text:
Endothermic reaction of solid ammonium nitrate

The above process is endothermic by mixing water. When the products (solution containing $\ce{NH4+}$ ions and $\ce{NO3-}$ ions) are heated up again, will it form back $\ce{NH4NO3}$ (solid ammonium nitrate) that is the original reactant?

I would be careful about heating an ammonium nitrate solution sufficient to drive off the water as solid ammonium nitrate is explosive. However, in principle, if you remove the water you get back solid ammonium nitrate

There is many salts with endothermic dissolution. If stable enough, they can be retained back as solid ionic matrix by spontaneous evaporation, providing the relative humidity is low enough. Heating speeds up the process but may affect the salt stability. In some cases it may lead to desintegration or explosion.

@Waylander  Hi, what about if I let the nitrate solution evaporate it in the sun, it will still give me back solid ammonium nitrate right? Thanks

@Waylander I am a high school student. Dont you think it is useful for designing an air cooling system. You just have to get some ammonium nitrate and water to cool the air. After that, the solution produced is evaporated in the sun to yield back the powder again.

@MaximusSu  Sure if you leave the solution in the sun you will get ammonium nitrate crystals back eventually. There are far more efficient ways of air cooling though.

In your high school gymnasium or medical office, they likely have packets of ammonium nitrate, with a sub-packet of water,  to make an "instant" cold compress for injuries. There are also ammonia absorption refrigerators: https://en.wikipedia.org/wiki/Absorption_refrigerator


text:
How to decide whether diborane undergoes symmetric or asymmetric cleavage?
How does one decide whether diborane undergoes symmetric or asymmetric cleavage, given the lewis base which attacks it. Is it possible to ascertain this theoretically?
Sterics is mentioned as the reason at some places, including some answers on this site. This reasoning does work when comparing between the cleavage products when ammonia and trialkyl amine are used as bases but does not explain the symmetric products when other bases like $\ce{CO, PF3}$ are used as bases as proposed in this unanswered question.
Therefore, is it really possible to determine the products just based on the lewis bases theoretically? If it is then what factors need to be accounted for?

Does this answer your question? Why is diborane cleaved symmetrically by trimethylamine but unsymmetrically by ammonia?

Nope, I've already read that, but the reasoning behind "sterics" does not answer why CO,PF3 cleave it symmetrically. I've already mentioned that in my question. Infact that is why I posted this to clarify what are the other factors that are affecting this.

CO and PF3 are good pi acceptors, NH3 not so much. Might the appropriate combinations of relatively high energy B-H bonding orbitals take notice?

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text:
Analysis of reaction mixture by TLC and finding column chromatography condition
I am conducting an organic synthesis. Below is the TLC image.

I used dichloromethane:methanol:triethylamine=10:1:0.15 as eluent, but I stored it during 3 days, so it may be a little bit concentrated.

Question 1. I used toluene to dissolve sample. If I spot only toluene and run TLC, and stain the plate, then there emerges brown color near the solvent front. I waited enough time to let toluene evaporate before staining, but are there chances that spot 1 is of toluene?

Question 2. I am assuming that my desired product is basic, so I used triethylamine. When I fixed ratio of dichloromethane and methanol and changed only ratio of triethylamine, only the position of spot 7 changed and the others are unchanged. Then can I conclude that spot 7 is not part of spot 6 or tailing 2?

(sample is more concentrated than first image)

Question 3. There are some tailings. Does existence of tailing always mean the band widening when I run the column chromatography? I heard that when I do the column, I had better use as minimum solvent as possible. But to prevent tailing, should I dilute the sample more? (I think the size of spot was small enough)

Question 4. In UV, spot 6 looks like one big spot, but after staining it seems that middle is empty. Does this mean there are two or more other compounds in same position?

Question 5. Is spot 4 and tailing 1 is same compound? Is spot 6 and tailing 2 is same compound? If I can't determine in this result, what should I do?

Another TLC image. Spot 4 is more identifiable. (sample is more concentrated than first image)

Question 6. I assume that the longer distance spot moves, the more nonpolar it is. Even in very polar solvent system (i.e. dichloromethane:methanol=10:1) many spots are close to solvent front. To separate spot 1, 2, 3, and 4, how can I change the solvent system?

additional information

starting material : aza-crown ether, bromothiophene, Pd catalyst, P ligand.
expected product : N-substituted thiophene.
aza-crown ether and spot 7 is not drawn up by solvent without (aza-crown ether is detectable only by ninhydrin) So I think I should use Et3N to separate these two.
spot 3 is assumed to be Pd catalyst. bromothiophene and P ligand is not detectable by formaldehyde/H2SO4 staining.

Better you refer to some senior colleague at your place. I won't start any column using those conditions but what is the crude the material the amount the columns around etc it is crucial, togheter with possible major changes of the eluent. As for 1 answer should be no. #2 not really important #3a likely #3b no 4# likely but it might depend on the handling of the plate. #5&6 reduce polarity. Note that this is an attempt to help,only. Otherwise I would have answered. Also one should know if the reactants are there. The entire tlc is messy. Tailing is due to Et3N.

Finally the start line might have been to far from the eluent surface level. Togheter with a not well dried plate this means bigger spots. And to all: the capillaries for TLC should be extremely thin, not those called "for ematocrito".

@Alchimista Oh Is that start line too far? I line position at 1 cm from the bottom of the plate.

I have just guessed from the pics. And this would be important only if one doesn't dry the plate after deposition (and not existing with volatile eluents). But it is the last problem with your TLC

@Alchimista I used heat gun after developing. But is it okay to use heat gun if some compounds on TLC is volatile?

think about it. At least they won't be gases. The plate must be dry even after deposition. Otherwise the spot starts the process.

@Alchimista Okay. I would try and compare the results again.

Any aminothiophene derivative is not going to be very basic - similar to an aniline but will show up on UV. Any residual thiophene SM or be-brominated Sm will be at solvent front. My guess is spot 6 is product. I would run a gradiented column starting with DCM and going to 5% MeOH then strip with 10% MeOH and added Et3N.

You should have idea of which the product is. A synthesis does not come out of nothing. That is why you check it while is taking place. Most of the time you are not interested in the isolation of everything. Etc. Try to discuss also with a senior student and the friendly people around. If any.

@Alchimista Sorry, organic synthesis is not main job in my lab, so there are no senior to discuss about synthesis, actually I'm doing this nearly alone. (seniors mainly work for electronics) I major in chemical engineering, not chemistry, so there are no colleague to discuss, too.

@Alchimista I know that I don't have to isolate all the things, but all spots except one (spot 3) are not matched with starting materials so I was confused about what the product is.

@Waylander For some reason my bromothiophene starting material is not detected by UV. What do you think about question 4?

It could be co-eluting with the phosphine oxide from the ligand. Did you try the permanganate dip for visualisation?

@Waylander I have image with permanganate dip, it has a change near solvent front and middle of the plate, but it is not unclear that this middle position is exactly spot 6. I would try again. When I spotted only P ligand, after permanganate dip, it was near the solvent front.

The oxide of the P ligand will run lower than the ligand and have a good UV spot

@KrangLee I see. Welcome to the "materials for electronics and the like" science world ;)

By the sheer size of the «blob» on the top of the TLC it is possible that your analyte solution either is too concentrated, or that you put the capillary with the analyte solution too often on the TLC prior running the TLC.  So you can't say if the different colours by staining are just by different concentration of the analyte, or by different compounds (well, looks like the one eluted first does not absorb much UV).  But it is better to ran with two lanes simultaneously, starting material and reaction (mixture), or three (co-spot), to ease assignment of the spots.

(Ctd.)  See, e.g., https://chem.libretexts.org/Ancillary_Materials/Demos_Techniques_and_Experiments/General_Lab_Techniques/Thin_Layer_Chromatography or https://chemistryhall.com/thin-layer-chromatography/.  But what excellent school lets you hang running these experiments in the dark with so little chemical supervision?

@Buttonwood Thank you for answers, I would try more experiments..

@Buttonwood likely real research rather than schooling....

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In electrophoresis of amino acids and dipeptides, how can I rationalize the observed mobilities?

For part b(i) why is the spot tyrosine? I know tyrosine has an overall -2 charge while phenylalanine has an overall -1 charge. However, phenylalanine has a smaller Mr so shouldn't it travel faster? Also, I don't understand (ii)... why are they close?

So you figured out that the mobility depends on charge and on size. Make a table of the three species and their charge and size, and see if you can predict which travels the fastest.

Species
Charge at pH 12
relative size

Dipeptide
-2
Large

Tyrosine
-2
Small

Phenylalanine
-1
Small


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Quantitatively Calculate Dipole Moment
Recently, I have been reading up on articles relating to the dipole moments of different molecules (specifically this). I see how they can get bond lengths and experimental dipole moments with reasonable accuracy and compare that to the theoretically predicted values.
What I don't get is the how theoretical prediction of dipole moment happens. I know that dipole moment equals the charge times the separation distance (u=q*d). So I must be able to predict the dipole (making sure its a vector) of any molecule given its distance of separation and partial charge on the molecules. The partial charge on the molecule depends on the electronegativity values. But then you can't put the electronegativity values (Pauling or Sanderson or others) directly as charge in the formula. Now, I know that Pauling or any other electronegativity scale is only arbitrarily chosen and is relative to the most electronegative element.
But then, how do we make theoretical estimates of dipole moments? Is there any way of relating the electronegativity values to partial charge on each atom? Or is there an entirely different approach?
I have already read these StackExchange answers -  this which answers the 'how' but not by 'how much'

:/ Plug your molecule in some serious software that gets distances and partial charges right, and lo and behold...

Not possible to make a connection between electronegativity and dipole moments. Forget it. Dipole moments are theoretically calculated by advanced quantum mechanical calculations (which I cannot do), and experimentally determined by microwave spectra (again not a trivial exercise) and by various methods. Still experimental methods give a reliable value. Theoretical calculations are pretty close.

@Mithoron I have tried this method using ChemAxon's MarvinSketch. I would like to know how reliable are these estimates, and possibly verify it myself.

@M.Farooq Very well, if the method of theoretical calculation of dipole moments is a complex thing, then I would be happy to see a high level overview of the process even if it skips some details without obstructing the understanding of the process. I have not found any resource online explaining the process which I would be more than happy to receive if somebody points that out.

@PrajvalK There is a software called GaussView which allows you to calculate dipole moments for a given structure. You have look up its documentation. Please note not everything is available on the internet and most of the stuff is behind paywalls.

@PrajvalK, And here it is. https://gaussian.com/force/, I have never used it but my collaborators use it exactly for determining dipole moments.

@M.Farooq GaussView is just a GUI for Gaussian, which does the actual calculations. Funny that you are promoting closed-source software company with shady practices and mentioning paywalls when there are a free alternatives such as ORCA with Avogadro as graphical interface.

@PrajvalK I highly doubt ChemAxon's MarvinSketch could be considered serious in the context of Mithoron's comment and your use case.

@M.Farooq Yeah, I have heard about Gaussian but I do not have the pockets for it

@andselisk I might check with Avogadro or ORCA for its predictions. The whole reason this came up was that there was an anomaly between values that some websites gave and what the software gives which I will be addressing this in another post. For now, though I will leave this question open for any person in the future who might know the actual process.

@andselisk Most people in computational chemistry I have met seem to only know about and use Gaussian, probably because it was developed a long time ago, and there is inertia in adopting new softwares. Orca is a good software and there is also GAMESS, NWChem etc. which are all free.

@Mithoron The software does not usually calculate the partial charges, the dipole moment can be directly calculated from wavefunction. Partial charges (e.g. ESP charges) are generated by a separate fitting procedure, which is not required for dipole moment calculation. OP, take a look at this page: https://chemistry.stackexchange.com/questions/31075/how-to-calculate-molecular-dipole-moment-from-a-known-wavefunction

@SRMaiti Thanks. That was exactly what I was looking for.


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For home chemistry purposes can you treat sodium percarbonate as sodium peroxide?
It is easy to find home chemistry experiments for sodium peroxide but not so much sodium percarbonate. Industrially sodium percarbonate is used a sodium peroxide source- it is close enough to used as a substitute for common home experiments?

you really need to specify what experiment you want to do.

Sorry my addled brain thought I was writing hydrogen peroxide

Then fix it in the question title/body, otherwise readers will take it as the valid content. (Fixing it on the author behalf is always a risk not to match what the author has in mind .). // In the current question wording, the answer is it cannot be said until the particular scenario is specified.

It is not a true peroxoacid anion like persulphate or peracetate. It is a peroxohydrate, releasing free hydrogen peroxide when dissolved, or at contact with suitable solvent.

These behave fairly differently, in most cases. Look at the Safety Data Sheets, first:

$\ce{Na2O2}$, sodium peroxide:
This chemical is considered hazardous by... 29 CFR 1910.1200
Skin Corrosion/Irritation Category 1A
Oxidizing solids Category 1

$\ce{Na2CO3.1.5 H2O2}$, sodium percarbonate:
This chemical is considered hazardous by... 29 CFR 1910.1200
Acute oral toxicity Category 4
Oxidizing solids Category 3

$\ce{Na2O2}$ is a more powerful oxidizer, more likely to start combustion in contact with organic substances, can cause skin corrosion, and is generally not very stable.
$\ce{Na2CO3.1.5 H2O2}$ is a less powerful oxidizer, a bit more stable, less toxic orally, and is not listed as corrosive to skin (caustic), just possibly irritating.
Yes, both can be used as a source of oxygen, $\ce{O2}$, but $\ce{Na2O2}$ can react abruptly, perhaps causing a conflagrating explosion. In some use, the water and $\ce{CO2}$ that accompany evolution of oxygen from $\ce{Na2CO3.1.5 H2O2}$ might interfere with a reaction.
Caveat pharmacopolium: It is possible, if you have the chemical and engineering knowledge, to use one in place of the other in some situations... but you'd need to be specific to get any assurance.


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What do the suffixes on Y-90s and Y-90v mean?
I'm writing software that needs to differentiate the following "isotopes" for a dose calibrator (as used in nuclear medicine). My software needs to normalize various inputs for "isotopes", so I'd like to parse the suffixes provided and have names for what they mean, so I can document this in the code.
I use the term "isotopes" because that's how they are being referred to in my organization, but Wikipedia says that the difference between Tc-99 and Tc-99m would be termed an isomer.
I have the following examples of isotopes with and without suffixes.

Ir-196
Ir-196p
Y-80
Y-90s
Y-90ss
Y-90v
P-32s
Tc-99m

So the suffixes I have to work with are

m: I know this means metastable.
p
s
ss: I think this means "sphere sources".
v

What do these various suffixes mean? Is there a general term to group their meanings (eg: metastable is a "state")?

Where, and in what context, have you seen those?

These might be calibration used for a specific container... my research indicates Y-90s might refer to measuring the source inside a syringe, and Y-90v, inside a vial. Atomlab lists calibration sources by index: http://www.biodex.com/sites/default/files/086330man_17355_0_opt.pdf , but there is no explanation. In this case, it should not be an embarrassment to ask your employer for more information; this is not readily found information.

@DrMoishePippik Great find! A little tip: a multipage PDF might be a pain to browse for the keyword, so one can append #page=<page_number> to the URL for PDF, e.g. https://www.biodex.com/sites/default/files/086330man_17355_0_opt.pdf#page=195 will open the file on the Appendix H: Atomlab 500 Isotope Index list where the aforementioned symbols are located.

@andselisk, thanks for the tip! Until now, I just used Ctrl-F to find a key word - but Y90 might have been written 90Y, requiring multiple searches.

@ DrMoishe Pippik: As  andselisk commented, Appendix H has listed about 98 isotopes, including Y-90s (syringe), Y-90v (vial), and Y-90s-s (sir-spheres). No other ways are written.

@DrMoishePippik I actually did ask my employer, but he wasn't sure their meanings.
I believe you have the answer I'm looking for. AtomLab is one of the brands of dose calibrator the software is communicating with, and the isotopes in the PDF match well with the full list of examples I have. If you'd like to post your response as an answer, I'd be happy to accept it,.

@duckbrain, I'm glad that was of use, and as you suggest, it's moved to an answer, with some explication. Kudos to Mathew Mahindaratne, Andelisk and you for filling in gaps!

These might be calibration used for a specific container... my research only found mention of those items in an Atomlab manual for radiation calibration sources by index, starting on p. 195, and show on p. 190 how to enter that ISOtope index. Appendix E, p 183, lists the following formats for 90Y sources:
Bremsstra[h]lung only, 10cc plastic syringe
Bremsstra[h]lung only, NIST vial
Bremsstrahlung only, [SIR]-Spheres

(Typos corrected above, in "SIP-spheres" and in Bremsstrahlung (or Bremßtrahlung, if you prefer).
So it appears that:

Y-90s is calibrated for (plastic) syringe,
Y-90v is calibrated for vial, and
Y-90ss is calibrated for SIR-Spheres, i.e., encapsulated in Sirtex resin microspheres.
And all the calibrations appear to be for the X-rays produced by Bremsstrahlung from β− decay... [with] energy of 2.28 MeV with an average beta energy of 0.9336 MeV. This is likely because the container blocks the β− radiation (that's beta particle, electron, not  eszett ;-), so only X-rays escape the container. Since some containers are plastic, containing no heavy elements, likely the X-ray spectrum is towards the low end of the spectrum, because electrons penetrate farther, spreading their energy loss over multiple collisions.

However, since you are working on an interface for a device used in nuclear medicine, be very cautious about any assumptions. Deaths have been attributed to computational errors leading to radiation overdose, such as with the Therac-25 linac. Therefore, it should not be an embarrassment to ask your employer for more information. This is not readily found information, my answer is imprecise, and human safety is involved.

+1 For words of caution. For this particular question, the meanings don't affect the behavior of the software (just comments in the code), but duly noted.

Apparently there are people who write "Bremßtrahlung"...? I wonder where that came from. This spelling would be >100% orthographically wrong in the German language where the word originates.

@Antimon, yes, the  eszett or scharfes S (a double "s"), I was taught, was the original spelling. However, languages change - Neanderthal has lost its "h", also. In any event, it's the electromagnetic radiation resulting from acceleration of a charge, as when an electron in an X-ray valve smacks into the tungsten anode. So Röntgen was making heavy-metal vibrations many years ago.

I highly doubt that the ß version would ever have been correct in German. It's a compound term and those two "s" belong to different words within it. In addition, the letter ß is a sharp "s" sound and that's not how you pronounce that word. I'm really confused where this came from... if you have any more historic pointers, I'd be really interested.

@Antimon, just search "Bremßtrahlung" online - it's in use by NASA, Elsevier, AAAS, and numerous other authoritative sites. Did they all make up the word?

They certainly didn't make it up themselves, but someone at some point in the past obviously did and I'm wondering who it was, seeing how illogical that spelling is. But with Google indicating only a few hundred hits (vs. >1 mio for "Bremsstrahlung") it seems to me like a fringe phenomenon anyway.

Perhaps a book on Deutshce etymology would help. As mentioned, spelling has been simplified to avoid special characters, and to make pronunciation easier for non-German speakers (as in "Neanderthal"), so my guess is that the original orthography used the special character... but that's a guess. The roentgen is the unit of radiation, but Wilhelm his family name, Röntgen.


text:
A problem on Intramolecular Cannizzaro's Reaction
My textbook has the following reaction given as a special case of Intramolecular Cannizzaro reaction:

As far as I understand, here one Phenyl group migrates (instead of hydride ion) from the carbonyl getting oxidized to that getting reduced.
My question is, what is making the migration of the phenyl group possible?

What makes it possible? Ultimately because it is energetically favourable. Plus it is irreversible, so if it occurs there's no way for the product to revert to starting materials

Well, as you noticed it's basically a Cannizzaro reaction - reasons are the same, but what you want is rather some kind of rationalisation and should specify which.

This is known as the benzil-benzilic acid rearrangement. As Waylander noted it's energetically favourable. Phenyl group acts as a group with high migratory aptitude in many other reactions too. One can again question as to what makes the migration of hydride possible in the main Cannizzaro reaction.


text:
If I have an oil that contains an impurity that's more soluble in water than in oil, is there a way to "wash out" the impurity with water?
Maybe this question is better classified as "physics" than chemistry, but I'm trying it here.  Suppose I'm concerned about a substance S that is highly soluble in water and less soluble in oil.  I have a batch of oil that is nonetheless contaminated with traces of S, and I want to remove as much S from it as feasible.
I suspect that at an interface between this oil and water, S would diffuse from oil into water, since it's more soluble in water than in oil.  I also suspect that diffusion of S across the oil-water interface would be slow, and I'd like to speed up the process if possible.  In principle, I can think of three ways of doing that:

Heating a vessel of oil and water to speed up diffusion
Increasing the surface area of the oil-water interface by spreading the oil thinly over a large area of water
Making an emulsion, either by mechanical agitation or with the addition of some emulsifying agent, and then breaking the emulsion later, hoping that S breaks out preferentially into the water fraction.

The last option seems most promising to me, but I don't know if it really works the way I suspect it does.  Any pointers on how I might "wash" this oil?  I am relatively unconcerned about traces of water remaining in the oil.

Option 3 is routinely done by agitating during liquid-liquid extration. Using emulsifiers would be wrong move, as you can get into troubles to separate oil and liquid later and oil would get contaminated by the emusifier. Getting fine emulsion is not really necessery. Extraction may or may not to be done repeatedly.

This is performed on a daily basis in organic synthesis laboratories - what you're describing is liquid-liquid extraction, which is commonly performed inside a separatory funnel. See, for example, this video as one out of many available on YouTube.

Goes to show that I don't know jack about the practice of chemistry -- thanks both for pointing me in the direction of liquid-liquid extraction!  I'm sure I'll be able to find more info now that I know the right term.

The comments above are somewhat correct but we don't treat oils like that. Rather your oil should be in solution. After extraction and separation the organic phase is dried and the solvent removed. Otherwise you run in the problem of your point 3, and assuming you can break the emulsion or wait enough, you might end up with less S but surely more water instead. It will very much depend on what the oil is, what would be its use, and its amount.


text:
If a company advertises a solution as 20% on their web-site, will it be 20% w/w?
I am trying to buy some ammonia solution from a company and I have found the product I want but the product particulars just tell me it's 20%.  They don't say if it's 20% w/w or w/v?  Their technical support people don't reply to queries (so far).
If I order it, will it have the product particulars on the bottle?
I think I may have to just risk it and buy it and hope the label makes it clear!

They seldom sell 183% H2SO4.

It should be labeled to be compliant with proper regulations. If its w/w it should say that and if its v/v it should say that.

Thanks Harley.  I was hoping that would be the case

This is a common issue with vendors and websites. Find the particular vendor's SDS, that has the best chance of having information as to the concentration units. Anecdotally, w/v is common for vendors that aren't a focused scientific laboratory supply vendor.

Most likely it's w/v. This table of ammonia solution densities* gives a specific gravity close to 0.92 for a solution with 200 g ammonia per liter, meaning 20% w/v, and Fischer Scientific lists their 20% ammonia solution as having a matching specific gravity of 0.92.
*Sorry for the long link. Attempting to shorten it gave me a 404 error.

I guess you mean a solution of ammonium hydroxide?
Maybe you could get some clues by having a look at the density / melting point, etc. of the sold solution if they are available, and compare them with some information available here. In particular, it seems that the density or melting point may be of great help, if it is given to you with a sufficient accuracy. Unfortunately, there are only few general rules about specifying a concentration in w/w or w/v (see here).
Otherwise, you will have to weight it or measure its melting point once you receive it. Indeed, according to this site, the melting point seems to be a fairly accurate mean of finding the concentration of an ammonia soliution, and it is not too difficult to measure if you have a laboratory freezer and the associated thermometer at hand.

Ammonium hydroxide is a decades old myth.

Not entirely. Ammoniun hydroxide may form at high pressure. Quoting: "We show that compression facilitates proton transfer from water to ammonia in all relevant mixtures". Such modeling is done in connection with the internal structure of ice-giant planets.

@Poutnik, three decades : https://pubs.acs.org/doi/10.1021/ed066p310 ^^ More seriously if you have other robust sources proving your sayings, I would be interested in reading them! I must admit that I just wanted to be sure that we were talking about this very chemical species. (Since no formula was given in the question, I didn't wanted to make a misinterpretation of Naz issue)

@OscarLanzi Correction: I do not say that ammonium hydroxide cannot exist - in fact I do know it does. The myth is about considering it at usual laboratory scenarios, like ammonia solution at room pressure and temperature or even selling it as the commercial product. As less than 1% of ammonia is present in the commercial ammonia solution as ammonium. By the similar principle, $\ce{Na3PO4}$ solution could be sold as "phosphoric hydroxide", due $\ce{PO4^3- + H2O <=> HPO4^2- + OH-}$

Hello.  I meant ammonia (aq).
That website Gestis is very useful. It has the concentration I would like but again describes the concentration as 20%.  So is that 20% w/w?  There is a reference document but I couldn't find it on the web.  If it's not clear when I receive it, I could indeed freeze it and find it's melting point and use the referece materials.  Thanks for the tip!

It may be easier to perform titration by HCl, after dilution.


text:
How to find standard electrode potential of a reaction resulting from two different reactions?

Q1) Calculate $E_{\ce{Cu+}|\ce{Cu}}^o=E_o$. Given that- $E_{\ce{Cu^2+}|\ce{Cu}}^o=E_1$ and $E_{\ce{Cu+}|\ce{Cu^2+}}^o=E_2$

Method 1- $$(i)\quad\ce{Cu^2+} + 2e^-\to \ce{Cu}  \quad... \Delta G_1=-2FE_1$$
$$(ii)\quad\ce{Cu+}\to\ce{Cu^2+}+e^-\quad...\Delta G_2=-FE_2$$
Required reaction is the "addition" of these two so-
$$\ce{Cu+}+e^-\to\ce{Cu}$$
$$E_o=2E_1+E_2$$
Method 2- Just directly add the two reactions, which gives us $E_o=E_1+E_2$

Q2) Calculate standard cell potential of the following reaction-
$$14\ce{H+}+6\ce{Cl-}+\ce{Cr2O7^2-}\to 3\ce{Cl2}+2\ce{Cr^3+}+7\ce{H2O}$$
Given that $E_{\ce{Cr2O7^2-}|\ce{Cr^3+}}^o=E_1$ and $E_{\ce{Cl-}|\ce{Cl2}}^o=E_2$

Method 1- Use the same process as done in the above example i.e. write expressions of $\Delta G$ of reduction of dichromate and oxidation of chloride, multiply them by suitable coefficients so as to get the desired chemical equation. This gives us $E_o=\frac{E_1+3E_2}{6}$.
Method 2- Just directly add the two reactions, which gives us $E_o=E_1+E_2$
According to my textbook Method 1 is correct for the first question while Method 2 is correct for the second. Why?

Arithmetics with E is the same as with DeltaG, except the extra multiplicator n (resp. nF ). so if you are going to honor the energy conservation law, when doing additions and subtractions with potentials, you always have to implicitly or explicitly involve the n multiplicators. As energy = charge . potential.

@Poutnik yeah then why does that not work with the second question? Method 1 should have been correct according to what you say but actually method 2 is the correct one.

There is another factor that you use another multiple of Delta G or of nFE, if you use multiplied chemical equation, like $\ce{2 Cu+ + 2e- -> 2 Cu}$  instead of $\ce{Cu+ + e- -> Cu}$

Define $E_2$ for Q. 2. Standard electrode potential is an $intensive$ property.

@Apurvium Sorry, that was a typo, corrected.

Q2) Calculate standard cell potential of the following reaction-
$$14\ce{H+}+6\ce{Cl-}+\ce{Cr2O7^2-}\to 3\ce{Cl2}+2\ce{Cr^3+}+7\ce{H2O}$$
Given that $E_{\ce{Cr2O7^2-}|\ce{Cr^3+}}^o=E_1$ and $E_{\ce{Cl-}|\ce{Cl2}}^o=E_2$

The half-cell reactions are:$$14\ce{H+}+6\ce{e-}+\ce{Cr2O7^2-}\to 2\ce{Cr^3+}+7\ce{H2O};~~~~~~~~~~n_1=6$$
$$3[\ce{2Cl-}\to \ce{Cl2}+\ce{2e-}];~~~~~~~~~~n_2=6$$
Note that, before arriving the final equation, you have to balance the charges and then cancel out the electrons by multiplying with appropriate coefficient. As $G$ is a state function:
$$\Delta G^0_1+\Delta G^0_2=\Delta G^0$$
$$\implies -n_1FE^0_1+(-n_2FE^0_2)=-nFE^0$$
$$\implies E^o=\frac{6E^0_1+6E^0_2}{6}=E^0_1+E^0_2$$

Q1) Calculate $E_{\ce{Cu+}|\ce{Cu}}^o=E_o$. Given that- $E_{\ce{Cu^2+}|\ce{Cu}}^o=E_1$ and $E_{\ce{Cu+}|\ce{Cu^2+}}^o=E_2$

Similarly, the half-cell reactions are: $$(i)\quad\ce{Cu^2+} + 2e^-\to \ce{Cu}  \quad... \Delta G^0_1=-2FE^0_1$$
$$(ii)\quad\ce{Cu+}\to\ce{Cu^2+}+e^-\quad...\Delta G^0_2=-FE^0_2$$
Required reaction is the "addition" of these two so-
$$\ce{Cu+}+e^-\to\ce{Cu};~~~~~~~~~~n=1$$
$$\Delta G^0_1+\Delta G^0_2=\Delta G^0$$
$$\implies -2FE^0_1+(-FE^0_2)=-FE^0$$
$$\implies E^o=\frac{2E^0_1+E^0_2}{1}=2E^0_1+E^0_2$$
Therefore, we can always get the correct result by equating the values of $\Delta G$.

Oh! I made a silly mistake. Sorry for wasting your time


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NaOH + atmospheric CO2: Unexpected results troubleshooting
I've conducted an experiment where a solution of $\pu{4.2 M}$ $\ce{NaOH}$ was left exposed to atmospheric $\ce{CO2}$ for a period of time and titrated using potentiometric titration and potassium hydrogen phthalate ($\ce{KHP}$) to investigate the degradation of $\ce{NaOH}$ over time. I was under the impression that the reaction for such a high pH solution would look like this:
$$\ce{2NaOH(aq) + CO2 <=> Na2CO3(aq) + H2O} $$
But the results I have got don't fit this. For example the rate of change of $[\ce{NaOH}]$ and $[\ce{Na2CO3}]$ should be a 2:1 ratio but I got  0.849:1. This is a significant difference. The  data I have seem accurate and the titration curves lined up well on each run, I don't think this is a significant error on my part so I'm now wondering if instead of the above reaction, $\ce{NaHCO3}$ was being formed:
$$\ce{NaOH(aq) + CO2(g) <=> NaHCO3(aq)}$$
This would fit the ratio I got much better but I thought this wasn't possible at a pH of 10 or above?
Here are my titration and Concentration vs time graphs. I had previously reasoned that the fact I was getting only two equivalence points was due to the third step being a weak neutralization with a pH of about 7 that would barely show but now it seems like it could be because $\ce{NaHCO3}$ is the reacting not $\ce{Na2CO3}$.

Any help working out what is happening here would be great...

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

With the initial pH, it is obvious the $\ce{HCO3-}$ content must be almost negligible. But in the section between both steeper segments, $\ce{CO3^2-}$ is being converted to $\ce{HCO3-}$. On the plateou near 105 mL, the ratio $\ce{CO3^2-/HCO3-}$ is 1:1 - carbonate/bicarbonate pH buffer in action.

@Buttonwood Thanks for that, I haven't seen that formatting before so cheers.

You add acid, right ? Does an acid convert carbonate ( in excess ) to bicarbonate ? $$\ce{NaOH + Na2CO3 ->[acid] Na2CO3 ->[acid]NaHCO3 + Na2CO3 ->[acid]NaHCO3 ->[acid]NaHCO3 + CO2 ->[acid] CO2}$$

@Poutnik Sorry mate I completely misread that. Yes I guess it could, so the reaction with $\ce{CO2}$ creates $\ce{CO3^2-}$ then the acid converts it back?

You do know about the concept of acido-basic equilibrium and acid dissociation constants, don't you ? $K_\mathrm{a2}=\frac{\ce{[H+][CO3^2-]}}{\ce{[HCO3-]}}$

@Poutnik Yeah I do, im just stuggling wrapping my head around all of it at once really. So given my titration stops when all of it is $\ce{NaHCO3}$ it will be a 1:1 with the $\ce{NaOH}$ but double the amount of $\ce{Na2CO3}$ that was present after exposure?

Let us continue this discussion in chat.

Real experiments and their outcomes are more interesting than expected so don't call it a trouble. First think of a couple of scenarios.
a) If you just leave a pellets of solid NaOH in open atmosphere. It would absorb water and form a protective coating of insoluble sodium carbonate. If you analyze the solution, that would be still be very high in NaOH conc.
b) If you leave concentrated NaOH (~ 50% w/w) in open atmosphere, you start to see a white crust and white powder settling to the bottom. Here again sodium carbonate is insoluble and acts like a protective coating. If you analyze the solution, you will see most of the NaOH is still intact.
c) Your NaOH solution is a relatively low concentration but its concentration is not changing as fast as you were expecting. The reason that you have a heterogeneous reaction. You have trace concentration of a gas reacting with quiet (unstirred) solution of NaOH. So two steps are involved:
(i) Absorption of carbon dioxide from a very dilute mixture (air) in NaOH.
(ii) Irreversible reaction of carbon dioxide and NaOH to form sodium carbonate.
All of this is mainly happening at the air-solution interface. Your bulk is still protected by very slow diffusion. So these two steps will not follow your expectations.
People have worked out the expected rates of absorption of carbon  dioxide in NaOH.
For example in Absorption of Carbon Dioxide into Aqueous Sodium Hydroxide and Sodium Carbonate-Bicarbonate Solutions, The Chemical Engineering Journal, 11(1976), 131-141.

Then the absorption of solute gas $\mathrm{A}$ is accompanied by an
irreversible chemical reaction, the rate of absorption
$N_{\mathrm{A}}$ is represented by $$ N_{\mathrm{A}}=\beta\left(2A_{i} \sqrt{D_{\mathrm{A}} / \pi t}\right) $$ where $A_{\mathrm{i}}$
and $D_{\mathrm{A}}$ are the interfacial concentration and the
liquid-phase diffusivity of the solute gas, respectively, $t$ is the
exposure time and $\beta$ is the reaction factor.

Another, point is that I did not mention sodium bicarbonate anywhere. The reason is that your NaOH concentration is pretty high, as long as you do have NaOH, sodium bicarbonate will not form. So seven days are not enough. Wait for months perhaps!
If you were stirring everything for 7 days, then the rate of carbon dioxide will be faster but still you are limited by diffusion of the gas into the solution.

Thanks for the answer, if i understand you correctly youre saying that the rates will change based on many factors which i totally agree with - shouldnt the rate that the $NaOH$ decreases at still be related to the rate that the $Na_2CO_3$ increases at? Regardless of how much reacts, if i can work out how much $NaOH$ is gone i should be able to work out how much $Na_2CO_3$ should be gone and the rates should follow the same ratio?

Also i think i possibly didnt phrase my question well, my problem is not that the concentration is changing slowly, it is that it is changing out of sync with the reactions products and therefore violating the stoichiometric relationship.

The simple rates of reactions apply when the reaction were homogeneous. Your bulk is still NaOH, but sodium carbonate is being formed is only at the surface.

Regarding the stoichiometry issue, do you stir the solution before sampling for titration?

And regardless, if the analysis is done right, your mass balance must remain correct. Forget about the rates. Think about the moles of products and reactants only.

Yes I stirred the soution before I titrated, and the overall mol when I add together the mol of $NaOH$ and the mol of what I assume is $Na2CO3$ remains constant (within uncertainty). The problem is that there should be half as much $Na2CO3$ as there is because it should be a 2:1.

Let us continue this discussion in chat.


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Acidity of alcohols
So everywhere that I've read, there's the reasoning that the acidity of a compound is determined by how  stable the conjugate base is, this should be the same for alcohols too, but on Chemistry LibreTexts there's an article which says otherwise, and the reason being that if the electron donating ie. inductive effect was responsible for this then we would have obsereved the same acidity order of alcohols in the gaseous phase too, where the relative acidity trend is quite opposite of what it is in solution.
The rationale behind this as is given in the article is:

Polarizibility almost completely accounts for the trend in gas-phase acidities. As the size of the substituent increases, the acid becomes stronger due to the ability for the charge to be distributed over a larger volume, thereby reducing the charge density and, consequently, the Coulombic repulsion.In solution, however, the ions can be stabilized by solvation, and this is what leads to the inversion of acidity ordering. Brauman and Blair showed that smaller ions are better stabilized by solvation, which is consistent with the Born equation.

Now there are these two things that are not clear to me

If the substituents are almost similiar in size such that the charge almost gets a similiar size to spread, let's say they are amine and methyl groups on the beta carbon, will we then take the inductive effect into account to determine the acidity of the alcohol?

Why is this then only applicable to alcohols? Why not the same reasoning for determing basicity and acidity of other functional groups?

That's just a thing to show whys and hows of "inductive effect" of alkyl groups vs hydrogen. See, carbon is actually more electronegative then hydrogen, and yet ethanol is weaker acid then water in aqueous solution. In gas phase, not so much.  And, yeah, polarisability is probably more adequately correlated parameter for gas phase in such situation.

@Mithoron Could you expand on the whys and hows, I'm not sure how electronegativity and inductive effect correlate.

As mithoron has said in the comments, alkyl basically acts as an EWG in the gas phase, as shown e.g. by Boand et al. It's easily seen that it is very very weak, which could be easily explained, as there's not much of a difference between the electronegativity of C and H, and the existence of hyperconjugation "donates" some electrons.
But, another common fact is that alkyl substitutions on ammonium ions decrease their acidity, as is shown by Aue et al., which means that it should be regarded as an EDG  for ammonium ions.
So, what is really happening here is that alkyls are highly polarizable compared to hydrogen, which means that whenever an alkyl is near a charge, no matter positive or negative, it will create a dipole that disperses charge, which would act as a stabilizing force.
In this way, we could explain the stability of carbocations, carbanions, etc.
Boand, G.; Houriet, R.; Gaumann, T. Gas-Phase Acidity of Aliphatic Alcohols. J. Am. Chem. Soc. 1983, 105, 2203-2206. DOI 10.1021/ja00346a600
Aue, D. H.; Webb, H. M; Bowers, M. T. Quantitative relative gas-phase basicities of alkylamines. Correlation with solution basicity. J. Am. Chem. Soc. 1972, 94, 4726-4728. DOI 10.1021/ja00768a049


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Ferrous chloride solution coming out purple rather than green?
I've got some steel wool in hydrochloric acid to attempt to make ferrous chloride and the whole solution has turned purple. From everything I've seen it's supposed to be green. I have no idea what's happened and if anyone could give any ideas that'd be great.

The steel wool has no soap
The glassware is clean of phenols/enols/anything that would give a positive result in the ferric chloride test
The HCl has 1% titanium dioxide for some reason (store bought in Australia)

https://i.stack.imgur.com/r394V.jpg

Chromium has a rich color chemistry. Steel wool is not iron wool.

HCl with titanium dioxide? This is strange.

Your observation with steel wool is somewhat perplexing because your HCl is an odd solution of titanium chloride expressed as 1% titanium dioxide. Initially, chromium complexes could be blamed for this unusual violet color after dissolving steel wool or iron wool. However it seems that something fancier is happening.
Let us assume your store HCl has dissolved titanium dioxide for whatever magical reasons (no idea why this is so). Titanium could be in $\ce{Ti^{4+}}$. When you add iron/steel wool,  $\ce{Ti^{4+}}$ begins to be reduced by hydrogen evolution reaction to $\ce{Ti^{3+}}$. Fifty years ago, we could have called that nascent hydrogen is doing this magic, but iron powder in HCl can reduce the +4 oxidation state of Ti to +3 by some mechanism.
$\ce{Ti^{3+}}$ chloride solutions are beautiful violet/purplish (source: Wikipedia):

With some chromium, iron, and titanium, you have a dirty mix of purplish color rather than the expected green color.


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How do I find an atom in VMD by using its index
For my work I was asked to run a computational NMR of a mostly water structure and then asked to show which hydrogen in this structure correspond to what chemical shifts
I am visualizing the structure in VMD and by clicking on the hydrogens manually I get useful info like its index which corresponds 1:1 as the order they appear on my separate chemical shift data file
My problem is that although I can write down that Hydrogen number 84 has a chemical shift of XXX, I can not as easily find Hydrogen number 84 from VMD itself, in the graphical interface. I know there is a command called atomselect but I am having a VERY hard time getting it to work and was wondering if anyone can give me some guidance on how I can easily look up specific hydrogens  though Identifiable keywords like its "index" witout having to resort to random guess clicking until I find the right one
Thank you

You do not need atomselect to do this in VMD, you can do this from the GUI as well.
1) Open the file with VMD
2) From the menu bar go to Graphics > Representations...

I would recommend using Lines or CPK as drawing method for the main system (selected atoms = "all"), as this gives small atom size in display.
3) Click on "Create Rep" button and a new representation will be created with the atom selection "all"
Delete "all" and type "index 10" to select only atom 10 for the visualization, then click "Apply". Then change drawing method to "VDW" to highlight that atom (it will become bigger). Alternatively you can change the colouring scheme to highlight atom. [Keep in mind that in VMD, atom number counting starts from $0$]

The result would look like this:

You can see that atom number 10 is now clearly visible among all others. (I am using a large system with ~ 3000 atoms so there are a lot of atoms on the screen; with smaller systems, you would be able to see the highlighted atom even more clearly.)
VMD can also select atoms in many other ways. You can go to the "Selections" tab in the Graphical Representations window to see all of the options available.

As an aside, VMD's main strength is visualising a large number of atoms, it is mainly targeted at molecular dynamics. If you are calculating NMR, I assume you are using a small system (~50 atoms max) and using some quantum chemistry software like Gaussian or Orca. For those, there are many other visualisation programs which would be easier to use I believe. For example, Avogadro can read output files from many quantum chemistry codes and visualise the structure. In the "Display types" windows on the left, you can turn on Label which will show the atom numbers of each atom in the display:

Then you can see the atom number of each atom in one display, instead of highlighting one by one in VMD.


text:
Planarity at the Junction between multiple Rings

My Approach:
As cyclohexane should be in the chair form and cyclopentane should be in the envelope form, more the number of five-membered rings more should be its planarity, as envelope form is more planar than chair form but also due to angle strains the structure five-membered rings might not be rigid and hence might change shape then how will we be able to predict planarity.
P.S.: Answer is C

I was gonna about to argue on basis of hydrogen to carbon adjacent to junction but it turns out that for each case adj carbons are all tert

@Buraian Yes, I had thought about that too and also from hybridization but in this question almost everything is similar, the only difference is the size of the rings, so most probably angle strain and geometry is being applied but the concept is contradicting the answer.

In a and b, you can probably assume the adjacent carbon bonds are around 110 degrees away from each other. In C, it looks like the bonds to the atom (C) are around 90 degrees because those three atoms are connected in sequence (straight line). Therefore those bonds will, in reality be out of the plane to try and increase the bond angle. Because the bonds in c have smaller angles between them, they will be more out of plane to maximize the distance away from each other.


text:
Purpose of using K2O and Al2O3 in Haber's Process
Our text book states use of catalyst such as iron oxide with small amounts of $\ce{K2O}$ and $\ce{Al2O3}$ ( these act as promoters) increases the rate of attainment of equilibrium in Haber's Process.
Why do we want to attain equilibrium? We want more production of ammonia so attaining equilibrium will not be efficient right?
Haber’s process is done at high pressure and low temperature (Le Chatelier’s Principle) to produce more output and increase forward rate of reaction. Here we do not wish to attain equilibrium.
When equilibrium is attained the rate of forward and backward rate becomes equal. If we want to maximise the output then the reaction should not reach equilibrium. According to my reasoning I feel that the reaction should go in forward direction and produce more of the products and simultaneously the product must be removed so as to facilitate more production otherwise it will result in backward reaction and when equilibrium is attained no more product shall be formed.
Why should we add catalyst and help the reaction attain equilibrium faster?

Catalysis is about reaching equilibrium faster. Catalysis does not provide going behind the equilibrium. The equilibrium is chosen by reaction conditions, like p, T, composition. The chosen T is a trade off between reaction thermodynamics (prefers lower T) and kinetics ( prefers higher T).// Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own explicit  effort.  How do I ask a good question.

IF the reaction doesn't approach equilibrium, the products will be hydrogen and nitrogen, the same as the inputs.

In fact, the reaction does not reach equilibrium even with the catalyst due short contact time. NH3 is separated after the catalyst passage and H2+N2 returns by a loop to the reaction for another catalyst path. For chemical engineering is very applicable the Pareto "80-20' rule, saying the 80% of result is achieved by 20% of effort. This is applied repeatedly, instead going of behind 80% result by enormous effort increase.

Why should we add catalyst and help the reaction attain equilibrium faster? Because without catalyst, the reaction is very slow, At very high temperature, where it is not slow, equilibrium is shifted too much toward reactants. So you would need more energy with less yield.

Please correct me if I’m wrong. Does this mean that we add catalysts to allow this specific reaction to take place at higher temperatures with better efficiency and also adding of catalyst does not affect this reaction’s equilibrium point as it is never reached ?

Generally, catalysts never affect eq. point, they just speed up reaching it from either side. // For this specific reaction, it is advantageous to reach eq. as close as possible to get as high yield as possible at the single catalyst pass. Also advantage is to have eq. point shifted as far toward products as possible. This needs low temperature, where reaction goes very slowly, but is accelerated by the catalyst.

@Poutnik  Understood! Thank you. This should be the answer.

Generally, catalysts never affect the equilibrium point, they just speed up reaching it from either side.
For this specific reaction,
$$\ce{N2(g) + 3 H2(g) <=> 2 NH3(g)}$$
it is an exothermic process, so higher temperatures speed up reactions, but provide lower yield at equilibrium.
Considering the reaction yield, it is advantageous:

Reaching equilibrium as close as possible to get a high yield at the single catalyst pass.
Having the equilibrium point shifted toward products.

The latter needs relatively low temperature, where reaction goes very slowly, but is accelerated by the catalyst.


text:
Why does iodine trichloride dimerize?

In which of the dimerization process, the achievement of the octet is not the driving force

$\ce{2AlCl3 \to Al2Cl6}$

$\ce{BeCl2 \to BeCl2}$ (solid)

$\ce{2ICl3 \to I2Cl6} $

$\ce{2NO2 \to N2O4}$

Answer: (3) $\ce{2ICl3 \to I2Cl6} $

The answer was given without explanation from the problem book I was solving, so I ask then, what exactly is the reason that $\ce{ICl3}$ dimerize?
From this website , I find it is because it is odd electron and wants to complete the octet, but the question asks why it dimerizes disregarding that rule.
Related

Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own explicit  effort.  How do I ask a good question. // Why not to analyze, where reaching the octet is and is not achieved ?

I understood what you said now, the thing is they specifically asked which one is not driven by octet rule/ odd electron atom rule

P.S.: Note that I was refering just to the partial question, if reaching octet is the case or not. I was not referring to the second part what is the cause if it is not.

It's a dreadful question. Achievement of an octet is never the driving force. The driving force is that dimerisation leads to a lower free energy. Achievement of an octet is merely a hand-waving rationalisation of that observation, it is not the cause.

@Ianbush It may be perhaps said that at lower periods, reaching minimal energy as the driving force  is often accompanied with completing octet octets. So the latter is merely an indicator, not fully reliable.

The real difference is that the others dimerize/polymerize to form more bonds, whereas ICl3 dimerizes (which, by the way, is true) to strengthen its existing bonds (2c-2e -> 3c-4e).

I was thinking similar in dimer no. of bonds increases as a whole.

In Icl3, the lone pairs are at an 120° angle. It dimerises such that the lone pairs are then at an angle of 180° which causes less repulsion and brings more stability


text:
Is stoichiometric imbalance in polymerization important if the process has an evaporation step?
Polymerization texts stress the importance of stoichiometric balance of the monomers.  For example here:
https://www.e-education.psu.edu/matse202/node/606
I can understand how stoichiometric imbalance leaves some unreacted monomers in the product and this drags down the average molecular weight.  I think this is what the theory is describing.
But, what if the process has subsequent evaporation process to remove light ends?  It seems to me that if this is the case, stoichiometric balance isn't really an issue anymore.  However, maybe there's an another factor I'm not considering.  Appreciate any insight!

This belongs on the Chemistry Stack.

Polymers is more of a chemical engineering than a chemistry subject.  For example, the polymer synthesis class on MIT open Courseware is part of the chemical engineering curriculum.  https://ocw.mit.edu/courses/chemical-engineering/10-569-synthesis-of-polymers-fall-2006/

Engineers touch on many topics, but others do them in greater detail...

The question is if it is copolymerization ( like ABS ) or polycondensation ( like PET or PC )

@Poutnik it's specific to step / condensation polymerization where there are two different monomers.  Something like ARA + BRB -> AR2B + AB.  Examples are polyurethane or polyimides.  Stoichiometric imbalance refers to different initial concentrations of the two monomers.

Yes, in theory, an idealised step-growth AA/BB monomer polycondensation is exquisitely sensitive to an imbalance in the ratio of the functional groups A and B present in the reaction mixture, as is described in your link. You're basically exploring the properties of a function close to a vertical asymptote. Since your evaporation step takes place after the polymerisation, then it is irrelevant to the polymerisation reaction itself - the excess monomer will already have caused its damage by the time you decide to adjust the weight distribution by some fractionation process (e.g. distilling off monomers/short chains).
The problem is that in your mind, you're unintentionally imagining the weight distribution of a polymer from a balanced polymerisation and then adding a thicker tail at the low weight end (which can be "fixed" by removing the low weight tail altogether). You can't do this - the monomer imbalance will have severely affected the entire weight distribution to start with, and at every step of the reaction. This is how you get the Carothers equation showing how the theoretical degree of polymerisation can vary anywhere from 100 to infinity depending on a reagent imbalance of a mere 1% or less.  You may want to have a look at Paul Flory's 1953 book Principles of Polymer Chemistry to really help you absorb the underlying mechanics of a polymerisation reaction. I can definitely recommend it.
Now in practice there may not be much difference, though this is dependent on a wild number of real-world parameters (exact reaction kinetics, reagent purities, possible side-reactions, measurement errors, mixing dynamics...) which make it virtually impossible to predict. You're much better off just trying it out on a small scale and seeing what happens. That said, the higher your target degree of polymerisation is, the more fragile the process is. If all you need is an average of a few dozen repeat units per polymer chain, there might be some leeway. If you need thousands or more, then you're in trouble.

Thanks for the reply.  I'm still struggling.  If you have 1 mol/L of A and 1.1 mol/L of B compared to 1 mol/L of each, it seems the excess B would only add to the extent of reaction.  The total amount of chemistry that occurs from an A+B -> product perspective seems like it wouldn't be negatively influenced by the imbalance.  If damage is done by the excess B, what exactly is the nature of the damage?

Try thinking of a more extreme case, such as having 10 or 100 times more BB monomer than AA monomer. At any point in time, any molecule (monomer, oligomer or polymer) with an A functionality has a large chance of reacting with a BB monomer, simply because there's a lot more BB monomer present than any oligomer/polymer chains containing B functionalities. So almost all reaction events involve adding a single BB monomer to the chain, until all functional groups A are exhausted. Thus, even at high extents of reaction, almost all that happened was the addition of BB monomers to growing chains.

Now how does this differ from a balanced polymerisation? In this case, as the reaction progresses, the amount of both AA and BB monomers is mostly used up, but there are always an equal number of A and B functionalities still available for reaction. Now, at high extents of reaction, you begin to observe a significant amount of condensation between chains (not just a chain + a monomer). This leads to a sudden explosive increase in average molecular weight at the very end - as almost all functional groups are consumed, the chains are forced to join together.

Thank you Nicolau for the valuable discussion!  You helped me a lot.


text:
E1CB mechanism with replacement of hydrogen step?
In this question by Jay, one of the sub parts of the total question has the following reaction identified as $\ce{E1CB}$ pathway:
$$\ce{Ph-CH2-CH2Br} \text{ on treatment with } \ce{C2H5OD/C2H5O-}  \\
\text{gives } \ce{Ph-CD=CH2} \text{ as the major product.} $$
Now, I know $\ce{E1CB}$ mechanism to be an elimination mechanism which involves strong base and weak LG which I found well illustrated in the following example:

From Wikipedia

In extrapolating the above scheme into the question's case, I can't understand the reaction at all. Firstly $\ce{Br}$ is a good leaving group, if $\ce{E_1CB}$ is to happen, I think it must be taken that $\ce{Ph}$ is the leaving group. Secondly, let's leave the leaving group part and focus on the hydrogens, how did the hydrogen turn into a deuterium in the final product?
I have seen the user TRC's answer where they simply claim it is because of the strong base, I feel the logic is flawed, since it could have been $\ce{E_2}$ as well in that case due to the bromine existing. So, I don't find the argument fully convincing.

The question is ambiguous for lab purpose. But in theory you should read first 2 comments of TRC below his answer. E1cb mechanism can only produce carbanion intermediate.

I agree with @Jay - questions like these are only good in theory. In practice you will of course have E2 as well - which will give the non-deuterium elimination product. My answer was only in accordance to the scenario given in the question itself, the only possible reason to explain both the reactant, reagent and already given product.

This would be a good read to understand the difference.


text:
Can sodium metal deprotonate terminal hydrogen atom from propyne to form propynide?
I've read that sodium metal is used to form ethynides but for forming propynide, they have used $\ce{NaNH2}$. Is there some reason for not using sodium metal for propynides as well?
Source: Chemistry NCERT class 11 part-II page 394

Under what conditions? Using what solvent?

The comments of this previous question appear relevant https://chemistry.stackexchange.com/questions/45203/reaction-of-sodium-with-terminal-alkyne

related https://chemistry.stackexchange.com/questions/27959/reactions-of-alkynes-with-sodium-ammonia-vs-sodium-amide

Does this answer your question? Reactions of alkynes with sodium/ammonia vs sodium amide

Sodium metal can indeed be used to deprotonate propyne, although as we will see sodium amide would be more selective.
Addison et al. [1] report the reactions of melted sodium metal with both ethyne and propyne. In both cases a mixture of products is obtained; both displacement of the acetylenic hydrogen to form an acetylide salt and reduction of the alkyne to an alkene occur. This is in contrast with sodium amide, which has no reducing power and thus the acetylide salt would be obtained more selectively with the amide.
One difference between the two hydrocarbon substrates is the acetylide salt that is obtained. With propyne the 1:1 salt is found ($\ce{Na^+C3H3^-}$), as the methyl group is not deprotonated. With ethyne, however, the deprotonation goes all the way to the di-anion, forming $\ce{(Na^+)2C2^{2-}}$ as the salt product. Sodium amide would be expected to accomplish only one deprotonation with either ethyne or propyne.
Reference
1.
C. C. Addison,   M. R. Hobdell  and  R. J. Pulham, "The reactions of acetylene and propyne at the surface of liquid sodium ", Journal of the Chemical Society A (1971), 1704-8. https://doi.org/10.1039/J19710001704

Why would sodium amide deprotonate only from one side?

Would not be a strong enough base to remove the second proton from ethyne. Sodium metal can reduce both protons.

I think  the reason for using NaNH2 for propyne and Na(with suitable solvent) for ethyne has to do something with the acidity of ethyne and propyne.
While any terminal alkyne is generally very acidic,ethyne is more acidic than propyne.In propyne,there is a methyl group connected to the sp hybridized  carbon leading to a slight reduction in the acidic nature of the terminal hydrogen(+I effect of methyl group).Hence it needs a better base like NaNH2.


text:
How can PCl₆⁻ be formed
So my teacher told me that PCl6- exists, But how? What I don't understand is that Phosphorus has 5 valence electrons hence a covalency of 5 but this doesn't allow for 6 fluorine atoms to bond with Phosphorus. Can someone show me the hybridisation or the technical reason and my teacher left off by saying that P can have hypervalent due to d orbitals but if electrons arent present to bond then how does this happen?

Try to search this site for similar questions, as similar topics have been asked multiple times. If googling, insert the term "site:chemistry.stackexchange.com".

Related  for SF6 and false idea of involving of d orbitals in $\mathrm{sp^3d^2}$ hybridization.

the main doubt is on excitation P can only have an sp3d hybridization that will provide 5 hybridized orbitals then how can a sixth one appear out of nowhere?

@orthocresol is the author of the linked answer and has much better insight, ragarding what from the link is applicable and what is to be added or modified.

@Poutnik OP's name suggests he is from India, where these hypervalent hybridizations are widely and unambiguously taught throughout grades 11 and 12. I have recently finished my 12th, and before I joined this site I also believed this false myth. Most unfortunate - those who don't take up chemistry in higher studies never even get to know that this entire concept is flawed.

I checked JD LEE by Sudarshan Guha, it has nothing about this in it. So, seems to be a rare point.

@TRC I remember being taught that as well in 80s. // If there is some test/exam/homework question that insists on the wrong sp3d2 hybridization answer then both electrons on the 6th orbital would come from F- anion, forming 6 fully occupied sp3d2 orbitals. // BTW LiPF6 solution in dialkylcarbonate mix is the electrolyte in Li-Ion cells.

@Poutnik - Exactly. That's what we are taught. The point about lithium ion cells was a new info for me, so thanks :)


text:
Why doesn't hydrolysis of isocyanides take place in basic medium?
Acidic hydrolysis of alkyl isocyanide gives rise to primary amines and formic acid. But alkyl isocyanides do not undergo hydrolysis in basic medium. Why?

The main reason for the inability of alkyl isocyanides to undergo hydrolysis in basic medium is because the negative charge at the carbon atom $(\ce{R-\overset{+}{N}#\overset{-}{C}\!:})$ repels the nucleophile $\ce{OH-}$ due to both being negatively charged. See this answer for more information.


text:
How do two species from the same buffer system have different pKa values?

$$
\begin{array}{lr}
\hline
\text{Reagent}~(\pu{0.1 M}~\text{solutions}) & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Ammonia} & 9.25 \\
\text{Acetic acid} & 4.75 \\
\text{Ammonium chloride} & 9.24 \\
\text{Sodium acetate} & 4.75 \\
\hline
\end{array}
$$
Using the equation below, calculate the ratio of $[\text{base}]$ to $[\text{acid}]$ to get the buffer of desired $\mathrm{pH}.$
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\text{base}]}{[\text{acid}]}$$

I am tasked to create a buffer solution with a $\mathrm{pH}$ level of $9.02.$ Given a table filled with the $\mathrm{p}K_\mathrm{a}$ values of two different buffer systems, I'm not sure if I should either use the $\mathrm{p}K_\mathrm{a}$ value of ammonia or the $\mathrm{p}K_\mathrm{a}$ value of ammonium chloride for the calculation.
Why do ammonia and ammonium chloride have different $\mathrm{p}K_\mathrm{a}$ values despite being from the same buffer system?

9.24 is no different from 9.25.

The table is misleading. Ammonia itself as an acid would produce $\ce{H+}$ and $\ce{NH2-}$, with a very very low Ka. Ammonia actually acts as a base. The value quoted in the table is the pKa of ammonia's conjugate acid, ammonium ion

I actually had the same thought on why Ammonia had a pKa when it acts as a base. Anyways, thank you so much for everyone's help!

If ammonia bothers you, so should sodium acetate, as it is the same case. Both refer to pKa of their conjugate(d?) acid.( ammonium and acetic acid )

Yup, also Sodium Acetate, but for this problem, I didn't bother minding Acetic Acid and Sodium Acetate since it wouldn't help me in solving the problem.

In fact, it could help, as you could more easily see, why ammonia and ammonium chloride refer to ( practically ) the same value - the acidity constant of ammonium. The tiny difference is due different data source.

Even though both are practically the same value (9.24 & 9.25), would you consider one value to be "more" correct than the other? or are they both just as correct as one another?

Can a constant (for the same conditions) have 2 different, equally correct values ? (rhetorical question ). The answer to quour question is matter of short research, which I would let on you. It is on an edge on a rounding error. Another question is the true value versus equally confident different values, due statistical errors.

I see. It's more of a statistical thing rather than a chemistry thing. I'll look into it for more information. Thanks!


text:
Why is the Tolman electronic parameter defined the way that it is?
I'm currently writing up a report on Hydrocyanation and I've been reading up on parameters people regularly use to justify their observations in regard to the performance of certain catalyst systems without too much regard as to how they are defined.
That is how I came upon Chadwick Tolman's Electronic Parameter $\chi$, which I was told is defined as follows:
The electron donating or withdrawing effect of phosphorous ligands can be measures using IR-Spectroscopy by comparing the frequencies of the A1 C-O vibrational mode ($\nu(\ce{CO})$) of a complex $\ce{[LNi(CO)3]}$ and the reference complex $\ce{[Ni(CO)3P(tert-Bu)3]}$.
Ergo:
$\chi$ = $\nu$(CO,L) $-$ $\nu$($\ce{CO}$, $\ce{P(tert-Bu)3}$)
If $\chi > 0$, Ligand causes net withdrawal of electron density towards the metal center (in this case, Nickel) compared to the ligand $\ce{P(tert-Bu)3}$.
If $\chi < 0$, Ligand causes net donation of electron density towards the metal center.
The reason that we use $\ce{[Ni(CO)3P(tert-Bu)3]}$ as a reference and not $\ce{[Ni(CO)4]}$ is because we are interested in comparing it to other phosphorous ligands and not carbon monoxide, which isn't a very interesting ligand in homogenous catalysis.
So here are the two questions:
Is my formula correct? Because I cannot find a short formula that defines $\chi$ in a single paper.
Is the reason for $\ce{P(tert-Bu)3}$ as a reference correct?


text:
Do CHF3 and acetone form a hydrogen bond?
I haven't been able to find a reference confirming that fluoroform forms hydrogen bonds with acetone. Do they?

See https://digitalcommons.usu.edu/chem_facpub/198/ and https://socratic.org/questions/592bd4b8b72cff0244178cc7

Apparently, it depends on your definition of a weak hydrogen bond. @Avon97

See also Hydrogen-bond dynamics of C-H...O interactions: the chloroform...acetone case

Acetone's $\ce{H}$ and  fluoroform's $\ce{F}$ and Acetone's $\ce{O}$ and fluoroform's $\ce{H}$ can H-bond simultaneously (You can visualize this by aligning the molecules)

@KarstenTheis Can $\ce{F}$ and $\ce{H}$, H bond in here other and $\ce{O}$ and $\ce{H}$ ? Its confusing in here with both H bonds the intermolecular force is strong.

This might help in your problem


text:
Reduction of conjugated ketone with lithium aluminium hydride
I came across a question about reduction of a conjugated ketone with $\ce{LiAlH4}$.

Although it seemed quite easy at first sight, everything became cloudy as I dug deeper. My doubts regarding the question are:

I have heard that conjugated double bonds are reduced by $\ce{LiAlH4}$. Is there any chance that the Oxygen from the ether group takes part in conjugation?
If not, why?
Is there a chance that the ether is cleaved during the acidic workup?
What is the true mechanism of this reaction, and does it involve the following intermediate?
Please help me with this question. Thanks in advance.

The reagents  should read 1) LiAlH4, 2) H3O+. After step 1), "d" is the product. After hydrolysis, "a" is formed.

You might be interested in this: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not/87195#87195

Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

I'll take the opportunity to expand on the Comments and links I provided previously. The intermediate you proposed does not occur during the LiAlH4 reduction of the enol ethers of dihydroresorcinols such as enone 1. The LiAlH4 reduction is a straightforward 1,2-reduction of the carbonyl group. If this reduction is worked up with water, no aqueous acid, allylic alcohol 2 would be expected. This species is no ordinary allylic alcohol because loss of water from protonated 2 -- namely, structure 3 -- leads to a highly stabilized allylic carbocation 4 with participation of the oxygen electron pair. Addition of water and loss of ethanol from 4 via intermediates 5 and 6 leads to the transposed enone 7.
This methodology is exploited in one aspect of the Stork-Danheiser reaction wherein the enol ether of dihydroresorcinol 8 (R1=isobutyl, R2=alkyl) is converted into a 4-alkyl cyclohexenone.


text:
Can all acids be classified as a Lewis acid?
There are three definitions of an acid that I am familiar with: Arrhenius acids, Brønsted–Lowry acids, and Lewis acids. From left to right, each new definition of an acid is broadened to include a wider variety of chemical species.
If a Lewis acid is the broadest category of acids, which necessarily includes all Arrhenius acids and Brønsted–Lowry acids, can all acids be considered a Lewis acid?

Logically the answer to your question can only be a robust yes

@Alchimista And of course, by extension, all bases are Lewis bases?

@Alchimista OP would need to start with true statement. The answer is no and questions lacks research. Broadest definition is Usanovitch's, and Lewis acids don't include Brønsted acids, which are adducts, by Lewis definition.

@Mithoron Interesting, I was not aware of Usanovitch's definition. So a Lewis definition is not the broadest. And indeed, not all lewis acids include Bronsted acids. According to Wikipedia, "A Lewis base is also a Brønsted–Lowry base, but a Lewis acid doesn't need to be a Brønsted–Lowry acid." I should have been more careful. Also, would it be fair to say that all Lewis bases are Bronsted bases?

With bases, it's technically true that one equals another, but in practice not necessarily. Also it's more correct that things act as acid or base then are.

@Mithoron I know. That was the point. There was actually no question. Yes I could have said that the "if" isn't satisfied. Perhaps I've been not very correct with OP but I was sure that this would have emerged.


text:
Preparation of Ferrocene
Although many synthesis methods of ferrocene are discussed in literature, I want to know about the black precipitation which occurs when $\ce{FeCl2・4H2O}$ in $\ce{DMSO}$ solution reacts with cyclopentadienyl anion in ether solution. What’s this?
At first, crushed $\ce{KOH}$ ($\pu{20 g}$) and diethyl ether $\pu{100 mL}$ are added into a $\pu{300 mL}$ erlenmeyer flask, then the solution is stirred in $10$ minutes. In addition, cracked cyclopentadiene (monomer)  $\pu{5.5 mL}$ is added into this solution.
Crushed $\ce{FeCl2・2H2O}$ ($\pu{6.5 g}$) and $\ce{DMSO}$ $\pu{25 mL}$ are added into a  $\pu{100 mL}$ beaker, then the solution is stirred. This solution is added drop by drop into the flask. This is the time when the black precipitation forms.

The major solvent is dirthyl ether and the other product of the reaction is KCl so my guess is that the solid product is KCl. No idea why is it black, perhaps is contains some Fe(0).

Based on some some results I am seeing online (but I am struggling to find primary sources) the opposite effect seems more likely. The iron could be getting oxidized and thus precipitating Fe3O4, maybe also Fe(III) oxide/hydroxide species mixed in.


text:
How to determine chlorophyll content from spectrophotometry data (absorbance vs wavelength)?
I boiled spinach solution (spinach leaves and 90% methanol) in solutions of varying $\mathrm{p}H$. I used a spectrophotometer to obtain wavelength ($\lambda$) vs absorption, and want to investigate chlorophyll degradation between the different $\mathrm{p}H$ and determine which $\mathrm{p}H$ would be most ideal to minimize chlorophyll degradation. I want to check if I am analyzing my data right: Can I simply take the $\lambda_{max}$ for each measurement and use that in the Beer Lambert law to determine the concentration? I know the light pathway (cuvette width) and absorbance (from spectrophotometer), but do not know the molar absorptivity. I was thinking that since its value would be a constant, I would leave it as with whatever concentration I determine from the calculation, and it would still be comparable. However, if I use this methodology for analysis, I would be ignoring the fact that the wavelengths at which the $\lambda_{max}$ is observed is different for each sample. How would I incorporate that because that is surely important to mention and analyze as well.
Or could I just analyze the data without determining the concentration and simply use the absorption to compare results? Is absorbance the same as concentration when talking about chlorophyll?
I’m stuck analyzing the data I obtained from the spectrophotometer. For each sample, I have the absorption against wavelength from the spectrophotometer. What I am planning on doing is to take the $\lambda_{max}$ for each sample and use the absorption from there so that I can utilize Beer’s Law to determine concentration. (Also I want to double check if this law is fit for my investigation)
$\textrm{Absorbance} = \textrm{molar absorptivity} \times \textrm{length of light path} \times \textrm{concentration}$
I know the absorbance and length of light path (length of cuvette), but I am still unsure what molar absorptivity exactly is. How do I determine it? I read that this can usually be found in literature, but I’ve been unable to find it for chlorophyll. Therefore, I was thinking that as molar absorptivity should be a constant (is this true? I have also read that it depends on wavelength), I could leave it as is in the calculations. What I mean by this is:
If molar absorptivity = $\epsilon$, I could leave the concentration as, for example, $1.49/\epsilon$. Would that make sense in a lab report?
However, obviously, it would be much better if I could get precise molar concentrations for each sample.
My concern for this method of processing is that I will be ignoring how the wavelength $\lambda_{max}$ varies in every sample. I mentioned this earlier, but if what I read about molar absorptivity varying based on wavelength is true, wouldn’t that affect my concentration calculations? Does that make sense?
In short, if I want to measure chlorophyll degradation (or amount of chlorophyll present), do I need to:
a) Just compare $\lambda_{max}$ (wavelength where absorption peaks) between the different $\mathrm{p}H$ samples
b) Just compare absorption at lambda max between the different $\mathrm{p}H$ samples
c) Compare absorption at one specific wavelength (if so, how do I determine the wavelength if $\lambda_{max}$ differs for each measurement?)
d) Determine concentration from absorption based on $\lambda_{max}$ using Beer’s law, and compare that between the $\mathrm{p}H$ samples?

I strongly recommend giving this a look: https://en.wikipedia.org/wiki/Molar_attenuation_coefficient

How to determine chlorophyll content from spectrophotometry data (absorbance vs wavelength)?

You have to consider two pigments, chlorophyll A and chlorophyll B, along with other substances that absorb in the same range. This paper gives extinction coefficients for both pigments, and a recipe to measure at two wavelengths to determine both in a mixture. There are newer papers with extinction coefficents for other solvents that also discuss degradation product, e.g. this one (behind a pay wall).

In short, if I want to measure chlorophyll degradation (or amount of chlorophyll present), do I need to:
a) Just compare $\lambda_{max}$ (wavelength where absorption peaks) between the different $\mathrm{pH}$ samples
b) Just compare absorption at lambda max between the different $\mathrm{pH}$ samples
c) Compare absorption at one specific wavelength (if so, how do I determine the wavelength if $\lambda_{max}$ differs for each measurement?)
d) Determine concentration from absorption based on $\lambda_{max}$ using Beer’s law, and compare that between the $\mathrm{pH}$ samples?

I would keep things simple and choose a wavelength and stick with it, using the absorption as an estimate of the total concentration (option c). In the image below, you see a conceptual diagram of the absorption spectra. Your observed spectrum will be some summation of these and the spectra of other absorbing material in your sample. If you measure at about 650 nm, you will capture both A and B. In reporting, you should also show the complete spectrum of your samples to give the reader an idea how different the composition from sample to sample is.

Source: https://ib.bioninja.com.au/standard-level/topic-2-molecular-biology/29-photosynthesis/action-spectrum.html

Thank you for your answer. I still have a few questions:
So if I go with option C and compare absorption at a specific wavelength, I won’t need to utilize extinction coefficients, as that is only needed for Beer’s law, is that correct? Also, could you clarify why the other options would not be a good idea? I was thinking that I could still try comparing the data using one of the other options like b) where I could compare absorptions levels at each respective lambda max in addition to option C, but would that be redundant?

You don't know why lambda max shifts. It could be a different composition of the absorbing species, solvent or pH effects (you could minimize those by measuring at the same pH even when incubating at different pH values), or something else. The community could help you further if you posted examples of two spectra with different maxima. In the end, you have to acknowledge that the task is complex, and that a single claimed concentration of "chlorophyll" will not do it justice. Nevertheless, it is possible to learn something from the data you collect, and make more general claims based on it.


text:
Nucleophilicity of sterically hindered enolates
Problem

Answer

 C

Question
Here $\ce{KH}$ obviously acts as a base, and forms the enolate from the ketone, which subsequently act as nucleophiles causing an SN2 reaction on the $\ce{CH3I}$. When this happens twice, we get the compound in option A.
However, after the removal of a third proton from $2,6$-dimethylcyclohexanone, the following enolate is formed -

This enolate will attack through the $\mathrm{C-2}$ atom. This atom is sterically hindered as it is a $3^\circ$ carbon. Hence I reasoned that like tert-butoxide, this enolate wouldn't act as a nucleophile either, so the reaction effectively stops at A.
Since the answer is C, it turns out that the sterically hindered enolate still does react further, and in good yield. Why is my reasoning incorrect? Also, this question is not from a standard textbook, so I would appreciate any reference regarding if and why/why not the enolate will react further.

T-butoxide anion statically hinders only in elimination reactions. Eg, the Hoffmann product is formed when t-butoxide is used. I have never come across a retardation of the reaction due to steric hindrance of the nucleophile (to an appreciable extent) in substitution reactions like Sn2. Steric hindrance of the substrate generally decreases the rate of Sn2. I may be incorrect though.

@AdityaRoychowdhury - I meant that if we have, say ethyl chloride, then tert-butoxide ion prefers elimination reaction (to give ethene) over substitution, because it's sterically hindered. So the yield of substitution product decreases. I would definitely have understood if the product of option C had been given as one of the products, but with a lower yield. The fact that it's yield is 81% is something I felt doubtful about.

That's just a methyl substituent, if you were adding t-butyls to the ring then I suspect your reasoning would be true.

@Mithoron - I see... so since the methyl itself is small it allows the attack of a bulky nucleophile. These results are best determined experimentally, of course. I'll still wait though if someone can find a reference. Thank you for the explanation :)

@Rishi - it's from a Fiitjee AITS paper - a mock test for JEE Advanced, but then probably you already know what AITS it ;)

@TRC One could say you kinda tert-butylate MeI - there's no problem with introducing t-butyl for ex. in benzene.

The permethylation of cyclohexanone with KH/MeI is described here JOC paper. The yield quoted is 96% and other ketones are also exemplified.

Thank you. I visited the paper but could access only the first page. The table at the bottom had a "yield" column with two numbers - like for cyclohexanone, yield says 96(81). Could you please clarify what those two numbers are?

Also, in the remaining paper (which I could not access), have the authors discussed any effects of steric hindrance on the reaction?

@TRC you can access the full paper on SciHub (legality is questionable)

@TRC the first yield is GC yield, the second yield is product after distillation. I also can only read the first page. It is, I think significant, that the authors mention that the system needs to be refluxed to get the 4th Me to add in some cases, thoigh not cyclohexanone

I must seem like a complete fool now... what is meant by the GC yield?

Gas chromatograph yield

@TRC: The yield is actually based on analysis remaining $\ce{CH3I}$ present in the reaction mixture by oxidation to $\ce{CH4}$ gas.

One thing that will help you understand how this can happen at a position that otherwise looks hindered is to consider the geometry at that position. Carbon 2 is planar in the enolate anion, and the nucleophilic molecular orbital is more $\mathrm{p}$-like than $\mathrm{sp^3}$-like. The picture below highlights this geometry (note that the orbital shown is a simplification of the conjugated HOMO of the enolate anion):

Because the enolate anion is planar at carbon 2, the electrophile approaches from above or below. The methyl group at carbon 2 is not in the way.

It's important that the ring is not planar, only the O-C1-C2 motive is. You'd probably call this molecular shape half chair, or less specific twisted. But that is hard to draw with a schematic representation.

Agreed. The half-chair / pseudo-twist conformation representation adds cognitive load (and is not necessary to convey this point).

In my experience, while showing it is difficult and unnecessary, it's important to mention these simplifications. There are not few people who otherwise take the images at face value. This is not a trivial reaction and the other answer references a scientific paper. I also think that the way the question is posed, too much simplification isn't necessary.

Thank you. While I haven't learnt yet about the usage of molecular orbital theory in organic compounds, the structures and planarity at the enolate carbon helps to clear the point.

Although Ben Norris has given excellent explanation for why the steric hindrence is not playing the role in the reaction of cyclohexanone with excess methyl iodide to give tetramethylated cyclohexanone, there is an enother reason why reaction proceds so smoothly under given condition to give more than 80% yield of 2,2,6,6-tetramethylcyclohexanone. It is the exceptionally high reactivity of $\ce{KH}$ towards ketones compared to that of other strong non-nucleophilic bases such as $\ce{NaNH2}$, and even $\ce{NaH}$ and $\ce{LiH}$ (Ref.1):

As shown in the graph above, methyl tert-butyl ketone (pinacolone) has given 100% metallated enolate within $\pu{5 min}$ in the presence of $\ce{KH}$ at $\pu{20 ^\circ C}$ while it has taken more than $\pu{120 min}$ to give at least 20% metallated enolate with $\ce{NaH}$ under same conditions. For comparition, $\ce{LiH}$ is even less reactive than $\ce{NaH}$ (Ref.1). In a typical reaction ($\pu{25 mmol}$ substrate in $\pu{40 mL}$ of THF in the presence of $\pu{28 mmol}$ of $\ce{KH}$), cyclohexanone is enolated 100% within $\pu{1.5 min}$ and 2-methylcyclohexanone is enolated 100% within $\pu{6 min}$. In the latter case, which can give two enolates, the ratio of isolated enols (as silyl ethers) was found to be $33:67$ less substitured one to more substituted one indicating that reaction is more facile with making more stable enolate even in the presence of steric hindrence.
Thus, it is clear that the question in OP's post is directly related to the reserch done in Ref.2 (which is mentioned by Waylander in his answer. Accordingly, the authors in Ref.2 reported yield of 2,2,6,6-tetramethylcyclohexanone as 96% (GLC analysis) and as 81% (isolated yield by distillation), which is the exact yield given in the question. The procedure was as follows:

To a flask containing  $\pu{216 mmol}$ of $\ce{KH}$ in mineral oil $(\pu{40 mL})$, which is in a waterbath maintained at $\pu{25 ^\circ C}$, $\pu{220 mL}$ of THF was added first and then $\pu{5.2 mL}$ of cyclohexanone $(\pu{50 mmol})$ was added dropwise over a $\pu{5 min}$ period. The mixture was stirred for additional $\pu{5 min}$ and $\pu{13.5 mL}$ of $\ce{CH3I}$ $(\pu{216 mmol})$ was added dropwise over a $\pu{15 min}$ period. The reaction mixture was stirred for additional $\pu{15 min}$ before workup with water. The simple distillation of dried organic layer gave $\pu{6.25 g}$ of expected product (81% yield).

References:

Charles Allan Brown, "Saline hydrides and superbases in organic reactions. III. Facile reaction of potassium hydride with ketones. Rapid, quantitative formation of potassium enolates from ketones via kaliation," J. Org. Chem. 1974, 39(9), 1324–1325 (DOI: https://doi.org/10.1021/jo00923a042).
Alan A. Millard and Michael W. Rathke, "Procedure for the permethylation of ketones using potassium hydride and methyl iodide," J. Org. Chem. 1978, 43(9), 1834–1835 (DOI: https://doi.org/10.1021/jo00403a055).

Beautiful answer - I'm glad I posted this question on this site :) Is there a way I can mark all the three answers as accepted?

@TRC: As much as I know, you can't accept all answers. But, it is best if you can decide what answer gives you the best information on your question. It's up to you.


text:
Adding H2O2 in water does what to the pH of the water?
Not looking for numbers (unless required). I searched for this question on Google and found irrelevant results, with only the following link of some use but it doesn't appear to answer my question (unless I've misunderstood it).
https://www.researchgate.net/post/What_Chemical_interaction_occurs_when_hydrogen_peroxide_is_added_to_water
My previous understanding was that $\ce{H2O2}$ spontaneously dissociates into $\ce{H2O + O2}$, including in water. I had imagined this would have no effect on the $\mathrm{pH}$ of the water its added to.
The link above says if the water is acidic, $\ce{H2O2}$ will dissociate into hydronium + hydroperoxide ($\ce{H3O+ + HOO-}$). I imagine the production of hydronium will increase the $\mathrm{pH}$.
For example, if the water is slightly acidic, say $\mathrm{pH}$ $6.5$, will the addition of $\ce{H2O2}$ make the solution even more acidic?

How will production of more hydronium ions increase the pH?

(Rhetorical question) Is hydrogen peroxide a weak acid ?

@TRC I thought this is one indication of increasing acidity. That was a typo - I meant increasing acidity.

@Poutnik it is a weak acid but I am not certain if what I read in the link is correct.

The $\mathrm p K_\mathrm{a}$ of $\ce{H2O2}$ is $11.75$ in Wikipedia.  As a consequence, adding enough $\ce{H2O2}$ to make it concentration  $\ce{[H2O2]}$ go up to $\pu{1M}$ with change the original pH $6.5$ to $6.0$

Ok I can understand the pKa, but if all of H2O2 is disassociating into H2O and O2, why is the pH dropping?

Ordinary water in contact with air has a pH 5.5 ± 0.3, depending upon the quality of the air. This is due to some atmospheric  $\ce{CO2}$ that gets slowly dissolved into water. If your water had a pH equal to $6.5$, it means that it was kept out of contact with air. If it goes down to $5.5$ during decomposition of $\ce{H2O2}$, it may be due to some dissolution of $\ce{CO2}$ and it may have nothing to do with $\ce{H2O2}$.


text:
Why do some elements form allotropes and others do not?
I just studied in my chemistry class about the allotropes of carbon. But why does carbon form allotropes? Also why only carbon? Why do some elements form allotropes and others do not?

What kind of explanation do you expect? Elements are just different, much like people.

One can get some other allotrope of pretty much any element, if only try hard enough.

Your own reasoning – based on searching, reading and thinking – is supposed to be present to avoid the question closure for lack of own explicit  effort.  How do I ask a good question. // It seems you have skipped even basic searching.

https://chemistry.stackexchange.com/questions/48751/allotropes-of-phosphorus?r=SearchResults&s=4|39.3424 ... https://chemistry.stackexchange.com/questions/141580/which-allotropes-of-sulfur-exist-naturally ... https://chemistry.stackexchange.com/questions/28124/allotropy-in-nitrogen-and-bismuth-of-group-15


text:
Sodium acetate stoichiometry from baking soda and vinegar
I want to make sodium acetate from sodium bicarbonate and vinegar. I'm a bit stumped at calculating amounts of each reagent. I'm starting with 8% acetic acid vinegar. I was going to add, to 1 L of the 8% acid (80 mL acetic acid), 80 grams of bicarb soda. I came to this as the 80 mL acetic acid has a mass of ~83 grams, and I'm guessing I should use an excess of acetic acid. Is this right?

You should use excess of acid, but excess in terms of the number of moles. Reactions occur not by weight, but by moles. Weight of compound = number of moles x molar mass

The density of acetic acid is 1.053 g/ml, so you'll have 80 x 1.053g = 84.24g of acetic acid in 1L. The molecular weight of acetic acid is 60.052 so you have 80 x 1.053/60 = 1.404 moles.  Multiply this by the molecular weight of sodium bicarbonate which is 84.007 to get the weight of sodium bicarbonate required for equivalence = 117.946g.


text:
Effective nuclear charge on 3p electron of aluminum
Using Slater's rule I calculated the effective nuclear charge on 3p electron of aluminium as:
Config: $\ce{1s^2 (2s^2 2p^6) (3s^2 3p^1)}$
$S = 2(0.35) + 8(0.85) + 2 = 9.5$
$Z_{eff} = Z - S = 3.5$
The Wikipedia page states its value as 4.066.
Which one is correct? Did I go wrong somewhere?

You're doing it correctly. The value you cite is from E. Clementi and D. L. Raimondi. "Atomic Screening Constants from SCF Functions." Journal of Chemical Physics, volume 38, number 11, 1963, pp. 2686–2689. They used a very primitive (by today's standards) method for computing the screening constants. It is entirely possible that their number is off. It's also known that Slater's rule is good for the first 2 rows, then starts to diverge from experiment from there, so aluminum falls into that category.


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Chlorophyll content in relation to the pH of the solution in which it is boiled
What would be the expected trend of chlorophyll content in relation to the pH of the solution in which it is boiled? Would there be the least chlorophyll degradation in basic or neutral solutions? Logically, I thought that both extremes of pH 1 and 13 would cause chlorophyll degradation, so the ideal solution pH would be 7, neutral. However, I read that acidic solutions increase degradation due to the presence of hydrogen ions, so I thought that the higher the pH, the less degradation. What would be hypothesized?