Determination of hardness of water using EDTA Finding hardness of water by using EDTA. For determining permanent hardness we take the sample of about $\pu{50 mL}$ and then add Eriochrome Black T (EBT) indicator that results in wine red color due to calcium complex made by and also ammonia buffer. Then titrating it with ethylene diamine tetra acetic acid (EDTA) where change in color (i.e. sky blue) marks the end point of titration.

While carrying out experiment we add EBT as an indicator. Is there any criteria to add so in a fixed particular amount?

There really is no need to add EBT in a fixed amount. Calcium EDTA complex is far more stable at $\mathrm{pH} \approx 10$ (ammonia buffer) and will be the dominant Ca-compound in the solution even if there is a huge excess of EBT.

However, the amount of EBT is important to track the color change. If there is too much EBT, the color will be too dark, and one can easily miss the red-blue transition. If there are only trace amounts of EBT, the colors will be pale and again the transition is going to be barely visible.

The amounts of EBT given in the lab manuals are typically approximate and are estimated for the convenience of the personnel and to reduce the error in determining equivalence point due to unclear reading of the solution's color.


Extracting metallic magnesium from cholorphyll My understanding is that when chlorophyll is exposed to an increased concentration of hydrogen ions (or rather the pH is lowered), the magnesium is displaced by a hydrogen ion and chlorophyll is converted to pheophytin.

My question relates to the Mg2+ ion that is released in this process. Is it possible, naturally depending on the acid used to convert the chlorophyll to pheophytin, to isolate the Mg released?

Note my background in chemistry is rather limited. My guess is that the Mg will react to whatever acid (say sulfuric acid) to create a compound like magnesium sulfate which in turn can be processed to extract Mg in metallic form (electrolysis?).

Any clarity would be much appreciated as this has tickled my curiosity for a while now.

You can make the magnesium salt, but electrolysis of an aqueous solution of this salt will reduce the water, evolving hydrogen, not reduce the magnesium.  You can get metallic magnesium via electrolysis of the molten chloride or high temperature reduction of the oxide with carbon,  neither of these cheap or easy.  See here.


One-pot synthesis of pyrazole I am trying to follow a published paper. It is a one-pot synthesis of pyrazole from a ketone and and acid chloride via a diketone intermediate.

Here is the produre from the paper:

General Procedure: Ketone (2 mmol) was dissolved in 5 mL dry toluene in a screw cap vial (with septum) and then the solution was cooled to 0°C under nitrogen. LiHMDS(2.1 mL, 1.0 M in THF, 2.1 mmol) was added quickly via syringe with agitation stirring and the formed anion was allowed to sit for approximately 1 minute before the addition via syringe of acid chloride (1 mmol) in one portion with stirring. The vial was then removed from the ice bath and allowed to stand for 1 minute, then 2 mL AcOH was added with stirring. 10 mL EtOH and 5 mL THF was added to form a homogeneous
mixture, and then hydrazine hydrate (2 mL, 1.1 g, 34.3 mmol) or substituted hydrazine(5 mmol) was added and the mixture was allowed to auto-reflux and was held at that temperature for 5 minutes, when LCMS showed that all diketone had reacted. The resulting solution was added to 1.0 M NaOH solution and extracted with EtOAc. The organic fraction was then washed with brine, dried over Na2SO4, and evaporated under reduced pressure.See specific compounds for purification details.

My questions are:

What is the purpose of excess hydrazine hydrate, acetic acid, and enolate in producing the pyrazole?
How many molar equivalents of sodium hydroxide should I use at the end of the reaction?

Excess hydrazine hydrate to neutralise the AcOH and ensure complete reaction of the diketone.

Excess AcOH to protonate all anions arising from the LiHMDS, mostly to completely protonate the anion of the diketone.

Excess enolate (and LiHMDS) to ensure all the enolate reacts. The diketone product is more acidic than the enolate and will hence protonate the enolate, so you need 2 eqs or more of base/enolate to get complete reaction.

How much NaOH for the wash at the end? More than eqs of AcOH. Don't see why they specify a base wash as everything you want to remove into the aq phase would be aq soluble anyway.


Why do molecular orbitals in solids merge to bands? Why do molecular orbitals in solids merge to bands?

For example: In silicon every atom is sp3 hybridised, but when I merge two of these orbitals then it yields a bonding and an antibonding MO. When a third atom binds to one of the $\ce{Si}$ atoms in this configuration, then again we should get two MOs, separated by the same energy. This way we we don't get so called bands. How do the bands form?

I have tried wikipedia and other sites but they answer in brief.

Let's explain this with your example of silicon that every atom is sp3 hybridised. when you merge two of these orbitals, then it yields bonding and antibonding MO. But  when a third atom bounds, it yields  3 MO (not four as you mentioned) bonding, antibonding and non bonding MO. The number of MO equals the number of atomic orbitals constituting them, according to LCAO theory.
The energy of the third MO is in between the nonbonding and anti-bonding MOs. When you add a forth, a fifth and so on atoms, you will have more and more MOs very close in energy and you will form a band.
This is a qualitative description about the formation of a band.


what is the correct order of basicity It asks about the basicity. All have the same hybridization so through inductive effect, 3 should be the most acidic. I feel that my assumption is wrong. Can anyone help? The answer is 3>4>1>2.

$\ce{-CH3}$ group has +I effect; it has electron donating effect. It increases basicity and reduces acidity.
Moreover due to resonance in phenyl group, the availability of electrons over N decreases.


What is the difference between doxycycline HCl and doxycycline hyclate? I would like to ask what are the main differences between doxycycline HCl and hyclate?

I work in the pharmaceutical/veterinary raw materials industry and some clients are saying that they prefer hyclate because it is more concentrated than HCl, but I have yet to come on any articles confirming this. Is this true?

[...]they prefer hyclate because it crystallizes more readily than HCl salt

The principal component is doxycycline - in both cases. The active component is a metabolite.

Doxycycline is an antibiotic and belongs to the tetracyclines (= four rings).

In order to prepare nice ions that can traverse membranes and pumps by the body easily, pharmaceutics containing amino groups or 'alkaloids' (see the red nitrogen atom in the structure), are often converted to their hydrochloride, citrate, oxalate, bromide etc. salts.

These are salts with the (protonated) pharmaceutical as the cation and chloride as the anion. Salts are water soluble.

Here, the notation for this is doxycycline$\ce{.HCl}$.

The crystallisation is done in a solvent or a mixture of solvents.

When the doxycycline$\ce{.HCl}$ precipitates (=when it forms a solid), molecules of the solvent may get embedded, trapped, or entrained in the crystal structure (but the crystals gets pulverized anyway and re-crystallized).

Note that this is not just a wet solid! There would be little or no absorption in that case.

Exactly this happens when the hyclate is formed during the crystallization of doxycycline hydrochloride in a mixture of water and ethanol.

Hyclate is just an abbreviation for hydrochloride hemiethanolate hemihydrate, which means that each molecule of doxycycline comes with 1 $\ce{.HCl\ }$  and 0.5 $\ce{.CH3CH2OH}$ and 0.5 $\ce{.H2O}$ in the crystal.

Whether it is more concentrated (= more active component in one pill) only depends on the formulation of the respective pharmaceutical companies.


How do they ensure there is one electron on an oil drop in Millikan's oil drop experiment? Since air is ionized using X-rays, in the 'observation chamber', there must be several electrons in the chamber, so the oil drops are exposed to several electrons so it seems intuitive that a single oil drop should catch more than one electron.

Then why is it that there's just one electron on a droplet or do they selectively choose those droplets which have a single electron residing on them?

There does not have to be just one electron per drop.  Say you have a drop which, in reality, picked up four electrons, another with five and a third drop with seven.  None has one electronic charge, but when you measure charges you find they have a common factor; the first drop has four times that factor, the second has five times that factor and the third drop shows a multiplier of seven.  It was this common factor, not necessarily the charge on any specific drop, that was recognized as one electronic charge.


How did the temperature in the solution with a lower concentration of CuCl2 increased in a faster rate? My lab assignment is about the reaction between $\ce{CuCl2}$ and Aluminium and we tracked the temperature of the solution every 15 seconds and we needed to make our own experiment to check how different things affect the temperature of the solution every 15 seconds. So I chose the concentration of $\ce{CuCl2}$ and thus we had four cups, one with $\ce{0.1 M}$, one with $\ce{0.2 M}$, one with $\ce{0.5 M}$ and one with $\ce{0.8 M}$. Obviously the temperature in the cups with a lower concentration of $\ce{CuCl2}$ barely changed and throughout most of the experiment stayed at the exact same state.

Now, the cup with the $\ce{0.5 M}$ concentration of $\ce{CuCl2}$ changed temperature in a faster rate than the cup with the $\ce{0.8 M}$ concentration. Not only that, at the end of the experiment, the temperature in the cup with $\ce{0.8 M}$ was 28° C meanwhile the temperature of the cup with the $\ce{0.5 M}$ concentration of $\ce{CuCl2}$ was 29° C .

Can somebody explain to me how that happened?

The reaction $\ce{Cu^2+ + 4 Cl- <=> CuCl4^2-}$ may lead to slowing down reaction with aluminium, decreasing the equilibrium concentration of $\ce{Cu^2+}$ in $\pu{0.8 M}$ $\ce{CuCl2}$. That all would lead to lower peak temperature due bigger heat dissipation before reaching maximal temperature.

Contributing factors may be

decreasing of chemical activity of water for the electrode surface reaction $\ce{Al(s) + 6 H2O -> [Al(H2O)6]^3+(aq) + 3 e-}$. These  electrons, charging the electrode, then reduce copper: $\ce{[Cu(H2O)n]^2+(aq) + 2 e- -> Cu(s) + n H2O}$
increasing of viscosity, therefore decreasing of diffusion coefficients and another decreasing of the reaction rate.


Why doesn't propynal give cannizzaro reaction? Cannizzaro reaction is a disproportionation reaction which is given by aldehydes and ketones having no α-H, forming alcohol (reduction) and carboxylic acid (oxidation) as its products. The compound, Propynal has no α-H and it should be giving Cannizzaro reaction but, my textbook says that the reaction is not given by Propynal. Can someone please explain to me why Propynal doesn't give Cannizzaro reaction?

It is known that terminal alkynes reacts$1$ with $\ce{NaNH2}$ and produces salt, much like an acid-base reaction, hence it is a fast reaction.
Furthermore, the aldehyde group is an EWG, which makes it easy to lose a proton, hence drives the reaction!
Whereas, cannizzaro reaction happens through multiple reversible steps as seen in the following image,

Therefore, acid-base reaction in this case has much higher probability than Cannizzaro reaction.

References
$1$: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch09/ch9-2.html


Why can't this mixture be used as a buffer solution? Explain, using a chemical equation to aid your answer, why a mixture of $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ ammonia solution and $50~\mathrm{cm^3}$ of $0.10~\mathrm{mol\:dm^{-3}}$ hydrochloric acid cannot be used as a buffer solution.

So writing the equation down: $$\ce{NH4OH + HCl -> NH4Cl + H2O}$$

there are $0.005~\mathrm{mol}$ of each reactant. From my notes, I think there has to be double the amount of the weak base, but I don't really understand why. Even after googling explanations, I can't find one that answers my question. Why do you need double the amount (if this is correct) of the weak base? An answer to the question above would also be appreciated.

You have an example of an alkaline buffer solutions here. They are commonly made from a weak base ($\ce{NH3}$ aqueous solution) and one of its salts ($\ce{NH4Cl}$) and work in a little simplified picture as follows.

Ammonia is a weak base, so the position of the equilibrium for the following reaction is well to the left
$$
    \ce{NH3_{(aq)} + H2O_{(aq)} <<=> NH4+_{(aq)} + OH^{-}_{(aq)}} \, ,
$$
and once you add some $\ce{NH4Cl}$ it pushes the equilibrium of the reaction above even more to the left (see Le Chatelier's Principle). As the result you have both lots of ammonia (due to the equilibrium of the above reaction being well to the left) and of ammonium ions (coming from $\ce{NH4Cl}$). And you will still have some hydroxide ions due to the reaction above, so that solution will be alkaline. The actual $pH$ depends on in which molar proportions the ingredients were mixed. Usually, $\ce{NH3_{(aq)}}$ and $\ce{NH4Cl}$ mixed in equal molar proportions, the solution would have a pH of 9.25.

When some acid is added to such buffer it will mostly be removed by the following process
$$
    \ce{NH3_{(aq)} + H+_{(aq)} <=>> NH4+_{(aq)}} \, ,
$$
and when some base is added it will mostly be consumed by the following reaction with ammonium ions
$$
    \ce{NH4+_{(aq)} + OH^{-}_{(aq)} <=>> NH3_{(aq)} + H2O_{(aq)}} \, ,
$$
where in both cases the equilibrium is well to the right.

So, as you see, both the weak base and its salt should present in the solution. Thus, there should be some excess of weak base when you mix it with a strong acid, otherwise there will be no weak base at the end of neutralization reaction. The idea that one should have double amount of a weak base comes from the fact that, as I mentioned, usually a weak base and its salt are mixed in equal molar proportions.

$%edit$


Five-pointed crown alkane C25H40: does it exist? My 4 year old granddaughter, playing with molecular models, began making a cyclopentane ring, each of whose carbons bore a spiro-connected cyclopentane ring of its own , in the manner of spiro[4.4]nonane.  If she hadn't run out of carbon atoms first, the result would have been a pentagonal crown, with the five outer cyclopentane rings perpendicular to the central one.

I wonder if such an alkane has been synthesized. Its molecular formula would be $\ce{C25H40}$. I can't think of why the molecule would be unstable or sterically impossible, but I couldn't find it in a Google search on $\ce{C25H40}$.

In particular,  I wonder if the five outer rings would be exactly perpendicular to the center one (eclipsed), or somewhat staggered.

These polyspiro compounds such as the one you gave are known as [m,n]rotanes, where m is the size of the central ring and n is the size of the decorating rings. So, your compound would be called [5,5]-rotane.
A Reaxys search unfortunately doesn't show anything for [5,5]-rotane, but in general, many such compounds have been made. A lot of the articles are in German or French (there are quite a few such articles). However, I did find one in English: J. Am. Chem. Soc. 1998, 120 (2), 317–328.
The authors synthesised and obtained X-ray structures of a number of rotanes, among them [5,4]rotane, which is the closest to the [5,5] that you are looking for. I quote:

In [5.4]rotane (7), the cyclopentane ring adopts a slightly distorted envelope conformation: C(1), C(5), C(9), and C(17) form an almost perfect plane (mean deviation 0.021 Å), while the flap atom C(13) is located 0.684 Å above this plane. The cyclobutane rings at C(13) and C(9) are nearly planar (φ = 9.7°) and planar (φ = 0.2°), respectively. The remaining three cyclobutane rings are moderately puckered (φ = 23.1, 26.5, and 28.2°).

The torsion angle φ of cyclobutane refers to the angle between the two planes that define the molecule. A completely planar four-membered ring would have φ = 0°, and an increasing value of φ indicates greater deviation from planarity:

As you can see in the X-ray structure, the five attached cyclobutane rings do eclipse each other to a certain extent: the obvious exception is the ring fused to C(13). This is not a very surprising conclusion at all. The envelope conformation of unsubstituted cyclopentane is known to suffer from torsional strain (i.e. unfavourable eclipsing interactions between substituents). Carey and Sundberg's Advanced Organic Chemistry, Part A writes (pp 162–3):

There is minimal angle strain in cyclopentane, but considerable torsional strain is present. Cyclopentane is nonplanar and the two minimum energy geometries are the envelope and the half-chair. [...] The planar portions of both conformations have torsional strain owing to C−H and C−C bond eclipsing.

There's a ton more information in the paper, and I'd recommend giving it a read if you are interested - I just tried to pick out the parts directly relevant to your question here, since you indicated that you were interested in the conformation of such a compound.


Is there a way to determine if a compound is oxygen or water sensitive? The synthesis of $\ce{CoH[P(OPh)3]4}$ requires the use of an inert atmosphere e.g. nitrogen as the compound is air-sensitive. Is there any way to tell whether it is the oxygen or water vapour in the air specifically? I thought it might be water sensitive seeing as metal hydrides are basic and may therefore react in an acid-base reaction:

$$\ce{CoH[P(OPh)3]4 + H2O ⟶ Co[OH][P(OPh)3]4 + H2}$$

Is rationalisation correct/is there more to it than this?

I think that this will be water sensitive for the reasons you present but also oxygen sensitive for two reasons:

Due to the relatively low oxidation state of $\ce{Co(I)}$ in the compound which will react with $\ce{O2}$ towards formation of $\ce{Co(II)}$.
The $\ce{P^{III}(OPh)3}$ ligand is air sensitive as well and can oxidize to $\ce{OP^V(OPh)3}$ as well as reacting with water: https://www.alfa.com/en/catalog/A18662/.


Can the reaction intermediate be lower in energy than the reactants or products? It makes sense to me that reaction intermediates exist transiently if they are of higher energy than the reactants and the products, my organic chemistry textbook seems to give me information that doesn’t seem to match up.

My knowledge is that the overall activation energy of the reaction is the energy change between reactants and the highest-energy transition state and the overall energy change is he energy difference between the reactants and the final products. If intermediates could be of lower energy levels than either the reactants or products, would this definition change?

I am confused because the solutions manual to one of the problems gives me an energy diagram with an intermediate that is of lower energy than both the reactants (but not the products), and labels the overall activation energy as the energy change between the intermediate state and the highest-energy transition state. The textbook also gives me an energy diagram of a catalyzed reaction that has several intermediates of lower energy than the reactants, but not the products.

(The textbook I’m referring to is Organic Chemistry. 8th edition by John McMurry)

Would it make sense for intermediates to only be lower in energy than the reactants but not the products since the products would still be more favorable and stable than any of the intermediates? Would this only happen if there is a catalyst?

The intermediate is a proper chemical species but a reactive one under 'normal' conditions and is not a transition state, i.e. it has a short, but measurable, lifetime. The intermediate can change only by crossing one of two transition states, one back to reactants and the other to another intermediate or to product.

Any intermediate must not be lowest energy species, including reactant and product, but can be lower or higher in energy than the reactant (but less than any transition state) and higher than product. The limiting activation energy will be the largest one in the whole scheme, so may be that between reactant & intermediate or between intermediate & product. Which it is depends on the exact species involved.


Why is Cu2HgI4 filtered while it is hot? Why is it required that the solution of copper (I) tetraiodomercurate (II), which has a thermochromic property, be filtered while it is hot?

Copper tetraiodomercurate, $\ce{Cu2HgI4}$, is typically prepared by heating combined suspensions of copper(I) iodide, $\ce{CuI}$ and mercury(II) iodide, $\ce{HgI2}$.

The iodides are prepared by (a) reacting copper(II) sulfate, $\ce{CuSO4*5 H2O}$ with potassium iodide, $\ce{KI}$, and (b) mercury(II) nitrate, $\ce{Hg(NO3)2}$ with with potassium iodide.

In order to avoid contamination of the final reaction product with one (or both) of these iodides, it is essential to filter off while the reaction mixture is still hot.


Why does a carbonyl reform in ester reactions I have noticed that frequently whenever esters react with a nucleophile, the nucleophile will attack the carbonyl carbon and eventually in order for the carbonyl to reform the other oxygen will leave for example the first addition of the grignard reagent to an ester:
$\ce{PhMgBr + Ph-(C=O)-O-R}$
will result in
$\ce{(Ph)2-(C-O^{-})-O-R}$
as an intermediate, and then rather than the
$\ce{O^{-}}$ simply getting protonated with water for example, its favorable for the $\ce{O-R}$ to leave and the $\ce{C-O^{-}}$ to become $\ce{C=O}$. Why is the carbonyl reforming more favorable, isn't $\ce{O-R}$ a bad leaving group?

Alkoxide ion is only a "bad leaving group" in a relative sense (say compared to chloride). In this case, the tetrahedral intermediate resulting from the initial reaction between the Grignard reagent plus the ester is itself an alkoxide ion, and when the OR group leaves, that then becomes an alkoxide ion - so that step should be close to energetically neutral.

Saponification of esters by sodium hydroxide is another situation where alkoxide anion is a leaving group - again because alkoxide and hydroxide ions are close in energy.


Common flame-retardant plastic film materials? More questions for my DIY hot air balloon:

What are some very common (i.e. easy to purchase) plastic film materials that have very good flame-retardant properties?

How is the "degree of flame retarding" of a plastic film material rated, so that I can compare materials by their manufacturer's rating?

Are there any common "spray on" flame retardant treatments that could work? Or is a flame retardant plastic film the only way to go to really stop it from bursting into flames?

What have you found so far? In a situation like this honestly I would go ahead and research what hot air balloons are commonly made with already. Find somebody more experienced in the hobby and ask them if you can, or even an experienced fire fighter you can get in touch with.

From wikipedia: "For modern sport balloons the envelope is generally made from nylon fabric and the inlet of the balloon (closest to the burner flame) is made from fire resistant material such as Nomex"

The type of measurement you are asking for seems a bit more complicated than something like measuring mass or volume, it seems to be commonly represented with images, videos of tests, graphs, etc. You may feel more at ease by testing how useful materials are before and after flame testing them yourself, that is what I would do anyways. I hope this helps you.


What exactly is meant by 'multi-configurational' and 'multireference'? Some topics here have touched on this before (see 1, 2, 3), but I haven't found a clear definition yet.

I would like to know what exact property of the wave function these terms refer to. It would also be helpful to have a clear definition of 'reference' and 'configuration'. I'll try to explain below where my problems are in clearly understanding/defining these terms:

Starting with Hartree-Fock, it is obvious that the wave function in HF is both single-reference and 'single-configurational': there is only one Slater determinant.

Going to configuration interaction methods, the wave function now becomes a linear combination of several Slater determinants. The additional Slater determinants are excitations of the ground state determinant: virtual orbitals from HF are taken and replace previously occupied orbitals in the determinant. However, these orbitals still have the same coefficients as in HF - only the coefficients in front of the Slater determinants are optimized for the linear combination. If I'm correct, 'configuration' here refers to one particular Slater determinant - a method is, therefore 'multi-configurational' if the wave function that is used has two or more (different) Slater determinants, correct? That also means that none of the CI methods (be it CIS, CISD,..., or Full CI) are multireference methods?

Continuing with CASSCF, this method is basically Full CI limited to the chosen active space of orbitals. It is therefore multi-configurational. At the same time, it is also often referred to as being 'multireference'. The only difference to CI, however, seems to be the optimization of the coefficients in the Slater determinants themselves, hence, this must be the defining criterion for 'multireference'? What does 'reference' here refer to?

Now there is also multireference-CI. From the above definition, I would expect this to be a form of CI where I also optimize the orbitals, but that does not seem to be the case. The Wikipedia article on MRCI starts with:

  In quantum chemistry, the multireference configuration interaction (MRCI) method consists of a configuration interaction expansion of the eigenstates of the electronic molecular Hamiltonian in a set of Slater determinants which correspond to excitations of the ground state electronic configuration but also of some excited states. The Slater determinants from which the excitations are performed are called reference determinants.

This is confusing to me: Is 'excitation of the ground state electronic configuration' vs. 'excited states' referring to the optimization of the Slater determinants themselves? Or does 'excited state' refer to configurations with different total spin? That would be a different definition of 'reference', but then CASSCF would only be a multireference method if it uses the corresponding SA-CSFs, regardless whether the Slater determinants are optimized or not?

Your problem seems to be with the terminology used in CI methods, so let me go through the different terms you mentioned:

A configuration is a certain occupation of (molecular) orbitals. Mathematically configurations can be represented in 2 ways. The first one is the Slater Determinant (SD), an anti-symmetrized product of spin-orbitals. Slater Determinants, however, are not eigenfunctions of the spin operator $\hat S^2$, but the electronic wave function needs to fulfill this requirement. Therefore one constructs spin-adapted Configuration State Functions (CSF) as certain linear-combinations of SDs. Using CSFs instead of SDs usually makes a calculation more stable. "Configuration" is a more general term, which does not explicitly say whether an SD or CSF is considered.
Multi-configurational just means the method considers more than one configuration.
Reference means we have a designated configuration from which the excitations are generated. Single-reference thus means we only have one such configurations (usually the HF configuration), e.g. in CISD or CCSD. Multi-reference means we have more than one configurations to generate excitations from.

So multi-configurational just means we have many configurations, while single-/multireference says something about how those configurations are selected/generated.

How does single-/multireference apply to the different methods?

In HF we do not generate any excitations, therefore this concept does not really apply here. But I think one could still call it single-reference and single-configurational.
CISD and CCSD are the typical examples of single-reference methods. Nothing special here.
In FCI and CASCI we don't restrict the excitations to certain degrees (Single, Double, etc.), instead, we just take all of them. The concept of having one (or more) reference configurations is a possible perspective, but it is not really necessary. FCI can be viewed as having one reference configuration and taking all excitation degrees (up to the number of electrons available), which would make it single-reference. But from another perspective, we can argue that every MRCI wave function is just a truncation of the corresponding FCI wave function. So if FCI covers everything and more than MRCI does, would it not be multi-reference as well? Again, I would just say the concept does not apply here.
MCSCF (CASSCF) is a combination of CI (CASCI) and SCF, where SCF means we optimize the orbitals as well. This is not done in CI(SD...), CC(SD...), MRCI, etc. Other than that, the above arguments apply in the same way to the CI space as before.

Multi-Reference CI and CASSCF

As argued above, personally I would not consider CASSCF to be multi-reference, as the concept does not really apply. But I can see why people would consider it as such.

One usually does CASSCF because single-reference methods fall short of describing the wave function correctly, in a qualitative way. Such systems are then called strongly correlated and CASSCF can treat that strong correlation (also called static correlation) which is missing in single-reference calculations. In turn, however, CASSCF is missing that kind of correlation single-reference method can treat well, called weak or dynamic correlation.

Multi-reference is now the approach to combine both, strong and weak correlation, by doing first a CASSCF, and then using those configurations as a reference space, to generate all the excitations from.

Orbital optimization, however, is usually not feasible for dynamic correlation, because dynamic correlation usually means to have a lot of configurations and therefore huge CI space. With static correlation there are much fewer configurations, so doing orbital optimization additional to CI optimization is feasible and yields qualitative improvements of the orbitals. Therefore this is done in CASSCF, but not in MRCI.

"Excited states" in MRCI

What is meant here by "excited states" are actually "excited configurations" (with respect to the HF configuration). If you, for example, choose a CAS as your reference space, then all configurations, except for the HF configuration itself, are excited configurations.

Please note, that "excited state" is not a good choice of words here since it is actually something different: a state is a result of a CI calculation and is physically observable. On the other hand, a configuration is a basis function you put in a CI calculation and therefore is more of an abstract mathematical tool. However, you can approximate a state as a configuration to some extent, this is for example done in HF.


How can you detect iron hydroxide or iron sulfide in quarried rocks? We have a quarry that produces some gray rocks that sometimes turn green, then orange and the color bleeds, making the rocks unsuitable for our purposes. But it can take up to 6 months out in the weather to see if the gray rock will end up having this problem. It's been suggested that the compounds causing the problem in the rock are iron hydroxide and iron sulfide. Is there a way to detect these chemicals sooner, or the chemicals that precede them, so that we don't have to wait 6 months to weed out the bad rocks? Would also appreciate knowing any resources that might help figure it out.

Rocks that turn green and then red seems to me like you might have some copper sulfides, like chalcopyrite (CuFeS2). At first, the copper reacts with the atmosphere to form something like malachite (similar to the green patina observed on things like the Statue of Liberty). Next, the iron oxidises to form iron oxy-hydroxides (as you correctly identified).

In order to detect these things rapidly, you have several options:

Use expensive methods like XRD or XRF. I would recommend against, because these only test the very thing you are analysing, and if you would like to test a whole rock face, this would be extremely expensive. It would however be beneficial to give one of the stained samples to a nearby research institute to test what the material is, either using SEM-EDS or XRD. I might be wrong and it would not actually be copper.
Just look at the rock very close, preferably with a hand-lens. Chalcopyrite (and other copper/iron bearing sulfides) and shiny, usually golden or slightly green-purple when tarnished. You can look at a gallery of chalcopyrite to have a rough idea of how it looks like. Note that the gallery has nice specimens, which yo clearly don't have, but try to see how the lustre and colour look like.
Spray bleach (sodium hypochlorite). This would speed up the reaction, better than acid. Note that this could affect other things in the rock and cause staining or dissolution of things that wouldn't have without the bleach.
Try to smell the rock, especially around the places where you see incipient greening. Can you smell a sulfurous smell? Our noses are surprisingly extremely sensitive to sulfur, and can detect it even when some instruments can't.


When water boils at its boiling point how do we call it as an equilibrium? On boiling, water becomes a vapor, liquid volume decreases and vapour volume increases.  Since volumes are not constant, how do we call it as equilibrium?

An isolated system consisting of liquid water and vapour is always in equilibrium.

If you are applying heat to a system consisting of liquid water and vapour the system is not in equilibrium.


Properties of powdered metals I have seen certain metals such as Aluminium used in a situation where it is in a powdered form and "compacted" into a shape under intense pressure (in a mould), and it ends up looking mostly like solid metal. It is also heated afterwards, I assume to form a complete metallic bond.
I have a few questions on this:

How are oxide layers on the powder affected when it is compacted? Eg. Can the $Al_2O_3$ layer, or at least the O atoms from them, be "inside" the moulded object when it is compacted?
Metals in a powdery form aren't as reflective as in their "solid" form (also please let me know the right word for that). Why is this?
Are most metals in their powdered form pure, eg. Not like a salt, but a single element?
I know gold powder, obtained from chloroauric acid, becomes gold in its metal form when melted down. It looks brownish and takes up far more volume than the gold it contains. Is this pure Au or a compound/salt?
Also, in 4. with the gold powder, it actually forms rather large clumps from the chloroauric acid. How is this possible assuming it's a pure element?

Thought I better get all the questions done in one go. Thanks in advance

Let me try to answer the part 1. of your question.
The process of compacting powdered Aluminium is also called Sintering. And as you can guess, the thin oxide layer of Alumina inhibits the binding of Aluminium grains between the powder particles.

So, while sintering, we need to either:
1) Break part of the Alumina film or,
2) Reduce part of it to Aluminium,
 in order to facilitate formation of a metallic bond between the powder particles.

For 1), we use a nitrogen atmosphere as it enhances the destruction of Alumina.
For 2), we use because magnesium as a alloying element because magnesium, even in small amounts, concentrates on the surface of powder particles
and supports sintering by local reduction of Alumina.

In conclusion, I would say that compacted Aluminium does have Alumina inside the molded object. But metallic bonds between powders are formed after partly removing the oxide film by the above mentioned methods.


Are there different types of melamine resin? Background

I have a little knowledge of chemistry.

I was talking to a sales rep from a cleaning supplies company. They sell a scrub sponge that looks and functions just like Mr. Clean Magic Eraser. The sales rep was trying to scare me by saying that magic erasers have formaldehyde.

I know that the properties of a chemical compound are different than the properties of each component. I suspected that their product is actually the same as the magic eraser so I contacted the company directly, but got a less than satisfactory answer.

I asked if their sponge was made of formaldehyde-melamine-sodium bisulfite copolymer. They did not answer yes or no. They told me their sponge is made of melamine resin.

According to Wikipedia, melamine foam (a foam-like material consisting of a formaldehyde-melamine-sodium bisulfite copolymer) is a special form of melamine resin that is used mainly as an insulating and soundproofing material and more recently as a cleaning abrasive.

I would like to be more knowledgeable next time this sales rep contacts me. Hence the question.

Questions

Is there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer? Is there a way to make a melamine resin without using formaldehyde?

Certainly the are many different types of melamine resin.

Is there a way to make a melamine resin without using formaldehyde?

Yes, a different aldehyde could be used such as acetylaldehyde.  Glucose, benzaldehyde, acrolein and other alternatives to formaldehyde have been studied also (see page 33 of Polymer Synthesis by  Sandler et al.).

Is there another way to make an abrasive foam sponge out of melamine resin that is not formaldehyde-melamine-sodium bisulfite copolymer?

Good sources of information about formaldehyde-melamine-sodium bisulfite copolymer and possible alternatives are the patent Resilient foam based on a melamine-formaldehyde condensate and related patents.

First the introductory portion of the patent explains:

  It is true that foams based on melamine-formaldehyde condensates have been described in various patent publications, but they have not hitherto been employed industrially for heat or sound insulation in the construction industry. They are produced by stirring air into an aqueous melamine resin solution which contains an emulsifier and a curing agent. Such as described, for example, in German Pat. No. 870,027, have the serious disadvantage that they are very hard and brittle, and easily break or crumble when handled. It is alleged in French Pat. No. 1,073,642 that melamine resin foams can be produced by heating a resin powder in a mold under reduced pressure. This process, however, does not give useful resilient foams. U.S. Pat. No. 3,093,600 describes melamine resin foams which are said to have improved resilience and resistance to crazing as a result of the incorporation of triols, e.g. trimethylolpropane. However, it has been found that the resilience, and especially the recovery after compression, of such foams is inadequate for many applications.

Then in describing what is considered new about the invention, many different alternatives are included. Such as adding aldehydes other than formaldehyde, urea, and many other substances.


Mechanism of the reaction of an alkene with excess HI in presence of CCl4 Problem

Answer

 2-iodopropane

Question
I thought of Markovnikov's rule and suggested 1,2‐diiodopropane as the the final product:

Apparently, this is wrong. I can not figure out the mechanism of this reaction (will $\ce{HI}$ dissociate in the presence of $\ce{CCl4}?)$ And what is the significance of excess $\ce{HI}$ here?

According to [1], 1,2-diiodopropane reacts with hydrogen iodide to give 2-iodopropane.

So the reaction path might be

$$\ce{CH2=CH\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH2\bond{-}I}$$

$$\ce{CH3\bond{-}CHI\bond{-}CH2\bond{-}I + HI -> CH3\bond{-}CHI\bond{-}CH3 + I2}$$

The mechanism of the second part is unclear to me. It may be elimination of $\ce{I2}$ followed by addition of $\ce{HI}$.

1 Th. M. Schmitt, Analysis of Surfactants, Second Edition, p. 67


Negative ions affecting galvanic cell voltage I don't see why negative ions don't affect the voltage. I mean the electric field of a negative ion like $\ce{SO4^{-2}}$ is not like what $\ce{NO3-}$ produces that means that it should affect the overall voltage.

This is the same reason why for example for $\ce{Cu+^2}$ the reduction electrode potential is higher than for $\ce{Zn+^2}$

Well, assuming that you're talking about an aqueous cell, in order to get the anion into solution, you need a salt: eg $\ce{CuSO4}$ or $\ce{Cu(NO3)2}$. Yes, the local electric field around a single $\ce{SO4^2-}$ anion would be stronger than around a single $\ce{NO3^-}$. However, at the same concentration of $\ce{Cu^2+}$ you would have twice as many $\ce{NO3^-}$ ions so the overall solution would still be neutral.

In a microscopic level, each ion is surrounded by a solvation shell of water molecules that all have their partial positive charge on their hydrogens pointing at the anion (or oxygen towards the cation). For this reason they are also shielded from the electrode. Speculation: double charged anions would probably have a thicker solvation shell.

I don't see how this explains the difference in potential of $\ce{Cu^2+/Cu}$ vs $\ce{Zn^2+/Zn}$ since both have the same charge.


pH calculation of HK2PO4 I am struggling with a calculation I should know how to do, but I feel like I may be missing crucial information. Here's the outline:

I dissolve $\pu{6.0 g}$ of $\ce{HK2PO4}$ in $\pu{20 mL}$ of water. I then add $\pu{8.0 mL}$ of $\pu{4 M}$ $\ce{HCl}$ and dilute to $\pu{30.0 mL}$. I take $\pu{25.0 mL}$ of the resulting solution and adjust the pH to 3.5 with $\pu{1 M}$ acetic acid.

What volume of acetic acid is theoretically necessary to reach pH 3.5?

Experimentally, I attempted this and the initial pH of the $\pu{25.0 mL}$ aliquot was 5.71. Technically, I should have been able to reduce the pH to 3.5 using less than $\pu{15.0 mL}$ of $\pu{1 M}$ acetic acid. However, I observed buffering effects around pH 4.0 and the total volume required was $\sim\pu{50.0 mL}$ of $\pu{1 N}$ acetic acid.

In complex analytical chemistry problems like this, it's useful to consider a stoichiometric matrix of the reaction network.

                               | H2PO4- | H3O+ | H3PO4 | CH3COOH | HPO4-2 | PO4-3
------------------------------ | ------ | ---- | ----- | ------- | ------ | -----
1st phosporic aciddissociation | 1      | 1    | -1    | 0       | 0      | 0
2nd phosporic dissociation     | -1     | 1    | 0     | 0       | 1      | 0
3rd phosporic dissociation     | 0      | 1    | 0     | 0       | -1     | 1
Acetic acid ionization         | 0      | 1    | 0     | -1      | 0      | 0

Each score represent the stoichiometric coefficient of that species in the given reaction, with a negative sign when it's a reactant and a positive value otherwise. I am not using $\ce{KH2PO4}$ or $\ce{HCl}$ as species because they are always fully dissociated. It's better to think of them as $\ce{H2PO4-}$ and $\ce{H3O+}$ respectively in regards to their molar concentration.

We can approach the problem by consider a "mixing $\to$ reaction" two-step process. Because equilibrium is achieved regardless of the path chosen, we can do this freely, that is, consider the "initial" concentration as though we mixed all parts together with no reaction occurring.

The mixing step mass balance will be $C_{\textrm{initial}} \times V_{\textrm{pre-mixing}} / V_{\textrm{solution}}$. Here's the full table:

                               | Solvent | H2PO4-       | H3O+         | H3PO4 | CH3COOH      | HPO4-2 | PO4-3 | CH3COO- | Total
------------------------------ | ------- | ------------ | ------------ | ----- | ------------ | ------ | ----- | ------- | ------------
Initial Concentration (mol/L)  | 0       | 2,2044883383 | 4            | 0     | 1            | 0      | 0     | 0       |
Initial volume (L)             | 0,03    | 0,02         | 0,008        | 0     | 0,5614062187 | 0      | 0     | 0       | 0,6194062187
Solution concentration (mol/L) | 0       | 0,0711806976 | 0,0516623809 | 0     | 0,9063619346 | 0      | 0     | 0

Since we are basically solving a nonlinear problem, I took the liberty to assign a random value to the initial acetic acid volume, which we will later determine by minimizing an objective function.

Let's call the stoichimetric matrix $A$, and the solution concentration (post-mixing, pre-reaction) vector $C_{\textrm{post-mix}}$. Let's define the extent of reaction vector $\delta$ as the array of values representing the amount of mols (per unit volume) produced or consumed by each reaction.

Reaction                   | Extent (mol/L)
-------------------------- | --------------------
1st phosporic dissociation | 0,0034158968
2nd phosporic dissociation | 1,46105223300036E-05
3rd phosporic dissociation | -2,218252003941E-14
Acetic acid ionization     | -0,0547767774

Once again, we do not know these values a priori. They will be optimized later on. The amounts of each reactant that is produced or consumed ($C_{\textrm{reacted}}$) is then given by the matrix product

$C_{\textrm{reacted}}^T = A^T \delta$

The equilibrium concentration is then the post-mixing concentration plus the reacted concentration,

$C_{\mathrm{eq}} = C_{\text{post-mix}} + C_{\textrm{reacted}}$

Now consider the vector of potentials of the equilibrium constants fpr each reaction ($\mathrm pK = -\log(K)$),

Reaction                       | pK
------------------------------ | -------------
1st phosporic aciddissociation | 2,1611509093
2nd phosporic dissociation     | 7,2076083105
3rd phosporic dissociation     | 12,3187587626
Acetic acid ionization         | 4,7447274949

For the equilibrium step, equaling the equilibrium expression to the $K$ of each reaction is the same as imposing the following matrix product (think of why?)

$A (\mathrm pC_{\textrm{eq}})^T = \mathrm pK$

Where the vector $\mathrm pC$ is calculed by taking the negative log of the species concentrations vector $C_{\mathrm{eq}}$. One of these values is the $\mathrm pH$, which is given by the problem.

We are then left with the task of finding the initial acetic acid volume by solving:

Find $V_\ce{CH3COOH}$ and $\delta$ (remember, this is a vector) that minimize the residuals in $A (\mathrm pC_{\mathrm{eq}})^T = \mathrm pK$ and $\mathrm pC_{\mathrm{eq}} \Bigg|_{\text{species}=\ce{H3O+}} = (\mathrm pH)_{\text{given}}$.

I used libreoffice-nlpsolver to yield

$$V_{\ce{CH3COOH}} = 0.5614062187\ \mathrm L$$

But, wait, this is the volume of acetic acid we'd employ to react with the entire solution volume. But the problem asks about only $25\ \mathrm{ml}$, it is thus just a matter of scaling the proportion using the total volume ($V_\mathrm t = 0{,}6194062187$),

$$V = 0{,}5614062187 L \times \frac{0{,}025 L}{0{,}6194062187} = 0{,}022659048 L \ \mathrm L$$

Or about $22{,}7\ \mathrm{ml}$

I purposely did not take into account the self ionization of water, because of the low pH in question. But it could have been done nonetheless. The only changes would be the need to include $\ce{OH-}$ in the stoichiometric matrix and take $\mathrm pW$ into action.


Is it correct to say that “some degree of pollution is an inescapable consequence of the Second Law of Thermodynamics”? The quote above is from ‘Enlightenment Now’ by Steven Pinker, I was just wondering if this was true or if it misrepresented the Second Law.

The full quote below...

  ...some degree of pollution is an inescapable consequence of the Second Law of Thermodynamics. When people use energy to create a zone of structure in their bodies and homes, they must create entropy elsewhere in the environment in the form of waste, pollution, and other forms of disorder

(Page 123)

Yes, this is true. However, the question is where that increase in entropy occurs.

Since the industrial revolution, life on earth is a bit one-sided, with humans controlling the chemistry going on, and a lot of it. When we burn fossil fuels, there is an increase in entropy on earth. Before that, one organism's waste was some other organisms food, and the entropy increase occurred mostly in the sun (giant nuclear reaction slowly going towards equilibrium).

Humans (and all other animals) benefit from the $\ce{O2}$ "pollution" from cyanobacteria and plants. Plants benefit - to a degree - from the $\ce{CO2}$ "pollution" from animals. However, when global concentrations change too fast, some species might benefit a lot, but many other species will go extinct.

If the "plastic ocean" had built up over millions of years instead of 50, some microorganism that uses the polymers as its primary source of food might have evolved, and there would be a cycle instead of a build-up.


Regioselective enolate formation, choosing major and minor products This question came in my organic chem assignment. The options highlighted are the answers.
The end carbon(adjacent to the carbonyl group) should have more acidic alpha-hydrogen and the alkene formed is more substituted (as compared to the one in F). Still it is given as the minor product. Could someone explain why?

A major factor contributing to the stability of the enolate ion formed as the major product (F) is the stability associated with extended conjugation or extended delocalisation via resonance. Due to this, terming (F) as the major product seems justified.
If the product were to be a kinetically-controlled one (KCP), then the presence of three acidic alpha hydrogen atoms could've caused faster formation of product (E) ; it still would not have been the more stable product, but since the major product in KCP-type reactions is controlled more by the kinetics , rather than the thermodynamics, we would possibly expect to see (E) in a larger yield in that case.
Note: Strictly speaking, the problem should mention conditions that can help us to unambiguously determine which pathway (KCP vs. TCP) should be chosen. But a practical advice, garnered from the experience of working problems which seldom perfectly mention the reaction conditions, is that unless mentioned otherwise, it is always wise to first consider the thermodynamically stable product.


How can we tell the particle size of solutes in solutions? This is a general question in regard to solutes dissolving into solvents. I am interested in the reality of the particle size of the solute during dissolution.

For example, take $\ce{NaCl}$ and dissolve it into water. Are the salt particles, which are now invisible to the eye, truly dissolved? Meaning have they turned into single atom particles of $\ce{Na+}$ and $\ce{Cl-}$ clinging to $\ce{H2O}$ water structures?

Chemical equations seem to imply this. However, when you look at colloids, then particle size matters and can vary. Do ionic salts still have large clusters of their cations/anions upon dissolution?

I am really interested in this topic and would love someone to give me a good explanation.

I assure you there are many reasons to believe that the ions really are solvated individually. The first thing to look at is the relative strength of interactions between the ions in a lattice and the ions when in a solution. For water, these interactions must be very strong because a number of hydrogen bonds must be broken, the interaction between ion and water must overcome this initial deficit and the loss of some entropy. So, the interaction is surely quite strong.

It is of course possible for two ions which dissolve to get back together and interact with each other. They are oppositely charged ions after all. By looking at the relative size of equilibrium constants for these two processes, however, I think you'll be convinced it is far preferable for the ions to be solvated. For instance, a quick google search will tell you that the $K_{sp}$ of $\ce{NaCl}$ is around $36$. That's an astronomically large solubility product. This means we dump sodium chloride into water until the concentration of both ions is around $6\ M$.

Thus, the easiest experiment to do is to just saturate a solution of water with salt and write the mass of salt added in molarity units. This doesn't really answer your question though because we assumed that the salt existed as two ions and then made a correspondence between that concentration of ions and the concentration of water which each ion is occupying. Plus, we would generally be interested in more dilute solutions than this.

This is why people have used methods which probe the structure of ions in a solution directly. Namely, using x-ray diffraction and neutron scattering experiments, it is possible to determine the local environment around water (or something else solvating an ion) and calculate the hydration number. This is the number of ions, of a specific ion, solvated per molecule of solvent.

Here is a review which discusses x-ray diffraction and neutron scattering experiments and one which discusses coordination numbers of ions solvated in water.

Using these structural methods, which show it is extremely unlikely there are chunks floating around in solution, the following data can be found for hydration numbers of ions in water:
$$n_{\ce{Li+}}=4$$
$$n_{\ce{Na+}}=5$$
$$n_{\ce{K+}}=6$$

Here is also an old review which describes many simple but outdated methods for determining the solvation/hydration numbers of ions in a solvent.

Hopefully some of this is helpful and convincing. The experiments are quite definitive, but if you believe the equilibrium constants, then it is very hard to believe these lattices would remain intact in solution.


Magnetic moments of tetrahedral Cobalt (II) and (III) complexes From the spin-only formula we can predict that for tetrahedral cobalt (II) complexes $$m_{eff} = 3.87 \mu_B $$

This ignores orbital angular momentum effects, which result in higher magnetic moments for the tetrahedral halide complexes (Hund's 3rd rule results in spin-orbit coupling "together"). Why do the Co(II) cyanide and Co (III) alkyl complexes have lower magnetic moments than the spin only formula predict ? I have quoted values of 2.15 $\mu_B$ and ~3 $\mu_B$ for these cases.

The question isn't really clear exactly what complexes you are referring to.  However, cyanide is quite a high field ligand, so perhaps the Co(II) cyanide compound is low spin  while the 3.87 value is for high spin (3/2).  I don't see how it could be low spin if it is really tetrahedral, but isn't $\ce {Co(CN)_4^2-}$ square planar?

For Co(III) alkyl, alkyl ligands can also be relatively high field, and it is at least theoretically possible to have low spin (s=1) Co(III) tetrahedral complex. However, it is rare for tetrahedral complexes to be low spin because the ligand field splitting energy is lower than for octahedral.

If those are not the reasons, see Iron, Cobalt, and Nickel Complexes having Anomalous Magnetic Moments Quarterly Review Chemical Society volume 22, pp 457-498.


How come ∆G ≤ 0 is used for spontaneity in electrochemistry, not ∆G ≤ W(other)? The spontaneity condition for a system at constant temperature and pressure in which the only type of work accomplished is of the $pV$ type can be expressed as:

$$\Delta G\le0\tag1$$

In case of other types of work (such as electrical, friction, etc.) there is an additional term and the previous equation becomes:

$$\Delta G\le W_\text{other}\tag2$$

In case of an electrochemical process the maximum work is the electrical work and can be expressed as:

$$W_\text{other}=-nFE\tag3$$

Question: Why it is still said that the electrochemical reaction is spontaneous if the change in Gibbs free energy is lower than zero, thus using Equation $(1)$, even though the proper equation to be used is number $(2)$?

Most likely the source is wrong, or it has purposfully omitted the correct equation to pseudosimplify a problem. It might be that an electrochemical reaction was under discussion, not a cell. A further possibility is pursued in  the appendix. I recommend adding a source where you came upon the quoted exchange to avoid a straw man argument; even though in this case  I have observed (and possessed) the misconception first hand.
One definition of spontaneity
I define the true spontaneity condition as such when there is a tendency toward net 'something different' (chemical reaction, expansion etc) to establish state 2 instead of some original state 1. An example could be a chemical reaction where there is a higher reaction extent than at equilibrium, so a net reverse reaction takes place.
You are correct that the true spontaneity equation (from Clausius's inequality) in the thermodynamic limit is, for net-constant pressure and temperature in the system,
$$\operatorname{d} G-\delta w_\pu{other}\leq 0.\tag1\label1$$
When electrochemical work is the only component besides expansion work, this implies (due to $|\operatorname{d} n_\pu{e}|=|\nu_\pu{e}|\operatorname{d}\xi$)
$$\Delta_\pu{r} G_{T,P} +|\nu_\pu{e}|FE\leq0\tag2\label2$$
where $\Delta_\pu{r} = \partial/\partial \xi$; here $\xi$ is extent of the reaction. The Greek $|\nu_\pu{e}|$ signifies an absolute value of the stoichiometric coefficient of an electron in some half-reaction. Equation $\eqref2$ assumes that only one net reaction occurs. The term $|\nu_\pu{e}|FE$
should be a good indicator of electrical work.
It might also be that they are discussing an electrochemical reaction, not the cell itself. We can have the process
$$\ce{Zn(sln) + Cu^2+(sln) -> Cu(sln) + Zn^2+(sln)}\tag3\label3$$
without harnessing its electrical work via an external circuit. So $\operatorname{d} G\leq 0$ would hold for spontaneity.
Appendix: An alternative
A different definition of spontaneity might be in play. Namely, the term spontaneity is also used to mean a large enough standard equilibrium constant (especially in biochemistry).
$$RT\ln \frac1K = \Delta_\pu{r} G^\circ_{T,P} \tag4\label4$$
Equation $\eqref4$ is technically a definition of the standard equilibrium constant. When non-negative absolute temperatures are quaranteed, the LHS of equation $\eqref4$ will become non-positive for all $K$ big enough, i.e., $K\ge1$ . That implies
$$\Delta_\pu{r} G^\circ_{T,P} \leq 0. \tag5\label5$$
Note, however, that $K\ge1$ doesn't necessarily imply much about the reaction balance itself because the equilibrium constant comprises of activities (not concentrations), and because the stoichiometric coefficients in the denominator may drown out the coefficients in the numerator. (Equation $\eqref6$ assumes that surrounding fugacity is equal to an agreed fugacity in the standard state, denoted here and elsewhere by '$^\circ$'.)
$$K(\pu{in solution, solvent \ce{A}}) = \left[a(\ce{A})_\pu{eq}\right]^{\nu_\ce{A}}\prod_i \left[a(\ce{B_$i$})_\pu{eq}\right]^{\nu_i}. \tag6\label6$$
Reactants–products (other than solvent) are designated by $\ce{B_$i$}$. But still, for $K$ big enough (2nd definition of spontaneity), equation $\eqref5$ will hold by definition. Also keep in mind $\Delta_\pu{r} G^\circ_{T,P} \neq \Delta G$ (even their dimensions are different!). So the source is presumably still at fault for poor notation.


Ozone in ozone layer If ozone is a gas, and gases have the highest entropy, then how does the ozone gas stay within a few layers of the atmosphere, even though they span for kilometers?

It does not have much to do with entropy, rather with the way and place of ozone creation.

Stratospheric ozone is produced typically at altitude 20-30 km by UVC radiation with $\lambda \lt \pu{280 nm}$:

$$\ce{O2 + \nu -> 2 O}$$
$$\ce{O + O2 -> O3}$$

See ozone cycle as courtesy of @M. Farooq.

Ozone in lower troposphere troposphere is created by UVB (280-320 nm)  (+UVA >320 nm ??) mostly due catalytic effects of nitrogen oxides that come typically of the oxidative smog of the Los Angeles type:

$$\ce{NO2 + \nu -> NO2^{*}}$$
$$\ce{NO2^{*} + O2  -> NO + O3}$$
$$\ce{2 NO + O2 -> 2 NO2}$$


Why doesn't velocity enter into Heisenberg's Uncertainty Principle? So I came across Heisenberg's uncertainty principle in my 11th grade book, and I'm not convinced. It says $\Delta x \Delta p \gt \frac{h}{4\pi}$. Okay, but how does velocity even matter to determine the path?

Now suppose we have, for the case of my argument, 100 exactly identical atoms of hydrogen. Now we find the positions of the lone electron in 100 different cases, and we can do it with 100% accuracy. Now, if we join the dots, won't we get the path of the electron? Why not? Please explain in simple words if you can.

Some problems with your method: you mention taking 100 copies of the same atom of hydrogen and finding the positions at 100 different times. (i) You actually can't make a copy of a quantum system without knowing its quantum state (the no-cloning theorem), (ii) your measurements of position will disturb the "trajectory" of the electron, so your 100 cases are distinct and cannot represent the path of a single electron, and this is especially true the more accurate you make your measurements.

You should realize, however, that the bound placed by the Heisenberg uncertainty principle is really quite small, and on a macroscopic scale we can essentially determine the trajectory of an electron. A bubble chamber or cloud chamber is essentially a major revision of your proposed method: we fill a vessel with supersaturated vapor, wait for charged ions to pass through the vapor and cause it to condense, and observe the resulting tracks. This describes the ion's motion with a resolution on the order of microns.


Will the electrolyte surface become charged if one electrode is raised out of solution? What would the potential distribution be if I were to immerse a single inert (graphite) electrode in an aqueous solution with good ionic conductivity and hold it at a high (ca. $\pu{24V}$) potential vs. an electrode that would remain in the air above the solution's surface?

I don't expect anything to happen at the immersed electrode interface because the solution should remain equipotential.

But, what about the air-water interface? Would the electric field induce a surface charge?

But, what about the air-water interface? Would the electric field induce a surface charge?

Yes.  Basically, by locating one of your electrodes out of the solution, you've assembled an air-gap capacitor, though a fairly inefficient/low-quality one.

Per that above link, typical metal-plate "[a]ir capacitors have a small capacitance which usually lies between $\pu{100pF}$ and $\pu{1nF}$". High-level, the definition of capacitance is:

$$
C = {q\over V} \quad\quad\Longrightarrow\quad\quad q = CV
$$

Assuming that this water/electrode air capacitor has similar properties to the above-referenced metal-plate air-gap capacitor, the charge accumulated at the electrolyte surface under a $\pu{24V}$ applied potential should be between $2.4$ and $\pu{24nC}$. A similar amount of charge, but of opposite sign, should of course accumulate at the surface of the elevated electrode.

  I don't expect anything to happen at the immersed electrode interface because the solution should remain equipotential.

This will be true at steady-state, as the entirety of the potential drop is going to be supported by the air gap.

However, you will have some minuscule amount of transient electrochemical activity at your submerged electrode right when you turn on your power supply, as the system adjusts from the $\pu{0V}$ to the $\pu{24V}$ applied voltage.


Haemoglobin literature spectra does not match I am studying haemoglobin absorption spectra in its oxygenated and deoxygenated state and I compiled values from different sources including:

A PhD thesis by van Assendelft [1], using extinction coefficients in (mmol/L)-1/cm

A book compiling almost one century of research on the topic by the same laboratory that helped standardizing the haemoglobin cyanide standard [2]. Extinction coefficients in the same unit as [1], numerical values are very close.

Coefficients compiled by Scott Prahl[3]. The unit is the same but there is a factor 3.7 up to 4.0 between his data and that of Assendelft and Zijlstra.

Coefficients measured by Kolyva[4].

All the published coefficients are defined as base 10 extinction coefficient. That is if I
is the light intensity, they are expressed in Beer-Lambert law as:
$I = I_0 \cdot 10^{-c\cdot \varepsilon \cdot l}$
Where $l$ is the light path length (cm), $c$ the concentration in absorbing compound (mol/L) and $\varepsilon$ the molar extinction coefficient (L/mol/cm).
How comes there is a 3.7-4.0 factor between Prahl and Kolyva data on the one hand and Assendelft and Zijlstra on the other hand?
At first I thought of some $log(10)\approx2.3$ difference, as if some of the author forgot to precise that they used base $e$ extinction coefficient at some point but it does not fit. If anyone has an idea on the topic, the help would be most appreciated.

Further information
The cyanmethaemoglobin method is often used for the determination of the haemoglobin concentration of a solution. In particular, its absorbance at 540nm is known to be 11.00L/mmol/cm[5] which is in agreement with Assendelft and Zijlstra published data.
References

O.W. Assendelft, Spectrophotometry of Haemoglobin Derivatives, 1970. ISBN: 902320560X.
W.G. Zijlstra, A. Buursma and O.W. van Assendelft, Visible and Near Infrared Absorption Spectra of Human and Animal Haemoglobin, 2000. ISBN: 9067643173.
Scott Prahl, Optical Absorption of Hemoglobin, 1999, https://omlc.org/spectra/hemoglobin/
C. Kolyva, Systematic investigation of changes in oxidized cerebral cytochrome c oxidase concentration during frontal lobe activation in healthy adults, 2012. DOI: 10.1364/BOE.3.002550. Coefficients are in MEDIA 1.
F. Meng, Determination of extinction coefficients of human hemoglobin in various redox states, 2017. DOI: 10.1016/j.ab.2017.01.002. Table 1.

As stated by Scott Prahl :

  One of the most confusing things about looking at hemoglobin (Hb)
  spectra is that the values are typically tabulated in equivalents. The
  term equivalent is used to indicate the amount of hemoglobin which
  contains 1 gm atom of Fe and combines with 1 gm molecule of O2 or CO.
  One equivalent of hemoglobin is assumed to be 64,500/4 or 16,125 gm. A
  concentration of 10$^{-6}$ equivalent is 16.125mg of hemoglobin per cc.

It seems that Assendelft and Zijlstra used the equivalent quantification of haemoglobin, meaning that all their extinction coefficients should be multiplied by 4. This leads to the following figure:

Sorry for not having read again Scott Prahl's description before posting. I read it several months ago, as I was gathering data from different source and forgotten it since...


Mechanism of the reaction of cyclohexanone and nitric acid to form Adipic acid Can I get some help in figuring out the mechanism for this reaction.
Adipic acid is a molecule with two carboxylic groups,but I can't see how that happens. My initial thoughts involved that the carbonyl group in cyclohexanone is protonated,but I couldn't get any further as I couldn't figure out what would happen after that.

It think this reaction goes via following mechanism.


Making carbon powder using fireplace and weak acid I am doing a little home experiment trying to make as pure carbon powder as I can. I plan to then used it to eat up oxygen from a container where I want to try to melt aluminum.
My process was following:

I have put some sawdust in an iron can.
I covered the opening of the can with clay, leaving only a small opening.
I put the can in a fireplace and let it heat to about $\pu{800 ^\circ C}$.
After removing, I ended up with about 1/3 volume of black powder.
I washed the powder with water and filtered the water out.
I placed the wet powder in a beaker and added more water.
I added about 12% hydrochloric acid to dissolve calcium and metal oxides and hydroxides.
The solution started bubbling slightly, so I left it for several hours with occasional mixing.
I washed the resultant powder with water twice and left it to dry.

Did I manage to purify the carbon in the powder somewhat? Did the use of acid bring any benefit?

The result of your first pyrolysis must be charcoal plus the mineral constituants that are present in wood ashes. Now the wood ashes are usually a mixture of : $40 - 70$ % $\ce{CaCO3}$, $5 - 10$ % $\ce{Na2CO3}$,  $5 - 10$ % $\ce{K2CO3}$, $2 - 3$ % $\ce{K2SO4}$, $2 - 5$ % $\ce{Ca3(PO4)2}$, $5 - 10$ % $\ce{MgCO3}$, $2 - 10$ % $\ce{SiO2}$, < $1$ % $\ce{FePO4}$. Washing your charcoal with water and hydrochloric acid must remove all these substances, except $\ce{SiO2}$. This impurity cannot be removed.


How to select the parent chain in mechlorethamine? Here is the structure of mechlorethamine:

Although the longest carbon chain is ethyl, why does Wikipedia name the compound as bis(2-chloroethyl)methylamine (or N,N-di(2-chloroethyl)methanamine)?

It seems that that is the preferred name, rather than 2-chloro-N-(2-chloroethyl)-N-methylethanamine which does give a result on ChemSpider.

I understand that the longest carbon chain does not necessarily yield the parent chain, as with cyclic compounds (where the chain lies within the ring) and compounds with multiple bonds (where the longest chain with multiple bonds is preferred over a longer one with no multiple bonds). I am guessing that perhaps the rule of more substituents is used to determine the parent chain here, but the name on Wikipedia seems to have two substituents compared to 2-chloro-N-(2-chloroethyl)-N-methylethanamine which has three. What is happening here?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), secondary and tertiary amines can be systematically named in various ways similar to the methods for primary amines:

  P-62.2.2 Secondary and tertiary amines

  P-62.2.2.1 Symmetrical and unsymmetrical secondary and tertiary amines are named only by the same methods described in P-62.2.1.2.

  (1) substitutively using the retained name ‘aniline’ or the suffix ‘amine’ and the name of a parent hydride with further N-substitution;

  (2) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘azane’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.

  (3) substitutively, by prefixing, in alphabetical order when required, the name(s) of the substituent group(s) R, R' or R" to the parent hydride name ‘amine’. In order to avoid ambiguity, the second prefix in a secondary amine, and the second and the third prefixes in a tertiary amines must be enclosed in parentheses when these prefixes denote simple substituents.

  Method (1) generates preferred IUPAC names. Functional parent names like diethylamine and triethylamine are deprecated. The prefixes in names of such secondary and tertiary amines formed by method (2) are set off by parentheses to distinguish them from these deprecated names.

Therefore, the preferred IUPAC name for the tertiary amine given in the question is generated according to method (1) using the suffix ‘amine’, the name of a parent hydride, and further N-substitution. (Other methods may be used in general nomenclature.)

Concerning the choice of the principal chain, the rules for amines state:

  P-62.2.2.2 Selection of the principal chain or senior ring system in secondary and tertiary amines.

  Names of amines formed substitutively by using the retained name aniline or the suffix ‘amine’ are based on a principal chain and a senior ring system (see P-44.1). When a choice for parent hydride is possible between a ring and a chain, the ring is preferred. In names using ‘amine’ as a parent hydride, substituent groups expressed as prefixes are cited in alphanumerical order; the prefix(es) immediately preceding the term ‘amine’ is (are) enclosed in parentheses.

Thus, the rules essentially refer to the usual seniority order for parent structures that is stipulated in P-44.1 and the following subsections. The relevant criteria for the compound given in the question are

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (…) and the seniority of suffixes (…).

and

  P-44.3.2 The principal chain has the greater number of skeletal atoms (…)

In case of the compound given in the question, both methane (methylamine) and ethane (ethylamine) are parent structures that have the maximum number of substitutents corrsponding to the principal characteristic group (i.e. the suffix ‘amine’) since the compound contains only one amine group. The principal chain for the compound given in the question is ethane rather than methane since it has the greater number of skeletal atoms (i.e. the longest chain).

In accordance with the above-mentioned method (1), using the name of the parent hydride ‘ethane’ and the suffix ‘amine’ with the appropriate locant yields ‘ethane-1-amine’. The chloro substitutent is placed at atom ‘2’ of the parent hydride ‘ethane’, which leads to
‘2-chloroethan-1-amine’. Further N-substitution with the substituents ‘methyl’ and ‘2-chloroethyl’ results in the preferred IUPAC name
2-chloro-N-(2-chloroethyl)-N-methylethan-1-amine.

Note that the locant ‘1’ is omitted in ‘ethanamine’; however all locants must be cited in ‘2-chloroethan-1-amine’ and ‘2-chloro-N-(2-chloroethyl)-N-methylethan-1-amine’.

The alternative name bis(2-chloroethyl)(methyl)amine can be generated according to method (3). This is not the preferred IUPAC name, but it is a correct systematic name that may be used in general nomenclature.

Note that, according to method (3), the parent hydride in this case is ‘amine’ (i.e. neither ethane nor methane). The names of the substituent group(s) R, R' or R" are added as prefixes in alphabetical order, for example: ‘ethyl(methyl)(propyl)amine’. The multiplicative prefixes ‘di’ or ‘tri’ are added as usual, for example: ‘diethyl(methyl)amine’. However, the multiplicative prefixes ‘bis’ and ‘tris’ are used for substituted substituents (such as ‘2-chloroethyl’) and other compound or complex entities, which results in the name bis(2-chloroethyl)(methyl)amine for the compound given in the question.

Note the correct use of parentheses according to Subsections P-62.2.2.1 (3) and P-62.2.2.2.

The name N,N-di(2-chloroethyl)methanamine, which has been proposed in the question, is simply wrong. It violates multiple nomenclature rules.


Hypervalency of elements I know that elements of the 3rd period and beyond can use their empty 3-d oribtals for bonding. But why is there a limit on their covalency?
eg. Al has 5 3d orbitals. And it has an empty 3p orbital. Which means that it's maximum covalency should be 9. But it is 6. Why is it so? Is it because of the small size of Al leading to increased repulsions? Also, why do the elements of the 3rd period use 3d before 4s for hypervalency? 4s has less energy so shouldn't it be used first?

First off, as suggested in the comment above, the contribution of 3d orbitals in hypervalency is probably minimal.

It's clear from sampling the inorganic literature that "beyond the octet," the size of the atom matters, as well as the shape of the ligands, size of the surrounding coordination sphere, etc. To some degree, it's a matter of fitting X atoms around the central atom.

If you have a 5th or 6th row element (2nd or 3rd row transition metal), lanthanide, actinide, etc., higher coordination is relatively common. For example, for lanthanide ions, 8-coordinate is the most common coordination number. Still, 8-coordinate iron complexes have been found, e.g. "Combined Mössbauer Spectral and Density Functional Study of an Eight-Coordinate Iron(II) Complex" Inorg. Chem., 2015, 54 (17), pp 8415–8422 or "Square-Antiprismatic Eight-Coordinate Complexes of Divalent First-Row Transition Metal Cations: A Density Functional Theory Exploration of the Electronic–Structural Landscape" Inorg. Chem., 2015, 54 (4), pp 1375–1383.

I'm not aware of a general overview of hypervalency and "what coordination numbers can you get", but there's a nice overview on modern descriptions using multi-center bonding.


Introductory chemistry book with calculus prerequisite I understand that the policy here is that book requests are allowed, as long there is an effort to narrow it down. This question is similar to Can someone recommend me a book ( preferably written long ago) that includes all the details of high school chemistry?.

I am trying to find a book which tries to approach chemistry from the perspective of someone who has a higher-than-high-school-math background, i.e., they're proficient in calculus, but they aren't familiar with chemistry. Does any such book exist?

Reason why I'm asking for this is that mathematics undergraduate students, in my experience, get very little experience applying their problems outside of mathematical theory. Chemistry, in my years of tutoring and teaching, gets virtually no mention in calculus courses unless a math professor decides to discuss balancing equations using techniques in linear algebra. It has been years since I've seen high-school AP chemistry, but I recall rate laws being a consequence of integration, although this was never actually shown in AP chemistry. I made an effort, when I was a Calc. I TA, to show a class of mostly chemistry majors how the first-order rate law is derived from a differential equation.

Most introductory physical chemistry courses and books don't rely very heavily on prior chemistry knowledge, and there is quite some math to do there. I think you can start most subjects there independently from scratch (QM, thermodynamics, kinetics).

To better prepare us for quantum mechanics, my (chemistry) faculty offered a two-semester lecture for "freshmen" where the professor derived the Schroedinger equation with us from first principles. The lecture was basically self-contained, both mathematically (introducing analysis) and chemically (first-years do basic chemistry and qualitative inorganic analytics, there is hardly any overlap with QM).

With students that are better in math than we were, this could work very well, and faster. :-)

I'm afraid I have no english book for this. The regular physical chemistry books don't dive into the actual mathematical derivation so much. We used "Wedler, Lehrbuch der physikalische Chemie".

Btw.: Students that are good in math AND chemistry are very valuable. Be prepared for your colleagues from the chemistry department to be after them. ;-)


Triflate use in metathesis reactions Why are triflates so widely used in coupling reactions, but not tosylates, mesylates nor nosylates?

As a result of the -I effect of the trifluoromethyl group, triflate is a better nucleofuge than mesylate and tosylate.
As compared to tosylate and nosylate, triflate is rather small. This might be an advantage in reactions at "crowded" centres.
Triflates do not undergo side reactions, such as hydrogen abstraction reactions, etc. A triflate, once introduced, can easily survive several steps in a synthesis.


Nitric acid decomposition I have a problem with the decomposition reaction of Nitric acid($\ce{HNO3}$) by light.
My problem: is that reaction is a redox reaction?
If it is please tell me the oxidation and reduction (balanced) reactions seperately.

Generally, nitric acid is a strong oxidizing agent which get reduce to nitrogen oxides on reacting with metals which act as reducing agent thus forming a redox reaction. But in this case, only nitric acid is involved which get reduced to nitrogen oxides on reaction with air/light. So, this may be called oxidation reaction where nitric acid gets oxidized. From wikipedia article of nitric acid:

  Nitric acid is subject to thermal or light decomposition and for this reason it was often stored in brown glass bottles:

  $$\ce{ 4 HN^{V}O3 → 2 H2O + 4 N^{IV}O2 + O2}$$

  This reaction may give rise to some non-negligible variations in the
  vapor pressure above the liquid because the nitrogen oxides produced
  dissolve partly or completely in the acid.

  The nitrogen dioxide ($\ce{NO2}$) remains dissolved in the nitric acid
  coloring it yellow or even red at higher temperatures. While the pure
  acid tends to give off white fumes when exposed to air, acid with
  dissolved nitrogen dioxide gives off reddish-brown vapors, leading to
  the common name "red fuming acid" or "fuming nitric acid" – the most
  concentrated form of nitric acid at Standard Temperature and
  Pressure (STP). Nitrogen oxides ($\ce{NO_x}$) are soluble in nitric
  acid.

There are some old papers(here and here) which discuss the mechanism of nitric acid decomposition. Give a quick read on the papers and you will understand the mechanism of the reaction.


Understanding the mechanism of Blanc chloromethylation I have seen this question: Mechanism for chloromethylation of benzene.

The answer there uses a picture from wikipedia which confuses me. It gives $$\ce{H2C=O <=>C[HCl] H2C+-OH <-> H2C=O+H}$$ and the benzene attacking the $\ce{C}$ from $\ce{H2C=O+H}$, not the $\ce{C}$ from $\ce{H2C+-OH}$ as I would have expected.

Is it only a quirk in the picture or do I have some fundamental misconceptions?

The result is the same no matter which resonance form you use. Resonance forms are only formal representations, extreme forms of the true structure which is somewhere in between.

Note that to get from the blue resonance form to the red resonance form, you'd need to push an arrow:

So there is no real difference which resonance form you use - in the end you would draw exactly the same arrows. In the blue reaction, you draw two arrows. In the red reaction, you only draw one arrow. But, to get from the blue resonance form to the red resonance form, you have to draw one arrow anyway.

Don't read too much into the number of arrows or whatever - the only point I am trying to make is that from a mechanism-drawing perspective, it does not matter which resonance form you use, because all roads lead to the same product anyway. All those arrows are just formal bookkeeping.


Why is the cyclic voltammetry of my ferrocene compound not showing a standard shape? I am trying to set-up a new cyclic voltammetry measurement station in my lab but the results are very poor. I am using the following parameters:

working electrode: platinum wire, 0.25mm diameter
counter electrode: platinum wire, 0.25mm diameter
reference electrode: none
solvent: acetonitrile
electrolyte: ferrocenecarboxaldehyde
support electrolyte: tetrabutylammonium hexafluorophosphate
scan rate: ~600mV/s

Platinum wires each have a submerged length of about 2mm to 5mm and are separated by about 2cm. I was previously using tetrabutylammonium iodide, but that is probably not a good support because the iodide ions will become iodine and forming a blocking layer on my electrode?

I also tried using graphite rods as the electrodes but obtained similar poor-looking CV shapes.

What should I change to obtain the standard CV shape of an outer-sphere redox species?

Without a proper reference electrode, knowing what potential your ferrocene/ferrocenium couple will show up is quite impossible. You should sweep a larger potential window, until the solvent breakdown potential on either end.

As Michael Dryden suggested, you should sweep slower as well. That will help with the signal getting resolved easier.


Hydrogen atom transfer between enantiomers of rac-1-phenylethanol (Taken from Dalton Trans., 2012, 41, 13423–13428. DOI:10.1039/c2dt31781h)

I have difficulties understanding the hydrogen transfer in the image above. I always thought that hydrogen transfers followed the following formula:

$$\ce{X^. + H-Y -> X-H + Y^.}$$

I can't really see the connection between the picture and the formula. Is it the whole ethanol group that is transferred?

Did you continue reading the paper? :) The source is important and is what was needed to answer the question!

I'm just going to quote from it.

  In DKR [dynamic kinetic resolution] a chemical catalyst is added in order to racemize the enantiomers, ensuring that all the substrate can be utilized. [...] Highly successful catalysts are often Ru(II) complexes [...] The metal complex catalyses racemization by a hydrogen transfer mechanism (Scheme 2).

So, the supposed hydrogen transfer occurs via coordination to a metal, followed by β-hydride elimination. Readdition of the metal hydride across the opposite face of the resultant ketone (acetophenone in this case) produces the other enantiomer.


How to name molecules with a hydroxyl group in a side chain? TL;DR: Are there clear nomenclature rules for naming the below molecule?

Today in school we learned about the nomenclature of alcoholes. Unfortunately, it was not mentioned what to do with hydroxyl groups in side chains.
With the rules we learned I would guess the name of this molecule is something like 2-(methylol)-propane-1,3-diol. But I could not find any specific rule telling me how to name hydroxy groups in side chains. 1

A free molecule builder with integrated naming I found names the molecule as follows:

  2-(hydroxymethyl)propane-1,3-diol

But I'm still not sure if this is the right way to do it.

Are there any precise IUPAC nomenclature rules that say that you use "-ol" in the main chain and "hydroxy" in side chains? 2

1 I did some research with no solutions: our schoolbook, some websites, Chemistry Stack Exchange...

2 I mean, isn't this IUPAC's aim? To provide (relatively) clear nomenclature rules?

The proposed name 2-(hydroxymethyl)propane-1,3-diol is correct.

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) includes three different methods for systematic names of alcohols. The preferred names are formed using the suffix ‘ol’ and (if necessary) the prefix ‘hydroxy’.

  P-63.1.2 Systematic names of alcohols, phenols, enols, and ynols

  Hydroxy compounds are named in three ways:

  (1) substitutively, using the suffix ‘ol’ and the prefix ‘hydroxy’. The presence of several ‘ol’ characteristic groups is denoted by the numerical multiplying prefixes ‘di’, ‘tri’, etc.; the final letter ‘a’ in a multiplying prefix is elided before the suffix ‘ol’. Rule P-44 is applied when a principal chain or a senior ring system must be chosen. When there is a choice for numbering, the starting point and the direction of numbering of a compound are chosen so as to give lowest locants to the ‘ol’ suffixes;

  (2) by functional class nomenclature and the class term ‘alcohol’;

  (3) as assemblies of identical units by multiplicative nomenclature when the conditions for its use are fulfilled (see P-51.3)

  Method (1) generates preferred IUPAC names. Names as assemblies of identical units, method (3), are preferred to those that are formed by simple substitution (P-51.1.5).

The first step is the selection of the principal chain, which will serve as the parent structure. The relevant general criteria for the choice of a principal chain in such compounds are:

greater number of groups that are expressed as suffix
longest chain
lower locants for groups that are expressed as suffix

  P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES

  When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.

  P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

  (…)

  P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].

  (…)

  P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.

  The senior ring, ring system, or principal chain:

  (…)

  (h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);

  (…)

For the compound that is given in the question, the principal characteristic group is $\ce{-OH}$, which is expressed as suffix ‘ol’. Whichever way you look at it, the principal chain can only have two of these groups. Therefore, the principal chain is a propane; the parent structure including the substituents corresponding to the principal characteristic group is propane-1,3-diol.

The compound that is given in the question actually has a branched chain structure. Therefore, the next step is the identification and naming of side chains, which will be treated as prefixes. In this case, the resulting name is 2-methylpropane-1,3-diol.

The third $\ce{-OH}$ group cannot be expressed as suffix ‘ol’ since it is not situated on the principal chain. Therefore, it has to be indicated by the prefix ‘hydroxy’.

  P-63.1.4 Substitutive nomenclature, prefix mode

  Hydroxy groups are indicated by the prefix ‘hydroxy’ when:

  (1) a group having priority for citation as the principal characteristic group is present; or

  (2) a hydroxy group cannot be denoted by a suffix.

In this case, the hydroxy group is placed on the methyl side chain, which results in the compound substituent group ‘hydroxymethyl’.

  P-29.4 COMPOUND SUBSTITUENT GROUPS

  P-29.4.1 Compound substituted substituent groups

  A compound substituted substituent group is formed by substituting one or more simple substituents into another simple substituent that is considered as the principal chain. The choice of the principal chain is fully discussed in Section P-44.3. The first criterion to be applied is that the principal chain is the longest chain; it will be applied in the following examples of acyclic compound substituent groups.

  (…)

Parentheses are used around such compound prefixes in order to avoid ambiguity.

  P-16.5.1 Parentheses (also called curves or round brackets)

  P-16.5.1.1 Parentheses are used around compound (see P-29.1.2) and complex (see P-29.1.3) prefixes; after the multiplicative prefixes ‘bis’, ‘tris’, etc.; to enclose a multiplied parent compound consisting of a parent hydride and a substituent suffix or a multiplied substituent prefix even though preceded by a basic numerical prefix, such as ‘di’, ‘tri’, etc.

Therefore, the complete systematic name for the compound that is given in the question is 2-(hydroxymethyl)propane-1,3-diol.


Why does the cathode have to be made out Cu in a Zn-Cu voltaic cell? The classic voltaic cell has a $\ce{Zn}$ anode and $\ce{Cu}$ cathode. The reduction half reaction occurring at the cathode combines aqueous copper ions and electrons supplied by the anode to form solid $\ce{Cu}$. So what is the purpose of the $\ce{Cu}$ cathode? $\ce{Cu}$ is produced anyways from the electrons supplied by the anode. Why can't we have instead of the $\ce{Cu}$ cathode, simply the wire connecting to the anode which will supply electrons to fuel the reduction half reaction?

You could certain have a zinc-copper cell in which copper was not the cathode material.  Graphite or silver or whatever could be the electrode material instead.   But once you close the circuit:

Current flows
Cu is deposited as a metal solid on the electrode surface, whatever it is made out of.
Copper metal is, as it always is, conductive.
As a result of steps 2 and 3, there is now a copper metal cathode instead of a cathode made of another material.

So even if you start without a copper cathode you wind up with one.


Computational chemistry software to generate files of cartesian coordinates and zmatrix of molecules Is there any software that I can provide bond lengths, bond angles and torsion dihedrals of a molecule and generate an output file in the format of cartesian coordinates or zmatrix?
Example:
I would have input similar to the one below (which i provide all bond lengths, bond angles and torsion dihedrals of a molecule)
BOND LENGTHS
     2 1 Si O2 1.63
     3 1 Si O2 1.63
     ...
BOND ANGLES
     2 1 3 Si O2 Si 148,482
     1 2 4 O2 Si C3 107,426
     ...
TORSION ANGLES
     3 1 2 4 -161,552
     3 1 2 8 -41,721
     ...

And I would like to have an output like this:

My idea is to vary some of this information and generate several Gaussian software inputs to calculate energy points at specific bond lengths, for example, without having to manually edit the molecule structure in this software.
I know that Open Babel software can do something like that, but I don't know what type of input to use.

This is not the answer to your question, but it is a comment about an approach to the aims, based on the comments. If coding is an option, RDKit is a good option, specifically adding constraints (restraints) to the ForceField and optimising. Given a molecule:
from rdkit import Chem
from rdkit.Chem import AllChem

acetate = Chem.MolFromSmiles('CC(=O)C')

Normally the 3D conformer is generated with:
AllChem.EmbedMolecule(acetate)

And it stops there. However, then you can manually control the force field —either UFF (more elements) or MMFF (classic choice).
ff = AllChem.UFFGetMoleculeForceField(mol)
ff.Initialize()
print(ff.Minimize()) # this tells you if it failed.

Within this, you can add constraints, such as:
ff.AddDistanceConstraint(0, 1, 2., 2.1, 1_000_000_000)

Note that the weight needs to be set absurdly high for absurd lengths. To check it work you could do:
dm = Chem.Get3DDistanceMatrix(acetate)
print(f'{dm[0,1]} Å. {ff.CalcEnergy()} kcal/mol')


Are homemade batteries a practical alternative during a blackout? If your home lost power without warning; are homemade batteries (such as the lemon/potato variety described in this question Are homemade batteries economically practical?) a practical alternative for powering your appliances while you wait for the electrician or what have you?

Are there other, better, non-produce-dependent methods for producing use-able electricity from household materials and minimal expertise?

... are homemade batteries ... a practical alternative for powering your appliances ...?

No, those type of batteries don't have enough energy to power more than a small light bulb or a transistor radio. If fact the light bulb probably won't stay lit for more than 10 minutes. An LED would be much better.

You need at least 30 car batteries to power your home for a significant amount of time; this is why most people going off the grid have a significant infrastructure cost.

  Are there other, better, non-produce-dependent methods for producing use-able electricity from household materials and minimal expertise?

Other than the alkaline batteries you buy in stores, not that I can think of.


Use the given information to write the formula of this compound the question says :

Two elements, A and B, combine to form two binary compounds. In the first
compound, 3.5 g of A combines with 8.00 g of B. In the second compound,
5.0 g of A combines with 17.1 g of B. If the formula of the first compound is
AB2, then the formula of the second compound would be ?

I can't even make the first step :(

For a molecule AB2, 3.5g of A represents one molar fraction, and 8.00g B represents 2 molar fractions (or 4.0+4.0). Therefore, a direct ratio can be given as 3.5:4.0, or 1:1.14.  This means a molecule AnBm will give a mass ratio for A:B of n:1.14xm

For a molecule AB, for every 1g of A, you will have 1.14g of B.

For a molecule AB2, for every 1g of A, you will have 2.28g of B.

For a molecule A2B3, for every 1g of A, you will have (1.14x3/2) 1.71g of B.

etc etc

So for 5.0g of A, in order to have 17.1g of B, you will have 1.14 x (m/n) = 17.1/5

=> m/n = 17.1 / 5 / 1.14 = 3

This means m/n=3, or AB3 (alternatively it could also be A2B6)


Determining the mass of a substance using calibration curve produced by HPLC I hope everyone is doing well and safe! Yesterday, I was fortunate enough to be taught about how to read a chromatography calibration curve through this stackexchange! Special thanks to those who explained to me patiently and meticulously. Anyways, I've tried to solve some questions regarding this field but ended up no-where with this one. Could someone run me through the procedure of how to solve this question? Huge thanks in advance!!! <3 <3

Info: A $\pu{1.0mL}$ aliquot of a drink was diluted to $\pu{50.0mL}$ with deionized water. A sample of the drink was run through the HPLC to produce the calibration curve below. In this experiment, the mass of caffeine in the drink was determined by HPLC. The peak area obtained for this diluted sample was $2800$ arbitrary units.

Question: What is the mass of caffeine in grams, in a $\pu{330mL}$ can of this drink?

Answer = $\pu{825mg}$
P.S.: Could someone explain to me the significance of the peak area ($2800$ arbitrary units) that is obtained? And thank you for the constant support and help. Much love to this community!!!

Warning: This is also a homework question. However, this is more like mathematics (geometry) than chemistry, I decided to help you. Nonetheless, please keep in mind that we won't help you in all future perspectives, if you do not show your effort to solve the problem, regardless more mathematics involved.
As  M. Farooq pointed out that this problem can be solved using the answers to your previous question, specially my old-fashioned approach. Here is how you could do it. Analyze your graph thoroughly to find two different points of graph where it perfectly crossed the cross-lines:

For me, these two points are: $(35,2000)$ and $(60,3500)$. Therefore the tangent of the slope ($\alpha$) of line is:
$$\tan \alpha = \frac{3500-2000}{60-35} = \frac{1500}{25}=60$$
Now, take another arbitrary point in the graph. Say it is $(x,y)$. Then:
$$\tan \alpha = \frac{y-2000}{x-35} = 60 \ \Rightarrow \ y = 60x - 100 \tag1$$
The equation $(1)$ is the straight-line equation of the graph.
When $y = 2800$, $x= \frac{2800+100}{60} = \pu{48.33 mg/L}$. This is for the diluted sample. I think it is fair to leave you some work to do. So you can back calculate to get your answer. When I did, it came close to $\approx \pu{798 mg/can}$.
Note: The error ($3.27\%$) is due to the finding of the slope. Also there are few flaws in the question. The calibration curve is made with HPLC traces of known standard caffeine solutions (not using diluted samples of the drink). The peak area of $2800$ arbitrary units was obtained by the injection of diluted sample of the drink ($1$ to $\pu{50 mL}$ dilution).