of elements "share" electrons with each other to have an octet or duplet. But, I also have been taught that electrons keep on revolving around the nucleus of the atom, much like the planets around the sun. But, if this is true, how are the able to easily share electrons if these electrons keep on moving around their respective nuclei? I hope I have made myself clear. If not, then please tell in the comments.

electrons keep on revolving around the nucleus of the atom, much like the planets around the sun.

Electrons are quantum particles. Trajectories are not applicable for them. One have to use quantum wave equation to explain "movement" (note the quotes, you cant say anything about electron movement, you can only determine the probability of electron presence in some place) of electron. Planetary (Bohr) model of the atom can help to describe some effects for isolated atoms using Newtonian physics' equations with some amendments (Bohr quantum rules), but it does not describe bonding.

So for bonding you can use valence bond theory (which exactly the case with sharing electrons you described) or some more complex theories (molecular orbital (MO) theory, as widely used example) which consider quantum wave equations for describing electrons. But you surely can't mix in Bohr atom model in this case, because it is too simple to describe bonding.

Hope my explanation helps.


How would one correct for a blank in a potentiometric titration? I want to do a potentiometric titration with perchloric acid of a sample consisting of active drug (weak base) and two other components. Based on theory, I already know that these two other components should not interfere with my titration.

However, I still want to do a blank titration with just these two components to make sure. What I would expect with this blank titration is just an exponentially decreasing curve. If this is indeed what I observe, would I have to correct my sample for this blank titration? If so, how would one do this?

Even though the other compounds should theoretically not interfere with your titration, a method/matrix blank like you've proposed is always a good idea.

What I would do is simply subtract the blank potential value from that of the sample for each volume addition, then plot and find the inflection point as you would usually do (i.e. from the derivative plot, etc.).

If you are correct in your assumption that the other compounds won't interfere, then the position of the inflection point should be the same as that calculated without the blank, within experimental error (i.e. within a standard deviation or two or three, etc. of a set of replicate titration values.).  If you are incorrect and there is some interference, then the blank subtraction will correct for it and the blank corrected value is what should be used/reported.


Which conformer of beta D-glucose is more stable? I drew the two chair conformers of  $ \beta $-D-glucose:

Which one is more stable?

I think it's the one with all of the OH groups in the equatorial position because of less steric hindrance, but I also know that the anomeric effect makes the axial one favored as well.  So I'm unsure which is more stable.

I also know that the anomeric effect makes the axial one favored as well.

The anomeric effect does stabilize an axial group, but only at the anomeric position. Other axial groups are not similarly stabilized. Thus $\alpha$-D-glucose is more stable than you think it should be based on steric considerations alone.

Knowing this, which is the more stable?

PS - very pretty chairs.


Why does the proton NMR of ferrocenecarboxaldehyde appear to be four singlets? One singlet stems from the aldehyde H, and another from the Cp ring with 5 hydrogens attached.

However, the ring with the aldehyde group attached has hydrogens in two chemical environments which are just three bonds apart from each other. Why do they not couple? (400 MHz, CDCl3)

I like to address some interesting NMR features of ferrocene derivatives. Try Hard's question about $\mathrm{^{13}C-NMR}$ of ferrocenecarboxaldehyde showing one signal for the substituted cyclopentyl ring prompt me to find some solid evidence. However, it seems like most of ferrocene derivatives give similar results regardless of how strong magnetic field you used. Therefore, Aldrich NMR spectra may not be the best spectra with high definition (e.g., they use $\lt \pu{300 MHz}$ machine for $\mathrm{^{1}H-NMR}$).

Following $\mathrm{^{13}C-NMR}$ values from very recent paper, which published NMR data of few ferrocene derivatives using a $\pu{400 MHz}$ (but $\pu{100 MHz}$ for $\mathrm{^{13}C}$) machine (Ref.1):

The values in left hand side is for $\mathrm{C_p}$ ring (substituted) while single value at the bottom is for all 5 $\ce{C}$s of unsubstituted  $\mathrm{C_{p'}}$ ring. If you look closely, you'd see monosubstituted rings have so close values that they may overlap in spectra taken in low resolution machines.

Reference:

N. D'Antona, R. Morrone, G. Gambera, S. Pedotti, “Enantiorecognition of planar “metallocenic” chirality by a nitrile hydratase/amidase bienzymatic system,” Org. Biomol. Chem. 2016, 14, 4393-4399 (DOI: 10.1039/C6OB00689B).


Can a high-spin complex be diamagnetic? There are two nickel complexes: one is octahedral and the other square planar. One complex is diamagnetic, the other is paramagnetic. Both are high-spin.

Now, $\ce{Ni^2+}$ that is octahedral should be paramagnetic, as should $\ce{Ni^2+}$ that is square planar. Actually any high-spin complex should be paramagnetic. Is it possible that the octahedral $\ce{Ni^2+}$ is paramagnetic, and square planar $\ce{Ni^2+}$ diamagnetic (achieved by not filling the last orbital even tho that goes against high-spin concept)?

Both are high-spin.

No, square planar complexes are nearly always low-spin. With a $\mathrm{d^8}$ configuration, the lowest four orbitals would be filled. That would make the complex diamagnetic.

For the octahedral $\mathrm{d^8}$ complex, regardless of whether it's high- or low-spin, it will have the same electronic configuration $\mathrm{(t_{2g})^6 (e_g)^2}$. By Hund's first rule, the two electrons in the $\mathrm{e_g}$ orbital will both be unpaired, so the complex will be paramagnetic.

Here's a quick sketch (energy levels not to scale):


Epoxide ring opens, or not? Does the epoxide ring open in this case? I guessed it does, but it really doesn't. The answer given (among the four options) is another epoxide, with some sort of substitution. Could someone please explain the mechanism for this reaction?

Also, what exactly is the condition for ring opening of epoxides? Thanks for the help!

Epoxide gets opened at the least hindered side by the ethoxide (acting as a nucleophile rather than as a base in this case).
The resulting alkoxide intermediate can itself act as a nucleophile. In the middle intermediate could either kick out ethoxide (which would give back the starting material), or kick out the chloride (to give a different epoxide). Since the chloride is a better leaving group, it is this one that probably leaves (theres an element to which the whole thing is reversible away, and if it does kick out ethoxide, the ethoxide can just attack again to re-form the middle alkoxide intermediate)

The product on the right hand side is probably what the mixture would converge upon, given that chloride is  a better leaving group than ethoxide.


Why are noble gases 'gases'? Why do group 18 elements exist as gases, why are they not found in any other physical state ?

Liquids and solids form because intermolecular forces hold together different molecules/atoms/ions/biscuits in a strong enough fashion to prevent them from drifting away from each other. In a very basic sense you could say that gases are non-ordered, liquids have near-range ordering but not far-range, and solids are well ordered into crystal lattices.

Noble gases have very little intermolecular forces acting between them since they are monoatomic and unpolarised. Thus, only London dispersion forces act and these directly depend on the number of electrons in a compound. If you compare each noble gas with the element to its left in the periodic table (i.e. to the neighbouring halogen; hydrogen for helium) and take $n$ to be the number of electrons the noble gas has, then the next-door neighbour will always form a diatomic molecule with $2(n-1)$ electrons. In every case except for hydrogen, this is substantially more than in the noble gas, thus there is a substantial increase in melting/boiling point when going from the noble gas to the corresponding halogen.

Note that radon’s boiling point is already at $-61~\mathrm{^\circ C}$, higher than that of the highly polar molecule $\ce{HCl}$ ($- 85~\mathrm{^\circ C}$). And note that ununoctium has been predicted to have a boiling point of $(+80 \pm 30)~\mathrm{^\circ C}$ and would thus better be termed a noble liquid. Unfortunately, ununoctium has a half-life of about $0.89~\mathrm{ms}$ so actually observing the noble liquid could prove tricky.

(All values taken from the respective gases’ Wikipedia articles.)


How does higher carbonyl compounds disolve in water? We know that higher the carbon, the more is hydrophobic the compound nature is, Still when we take a compound like sugar , a hand of sugar, it readily dissolves in water? can someone explain me reg this

Sugar contains a lot of hydroxyl groups. So due to hydrogen bonds, sugar can readily dissolves in water.


Can not find structural formula with compounds' IUPAC name I currently need to read a paper. In that paper the following compound was mentioned:

1,2-benzo-6-(N-ethyl-N-isobutylamino)fluoran

I tried to find the corresponding structural formula using the OPSIN parser at: http://opsin.ch.cam.ac.uk/
but I got the error message:

  1,2-benzo-6-(N-ethyl-N-isobutylamino)fluoran was uninterpretable due to the following section of the name: 1,2-benz

I also tried just googleing the compounds name but the only thing I found where patents which weren't very useful.

I hope somebody can help me find the structural formula.

The name given is not standard.

Fluoran itself is a spiro compound with two components:

(it's strictly an isobenzofuranone, but I'm lazy to type that multiple times.)

Because of the nomenclature rules of spiro compounds (which I'm sure someone will come along and tell you all about), the atoms in the benzofuran system are labelled 1-7 and those in the xanthene system are labelled 1'-10', with an apostrophe.

Atoms 1 and 2 of the benzofuran system are both within the furan ring. If I am not wrong, atom 1 is the carbonyl carbon, and atom 2 is the ring oxygen. The given name, that starts with "1,2-benzo", therefore makes no sense as you cannot fuse a benzene ring to atoms 1 and 2 of the benzofuran system.

After a little bit of Googling, I came across this compound: 6'-(Diethylamino)-1',2'-benzofluoran. I think with the apostrophes, it makes much more sense.

The numbering within the xanthene system parallels that of anthracene:

Having "1',2'-benzo" means that you have an additional fused benzene ring to carbons 1' and 2'. The 6'-substituent also makes sense. Chances are if the paper you were reading labelled the xanthene atoms with 1-10 without apostrophes, the 6-amino group should be a 6'-amino instead. Your desired structural formula is probably

ChemDraw's name to structure generator gives this molecule when you input 6'-(N-ethyl-N-isobutylamino)-1',2'-benzofluoran. The amino group, which is just an ordinary substituent, should probably be a prefix to "benzofluoran" which is the parent compound.

I am sure in due time someone will come along and give you all the official IUPAC rules but I figured that if you just wanted the structural formula, my not-very-rigorous explanation would suffice.


Covalent and Ionic Bonds Is sodium nitrate a covalent or ionic compound? This is a test question. I put the answer ionic compound but my teacher says that statement is false because it is only partially ionic as there is covalent bond between nitrogen and oxygen in the nitrate ion. Is that true?

Bonds are usually used to describe how electrons are shared. "Partially ionic bonds" are polar covalent bonds. Which is not the same as saying that a compound has different types of bonds in the whole compound. I think that your teacher gave you a bad question and a bad answer. If the question had asked whether all the bonds in sodium nitrate are ionic, then it would be a different story.

However, it is true that the bonds between the nitrogen and oxygen are covalent. Nevertheless, stating that sodium nitrate is ionic shouldn't be wrong.


What is the mechanism of alkane combustion? I understand that combustion of hydrocarbons (such as alkanes) produces energy by breaking bonds:

$$\ce{C_nH_{2n} + $\frac{3n}{2}$ O2 -> nCO2 + n H2O}$$

And I can find dozens of explanations of how combustion works in an engine.

What does the chemical reaction (not equation) of hydrocarbon polymer combustion actually look like?

In other words, how is a hydrocarbon polymer actually broken down to form energy?

What process actually breaks up the polymer into its respective monomers (methane?) for combustion to take place?

Please explain. I would also love a visual (like this or this ) if someone could find one!

The following, very general radical reactions apply to any common combustion reaction of organic material. Details vary largely.

Step 1

An energy spark of whatever kind turns triplet oxygen into its high energy singlet state$^*$, which typically strips a hydrogen atom off of some organic molecule
$\ce{R-CH3 + O2^* -> R-CH2. + HOO.}$
Now you have two radicals.

Step 2

The perhydroxy radical strips another hydrogen off somewhere,
$\ce{R-CH3 + .OOH -> R-CH2. + HOOH}$
while the alkyl radicals react with more oxygen (self-catalysed)
$\ce{R-CH2. + O2 -> R-CH2-OO.}$
$\ce{R-CH3 + .OO-CH2-R -> R-CH2. + HOO-CH2-R}$
, this incorporates $\ce{-OOH}$ hydroperoxide motives into more and more molecules. This is in itself a linear process, with a constant number of radicals. Very similar to free-radical polymerisation or organic radical-substitution reactions, and very much the same as formation of the infamous ether hydroperoxides.

Step 3

These peroxides are not stable at elevated temperature, and their O-O bond tends to split homolytically
$\ce{R-OO-H -> R-O. + .OH}$
Now the number of radicals grows rapidly, the combustion is gaining speed.

Step 4

As temperatures rise (see step 5!), esp. larger organic molecules start to break up also by themselves
$\ce{R-CH2-CH2-R -> R-CH2. + .H2C-R}$
(That's also a possible initiation step.) You get more radicals, which can react with more oxygen to form peroxides. Now the chain reaction snowballs!

Step 5

$\ce{HOOH -> HO. + .OH}$ gives more hydroxy radicals (also from step 3), which turn into water
$\ce{HO. + H3C-R -> H2O + .H2C-R}$
leaving another radical behind. Producing water is highly exothermic, this fuels the reactions creating radicals (in steps 3+4+5). Making radicals costs a lot of energy.
Steps 2-5 (it's really not a sequence) run in parallel and repeat as they produce more radicals and peroxides all the time. The reaction is usually limited by the inflow of fresh oxygen.

Step 6

Termination: Two carbon radicals can recombine, but at high temperature this does not really happen, because the energy from the recombination has to go somewhere. When temperatures are too low, this makes coke and smoke.
This picture is by no means complete. $\ce{CO2}$ would likely occur via decarboxylisation of $\ce{R-COO.}$ radicals, etc.  Organic radicals at high temperatures have more internal reactions where they split off smaller molecules, e.g. $\ce{R-CH2-CH2. -> R. + H2C=CH2}$. (The latter can be seen as depolymerisation of polyethylene.) A radical in the middle of a chain splits it in half ($\ce{R-CH2-CH2.-R -> R. + H2C=CHR}$, increasing the volatility.
And remember that this picture is terribly simplified. I'm by no means an expert on this subject, and the experts also have not agreed on all details yet, especially since it varies a lot.


Faraday's Law of Electrolysis problem Question:

  How many moles of iron will by passage of $\pu{4 A}$ of current through $\pu{1 L}$ of $\pu{0.1 M}$ $\ce{Fe^{3+}}$ solution for $1$ hour. Assume $\ce{Fe^{2+}}$ is not present in the solution initially.

Attempt:

Using Faradays' Law, we have:

$\text{mass delivered = } \mathrm {Z\times I\times t} = n\times M$  where $\mathrm Z = \dfrac{\text{Equivalent mass }}{\pu{1F}}$

Equivalent mass of Iron here would be $\dfrac{M}{3}$ where M is molar mass.

So,

$n \times M = \dfrac{M}{3}\times \dfrac{\pu{1 mol}}{\pu{96500 C}}\times \pu{4 \frac{C}{s}} \times \pu{60 \frac{s}{min}} \times \pu{60 \frac{min}{hr} \times \pu{1 hr}}$

$\implies n = 0.049$ moles.

But answer given is $\pu{0.0245 moles}$

Where have I gone wrong? I can't figure out why there's a difference of factor of 2 in the answer.

I think the key part of the answer is

  Assume $\ce{Fe^{2+}}$ is not present in the solution $\color{red}{\underline{\text{initially}}}$.

Since iron reacts with $\ce{Fe^3+}$ to form $\ce{Fe^2+}$ via:

$$\ce{Fe^0 + 2 Fe^3+ -> 3 Fe^2+}\tag 1$$

You have to account for this additional reaction. If the reaction did not produce any $\ce{Fe^2+}$ as you assumed then of the $\pu{0.100 mol}$ of $\ce{Fe^3+}$ in the solution then $\pu{0.051 mol}$ would remain after and you would be correct, but it does react. Now, my math says that the answer is more like

$n \times M = \dfrac{M}{3}\times \dfrac{\pu{1 mol}}{\pu{96500 C}}\times \pu{4 \frac{C}{s}} \times \pu{60 \frac{s}{min}} \times \pu{60 \frac{min}{hr} \times \pu{1 hr}}\tag 2$

$\implies n = \pu{0.0497 mol} = \pu{0.050 mol}$

Which means $\pu{0.050 mol}$ of $\ce{Fe^3+}$ remains which must be neutralized by the $\pu{0.050 mol }\ce{Fe^0}$. From equation 1 we can see that it takes one $\ce{Fe^0}$ for every 2 $\ce{Fe^3+}$ which menas $\pu{0.025 mol}$ of $\ce{Fe^0}$ is needed to react with $\pu{0.050 mol } \ce{Fe^3+}$ and your remaining $\ce{Fe^0}$ is $\pu{0.025 mol}$ or $\pu{0.0245 mol}$ in the case of $n = \pu{0.49 mol}$.


Ponderomotive Effect on Ionic Liquids Would an alternating electric field cause net motion of ions in a liquid in the case of the electrodes being placed outside the liquid?

Generally speaking, what happens if you apply an external electric field to an ionic liquid, taking into account that there are no electrodes places inside the liquid itself, but are separate from the liquid container in the same manner you have a parallel plate capacitor.

Yes, due to electrostatic coupling. The conductors and the liquid would form a capacitor, causing back-and forth motion of the ions (and in polar molecules). This is an effective way to heat something without contact, as in dielectric heating.

This capacitance enables liquid crystal displays and E-Ink to display images using little or no current when not actively switching.


Where do things connect in EAS reactions? If you have an aromatic ring that requires substitution of some molecule, how do you know where this molecule is supposed to substitute on the ring?

In (i) and (ii) how do we select the ring and position that will receive the molecule (HSO3 in (i) and Cl in (ii))? In (iii), how do we know where to attach the I?

The position of substitution in EAS is determined by the electron-donating/-withdrawing effects of existing substituents. The effects can be inductive (the electronegativity of heteroatoms polarize the bonds), or they can be resonating (free electron pairs are donated to a double bond in the transition state).

1. rule; resonance effects overrule inductive effects.

If a ring has both a resonating and an inductive substituent, or one substituent with both effects, the resonating effect will control the reaction.

2. rule; donating effects activate the ring (more electrons for the electrophile), withdrawing effects deactivate the ring (less electrons for the electrophile).

Strong donating groups: -OH, -OR, -NH2, -NHR, -NR2, -SH, -SR, -F, -Cl, -Br, -I, -CH=CH-X (X is any donating group,

Neutral/weak donating groups: -CH3, -(CH2)R, -CHR2, -CR3 (R is any hydrocarbon), Ph.

Strong withdrawing groups: -(C=O)R, -NO2, -SO3,

Weak withdrawing groups:  -(CH2)X (X is any atom more electronegative than carbon),

3. rule; activated rings favor para and ortho positions, deactivated rings favor meta position (para and ortho are deactivated).

Deactivated rings can still react, but much higher temperatures are required, and the reaction is much slower.

4. rule; para is favored over ortho.

Because the existing substituent is "in the way" of the attacking electrophile in ortho position.

An Example

In electrophilic aromatic substitution the aromatic ring has a positive charge in the transition state (because an electrophile (E+) has "borrowed" to two electrons). The electrophile can, and will, attack all three possible positions, but this first step is reversible, and any charge on a carbon is highly unfavorable. Therefore, if the charge cannot be placed on in the bottom path in the example below, the electrophile will be ejected quickly, while the other two paths are stabilized long enough for a base (B-) to extract the proton, leaving the electrons used to bond with the proton to the ring.

rule; Oxygen is inductivly withdrawing, but at the same time donating by resonanse. => donating.
rule; Donating => activated ring.
rule; Activated => ortho/para.
rule; both ortho and para are available. => para favored

Your example

Question (i), the -(C=O)R is a withdrawing group on the left ring, while the RO- group has two electron pairs that it can share with the right ring. ==> para position on the right ring.

Question (ii), the -NO2 on the left ring is deactivating for that ring, while -CH2R is weakly activating for both ring. ==> para position on the right ring.

Question (iii), the ring has a strong and a weak withdrawing group, and is thus highly deactivated. Also, as ssavec pointed out, NaI is no electrophile.

Still let us assume that you had some strong electrophile; Both substituents deactivate their ortho positions, while having little effect on their meta positions. Since -NO2 is a stronger withdrawing group, it rules, and the reaction would happen meta to the -NO2 group (ortho to the -CH2Br).


Are there carbon-silicon alloys? Both carbon and silicon expose a lot of metallic properties like metallic lustre, electrical and heat conductivity, etc, so often considered metalloids. The both form alloys with other metals such as iron.

On the other hand, the usual silicon carbide is not an alloy at all. It is tranparent, electrical insulator etc.

I wonder whether there is a different carbon-silicon compound that has more metallic properties?

Carbon (graphite):

Silicon:

Silicon carbide:

Here you can find a phase diagram for $\ce{Si/C}$ system. It does not have zones with homogeneous non-stohiometric solids. So, there is not thermodinamically stable Si/C isomorphic alloys. However, since the liquid likely to be homogeneous, it is likely for fast cooled liquid to form amorphous alloys. Indeed, google search provides plenty of links for amorphous $\ce{Si/C}$ alloys. However, it is also possible for cooling liquid to form polycrystalline alloy, say, with $\ce{SiC}$ and $\ce{C}$ grains, similar to pig iron.


What will be the product ratio if I do chlorination of p-toluic acid? If I do chlorination of p-toluic acid using Lewis catalyst, e.g. $\ce{FeCl3},$ what will be the ratio of major product to side product? Main product shall be 3-chloro-p-toluic acid and other product may be 2-chloro-p-toluic acid for monochlorination.

So, what will be approx ratio of these two?

I couldn't find any literature evidence for the nuclear chlorination reaction of 4-methylbenzoic acid (p-toluic acid). Instead, I found a patent in Japanese for the nuclear chlorination of 4-methylbenzoic acid chloride, of which, I believe close enough to your question (See below the translation given in Google).

According to the patent, when mixture of 99.7% pure 4-methylbenzoic acid chloride ($\pu{927.6g},\; \pu{6.0mol}$) and ferric(III) chloride ($\pu{2.9g}, 0.3 \:wt\%$) was heated at $\pu{50\!-\! 55 ^\circ C}$ for $\pu{22 h}$ in the presence of $\ce{Cl2}$ gas. After completion of the reaction, 84.9% of 3-chloro-4-methyl benzoic acid chloride was isolated. In addition, 0.1% of 2-chloro-4-methyl benzoic acid chloride and 2.2% of dichloro-4-methyl benzoic acid chloride were also isolated (dichloro-4-methyl benzoic acid chloride was not identified, but it should be 3,5-dichloro-4-methyl benzoic acid chloride, judging from the ratio of 3-chloro-4-methyl benzoic acid chloride to 2-chloro-4-methyl benzoic acid chloride).

Reference:

Inventor: 智規 加藤亮太 甲斐幸基 小林, "Isopropyl 3-chloro-4-methylbenzoate and method for producing same." Patent application # (2013): WO2013015203A1 (https://patents.google.com/patent/WO2013015203A1/en).


Justification for Freezing Point Depression & Boiling Point Elevation in Solutions? I was wondering if the following justification for freezing point depression and boiling point elevation are conceptually correct. The reason why I ask this question is because I have been self studying chemistry for a course I will be taking this fall, and I don't have a human reference to check with. My book is also very confusing on this subject, and I don't have a world renowned memory (meaning I'd rather understand than memorize the equation), so I tried to think about it and came with the following conclusions:

-Freezing Point Depression: When a solvent is not pure and has particles dissolved in it, there are constituents (be it molecules, ions, etc.) that take up volume in the solution. For freezing to occur, these constituents need to aggregate to form a lattice, bonds, etc. Say we have a pure solvent X. This process occurs at a temperature Y. However, when this solvent is not pure and has dissolved solutes, there is "blockage" impeding these constituents from aggregating and bonding. Therefore, the temperature Y needs to be reduced further to slow down the kinetic energy of all constituents in solution (solvent and solute), which gives a greater statistical probability of intermolecular forces getting an opportunity to bind solvent-solvent constituents.

-Boiling Point Elevation: Boiling, if I'm not mistaken, occurs when the partial pressure of a vapor exceeds atmospheric pressure. For a gas to enter the vapor phase, molecules in solution need to have a certain kinetic energy (as specified by a Boltzmann distribution) to free themselves of the intermolecular forces in the solution phase and become a vapor. When a pure solvent is now diluted with other particles, the new solute-solvent bonding energy needs to be overcome to allow a solvent molecule to exist as a vapor. One way to overcome this energy is to increase the average kinetic energy of all molecules, and this is done through a temperature increase.

This is for an introductory chemistry course, and I'm not sure if my explanations are sound. I just want a yay or a nay in terms of my logic, because I don't have the most intimate background in chemistry and I'm trying to conceptually understand things.

I think you are looking at the problem from slightly the wrong angle. The central quantity when dealing with colligative properties is entropy and not solute-solvent or solvent-solvent molecule interactions. Of course the interactions are important in the sense that they affect the entropy but since we are dealing with thermodynamics here the way you think about it at the moment clouds the main effect at work here imo.

So, when trying to picture how colligative properties work on a microscopic level think about the effect of the solute on the entropy. Boltzmann's entropy formula tells you that the entropy $S$ is proportional to the logarithm of the number of microstates available to the system. The solute is present in the solution in small quantities and the molecules are really dissolved and don't form a different phase. By their presence the solute molecules disturb the local order of the solvent molecules and introduce new available microstates into the system. As an extremely simplified example you can think of it like this: If you want to place 10 identical blue balls (solvent molecules) on 10 fixed spots there is only 1 way (microstate) to do this, since all the balls are alike. But if you replace one of the blue balls with a red ball (solute molecule) you suddenly have 10 possible distiguishable ways to arrange your balls on the 10 spots. So, when comparing the initial situations of a solution of pure solvent and one of solvent plus solute, the second solution will have a higher entropy to start with, i.e. $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$.

Now, what happens when you want to freeze both solutions? In order to freeze the Gibbs free energy of the frozen (solid) phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{s}} \leq G_{\mathrm{l}}$ or put differently $\Delta_{\mathrm{m}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{l}} - G_{\mathrm{s}} \geq 0$. The Gibbs free energy is given by

\begin{equation}
 G = H - TS
\end{equation}

where $H$ is the enthalpy and $T$ is the temperature, so that

\begin{equation}
 \Delta_{\mathrm{m}} G = (\underbrace{H_{\mathrm{l}} - H_{\mathrm{s}}}_{\Delta_{\mathrm{m}} H}) - T (\underbrace{S_{\mathrm{l}} - S_{\mathrm{s}}}_{\Delta_{\mathrm{m}} S})
\end{equation}

Since there is only very little solute present in the second solution you can assume that $\Delta_{\mathrm{m}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{m}} H^{\mathrm{pure}}$. So, the difference between $\Delta_{\mathrm{m}} G^{\mathrm{impure}}$ and $\Delta_{\mathrm{m}} G^{\mathrm{pure}}$ will mainly be caused by the entropical term. Let's have a look at that: the solid phase is assumed to be pure in both cases, i.e. consisting only of solvent molecules, and thus $S_{\mathrm{s}}$ must be the same in both cases. Furthermore, you generally find that the entropy of a solid phase is lower than the entropy of a liquid phase, i.e. $S_{\mathrm{s}} < S_{\mathrm{l}}$. Above I established that $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$. It follows that $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Putting all this together gives you the following picture: For $\Delta_{\mathrm{m}} G$ to become equal to zero the entropical term $T \Delta_{\mathrm{m}} S$ has to balance $\Delta_{\mathrm{m}} H$, which is assumed to be about the same for the pure-solvent and the solvent-plus-solute case. Since $\Delta_{\mathrm{m}} S^{\mathrm{impure}} > \Delta_{\mathrm{m}} S^{\mathrm{pure}}$ the temperature at which freezing becomes possible must be lower in the solvent-plus-solute case, i.e. adding the solute leads to a freezing point depression.

An analogous argument can be made for the Boiling point elevation. The focal point is again that the solute raises the entropy of the liquid phase, thus affecting the entropical term in the Gibbs free energy equation for the phase change. For the boiling process to happen the Gibbs free energy of the gaseous phase must be lower than or equal to the Gibbs free energy of the liquid phase, i.e. $G_{\mathrm{g}} \leq G_{\mathrm{l}}$ or $\Delta_{\mathrm{vap}} G \substack{{\scriptsize \mathrm{def}} \\ =} G_{\mathrm{g}} - G_{\mathrm{l}} \leq 0$.
Requiring the Gibbs free energy of vapourization

\begin{equation}
 \Delta_{\mathrm{vap}} G = (\underbrace{H_{\mathrm{g}} - H_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} H}) - T (\underbrace{S_{\mathrm{g}} - S_{\mathrm{l}}}_{\Delta_{\mathrm{vap}} S})
\end{equation}

to be equal to zero gives

\begin{equation}
 \Delta_{\mathrm{vap}} H = T \Delta_{\mathrm{vap}} S
\end{equation}

Again, we can assume that $\Delta_{\mathrm{vap}} H^{\mathrm{impure}} \approx \Delta_{\mathrm{vap}} H^{\mathrm{pure}}$. Since the entropy of a gaseous phase is generally higher than the entropy of a liquid phase, i.e. $S_{\mathrm{g}} > S_{\mathrm{l}}$, and since $S_{\mathrm{l}}^{\mathrm{impure}} > S_{\mathrm{l}}^{\mathrm{pure}}$, it follows that $\Delta_{\mathrm{vap}} S^{\mathrm{impure}} < \Delta_{\mathrm{m}} S^{\mathrm{pure}}$. Thus the boiling temperature must be higher in the solvent-plus-solute case, i.e. adding the solute leads to a boiling point elevation.


What physical/chemical reasons could have caused Dorian Gray's picture to degrade? In the novel 'The Picture of Dorian Gray' by Oscar Wilde, Gray locks away a portrait of his in the attic for 18 years and over that time the picture degrades to an almost unrecognisable version of its former self. The novel explains this as the fulfilment of a wish by Gray to have the portrait age instead of himself but I have been wondering recently, are there be any natural reasons that could explain this transformation if a painting is locked away for a large period of time?

Preface: So far, I do not know the novel you refer to, and assume the portrait is a classical oil painting.

Such a painting consists of several layers of products applied.  Similar to a wall in your living room, the canvas is prepared to accomodate the pigments dispersed in linseed oil, what in other fields of application would be called a "primer".  Then there are multiple layers of the painting itself, topped by the varnish.  To mention a few of the potential issues,

Pigments may change their colour upon exposure to air. Pigments like lead white turned over time black, for example.
Pigments may interact with the oil, forming complexes, of different mechanical and complexation strength:

(source)

Components may migrate across the picture and locally, and reactions may occur.  The formation of soaps of lead and zinc are the most prominently known ones, for example based on palmitic, stearic, or azelaic acid.  These processes may alter colors, even rendering them transparent enough to recognise the support of the paint, and they may eventually protrude the surface of the paintings as shown from a topview:

(source, p. 14)

or cross section:

(source)

These may not simply be "washed off" as a normal cleansing would do, as water and humidity in general amplify the problem.  If left untreated, the painting however will age and degrade.

Further reading may be found, for example, here, here, here, here/here.


Wittig reaction in presence of carboxylic acids The following reaction came in a recent test:

According to this reaction, the acidic hydrogen would protonate the ylide, and the standard wittig reaction would not take place. This source agrees with the same results.
However, according to El-Batta et al.1, the Wittig reaction in water can take place even in the presence of acidic groups, such as carboxylic acids and phenols:

It is also interesting that the Wittig reaction in water works well with substrates having acidic unprotected functional groups present; hence the presence of a carboxylic acid group (vide infra) or phenolic groups, such as p-hydroxybenzaldehyde, gives the corresponding cinnamate product 16 in 92% after heating the reaction for 1 h at 90 °C (entry 16). In sharp contrast, the same reaction in refluxing CH2Cl2(4 h) has been reported to provide only 8% of 16

Which one of these references is correct? Does the Wittig reaction really proceed in the presence of acidic groups? I'm looking for a good reference, along with an explanation on what occurs when a carboxylic acid group is present on the substrate.
References:

El-Batta, Amer, et al. “Wittig Reactions in Water Media Employing Stabilized Ylides with Aldehydes. Synthesis of α,β-Unsaturated Esters from Mixing Aldehydes, α-Bromoesters, and $\ce{Ph3P}$ in Aqueous $\ce{NaHCO3}$.” The Journal of Organic Chemistry, vol. 72, no. 14, July 2007, pp. 5244–59. doi:10.1021/jo070665k.

This works because the acidic starting material is going to be deprotonated as there is excess bicarbonate present. Because of the adjacent carboxylate group, the pKa of the phosphonium species is going to be around 6-7  pKa Table here so within the range where bicarbonate can deprotonate it. The reaction is between the anion of the starting material and the anion of the phosponium species in aqueous phase. This will only work for readily deprotonated phosphonium species and aqueous soluble aldehydes in the presence of excess base.


Water color additive that will maintain color when water is a thin film I'm producing thin sheets of liquid (50 to 100 microns) from a nozzle and trying to photograph them.  I think it would help if I could color the water. However, typically as water becomes more of a thin film it tends to lose color.  I'm wondering if anyone has any thoughts on this.  I think white would be a particularly good color.   Food coloring containing titanium dioxide might work.  I know that just using milk will work, but it's a mess to deal with because I am dealing with sprays traveling at a minimum 5 meters per second.

Use fluorescein or other fluorescent dye, along with an effective UV filter (AKA haze or skylight filter) on the lens and an ultraviolet source, such as a xenon strobe (speed light) or UV flashlight. A UV-transmission filter on the strobe can increase contrast, and can be made cheaply, but might not be needed. Because fluorescence makes the fluid become a source of light, it is quite obvious against a dark background. You can even use most laundry detergents, which contain blue-fluorescing dyes -- though foaming might be a nuisance.

See this image of fluorescent soap bubbles or this one. Soap bubble film may be just a few hundred nm thick, yet show clearly with fluorescence. Also watch a demo of strobe illumination of water droplets using sodium fluorescein for contrast. In the video, notice, in particular, that the stream tends to break into pairs of droplets, one perhaps a millimeter in diameter, and the sibling, just a few microns across, but all clearly visible!

Choose a near-UV light source and fluorescent dye for safety, since shorter wavelengths are more damaging to eyes and skin. A search for UV flashlight returned items from US$10 and up, some of which might be suitable. Most were in the 395 to 365 nm spectrum, relatively innocuous.


Why do resorcinol and iodine react to give 2-iodobenzen-1,3-diol? I am supposed to perform the above shown reaction in the lab soon, but I don't really understand the mechanism behind it. I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which resorcinol (benzin-1,3-diol) acts as the electrophile and iodine as nucleophile. I imagine the resorcinol also electron-poor because of the -I effect caused by the electronegative Hydroxide substitutes. But isn't the +M Effect stronger? Then why should resorcinol act electrophilic?
Why is Iodine nucleophilic when there is no Lewis-acid present?
And last but not least: why is the Product 2-iodo-1,3-diol and not 4-iodobenzen-1,3-diol?

Reading this question, I realized that OP is very new to organic chemistry, and in need for learning a lot about electrophilic aromatic substitution reactions. Thus, I recommend that OP should concentrate on the electrophilic aromatic substitution reactions and read the chapters of OP's textbook dedicated to that subject. Said that, I'm going to answer plethora of subjective questions, which should be answered and cleared to show the way to right direction.

I got told by my lab assistant that the mechanism is an electrophilic aromatic substitution in which the Resorcinol (Benzin-1,3-diol) acts as the electrophile and Iodine as nucleophile.

First, resorcinol is not called benzin-1,3-diol (may be in an old book). It's correct name is 1,3-dihydroxybenzene. It is true that the reaction is an electrophilic aromatic substitution. However, the electrophile here is polarizable $\ce{I2}$ molecule because it exsists in $\ce{I^{\delta +}-I^{\delta -}}$ form under reaction conditions (where $\ce{I^{\delta +}}$, which lacks electrons is the reactive center). Aromatic nucleus with two hydroxy group is electron rich (hydroxy groups activate the ring), hence acts as the nucleophile.

But isn't the +M Effect stronger?

Yes, mesomeric effect (old resonance effect) is stronger than inductive effect. Here you have two hydroxy groups to give strong $(+M)$-effect to activate the aromatic nucleus. Hence the reaction is o,p-directive. As a result, you can expect either 2-iodo-1,3-dihyroxybenzene or 4-iodo-1,3-dihyroxybenzene or 2,4-diiodo-1,3-dihyroxybenzene or etc. as your product or mixture of them based on the reaction condition used (Ref.1 & 2).

Why is the Product 2-Iodo-1,3-diol and not 4-Iodobenzen-1,3-diol?

As I explained before, the reaction conditions govern the product ratio (see the PDF of Ref.1 I attached below for better understanding). The iodination of resorcinol is not a straight forward reaction. Slight changes of reaction conditions gives you different product. For example, the major product given in the reaction using $\ce{I2/NaHCO3}/\pu{0 ^\circ C}$ condition is 2-iodo-1,3-dihydroxybenzene as indicated in the OP's question. In a large scale reaction, the yield obtained was only 66% (Ref.3). Using the same conditions, Ref.1 reported 77% isolated yield. Keep in mind that, some of 2-iodo-1,3-dihydroxybenzene isomerized to 4-iodo-1,3-dihydroxybenzene (major) and 4,6-diiodo-1,3-dihydroxybenzene in an acidic medium at high temperature (Ref.1). Thus, temperature plays an important role in percent yield in the reaction.
On the other hand, when the electrophile ($\ce{I+}$ source is changed to $\ce{ICl}$, the main product is 4-iodo-1,3-dihydroxybenzene (4-iodoresorcinol) with 4,6-diiodo-1,3-dihydroxybenzene (4,6-diiodoresorcinol) as minor product (Ref. 1 & 2). Ref.2 also stated that use of nascent iodine $(\ce{KIO3/KI/HCl})$ has resulted in the regiospecific 2,4-diiodonation of resorcinol at room temperature (which is a rapid reaction). The $\ce{^1H}$-NMR spectrum of this reaction mixture has revealed only two products, 2,4-diiodoresorcinol as major product (56% isolated yield) and 2,4,6-triiodoresorcinol as minor product in $89:11$ ratio.
References:

Ib Thomsen, and Kurt B. G. Torssell, "Iodination of Resorcinol, 5-Methoxyresorcinol, Phloroglucinol and Resorcyclic Acid," Acta Chemica Scandinavica 1991, 45, 539-542 (DOI: 10.3891/acta.chem.scand.45-0539)(PDF).
Frederick L. Weitl, "Regiospecific 2,4-diiodination of resorcinol with nascent iodine," J. Org. Chem. 1976, 41(11), 2044–2045 (DOI: https://doi.org/10.1021/jo00873a040).
Shin-ichiro Tsujiyama and Keisuke Suzuki (submitted); David B. Guthrie, Heather M. Gibney, and Dennis P. Curran (checked), "Preparation of Benzocyclobutenone Derivatives Based on an Efficient Ggeneration of Benzynes [Bicyclo[4.2.0]octa-1,3,5-trien-7-one, 5-(phenylmethoxy)-]," Org. Synth. 2007, 84, 272-284; Coll. Vol. 11 2009, p. 488-497 (DOI: 10.15227/orgsyn.084.0272).


Molecular formula of a hydrocarbon A is a non cyclic hydrocarbon chain with $\ce{C_{3x-1}H_{6x}}$ as a molecular formula which is mainly used in cars as fuel, its cracking leads to hydrocarbon (B) which belongs to the same family of hydrocarbon A, also two molecules of ethene form the second product in the above reaction.

My effort:
To determine if it is an alkane, we compare the given subscripts on $\ce{C}$ and $\ce{H}$ ($3x-1$ and $6x$, respectively) to those in the general formula for an alkane, $\ce{C_{n}H_{2n+2}}$. Let $n=3x-1$ where $n$ is the number of carbon atoms.

$$6x\buildrel{?}\over{=}2(3x-1)+2=6x$$

and therefore A is an alkane, with the molecular formula satisfying the form $\ce{C_{n}H_{2n+2}}$.

You're right in thinking that it's an alkane. Now I think I shouldn't blame you for getting confused by

  $\ldots$ as a molecular formula which is mainly used in cars as fuel $\ldots$

Because

First off, we don't know if $x$ in the molecular formula is an integer or a decimal number.
Second, we don't know what fuel is meant. It's just supposed to be a fuel used in cars. And what are the fuels used in motor vehicles again?

  Currently, the majority of motor vehicles worldwide are powered by gasoline or diesel. Other energy sources include ethanol, biodiesel, propane, compressed natural gas (CNG), electric batteries charged from an external source, and hydrogen.

What are these?

  The bulk of a typical gasoline consists of $\mathrm{\color{red}{hydrocarbons~with~between~4~and~12~carbon~atoms~per~molecule}}$ (commonly referred to as C4-C12).

  Petroleum-derived diesel is composed of about 75% saturated hydrocarbons (primarily paraffins including n, iso, and cycloparaffins), and 25% aromatic hydrocarbons (including naphthalenes and alkylbenzenes). The average chemical formula for common diesel fuel is $\color{red}{\ce{C12H23}}$, ranging approximately from $\ce{C10H20}$ to $\ce{C15H28}$.

Ethanol is an $\color{red}{alcohol}$.

  Biodiesel refers to a vegetable oil - or animal fat-based diesel fuel $\mathrm{\color{red}{consisting~of~long-chain~alkyl~(methyl,~ethyl,~ or~ propyl)~ esters}}$.

Propane is an $\color{navy}{alkane}$.

  CNG is made by compressing natural gas (which is mainly composed of $\color{red}{methane}$, $\ce{CH4}$), to less than 1 percent of the volume it occupies at standard atmospheric pressure.

Electric batteries and hydrogen are $\ldots$ a bit irrelevant. Let's not make this answer longer.

Thus, the answer would be propane, since it's the only alkane among the common car fuels. But is it? Of course not; since propane cracking isn't in high school syllabuses anywhere in the world. There is a very very good chance the querent had gasoline mind.
Now that we trust our natural instincts (!) that they meant gasoline when they were referring to the fuel, we should remain perplexed and run in circles. Because gasoline isn't consisted of one or two distinctive and well-known substances, and it's a mixture of  It is a mixture of paraffins (alkanes), cycloalkanes (naphthenes), and olefins (alkenes), where the usage of the terms paraffin and olefin is particular to the oil industry. To make matters worse, gasoline composition is obviously not the same around the world.

$\hspace{25ex}$
$\hspace{25ex}~\small{\mathrm{Some~of~the~compounds~commonly~found~in~gasoline}}$
$$\small{source}$$

So, let me rephrase your question into something actually meaningful:

  $\ldots$ as a molecular formula which is mainly known to be a really smooth-burning species in gasoline $\ldots$

Now we're talking. The best known gasoline would be one that consists of 100% isooctane, and that's why we have octane rating. It's not a real fuel, just the material in an imaginarily ideal mixture that would burn "real smooth". That's why your question is problematic and scientifically incorrectly asked.

Back to the question, 2,2,4-Trimethylpentane must have been the compound the querent had in mind. Now we have the answer for A.

$\hspace{25ex}$
$\hspace{28ex}$ Isooctane (PIN: 2,2,4-Trimethylpentane)
$\hspace{43ex}$ $\small{source}$

But what would the products of the cracking be?

According to the Wikipedia article on cracking the cracking process in entropy-driven and the radicals formed in certain conditions usually move the reaction forward. But the final products (and even which radicals form) highly depend on the methods of cracking (thermal, steam etc.) and/or the catalysts. There are numerous catalysts which demonstrate the diversity in the mechanism of cracking.

For example, take a look at this paper (It's just an abstract, but it gives you an idea)

  $\ldots$ The main reaction route on FAU is hydride transfer followed by β-scission leading to mainly C4 species, while on MFI protolytic scission is responsible for the formation of high amounts of C1−C3 species. $\ldots$

So, we're left with speculations. This isn't a real question. The only answer the examiners want is B's most probably butane, as it must be an alkane.

As this also demonstrates, a very simplistic view is that an octane can break into two ethene molecules and one butane molecule. So, the reaction they must have had in mind is
$$\ce{C8H18 -> C4H10 + 2C2H4}$$
Pfft.

TL;DR:

This isn't a good or correct question to ask at all; at the very least, it's very poorly worded. The answers are supposedly isooctane for A and butane for B, but the OP's confusion was innocent.


Can the inverse of a micelle be formed with non-polar solvents? A micelle is formed due to the clustering of soap molecules wherein the interior of the cluster is formed by the hydrophobic end and the ionic ends are on the surface of the cluster.

So, if a non-polar solvent is used (Assuming non-polar solvents dissolve non-polar solvents), can the reverse of micelle be formed? That is, wherein the ionic end forms the interior and the hydrocarbon end the exterior?

Yes, they are aptly enough called inverse micelles, eg such as formed by the detergent AOT. While you retain the hydrophobic/philic balance typical of a water-soluble detergent, wherein you have a long hydrophobic tail and hydrophilic headgroup, you generally have to alter the headgroup area (making it smaller relative to the tails) to favor inverted phases of this type. In addition if charged the headgroups will strongly repel, requiring a cosolvent or some counterions to reduce such interactions.


Purification/Extraction of Sodium Permanganate I read this question and answer, and it seemed like a rather nice way to make sodium permanganate.  Using the method with sodium hydroxide, I produced a solution. Once filtered and decanted, it was dark green.

At first this seemed somewhat strange, as I would expect to see a red color for sodium permanganate. Though, after a bit of research it seems this is actually sodium manganate. The color matches, and considering it only exists in very basic conditions, and since there is a lot of sodium hydroxide in the solution, this would make sense. To test this further, I added some sodium bicarbonate to, oddly enough, act as an acid. The sample of solution quickly turned the pink color I had expected.

So, I currently have some amount of sodium manganate, sodium hypochlorite, sodium hydroxide, and sodium chloride in this solution.

At first I hoped I might be able to extract this using organic solvents, but unfortunately wikipedia claims potassium permanganate decomposes in organic solvents and alcohol. I assume the sodium permanganate or manganate would do the same.

Upon boiling down the solution I would end up with a mess of crystals of various shades of blackish and whitish.

So, my real question here is, if there is a way or ways to remove the other substances (sodium hypochlorite, sodium hydroxide, sodium chloride) and attain more or less pure sodium permanganate crystals.

You are correct. First step produces $\ce{K2MnO4}$, then you need to oxidize it.

My first guess was that adding $\ce{H2SO4}$ will cause disproportionation of $\ce{K2MnO4}$ into $\ce{KMnO4}$ and $\ce{MnO2 + K2SO4}$. Filtering removes $\ce{MnO2}$. Crystallization will give you pure product. Or you can use a mix of
$\ce{KMnO4 + K2SO4}$ for many purposes.

Industry uses more efficient method: electrolysis of $\ce{K2MnO4}$.


Very elementary question about valence electrons of ionic compound In a university entrance question, I found this question:

  What is the number of valence electrons of $\ce{Mg(OH)2}$? Answer = 24

I know that Mg has 2 valence electrons and O has 6. But I don't know what I must do for a complete ionic compound. If I add them up I get 2+6+6+1+1 = 16, but the answer of this question was 24. Is the answer right? If yes, then how to get it?

You get 24 if you say Mg2+ has 8 valence electrons (its electron configuration is 1s22s22p6) instead of zero.

In that case, there are 8 in Mg2+, and 8 in OH-  (6 from O, 1 from H, and and extra 1 because of the -1 charge).  Therefore there are $8 + 2 \times 8 = 24$ per formula unit made of one Mg2+ and two OH-.

Some people do argue ionized Group 1A and 2A metals (above Period 2) have 8 valence electrons instead of zero, because technically there are indeed 8 electrons in the outermost electron shell and they are isoelectronic with the noble gas at the end of the previous row (e.g. in your case Mg2+ has the same electron configuration as Ne).

But I think most chemists would prefer to say Group 1A and 2A cations have zero valence electrons, that is, that the valence shell in these ions is the shell corresponding to their Period that used to have the outermost electrons.  In that case, the answer to your question is that there are zero valence electrons in Mg2+ and 8 in OH-, for a total of 16 in Mg(OH)2.  Note that this is slightly different from your reasoning, however.

An argument for saying Mg2+ has zero valence electrons is that if we draw the Lewis dot structure for the formula unit, I don't know anyone who would put any dots around the Mg.  On the other hand, an argument for saying that it has 8 is that we frequently argue that Mg forms the Mg2+ ion because that gives it a complete octet in the outer shell.  We rarely present the octet rule as requiring 0, 2 or 8 electrons in the valence shell.

I don't think the correct answer can be found in any mathematically rigorous sense, it depends on your preference in labeling, what you want to call "valence" electrons, and there are nontrivial arguments both ways.


IUPAC Name clarification $\ce{CH3-CH(F)-CH2-CH2-CH(C2H5)-CH2-CH3}$

Will the correct IUPAC name be 3-ethyl 6-fluoro heptane or 5-ethyl 2-fluoro heptane?

The one with the lowest set of locants is correct. I.e. the one with lower first number in the locant set; if they are same, then the second one, etc. But, it is done on the ordered sets of locants, (3,6) vs (2,5) not (5,2). The substituent prefixes are then sorted alphabetically. So, the correct name is 5-ethyl-2-fluoroheptane.


Reaction of a diketone What is the product when ammonium bicarbonate reacts with hexa-2,5-dione?

I thought of formation of 2 moles of NH3, water and carbon dioxide. And then the usual addition followed by elimination to the carbonyl group. But is it the case? Or will some cyclic compound result?

Ketones react with NH3.

R2C=O -> R2C=NH. Hexa-2,5-dione has two carbonyl groups so it leads to cyclic compound. Check Paal–Knorr synthesis in Wikipedia to know the answer.


Scale to be considered for comparing electronegativities of nitrogen and chlorine The Pauling scale gives the $\chi$ values of $\ce{N}$ and $\ce{Cl}$ to be $3.04$ and $3.16,$ respectively, but the Allen scale gives the $\chi$ values of $\ce{N}$ and $\ce{Cl}$ to be $3.066$ and $2.869,$ respectively.

Another point that should be made note of is that the most electronegative element as per Allen scale is $\ce{Ne}$. So, why is it accepted in the first place? And whose electronegativity is actually greater?

Please take into account that there are a lot of different ways for calculating electronegativity.

1) Pauling obtains it by comparing the energies of dissociation of the three diatomic molecules AA, BB, and AB, stating that H electronegativity  is 2.1 !  Unit : √energy

2) For Mulliken, it is the average value between the electronic affinity and the first ionization energy of the atom. Unit : energy

3) For Allred - Rochow, it is the force acting between a valence electron and the charge of the atom when the other valence electrons are removed. Unit : force

4) For Sanderson, it is the average electronic density of an atom divided by the average electronic density of the corresponding atom of inert gas.No unit.


Deviation angle for 1st order diffraction (i) A diffraction grating has 560 lines per mm. Calculate the grating spacing. If the total length of the grating is 2 cm, how many lines are there?

  (ii) Light of wavelength 500 nm is incident on the grating with this spacing. Calculate the deviation angle for 1st order diffraction.

I am having difficulty with the (ii) part. For my spacing I got 0.0018 mm. So, then I used

\begin{align}
n \lambda &= d \sin{\Theta} \\
\frac{n \lambda}{d} &= \sin{\Theta} \\
0.27 &= \sin{\Theta} \\
\Theta &= \sin^{-1}{(0.27)} = 15.7^\circ
\end{align}

The grating equation you've used is correct, and in order to find the first order diffraction angle ($n = 1$) you need to use arcsine trigonometric function and carefully place data with respective units:

\begin{align}
n \lambda &= d \sin{\Theta} \tag{1}\\
d &= \frac{\pu{1 mm}}{N}~[\pu{mm}] \tag{2}
\end{align}

where $N$ is number of lines per unit length. Combining (1) and (2) and converting the wavelength from nm to mm:

$$\Theta = \arcsin{\left(\frac{n \lambda}{d}\right)} = \arcsin{\left(\frac{n \lambda N}{\pu{1 mm}}\right)} = \arcsin{\left(\frac{1 \cdot \pu{5e-4 mm} \cdot 560}{\pu{1 mm}}\right)} = 16.26^\circ \tag{3}$$


Difference between state of matter and phase What is the difference between state of matter and phase?
 This site  said that:

  Phases are different from states of matter. The states of matter (e.g., liquid, solid, gas) are phases, but matter can exist in different phases yet the same state of matter.

I don't really understand this. Could anyone start from scratch and explain this? Or send me a link to go to?

A phase is a region of space where all physical and chemical properties are uniform. This means that for example the density, the chemical composition and the temperature is equal everywhere in the region. This will also mean that the state of matter (solid, liquid, gas, etc.) is equal everywhere.

A state of matter is a form that matter can take. Examples are solid, liquid, gas, plasma. There are some others that appear under extreme conditions.

For example, nitrogen and water at 50K are both solid, so they have the same state of matter. However the chemical composition is not the same and hence they don't have the same phase.


Comparison of chemical shifts among alkanes in NMR spectroscopy I was reading through the factors influencing chemical shifts in proton NMR. I found the order of chemical shifts in the image above. Why is this order supposed to be as it is? I thought it would be the reverse as the inductive effect is the maximum in tertiary carbon, so the hydrogen attached to that carbon is shielded the most and thus has the least chemical shift. But why isn't that true?

A carbon atom is slightly more electronegative than a hydrogen atom. For tertiary hydrogens, the extra bonded carbons will inductively withdraw more electron density from the hydrogen. This means the hydrogen will be deshielded slightly more, and resonate at a higher ppm.

I would classify cyclopropane in its own class, as its molecular orbitals make it somewhat unique. The C-C bonds do not exhibit exclusive sigma bond characteristics, and due to that phenomenon the hydrogens will hold slightly more electron density.


Why is formyl chloride unstable while higher acyl chlorides are stable? Source: Paula Bruice Ed. 2017 Pg.877

From this excerpt, we infer that formyl chloride is unstable.
But how does that affect the reaction in any way? Is it just mentioned as a point to note or does it have an effect on the reaction. And if acetyl chloride is unstable why is acyl chloride stable?

Also note that I do realise that this is a duplicate of
Stability of formyl chloride

But we’ve not used water in Gatterman Koch reaction as reasoned by @SimpliFire in the thread. So how does it remain unstable even when water is not involved.

The simplest stable acyl chloride is ethanoyl chloride or acetyl chloride; methanoyl chloride (formyl chloride) is not stable at room temperature, although it can be prepared at –60 °C or below. (Wikipedia)

The instability of $\ce{HCOCl}$ is caused by ease of elimination of HCl from its molecules. Cl is decent leaving group and after it's gone, remaining acylium cation has a very acidic hydrogen instead of alkyl present in other acyl halides. Thus easier breaking of C-H bond vs C-C bond causes drastic difference in thermal stability.

Note also the formyl chloride can be considered not only the derivate of formic acid, but also a derivate of formaldehyde.


When an inert gas is added to a system of constant volume, equilibrium is not effected? But pressure of the system increases. Because of this increase in pressure, wouldn't (according to Le Chatelier's principal) equilibrium shift towards the side with less moles of gas?

Addition of an inert gas at constant volume:
When an inert gas is added to the system in equilibrium of constant volume though the total pressure exerted by the gases increases there is no shift in the equilibrium.

There is no shift in the equilibrium since there is no change in the molar concentrations(kept at constant volume) of each of the reacting components.

BUT

If an Inert Gas is added to a mixture kept at constant pressure there is a change in volume of overall mixture and hence the molar concentrations of the reacting components also decreases.
Hence according to Le Chatelier’s Principle the reaction shift in direction of lower molar concentration.

Basically - ((Inert/Non Reacting) Gases)) do not affect the original concentrations and hence do not affect the equilibrium in case of constant volume situations


How can a conductivity experiment have higher voltage than battery? I did an experiment where I tested the conductivity of different solutions and some substances.
We connected a 9 bolt battery, some aluminum foil sticks, and a multimeter that reads the voltage.
It connects like  liquid --> aluminum --> multimeter --> battery --> aluminum --> liquid.

My understanding is that if a liquid contains more ions, it conducts electricity better.
Here's my question. We tested the battery just by itself and it read about 9 volts. When we tested with baking soda, it reads about 9.2 volts.
How can it be higher than the energy source? Does baking soda produce its own electricity?
Also, when we tested deionized water, it came out to be about 8.5. It seems close to 9, but it was one of the lowest of the tested solutions.
Why is it like this? If it's deionzed, shouldn't it be nearly zero? since it doesn't conduct electricity at all? (although there can be impurities so it can have some, but I think 8.5 is still too much).
I don't really understand. Can someone explain why the results came out like this?

A voltmeter is a zero current(i.e. very high resistance) device. Therefore in your scheme, the readings you are getting are not reliable. The way to make that measurement is by either switching your multimeter into an ammeter in the configuration you have and measure the current, or adding a resistor to the circuit in place of where you have the multimeter now and connect the multimeter in parallel to that resistor. In either case, you will be measuring the current going through the solution, which will tell you about the conductivity.