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Synthesis Golf XI: Hyperielliptone HA
It has been a little over a year since the last synthesis-golf was posted. I have talked to orthocresol and would like to help revive the tag!
A full FAQ post has been written on chemistry.meta.SE, explaining the premise of synthesis golf and the 'rules'. Please take a look at this before answering (if you haven't already).

Hyperielliptone HA is a phloroglucinol extracted by Wu et al1 in 2008. The compound is of particular interest because of its highly unique structure and demonstration of  DNA oxidation inhibition.

Rules:

Your synthesis must involve a method for the preparation of the phloroglucinol core.
Your synthesis must be stereoselective; however, the stereocenter attached to the phloroglucinol core can be racemic. Bonus points if you can develop a stereoselective synthesis for this moiety!
All reagents must be available from Sigma Aldrich.
If you do not have time to complete the entire synthesis but have a clever idea for a specific moiety, feel free to share your answer.

References

Wu, C.; Yen, M.; Yang, S.; Lin, C. Phloroglucinols with Antioxidant Activity and Xanthonolignoids from the Heartwood of Hypericum geminiflorum. J. Nat. Prod. (2008) 71 (6), 1027-1031. DOI:10.1021/np8001145.

HINT: This compound exist in two tautomeric forms. This might make conceptualizing this molecule a bit easier -- it's good to tackle a synthesis by looking at the molecule from multiple angles!

If you have not yet thought about it; maybe you would like to submit a synthesis.

I'm sure there are a lot of issues  with my attempt I haven't spotted (or maybe some of the ones I have are in fact lethal), but I saw the question and thought I'd give it a go.
Retro:
Went straight for the Diels-Alder with something resembling Danishefsky's diene. Disconnection to reveal the species I named A and B comes from a Baylis-Hillman reaction, which works reliably with α,β-unsaturated esters (unsure whether the attached enol-looking oxygen would have anything to say about this). Asymmetry in A is provided by a Sharpless epoxidation, and in B by a bit of a longshot centred on MacMillan imidazolidinones.
Forward synthesis:

Notes:
TMS groups are used throughout but could always be beefed up to TBS as they're removed with TBAF at the end anyway.
I'd welcome suggestions of a better way of achieving step IIIa (formation of the enol-ether-like structure in A).
MacMillan catalyst sequence (steps VI to VII):
I'm aware that these catalysts (R = Bn, BnIndole etc.) are designed with one-pot systems using poor nucleophiles and electrophiles in mind, and hence expect this to be a bit of a stretch. I'm wondering if a more robust method might be to go with stoichiometric addition of the imidazolidinone before addition to THF containing the lithium alkyl cuprate shown to the left of step IV, rather than simultaneous addition of a Grignard and copper(I). The cyclisation seems like it would follow naturally and the correct stereochemistry would be provided by the imidazolidinone in accordance with MacMillan's models.
Not sure how big of a concern rearrangement during the Barton McCombie reduction step (IXa) is, but perhaps it would be better to conduct after the Diels-Alder; I don't know.
My hope is that the Diels-Alder step would get some stereocontrol from the 'B' section of the molecule folding down underneath the diene section, but expect this to be the sort of thing that has to be tried out in the lab.

For now, I'll just comment on the preparation of A. Very interesting use of what I assume is double enolate in step II. I haven't seen this used before, but if you have any references, I'd love to read up on it. For step IIIa, I have a few concerns. The enolate will be a resonance hybrid of both of the carbonyl groups, but the majority of the enolate character will reside on the ketone (since ketone alpha protons are more acidic), making it relatively nucleophilic. Even if you did deprotonate the alcohol, I still don't think it would be able to attack the nucleophilic enolate form.

For step IIIa, I thought about working the molecule up to its neutral form (you would need a less labile protecting group like TBDMS) and then adding a Lewis acid like Al3+. Since five-membered rings form faster than seven-membered ones, the OH should preferentially attack the ketone in the presence of the Lewis acid. As for the elimination, hopefully, the Lewis acid would be enough to make the -OH of the hemiacetal a good leaving group for an E1 elimination. As long as you use the same alcohol ROH, ester hydrolysis shouldn't be a problem.

@EliJones On step II; as far as I'm aware it's fairly standard practice - if you've got Clayden 2nd ed. the method is covered briefly on page 601, but you can also find a fair amount about it by searching online for the acetoacetate dianion. Somewhat hilariously, I've just found this: https://books.google.co.uk/books?id=c8U0HhptJ2QC&pg=PA37&lpg=PA37&dq=acetoacetate+dianion&source=bl&ots=A6WcJaB-sj&sig=ACfU3U1pNQ8cocfwpMAockZw4a2aqTpHWw&hl=en&sa=X&ved=2ahUKEwjU_cHWo_HvAhXK_rsIHVV3DfQQ6AEwEXoECAsQAw#v=onepage&q&f=false which continues with a similar transform to my IIIa!

My initial thoughts about IIIa were that simple dehydration under slightly acidic conditions would probably get the job done because the non-enol form would still exist in high enough equilibrium concentration. This seems to be exactly what they were going for in the case I just linked, but I was as dubious as you until now as I'd never seen this done before!

Thanks for the reference! I also noticed a possible issue in step V. You may get polymerization and a mixture of products since the Wittig reagent might just keep reacting with the product, considering the carbonyl group gets more reactive with each Wittig addition. You could try protecting the Wittig reagent's formyl group first?

Most welcome. Good thinking, though this ylide is often used to make enals quite successfully - maybe this is a thermodynamic effect as the ylide itself is so stable in solution: It's available from Sigma and I believe it can even be recrystallised from water. I'd guess extending the π-system on one enal is less favourable than making a new π-system in the form of another enal. You can find an enal-forming example on page 628 of Clayden, or here's an old paper featuring them: S. Trippett and M. Walker, J. Chem. Soc, 1961, 1266. (Found this in Wyatt and Warren's Strategy and Control, 232).


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Why is the solubility of calcium sulfate so low even though it is a salt?
What is the reason the solubility of calcium sulfate in water is so low? Compared to other salts like potassium sulfate (110 g/L) or calcium nitrate (1470 g/L) the solubility of calcium sulfate seems to be very low (2.4 g/L).

Being a salt doesn't mean it has to be soluble

That's a pretty good equivalent of "Why is Donald Trump so poor, even though he is American?"

Solubility of calcium sulphate is quite high, compared to many much less soluble salts, like barium sulphate, calcium phosphate or calcium fluoride.
Most of insoluble minerals are salts.
Insoluble/limited solubility salts have as crystals lower Gibbs energy than dissolved, leading to solution being thermodynamically  unfavourable.
That is related to mutual relation of energy of crystal lattice and of hydration of ions, and also to dissolution entropy change.

@ Poutnik. You are right. But the question should be rewritten so : How to explain why insoluble salts have as crystals a lower Gibbs energy than dissolved ?

There is a similar situation with halogenides. Calcium iodide, bromide and chloride are extremely soluble in water. Surprisingly, Calcium fluoride is absolutely insoluble in water. Why ?

Only if the opposite has to be explained as well, i.e.why soluble salts have lower G if dissolved. :-)

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Mechanism for the ring contraction
In this reaction I am unable to understand its mechanism. How the peroxide HCCL3 is causing the ring contraction. I know that the 8 member ring is unstable due to strain and hence it will try to contract. But I am not able to initiate the mechanism due to reagents present. Please can someone explain me how the contraction of 8 members ring to 2 five member ring is taking place?

Peroxide is acting as a source of radicals.

@Waylander what will be the first step?

Creation of the CCl3 radical and addition to one of the double bonds. See this for a summary  https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemistry_(Smith)/Chapter_15%3A_Radical_Reactions/15.13%3A_Radical_Addition_Reactions_to_Double_Bonds

@Zenix: This is a radical reaction. Peroxide initiate the radical formation by abstracting a hydrogen from $\ce{HCCl3}$ to form $\ce{^.CCl3}$ radical, which attacks one of the double bond first, and rearrangement happens there after to give $2^\circ$ carbon radical. It abstract hydrogen from another $\ce{HCCl3}$ to give the product and another $\ce{^.CCl3}$ radical, so reaction progress to the end.

As has already been said in the comments, this is a free radical-based reaction. The peroxide $\ce{R1-O-O-R2}$ acts as a radical initiator( where $\ce{R1}$ and $\ce{R2}$ represent alkyl groups). These peroxide radicals generate $\ce{^.CCl3}$ which attacks the double bond and completes the reaction as shown.
An important point to note is that $\ce{^.CCl3}$ is regenerated at the end of the reaction, so it can be described as a catalytic intermediate for this mechanism

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What is the product formed after nitrosation of ketones (HNO2 + ketone)?

On C-nitrosation of ketones we get oximes. After this, the oxime formed is hydrolyzed ($\ce{H3O+}$ is added). What will be formed in the end? Does it follow Beckmann rearrangement or what?

Two answers here: https://chemistry.stackexchange.com/questions/86061/what-is-the-mechanism-of-the-reaction-between-hexane-2-5-dione-and-nitrous-acid/86090#86090

Hydrolysis of the ketone oxime gives the 1,2-diketone as reported in this Synlett reference:

Treatment of alpha-methylene ketones with excess sodium nitrite and
  aqueous HCl in THF at reduced temperature provieds an effective tool
  for the preparation of a variety of 1,2-diketones


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Vapor Pressure and Boiling Point
I heard that the boiling point of a solution is the temperature where its vapor pressure at that temperature equals the atmospheric pressure at that temperature. Thus at higher altitudes, as the atmospheric pressure at that temperature is lower, the boiling point is also lower.
Why is the boiling point the temperature where the vapor pressure equals the atmospheric pressure?

Related: https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers/38654#38654

When a substance's multiple phases are in thermodynamic equilibrium with each other the vapor pressure is the pressure exerted by a vapor existing above a liquid surface. Vapor pressure is related to volatility; the greater the pressure above the liquid the easier it is for vapor molecules to escape the equilibrium and transition to the gas phase.
A simple analogy would be to relate the vapor pressure to the vapor molecules' desire to break free from the equilibrium (multiple reasons for this, more complicated than the question though.) The force of the atmospheric pressure acts as a permeable barrier that can prevent the vapor molecules from escaping (by this I mean transitioning to the gas phase;) when the atmospheric pressure is greater than the vapor pressure the atmosphere is able to contain most of the vapor but when the pressures are equal the vapor is trying to escape just as much as the atmosphere is preventing it from escaping. This is the boiling point; when any more liquid in the equilibrium transfers to vapor the pressure of the vapor is too much to contain and molecules freely convert to gas.
Does that make sense to you? I can put together something more detailed and maybe file some diagrams/pictures that will help if necessary.
Edit: I think this image might help:

Pay attention to the arrows which indicate the opposing forces of the vapor pressure and the atmospheric pressure. When the temperature reaches the boiling point the force of the vapor pressure equals that of the atmosphere's and liquid can freely transition to the gaseous phase.
This is also good


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Why is the crystal field splitting energy larger for square planar than octahedral complexes?
I recently came across a fact that for the same combination of metal and ligand, the crystal field splitting energy for square planar complexes is larger than that of the corresponding octahedral complex.
In fact, I came to know that

Δsp is greater than Δo by a factor of 1.3 to 1.7.

Is it possible to explain why it is greater theoretically or was it just observed through experiments?
And can that factor of 1.3 to 1.7 be justified?
I think in octahedral field placement of ligands both $\mathrm d_{x^2-y^2}$ and $\mathrm d_{z^2}$ orbitals of the metals face direct contact with the ligand (when compared to only $\mathrm d_{x^2-y^2}$ orbital in the corresponding square planar complex) and hence they must experience more repulsion. As the result, there must be greater enhancement of the energy levels and the splitting energy of the octahedral complex must be higher than the corresponding square planar complex. What is wrong in my reasoning?

I try to guess the answer with this reasoning. When four ligands L are at the corners of a horizontal square and around the metal M, the bond length ML takes what I call a reference value. If I add a 5th and a 6th ligand above and under the metal, they may interact with the four first ligands, and repel them a tiny bit, so that the bond ML is also increased. And the intensity of the $\Delta$ effect is slightly decreased

Think of square planar as if two ligands from octahedral have been removed (the ligands above and below the plane). Now, the energy of the $\mathrm d_{z^2}$ orbital decreases as there are no ligands at the axial position. So, the order of energy changes too:
$$\mathrm d_{x^2-y^2} > \mathrm d_{xy} > \mathrm d_{z^2} > \mathrm d_{xz} = \mathrm d_{yz}.$$
The $\mathrm d_{z^2}$ which was once with $\mathrm d_{x^2-y^2}$ and $\mathrm d_{xy}$ which was once with $\mathrm d_{xz}$ and $\mathrm d_{yz}$ in octahedral splitting, has now split further which has increased the gap between the orbitals $(\mathrm d_{x^2-y^2})$ and $(\mathrm d_{xz}$ and $\mathrm d_{yz}),$ which means the splitting has increased.

Please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown.


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less toxic chemical laser chemistry
While most common lasers are electrically powered, there are also chemically powered lasers. These are high power systems, since for lower power systems electrically powered lasers (including optically pumped) are more practical. The chemistries I'm aware of (from wikipedia) use highly toxic fuels or have highly toxic reaction products. (Iodine based lasers that I could find use chlorine gas as an input, hydrogen fluoride lasers result in hydrogen fluoride.)
Are there any chemistries that could conceivably be used as a chemical laser that do not require handling of such dangerous compounds? (That preferably are usable at a smaller scale, although I am not aware of any reason why existing chemical lasers could not be used at small scales.)
edit: The question is just out of curiosity, for practical applications other types of lasers would be more useful.

If there were numerous reactions that produced excited states chemically then these might be of some use, but reactants are needed in huge quantities and so these types of lasers are v. expensive to run and of limited use anyway since there are alternative solid state lasers for most tasks nowadays. If you want to shoot down planes or satellites then multi-kw chemical lasers may be your only choice. (btw most lasers are optically pumped say from another laser, laser diode or flash lamps. Argon ion (gas) lasers are still electrically excited but these are rarely used nowadays due to cost).

@porphyrin Yes electrically powered (including with diodes, flash lamps) lasers are more practical at small scale, I'm just wondering if these chemical lasers need to use dangerous chemicals or if more safe-ish reactions could in theory also work. For a smaller laser you'd also need less chemicals. Or are you saying that the amount of reactant doesn't scale more or less linearly with the output power of a laser?

@EdV What do you base that conclusion on? Lack of googlable alternatives? I also considered that line of reasoning, but on the other hand, chemical lasers have only really been considered for the highest power regimes, where more energetic and thus dangerous compounds are preferable and the dangers can be managed by professionals. So there probably just hasn't been much research into safer chemistries. I'm also happy with answers that just 'in principle' could lead to laser action but which haven't been demonstrated.

I don't doubt their competence, but what did their design space and trade offs look like? My guess is that concerns like high output power were more important to them than chemical toxicity.

Existing chemical lasers work fine on small scales. Usually they were developed at small scales and later scaled up.
Chemical fuels for lasers need high energy reactants in order to create excited states. High energy substances are often reactive, which makes them toxic as they will also react with human body components. Stable laser fuels also exist, in the form of regular hydrocarbon fuels. Many hydrocarbon fuels can be used to power CO₂ gas-dynamic lasers.
The reason iodine and hydrogen fluoride lasers are used is that these are the few types of chemistries that produce relatively short wavelength output. Most chemical reactions deposit their energy into rotational-vibrational exitation modes. These have a much lower energy than electronic exitation modes and will lase in the deep infrared. CO₂ gas dynamic lasers emit light in the 10 µm range. There has been research into other chemical laser pathways resulting in visible or near visible output, but so far without result as far as I was able to find. Some (old, 1980s) background: Technical Report JSR-80-14: Visible Chemcial Lasers


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Does the amount of solvent interaction energy depend on the absolute gas phase energy?
I've been performing Gaussian 16 DFT optimization and single-point energy calculations on different conformers of a rather large organic molecule. My first series of calculations was without solvent model, the second one was with dichloromethane solvent effects included via SMD model. When I analyzed the results, I saw that the absolute energy of the conformers became lower in solvent (as expected due to solvent interactions) but also the energy differences between the conformers became smaller. The order of conformers according to energy however did not change a lot. This would mean that conformers having a higher energy in gas phase than other conformers would experience more stabilization by the solvent. Is this possible? How could this be caused?
If more information is needed about my problem, feel free to ask.
Thanks in advance

One way this can happen is if the different conformers have different dipole moments. Those with largest dipoles might have higher energy due to greater charge separation. However those conformations with larger dipoles will also interact most strongly with the surrounding solvent medium (dielectric), and thereby the solvent will reduce their energy by a greater amount.

Thank you for your answer, I will check the dipole moments and see if this is indeed the case.

@Bub note that my answer might be an oversimplification given that you speak of a "rather large organic molecule" in which case other moments and local charges might matter. Your focus is on how a simple continuum dielectric model of the solvent can have this effect, I provided one simple explanation.

For my data, there is no visible relation between the dipole moment and the gas phase energy so probably other properties of the conformation indeed matter

In any case you can be certain that the effect of the solvent is purely electrostatic (and not H bonds say), so focus your attention on the charge distribution in the conformers. If charges become more exposed to the solvent (rather than being buried) that is liable to lower the energy as well.


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Cross checking the fact that gas solubility decreases with temperature increase and Henry's law's validity
It is known that the solubility of gas decreases with increase in temperature which is due to the kinetic energy freeing the solute particles into the vapor phase. Henry's law states that the partial pressure of a gas is directly proportional to the mole fraction of that gas in the solution. So that must imply that with increased vapor pressure, the mole fraction of the gas in the solution must increase, but quite the opposite is true. I know that the Henry's constant changes with temperature too.
Is it the case that the growth the Henry's constant outgrows the increase in partial pressure of the gas, so that the mole fraction of the gas in the solution decreases?

So that must imply that with increased vapor pressure, the mole fraction of the gas in the solution must increase, but quite the opposite is true.

Vapor pressure, or saturation pressure, $P_{sat}$, is different than system pressure, $P_{sys}$. $P_{sat}$ is an intensive property, relative to the compound of interest, while $P_{sys}$ is an extensive property related to the system in which compound resides.
For Henry's Law, $$c_i = H(T)p_i$$ the pressure of interest is not the vapor pressure, but the system pressure. The partial pressure in Henry's law, $p_i$, is related to the system pressure, $p_i=y_iP_{sys}$.
Your intuition about increased vapor pressure leading to a decrease in the solubility of a component in solution is correct. Generally, if $P_{sat}(T) > P_{sys}$, then the component of interest will reside in the gas phase for a two-phase system (liquid & gas). Generally however, for gases like $\ce{N_2}$, $\ce{O_2}$, $\ce{CO_2}$, etc. the vapor pressure of the compound is already sufficiently high to begin with, i.e $P_{sat}(T) >> P_{atm}$. For example, the vapor pressure of $\ce{CO_2}$ is near $60$ atm at $25$ degC. This is where the role of the system pressure, $P_{sys}$, plays an important role: as you increase $P_{sys}$ sufficiently, you lessen the gap between $P_{sys}$ and $P_{sat}(T)$; this meaning that you will increase the prevalence of a component in solution.

Is it the case that the growth the Henry's constant outgrows the increase in partial pressure of the gas, so that the mole fraction of the gas in the solution decreases?

The Henry's law coefficient's relationship for temperature (for a limited temperature range) can be described by the following Van't Hoff relationship,
$$H(T) = H_oexp[\frac{-\Delta_{sol}H}{R} (\frac{1}{T}-\frac{1}{T_R})]$$
Where, $\Delta_{sol}H$, is the Enthalphy of dissolution (generally a negative term), $H_o$, is the Henry's constant at STP, and $T_R$, is the reference temperature (usually 298.25 degC). With this, we can re-write Henry's equation as,
$$c_i = H(T)p_i = p_iH_oexp[\frac{-\Delta_{sol}H}{R} (\frac{1}{T}-\frac{1}{T_R})] = y_iP_{sys}H_oexp[\frac{-\Delta_{sol}H}{R} (\frac{1}{T}-\frac{1}{T_R})] $$
We can see here that it is not that the Henry's constant outgrows the partial pressure; it is actually quite the opposite. The effect of increasing temperature actually lowers the value of $H(T)$, thus lowering $c_i$.


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Product operator analysis for CHn groups in HSQC and HMQC
I'm taking and advanced NMR course this semester and we are learning about product operator formalism. I have a homework where I'm supposed to apply this formalism to an HSQC and HMQC experiment. The problem I have is that I should use a 3 and 4 spin systems $(\ce{CH2})$ and $(\ce{CH3})$ groups instead of just simply $(\ce{CH})$ and we only did two spin systems in class. I understand how to use the formalism and how to apply it to the experiment, but I need let's say a concept clarification. We use basic pulse sequences for HSQC and HMQC.
In the exercise, we assume that the hydrogens do not couple to each other, only to the carbon. I did the example a found out that the result (in the product operator formalism it was the same observable term which with I was left at the end) was the same for $\ce{CH3}$ group as for $\ce{CH}$ after optimization of delay time. After I thought about it, I feel it makes sense. Since the three hydrogens will do the same as one hydrogen, only I will have a different intensity afterward in the spectrum and of course different shifts. I even calculated that the optimal tau delay is the same for both $\ce{CH}$ and $\ce{CH3}$ groups.
And since HMQC basically gives the same info, I assume it will work the same, and results for $\ce{CH}$ and $\ce{CH2}$ group will also give me the same.
Is this correct? Is there any other difference in what happens during the experiment to the $\ce{CH}$ and $\ce{CH3}$ group (or $\ce{CH2})?$
Just to clarify, it is not a multiplicity editing experiment, just a simple HSQC/HMQC applied to $\ce{AX2}$ and $\ce{AX3}$ spin systems.

Your analysis is generally correct. In the case where all the protons are equivalent, you don't need to worry about proton–proton coupling.
For equivalent protons, the main thing you might need to worry about is that when you have coherences on carbon, such as $S_x$, you will get evolution of multiple proton–carbon couplings, represented by (e.g.) the Hamiltonian $2\pi J_{IS} I_{1z}S_z + 2\pi J_{IS}I_{2z}S_z$ for a $\ce{CH2}$ pair. This could, in theory, generate more complicated product operators such as $4I_{1z}I_{2z}S_x$. To analyse this, you can separately evaluate the effects of the two J-couplings. The first one goes as:
$$S_x \xrightarrow{2\pi J_{IS} I_{1z}S_z \tau} (\cos \theta) S_x + (\sin \theta) 2I_{1z}S_y$$
where $\theta = \pi J_{IS}\tau$. To evaluate the second coupling, the $S_x$ term becomes
$$(\cos\theta)S_x \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos^2 \theta) S_x + (\cos\theta \sin \theta) 2I_{2z}S_y$$
and for the $I_{1z}S_y$ term, we can just "carry through" the $I_{1z}$ terms because they don't evolve under a Hamiltonian on different nuclei*:
$$(\sin\theta)2I_{1z}S_y \xrightarrow{2\pi J_{IS} I_{2z}S_z \tau} (\cos \theta\sin\theta) 2I_{1z}S_y - (\sin^2 \theta) 4I_{1z}I_{2z}S_x$$
So all in all, single-quantum coherence on carbon evolves under two different couplings to give four terms:
$$S_x \xrightarrow{2\pi J_{IS} I_{1z}S_z \tau} (\cos^2 \theta) S_x + (\cos\theta \sin \theta) (2I_{1z}S_y + 2I_{2z}S_y) - (\sin^2 \theta) 4I_{1z}I_{2z}S_x.$$
Thankfully, you do not need to worry about this, because the only times that single-quantum coherence on carbon exists is during the $t_1$ periods of the HSQC. Conveniently, the HSQC also has a proton 180° pulse in the middle of $t_1$, which refocuses all C–H couplings, so this whole evolution of $J_{IS}$ can be neglected. Outside of the $t_1$ periods the magnetisation of interest is single-quantum on proton, so the only coupling that evolves is that one proton–carbon coupling, which behaves in exactly the same way as for a CH system.
For a more detailed treatment I suggest consulting Chapter 12 of James Keeler's Understanding NMR Spectroscopy, 2nd ed. (2010).

If the protons are equivalent, then the above also applies to the HMQC: the only time when there are coherences on carbon is during $t_1$, and there is also a 180° pulse in the middle of $t_1$ which removes the effects of carbon–proton couplings.
However, if the protons are not equivalent, then you will start to get evolution of H–H homonuclear couplings, which can't simply be refocused by a 180° pulse. For the HSQC this is less important as it just leads to a reduction in intensity from magnetisation that goes down non-useful coherence pathways. Because the magnetisation during the HSQC $t_1$ is single-quantum carbon, it does not evolve under homonuclear couplings. However, for the HMQC which has multiple-quantum coherence present during $t_1$, the proton component will evolve under homonuclear couplings, which means that you will get multiplets in the indirect dimension after Fourier transformation. In practice it is difficult to resolve these splittings and so it is just manifested as line broadening.
There are also considerations to be made when using a sensitivity-enhanced HSQC sequence, but I assume those are not relevant for your current situation. The interested reader is directed to J. Biomol. NMR 1994, 4, 301–306.

As a final point, in the HSQC above, notice that when we allow single-quantum carbon coherence to evolve under $J_{IS}$, the $\ce{CH2}$ group obtains a phase factor of $cos^2\theta$, whereas a $\ce{CH}$ group would only have a factor of $\cos \theta$.
This is the basis of multiplicity editing in the HSQC: immediately after $t_1$, a spin echo of total duration $\tau = 1/J_{IS}$ is added in order to allow $J_{IS}$ to evolve. This choice means that $\theta = \pi$, so $\cos\theta = -1$: thus, $\ce{CH}$ groups are inverted and $\ce{CH2}$ groups are not. The resulting spectrum will therefore have different signs for peaks belonging to $\ce{CH}$ and $\ce{CH2}$ groups.
($\ce{CH3}$ groups get a factor of $\cos^3\theta$, so have the same sign as $\ce{CH}$ groups.)

* To be more precise, this is because $I_{1z}$ commutes with the Hamiltonian. So, it also commutes with the unitary propagator $U = \exp(-\mathrm i H\tau)$, and we can write
$$2I_{1z}S_x \xrightarrow{H\tau} U(2I_{1z}S_x)U^\dagger = 2I_{1z} \cdot US_xU^\dagger$$


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Do all metals expand on heating?
This question came up when I was reading about substances with negative thermal expansion. The article by Takenaka [1] gives a good list of materials displaying negative thermal expansion. I noticed that almost all the substances mentioned are alloys or crystals, with not a single pure metal included.
This raised the question: Do all metals expand on heating? If so, why? If not, give an example of a metal that contracts on heating.
References

Takenaka, K. Negative Thermal Expansion Materials: Technological Key for Control of Thermal Expansion. Sci. Technol. Adv. Mater. 2012, 13 (1), 013001. DOI: 10.1088/1468-6996/13/1/013001.

@andselisk how do you get that "reference" text? Do you copy it from somewhere or do you type it out manually?

There sure are also pure metals like that, as mentioned in earlier answer.

@AniruddhaDeb I use Zotero and ACS style file. There is also a simplified online version, ZoteroBib, if you don't feel like installing and configuring stuff.

Depends on the temperature range. Many pure metals do some odd things at low temperature.

@Mithoron no. My question specifically targets metals and not materials. Please suggest an edit if that is not clear in the question.

From andselisk's answer^ "(...) certain liquid-solid phase transitions are accompanied by negative thermal expansion (NTE). This also includes elements (Si,Ga,Ge,Sb,Bi,Pu) - your question is subset to earlier one and metals are also "materials".


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Which is a dye: Ferric Acetate or Ferrous Acetate?
I’ve been trying to make Iron Acetate to dye leather and I cannot, for the life of me, figure this out experimentally.
I’ve been allowing steel wool to react with pure white vinegar in a sealed glass vessel with a pressure release for about a month. It turns dark black to green which I assume is due to the ferrous acetate ($\ce{Fe(C2H3O2)2}$). It also contains orange particulates that I assume are the ‘insoluble in water’ salt complex basic iron acetate, Ferric Acetate $\ce{[Fe3O(OAc)6(H2O)3]^+}$.
The solution has dyed leather black and acted as a mordant, and I assume it is a reducing agent. I’ve been trying to separate the iron acetates but I don’t know what properties they each have. Here’s the question:

Which is the dye? Which one turns leather and wood black? Any information anyone has on the properties of either one would be immensely helpful.

The gist is in the solubility products for the hydroxide precipitates of $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$.

$\ce{Fe(OH)3}$ doesn't really exist but is rather $\ce{FeO(OH)}\cdot x \ce{H2O}$. However the $K_\mathrm{sp}$ is calculated as $\ce{Fe(OH)3}$.

Now for the vinegar solution let's consider a somewhat "spent" solution where a lot of the acid has reacted. So consider a nominal pH of 5.5, so pOH = 8.5 and $\ce{[OH] = 3.16\times 10^{-9}}$. Let's calculate the equilibrium molarity of Fe species as "Mol(Fe) @ pH=5.5" in the table below.
\begin{array}{|c|c|c|}\hline
 & K_\mathrm{sp} & \text{Mol(Fe) @ pH=5.5}\\ \hline
\ce{Fe^{2+}} & 8.0 \times 10^{−16} & 80.12 \\ \hline
\ce{Fe^{3+}} & 2.79 \times10^{−39} & 8.84\times 10^{-14}\\ \hline
\end{array}
A $\ce{Fe^{2+}}$ solution can't be 80 molar of course, but what that does mean is that all the $\ce{Fe^{2+}}$ formed will stay in solution.
The calculations also show that a substantial amount, if not all, of the $\ce{Fe^{3+}}$ created will precipitate. The precipitation does lower the pH, so the equilibrium values are indeterminate without knowing the concentrations of the acetate anion and the nominal concentration of $\ce{Fe^{3+}}$ (amount dissolved plus amount in the precipitate). However looking at the overall reactions, it wouldn't seem that the pH could be lowered much.
\begin{array}{c c c c}
 & \ce{2Fe + 6H+} & \ce{<=>} & \ce{2Fe^{3+} + 3H2} \\
 & \ce{2Fe^{3+} + 6OH- } &\ce{<=>} & \ce{2Fe(OH)3} \\
 & \ce{6H2O} & \ce{<=>} & \ce{6H+ + 6OH-} \\ \hline
\text{Overall} &\ce{2Fe + 6H2O} & \ce{<=>} & \ce{2Fe(OH)3 + 3H2}\\
\end{array}
A further consideration here is that oxygen from the air, which dissolves into the solution, can oxidize $\ce{Fe^{2+}}$ to $\ce{Fe^{3+}}$.
Now I'd assume that it is the $\ce{Fe^{2+}}$ salt that you'd really want since that salt could diffuse into (soak into) whatever you're coating the solution with. The $\ce{Fe^{2+}}$ would then be oxidized to $\ce{Fe^{3+}}$ in the interior of the item.  The $\ce{FeO(OH)}\cdot x \ce{H2O}$ on the other hand would just sit on the surface of the item like a paint.

When the metallic iron reacts with an acidic solution, it does not produce the ferric ion $\ce{Fe^{3+}}$ as MaxW states. It always produces the ferrous ion $\ce{Fe^{2+}}$ according to $$\ce{Fe + 2 H^+ -> Fe^{2+} + H2}$$ The ferric ion is produced by air oxidation of the ferrous ion.

Without getting into the chemistry which has been so clearly discussed by @MaxW, from my personal experience in "ebonizing woods" [particularly Oak and Black Walnut due to their tannin content] the light yellow solution you get as soon as the iron has dissolved, works the best.  I am assuming that the red solution obtained after allowing the iron acetate to set for a month is iron in a higher oxidation state and possibly trimeric [Cotton and Wilkenson, 1st edition], and hence more stable or slower to react with tannins.
Bottom line: Use the light yellow iron-acetate on the leather rather than the dark red solution.


text:
thinking about osmotic pressure when liquid is replaced by gas
An option in my test says:

"The osmotic pressure of a dilute solution is the same as it would exert if it exists as a gas in the same volume of the solution and at same temperature."

I'm not able to think how should I relate between a solution and a gaseous mixture. Am I missing something?

The equations are analogous but in my opinion that is a misstatement. The entire solution, if evaporated, would not exert the same pressure. It is more subtle than that. It is the solute that would exert an equivalent pressure if a gas.

thank you ill check the derivation of the osmotic pressure

Dah.. deleted my earlier comment. // Still believe... The sentence doesn't seem to be readily comprehensible to me. It might make more sense if the full context was known. // I think what the sentence is trying to point out the association of gas pressure and osmotic pressure. Think of a setup to measure osmotic pressure. Initially the opposite sides start off with the same water level. Think of solute side as a gas with  $V_0$ and $P_0$. If the solute side liquid had been a gas and confined to its original volume, then the pressure would go up to $P_1$, where $P_1$ is the osmotic pressure.

my teacher had said that in calculation of osmotic pressure we consider the solution to be ideal so when it is ideal the state , that is gas or liquid doesnt matter. Does this seem right? I think you are also trying to tell the same thing

@charansahith The condition of ideal behavior (essentially ignoring interactions between has molecules or solute molecules) is a key condition during derivation of an expression for the osmotic pressure. The underlying reason for the similarity has presumably to do with entropy associated with a volume expansion (rather than enthalpy changes) playing a central role in both gas and osmotic expansion work.

In both the case of the osmotic pressure of a dilute solution and the case of the pressure exerted by an ideal gas, the solute or gas may be described as composed of non-interacting (ideal) particles, and the mathematical expressions (equations of state) describing the two situations are very similar (in one case $p=cRT$, in the other $\pi = cRT$)$^\dagger$. However it might be less confusing if equivalent to state that in both cases the equations describe similar relationships between the work required to change the volume of the system and the accompanying change in the concentration of gas or solute. In both cases work can be done by the system through an expansion, but in one case the expansion results from pressure exerted by the gas, while in the other it results from pressure exerted by the solvent. In the case of osmotic pressure, since the chemical potential of the solvent is coupled to that of the solute (as described by the Gibbs-Duhem relation) it is possible to relate the osmotic pressure to the solute concentration  (in the limit of an ideal solution as described by Henry's Law).
$^\dagger$As I commented, the equations are analogous but in my opinion, "The osmotic pressure of a dilute solution is the same as it would exert if it exists as a gas in the same volume of the solution and at same temperature." is a misstatement. The entire solution, if evaporated, would not exert the same pressure. It is more subtle than that. It is the solute that would exert an equivalent pressure if a gas.


text:
What is the precise meaning of "in-phase" (and "out-of-phase") in this context?
I am currently studying the textbook Infrared and Raman Spectroscopy, 2nd edition, by Peter Larkin. In a section entitled Symmetry: Infrared and Raman Active Vibrations, the author says the following:

In a molecule with a center of symmetry, vibrations that retain the center of symmetry are IR inactive and may be Raman active. Such vibrations, as shown in Fig. 2.12, generate a change in the polarizability during the vibration but no change in a dipole moment. Conversely, vibrations that do not retain the center of symmetry are Raman inactive, but may be IR active since a change in the dipole moment may occur.

For molecules without a center of symmetry, some vibrations can be active in both the IR and Raman spectra. Molecules that do not have a center of symmetry may have other suitable symmetry elements so that some vibrations will be active only in Raman or only in the IR. Good examples of this are the in-phase (symmetric) stretches of inorganic nitrate and sulfate shown in Fig. 2.13. These are Raman active and IR inactive. Here, neither molecule has a center of symmetry but the negative oxygen atoms move radially simultaneously resulting in no dipole moment change. Another example is the 1,3,5 trisubstituted benzene where the C-Radial in-phase stretch is Raman active and IR inactive.

What is the precise meaning of "in-phase" (and "out-of-phase") in this context? I understand the concept of phase in the context of waves (physics), but their use here, as related to molecules and vibrational spectroscopy, is unclear to me. I would greatly appreciate it if people would please take the time to explain this, including the associated mathematics, so that I have a more substantive understanding of what this means in this context.

It is just a different way of calling symmetric (in phase) and asymmetric (out of phase) vibration modes. One may see "in phase" vibrations this way: take in every moment an t-y(n) plot, where t is time, and y1,y2,yn is the bond length between the central atom and the surrounding ones. In a in-phase (Symmetric) vibration mode, the approximately harmonic vibrations for the plotted bonds show the same phase (ie: maxima at the same t, minima at the same t)

@The_Vinz What you're describing is not clear to me.

Take $\ce{CO2}$ as an example. The symmetric vibration has the $\ce{C=O}$ bonds moving in-phase with each other: when one stretches out, the other stretches out at the same time. For the antisymmetric stretch, it is the opposite, so as one bond stretches out, the other compresses.

@Tyberius Ahh, so that's all it means? Sounds simpler than I thought.

Yes, at a physical level it's just that simple. The_Vinz gives a pretty decent overview of the math and what the phase is actually referring to (the offset in time between when one bond reaches it max/min length and when the other does).

@Tyberius Ahh, ok, now I understand what The_Vinz is saying. Thanks for the clarification.


text:
Why is maximum work done for reversible paths in isothermal expansion and not minimum work done?
Work done is given by $W = -P_{ext}dV $ according to convention.
Which should mean that in cases of expansion, work done on the system would be negative. If work done is negative, isn't it wrong to say that work done in isothermal reversible process is maximum? Because in irreversible expansions, area covered is smaller but then, if the value is negative, then wouldn't it be that minimum work is done in case of reversible isothermal expansion and maximum for irreversible case?

Image for clarification. This image is for expansion. The dark gray area is for irreversible process and the light gray + dark gray is the work done by reversible process. Clearly reversible process has higher area and hence higher negative value.

What is the maximum of -5 and -8?

The work done by the system is maximum.

-5 is bigger than -8. Magnitude wise, 8 is bigger than 5.

@BuckThorn Ahh, is that what they actually mean?

Well, in the opposite direction (compression) you would say that the work done by the surroundings is least for a reversible process.

@BuckThorn No, I think that work done on the system for compression would still be minimum for reversible process because for irreversible process, the higher pressure curve would cover a bigger area than reversible process curve.

@BuckThorn Apologies, I thought you were saying that work done on system was maximum for reversible isothermal process.

The equation you gave  is for the work done on the system.  The work done by the system is $+P_{ext}dV$.

@ChetMiller my teacher told me that work done is maximum in reversible isothermal process compared to any other process. is this correct?or it needs more conditions to be included?

For the same volume change, more work is done at constant pressure.

When we talk about the work done in the above equation,

$W=-P_{ext}dV$ ,

the negative sign only tells us about whether the work is done by the system or work is done on the system. So during expansion the work comes out to be negative indicating the work is done by the system. It is just a physical notion and has nothing to do with the actual magnitude, as we have for acceleration or velocity where negative indicates the opposite direction, just like that.


text:
How to properly clean spilled battery content?
I was cleaning my wardrobe when I found some huge white-brown blots where there was a lot of batteries that I have use. I was keeping them for properly disposal but I forget to do that.
I accidentally touch one of them and noticed a incolor liquid. I just washed my hands with soap and put the batteries outdoor in a disposal bag. There was not any rust or white mass, just blots on the wood of the shelf and maybe a little more of that liquid (I guess the wood absorbed most of it).
Most of the batteries was Duracell LR6 alkaline but there was a few of  Eveready carbon zinc R6 too. So, I would like to know how to clean up the possible remaining liquid and if I need to worry about vapors or something.

Carbon-zinc (Leclanché ) chemistry is comparatively innocuous, but manganese-alkaline cells contain potassium hydroxide, $\ce{KOH}$, which is caustic (can dissolve skin). Some older alkaline cells contained small amounts of mercury to improve shelf life, but that has been phased out. Mercury compounds are very toxic, but there should be little even in old cells.
The main damage is caused by the electrolyte, which contains some water, and for alkaline cells, $\ce{KOH}$. The liquid does not create toxic fumes,  but it can damage paint, wood surfaces, and skin if left in place for a long time. Just wash with plenty of water, scrub, and wash with more plain water. If your hands have a "soapy", slippery feel from the caustic $\ce{KOH}$, after rinsing with water, you can rinse with a little bit of vinegar. Vinegar may also help remove some stains, idf the surface is not too badly damaged.
Both cell types are very messy if broken open, because the black manganese dioxide, $\ce{MnO2}$, is finely divided and hard to remove from hands, cloth and carpet (as I discovered in my misspent youth).

Per this reference, there is a range of metals employed in battery technology.
Some problematic metals salts likely found in select batteries include ions of Hg (mercuric oxide battery), Cd and Ni (in the nickel and cadmium battery, nickel–hydrogen battery and nickel–zinc battery).
I have listed them in decreasing order of which metal salts for which I would prefer to avoid direct contact.
I would recommend proper (as in professional) disposal of Mercury and Cadmium salts.
For cleaning up leakage of battery contents, I recommend Seltzer water (aqueous CO2) and/or Washing Soda (Na2CO3).
Do not add complexing agents (like acetic acid) increasing solubility and the likelihood of skin absorption of toxic metals and possible further introduction into the water table.


text:
Drying sodium tetraborate for use as a standard
I am currently attempting to purify technical grade sodium tetraborate decahydrate by recrystallization for use as a titration standard. I don't have a requirement for a specific grade or purity. I'm just looking to improve my results a bit.
I understand that one of its qualities for use as a primary standard is the stable hydration number. However, if I'm drying wet crystals, what would be the best way to ensure that there is no extra water without affecting this? Could heating produce the pentahydrate or anhydrous form?
This is a separate question really, but would it be better to convert to boric acid, recrystallise and then neutralise with a sodium base?

It is generally not a good idea to consider salts with crystal water as standards. As, with possibly rate exceptions, you cannot guarantee sufficient moisture elimination and defined composition at the same time.

I'm a little confused. Yes, they would need to be dry, but not anhydrous. Sodium tetraborate decahydrate is considered a good primary standard. Is this dried in some way not accessible to a lab? Or purified in some way that doesn't involve water?

I suppose borax may be within those exceptions. But the result should be controlled by a commercial primary standard. BTW it is not true decahydrate, as there are 4 OH groups bound to boron.

Yes, I could just buy a higher purity commercial primary standard, but I'm interested to find out how much closer to that I can get with the supplies I have and basic methods and equipment. The question is what drying method to use. Would there, for instance, be a temperature to stay below to avoid driving of the water of crystallisation?

Unless some differential gravimetry data are available, you may need to experiment to get m=f(T).

Do you mean that comparing to a mass/temperature curve could indicate when the sample has dried, or something like that, thermogravimetry?

Whatever your instrumentation allows. There are 2 constraints: T high enough to dry it well T low enough not causing releasing crystal water. I would use known masses from the same lot, dried at different temperatures, plot the chart and decide. Or search for online date before that.

recrystallise below 50C to ensure the decahydrate forms, don't heat it to dry it.

Probably also aim to obtain small and equally sized crystals, and spread them out in a thin layer - or use a rotary evaporator to constantly agitate, with a heating bath and not much vacuum (or a gentle air flow).


text:
How come we can't use the equivalence point equation of NV=NV in this problem?
So, the problem is you prepare to standardize a $\ce{Na2S2O3}$ solution. $\pu{32 mL}$ of $\ce{Na2S2O3}$ solution is titrated into $\pu{50 mL}$ of a $\pu{0.01 M} \ \ce{KIO3}$ solution to reach the equivalence point. They first titrate the $\ce{KIO3}$ solution until it loses color, then add a starch indicator until the reaction is complete. The reaction proceeds in these two steps:
$$\ce{IO3- + I- + H+ <=> I3- + H2O} \tag{1}$$
$$\ce{I3- + S2O3^2- <=> I- + S4O6^2-} \tag{2}$$
Determine the concentration of $\ce{Na2S2O3}$ solution at the beginning of the experiment.
The answer is $\pu{0.094 M}$, and basically you need to first balance out the two formulas to create a net ionic equation for the two reactions, and then you use the moles of $\ce{KIO3}$ (it's molarity times it's volume) and then the balanced stoichiometric ratio ($\pu{6 mol}$ of $\ce{S2O3^2-}$/$\pu{1 mol}$ of $\ce{IO3-}$) to get the moles of $\ce{S2O3^2-}$, which is the dissociated version of $\ce{Na2S2O3}$ in a $1:1$ ratio (dissolved $\ce{Na2S2O3- -> 2Na+ + S2O3^2-}$). Then you just divide the moles of $\ce{S2O3^2-}$ by it's volume converted to liters to get $\pu{0.094 M}$.
My question is, since we are looking at the equivalence point in a titration, why can't we use $N_\mathrm{acid} \times \pu{Volume_{acid}} = N_\mathrm{base} \times \pu{Volume_{base}}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it?

Pay more attention to enumeration of chemical reactions, there are errors (or they are intentionally incomplete in task description )  in both (1) and (2).

My question is, since are looking at the equivalence point in a titration, why can't we use $N_\text{acid} \times \text{Volume}_\text{acid} = N_\text{base} \times \text{Volume}_\text{base}$, since equivalents of acids and bases are the same at the equivalence point? If we can, how would we do it?

First of all, I would suggest that you stick to the original method using moles and proper mole ratios.
Please forget about this normality equation because it is only historically important and rarely anyone uses this concept in 2020 for research or academic work. Note, industrial analytical calculations still use this equation for very routine work. Still if you want to know why this normality equation fails in your case is as follows.
0) The $N_aV_a = N_bV_b$ relation holds if and only if all the concentration units are in normality, where $a$ and $b$ can be acid-base or oxidizing-reducing agent pair. It will also work for precipitation titrations etc. Corollary: This equation fails in titration calculations when the units are in molarity.
(i) This is a redox titration not an acid-base titration.
(ii) Let us change the subscripts for the sake for indexing, $$N_\text{ox}V_\text{ox}=N_\text{red}V_\text{red}$$
You have $\pu{0.01 M}$ $\ce{KIO3}$, what will be its concentration in normality units?
For that you need to see how many electrons are gained in the half-reaction...
$\ce{KIO3}\longrightarrow\ce{I2}$; Balance the equation and see how many electrons are involved in the redox process. See the example here
Balancing this reaction
Multiply the number of electrons involved in the half reaction by the molar concentration to get normal concentration of potassium iodate.
Apply the above equation and you will get the correct result but in normality units i.e. thiosulfate concentration in normality units.

I'm still a little confused on how to find the equivalent number of electrons. This seems tricky to balance out both equations into half reactions. Would IO3 have 1 electron equivalent while S2O3 (2-) has 2 electrons equivalents (normality being equal to moles times equivalents)? Also, for the equation above, where did you get the I2 for the KIO3 - > I2 in the equation above?

Write half reduction of KIO3 to I2. I2 reacts with I(-) to form I3(-). Forget about balancing. You deal with half balanced reaction in order to convert molarity to normality.  Read this first carefully, you will understand normality http://spots.gru.edu/smyers1/Normality%20Primer.pdf.

If there still a query after reading the pdf, post a new question

This is a note to M. Farooq's answer: This titration consists of two back to back redox reactions. As Farooq pointed out, your first reductive half reaction is actually $\ce{IO3- <=> I2}$ (or $\ce{IO3- <=> I3-}$). In either way, generated $\ce{I2}$ in the presence of extra $\ce{I-}$ would become $\ce{I3-}$. The presence of $\ce{I3-}$ can be detected by addition of starch, which gives a deep blue color. Thus, you must have initially added excess $\ce{KI}$ solution to your $\ce{KIO3}$ solution so that this redox reaction happens spontaneously before you titrate it with thiosulfate solution. That is the second redox reaction ($\ce{I3- <=> I-}$ and $\ce{S2O3^2- <=> S4O6^2-}$).
Your balanced ionic reductive half-reaction of first redox reaction is:
$$\ce{IO3- + 6 H+ + 6 e- <=> I- + 3 H2O} \quad E^\circ = \pu{1.085 V} \tag{1}$$
I put $E^\circ$ value to show the reaction is spontaneous under standard condition. Your balanced ionic oxidative half-reaction is:
$$\ce{3I- <=> I3- + 2 e-} \quad E^\circ = \pu{-0.536 V} \tag{2}$$
To get the first complete redox reaction, you must add equations $(1)$ and $(2)$ in order to cancel out electrons. Doing so you'd see: $E_\mathrm{rxn}^\circ = E_\mathrm{red}^\circ + E_\mathrm{oxi}^\circ = \pu{0.549 V}$, indicating a spontaneous reaction. After the initial redox reaction has completed, you have $\ce{I3-}$ in your solution. That's when your second redox reaction kicked-off as you add thiosulfate solution. The two relevant   balanced ionic reductive and oxidative half-reactions are:
$$\ce{I3- + 2 e- <=> 3I-} \quad E^\circ = \pu{0.536 V} \tag{3}$$
$$\ce{2 S2O3^2- <=> S4O6^2- + 2 e-} \quad E^\circ = \pu{-0.08 V} \tag{4}$$
To get the complete second redox reaction, you must add equations $(3)$ and $(4)$ in order to cancel out electrons. This is easy, because you have only $\ce{2e-}$ in either side of each half-reaction. Thus, net redox reaction is:
$$\ce{2 S2O3^2- + I3- -> S4O6^2- + 3I-} \quad E_\mathrm{rxn}^\circ = \pu{0.456 V} \tag{5}$$
Using these two balanced two redox reactions, you may able to use calculations as M. Farooq explained. However, do not forget to include correct stoichiometric mole ratio $\left(\frac{\pu{mol}_\text{oxi}}{\pu{mol}_\text{red}}\right)$ in each calculation.


text:
On the assignment of axial and equatorial to substiuents of polyhexacyclics
Must the assignment of axial and equatorial substituents of polyhexacyclics always be with respect to each six-membered ring or is there a shorthand?
For example, how would X, Y, and Z be assigned axial/equatorial in the following substituted adamantane?

It depends upon which ring is the reference. Equatorial to ring "A" can be axial to "B".

What would Z be relative to the ring that has both X and Y as substiuents (the one at the bottom)?

The perspective of Z is not drawn well but it appears to be axial to the ring bearing X in which case both X and Z are axial to their common ring.

The structure you have drawn where X = Y = Z = H is known as adamantane, C10H16. It has 3 degrees of unsaturation but visually appears to have 4 rings. Once 3 rings have been defined (rings A, B and C), then all C-C bonds have been accounted for. Put another way, you only need to erase three C-C bonds and add hydrogen to obtain a saturated, acyclic hydrocarbon, C10H22.
In structure 1 ring A is highlighted in red with X and Y equatorial to ring A. In structure 2 substituents X and Z are both axial to ring B. Structure 3 has the configuration of the Z substituent changed to equatorial relative to ring B. Ring C bears the same relationship to ring A as ring B does to ring A. Therefore in structure 4, substituent Y is axial to ring C. Now locate ring "D" and define equatorial/axial relationships of the Z substituent to ring "D" in structures 1 - 4.

I used the URL of the image from your first version.


text:
Getting warnings during SDF to PDB conversion using obabel?
I did batch conversion of SDF to PDB for the Maybridge Screening library (containing ~50000 ligands) using obabel batch conversion tool using the following code.
 obabel *.sdf -opdb --gen3d -m

When I did it, I got successfully the ligand converted but I am getting a few warning referring to incorrect stating that Stereochemistry is wrong, using the distance geometry method insteadand wish to clarify the meaning of those warnings
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 2 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 1 5
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 2 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 2 4
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 3 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 1 3 4
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 4 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0 2 3 4
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 2 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 1 12
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 1 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 1 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 1
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 2 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0 2
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 2 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0 11
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 3 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0 2 13
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 1 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 1 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 1 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 1 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 0
Warning: Stereochemistry is wrong, using the distance geometry method instead
==============================
*** Open Babel Warning  in CorrectStereoAtoms
  Could not correct 5 stereocenter(s) in this molecule ()
  with Atom Ids as follows: 8 9 30 31 52
Warning: Stereochemistry is wrong, using the distance geometry method instead

What do these warnings stand for and how to handle these warnings?

Since you are using batch conversion you are getting warnings for more than one converted molecule. The software is apparently circumventing the encountered problem by applying an alternate algorithm to determine stereochemistry. While someone else here may know from experience what exactly the program is complaining about, you could help resolve the problem by narrowing it down to a specific example of a sdf formatted structure that triggers the warning. In other words, run the software on subsets until you find one structure that triggers the warning. Then post the sdf for that structure


text:
Problem with the solubility and Ksp of zinc hydroxide
My book gave me that solubility (g/L) is $4.2\times 10^{-4}$, so solubility (mol/L) is $4.2\times10^{-6}$. Then, using the chemical formula $\ce{Zn(OH)2}$, I know that the solubility product is $K_\mathrm{sp} = \ce{[Zn^{2+}][OH-]^2}$, so:
$$K_\mathrm{sp} = (4.2\times10^{-6})(2\times(4.2\times10^{-6}))^2 = 3.0\times10^{-16}$$
However, the book says that the $K_\mathrm{sp} = 3\times10^{-17}$. Where did I go wrong?

Direct relying of solubility and Ksp implies that 1) All Zn(aq) is as Zn^2+ 2) All OH- remains as OH- 3) Water autoionization plays no role 4) Data are for the same temperature and the best from the same source. If e.g there is present ZnOH+(aq), data cannot match well.

Also from Maurice's answer and comments (5) activity coefficients should be used instead of concentrations (6) ion pairing is negligible.

Down the rabbit hole "Zinc Hydroxide: Solubility Product and Hydroxy-complex Stability Constants from 12.5-75 C", RA REICHLE, KG MCCURDY, LG HEPLER  CAN. J. CHEM, V53(1975) p 3841

There are no simple relation between the solubility and the solubility product when a doubly charged ion is involved. The measured solubility is always much bigger than the value obtained from the solubility product. This is due to the doubly charged ion.
Doubly charged ions like $\ce{Zn^{2+}}$ are usually hydrolyzed in water, and are partly transformed into basic ions like $\ce{[Zn(OH)]^+}$ by the following reaction : $$\ce{Zn^{2+} + H2O -> [Zn(OH)]^+ + H+}$$ If $\ce{Zn(OH)_2}$ is dissolved in water, it will not only produce $\ce{Zn^{2+}}$ ions but also basic ions by the following equation :
$$\ce{Zn(OH)_2 -> [Zn(OH)]^+ + OH^-}$$
As a consequence, the concentration of non complexed $\ce{Zn^{2+}}$ may be calculated from the solubility product of $\ce{Zn(OH)_2}$ but this $\ce{[Zn^{2+}]}$ is much smaller than the measured solubility of $\ce{Zn(OH)_2}$, because the measured solubility of $\ce{Zn(OH)_2}$ is the sum of $\ce{[Zn^{2+}]}$ and $\ce{[Zn(OH)]^+}$. It is even possible that ion pairs like  $\ce{Zn^{2+}(OH^-)_2}$ are dissolved in water without being dissociated.
In the case of $\ce{Zn(OH)_2}$, the measured solubility is $4.2 ·10^{-6}$ M. And the concentration of $\ce{Zn^{2+}}$ taken from $\ce{K_{sp} = 3·10^{-16}}$ is $\ce{[Zn^{2+}] = 1.95·10^{-6}}$ M. This is much less than the measured solubility. It means that the total concentration of the dissolved ion pairs $\ce{Zn^{2+}(OH^-)_2}$ and of the basic ions $\ce{Zn(OH)^+}$ is equal to the difference $(4.2 - 1.95)·10^{-6} = 2.25·10^{-6}$ M. Only $46.4$% of all the dissolved $\ce{Zn(OH)_2}$ produce  simple $\ce{Zn^{2+}}$ ions (which may be $\ce{[Zn(H_2O)_6]^{2+}}$ by the way).

Magnesium and calcium hydroxides do seem to have solubilities close to the prediction for full ionization because both are sufficiently strong bases relative to their solubilities.  But they are just about the only divalent metal hydroxides with both limited solubility and sufficient basic strength.

(1) Agree that overall notion is a possible explanation. (2) wouldn't say "There are no relation between the solubility and the solubility product when a doubly charged ion is involved." but rather the relationship can be complicated. (3) Not "doubly charged ion" but an ion that forms complexes. $\ce{Cu+}$ forms complexes, $\ce{Ca^{2+}}$ doesn't. (4) I'd expect $\ce{Zn(H2O)6^{2+}}$ in aqueous solution not "simple $\ce{Zn^{2+}}$ ions".

oh ok, so my understanding is simplified :)

@MaxW. I agree with your remarks, except (3). I agree that $\ce{Cu^+}$ forms complexes. But the formation of a complex is not necessary to explain why the measured solubility is higher that the solubility calculated from $\ce{K_s}$. To take your example, namely Calcium, which does not form complexes, the solubility of $\ce{CaSO_4}$ is $4.7$ mM when calculated from $\ce{K_s}$ and $18$ mM when measured. Check : J. Chem. Educ. 7, 12, Dec. 2000, p. 1558. It is exactly like the case of $\ce{Zn(OH)_2}$

article link is https://doi.org/10.1021/ed077p1558.1 // Article discusses $\ce{CaSO4}$. // Not surprised that the activity coefficients have an effect. The ions parring is more influential due to a doubly charged cation and a doubly charged anion.  I'd expect that ion pairing would affect $\ce{CaCl2}$ less.

The molar mass of zinc hydroxide is 99.424, so all your math checks out. Wikipedia also notes for zinc hydroxide that the $K_\mathrm{sp} = 3.0\times 10^{-17}$,  so there is an inconsistency between the $K_\mathrm{sp}$ and the solubility data. Based on the $K_\mathrm{sp}$, the solubility should be $\pu{1.9\times10^{-4} g/L}$.
Obviously one source of the inconsistency is just different sources for the data. Another potential source of the is the temperature. It would take a careful literature search to find the original data sources and evaluate their relative quality to sort this out.

ok thanks! so the two numbers dont "fit" together?

@12345bird - Impossible to say without a careful literature search. The Ksp could have been taken at $\pu{15 ^\circ C}$ and the solubility data at $\pu{25 ^\circ C}$. So in the absence of any other knowledge about the data all that can be said is that the values seem to be inconsistent.

Another possibility is that $s = 2.1 \times 10^{-4}$ instead (half the original value). Thus, $K_\mathrm{sp} =  3.8 \times 10^{-17}$.

@Maurice - Your last paragraph is inconsistent with $K_\mathrm{sp} = 3.0\cdot10^{-17}$. You gave:
$$\ce{[Zn^{2+}] = 1.95\cdot10^{-6}}$$
$$\ce{[Zn(OH)+] = 2.25\cdot10^{-6}}$$
Neglecting the autopyrolysis of water which is really an insignificant correction...
$$\ce{[OH-] = (2\times1.95 + 2.25)\cdot10^{-6}} = 6.15\cdot10^{-6}$$ $$\therefore K_\mathrm{sp} = \ce{[Zn^{2+}][OH-]^2} = 7.4\cdot10^{-17}$$
Neglecting the the autopyrolysis of water means that the following approximation is used
$$\ce{[OH-] \approx 2\times [Zn^{2+}] + [Zn(OH)+]}$$
instead of the exact formula:
$$\ce{[OH-] = 2\times [Zn^{2+}] + [Zn(OH)+] + [H+]}$$
From solubility value $\pu{4.2\cdot10^{−6} mol/L}$, the $K_\mathrm{sp} = 3.0\cdot10^{-17}$, and including the autopyrolysis of water:
$$\ce{[Zn^{2+}] = 1.079\cdot10^{-6}}$$
$$\ce{[Zn(OH)+] = 3.121\cdot10^{-6}}$$
$$ \ce{[OH-]} = 5.272\cdot10^{-6}$$
The pH would be 8.72 and only 26% of the zinc is $\ce{Zn^{2+}}$.
From the paper

"Zinc Hydroxide: Solubility Product and Hydroxy-complex Stability Constants from 12.5-75 C", RA REICHLE, KG MCCURDY, LG HEPLER CAN. J. CHEM, V53(1975) p 3841

the following image shows that three species ($\ce{Zn^{2+}, Zn(OH)+, Zn(OH)2(aq)}$) would have to be considered in this pH range. Also the activity coefficients and ion pairing would have some yet unknown effect.


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Creating a 10 ppb Lead Nitrate Solution
I’m trying to make a 10 ppb Lead Nitrate solution as part of a concentration calibration step for a Beer Lambert’s Law experiment. I plan on performing a dilution in three steps. The calculations have been evaluated using:
$c_{1}v_{1}=c_{2}v_{2}$
Step 1: Creating a 1000 ppm (=1000 mg/L) stock solution (let’s call it A) by dissolving 1 gram (1000 mg) of Lead Nitrate in 1 L of distilled water [creating 1L of stock solution A]
Step 2: Creating an even more dilute  stock solution B of concentration 1 ppm (=1 mg/L = 1000 ppb) by diluting 1 mL of stock solution A with 1999 mL of distilled water. [creating 2L of stock solution B]
Step 3: Creating the desired solution C of concentration 10 ppb by diluting 0.25 mL of stock solution B with 24.75 mL of distilled water. [creating 25 mL of desired solution C)
Does this seem like an adequate method to obtain the desired solution? Should I add 1% v/v of Nitric Acid between every dilution step as a stabilizer to keep the lead ions in solution for as long as possible?
I only plan on using these solutions for about 10 days, so I’m not too concerned about the longevity of the solutions.

Your plan makes sense. And the idea of adding dilute nitric acid is also a good idea.

Note that, for ppb-level concentrations, you should prepare a blank of your nitric acid and water in the same way.

I wasn't sure how accurate you want your final solution. Serial dilution is always risky. But, I'd recommend do the last dilution 1 mL to 100 mL instead of 0.25 mL to 25 mL. Measuring 0.25 mL makes a lot of error.

@MathewMahindaratne That makes sense. Is there perhaps a less risky alternative to serial dilution?

You need it done by 3 steps, that restricts a lot of options.

@MathewMahindaratne It doesn’t by necessity need to be done in three steps. I’m definitely open to suggestions if there are better methods out there!

Step 2 has an error. For 2L of 1ppm, you might want to mix 2mL of 1000ppm with 1998mL diluent. In general, the use of a lab balance for measuring the volumes by weight might be advisable to increase accuracy, as ppm etc actually are by weight and balances give you more accuracy than pipets.

In general, to obtain a reliable 10ppb solution of lead nitrate (do you want 10ppb lead or 10ppb lead nitrate, actually?), you'll need water which has been certified for trace analytics regarding lead content. The other thing that might go wrong is adsorption of lead ions to your container (glass?), esp. upon storage of the solution. You might be better off using a suitable ionic matrix to prevent that, e.g. 100mM (or so) sodium nitrate, possibly in dilute nitric acid, to prevent the solution being alkaline and extracting stuff from the glass containers.

agree with imalipusram -- 1000 ppm $\ce{Pb^{2+}}$ = 331.2/207.2 = 1.598 g/L $\ce{Pb(NO3)2}$

1 ml of stock solution at 1000 ppm diluted to 2 liters gives 0.5 ppm, not 1 ppm.

Your dilution approach can be heavily improved.
Rule no. 1: 2+2 $\neq$4 in solution preparation from an analytical chemist's perspective. An extreme example is that you add 50 mL of water+50 mL of ethanol $\neq$ 100 mL solution. If you add a solid NaCl to 1 L of water, the volume may not remain 1 L. It may decrease!
Analytical dilutions are always prepared in highest quality volumetric glassware (Called class A glassware), which comes in fixed volumes. You might have 2 mL, 5 mL, 10 mL, 25 mL, 50 mL, 1000 mL flasks in the university. Similarly, pipettes comes in fixed volumes. The graded ones (so called Mohr's) are not that great.
With these limitations, you design dilutions, in such a way that volumes are not fractional so a volume of 24.75 mL for dilution is a no-no.
The dilution relation is handy $$\mathrm{C_iV_i=C_fV_f}$$, where i and f indicate initial and final concentrations (C) and volumes (V).
So think about it now in order to prepare 1000 mg/mL you would like to dilute it 1000-fold in steps to make it a 1 ppm solution or 1000 ppb $\ce{Pb(NO3)2}$. From there you would like to dilute it 100-fold to make it 10 ppb.
The first question you would like to think now, is what volumetric glassware is available.

Thank you for the clarification! Is it then that dilutions should only be performed with whole number volumes of diluent? Further, I only have access to standard glassware and was wondering if there was a way to quantify uncertainty on the concentration (because I don’t think my procedure will be near 100% accurate due to the materials I have available to me)? I’ve also looked into purchasing different micro-pipettes in the 0.1-2 micro liter range, so I don’t think the pipetting should be too huge a problem.

Smaller the volumes, larger the errors. What are you trying to do with 10 ppb $\ce{Pb^2+}$?

I was trying to create a calibration curve for Beer’s Law to determine Lead Nitrate concentration in solution, using 1,3,4 dimercapto 2,5 thiadiazole as the colorimetric complexing agent. The standardizing Lead Nitrate solutions are of concentrations 10 ppb, 20 ppb, 30 ppb, and 40 ppb. There is a calibrated curve published already, however. Would it be wise to simply use that calibration curve given the large margin of error in trying to create these select, very dilute solutions?

No, you should not use a published calibration curve.


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Why does the pH before the equivalence point of a titration depend on the initial concentration only when the acid is strong?

SO, in this picture, two different titration curves are shown: one for a weak acid, one for a strong acid. I'm confused as to why for a strong acid, if you vary the concentration of the acid, (and the concentration of the acid and titrant are the same) the pH lines are spread apart below the equivalence point, but for a weak acid, where the pHs are initially slightly different, the lines converge to a common path

It is due to the fact that at half equivalence point, the pH of the solution is equal to the pKa value of the weak acid. And this pH does not depend on the initial concentration of the acid. You should take into account something that does not appear on your diagrams. The concentration of the strong base (used on the abscissa) is not given ! They are not the same in the three titrations. They are equal to the concentration of the weak or strong acid.

@Maurice I agree. But you should turn your comment into an answer :)

Consider the situation exactly halfway to the equivalence point. For the weak acid cases,  the pH equals the pKa in all three cases: this is the center of the buffer region. For the strong acid cases, the added NaOH was completely neutralized, so the hydrogen ion concentrations decrease by a factor of two (because of the neutralization) and also by the dilution caused by adding the NaOH solution. So the pH values halfway to the equivalence are still separated by a pH unit for values in your example.

Please consider giving the green checkmark to the most helpful of the posted answers. It encourages people to put some thought and time into crafting answers that are factually correct, relevant, understandable and likely to be of benefit to those, in future, who encounter the question and accepted answer. It is a small reward for those who volunteer their considerable time, effort and experience to aid others and they might well look favorably at future questions from the same person. Thanks for considering this!

To understand what is happening with the two sets of titration curves, start by considering the titration of acetic acid ("$\ce{HOAc}$" for short) with sodium hydroxide solution. The titration curves are shown below:

In the figure, the mutual concentrations are simply the concentrations of the reactants. Thus, $\pu{0.1 M}$ $\ce{HOAc}$ is titrated with $\pu{0.1 M}$ $\ce{NaOH}$, $\pu{0.01 M}$ $\ce{HOAc}$ is titrated with $\pu{0.01 M}$ $\ce{NaOH}$, and so on.
Acetic acid is a monoprotic weak acid, with $\mathrm{p}K_\mathrm{a}=4.756$. The titration curves for the three highest mutual concentrations almost coincide at the half-equivalence points, i.e. at the center of the buffer region. This is as expected from the Henderson-Hasselbalch equation, which ignores the autoionization of water. However, the three lower concentrations demonstrate that the titration curves do not coincide at the half-equivalence points: as the concentrations decrease, the $\mathrm{pH}$ values at the half-equivalence pints move toward the assumed $\mathrm{pH}$ of 7 for pure water.
As the concentrations are lowered, the effect of water's auto-ionization can no longer be neglected and the Henderson-Hasselbalch approximation is no longer viable. The calculated $\mathrm{pH}$ values at the half-equivalence points are shown in the second column of the table below:

Note that for $\pu{0.1 M}$, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $, as expected. Even at $\pu{0.001 M}$, the approximation is not bad. However, as concentrations continue to decline, the $\mathrm{pH}$ at the half-equivalence points simply move even closer to that assumed for pure water.
Now consider the titration of hydrochloric acid with sodium hydroxide solution. The titration curves are shown below:

As shown in the figure and in the third column of the table, the $\mathrm{pH}$ values at the half-equivalence points are one $\mathrm{pH}$ unit apart until the concentration is down around $\mathrm{1 \mu M}$.  Then, as concentrations continue to decline, the $\mathrm{pH}$ at the half-equivalence points must invariably move toward that of pure water.
For hydrogen ion concentrations high enough to neglect the auto-ionization of water, the $\mathrm{pH}$ at the half-equivalence point is easy to calculate. First, divide the initial hydrogen ion concentration by 2, to account for the fact that half of the number of hydrogen ions, from the 50 mL of starting $\ce{HCl}$ solution, were neutralized by addition of $\pu{25 mL}$ of equimolar $\ce{NaOH}$ solution. Then multiply the hydrogen ion concentration by 2/3, because the solution volume increased from $\pu{50 mL}$ to $\pu{75 mL}$. The net result is that the hydrogen ion concentration, at the half-equivalence point, is 1/3 of the initial value. Then take the negative logarithm base 10, as usual:
$$\mathrm{pH} = -\log{\left([\ce{H+}] \times \frac{1}{2} \times \frac{2}{3}\right)} = - \log{\left(\frac{[\ce{H+}]}{3}\right)}$$
If $\ce{[H^+]} = \pu{0.1 M}$, then $\mathrm{pH} = 1.477$. N.B. Significant figures were ignored in this answer, so $\pu{0.1 M}$ is treated as exact, etc.

It is due to the fact that at half equivalence point, the pH of the solution is equal to the $\ce{pK_a}$ value of the weak acid. And this pH does not depend on the initial concentration of the acid. You should take into account something that does not appear on your diagrams. The concentration of the strong base (used on the abscissa) is not given ! They are not the same in the three titrations. They are equal to the concentration of the weak or strong acid.


text:
Prostaglandin Nomenclature
Prostaglandins are lipid substances derived from fatty acids, which act as powerful biochemical regulators.
In Wade's Organic Chemistry, they indicate that there are several families of prostaglandins (A, B, C, …) and that, according to this classification, a prostaglandin could be formulated as PGA1 or PGA2.
As an introduction it is very well explained, but I would like to go deeper into its nomenclature. I have searched on the Internet, but I have found nothing.
For example, the next prostaglandin is PGB1, but I don't know why. Nor would I know how to formulate one from its structure.

Here is a reference: https://www.lipidhome.co.uk/lipids/fa-eic/eicprost/index.htm All PG's have an E-allylic alcohol in the lower chain. If a Z-alkene is in the the upper chain, then it has a "2" in the name. Other letters, in the name refer to the functionality. Info not hard to find.

Each prostaglandin is named using the prefix 'PG' followed by a letter A to K depending on the nature and position of the substituents on the ring. PGA to PGE and PGJ have a keto group in respective positions on the ring which are further distinguished by the presence or absence of double bonds or hydroxyl groups in various positions in the ring(indicated by numbers or Greek letters*). PGF has two hydroxyl groups and PGK has two keto substituents on the ring. PGG and PGH are bicyclic endoperoxides. An oxygen bridge between carbons 6 and 9 distinguishes prostacyclin (PGI). Also, all prostaglandins have a hydroxyl group in the S-configuration on carbon 15 and a trans-double bond at carbon 13 of the alkyl substituent (R2).
*Numerical subscripts (1 to 3) are used to denote the total number of double bonds in the alkyl substituents, and Greek subscripts (α or β) are used with prostaglandins of the PGF series to describe the stereochemistry of the hydroxyl group on carbon 9.
So for example, if we write PGF2α, the F will indicate two hydroxyl groups in the cyclopentanone ring (F series), the 2 will indicate two double bonds, and the α will indicate that the hydroxyl grouping at carbon 9 is in the α configuration.
References

https://www.lipidhome.co.uk/lipids/fa-eic/eicprost/index.htm (@user55119's link)
https://www.glowm.com/section_view/heading/The%20Prostaglandins:%20Basic%20Chemistry%20and%20Action/item/312#2133
Norman A. Nelson, Journal of Medicinal Chemistry, 1974, 17 (9), 911-918
DOI: 10.1021/jm00255a001


text:
can you explain the structure of gemtuzumab ozogamicin? where are the linker boundaries?
could someone help me to decipher the structure of this molecule? I am trying to understand the structure of gemtuzumab ozogamicin: The anti-CD33 antibody is acylated with the bifunctional AcBut linker, which allows for hydrazone formation with the calicheamicin derivative. This derivative, N-acetyl-gcalicheamicin dimethylhydrazide (DMH), has a disulfide stabilized by both the two methyl groups and the calicheamicin bicyclic core, and an acetylated amino sugar. References: Joel C. Barrish, Percy H. Carter, Peter T. W. Cheng, Robert Zahler Accou... I would like to know the boundaries of AcBut linker. I got confused because in this picture it is depicted like that:  I thought that DMH was the part after the benzyl ring in the linker. Thank you in advance!

The trick is probably the disulphide bridge which will be cleaved by reduction, once the construct has been internalized into the target cell, liberating the drug. So I'd consider everything proximal from it towards the antibody as linker structure.

thank you for your answer!but this disulphide bridge does not belong to the DMH calicheamicin?

Calicheamicin (en.wikipedia.org/wiki/Calicheamicin) has a H3C-S-S-S- group. Apparently, this modification is ok, without losing its DNA binding and cleaving functionality which is exerted by the aryl tetrasaccharide mojety and the en-diine respectively.


text:
Fumes by Melting Aluminium Soda Cans
I am planning to melt and cast aluminium from soda cans. I went to quora page that deals with "burning" aluminium can, which can release toxic gases as the cans are coated with plastic inner lining. So my question is if melting aluminium has the same effect? If so can the affects be mitigated by melting aluminium in the open and wearing respiratory masks (not sure if they are rated for plastic related fumes)?

It is my understanding that all aluminium cans used to store carbonated beverages have a spray-on coating/liner to prevent the liquid and metal from interacting. As you heat the cans to melt them, you are likely to start burning that lining which may produce dark smoke, acrid odors, and many other things. The type of mask that would be suitable is hard to know just from internet research. If you do this in the open, don't you risk having some annoyed neighbors?

Metallic aluminum is always covered by an invisible, colorless, thin and continuous layer of aluminum oxide. If you heat soda cans above 660°C, the metal will melt, and not the alumina. So the liquid will stay imprisoned in a mantle of aluminum oxide. You may have difficulty to get it out of this protective varnish.

@J.Ari I am glad you mentioned If you do this in the open, don't you risk having some annoyed neighbors? I wont be doing this in vicinity of other people due to potential cancerogen fumes. Im just trying to protect myself and a helper who would be only one in the vicinity. I guess I really need to search for proper respiratory masks then.

@Maurice I am not sure I understood well. Dont people skim the alumina and other dross by simply using a spatula? I am more concerned in ensuring if the fumes can cause significant damage if I melt say 100 cans. My understanding is that since the protective film is very small and the hazard to being exposed to the fumes is negligible but I have to be sure as I will have a friend help me as well.

I have this shrewd suspicion that casting Al is nowhere near as simple as doing it with lead, zinc, or even iron.

@Karl it might not be as hard as it sounds but the real question is how big is the risk of being exposed to the fumes due to coatings being burnt. Check here

Youtube, honestly? ;-) (Btw. I meant tin, not zinc.)

@gfdsal. If you melt aluminum cans, you may get a lot of bags made of alumina and filled with liquid aluminum. You may have difficulty remove entirely these bags from the liquid. A lot of aluminum will remain in the bags if you skim the liquid with a spatula.

@Maurice allright. I am not understanding what you mean by "bags". You mean quantity? so large quantity of alumina $(Al_2O_3)$ will be produced and little elementary aluminium $(Al)$?

@gfdsal. Exactly ! I am afraid you may obtain a lot of alumina plus some drops of aluminium imprisoned in the bags here and there. I would be pleased to be wrong ...

There will be some organic coatings that will burn ,producing -who knows what? You will have little success melting aluminum cans in air , they will basically oxidize. Aluminum cans were not worth anything as scrap until years ago commercial melting began in inert ( non-oxidizing) atmospheres. I have seen the remains aluminum tank trucks ( gasoline) that burned , only steel is left , no melted puddles of aluminum. Apparently, the burning gasses carry away most of the alumium oxides produced.

This makes sense. I was gonna melt them properly in a small homemade furnace as here. Are you suggesting that most of the aluminium will oxidize and we will be left with uncommercially small aluminium?

They will if you have any oxygen present in the atmosphere. Aluminium will react with Nitrogen above 850C too.

@Waylander alright thanks for the tip. I did see too many tutorials on youtube where the yield from melting soda cans is sufficient. Roughly 50 cans yield 2kgs of Aluminium. I assume that the combustion of the fuel inside the furnace uses up all the oxygen.

please do care to look at this

In addition to the problems listed in the other answer (typically you get very low yield because most of the aluminum oxidizes, and you have to burn off a toxic plastic coating), aluminum cans are made out of aluminum sheet, which is rolled (hot/cold rolled) and not cast.  The alloys used are not especially suited for casting, (the obviously named) cast aluminum alloys are way better.  You find cast aluminum often in car parts such as pistons, other engine parts, and some wheels.  Since these are in big chunks instead of sheets, they won't oxidize nearly as badly, either.

This makes so much sense. Are you suggestion that aluminium alloy in sheets is not good for casting (apart from low yield and oxidizing) because the alloy has other elements that make casting less desirable? Someone suggested me exactly what you mentioned about car parts but I want to use what is avaialble to me.

Aluminum casting alloys usually contain silicon, up to 12 %.

@blacksmith37 yes and a quick search also reveals the soda can has 1% magnesium, 1% manganese, 0.4% iron, 0.2% silicon, and 0.15% copper along with aluminium. I am assuming they are there to make aluminium more malleable to be formed into cans using punching dye's as well as make them recycling friendly. source

It's not that you can't cast with sheet aluminum, just that cast aluminum alloys works better, it has a lower melting point, flows better, and has a lower tendency to 'hot crack' in the mold.  But aluminum alloy is cheap enough (or easy to find scrap for) that it might be better to work with a more suitable alloy. (ASIDE:   I'd also worry a bit about the coating on cans forming some aluminum carbide in the melt.)

please do care to check this out


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Addition of HBr on alkene
In 1-chloroprop-1-ene(E), if we add HBr, where will the proton add to, C1 or C2? I am asking this because I can't decide whether mesomeric as well as negative inductive effect of Cl will dominate or hyperconjugation of methyl group at C2 carbon, which in turn will decide the position of the formed carbocation.

Chlorine in chlorobenzene is an ortho, para-director. What does that tell you?

@user55119,But also, chlorine deactivates the ring. Therefore, it is a bit confusing that whether hyperconjugation (which only stabilises carbocation) will dominate, or +M and -I by chlorine.

The +M effect is dominant here over hyperconjugation. Hyperconjugation is dominant over +M when number of alpha hydrogen is more than 9. I think this gives you a hint what can be the answer to your question.

Simply follow the experimental results.


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What are the limitations of the Born-Lande' equation?
The Born-Lande' equation is used to theoretically calculate the lattice energy, $\Delta U$, of ionic compounds. It is often cited as such in literature,
$$\Delta U = -\frac{k_Az_1z_2Me^2}{4 \pi \epsilon_\circ r_\circ}\left(1-\frac{1}{n}  \right)$$
In which, $n$ is the Born exponent, $k_A$ is Avogradro's constant, $M$ is the Madelung constant, $e$ is the elementary charge, $\epsilon_\circ$ is the permittivity of free space, $z_1$ is the magnitude of the relative charge of the anion and $z_2$ is the magnitude of the relative charge of the cation  with $r_\circ$ being their closest distance apart.
However, the most apparent limitations of the Born-Lande' equation I'm able to think of are as follows.

The corresponding Born exponent, $n$, and Madelung constant, $M$, must be determined empirically or theoretically for each ionic compound.
The bonds between the cation and anion are assumed not to possess any covalent character. However, the difference in electronegativity between the cation and the anion results in covalent characteristics which I believe can be further explained through Fajans' rule. Moreover, I suspect it is the primary reason for the discrepancy between its calculated lattice energies and that of which are determined through the Born-Haber cycle.

Therefore, I would like to ask the following questions.

What other limitations does the Born-Lande' equation have?
Why are the lattice energies of ionic compounds determined from the Born-Haber cycle considerably higher than that predicted by the Born-Lande' equation?
How exactly is Fajans' rule related to the deviation from perfectly ionic behaviour?
Is it possible to quantify the discrepancy in the values of the lattice energies in terms of the covalent character of the ionic compound?
If electrons have particle-wave duality, how are we able to define ionic radii to determine $r_\circ$ if the location of an electron is described by its probabilistic wave-function?

Nice question. I agree with premise that partial covalent bonding could be a better explanation. // The Madelung constant, M, is determined by the crystal structure. So you could create a table of the appropriate values by normalizing to the nearest neighbor distance $r_0$. // The value of $r_0$ would have to be determined experimentally by say x-ray diffraction. // Wikipedia says that $n$ depends on "the compressibility of the solid" which makes no sense to me since $r_0$ would already have been determined experimentally. // basically the fudge factor is $n' = 1 - 1/n = (n-1)/n$

Wikipedia says that n is "typically a number between 5 and 12" so $0.9167> 1-1/n >0.8$ – Max

(a) The limitation arises from the assumption that the repulsive term is $\sim 1/r^n$. A different/ more exact expression could possibly be obtained depending on the type of atoms involved. This may need an accurate MO calculation. (b) The values for NaCl from both methods seem to be close.  Do you have examples where the comparison is poor and there are no reasons for this? (e) Atomic/ion  radii can be determined experimentally or from theory. Electrons in atoms are constrained by the positive nuclear charge so are 'local' in that sense.


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Are phase transitions reversible when performed at supercooled temperatures?
For the state transition of supercooled water to ice, is the process reversible? I wonder this because I want to know if I can use entropy change of the universe is 0. If it isn't reversible, how do you find the entropy change of the system? I know you can find the entropy change of the universe by calculating the heat of the process and divide by the temperature. Also how does the heat capacity factor into this problem? You can calculate the new enthalpy change at the new temperature but how does that help you with the entropy?

The entropy change for a system is the same whether or not the process that takes the system from its initial to final states is performed reversibly. That's because the entropy is a state function. The difference between a reversible and irreversible process, all else being equal with the terminal states of the system, lies in what happens to the surroundings.
Therefore, to find the change in entropy of the system you start by describing a reversible path for which you can compute the entropy change. Such a path might consist of the following steps:
(1) Warm supercooled water from the initial temperature ($\pu{-20^\circ C}$) to the normal freezing point ($\pu{0^\circ C}$)
(2) Freeze the water to ice at the normal freezing point ($\pu{0^\circ C}$)
(3) Cool the ice from the normal freezing point to the final temperature ($\pu{-20^\circ C}$)
The path involves reversible heating and cooling of supercooled water  and ice, respectively, and the intermediate step of freezing water to ice at the normal freezing point of water ($\pu{0^\circ C}$).
You can use the heat capacities to compute the entropy changes during heating and cooling of the system, respectively, from the initial and to the final states. For one mole of water:
$$\Delta S_{m,T_i\rightarrow T_f}=C_{p,m}\ln\left(\frac{T_f}{T_i}\right)$$
making sure to use the appropriate heat capacity (assumed constant over the temperature range) for water or ice.
For the first step, you warm up the supercooled water, and compute the entropy change by assuming that the supercooled state is stable. For the final cooling step, you compute the entropy change of the cold ice.
For the intermediate freezing step
$$\Delta_fS_m=\frac{\Delta_f H_m}{T_f}$$
The molar change in entropy for the system undergoing the irreversible freezing process is then just equal to the sum of the above changes for the three steps.
As for the computation of the entropy change of the surroundings: you compute the enthalpy change for the alternate path; then, since the enthalpy change for the actual irreversible process is the same as that for the reversible path, and the process is isothermal and isobaric (the intial and end states have same T and p), you can equate the enthalpy change with the heat, and can compute the change in entropy of the surroundings by dividing the heat by the temperature, making sure to use the correct value ($\pu{-20^\circ C}$).

Thanks for your help. I was still wondering then how do you go about calculating the entropy change of the surroundings? Do you adjust the enthalpy change to the new temperature and then divide by the temperature?

@NathanMatthieuTang You compute the enthalpy change for the alternate path; then since the enthalpy change for the actual irreversible process is the same, and the process is isothermal and isobaric, you can equate the enthalpy change with the heat, and do as you mention (divide enthalpy by temperature, making sure to use the correct T)


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Can one describe the bonding of Na to F in terms of molecular orbital theory? What about valence bond theory?
I would kindly appreciate an explanation in terms of the two accepted quantum mechanical theories -valence bond & molecular orbitals- for the electronic energy level structure in natrium fluoride $\ce{NaF}$.
I read here that from a non-quantum perspective the $\text{3s}$ electron of $\ce{Na}$ "jumps" to join the $7$ electrons of the $\text{2p}$ level of $\ce{F}$. This way, the $\text{Na}$ atom becomes the $\ce{Na+}$ ion, while the $\ce{F}$ atom becomes the $\ce{F-}$ ion. All nice until now but handwaving, or high-school level if you prefer.
So how is then the $\ce{Na-F}$ bond (thus the molecule) defined in terms of the electronic wavefunctions of the two original atoms?
How is this generalized to $\ce{MgF2}$ to explain a bond angle of 180 degrees (is this right?)? What about the crystal structure explained quantum mechanically?
Sorry if I ask too much in one shot.

Sooo, you're asking mostly about NaF and MgF2 molecules? Well, they are just that - normal molecules, albeit highly polarised, so ionic contribution to bonds may be higher then covalent. Also you need quite high temp. to make them. Even after vaporisation there are dimers in gas phase.

https://chemistry.stackexchange.com/questions/14174/what-is-sodium-chloride-like-in-gas-state https://chemistry.stackexchange.com/questions/2511/will-gaseous-ionic-compounds-be-free-moving-ions