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What type of reaction is this? (Modified organic nucleoside synthesis)
What type of reaction is shown in step F of this patent image?
I tried looking it up in SciFinder, but I didn't get any results, and I have not seen this before.

Thank you!
Edit 1: I'm also wondering if there is a specific name for this reaction, so that I can look it up.
Edit 2 Yes I'm sorry I forgot the reagents. The patent number is WO2014078463A1 (Bennett et al.) and I included a screenshot below of the relevant description.

Complement your query with Elsevier's «Reaxys» database.  Because its roots -- Beilstein (organic, overlap into organometallic) and Gmelin (inorganic chemistry)  -- followed a different approach to index literature references for reactions differ from the more bibliographic approach taken by Scifinder.  If your school has the money, Science of Synthesis (once Houben-Weyl, by Thieme), too.  None of the databases covers all and everything; the point is to complement their results, regardless of the obvious overlap of them indexing the literature.  And cite the source of the drawing shown.

What are the reagents and conditions? This is essential information which needs to be added into the question. If it is a patent the corresponding procedure may take a bit more time to find, but this is no excuse. Can you please also cite the patent from which this is taken as Buttonwood already suggested?

Thank you Buttonwood and orthocresol. I edited my post to include the patent number and reaction description.

This is an osmium dihydroxylation https://en.wikipedia.org/wiki/Dihydroxylation  The interesting part is the cyclisation through the carbamate NH

@Waylander: Now with the knowledge about excesi K osmate, there is no net oxidation or reduction. Is the "fancy" N-O bond somehow oxidizing the osmate ester? Or is the Na sulfite reducing the N-O bond? Interesting!

@user55119  something must be cleaving that N-O bond. My money is on the Na sulfite.

My thoughts are along the lines of @Andrew and @Waylander. Step F doesn't tell us much, @orthocresol! I prefer a cyclization, 1 $\rightarrow$ 2a, followed by hydrolysis of ester 2a to alcohol 2b. Alternatively, N-O bond homolysis with light would probably involve a cage mechanism. .

EDIT: I miss read the image and saw a carbonyl group where there was an alkene. This explanation has been changed.
This is my guess on what is occurring (an electrophilic addition to an alkene):

The nitrogen-oxygen bond on the amide is severed, forming a carboxylic acid an an electron deficient amide. The electron pair of this carboxylic acid is stabilized by resonance. What is left is a nitrenium ion ($\ce{NR2^{+}}$), which is described on Wikipedia as

Nitrenium species have been exploited as intermediates in organic reactions. They are typically generated via heterolysis of N–X (X = N, O, Hal) bonds. For instance, they are formed upon treatment of chloramine derivatives with silver salts or by activation of aryl hydroxylamine derivatives or aryl azides with Brønsted or Lewis acids. The Bamberger rearrangement is an early example of a reaction that is now thought to proceed via an aryl nitrenium intermediate. They can also act as electrophiles in electrophilic aromatic substitution.

The mechanism for an electrophilic aromatic substitution is similar to that of an electrophilic addition to an alkene.

The nitrenium ion then performs an electrophilic attack on the alkene (second carbon of the oxolane), with beta carbon undergoing a nucleophilic attack by either water in neutral solution or a hydroxyl group in basic solution, forming the primary alcohol. These two processes are likely concerted (otherwise, there would be the formation of a primary carbonation).

Below is a scheme for the predicted mechanism.

A concerted reaction seems more likely than stepwise

It looks a bit like a variation on a Zard radical cyclisation reaction https://warwick.ac.uk/fac/sci/chemistry/research/clark/clarkgroup/publications/tl_1998_1265_b-lactam.pdf

All good points--I have updated the answer.

How can you suggest a mechanism if you don't know what chemicals are used to effect this transformation? It's not even obvious whether if it's an ionic or radical process.


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How is electron-impact mass spectrometry data formally reported?
NMR data has the following format when reported in a paper:

How would I go about reporting my EIMS data formally? I tried looking for documents online but couldn't find anything.

Go to Journal of Organic Chemistry published by ACS. Open any synthesis related paper and click on the Supporting Information. You will see how the MS data is reported in text.

As so often: Check if there is a policy in your group (if you report in a masters / PhD thesis), or see the SI for typical synthesis publications in the journal you would like to submit to (JOC, OrgLet, BJOC, Synthesis, etc. pp.).  Especially for reporting the NMR data, there may be differences (left to right, right to left in ppm scale), but some NMR software (e.g., MNova) helps you to automatically format this by applying a style sheet.

Thank you! I was looking more for a specific style guide information rather than just copying other papers and taking it on faith that they are following the styleguide perfectly.  Thanks for the tip about MNova -- I will have to take a closer look at the documentation.

There is no law to present data in any specific way. These are style guides but not canonical books which should be followed in each letter and spirit. If you like a certain MS results presentation from a good reputable journal, follow it, as long as there are no requirements by the publisher, editor and reviewers.

Third edition (latest offline edition) of the ACS Style Guide (currently known as The ACS Guide to Scholarly Communication) has a section devoted to reporting analytical data, including mass-spectrometry and EIMS in particular [1, p. 275]:

Mass Spectrometry
MS $m/z$ (relative intensity): $238.2058~(44.8\%),$ $195.1487~(100\%),$ $153.1034 ~(21.2\%).$
GC–MS $m/z$ (% relative intensity, ion): $202~(9,~\ce{M} + 4),$ $200~(32,~\ce{M} + 2),$ $198~(23,~\ce{M+}),$ $142~(35,~\ce{M} - 2\ce{CO}),$ $321~(95,~\ce{M} - \ce{Me}),$ $415~(\ce{M+} - \ce{Cl}).$
HRMS–FAB $(m/z):$ $[\ce{M} + \ce{H}]^+$ calcd for $\ce{C21H38N4O6S},$ $475.259;$ found, $475.256.$
EIMS $(\pu{70 eV})$ $m/z:$ $\ce{M+}~420~(15),$ $241~(15),$ $201~(59),$ $135~(14),$ $69~(23).$
Abbreviations: $m/z$ is the mass-to-charge ratio, $\ce{M}$ is the molecular weight of the molecule itself, $\ce{M+}$ is the molecular ion, HRMS is high-resolution mass spec-
trometry, FAB is fast atom bombardment, and EIMS is electron-impact mass spectrometry.

The recommendations for ACS journals also align with the general recommendations given in RSC Experimental data policy — Guide to the presentation of experimental data (accessed 2021-04-22):

Mass spectrometry data
Given in the form: $m/z~183~(\ce{M+},~41\%),$ $168~(38),$ $154~(9),$ $138~(31)$ etc. The molecular ion may be specified as shown if desired. Relative intensities in parentheses (% only included once). Other assignments may be included in the form $m/z~152~(33,~\ce{M} − \ce{CH3CONH2}).$ Metastable peaks may be listed as: $\ce{M^*}~160~(189\to 174),$ $147~(176\to 161),$ etc. The type of spectrum (field desorption, electron impact, etc) should be indicated. Exact masses quoted for identification purposes should be accurate to within $\pu{5 ppm}$ (EI and CI) or $\pu{10 ppm}$ (FAB or LSIMS).

If you are a $\mathrm\LaTeX$ user, check out chemmacros package (TeX.SE) and its spectroscopy module (PDF) which assists with the input of various experimental data.
Reference

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.

Thanks! I am indeed using chemmacros, but this has no support for EIMS input specifically. I am however using it for my NMR data and I can configure it to use the EIMS in this format you described.
I can see why I couldn't find anything -- this ACS guide is behind a paywall! Thanks for the info. Could you link the section for NMR as well? I'd like to make sure my styling is correct.

Also would MP be reported as: "M.P.: 206—208ºC"?

Nevermind, I was able to find a PDF copy of the styleguide. Thanks for your help!

Yes I can, but it's better to ask NMR data reporting as a separate question. Fee free to post a new question if that hasn't been asked before. As for the melting point, the example from ACS Style Guide is "mp 175.5 °C (lit.$^{25}$ mp 175–176 °C)". Happy LaTeXing:)

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If all molecules have basically the same KE at a given temperature, why do lighter isotopes evaporate more readily?
One of the proxies used for paleoclimate is the ratio of oxygen-18 to oxygen-16 in ice cores and in sedimentary rocks.  The idea is that water molecules with oxygen-18 generally evaporate less readily than oxygen-16, and condense more readily;  and this proclivity varies with temperature.  One can then use the deviations of oxygen-18 concentration from a standard ratio to say something about Earth's temperature in the distant past.
Why do light water molecules evaporate more readily?  It makes some intuitive sense that lighter molecules would evaporate more easily.  But how does this square with the equipartition theorem and the idea that all molecules have the same amount of kinetic energy at a given temperature?
It would seem to me that because the intermolecular forces between two water molecules are independent of their mass, it should take a certain amount of energy to remove a water molecule from its neighbors in the liquid phase and move it far away.  This energy requirement doesn't depend on the mass of the molecules, only on how the force on it varies with distance ($W = \int \vec{F} \cdot d\vec{r}$.)  So by this argument, molecules with the same amount of kinetic energy would have the same likelihood of evaporation.  And by the equipartition theorem, the translational kinetic energy of a water molecule is only dependent on the temperature, not on the mass of the molecule.
EDIT:  Just to pre-empt the "it's velocity that matters, not energy" argument:  I agree that lighter molecules would, on average, have a higher velocity.  But from classical physics, we know that applying a fixed force to a smaller mass leads to a larger change in velocity, while a larger mass gets a smaller change in velocity.  That would imply that the "escape velocity" required for a heavier molecule would be smaller; since it's heavier, its motion would not be affected as much by the force it feels from the other molecules. This would imply that while heavier molecules are moving more slowly on average, they also don't need to be moving as quickly to escape.  In other words, the velocity threshold for evaporation should be lower for heavier molecules.  These two effects (lower speeds and lower threshold for evaporation) should cancel each other out.
Where does my logic fail?

Energy is not quite the same. The vibrational levels of the heavier molecule are deeper.

Related: https://chemistry.stackexchange.com/q/81289/72973

One of the lessons by statistical thermodynamics is that there is no fix, uniform kinetic energy for the molecules of a compound.  Instead, there is a distribution of molecules with higher, intermediate, and lower energy.  The shape of this distribution is temperature dependent, too (e.g., slide 32/36 here).  Depending on what you want to describe (and level of detail), this branch influenced by the work by Boltzmann offers a better description than the by classical thermodynamics.

@Buttonwood:  Yeah, I've taught "thermodynamics and statistical mechanics" in my physics department a couple of times, and I always enjoy teaching the "Statistical mechanics" part more than the "thermodynamics" part.

@MichaelSeifert I then assume you present a question like the one by you «about water and water» explicitly comparing molecules differing only by their isotopic composition (e.g., in kinetics when it comes to the deuterium isotope effect).  Because, within reason, intermolecular forces (e.g. cohesion) must stay the same for such a comparison; otherwise, acetic acid (60 g/mol, bp 118 C @ 1 atm) should evaporate at higher rate than benzene (78 g/mol, bp 80 C @ 1 atm).

Evaporation depends on molecular velocity not on the energy
The simple mistake in your assumption is that evaporation is driven by molecules having enough energy to escape the liquid phase. This is wrong. Evaporation depends on whether the molecule is travelling fast enough to escape the liquid.
The molecules in the liquid will have, on average, the same kinetic energy. But this means that, in practice, the molecules will have a range of different kinetic energies across a wide distribution. But, since kinetic energy is a function of both molecular mass and molecular velocity, there will also be a distribution of velocities. And, crucially, the more massive molecules (in this case those with 18O) will have slightly lower velocities at the same average kinetic energy. So the lighter isotopes will be slightly more likely to have a high enough velocity to escape the liquid phase.
You can work through the mathematical details (taking into account the statistical distribution of kinetic energy and converting that to the distribution of velocities and then calculating the proportion of molecules with a velocity above the threshold for evaporation) and check that the proportion of light molecules is higher than the proportion of heavy molecules.

Clarification: why is the force-based view wrong
The argument that applying a fixed force leads to a smaller effect on a heavier molecule so heavier molecules don't need to moves as quickly to escape the liquid is flawed for several reasons.
It sounds relevant but isn't because the vast majority of all collisions don't lead to escape from the liquid. There are so many momentum (and KE) redistributing collisions that all we have to consider is the distribution of energy/momentum across the ensemble of molecules in the liquid. And what we need to know is the average proportion of molecules having enough velocity to escape after their last collision.
Yes, the force argument matters for individual collisions (if they are not purely elastic), but, if you simulated a drop of liquid down to the molecule level, you would find that the distribution of energy/momentum would match the statistical one expected from theory and you would have gained no new knowledge. Follow an individual molecule and, yes, the thinking about force will apply. But follow the distribution of velocities across the ensemble of molecules and what matters is the proportion of molecules having a high enough escape velocity and that is governed by the statistical distribution. Think about the properties of the ensemble of molecules not the individual molecule.
And the force argument ("the escape velocity will also be lower") doesn't work at the ensemble level. All that matters is the velocity after the last collision and the only way to know that is to look at the statistics of the whole system not the individual forces on individual molecules.

To add a comment to your clear answer I would mention that as evaporation occurs the total mass of liquid diminishes and at the same time as only the quick molecules escape the kinetic energy of the molecules remaining in the liquid is continually being lowered. Thus cooling accompanies evaporation which explains the action of most refrigerators.

From physics classes, I know that applying a fixed force to a smaller mass leads to a larger change in velocity, while a larger mass gets a smaller change in velocity.  That would imply that the "escape velocity" required for a heavier molecule would be smaller; since it's heavier, its motion is not affected as much by the force it feels from the other molecules. This would imply that while heavier molecules are moving more slowly on average, they also don't need to be moving as quickly to escape, and the effects should cancel each other out.  What am I neglecting?

[edited original question to include the above argument explicitly]

@MichaelSeifert Your argument about force and mass is irrelevant as the average speeds are determined by the statistical distribution (guaranteed by equipartition). There is no "fixed force" governing evaporation or the speeds of individual molecules: there is only a vast number of momentum-redistributing molecular collisions which ensure the KE of all molecules is distributed in a particular statistical distribution.

@matt_black:  Are you saying that we can't model a molecule near the surface as a classical object being pulled back towards the liquid by the other molecules? I wouldn't think that collisions would play a huge factor for molecules that are "leaving" the liquid;  the only way such a molecule would be pulled back in would be via such attractive forces.

@MichaelSeifert see my additions to the answer. The key point is that–even if you did a complete molecule-level simulation including the detail of attractive forces–all that matters is the overall distribution of velocity after the last collision. equipartition in a large collection of molecules guarantees that the apparent logic of that individual molecule's view is irrelevant.

@matt_black Probably, I did not think through these lines of thought until reading the OP.  But, if the reasoning may be simplified (maybe I go too far here) to $$E_{kin} = \frac{1}{2} mv^2 \Leftrightarrow v = \sqrt{ \frac{2E_{kin}}{m}}$$ then, assuming the same  $E_{kin}$ per molecule of $\ce{H2^{16}O}$ and $\ce{H2^{18}O}$ would mean the ratio of their velocities were about 1.05 : 1.00. Which in turn reminds me about heavy water, $\ce{D2O}$, (of molecular mass more similar to the one of $\ce{H2^{18}O}$ than $\ce{H2^{16}O}$) and its higher boiling point ($\pu{101 ^\circ{}C}$) than $\ce{H2O}$.

Is there not a parallel to the escape velocity of a rocket leaving the gravitational potential of the earth? All rockets need to reach this value otherwise they return.

@porphyrin Only a little, as a rocket sees a force, gravity, that declines with the square of the distance and intermolecular forces often decline with the sixth power of the distance (though the reality in bulk materials is more complex). And also because it is the statistical distribution that matters not the individual force point of view, unlike the situation for a single rocket.

@matt-black, the nature of the force does not matter just its direction, and as for rockets just consider all those that could be launched all with various speeds (and masses) and the analogy still holds perfectly well.  The OP does not recognise that it is velocity that is important so I thought that analogy this would help.

@porphyrin:  The gravitational analogy isn't a good one because in that case, the gravitational force is is proportional to the mass, and so therefore the acceleration of the projectiles is the same.  My argument relied on the amount of force being the same for the heavier and lighter molecules.

As very simple model to find what properties are  important in evaporation consider a molecule that when close to the surface moves around in a cell of side $a$ and collides with frequency $f$ into its neighbouring molecules. There is an area $a^2$ through which it can escape the surface and in a second it moves on average a distance $v= 2af$ back and forth in the cell, which is the average velocity.
The surface presents a barrier $\Delta E_b$ to escape due to intermolecular forces and the chance of exceeding this is proportional to $\exp(-\Delta E_b/k_BT)$ at temperature $T$. Thus the number leaving the surface per unit area per second is proportional to the frequency of hitting a side of the cell at the top of the liquid, which is proportional to its velocity, and the chance of escape over the potential barrier or $\sim v\exp(-\Delta E_b/k_BT)$.
At a given energy the more massive molecule will have the smaller velocity but the barrier remains the same as this is determined by intermolecular forces and so the more massive molecule evaporates less.

The simplest and most direct explanation is that isotopically heavier molecules have higher boiling points. Regular water (R) boils at 100.0$^o$ C. Heavy water (H) boils at 101.4$^o$ C. A mixture of the two (R + H) can be separated by a distillation column. The heavier molecules require more energy to evaporate, to push away one atmosphere.
Any similarity to (astronomical) orbital mechanics is misleading. The orbital metaphor is like throwing a stone, involving one total energy. Liquid evaporation (even at lower temperatures) is like wiggling thru a crowd: an equilibrium of molecules leaving and returning to the liquid, with a net loss of liquid (like diving thru a gap in traffic).
Evaporation of either R or H at lower temperatures relies on diffusion of the escaped molecule to a distance so far away that it is no longer in equilibrium with the liquid. Here the higher velocity of an escaped molecule gets it farther away quicker, and less likely to return to the liquid. Heavier molecules with the same energy move slower, don’t get as far away as fast - thus evaporate slower.
Looking at the macroscopic systems (like pure R and H) rather than individual molecules focuses on measurable properties (bp). It all boils down to differences in boiling point.

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How to determine the concentration of oxygen in a nitrogen/oxygen mixture?
I need to prepare a 10% oxygen in nitrogen mixture for an aerobic reaction I'm developing in lab and I found an old instrument in someone else's lab that controls the flow of two gases. So I am able to control the "percentage" of both gases in the mixture using this manifold.
However, I want to test whether or not that thing is calibrated and how exact/precise it is since I don't want to report in a paper that I'm using 10% oxygen when it is actually 20 or 30 or 50%. What is the easiest way to determine the percentage/partial pressure of this gas mixture that will give me reliable results?
I thought about physical-chemical methods such as weighing the mixture and based on basic thermodynamic equations determine its partial pressures but I think a chemical method would be more interesting, such as reacting a known volume of gas with something that would consume the oxygen in a known stoichiometry. Any ideas?

Pyrogallol is your friend.

There are commercial oxygen sensors. Gas chromatographs can also be used but you need special columns and then gas standards. Pyrogallol will not help.

Intro lab: https://chem.libretexts.org/Courses/Westfield_State_University/Chem0103_Chemistry_of_the_Life_Sciences_(Theis)/05%3A_Labs/5.01%3A_Chemistry_labs/5.1.02%3A_Oxygen_in_air


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Dr. Stone making Acetone from Seashells and Wood vinegar
Could you make Acetone from boiling seashells and wood vinegar as said in Dr. Stone? I haven't found anything online that says whether or not it will work.

https://chemistry.stackexchange.com/questions/61408/chalk-and-vinegar  https://chemistry.stackexchange.com/questions/12250/why-is-it-acetone

Isn't that making acetone from acetic acid and calcium carbonate not methanol and calcium carbonate

No, it's not. https://chemistry.stackexchange.com/questions/143645/make-acetic-acid-from-wood

I think I'll try this in real life to test it out. I believe Dr. Stone uses the acetic acid contained in the wood vinegar and reacts it with the seashells to make acetone as you described. Thanks a lot @Mithoron.

Who is Dr. Stone and why is it related to Chemistry? Am I missing something?

Dr. Stone is a manga and now anime which has a lot of science in it

If calcium carbonate is reacted with acetic acid, this mixture produced a solution of calcium acetate. If now this solution is evaporated, the solvant water will disappear and pure calcium acetate is obtained. And if calcium acetate is pyrolized, ift produce acetone according to the following equation $$\ce{Ca(CH3COO)2 -> CH3COCH3 + CaCO3}$$. This is the way acetone was produced for the first time in History. And this is even the reason why acetone got this name instead of the correct propanone.  The word "acetone" comes from "acetate" in a time where the organic chemistry was in infancy.

Yea I know you can make acetone from acetic acid, but I didn't know there was acetic acid in wood vinegar so I thought the wood vinegar was just methanol


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Acid catalyst decarboxylation of 1-carboxyl-tryptoline
One of my labs had the following reaction:

I understand the Pictet Spengler mechanism. However, I am confused about the decarboxylation. I cannot come up with a way in which this decarboxylation would be acid catalyzed. Here is what I have for the Pictet mechanism:

I have been trying to figure out the acid catalyzed decarboxylation for several days now but I cannot. This is the closest I can come up with, but I'm not sure that this is correct:

Thanks!

If you protonate the beta position of the indole, the resultant immonium salt is isoelectronic with a a beta-keto acid, which readily decarboxylate.

Your suggested mechanism should be acceptable (need two more steps, protonation and deprotonation, see last two steps of my scheme below), if you have isolated the carboxylic acid product (compound $\bf{1}$ in my scheme). If compound $\bf{1}$ is not isolable, I suggest following path would be taken to decarboxylation:

Your reaction is an analog to Pictet Spengler reaction (Wikipedia). When the intermediate $\bf{I}$ is formed during the reaction as depicted in the mechanism, 1,2-hydride transfer is possible to give more stable carbocation $\bf{II}$, which is resonance stabilized $(\bf{\ce{II <-> III}})$. The intermediate $\bf{III}$ can easily undergo decarboxylation to give some what stable enamine product, $\bf{2}$. In acidic medium, this enamine $\bf{2}$ would undergo protonation to give the intermediate $\bf{IV}$, which would then deprotonated to gain aromaticity and give thermodynamically stable indole product, $\bf{3}$ (the intended product).


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Confusion in thermodynamically and Kinetically controlled reaction
I am confused in following reaction:
$\ce{(CH3)2CH-CHO}$ when reacts with conc. $\ce{NaOH}$, will it undergo cannizaro reaction?
According to our professor,  It can undergo both Cannizzaro as well as aldol condensation reaction.
If we consider aldol condensation,  we get tertiary carbanion as intermediate,  which is unstable and can be formed in thermodynamically controlled reaction, but the aldol is kinetically controlled .
Therefore, it undergoes Cannizzaro reaction also which is a thermodynamically controlled reaction.
Now, I know a thermodynamically controlled reaction means getting product that is stable and kinetically controlled means getting product which forms fast.
I am not able to understand how the Cannizzaro reaction being thermodynamically controlled helps.

The cannizaro reaction is only possible for molecules without an H-atom at the alpha position so an enol form is not possible.
Your molecule has a alpha-H atom so it should not undergo carnizzaro and instead it will form the enol tautomer and undergo aldol condensation.

@Waylander :
see this , https://chemistry.stackexchange.com/q/65089/108706 it undergoes cannizaro,  but what is role of thermodynamic and kinetic control has in it ?

Ok I see. So what exactly is your question? Two reaction pathes are avaible according to the link you provided (even thought it is more cannizaro-like because definition of cannizaro is explictly without a-hydrogen).
Usally when you have two reaction paths, one is thermodynamicaly controlled, one is kinetic controlled. You can influence which way the reaction takes by changing things like temperature, reaction time, concentration etc.

@Inselino : My question is how can we say a reaction for eg. Cannizaro reaction is thermodynamically controlled ?and how it being thermodynamically controlled makes its product major.

There is few good explanation given in here. Specially mechanism given by @user55119's answer.


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Oxygen manufacturing on Mars
I was reading about the Perseverance lander on Mars and was surprised to find that it has an oxygen manufacturing instrument on board. The oxygen is needed as a part of a binary rocket fuel to get back from Mars. It makes oxygen from carbon dioxide, a technology that would be very useful for carbon capture here on Earth. I can not find the chemistry or the technology that makes this work. Does anybody know the chemistry of this instrument?
The following is the text from nasa.gov explaining the mechanism of MOXIE:

Mars’ atmosphere is 96% carbon dioxide. MOXIE works by separating
oxygen atoms from carbon dioxide molecules, which are made up of one
carbon atom and two oxygen atoms. A waste product, carbon monoxide, is
emitted into the Martian atmosphere.

Carbon capture is not about chemistry at all. The real problem is "where is the money".

It would only be a useful process if the energy cost of doing it were remotely reasonable (heck, if we did it now we'd probably need to burn even more fossil fuel to generate the energy, releasing even more CO2 than we'd consume by the process!)

If you really wish to see the details, see this article: Mars Oxygen ISRU Experiment (MOXIE), it has all the pictures that you will ever need. https://link.springer.com/article/10.1007/s11214-020-00782-8

The abstract in this link has some details of the process  -  https://www.nature.com/articles/s41560-019-0457-4

On this site, Nasa give some closer information on the MOXIE system.
It splits electrochemically $\ce{2 CO2}$ into $\ce{2 CO}$ and $\ce{O2}$.

There must be a catalyst involved.

I think we are talking about the principle of a solide ocide electrolyzer which is roughtly explained here:
https://en.wikipedia.org/wiki/Solid_oxide_electrolyzer_cell

For this to be a good answer it should include the details of how the process works: link-only answers are frowned upon.

I am sorry but the question was, which chemical principal is behind it. The principal is electrochemical split via a solid oxide electrolyzer cell. This could be found with an easy online research. If he is looking for a detailed description of the process, which preasure temperature, which currency, voltage etc. than I expect to do some research before and then ask a more precise question.


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Henry's Law and "Gases in Vapour Phase"
I found the following definition of Henry's Law in my textbook:

"The partial pressure of the gas in vapour phase is proportional to the mole fraction of the gas in the solution."

My questions:
(a) Why should the undissolved gas be in vapour phase?
(b) As i understand it, a gas is called a vapour if it is below its critical temperature.
(c) But from other sources on the Internet (such as Wikipedia )there is no such restriction on the temperature for Henry's law to hold[I mean the temperature is not supposed to be below the critical temperature of gas to use Henry's law].
(d) Can a gas above it's critical temperature be dissolved in a liquid?
P.S:
(a) I am having trouble understanding why the word vapour phase is used in the definition. Although I do understand what Henry's Law is trying to say.
(b) I tried to search all over the internet but could not find any satisfactory answer.

Rather vapour/gas in gaseous phase.//1 Why should not ? //2 + usually if it is possibnle to have it as liqiud at give circumstances as well // 3 no such restiction // Even helium with crit. T few K dissolves in liquids

@Poutnik [1]I have no problem with the gas being a vapour but stating explicitly in the definition means that if it is not a vapour then we can't use the Henry's law and that is what bugs me.  [2] I can't find a satisfactory example of helium being dissolved in liquid at a temperature above its critical temperature on the net ; please clarify.

Henry's law is explictly about dissolving of gas in supposedly nonvolatile solvent. This law is closely related to the Raoult law, considering volatile liquid components and their vapours. // What example do you need ? All gases dissolve in liquids, regardless of their critical temperature, just their Henry constants differ.

@Poutnik [1]So the definition in the book (or rather the phrase "gas in vapour phase") is unnecessary? [2] I am confused as to why the word "vapour phase" has been used...(pls see ques part a,b and c)...[3]Any well established example of a gas dissolving in a liquid when the temperature is more than the gas's critical temperature.

1/ a gas is a component of gaseous phase, or part of mixed gas without mentioning the phase. 2/ There are sometimes disagreements when to call a gas vapour, but expression "vapour phase" is not used. 3/ The fact that you live is such an example. You do breathe oxygen, don't you ? What is its critical temperature ?

@Poutnik [1]So, in Henry's law , it is of no significance whether the undissolved gas above the liquid is vapour or not (means the temp. may be more or less than the gas's critical temperature)?   [2] I don't think there are any disagreements about whether a gas is vapour or not; it's termed a vapour if the gas is at a temperature lower than its critical temperature (src: Wikipedia).  [3] O2 is not a great example...most of it is transported by binding to haemoglobin and not simply dissolving in blood. [4]Thanks for such active replies to my querries :)

Calling gas a vapour or not is strongly context dependent and the boundary is fuzzy. CO2 in air is never called CO2 vapour, probably because you do not meet liquid CO2 at atmospheric conditions. Alternative approach is, gas is called vapour, if its boiling point at given pressure is above given temperature // Oxygen would not bind to haemoglobin, if did not previously dissolved in liquid. Fish breathe dissolved oxygen in water. Plus, you did hear about the diver keson disease, due releasing nitrogen bubbles in veins.

[1] Ok, agreed on O2 but what about this....{So, in Henry's law , it is of no significance whether the undissolved gas above the liquid is vapour or not (means the temp. may be more or less than the gas's critical temperature)? .....}

Yes, it does not depend on it. Or rather very indirectly. As high critical temperature often - not always - means high polarity , solubility and non ideal behaviour of the solution.


text:
Finding half-life of a nuclide based on gas release
Full Question: The hydride of an unstable nuclide of a Group IIA metal, $\ce{MH2 (s)}$, decays by alpha emission. A $\pu{0.025 mol}$ sample of the hydride is placed in an evacuated $\pu{2.0 L}$ container at $\pu{298 K}$. After 82 minutes, the pressure in the container is $\pu{0.55 atm}$. Find the half-life of the nuclide.
Here's what I have so far:

Find nuclear equation MH2(s) ---> He(g) + Noble gas + H2 (g) (not sure if this is right)

Find total moles of gas in container at 80 mins
$$n = \frac{PV}{RT} = \frac{(\pu{0.55 atm})(\pu{2 L})}{(\pu{0.0821 L atm mol-1 K-1})(\pu{298 K})} = \pu{0.045 mol}$$

Use integrated first rate law to calculate k and half-life.
Use $\ln N_t = -kt + \ln N_i$, where $N_i$ is the initial amount of moles of $\ce{MH2}$ ($\pu{0.025 mol}$) and $N_t$ is the moles of hydride remaining after 80 minutes.

Then find half life using k.
$$t_{1/2} = \ln 2 / k$$

Here's what I'm having trouble on: How to find $N_t$, i.e. the moles of metal hydride remaining. This part has been stumping me for quite some time, so help is appreciated! Thank you!

Welcome to Chemistry! You may be interested in: FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. See also my edit.

You need to figure out what the chemical process is when this nuclide decays. If it is in group II, and emits an alpha particle, what group is the resulting element in? You need to come up with a chemical equation for this.

@orthocresol  I understand that the MH2 would decay into helium, another noble gas, and then H2 (I believe?). I'm still not sure how to find the remaining moles of H2.

If the equation for the resulting amount of gas is correct (one mole of helium plus one of hydrogen per decay) then all you need to know the moles that have decayed is the number of moles of gas divided by two. Since you know the amount at the start you also know the amount remaining.

@matt_black I believe the products would be helium, hydrogen gas, and another noble gas. So why would we divide by 2?

@Emily Sorry, you are right I misread the equation, you should divide by 3 because 3 moles of gas are generated per decay.

For this we can first write the equation for the reaction again:
$$MH_2 => He_{(g)} + Noble gas + H_{2(g)}$$
Next we can find the moles of gas at 80 mins (I am assuming it is 80 minutes and not 82). You already did this and you got 0.045 moles. Since each mole of $MH_2$ forms 3 moles of gas, $\frac{0.045}{3}=0.015$ moles of $MH_2$ used in 80 minutes. Then we can use the first order integrated rate law now:
$$ln(N_f) = -kt + ln(N_i)$$
$$ln(0.025-0.015) = -k(80 mins) + ln(0.025)$$
$$k = 0.0115 min^{-1}$$
Now that we have k, we can solve for the half-life.
$$t_{1/2}=\frac{ln(2)}{k}$$
$$t_{1/2}=\frac{ln(2)}{0.0115min^{-1}} \approx 61 mins$$

Thank you! Now it just clicked! I appreciate it.


text:
Dilution factor and ICP
I did total digestion on a $\pu{0.05 g}$ sample and had a final volume of $\pu{10 mL}$ once it was digested. I then took $\pu{5 mL}$ of this $\pu{10 mL}$ solution and diluted it to $\pu{10 mL}$ with water for ICP analysis.
The ICP gives a concentration of $\pu{1.5E-3 mg L^-1}.$ I haven't accounted for the dilution factor yet. Would it be $\pu{10.00 mL}/(\pu{0.05 g}) = \pu{200 mL g^-1}$ and then   $\pu{10 mL}/(\pu{5 mL}) = 2,$ so the overall dilution factor would be $200\times 2 = 400?$
So the actual concentration would be $400\times\pu{1.5E-3 mg L^-1} = \pu{0.6 mg L^-1}?$

You just diluted by a factor of two, so just double the concentration from the ICP determination. What element are you looking at with the ICP, by the way?

Are you really trying to find the mass of your determined element of interest that is contained to your 50 mg sample? If so, then it is just 30 ng, i.e., 0.003 mg/L times 10 mL.


text:
Is there a particular definition for the partial derivative of internal energy with respect to volume at constant temperature?
If we have specific internal energy $u=u(T,v)$ then the total differential of $u$ is given by
$$du(T,v)=\left ( \frac{\partial u}{\partial T} \right )_vdT+\left ( \frac{\partial u}{\partial v} \right )_{T}dv$$
where
$$c_v\equiv \left (\frac{\partial u}{\partial T}  \right )_v$$
Is there any particular usage of $\left ( \frac{\partial u}{\partial v} \right )_{T}$? I know it can be expressed in the following form:
$$\left ( \frac{\partial u}{\partial v} \right )_T=T\left ( \frac{\partial p}{\partial T} \right )_v-p$$
I understand what it "means", but I don't know if it has a particular meaning that is significant.

Are you asking whether it has a "name?"

$(\partial U/\partial V)_T=0$ for an ideal gas and states  that its energy is a function of temperature only. The full eqn. is needed when working with  a real gas, such as described by van-der-Waals or similar eqn.

@Chet Miller Yes, essentially.


text:
Do "weak-field" and "high spin" complexes refer to the same thing?
I will keep it brief; I know that there are weak field, "moderate" field, and strong field ligands and that these might contribute to a complex being high or low spin overall. My question is, does a complex having high spin indicate a weak field complex, regardless of ligand "strength" (or vice versa)?
For example, $\ce{[Fe(H2O)6]^3+}$ has a moderate strength ligand ($\ce{H2O}$) but is overall a high spin complex, so would $\ce{[Fe(H2O)6]^3+}$ be considered a weak field complex, despite the ligand strength not being weak? If not, is it only appropriate to call it a high spin complex? Thank you.

@NisargBhavsar Sorry, that's not right. The complex $\ce{[Co(CO)5]}$ does not exist.

Also, @ajent what do you mean by "weak-field complex"? Ligands can be weak or strong-field and complexes can be low or high-spin. I have never seen complexes referred to as weak or strong-field.

There is no such thing as a weak-field complex. || @NisargBhavsar There is no such thing as [Co(CO)6] either. Cobalt forms a dinuclear carbonyl: Co2(CO)8. https://en.wikipedia.org/wiki/Dicobalt_octacarbonyl

@ShoubhikRMaiti sorry for the late response, I have been busy. I'm trying to understand a question on my homework for this class; the way it's worded implies that the complex itself be labeled as weak-field or strong-field, rather than merely the ligand. I assume this is just a grammatical problem if it is inappropriate to call the complex itself weak-field or strong-field.

The spectrochemical series only gives info on the relative strength of the ligand field. Whether or not a metal complex is high or low spin (assuming there is an appropriate number of d electrons that both are possible) is a function of both the ligand and the metal center.
Thus, it is possible for "weak field" ligand to be in a low spin complex if the metal center favors low spin. For example, almost all complexes of second and third series transition metals are low spin, independent of ligand identities.

Giving this the answer so that it doesn't remain unanswered for eternity. I think my confusion stems from the problem I am working with being worded unclearly. Thanks for your response though!


text:
What exactly is causing analyte protonation in electrospray ionization MS?
I commonly see explanations that the presence of acid in the mobile phase is a source of protons for the analyte, and that you should be multiple pH units below the pKa of the analyte, but in practice plenty of compounds that you end up detecting have pKa's that are well below the pH of any practical mobile phase or are not even protonatable - so how are these molecules being protonated?

Ionization in ESI-MS is not well understood as yet because it is all gas phase reactions in a strong electric field. There are two models. Ion evaporation model (IEM) and the charge residue model (CRM). See Wiki and start from the references therein.

Fenn, the Nobel laureate who invented ESI, said that he could make the elephants fly...but did not know how how the elephants were flying.

[ This is a short answer, as a complete answer would require going through multiple references and books on the topic. ]
First of all, the ESI process does not require the pH to be acidic : positive ESI is routinely done in ammonium acetate or ammonium bicarbonate media, which have a pH of 7.4 or 8.1. But indeed, to produce protonated analytes it indeed requires a source of protons. So what is the source of protons?
One should go back to the initial stage of ESI, which is the production of positively (or negatively) charged droplets under the influence of a strong electric field. Let's assume that we focus on positive ESI and that the droplets carry an overall positive charge. These charges are often simply represented as + signs, but their molecular nature is important. In the case were an electrolyte is present in high concentration and in its ionic form, the major charge carrier will be the positive charge of this electrolyte. So when considering an organic acid at low pH, the positive charges will be mostly solvated protons. If the electrolyte was ammonium acetate at neutral pH, it would be ammonium ions. Even at higher pH, if the electrolyte is ammonium hydroxide, the droplets would still bear a net positive charge borne by an excess of ammonium ions.
Then come the evaporation / Coulombic fission stages whereby smaller droplets are formed from the larger one, with an overall concentration step. The concentration effect of this stage will affect the overall pH of the droplet if initially acidic, but it is not required to lead to protonated species. The important point being that within the droplets, there is still an excess of positive charges, mostly in the form of solvated proton. At some stage, as concentrations of both the analyte and the solvated proton increase, the proton will be placed in close proximity with the analyte (both not being volatile do not evaporate). At this stage, the system is no more in the bulk solution thermodynamic conditions, but not really either in the gas phase thermodynamic conditions.
But we can still at this stage approximate the situation as "in the middle" by considering the thermodynamic parameters for both ends: on the bulk side, we have pKa of the analyte, with decreasing pH, and on the gas phase side, we have the proton affinity (PA) of the analyte that should be compared to the PA of the other species competing for the proton. The two scales are related but not strictly: for instance water has a quite low proton affinity (692 kJ/mole) compared to most polar molecules (ketones are in the 800 kJ/mole range). Thus is the proton located on the analyte is more stable than on molecules from the mobile phase, it will end up protonating the molecule, without considderation of the pKa.


text:
How to calculate the charge on silicates?
An isolated $\ce{SiO4^4-}$ ion has $-4$ charge. These discrete tetrahedral units form the ortho-silicates. As these tetrahedral units combine, they form various types of silicates.
How do we calculate the total charge on various forms of silicates?
Examples: $\ce{Si4O11^6-}$ has $-6$ charge and so does $\ce{Si2O7^6-}.$ But I am not able to understand how to calculate this charge. Is there any general rule?

Check the oxygens that are not shared each gives -1.

Ever heard of oxidation states? Si is +4, O is -2. Yes, it is as simple as that: sum up all oxidation numbers, and that would be your charge.

In this answer, the basic concept of how to understand what figure they are and how much charge they have would be discussed.
As OP has well stated, the basic unit of any silicate is the $\ce{SiO4^4-}$ tetrahedron which has four oxygens that are connected to one silicon atom.  Each unshared oxygen provides -1 charge and each shared oxygen contributes 0.5 oxygens to one tetrahedral unit.
Now, depending on how these units interconnect and how oxygens are shared, we get multiple different structures. I shall discuss three of them here since that should be enough to get the idea across.
Type 1: Chain Silicates

In this type of silicates, as you can see above, the tetrahedron shares one oxygen on either side and so the basic unit here would be the $\ce{SiO3^2-}$ unit, since two oxygens are being shared.
Since two oxygens are shared per tetrahedral unit, the basic unit would have 3 (2 unshared + 0.5 * 2 shared) oxygens and 2- charge.
Type 2: Pyro silicates

Here, the one oxygen in the middle is shared by the two tetrahedral units. This means that there is a total of 6 free oxygen atoms that are only bonded to the silicon with a single bond. Therefore the structure arises as $\ce{Si2O7^6-}$.
Since one oxygen is shared per tetrahedral unit, the basic unit would have 3.5 (3 unshared + 0.5 * 1 shared) oxygens and 3- charge. However since the values need to be integers $\ce{SiO_{3.5}^3-}$ would be written as $\ce{Si2O7^6-}$
Type 3: 3D Silicates
In this case, all the oxygen atoms are shared leading to the formation of $\ce{(SiO2)_n}$ molecules.
Further Reading
https://www.adichemistry.com/inorganic/p-block/group-14/silicates/silicates-1.html

Your "Type 3" description is incorrect - it wouldn't even be silicate!


text:
Why are Grignard reagents nucleophiles and not ambiphiles?
The way I see it, the -R (alkyl part) obtains a negative charge and the MgX obtains a net positive charge on bond dissociation. So a nucleophile and an electrophile both are obtained simultaneously. So why is Grignard reagent a nucleophile and not an ambiphile? Where am I wrong?

R- doesn't like being R- and MgX+ is fine with it.

There are some cases where the Mg cation binds to the Grignard substrate activating it or directing the nucleophilic attack, so you could say in these instances it is an ambiphile.

@Safdar Faisal would that mean R- would be a better nucleophile? Ready to get rid of that charge instantly?

@Waylander ah can you give me an example?

Exactly, the instability of R- is what acts as a driving force in most Grignard reactions

@Safdar Faisal oh I get it, thanks! Just one more doubt tho: by that logic , RCOO- is also a nucleophile, but the negative sign here is stable on the O, through resonance. So why does this act as a nucleophile?

@SaumyaChaturvedi  The Grignard addition to nitriles is arguably one.

@Waylander okay I'll check that out thanks!

In fact a Grignard reagent reacting with, let us say, a ketone, is ambiphilic. However, the electrophilic component of the reaction, in which the positively charged magnesium combines with the carbonyl oxygen, does not figure into the organic product on workup. See this answer.


text:
Molecular diffusivity of a three-component mixture
I usually used a following expression by Fuller & Schettler & Giddings to predict molecular diffusivities of binary mixtures:
\begin{equation}
D_i = \dfrac{1\cdot 10^{-4}\left(0.00143\cdot T^{1.75}\right)}{\left(P\cdot {M_{ib}^{1/2}}\cdot \left(\left(\sum \nu_i\right)^{1/3} + \left(\sum \nu_b\right)^{1/3}\right) \right)^2}
\end{equation}
(where $P$ is a pressure, $T ...$ temperature, $\nu ...$ molecular constant, $M ...$ molecular weight) as it gives a decent accuracy for most of the binary mixtures.
There are other formulas to predict a diffusivity, but to my knowledge, all them are for a binary mixtures.
What would you recommend for a molecular diffusivity prediction of a three component mixture of $\ce{N2}$, $\ce{CO2}$ and $\ce{O2}$?


text:
Do all metal nitrates give nitrogen dioxide on decomposition?
Some metal nitrates decompose to give nitrites: $\ce{2MNO3 -> 2MNO2 + O2}$

whereas some metal nitrates decompose to give $\ce{NO2}$. A possible reaction is $\ce{M(NO3)2 -> MO + NO2 + O2}$
Based on my observation my question is that, is it true that all metals (including transition metal) nitrates decompose to $\ce{NO2}$ apart from $\ce{[Na , K, Rb, Cs]}$ (alkali metals which give nitrites instead) ?

$\ce{NO2}$ is pretty much always there when a nitrate is decomposed (also for $\ce{MNO3}$). The question is, how much. And please try not to use chemical formulas in titles (especially those requiring MathJax), use chemical name instead.

Strongly related: https://chemistry.stackexchange.com/questions/51769/why-thermal-decomposition-of-ironii-nitrate-and-sodium-nitrate-yields-differen

and similar questions: https://chemistry.stackexchange.com/questions/50039/do-all-covalent-nitrates-on-heating-decompose-to-give-no2 ... https://chemistry.stackexchange.com/questions/19258/products-of-leadii-nitrate-decomposition ... https://chemistry.stackexchange.com/questions/122411/decomposition-of-metal-nitrates ... https://chemistry.stackexchange.com/questions/127735/decomposition-reaction-of-tinii-nitrate ... https://chemistry.stackexchange.com/questions/88173/decomposition-of-lino3

Thanks @NilayGhosh I just read all of your suggestions. They did answer my second potential question of whether the metals would exits in free state or as oxides after decomposition. Although for this question I was seeking a possible generalization and affirmation on this observation and to know any exceptions if they might exist.

TL;DR: Decomposition of nitrites always afford $\ce{NO2}$ if you apply the correct reaction condition to it.
Long answer
To generalize the decomposition of metal nitrates (I concluded from the article above, and hereby using a generic monovalent metal M for the equations), we can say this:

The first stage of decomposition is always the decomposition to nitrites: $\ce{MNO3 -> MNO2 + 1/2 O2}$

Note that nitrites are almost always (if any) less thermally stable than the corresponding nitrates. Thus, the formed nitrites from the reaction above quickly decomposes. I think that (and from the article), nitrites decompose to form the oxide and $\ce{N2O3}$, the latter disproportionates into $\ce{NO}$ and $\ce{NO2}$: $\ce{2 MNO2 -> M2O + NO + NO2}$.

Everything becomes interesting from here. It is due to the fact that the nitrite formed from the first step and not decomposed (kinetics matter) can be oxidized by either the $\ce{NO2}$ or $\ce{O2}$ back to the nitrate. This is especially significant when the nitrite is stable enough ($\ce{KNO2, NaNO2}$ etc). In fact, the decomposition of $\ce{KNO3}$ is reversible (equilibrium) at about 650 - 750 °C in the atmosphere. In the other hand, nitrites of transition metals (which are mostly covalent in character) are so unstable that they decompose rapidly and thus only appear as the so-called reactive intermediates.

Finally, the oxide can decompose into the metal and $\ce{O2}$. This is a characteristic of metals that lie in the less-reactive end of the reactivity series. If the decomposition temperatures of the oxide and the nitrate are different enough, you can control the temperature to get the oxide or the metal ($\ce{HgO}$ from $\ce{Hg(NO3)2}$ is an example.)

The re-oxidation of stable nitrites can be avoided by two ways: either do the reaction under a flow of inert gas to displace the gaseous products away, or crank up the heat more to increase the reaction temperature, which means faster decomposition of nitrites. For example, decomposition of $\ce{KNO3}$ goes to completion at about 800 °C.
Reference

High Temperature Properties and Decomposition of Inorganic Salts Part 3, Nitrates and Nitrites, Kurt H. Stern, Journal of Physical and Chemical Reference Data 1, 747 (1972); DOI: 10.1063/1.3253104

Apart from formatting, you should also learn to properly cite papers. See this meta posts on how to cite papers: https://chemistry.meta.stackexchange.com/questions/3529/should-we-enforce-including-actual-citations-instead-of-just-linking-to-sources ... https://chemistry.meta.stackexchange.com/questions/2944/what-is-the-recommended-style-of-citing-on-chemistry-se


text:
How can flammable gas become non-flammable when mixed with another gas with comparable percentage?
For example, R-410a is a 50-50 mixture of R-32 and R-125. R-32 is flammable (class A2L) and R-410a is not flammable (class A1).
Flammability of refrigerant is mainly a concern when it is leaked since there is no oxidizer when it is not leaked. But when leaked, it become gas in air. In the case of R-410a, it will become gas of 50-50 R-32 and R-125, which then diffuse and  mix with air. Since air in the room is far more than the leaked R-410a, the oxygen molecules will surround (i.e. collide with at some rate) R-32 molecules just like when pure R-32 is leaked. Then how does R-125 make R-32 inflammable under the same condition?
The only way that I can imagine for this to happen is that R-125 and R-32 molecules combine to form some aggregation of a couple of molecules in air which prevent the reaction with oxygen to happen - do they? What are other ways that can make this (i.e. the title) happen.

Aggregation is not a part of the picture, nor is it needed. Think of it this way: will the air with just a few molecules of R-32 be flammable?

For many compounds, there are tables about lower and the upper experimentally determined flammability limits.  That is, what is the minimal concentration / maximal concentration of a compound in air (typically) or a specific gas to be susceptible to burn / to oxidize / to explode.  It is not a simple scalar property you may add by mixing gas A and B to obtain a flammability.

There are many compounds that suppress burning even in low concentrations. The mechanism is that they interfere with the reactions that propagate combustion (often by terminating some radical process). Combustion is a bulk process so the presence of small amounts of molecules that interfere can stop the process happening.

Who else saw the title edit and cracked up?

I think there must be distinction between flammability and explosive limits. The former is about  composition of potential fuel in contact with air, while the latter about its gaseous mixture with air.   Like how much of N2 can be in CH4 to be still able to  light it up, versus how much of given CH4/N2 mixture, mixed with air,  can explode.


text:
Products of the reaction between cyclohexanone and N-bromosuccinimide
Problem

Treating cyclohexanone with N-bromosuccinimide in the presence of carbon tetrachloride gives A. A on treatment with aqueous KOH gives B. Deduce the structures of A and B.

Solution

 A is 2-bromocyclohexanone and B is potassium cyclopentane carboxylate.

Questions

In the first step, why did the bromine atom get attached to the second position? The carbonyl group, being electron withdrawing, should direct the incoming electrophile away from it.
In the second step, I am quite sure that the product mentioned is a result of Favorskii rearrangement. But there is also a possibility for the hydroxide ion to substitute bromine. Is this possible? If yes, why is the substituted product not a major product? If not, why?

With regard to your first question - consider the role of keto/enol tautomerism. There is an error in the stated answer; B is not cyclohexane carboxylate potassium salt.

Count the number of carbons!

@Waylander Could you elaborate your first statement?

It isn't the ketone that brominates, its the enol. Google is your friend! Search "enol bromination"

That explains the first part. Thank you!

I don't have a definitive answer for 2 but it is like a E2 vs SN2 for secondary haloalkanes. The solvent, temperature, and concentration will favour one or the other. See the last link below. However, it is also important to note that the hydrogen is rather acidic so it will likely favour deprotonation rather than substitution.                              https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkyl_Halides/Reactivity_of_Alkyl_Halides/The_Reaction_of_Alkyl_Halides_with_Hydroxide_Ions.

Also consider the steric factors; bromine is a large atom and the hydroxide ion must attack from the reverse face

thank you lots! That's convincing enough.


text:
Equilibrium composition of methane–water system under given conditions

Calculate the equilibrium composition of a reactive system consisting of an equimolecular mixture of $\ce{CH4}$ and $\ce{H2O}$ at $\pu{900 K}$ and $\pu{1 atm},$ considered to be an ideal mixture of ideal gases. Under these conditions, the equilibrium constants of the two possible reactions are
$$
\begin{align}
\ce{CH4 + H2O &<=> CO + 3 H2} &\quad K_{p,1} &= 1.307 \tag{R1}\\
\ce{CO + H2O &<=> CO2 + H2}&\quad K_{p,2} &= 2.210 \tag{R2}
\end{align}
$$

I have tried to solve it by expressing each $K_{p,i}$ independently, considering there are five constituents in this ideal mixture and that $\ce{CH4}$ and $\ce{H2O}$ are the starting compounds, with $n^0 = 1.$ I associated a different degree of advance, $\xi_i,$ to each of the reactions and solved the two-equation system, but it seems that this is not the proper solution because the reactions are not independent.
Can anybody figure out a solution?

Editing note: both $K_p$ values had a comma which I interpreted as a decimal separator, but it might as well be a thousands separator. I'm too lazy to look up a plausible $K_p$ value, so please be aware.

Vandalism is not accepted behaviour even over own posts.

Let
$$moles\ CH_4=1-e_1$$
$$moles\ H_2O=1-(e_1+e_2)$$
$$moles\ CO=e_1-e_2$$
$$moles\ CO_2=e_2$$
$$moles\ H_2=3e_1+e_2$$
where the $e_1$ is the number of moles of $CH_4$ destroyed and $e_2$ the number of moles of $CO_2$ that are produced.
So the equilibrium equations are going to read:$$\frac{(e_1-e_2)(3e_1+e_2)^3}{(1-e_1-e_2)(1-e_1)(2+2e_1)^2}=K_1$$and $$\frac{e_2(3e_1+e_2)}{(e_1-e_2)(1-e_1-e_2)}=K_2$$

@ShoubhikRMaiti. This is the start of how to solve it.  I don't want to reveal too much.  What part of this don't you understand?

Ok I see, you are giving a hint answer. I think you should change the wording of number of times each of the two reactions has been executed, I had to read it a couple of times to understand. Maybe change it to extent of reaction or something like that. Also, hint answers are discouraged nowadays: https://chemistry.meta.stackexchange.com/questions/4287/hint-answers-revisited

@ShoubhikRMaiti OK, I changed the wording to make things clearer.

The reactions don't each have separate numbers of moles in the denominator to get the mole fractions for each.  The total number of moles for both reactions is $(2+2e_1)$.  And, the two equations are coupled, so  you must solve these two non-linear algebraic equations simultaneously.

–1 for improper notations and terms like "$moles\ CH_4$", "number of moles" and also, this is a hint, not an answer. Please post suggestions and hints in the comments section shall you decide not to reveal the complete solution.


text:
Is it possible to distinguish between cyclohexanol and cyclohex-2-en-1-ol with simple chemical test?
Primary, secondary and tertiary alcohols can easily be distinguished by using reagent tests such as:

The triiodomethane test for $\ce{R-CH(CH3)OH}$ alcohols.
Lucas' reagent to distinguish between three alcohol groups.
A combination of reagents to distinguish between three alcohols. For example, dichromate to distinguish between tertiary and primary/secondary alcohols and subsequently Fehling's solution to distinguish between primary and secondary alcohols.

Would it be possible to distinguish between two secondary alcohols? For example, between cyclohexanol (1) and cyclohex-2-en-1-ol (2) using simple chemical tests?

There are reagents like MnO2 which will selectively oxidise allylic alcohols, and can therefore act as a basis for chemically differentiating the above two compounds. I'm not sure that's what you are looking for, though? Are you looking for something that can be done quickly in a test tube in a high school-ish lab?

An alternate approach for this specific case would be Br2 reaction with the alkene

Yes, it is possible to distinguish.

Yes. Try Br2 in CCl4/KMnO4 in either acidic or alkaline medium/HBr


text:
could potassium cyanide be turned into potassium nitrate in a single pot reaction?
Is there a way one could turn potassium cyanide into potassium nitrate (actually oxidize the cyanide anion)? It's fairly easy to convert it to ammonia by hydrolysis, but the conversion of ammonia to nitrate is difficult (bionitrification is out of question). Would peroxide be able to attack it?

You could try adding copper nitrate solution, the copper will precipitate out as cyanide leaving you with a potassium nitrate solution. Pour the solution in a dish and let the water evaporate and you will get potassium nitrate salt.

What is the background of this question ? Why do you want to do so ?

it's more of a thought experiment. I saw someone make cyanide from charcoal and pot ash in a furnace. I was wondering if it was possible to obtain gunpowder from it with equally primitive means.

@Francis L. Potassium cyanide cannot be made from charcoal and pot ashes, for lack of $\ce{N}$ atoms. Pot ashes are made of different substances, none of them contain $\ce{N}$ atoms. Some nitrogen derivate has to be added to the mixture.

I didnt mention its done under atmosphere of wood gas. It's called a bucher process

If we were to directly oxidise potassium cyanide, it would form potassium cyanate, which then oxidises to nitrogen gas and CO2. In this way, a one-pot reaction is unfeasible.
The viable method would be to dissolve excess potassium cyanide in water, before adding 1 Eq.(https://chemistry.stackexchange.com/a/64924/77157) silver nitrate. These reactants would form silver cyanide and aqueous potassium nitrate in a precipitation reaction.
The silver cyanide precipitate can be extracted to leave water and potassium nitrate. Of course, we can pour the solution into an evaporating dish, so the reactants form potassium nitrate.


text:
Calculating electric reduction(or oxidation) potential for complex ion formation equation?
I would like to calculate the standard electric potential($E°$) for the following equation.
$$\ce{Pt^{2+}(aq) + 4Cl-(aq) → [PtCl4]^{2-}(aq)}$$
$K_f = \pu{1.0 * 10^16}, T =\pu{ 298K}$
I figured out
$ΔG° = -nFE° = -RT\ln K$.
So, $E° = (\frac{RT}{nF})\ln K = (\frac{0.06}{n})\ln K (\text{approximated }\frac{RT}{F} = 0.06)$.
However, I'm having trouble finding $n$, or the number of moles of electrons transferred.
I'm not sure whether electrons are transferred in an complex ion formation equation.
Could you please help me with this confusion?


text:
How to evolve hydrogen from aluminum when the household drain cleaner is safe to use on aluminum?
I live in Japan. I would like to evolve hydrogen to demonstrate to my son buoyant balloons (we have helium aerosol cans here which barely fill one balloon). There are drain cleaners here with low concentration of sodium such as up to 4% and say safe for aluminum (probably the calcium hydroxide also contained). I have obtained some wire and batteries and I think electrolysis could help- what should I do  for the best chance of evolving hydrogen?

It depends on the nature of the sodium compound. If it is sodium carbonate, the reaction with aluminium is extremely slow. If it is sodium hydroxyde, a 4% solution will quickly react with aluminium and produce a lot of $\ce{H2}$ bubbles. But take care! The bubbles have a tendency to sweep and fill the flask. Take also care and  of the droplets emitted when the bubbles blow up. They contain $\ce{NaOH}$ which is corrosive.

Getting a little hydrogen at home is easy. Getting enough hydrogen at sufficient  pressure to inflate a balloon is another story, though.

@Ivan: Mylar balloons can be purchased without helium inside and can be inflated easily to the point where they will float in air. But they will still look floppy.

First, any drain cleaner that is at all effective is probably not safe for a child to use, even under adult supervision.
Second, you can easily produce enough hydrogen to inflate a plastic bag, perhaps one to four liters, using electrolysis. Unlike an elastic balloon, the bag will not resist inflation appreciably until full. Try carbon electrodes (e.g., from zinc-carbon cells) with $\ce{MgSO4}$ (Epsom salt) in water as the electrolyte.
You can also capture oxygen generated and test it with a glowing wooden splint.
It is an exercise for the experimenter to calculate how many ampere-hours it would take to inflate the specific bag.
BTW, you might want to perform this outdoors, in case a spark sets off the hydrogen!

How do I use a plastic bag to collect Hydrogen?


text:
Why is the variable on the y-axis of Maxwell's Speed Distribution Curve 'fraction of molecules'?
I came across the following graph in my textbook showing the kinetic energy distribution of the gas molecules in a gas sample. Now, the thing is the textbook is telling that the area under the curve gives us the fraction of molecules with energy E. Suppose $dN_{E}$ represents the number of molecules with energy E, $N_{T}$ represents the total number of molecules and $E$ represents the kinetic energy of the molecules. Then, the fraction of molecules with energy $E$ is given by $\frac{dN_{E}}{N_{T}}$, right? So this actually represents the variable in the $y$-axis. Now, let us take the area of a small strip under the curve. It is given by $\frac{dN_{E}}{N_{T}} × dE$ which is clearly not equal to the fraction of molecules with energy $E$. Then why is the book telling that the area under the curve gives us the fraction of molecules with energy $E$? It clearly does not. I just proved it. If the area under the curve really has to represent the fraction of molecules with energy $E$ then instead of fraction of molecules i.e. $\frac{dN_{E}}{N_{T}}$ on the $y$-axis, the variable in the $y$-axis should actually be $\frac{1}{dE}\frac{dN_{E}}{N_{T}}$. Now if we calculate the area of a small strip under the curve :
$\frac{1}{dE}\frac{dN_{E}}{N_{T}} × dE = \frac{dN_{E}}{N_{T}}$
The quantity $\frac{dN_{E}}{N_{T}}$ actually gives the fraction of molecules with energy $E$. Thus, my book is wrong, isn't it? Please can someone clarify if what I have done is correct or not. Can someone explain?

y represents the fraction density.  y . dE represents the fraction dN/N. Integral E0 to infinity of y.dE then represents fraction E>E0.  y cannot be 1/dE . dN/N, as the bottom dE does not follow the upper dE.

Please elaborate.

What part do you not understand ?

Is the variable in the y-axis $\frac{1}{dE}\frac{dN_{E}}{N_{T}}$ or is it $\frac{dN_{E}}{N_{T}}$ or is it something else?

The function on the graph is not the fraction, but the fraction density. The fraction is get by multiplying by dE, or integrating over E interval. So the former, but rearranged, as it is formally wrong. It is derivative of the cumulative function f(E)=N(E_mol<=E)/N_T. . Your original expression (1/dE).(dN/N).dE creates false impression both dE are the same, mutually clearable differentials.

The graph is drawn between 'fraction of molecules' and 'kinetic energy'. My final question is what exactly is 'fraction of molecules'? Can you please write 'fraction of molecules' in mathematical form? (For example, in the graph between speed and time, speed in mathematical form can be written as $\frac{dx}{dt}$ and time is $t$. So, graph is actually being drawn between $\frac{dx}{dt}$ and $t$). In this way, could you tell me what 'fraction of molecules' and 'kinetic energy' mean in mathematical form? Can you represent them in terms of variables like I did with speed and time?(Please help me)

Let us continue this discussion in chat.


text:
Why catalyst does not alter the entropy of a reaction?
In my textbook it is written that "Catalyst does not alter the entropy of a reaction". But why? There is literally no explanation given in my textbook as to why this is true. Could you please explain why this is true? I mean if a catalyst speeds up or slows down a reaction so shouldn't it affect the entropy of a reaction? I am really confused. Please explain. Would greatly appreciate it!

Entropy is a state function, what means its value depends only on the system state, not on the path, how the system got from its initial to the final state. So it does cause entropy change of the reaction, but may cause entropy change by its presence, independently on if it is a catalyst or not.

Errata: ..so it does not...

TL;DR Catalysts by definition do not alter the magnitude of changes in thermodynamic properties that accompany conversion of reactants to products, they only alter the conversion (reaction) rates.
The reason a catalyst does not alter the entropy of a reaction is that this is one part of the definition of a catalyst! Ideally none of the changes in thermodynamic functions accompanying a reaction are altered by the catalyst. However, that is a circular and not very helpful argument. The fundamental reason is that catalysts are not consumed in chemical reactions. Since the products and reagents remain the same, it follows from Hess's law that catalysts do not contribute to thermodynamic changes associated with the reaction.  Only the rate with which reactants and products interconvert (kinetics) changes.
Catalysts alter the reaction rate by altering the rate of encounter of the reagents or products or by assisting in formation of new lower energy (more stable) reaction intermediates. A catalyst may be as simple as a material that increases the proximity of reagents by increasing their local concentration, without otherwise altering their reactivity (bond enthalpies). That can be the case in either enzymes or solid supports. However catalysts often also alter reactivity by weakening specific bonds. An example of a catalyst that provides both functions is platinum.


text:
Measuring Cylinders -- small or big?
My teacher often recommends the usage of a smaller/more narrow measuring cylinder over a larger one while performing experiments but why is that? Does it help in accuracy in any way?

The relative cylinder resolution is given by the height resolution divided by the actual  height difference. Which of the narrower versus wider cylinder gives better value of this parameter for a given volume?

Related answer: https://chemistry.stackexchange.com/a/37788/101286

Adding an excess of ten drops in a narrow cylinder is visible : it does make a difference. It does not in a wide cylinder. The accuracy is better in a narrow cylinder.

That is why a volumetric container always has a narrow neck.

Well, that explains why my 1 cm long thermometer, with 10 cm diameter, is so insensitive and hard to read! ;-)


text:
Why can zinc porphyrins assume 5 coordinate geometries?
It has been reported that 5-coordinate geometries are possible and stable for Zn(II) porphyrin complexes. How do these complexes form and why are they stable when other 5-coordinate geometries of Zn(II) and other d10 metals are not stable?

First off, who reported it?  Second, is this really any different from PCl5 or SF6?

See for example 10.1021/ic50194a058

Honestly I don't know if its any different, I was hoping someone could give a quick rundown on how ligand field theory works in the d10 case where crystal field theory seems to suggest ligation would not occur, with a focus on how 5 coordinate geometries can emerge.


text:
Why on changing temperature of a reaction its activation energy and pre-exponenetial factor do not change significantly?
My teacher told me that on changing temperature of a reaction, its activation energy ($E_{a}$) and pre-exponenetial factor ($A$) do not change significantly and hence we can assume them to be constant on changing the temperature. But my question is why? Why don't they change significantly? Even in the graph given in my textbook (given below), on changing temperature they have assumed activation energy to be constant. Is it really true? Please clarify. Would greatly appreciate it!

The activation energy is determined by the electronic interactions between the reactants as they collide and form a transition state, and not on temperature. The pre-exponential  terms does depend on temperature but the effect is generally small, it mainly depends on how many times the reactants collide per second.

On what factors does the activation energy depend on? Could you please tell me?

This may help.

As I write in my comment, it depends on the electronic properties of the molecules involved, ie. the energy of the transition state complex  which is sort of half reactants and half  product. Calculating the energy of this is very hard particularly as there is little or no experimental data. If the reaction is AX+B $\to$ A-X-B $\to$ A+BX it is the energy of A-X-B.


text:
Difference between Tishchenko reaction and Meerwein-Ponndorf-Verley reduction
For the Tishchenko reaction, Al(OEt)3 is used as reagent where an aldehyde is oxidized as well as reduced, so it's a redox reaction.
Whereas in Meerwein-Ponndorf-Verley reduction, an aldehyde is converted to secondary alcohol and acetone with the same reagent and this reaction is reversible with the Oppenauer oxidation.
So, my question is what is the difference between these two reactions i.e what is the difference in mechanism of this reaction that lead to different products or there are some different conditions other than the reagents which changed their products?

Generally MPV reduction uses $\ce{Al(O^iPr)3}$.

Okay, but same can be used for Tishenko reaction also ? Is there any specific reason for it's use only in MPV.

See here in reactions

I am not sure, but I think the only difference is the presence or absence of acetone. In MPV reduction, acetone is present so it can go the reduction route, however, in Tischenko there is no acetone present so it goes to disproportionation route. See the mechanisms here and here

$\ce{Al(O^iPr)3}$ is used specifically in MPV reduction because $\ce{^iPrOH}$ is the reduced form of the acetone ($\ce{^iPr=O}$). If you look at the mechanism you will see why they do that. If you use something else, you would get a mixture of a lot of compounds. This way there are only 2 components in equilibrium $\ce{^iPr=O +carbonyl<=> ^iPrOH +alcohol}$


text:
Gibbs Free Energy Spontaneity Criterion Derivation
In Atkins's Physical Chemistry, spontaneity criterion is derived from Clauisius inequality as follows:
Firstly, system is initially at the same temperature T as surroundings. When a change in the system (its state) occurs, there is an corresponding entropy change in the system and a transfer of heat between system and surroundings and Clauisius inequality for this process can be written as: $$dS -\frac {dq}{T} \geqslant 0$$
If process is carried at the same temperature and pressure of a system than $$ dq = dH $$
If we plug dH in Clausius inequality, multiply by T and rearrange what we get is: $$ dG = dH - TdS \le 0$$
Which should prove Gibbs free energy spontaneity criterion for systems at constant p and T. Problem here is that T in the equation for G is of the surroundings not of the system and the whole point of G is to forget about surroundings. Answer you can give to this question is that surroundings and system remain at the same temperature during state change of the system, but if so how can there be any heat transfer between them, for they are already at thermal equilibrium and dq should is zero. When taking Gibbs free energy as spontaneity criterion, system needs to stay at the same pressure and temperature. To keep the system at the same temperature, heat generally needs to be exchanged with the surroundings and if so, heat exchanged with surroundings will change its temperature, so T in Clausius ineqaulity generally is not a constant, it can only be thought as constant if surroundings is uncomparably bigger than the system so that its heat capacity is uncomparably bigger than of system (which in most cases really holds) only in that case T in Clausius inequality can be taken as T of the system. Problem with this is that such an approach gives only approximate solution and equation for G as spontaneity criterion IS NOT AN APPROXIMATION, it holds always regardless of how big system is compared to the surroundings. So, I am not sure this derivation in Atkins's textbook is actually correct for all scenarios not only when surroundings is uncomparably bigger than the system. What are your thoughts?

There is a much better derivation of this in Denbigh's book, where the saturation is laid out much more precisely.

Will check it out, I actually have found better derivation in one thermodynamics textbook already. What are your thoughts on this derivation? Do you think it holds only when surroundings stay at the same temperature?

When they start using differentials to describe potentially irreversible processes, it makes me cringe, and I immediately discount it.

You mean exact differentials?

No, I mean any differentials.

What is a problem in using differentials for irreversible processes?

The problem is that the state functions are defined precisely for thermodynamic equilibrium states, and, in an irreversible process, the system passes through non-equilibrium states.  So one needs to make the additional approximation that, in an irreversible process (where conditions can vary with spatial position within the system), there is approximate local thermodynamic equilibrium and the state function variations can be integrated over the volume of the system.

Yes, but I looked at the derivation you mentioned (Denbigh) and he used both differentials and Clausius inequality to derive spontaneity criterion which refers to both reversible and irreversible processes. He used differentials to describe potentially irreversible processes to conclude.

To make it clear, I do agree with what you are saying, although I think I've never seen someone not using differentials for irreversible processes in any textbook (including Denbigh: Principles of Chemical Equilibrium), so yes they use a concept of approximate local thermodynamic equilibrium. If we suppose local equilibrium, do you think that this derivation in Atkins is wrong?

Well, I still don't like it, so I'm going to write out the development using deltas (between the initial and final states) and you can compare for yourself.  Stay tuned.

I see you are retired ChemE, I am student of ChemE at my last year. There is a lot to learn in thermo though.

So, you agree that this derivation is applicable only when surroundings is much bigger than the system?

We are talking about a system that is in contact with an ideal isothermal reservoir.  Such a reservoir would have an infinite mC and thermal conductivity so that its temperature could never change even at the interface with the system.  A good approximation to this might be one (like an ice bath) in which a solid phase and liquid phase are in equilibrium at the melting point.

Yes, so you agree that this is an approximation since our surroundings may be small compared to the system if our isolated system is not big. In that case, this derivation wouldn't work.

One reason I am opposed to using differentials for irreversible processes is as follows:  In an irreversible process where temperature and specific entropy both vary with spatial position within the system, what is TdS supposed to mean?  Also, since specific volume and force per unit area within the fluid both vary with spatial position, what is PdV supposed to mean; in an irreversible process, the force per unit area varies not only with spatial position but also with direction at each spatial location.

For a system in contact with a constant temperature reservoir at the same temperature as the system in its initial state, and also in contact with a constant pressure surrounding at the same pressure as the system in its initial state, $$\Delta U=Q-W=Q-P\Delta V-W_{npv}$$where $W_{npv}$ is the amount of non-PV work done by the system.  In addition, from the 2nd law of thermodynamics, we have $$\Delta S=\frac{Q}{T}+\sigma$$where $\frac{Q}{T}$ is the amount of entropy transferred to the system from the surroundings across the interface between the system and surroundings (recall that the latter is always at the initial temperature of the material comprising the system).  Note that this does not say anything about how the temperature may be varying with location and time within the internals of the system during the process.  In addition, the parameter $\sigma$ represents the amount of entropy generated within the system due to irreversibilities during the process. Note that $\sigma$ can only be positive (irreversible process) or zero (reversible process)
If we combine these two equations, we obtain:
$$\Delta U=T\Delta S-T\sigma-P\Delta V-W_{npv}$$or$$\Delta G=-W_{npv}-T\sigma$$ According to this, if no non-PV is done by the system on the surroundings during the process, $\Delta G$ must be negative (irreversible) or zero (reversible).

Yes, thank you for more detailed analysis. I think that Gibbs energy spontaneity criterion can be derived better without approximation of constant temperature surroundings which I found in Modern Thermodynamics by Prigogine and Kondepudi. This criterion is derived from three principles: 1) Definition of G 2) First Law 3) General entropy balance which is equation you've written in your answer. In that way, there are no restrictions of constant temperature surroundings.

I had an interested discussion with someone once about what about non - spontaneous processes like electrolysis which tend to increase G of closed systems at constant p and T. It may seem that such processes violate second law. I said that such process can only happen if somewhere in our closed system G decreases more than its increasing in electrolysis since in many cases electrolysis is powered by battery in which spontaneous reaction lowers G more than electrolysis increases it, so that spontaneity criterion is satisfied. Do you agree with me?

A small note, Chemists nowadays use the convention $\Delta U=Q+W=Q-p\Delta V$ for the first law, but physicists still use $Q-W$ as I understand it.

As an engineer, I was taught (and still use) Q-W.


text:
How is the atomic number of a beta particle zero?
In my textbook, $\ce{β-}$ particle is denoted as $_{-1}e^0$. This means that the atomic number of $\ce{β-}$ is $\ce{-1}$. But that doesn't make any sense. Also, how is the mass number of $\ce{β-}$ zero? I mean it must have some mass right? How can it be massless?

I will start you off: mass number is not mass. Mass number is total protons plus neutrons, so it is zero for an electron.

But what is the meaning of having atomic number equal to $-1$? Total number of protons equal to $-1$? That doesn't make any sense.

An electron has a negative charge, hasn't it ?

Get a better textbook, the authors of this one have no idea what they are writing about, or it's poor editorial job. Either way, beta particle is denoted with e, e⁻ or β⁻. All particle symbols are upright and the charge is denoted in a superscript.

Actually, it makes sense in a pedantic or slightly silly way. A proton can be defined as having atomic number (Z) of +1. After all, it is the nucleus of the simplest hydrogen isotope and atomic number is number of protons in an atom. So a neutron has Z = 0 because it has no protons. But, in free space, neutrons spontaneously decay, in roughly 15 minutes, to protons, electrons and electron antineutrinos. So, ignoring the neutrino, assigning the electron as Z = -1 keeps the bookkeeping honest, i.e., 0 goes to +1 plus -1.

Also, it is best not to dwell on such contrived stuff: people who write beginning level chemistry books often strive for an artificial consistency, leading to what you have encountered. I have seen it as well and been annoyed by it. So what @andselisk says is good advice: get a better book, if you have the opportunity.

@EdV you forgot the antineutrino. β⁻ decay is not only emission of an electron.

@IncnisMrsi I did not forget: read my comment again.


text:
How are intermolecular forces between non-metallic substances affected when the substance is heated to its melting point?
I have 3 main questions that I am unsure of:

Are intermolecular forces acting simultaneously causing attraction between particles in a sample of a substance?
Example: Hydrogen bonding and dipole-dipole attraction present in a substance both influencing the attraction between particles.

If there are multiple intermolecular forces present, would they be overcome in an order?

What actually happens to the intermolecular forces between particles as a non-metallic molecule is heated?

I'll try to answer this using $\ce{H2O}$ as example. 1) I didn't clearly understand what kinda overlap but in general sense will they overlap? Yes, a H-bond and any other intermolecular forces could overlap. 2) Yes, induced ones should be overcome easily. I'm not confident about the order so I'll not comment about that. 3) The forces become negligible because at higher temperatures the interaction becomes negligible due to high speeds.

Yup, just rewrote the question. Thanks for the answer.

From your first question I assume you're asking if the intermolecular forces are causing attraction between particles or can there be repulsion as well? For this you can find info on Wikipedia. There are attractive as well as repulsive intermolecular forces. But if as you mentioned the two simultaneous forces are only attractive, then yes only attraction is caused. But repulsive forces due to electron clouds always are in existence, we just neglect them for simplicity.

@Desay better speak of kinetic energy rather than speed, as it is the quantity of relevance.

First of all, multiple intermolecular forces could be present between molecules, and, yes, all forces influence attraction, binding and etc. However, if you study any chemical system, you will see that is usually dynamic.
It all depends on conditions like temperature, pH and especially compound properties, etc. For example, the drug captopril (an antihypertensive drug) could form multiple bonds with the protein angiotensin-converting enzyme (ACE).:

How about one specific atom of a molecule? In theory, one atom could form multiple intermolecular forces at a specific time (let's say picosecond), however, it is not practical - energetically unfavorable. Imagine that atom should be in such a position, an electronic configuration that reduces its potential energy and also form multiple intermolecular bonds. It is possible that such systems exist, but they are not common.
Second answer:
Yes, intermolecular forces have specific values of bonding energy:

For example, it takes less energy to break London dispersion forces than to break hydrogen bonds.
Third answer:
It all depends on the level you want to know... Basically, when you heat molecules (give them more energy) they absorb it and move faster.  Vigorous movements make bonds break. For example, water molecules at 100 °C start to move so fast that almost all its intermolecular forces are broken. Fun fact: you can smell lemons at room temperature because it has volatile molecules (molecules that form very few intermolecular bonds) called Limonene.

The forces among molecules don't change with temperature; the kinetic energy of the particles is what changes
All the intermolecular forces acting on molecules contribute to the net forces making a liquid hold together. They vary in strength but they all matter in the aggregate liquid.
But the reason liquids boil isn't because the forces get weaker or change with temperature: it is because, at higher temperatures, the average molecule has more kinetic energy (this is essentially the definition of temperature). At any given temperature the momentum or kinetic energy of the molecules shows a distribution of values (some move very fast; some slow). There are a very large number of molecular interactions happening all the time that redistribute that energy among the molecules to give a particular statistical distribution of speeds. Evaporation and, eventually, boiling occur because a proportion of molecules have enough speed to escape the liquid and that proportion depends on the temperature. The intermolecular forces don't need to change at all for this to happen.
So, to answer the questions in order:

Yes. All the forces contribute to the attraction between molecules
No. The strength order doesn't matter, just the net forces. And they are not "overcome" during heating it is just that a larger proportion of the molecules have enough speed to escape from the liquid.
Nothing. The net forces remain the same on heating. It is just the average kinetic energy that changes (and therefore the proportion of molecules in the liquid having enough energy to escape the net forces holding the liquid together.)


text:
Is the rate of a zero order reaction always independent of the concentration of its reactants?
Consider the following reaction :
$\ce{A + B}$ $\rightarrow$ $\ce{Products}$
where order of the reaction w.r.t. $\ce{A}$ is $-1$ and order of the reaction w.r.t. $\ce{B}$ is $+1$. So, overall order of the reaction $=(-1)+(1) = 0$. So, it is a zero order reaction. My teachers told me that in a zero order reaction, the rate of the reaction is independant of concentration of the reactants. Below I have derived the integrated rate law for the above reaction. You can clearly see that in the integrated rate law there are terms of concentration of $\ce{A}$ and
$\ce{B}$. Obviously, the integrated rate law is telling us that the rate of the reaction depends on the concentration of the reactants. Then why does everyone say that in a zero order reaction the rate of the reaction is independant of concentration of the reactants. Why?

Is it possible that a reaction could be of order $-1$ with respect to a reactant ? The reaction rate should decrease when increasing the concentration of the reaction. Does it make sense ?

It may occur for complex reaction with catalysts and much more.

I think your teachers were talking about elementary zero order reactions. If you increase the concentration of all the reactants in a complex zero order reaction uniformly, the rate doesn't change.

In this case it is important to differentiate between the overall order of the reaction and the order with respect to each species.  The overall order of the reaction is zero.  However, the rate law shows us that this is more complex that a simple unimolecular zero-order reaction, where rate = k.
As @M.L. pointed out, if you changed the concentration of both reactants by the same factor at the same time, then there would be no net change in the rate.  However, if you change the concentration of only one reactant at a time, then the rate would change according to the order with respect to that reactant.
Let's imagine the following situation for your reaction, where the rate law is $\ce{rate = k[A]^{-1}[B]^1}$ and k=0.500 mol/Ls.

Trial
A
B
Rate(mol/Ls)

1
0.10
0.10
0.50

2
0.10
0.20
1.0

3
0.20
0.10
0.25

4
0.20
0.20
0.50

You can see that when we double the concentration of B, the rate doubles.  This makes sense, since the reaction is first order in B.  In trial 3, we double the concentration of A and the rate decreases by half, since the reaction is -1 order in A.  If we double the concentration of both reactants, as in trial 4, the resulting rate is the same as in trial 1, because the change with respect to each reactant is cancelled out.

When you look at the overall reaction, it is if you have an equal concentration of each reactant, then by the reaction progressing or adding equal amounts of both, the reaction rate won't change. For example, if you have 2M of both A and B, $[A]^{-1}[B]^1 = 1/2*2=1$, and increasing the concentration to 3M for both won't change the rate since $1/3*3=1$. The differential rate law applies to only one single species, so unless you have equimolar amounts of each reactant or make an approximation using an excess of one, you cannot apply the differential rate law.


text:
Major product of the nitration of benz[cd]indol-2(1H)-one

Decide major product for given reaction of electrophilic aromatic substitution:

Is the major product a because of steric hindrance, or is it b because of hydrogen bonding?

According to this patent here, N-ethyl 1,8-naphtholactam nitrates in the 4 position i.e. structure a. It seems entirely likely the N-unsubstituted material will do likewise.

I Got the answer but I didn't understood "It seems entirely likely the N-unsubstituted material will do likewise." Meaning

Meaning I cannot find a reference to the nitration of  1,8-naphtholactam so I've quoted the nearest example. In my opinion 1,8-naphtholactam will also nitrate in the same position.

@Waylander Is iprdb.com a reliable source from your experience? Don't you think a conventional link to Google Patents (which they seem to copy content from without any quality control) would be better?

The reference is what Google found for me. It references the US patent so in this case it seems reliable. The result is in accord with my expectations


text:
Are there any actual uses of isodiaphers?
While studying atomic structure, I came across the terms isotopes, isobars, isoelectronic species, isotones and isodiaphers. While I can accept that the classification of isotopes, isobars, isoelectronic species and isotones may be useful, I do not understand what could be the use of species with the same difference between number of neutrons and protons (isodiaphers).
Is there any place in chemistry where they have a practical use (like in experiments, theories or laws), or is it just a useless classification term?

See: https://chemistry.stackexchange.com/questions/29152/what-are-isodiaphers

In practical radiochemistry, the term is rarely needed but it is not useless. In particular, isodiaphers are used in radiochemistry to describe chains of alpha decays.
I would expect, the relative frequencies of the practical use of the related terms is about
"isotopes" > "isobars" > "isodiaphers" > "isotones".
Isobars are used in radiochemistry to describe chains of beta decay, therefore their importance might be similar to isodiaphers for alpha decay. However, isobars are also very important to describe fission yields so that in total they might be used more often than isodiaphers.

The isodiapher concept provides a means of organizing stable nuclei.
In this corner...
An atomic nucleus with more than one nucleon (if it is smaller than a neutron star) essentially represents a battle between two forces through which the nucleons interact:

electrostatic forces, which tend to favor more neutrons for a given mass number due to repulsion between like charges

internuclear forces, which are more attractive between a proton and a neutron than between like nucleons and thus tend to favor equal numbers of neutrons and protons.

The neutron-proton difference thus represents a measure of the extent to which electrostatic forces intervene and prevent an atomic nucleus from reaching equal numbers of protons and neutrons.  Isodiaphers therefore represent a similar balance between these forces and thus have similar stability characteristics, at least if the atomic masses are close enough.
Parallel Lines
Perhaps the most prominent example of this similarity involves one family of isodiaphers, having a difference of zero.  These are atoms where electrostatic forces are small enough to allow the internuclear forces their way with proton-neutron balance.  Included in this series are the only four odd-odd nuclei that are fully stable against radioactive decay: deuterium, lithium-6, boron-10 and nitrogen-14.  In other words, odd-odd nuclei are stabilized by the specific combination of maximal internuclear attraction with small electrostatic repulsion.
Beyond nitrogen-14, this zero-difference family becomes more like other isodiapheric families.  Because the electrostatic/internuclear force balance shifts only slowly with increasing mass number, we see the fully stable nuclei arranged into parallel, overlapping families having a common nucleon difference.  With an even difference we see the isodiaphers at every other element, like neon-20, magnesium-24 and silicon-28 in the zero-difference series;  and titanium-48, chromium-52 and iron-56 in the +4 series.  This occurs because of the stability difference between even-even and odd-odd nuclei heavier than nitrogen-14.  When the nucleon difference is odd, this parity effect is absent and then we can see isodiaphers at adjacent elements, like neon-21, sodium-23, magnesium-25, aluminum-27 and silicon-29 (+1) -- which includes all five of those elements versus only three of the five for the heavier part of the zero-difference family given earlier.