text:
How do you derive the boiling point elevation equation using Raoult’s law and Clausius-Clapeyron relation?
I know that Raoult's law is $P_{solution}=X_{solvent}P_{solvent}$
and Clausius-Clapeyron equation is $ln(\frac{P_{2}}{P_{1}})=-\frac{L}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$
But how do you derive the boiling point elevation which is:
$\Delta T=K_{b}m i$

You may find useful these links for text and formula formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

p=f(T,x). dp=(dp/dT).dT + (dp/dx).dx. for dp=0,  (dT/dx)=-(dp/dx)/(dp/dT)

What you have done by adding an involatile solute to the liquid is to lower the activity of the solvent. This lowers the vapor pressure of the liquid.
Now if we assume that we live in a perfect fluffy nice world where all activity functions are equal to one. Then we need to merely consider the mole fraction of the liquid which is solvent (x1).
For example we can have water which when pure is 55.4 moles per litre.
If we assume that the molar volume (volume occupied by a mole of solute) is the same  no matter what the concentration is in the solvent, for arguments sake we have a sugar solution. For solution of glucose in water the glucose occupies a volume of 112 ml per mole of glucose. Thus if the mole fraction of glucose (x2) increrases then the number of moles of water in one litre will decrease which will make x1 drop.
If we assume at a constant temperture that the vapor pressure is proportional to x1, this is as we assume that activity of the water is proportional to x1. As for a dilute solution where x2 is small we can write
x1 = 1 - k[2]
Where [2] is the concentration of the solute in moles per litre and k is a constant.
Now as pressure is given by the equation
P = P* x1
Where P* is the vapor pressure of pure water then for our ideal system we can understand that there will be a linear decrease of vapor pressure as the concentration of the glucose increases.
To restore the vapor pressure back to the original value we need to increase the temperture.
Now we can go from the Clausius-Clapeyron equation to give us
P = k exp (-Delta H vapp / RT)
For the water to boil then we need the vapor pressure to be equal to the atomospheric pressure, let us assume that we have exactly 1 atmosphere pressure ansd the boiling point of water is 373 K.
We can then write the equation for the mixture
P = x1 k exp (-Delta H vapp / RT)
The gas constant R is 8.31
If Delta H vapp is 40.8 x 10^3
We can make some progress, our value of k will be 519695
I used excel to then consider the tempertures required to get a vapour pressure of one for dilute solutions of glucose and I found that the boiling point of the solutions was described by the line T = m[2] + c where m was 0.4921 and c was 373. I used the solver to obtain the boiling points for the solutions.
Thus I have shown that ΔT=Kb Mi is true at least for dilute ideal solutions.
Now this will only work for dilute solutions, when we get concentrated glucose which will be what I call "damp sugar" things are going to change. One of my professional interests is in very concentrated solutions and the way that the activity functions change. I have an interest in things like 5 moles per litre solutions of sodium chloride and mixtures of things like choline chloride and ethylene glycol.
Firstly for fun I considered "damp sugar" AKA 4 M glucose, for a range of glucose solutions where the concentration ranged upto 5 moles per litre I found that we best describe the system with a polynomial.
T = 373 + k1[2] + k2[2]^2 + k3[2]^3
Where k1 = 0.548, k2 = 0.0098 and k3 = 0.0186
For some systems things work better if we consider the amount of solute per kilo of solvent (m) which is different to the concentration per litre (M).
I recalculate for moles per kilo.
For very strong solutions concentrations expressed in moles per kilo of solvent tend towards infinity so I tend not to like this method of expressing concentration.
If we consider the polynomial which describes the boiling point of our ideal sugar solution then we get
T = 373 + 0.5101m - 0.00034(m^2)
When the concentration of sugar is low then regardless of if we express the concentration in moles per litre or moles per kilo of solvent the boiling point will increase in a linear manner. It is only when we get strong solutions that things will become more complex.


text:
Enthalpy contribution and heat capacity
Imagine having at the input of a burner 105 mol/h of $\ce{O_2}$ at 50 °C. The enthalpy contribution of this input mass is calculated using the Kirchhoff equation
$$\Delta H = \Delta H^° + \int_{25 °C}^{50 °C} C_p \,dT $$
where
$$ \Delta H^° = 0 \dfrac{\text{cal}}{\text{mol}}$$
$$C_p = (7.129 + 0.1407 \cdot 10^{-2} \hspace{3pt} T - 0.6438 \cdot 10^{-5} \hspace{3pt} T^2) \dfrac{\text{cal}}{\text{mol}  \cdot °\text{C}}$$
Compute the definite integral
$$ \int_{25 °C}^{50 °C} C_p \,dT = \Big( 7.129 \hspace{3pt} \Big[T\Big]^{50 °C}_{25 °C} + 0.1407 \cdot 10^{-2} \hspace{3pt} \Big[\dfrac{T^2}{2}\Big]^{50 °C}_{25 °C} - 0.6438 \cdot 10^{-5} \hspace{3pt} \Big[\dfrac{T^3}{3}\Big]^{50 °C}_{25 °C} \Big) $$
The result will be a number whose unit of measure will be $\dfrac{\text{cal}}{\text{mol}   \cdot °\text{C}}$. While enthalpy is measured in $\dfrac{\text{cal}}{\text{mol}}$, there is a discrepancy between the units of measurement.
Can you help me?

You have omitted the multiplication by T within  the integration. E.g. $\int{\vec F \vec {\mathrm{d}l}}$ does  not have dimension N, but Nm=J.  Similarly C_p.dT has dimension in your case cal/mol.

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text:
Can we find dipole moment of water from a familiar property?
There are many videos on Youtube showing deflection of a stream of water by a charged rod. My question is whether we can do a rough calculation of the dipole moment of water molecules using this deflection.
My reasoning is: we can approximate the water stream as a short cylinder which, as it falls, travels a measured distance horizontally. This gives us a horizontal force which we can relate to the charge and induced charge on the surface of the rod and cylinder of water.
I am omitting my calculations, which did not come close. I assumed a cylindrical stream of 3mm diameter, 10 mm long, 23 cm above a sink, displaced about 3 cm by a negatively charged rod placed 4 mm from the stream (near the top). I have some confidence in the force measured, little beyond that.
There are a lot of geometrical simplifications to be made but I may be missing subtleties about the way charge behaves on the surface of water and how to relate that to the dipole moment. If anyone has a plausible calculation (or just an outline, highlighting sources of potential error), or reasons why this cannot be done, either one would probably help my intuition.

Cool question! You might find this paper useful.

@Paul: This was an interesting article, thanks. I used a cylindrical stream so my  volume was larger but the deflection force was also comparable to the force of gravity.

The dielectric constant is fairly easy to measure accurately, though it might require a simple AC current meter and source.

@DrMoishePippik: Yes, don't see a way around that. Unless someone has a better idea I think this must be taken as measured/known.

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text:
Comparing stability of 1,3-dioxolan-2-ide and 1,3-dithiolan-2-ide carbanions
I'd like to compare stability of 1,3-dioxolan-2-ide and 1,3-dithiolan-2-ide carbanions:

I think there is no mesomeric or hyperconjugation effect and we should consider inductive effect.
Oxygen is more  electronegative compared to sulfur, and therefore should pull the electron cloud in a better way to stabilise the carbanion. Also, as sulfur is bigger in size, therefore I think it would also destabilise the ring creating some strain, resulting in 1 being more stable.
But the answer says that 2 is more stable. Where am I wrong?

While the acidity of 1,3-dithiolane is greater than 1,3-dioxolane at C2, it is nonetheless susceptible to fragmentation. That  is why 1,3-dithiane, the 6-membered ring is used as an umpolung  reagent.

sorry i am not familiar to umpolung reagent

Did you google "umpolung reagent"? Try this: https://en.wikipedia.org/wiki/Corey%E2%80%93Seebach_reaction

what happens if i take 1,3-dithiolane this as my reagent? will it not give hydrolysis or some other bond will be broken?

Hydrolysis of 1,3-dithiolane (Hg++, Ag+, etc.) will give formaldehyde and 1,2-ethanedithiol. What is unsuccessful is to attempt tp form the 2-lithio species and do an alkylation with, say 1-bromopropane, with the intent of forming n-butyraldehyde after hydrolysis. The anion fragments to ethylene and the "thioacetaldehyde" enolate. The fragmentation does not occur with 2-lithio-1,3-dithiane.

I almost understood everything ,but what is "thioacetaldehyde enolate".is it(https://pubchem.ncbi.nlm.nih.gov/compound/Thioacetaldehyde), if yes i am not able to figure out how ethylene forms, if possible please provide me with the reaction.Thanks!

My bad! The sulfur entity is not the enolate of thioacetaldehyde but rather H(C=S)S-. Just push the electrons from C2 around the ring and ethylene will pop out.

oh got it ,thanks for telling me that.

This is only a theoretical answer with electronic effects. Deciding stability involves on number of factors.

I think there is no mesomeric or hyperconjugation effect and we should consider inductive effect.

Your assumption was incorrect. Because sulfur has vacant d-orbitals so that electrons from carbon can be involved in delocalisation with these vacant orbitals. So here sulfur acts as an electron withdrawing group which stabilises the negative charge on carbon.
Whereas this effect was not observed in case of oxygen since it doesn't have vacant orbitals. So oxygen can only exhibit −I effect on carbon. Therefore 2 is more stable than 1.


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Does 'aromaticity' apply to individual resonance structures or the hybrid?
When we define the aromaticity of a compound, is it meant to describe the resonance hybrid of the compound, or for individual resonance structures of the compound?
I think the latter is wrong because in resonance structures all bonds are localized so there is no delocalisation of electrons, hence we cannot call it aromatic or non-aromatic. Also, resonance structures are hypothetical; they don't exist, so it doesn't seems reasonable to classify them as aromatic, non-aromatic, or anti-aromatic.
Yet, it's common to see some facts rationalised along the lines of "this resonance structure is aromatic and hence has a greater contribution in the hybrid". (For example, Is the aromaticity broken in some resonance structures of para-nitro-aminobenzene? where @ron begins his answer with the line some of the resonance structures in your figure are not aromatic. ) Is this usage of the term 'aromatic' correct / well-defined?

Aromaticity is nothing which is defined based on structure alone, especially not from diagrammatic structures. This includes, but is not limited to, resonance hybrid representations. As aromaticity is in itself a special form of resonance, and resonance structures (always more than one) are only (very, very crude) approximations to delocalisation, a single structure can never be called aromatic.

@Martin-マーチン Agreed, but (in practice) there is a general trend of calling certain resonance structures aromatic to rationalise structure or other properties. For example, bond lengths in naphthalene / anthracene. I'm pretty sure this is a cheap & technically incorrect way of saying something else; but I can't quite put my finger on what that something else is.

For example, I'm just not sure whether it's a cheap way of saying "this resonance structure must contribute more in order to reproduce the correct electronic structure of the compound, which in turn arises because it maximises stabilisation from aromaticity". My current thinking is something along those lines, but I'm not sure. [FWIW, I think this is actually a very good question.]

There is some argument to be made that near equal bond lengths indicate near equal bonding situation, which in itself does imply a higher degree of delocalisation, which could be seen as more stabilisation, which could be inferred to as 'more aromaticity'. That is a structural argument which does  bring molecular structure close to electronic structure. As far as approximations go, not a bad one. Anyway, for aromaticity (in resonance terms) you need at least two structures of (almost) equal weight, or one predominant structure with more structures of significant weight.

https://chemistry.stackexchange.com/a/154063/9961

This^ is how determining aromaticity is possible. Talking about resonance structures is kinda missing the point

And about resonance structures https://chemistry.stackexchange.com/questions/51632/what-is-resonance-and-are-resonance-structures-real

https://chemistry.stackexchange.com/questions/162792/does-aromaticity-apply-to-individual-resonance-structures-or-the-hybrid


text:
Making C14 or C13 chlorobenzene site specific label
Imagine that you want to make chlorobenzene with a carbon-14 label in one site only, the only site you want it to be in is on the carbon bearing the chlorine. Can you think of a better and shorter synthesis from a labelled C1 compound such as methyl iodide or barium carbonate ? The literature method is
BaCO3 --> CO2 --> Acetic acid --> Ethyl acetate
Next react with pent-1,5-diyl magnesium bromide to form 1-methylcyclohexan-1-ol
Dehydrate to 1-methylcyclhex-1-ene, dehydrogenate with palladium to toluene, convert to benzoic acid with an oxidant, use sodium azide and acid to make aniline. Last use nitrous acid and copper(i) chloride on the aniline to make the chlorobenzene. I count ten reactions.

My suggestion is to react labeled nitromethane with formaldehyde
Then dehydrate to make nitroethene
Use the nitroethene for a Diels-Alder reaction with furan
Dehydrate with an acid to make labeled nitrobenzene
Reduce with hydrogen and palladium on carbon to make aniline
React thw aniline with nitrous acid to form the diazonium salt
Convert the diazonium salt with copper(I) chloride to the target molecule.
This is seven steps, can anyone come up with a shorter synthesis ?

Loosely related: How are organic compounds with radioactive atoms synthesized?

Reaction of dichlorocarbene with cyclopentadiene is reported to produce 6-chlorofulvene and\or chlorobenzene (58% yield) depending upon carbene precursor and reaction conditions. Note that the fulvene could likely be thermally rearranged to the chlorobenzene. The reaction producing chlorobenzene also seems site specific as the same reaction with indene is reported to produce only 2-chloronaphthalene (69% yield).

(Cont.) Since $\ce{^{13}C}$ labeled chloroform is commercially available, this small-scale route could likely be optimized to produce the desired $\ce{^{13}C}$ labeled chlorobenzene in one or two steps depending how you count.

Did you consult a database (like Reaxys [Elsevier], or Scifinder [The American Chemical Society]) for already known and published protocols?  Because how else entered #587958 in the catalogue of Sigma-Aldrich ... (to mention one commercial supplier [though synthesis on demand]).

I did consult Reaxys for the literature method, see J. Am. Chem. Soc. 1968, 90, 6, 1601–1605
Journal of the American Chemical Society, 1973, vol. 95, p. 8358,8363
Journal of labelled compounds and radiopharmaceuticals, 2004, vol. 47, # 4, p. 266 – 267 for the literature met hod

Regarding Ron's method, maybe it could work. One problem I see is that C14 chloroform sounds like a bit of a nightmare, I suspect that compared with the intermidates in the literature method it would be more able to difuse through plastic bags and gloves thus making it a greater radiological hazard.


text:
What is heat bath here?
At the beginning $\pu{1 mol}$ of air is compressed to $\pu{3 atm}$ at a of temperature $\pu{523 K}$. After a random process, the pressure of the air equates to $\pu{1 bar}$ at a temperature of $\pu{353 K}$. A amount of heat is transferred to a bath of $\pu{303 K}$.
Where does the heat come from? From the gas? Or the bath is the gas?

The bath is envisioned as a liquid at a constant temperature of 303 K.  The heat flows from the gas to the bath, driven by the higher temperature of the gas.

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.  It includes an option to describe measurements (like $T = \pu{300 K}$ by $T = \pu{300 K}$), so unwanted line breaks between the number and the symbol of the unit may be prevented, too.

In voluntary / spontaneous processes, heat is passed from the reservoir of higher energy to the one of lower energy.  Thus, heat from the reservoir of higher temperature transfers to the reservoir of lower temperature, until both share the same temperature.  The reverse is possible only if you invest work.
As an analogy, compare the situation with a pumped-stage hydroelectric plant, where energy is stored in the potential energy of the water in the upper basin.


text:
When does the Mukaiyama-Addition follow which transition state and how does it affect the stereochemistry?
in the course on enantioselective organocatalysis, we discussed the route to hexoses via organocatalytic aldol addition and subsequent Mukaiyama addition. The use of either TiCl4 or MgBr2 leads to different products. Apparently this is due to either the cyclic or open transition state. However the information I can find on the transition state of the Mukaiyama is unsatisfying, and it is impossible for me to answer the following questions: Why are there different transition states for MgBr2 and TiCl4? How do they look? How is the stereochemistry affected, and how can I see this in the transition state?
I attached a screenshot of the presentation slide regarding this problem. Thanks for any help in advance!
Best regards. :)


text:
Why is HONH2 basic?
I've been told that $\ce{HONH2}$ is a weak base because of its $K_b$ value of $\pu{1.1E-8}$. However, the $K_a$ value is larger ($K_a=K_w/K_b=9.9\times10^{-7}$), which I think would make it more likely to lose rather than gain a proton. What is the error in my understanding?

The Ka belongs to NH3OH+, not to NH2OH.(frequent mistake)

You are comparing the $K_b$ of $\ce{H2NOH}$ with the $K_a$ of its conjugate acid $\ce{H3NOH^+}$. To assess
whether $\ce{H2NOH}$ has a net acidic or basic character in water you have to compare its $K_b$ with the $K_a$ of $\ce{H2NOH}$ itself, not with the $K_a$ of its conjugate acid.
The autodissociation product $K_w$ relates the $K_b$ of a solute with the $K_a$ of the conjugate acid, not with the $K_a$ of the same solute species. We can only suppose, generally, that the proper $K_a$ for comparison is less than the conjugate-acid $K_a$ you get using the ion product.
So how much less? That varies from solute to solute. In the case of $\ce{H2NOH}$ the $K_a$ of the neutral molecule is a lot less than that of the conjugate-acid cation, too small even to be evident in aqueous alkali. With the neutral $\ce{H2NOH}$ molecule, as opposed to the cation, $K_b$ wins hands down and thus $\ce{H2NOH}$ will act in water as a weak base.

Care to produce an alternativecanswer maybe?


text:
A Multistep Reaction Provided with IR Spectra (Guessing Started with Esterification)

Below, IR spectra for this sequence are provided:

For compound E, I suggest the formation of a conjugated ester with the elimination of $\ce{-OH}$ group. The IR peak at $\pu{1732 cm-1}$ is a reasonable value for $\ce{C=O}$ stretch of ester.
Subsequently for F, I suggest the presence of a hydroxyl group due to the strong peak at $\pu{3400 cm-1}$. The reformation of hydroxyl group from $\ce{C=C}$ is unlikely due to the conjugated system. Plus, hydrolysis of ester will give broad $\ce{-OH}$ peak which is unlikely from the spectrum. Hence, I am inclined to suggest  vinylogue Michael addition of compound D to yield compound F. However, I am hesitant because of the possibility of dehydration.
For compound G, I would like to suggest the presence of alkyne yet the peak is strong. Possibly, would it be $\ce{CO2}$? If decarboxylation occurs, compound F must be β-keto-acid which is generated after hydrolysis of ester. The nucleophilic attack from compound D would also attack the carbonyl group instead of the $\ce{C=C}$ bond of compound E.

Compound
Suggestion

E

F

F'

F''

Possibly, my reasoning is wrong at the first place. I am confused currently. Any suggestion is really appreciated.
After reading the suggestion, I consider the potential of O-nucleophile. Reversible acidic hydrolysis of ester, E can form acrylic acid. Then, it will  form acrylic anhydride. The conjugated system might lower the C=O stretch to lower frequency. Is this plausible? Is the condition suitable for the formation of acid anhydride without phosphorous pentoxide? However, I am still clueless for compound G.

consider the possibility of lactone formation

The suggest of the metharcylate E makes sense given the reaction conditions and comparison of the IR spectrum given to you with an experimental spectrum by AIST/SDBS (entry SDBS 1220).  Does the original question really truncate provision of the conditions/reagents for $\ce{E -> F -> G}$?  Or, does it state that $\ce{D -> G}$ proceeds by telescoping/a one-pot procedure?

@Buttonwood The original question only states that this is a multistep reaction sequence. I am unsure about the possibility of one-pot procedure. However, only conditions/reagents are shown for the first step. I believe the subsequent steps will follow the similar conditions.

@Waylander Except for cropping the header, replacing English transmittance and wavenumber by Chinese, this is copying the work by AIST/SDBS (entry 1220 mentioned earlier) without proper credit where (to whom) is credit is due.


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Selective Removal of Sodium Acetate
I am currently working on an extraction and synthesis of sodium salicyate from acetylsalicylic acid and sodium hydroxide. After doing a liquid-liquid extraction from DCM, using a sodium hydroxide solution, I am left with a solution containing sodium salicyate and sodium acetate. I need a way to remove the acetate while leaving the salicyate alone. Thank you!

Considering limited solubility of salicylic acid in water and good solubility in some less polar solvents, I would try acidification and either crystallization at low temperature, either extraction. Probably the former.

Be sure that hydrolysis is complete (e.g., TLC with a lane of your starting material, a lane with your raw (but quenched) product, and a lane for co-spotting).  When then performing a back-extraction to yield salicylic acid, don't forget the addition of acid or a good amount will remain in the aqueous phase.  You go for the sodium salt after its acid was isolated.


text:
Comparing acidic strength of acetic and p-anisic acids
In p-anisic acid the phenyl ring with methoxy group acts as strong electron donating group by mesomeric effect (+M), whereas in acetic acid there is only an electron donating nature by inductive effect.
According to this, acetic acid should me more acidic compared to p-anisic acid, but it turns out that p-anisic acid is significantly more acidic:

Acid
Structure
$\mathrm{p}K_\mathrm{a}(\pu{25 °C})$

Acetic acid

4.76

p-Anisic acid

4.47

Why is p-anisic acid more acidic and what causes a significant difference in the $\mathrm{p}K_\mathrm{a}$ values?

An sp^2 hybridized carbon atom is more electronegative than an sp^3 hybridized carbon atom. The carboxyl group is attached to sp^2 carbon in p-Anisic acid whereas it is attached to an sp^3 hybridized carbon in acetic acid which can be a contributing factor in p-Anisic acid being more acidic.

A better way to interpret this would be that a stronger acid yields a more stable conjugate base upon losing an H+ ion. In the C.B. of these acids, the carboxyl group gets converted to COO-. In 4-methoxybenzoate(C.B of p-Anisic acid), this negative charge on COO- undergoes resonance thereby distributing electron density throughout the ring. In acetate ion, the negative charge remains localized and the +I effect from the methyl group makes it more unstable. Therefore, the C.B of p-Anisic acid is more stable than that of acetic acid and thus, p-Anisic acid is more acidic than Acetic Acid.

@Krish Ratta: Understandably, benzoic acid (4.20) and p-nitrobenzoic acid (3.41) are less acidic than p-anisic acid (4.47). What one doesn't know is the resonance effect of the MeO/NO2 groups relative to the sp2 vs. sp3 argument. Your argument holds because you were given the answer. A priori, p-anisic acid might have been argued to be less acidic than acetic acid (4.76).

@user55119 Agreed. No theoretical argument can enjoy precedence over pKa values and the best evidences infact are the experimental observations and pKa values. Theory is just an explanation to satisfy ourselves as to why we got the observed result. So it is always fair to argue in this domain and to get confused between acidities or such factors because theoretically, we do not know which effect dominates and why just by only seeing the structure of the compound. And yeah, knowing the answer surely helped me plot out a reason for p-Anisic acid being more acidic.

@user55119 I would also like to inform you about a slight error that creeped into your comment. As per the values furnished by yourself, Benzoic acid and p-nitrobenzene are more acidic than p-Anisic acid and not less acidic and the sp^2 or sp^3 argument doesn't hold here because all these 3 compounds that you are comparing have a benzene ring and a COOH attached to it. Only the groups attached at the para position are different which would make comparison of their acidities(by M and I effect) much simpler compared to the one in the O.P's question

@Krish Ratta: Yes, they are more acidic. My error in not proofreading more carefully. Thanks for catching that.


text:
Does the same name represents all resonance and hybrid structure of a molecule?
The name chlorobenzene represents the following molecule.

But the molecule shows resonance and its resonating structures are

So does the same name represents all the resonance structures? Does it also represent the hybrid structure of these structures? When I used this service to find the name of the last structure I found its name to be Chloraniumbenzenide. See this image. But interestingly when I used the same service to get back the structure from the name it gave me an error. Google search also didn't give good results.

Please note: The molecule shows resonance and therefore it has resonance structures. It has nothing to do with the classical concept of resonating, therefore the term you are using (while unfortunately get popular) conveys the wrong meaning. For more on resonance, I highly recommend https://chemistry.stackexchange.com/q/51632/4945

@Martin I know that. I know that only resonance hybrid is the correct lewis structure of a molecule and not the resonance structures, in the sense that it helps use explain observations. But that doesn't stop us from naming the resonance structures. Zeus is an imaginary character but it still has a name. In many places for example on wiki it can be seen that the name chlorobenzene is used for the first structure and resonance hybrid structure (3d model). My question was whether the same name represents all other resonance structures as well or do they have different names of their name.

https://www.wikiwand.com/en/Chlorobenzene

My comment was not about naming hypothetical structures, or even about whether or not they are correct (they are valid Lewis structures after all, a resonance hybrid technically is not as it is an extension to the original concept). These are concepts with limitations, and for nomenclature there are different rules or conventions that may or may not convey these technicalities. I was merely pointing out, that they are called resonance structures and not resonating structures. This might seem nitpicky, but I still think it's important.

One should be careful with the structures to be submitted to a structure2name conversion.
The one you showed for chlorobenzene are formally correct

however the first one is the preferred one because it describes a structure lower in energy than the one with separated formal charges and charge separation costs energy.  In addition to this, specific to this example, recall chlorine is more electronegative than carbon; in Pauling's scale (reference), EN(Cl) 3.16, EN(C) 2.55 - thus, it is not that likely chlorine would donate valence electrons to the more electropositive carbon.
Grossman condensed the (empiric) rules about resonance formulae to a set of four of decreasing order of importance.  To quote (emphasis in the original):

No first-row atom (B, C, N, O) can have more than eight electrons in its valence shell.  (The octet rule is less sacred for heavier main group elements such as P and S, and it does not hold at all for transition metals.)

Common error alert: Resonance structures in which all atoms are surrounded by an octet of electrons are almost always lower in energy than resonance structures in which one or more atoms are electron-deficient.  However, if there are electron-deficient atoms, they should be electropositive (C, B), not electronegative (N, O, halogen).

Resonance structures with charge separation are usually higher in energy than those in which charges can be neutralized.

If charge is separated, then electronegative atoms should gain the formal negative charge and electropositive ones should gain the formal positive charge.

Now you may set the description of e.g., DMSO

into context, too.
Reference:
Robert B. Grossman.  The Art of Writing Resonable Organic Reaction Mechanisms. Springer 2003 (second edition), p. 5.  He moderates a companion web-site pointing to all (currently) three editions, supplementary material, and errata, too.

I disagree with this answer. Resonance structures cannot be separated from each other, they cannot be measured, they do only formally exist. The notion that one structure is lower in energy than others of the same set implies a stability that doesn't exist. While I can guess I know what you are actually trying to say, is only my interpretation, and in that context, I think this is wrong. And in DMSO there are two electrons omitted, which are important. Arguably, the second structure is the main contributor, which contradicts the set of rules stated before. Details matter.

@Buttonwood so the same name represents all the structures?


text:
Is an electron's charge arbitrarily called negative?
In Cosmos Carl Sagan writes

Electrons are electrically charged, as their name suggests. The charge is arbitrarily called negative.

Is this true (I know the book is out of date, like how Titan is called the largest moon in the solar system)? If so, why is it negatively charged? Was the proton discovered first and then they just made an opposite, because now that I think about it, a negative charge doesn't make any sense.

I guess it makes more sense to put something positive in the center of the atom, and something negative around this center.

Formal polarity and potential scale orientation in macroscale electrotechnics has been established before electron was discovered as the first known elementary particle  and before foundations of quantum physics has been laid.

Obligatory xkcd: https://xkcd.com/567/

I’m voting to close this question because it has been asked and answered several times on Physics.SE (https://physics.stackexchange.com/questions/142150; https://physics.stackexchange.com/questions/176589; https://physics.stackexchange.com/questions/377036; https://physics.stackexchange.com/questions/475823). Including @Loong's XKCD 567 reference, too.

Yes, it is arbitrary in the sense that you could redefine everything that relates to the charge of electrons and protons. Once you call one charge negative or positive (like that of the electron), all other charges (proton, positron, cation, anion, cathode, anode, muon, up quark, down quark, glass/fur, glass/silk) follow.
However, the electron is named after amber (from Greek ēlektron "amber"):

Thus, when amber is rubbed with fur, electrons are transferred from the fur to the amber, giving the amber an excess of electrons, and, hence, a negative charge, and the fur a deficit of electrons, and, hence, a positive charge.

Source: https://farside.ph.utexas.edu/teaching/316/lectures/node12.html
So it makes sense to use the name electron for the particle that has the same charge as amber rubbed with fur. Whether you call it negative or positive is arbitrary, but it makes sense to stick with negative with all the publications that came out in more than a century since it was discovered.


text:
Chlorinated hydrocarbon destruction
What's a convenient method for lab-scale destruction of chlorinated hydrocarbons?
Reductive dechlorination, total oxidation are fine as well as other routes you want to point out. If specific chlorocarbons are needed, consider 100ml of dichloromethane or tetrachloroethylene. Ideal end products would be $\ce{H_2O}$, $\ce{CO_2}$ and $\ce{HCl}$; formic acid and other organic compounds are fine as long as they are not chlorinated.
So far I've been looking at oxidative species such as $\ce{O_3}$, $\ce{OH^{.}}$ or $\ce{SO_4^{.-}}$ via Fenton's reaction, UV/ozone, UV/$\ce{TiO_2}$, Persulfates. Would $\ce{H_2SO_5}$ oxidize effectively?
I am open to completely different routes. Reference to papers are much appreciated.


text:
Comparison of hydrogen bond strength

Consider the following sets of H-bonds:
$$
\begin{align}
&\text{P:} &\ce{-O-H\bond{....}N} \\
&\text{Q:} &\ce{-O-H\bond{....}O} \\
&\text{R:} &\ce{-N-H\bond{....}N} \\
&\text{S:} &\ce{-N-H\bond{....}O}
\end{align}
$$

Since oxygen is more electronegative $(\mathrm{EN})$ and polarises the hydrogen more than nitrogen, I think the order of decreasing strength is Q > P > S > R.
But the correct answer is P > Q > R > S. The reason given is that for H-bonding in
$$\ce{-A-H\bond{....}B}$$
to be strong, $\mathrm{EN}(\ce{A}) > \mathrm{EN}(\ce{B})$ and $\ce{B}$ should be electron-rich.
I am not able to understand where I went wrong.

A hint: Which one is a stronger base, $\ce{NH3}$ or $\ce{H2O}$?

@Poutnik NH3? I still don't get it,

Which ( Broensted-Lawry ) base is supposed to be more willing to bond to hydrogen by its free electron pair? A stronger one or a weaker one?

@Poutnik Thank you! I understand it now.

@Intermechanic If you got your answer, feel free to self-answer it.

One way to understand this is through partial charges. A-H in −A−H⋅⋅⋅⋅B is acting as a hydrogen donor and B is acting as a hydrogen acceptor. H is partially positively charged in −A−H⋅⋅⋅⋅B. The more electron is withdrawn from the H, the more positively charged it is, and the more attraction there will be between H and B. Again, the H in −A−H⋅⋅⋅⋅B is partially positively charged, lacks electron density or is electron-poor, therefore it is attracted towards an electron-rich B.
Oxygen is more electronegative than nitrogen, it can make H to have a bigger partial positive charge than nitrogen can.  Nitrogen is less electronegative than oxygen, can put its partial negative charge into H⋅⋅⋅⋅B more readily than oxygen can.


text:
Question regarding chain silicates
In my textbook, $\ce{Li2SiO3}$ has been given as an example of single chain silicate. It is represented by general formula $\ce{(SiO3)^{2-}_n}$. In $\ce{Li2SiO3}$, since there is only one $\ce{SiO3^2-}$ ion, it means n=1 and therefore only one unit of $\ce{SiO3^2-}$ which combines with two lithium ions. But if there is only one silicate unit the number of oxygens should be four and the charge should be -4 and basically be a single isolated tetrahedron as $\ce{SiO3^2-}$ on its own will not be stable because Silicon needs 4 bonds.
Please clarify my doubt. Here is my textbook for reference.

I am not quite getting your question. Are you asking why lithium silicate is formulated as $\ce{Li2SiO3}$ (n=1) although it is a chain silicate?

yes that's my question

See Wikipedia: "Silicate anions are often large polymeric molecules with an extense variety of structures, including chains and rings (as in polymeric metasilicate $\ce{[SiO3^{2-}]_n}$), double chains (as in $\ce{[Si2O5^{2-}]_n}$, and sheets (as in $\ce{[Si2O5^{2-}]_n}$." See the anion is a polymeric complex, they are formulated in simplified way as compounds with n=1 to avoid confusion (although they are classified as what type of silicate it contains).

It is true that each tetrahedron has four oxygen atoms. But adding up the separate tetrahedra double-counts some oxygen atoms that are shared between two of them.
If two vehicles approach an all-way stop from different roads at the same time, laws (which may vary by state and by the type of vehicles) specify which one has the right-of way. Similarly, we can make a "law" saying, for instance, that when two of the tetrahedra in a chain from left to right have a bond to one oxygen atom the tetrahedron on the right effectively has the right to it for the purpose of single-counting.
Thus each tetrahedron gets only one of the shared oxygen atoms in this count along with the two terminal oxygens, therefore $\ce{SiO3}$ instead of $\ce{SiO4}$ (we do get the latter in compounds where the tetrahedra are completely apart, such as $\ce{Mg2SiO4}$). Likewise the oxygen we counted with the neighboring tetrahedron effectively takes the electron pair polarized towards it, so the $\ce{SiO3}$ unit we identified in the chain is left with two negative charges rather than four.
The lithium ions are more straightforward. No such ion is bonded to a shared oxygen, so each one is associated with only one of the tetrahedra. It is thereby easy to see that there are two lithium ions to go with each $\ce{SiO3}$ unit, thus the overall empirical formula $\ce{Li2SiO3}$ (per tetrahedron).

my question is why are we not representing other combined tetrahedra in the formula Li2SiO3 why take only one combined tetrahedra in the formula?SINCE ITS IS COMBINED IN A CHAIN WHY NOT Li2(SiO3)2 or Li2(SiO3)3 ETC?

@AJknight There's absolutely no need for you to capitalise your text like that. Don't do that again on SE.

@AJknight because chain silicates are not made of specific discrete molecules with known numbers of tetrahedra. They are long chains and the formula represents their average composition.

Look at the figure: for the chain silicate (inosilicate) the structural unit is constituted of two neighbouring tetrahedra. That is, you have 2 Si atoms + 7 O atoms. Anyway, only 5 atoms are fully contained in the structural unit, whereas 2 O atoms are linked with neighbouring structural units. Hence these 2 O atoms are half-contained in each structural unit and so they totally corresponds to 1 O atom. So the structural unit is consituted of 2 Si+ 6 O, corresponding to [SiO3]2-.

my question is why are we not representing other combined tetrahedra in the formula Li2SiO3 why take only one combined tetrahedra in the formula?SINCE ITS IS COMBINED IN A CHAIN WHY NOT Li2(SiO3)2 or Li2(SiO3)3 ETC?

@AJknight There's absolutely no need for you to capitalise your text like that. Don't do that again on SE. (Just like orthocresol says.)

@AJknight consider for example your proposed Li2(SiO3)2 formula: you have 2 Si+4 ions (giving +8 charges) and 6 O-2 (giving -12 charges). So you cannot balance the remaining -4 charges (8-12) with only 2 Li+ ions. So your simplified chemical formula corresponds to Li2SiO3.

I'm sorry I didn't mean to capitalise I might have pressed on accident.So even though there may be a chain of tetrahedra a compound will be formed only when the cation is able to neutralise the charge on anion which in this case is SiO3 2-?

Yes, since each tetrahedron is $\ce{SiO3^{2-}}$ when we properly count oxygen atoms, two $\ce{Li^+}$ ions will attach to each tetrahedron to neutralize it, which matches the unshared oxygen atoms where the net negative charges reside. Thus as I say, $\ce{Li2SiO3}$ is the formula per tetrahedron, thus also the overall atomic ratio.


text:
How holes in cooking burner help it to perform complete combution but candle thread can't by providing sufficient oxygen supply?
Recently I was introduced to organic chemistry for the first time.
In the section of complete and incomplete combustion of carbon compounds,
they provide example of two different objects, namely a cooking gas burner and a candle, and said that "complete combution can be seen in burner but incomplete combustion can be seen in candle"
The explanation was

Apart from unsaturated carbon compound present in candle while saturated carbon compound present in burner's gas help in complete combution as less energy is required to break single bond in gas  than double and triple bond in candle wax"

But also that

the holes in burner help in mixing of oxygen more to perform complete combution but candle's thread dont allow much easy mixing of oxygen thus insufficient oxygen supply cause incomplete combution"

Can anyone explain the second reason in more detail?
I can't understand how holes help in mixing oxygen thus providing sufficient oxygen supply for the gas to combust?  Why can't a candle thread provide sufficient oxygen?
One more thing , where does the reaction between oxygen and gas take place in the burner?  Inside it or outside the burner once the gas comes out of the holes?

Is has very little to do with saturated/unsaturated nature of hydrocarbons.
The gas burner has gas and air premixed, similar as Bunsen and other lab gas burners. If you did not premix gas with air, there would be incomplete combustion with yellow shining flame, like when you close the air input of a Bunsen burner. The holes help to prevent premature ignition of the gas/air mixture inside the burner.
Candles evaporate the paraffinic wax hydrocarbons. These vapors are not premixed with air and burning happens in transition region where is the hydrocarbon/air ratio range supporting burning. There is no burning in outer regions that have too little of vapors not in the inner regions that have too little of air. If you evaporated wax by electric heater, mixed it with air in ideal ratio and high enough mixture temperature, there would be complete combustion.
Additionally, the size of  molecules and carbon chain of long n-alkanes play its role in the burning rate.


text:
Salt extraction from water without affecting mercury concentration
I am looking to theoretically bio-analyze a water sample regarding its mercury concentration from both sea and lake water using a biosensor with an enzyme biorecognition element. However, a big issue is salt, which can negatively affect the enzyme into the sensor, thus rendering it ineffective. As a result, I would like to filter the salt ions while at the same time keeping the mercury in the water. Unfortunately, I am finding it extremely difficult to find a way online of doing precisely this. The relevant papers are either about removing heavy metals like mercury, filtering saltwater, or removing both. Please keep in mind that this is a theoretical exercise and not a real-life experiment, so there's no need to take practical limitations into account. I am not sure this is the right substack to ask this, so thank you for your patience and your answers.

Aside of the main topic, what form of mercury is to be detected ? As e.g. free mercury, mercury ions and organomercury compouds like $\ce{(CH3)2Hg}$ have very different chemical properties and biological impact.

You have to think the otherway round and simplify your problem. You have to isolate mercury from a very complex matrix like seawater or lake water because no "sensor" can be so selective that it will respond to parts per billion level of mercurcy and ignore the rest of the junk. For that read about cold vapour mercury analysis, how Hg is isolated.

As @M.Farooq states, it's far easier to separate the Hg, then use your enzymatic sensor on the Hg-enriched portion.


text:
Can Electron Repulsion Integrals be negative?
I am trying to write a HF code that can calculate the orbital energies for a given molecule at RHF level. I got upto calculating electron repulsion integrals (ERI) and kinetic energy (KE) integrals, and found that many values in the ERI where negative numbers. The results for the KE calculation is verified to be correct.
Following this question, I understand that a positive value means the interaction is destabilizing, while a negative value means it is stabilizing in the case of KE integrals. Is it the same for ERI also? Or am I missing something serious here?
The code can be found here. I used water as a test case with 3-21g basis set.

All negative energies are stabilizers.
The ERI integrals are a sum of the Coulomb operator ($\hat J_i$) and the exchange operator ($\hat K_i\{J\}$)
$$\hat J_i+\hat K_i\{J\}$$
Since the Coulomb operator is a repulsive energy (so a positive energy) and the exchange operator is a stabilizer, it's a case-dependent answer.


text:
Important differences for the hydration free energy of dioxygen in the literature
I am trying to find a reliable source for the value of $\Delta G^{\circ}_{solv}$ of $O_2$ in $\ce{H_2O}$ at $T = 298.15 K$ to compare with my DFT calculation.
I get $\Delta G^{\circ}_{solv} = 6.33 \text{ kcal/mol} $ from: Abraham, Michael H. Journal of the American Chemical Society, vol. 104, no. 8, Apr. 1982, pp. 2085–94. https://doi.org/10.1021/ja00372a001
More recently, you can find $\Delta G^{\circ}_{solv} =  3.96 \text{ kcal/mol}$ from: Plyasunov, Andrey V., and Everett L. Shock. Geochimica et Cosmochimica Acta, vol. 67, no. 24, Dec. 2003, pp. 4981–5009, https://doi.org/10.1016/j.gca.2003.08.003
The chemistry handbook (95th edition, 2015) only gives enthalpy of hydrations of gases, $\Delta H^{\circ}_{solv} = -0.29 \text{ kcal/mol}$ but still very different from the first source which gives $\Delta H^{\circ}_{solv} = -2.88 \text{ kcal/mol}$
The standard state for the first source is 1 atm, the second 0.1 MPa, which I think is pretty close. Why such big difference?

Instead of writing $H_2O$ to get  $H_2O$, write $\ce{H2O}$ to get $\ce{H2O}$. Ch. formulas are written upright. // You may find useful these links for text and formula formatting (not to be applied to titles):
Notation basics ,  Formatting of math/chem expressions  and
upright vs italic

For what it's worth, if I calculate based on the Keq derived from reported values of the Henry's law constant, I get something close to the 3.96 kcal/mol value. The Henry's Law constant values given in the wikipedia article on Henry's Law are consistent with the higher end of the solubility values I've found for O2 in H2O at 298 K, which seem to cluster between 1.2 and 1.3 mmol/L


text:
Is acetone an effective solvent for bitumen?
Japanning is an age old process of protecting metal surfaces by applying several successive layers of a syrup-like black coating and then baking it to a hard and durable finish. It was often applied to vintage machines and tools before the advent of petrochemical based coatings. Although there's debate on its exact recipe among vintage machine restorers, the coating was made by dissolving crushed, or powdered ashphaltum (bitumen) in turpentine and then combining it with boiled linseed oil.
While turpentine is a fairly good solvent for ashphaltum, it doesn't fully dissolve it, leaving large sized particulates suspended in the mixture. Getting decent results requires allowing the large particles to settle to the bottom over several hours and then filtering it. Even when filtered, it still tends to clump in regions when applied and doesn't consistently produce the highly-prized smooth satiny finish.
I suggested using acetone as an alternative solvent to another restorer, but he claimed the Ashphaltum only stayed in solution for 30 minutes before settling to the bottom, while turpentine action was faster. I'm surprised since acetone is very good at dissolving greases, oil, waxes and other organic substances. Why would it be less effective than turpentine?

My question is what is in the "ashphaltum". I am a metallurgist but the materials section at Amoco also included the asphalt experts; Amoco did more R&D on asphalt pavements than any other oil company. I heard many times that asphalt is a blend of asphaltenes , resins , and oils , giving a wide variety of properties . Asphalt is fully soluble in acetone, I believe they usually used MEK to clean lab equipment. So if it won't dissolve, your material is not all asphalt.

Thanks. It could be that acetone may be an excellent solvent, but isn't good at forming a suspension with certain materials?  Oil doesn't dissolve in water, but it can be suspended in water. Milk is an example. Does the polarity of acetone or higher density of the ashphaltum make it difficult for acetone to hold it in suspension?


text:
Is the graphene sheet in graphite considered molecule? What is considered molecule?

molecule, a group of two or more atoms that form the smallest identifiable unit into which a pure substance can be divided and still retain the composition and chemical properties of that substance. -source Britannica.
I am not sure if this is the answer to my question quote wiki
"molecule is an electrically neutral group of two or more atoms held together by chemical bonds.[4][5][6][7][8] Molecules are distinguished from ions by their lack of electrical charge."
Does that mean the graphene sheets are not molecule because they have electronic charge? what about diamond and quartz? or are they structurally different from molecules like ethanol or water?
Thank you in advance!

I suggest reading about network covalent bonding (and format references into proper functional ones while you are at it). The question is likely a duplicate of https://chemistry.stackexchange.com/questions/8068.

oh! Thank you! It's the exact same question I have! sorry for posting another one


text:
Are there any papers illustrating carboxylic acids displaying J-coupling of CH to the OH protons?
I understand that ethanol will display a J-coupling to the OH proton if observed in the absence of water, but are there any papers showing how to find this J-coupling (a 3-bond H-H J-coupling between $\ce{CH}$ and $\ce{COOH}$) in carboxylic acids (specifically formic acid)?

Generally, coupling to OH groups is quite hard to observe, mainly because of chemical exchange. I did find some examples of coupling to carboxylic acids, though.
In this first example, the authors made complexes of formic acid with hydrogen bond acceptors, leading to complexes of the form $\ce{HC(O)O-H...B}$. The CH and OH doublets, split by $^3\!J_\ce{HH}$, are visible in the spectrum below. Note that the exact coupling constant depends on the choice of the hydrogen bond acceptor, though.
Golubev, N. S.; Denisov, G. S.; Koltsov, A. I. Proton spin-spin coupling in complexes of formic acid with proton acceptors. Journal of Molecular Structure 1981, 75 (2), 333–337. DOI: 10.1016/0022-2860(81)85247-7.

In the spectra of solutions containing 0.02–0.05 mol l-1 HCOOH and 5–8 fold excess of THF or HMPT at 110–150 K, two well resolved doublets with the same splitting values (11.6 cycles) were observed. As one of these
doublets coalesced on RF saturation of the other, the splitting must be
caused by spin–spin coupling. The high value of $J$ confirmed the non-symmetrical structure $\ce{A-H...B}$.

Later, $^4\!J_\ce{HH}$ in similar complexes with acetic acid was also observed. Note in this case that acetic acid was labelled with $\ce{^13C}$ so the carboxylic acid OH group is split, both by $^2\!J_\ce{CH}$ as well as $^4\!J_\ce{HH}$.
Tolstoy, P. M.; Schah-Mohammedi, P.; Smirnov, S. N.; Golubev, N. S.; Denisov, G. S.; Limbach, H.-H. Characterization of Fluxional Hydrogen-Bonded Complexes of Acetic Acid and Acetate by NMR:  Geometries and Isotope and Solvent Effects. J. Am. Chem. Soc. 2004, 126 (17), 5621–5634. DOI: 10.1021/ja039280j.

The carboxylic proton signal resonates at $\pu{10.754 ppm}$. It is split into a doublet arising from scalar spin−spin coupling with the carboxylic carbon nucleus, with $^2\!J(\ce{CH3\mathit{C}OO\mathit{H}}) = \pu{6.7 Hz}$. This assignment is supported by $\ce{^13C}$-decoupling. Each component of the doublet is furthermore split into a barely resolved quartet, arising from long-range scalar coupling with the methyl group protons, where $^4\!J(\ce{C\mathit{H}_3COO\mathit{H}}) \approx \pu{0.85 Hz}$.


text:
What makes ammonia to dissociate when volume is increased?
What makes ammonia dissociate to $\ce{N2}$ and $\ce{H2}$ when the volume is increased (pressure is reduced)?
I understand how pressure affects equilibrium. What happens after the $\ce{NH3}$ is being formed, then the pressure is reduced, for example, does the existence of a certain value of pressure is necessary for the formation of the $\ce{N-H}$ bond?
Also, the dissociation reaction is endothermic, where does the $\ce{NH3}$ get the energy to absorb so it gets dissociated?

The system moves in the direction to remove stress. After NH3 is formed, pressure is reduced but the presence of heat can be reduced also by moving back to the endothermic reaction recreating some N2/H2.

Without energy input, whether from heat, ultraviolet light, ionizing radiation or other source, NH3 is stable. If you're thinking of its synthesis, consider the starting temperature in the Haber process.

This your statement I understand how pressure affects equilibrium. kind of contradicts the whole question. Equilibrium means both forward and backward reactions are ongoing by the same rate ( which can be very low at low temperature a/o without catalyst). Decreasing pressure makes ammonia decomposition going faster than synthesis.

The reaction of molecular nitrogen and hydrogen would not lead directly to formation of ammonia, the mechanism resulting in ammonia is more complex than a bimolecular process. However, if you consider the encounter of nitrogen and hydrogen only as the first step in the forward reaction to ammonia, you can explain the effect of increased pressure by understanding that increased pressure correlates with increased concentration at constant T, and therefore increased collisions.

The equilibrium constant $K$ is, well, constant at fixed temperature.  By working out the degree of dissociation $x$ of hydrogen which is also degree of ammonia formation we find that $K=f(x)/p^2$ where $f(x)$ (complicated polynomial in $x$) increases with $x$. So, if $p$ decreases so must $x$ (degree of dissn) decrease to keep $K$ constant and maintain the equilibrium. Even in an endothermic reaction there is a small but finite chance of crossing the energy barrier.


text:
Should the principal carbon chain contain maximum number of functional groups or just the principal functional group?

My answer was the 6 carbon chain, i.e. the name would be 2-(2-hydroxyethyl)-4-methylhexanoic acid.
But the given answer is the 4 carbon chain, i.e. the name would be 4-hydroxy-2-(2-methylbutyl)butanoic acid.
The reasoning given is that the principal carbon chain should contain the maximum possible number of functional groups.
However, I was (maybe wrongly) thinking that the principal chain should contain the principal functional group (-COOH), and then the maximum number of carbon atoms, and then the maximum number of substituents.
So, should the principal carbon chain contain maximum number of functional groups or just the principal functional group?

Feel free to have a look at Asking nomenclature questions and add a reference for your assignment (i.e. the "not to be" screenshot).

ChemDraw 20.1 agrees with you (2-(2-hydroxyethyl)-4-methylhexanoic acid) but the rules are apt to change. Wait for @Loong.

Your rationale is correct.
The relevant (in this case) simplified criteria for the choice of a principal chain in such compounds are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
lower locants for suffixes
greater number of prefixes (e.g. “chloro-” or “hydroxy-”)
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the nomenclature rules (taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)) reads as follows:

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

You have correctly identified the principal characteristic group as carboxylic acid.
According to Rule P-44.1.1, the principal chain has to include the carboxylic acid. Therefore, the option labelled “7 carbon chain” can be ruled out. However, since there is only one carboxylic acid group in the given compound, this rule doesn't help you any further.
The next relevant rule is P-44.3.2, i.e. the principal chain is the longest chain. Therefore, the option labelled “6 carbon chain” is the correct choice. The resulting name is 2-(2-hydroxyethyl)-4-methylhexanoic acid.

Only if there is still a choice, the further rules would be considered. Thus, the mentioned “maximum possible number of functional groups” (other than the principal characteristic group) would be considered later in Rule P-45.2.1. Even in this case, however, the given explanation would not help since both names “2-(2-hydroxyethyl)-4-methylhexanoic acid” as well as “4-hydroxy-2-(2-methylbutyl)butanoic acid” have two substituents cited as prefixes.


text:
Are good leaving groups good in both directions?
In biology, phosphate groups are often used, good leaving groups.
If I had an opposite case, where a phosphate group remains bound to an enzyme while the main substrate is removed (e.g. E-P04-S => E-PO4 + S) is the substrate considered a good leaving group?
Put another way, does the fact that Phosphate is a good leaving group affect the ease at which S is removed?

Can you specify whether the whole PO4 is coming off the substrate or if just PO3 is removed? In the former case, something needs to add to the substrate to replace the PO4 or there needs to be deprotonation/elimination

@Andrew Sorry, I may have been too vague in my example. To clarify, I am thinking of a pTyr phosphatase in this example. This enzyme has an intermediate where after the phosphate is removed from the tyrosine, it is covalently bound to a Cys in the enzyme active site. I am wondering if the elimination of Tyr from the Tyr-PO4-Enz complex is energetically favorable as normal PO4 hydrolysis. As you're still breaking the S-OPO3 bond. Hopefully this helps!

Based on your description, you are breaking the SO-PO3 bond not (as you wrote) the S-OPO3 bond, right? That is what happens in normal hydrolysis of a phosphate group.

While you are correct that phosphate is a good leaving group, it does not play that role in the reaction you described.

Source: https://doi.org/10.1074/jbc.273.17.10454
The reaction catalyzed by protein tyrosine phosphatase is an example of a "phosphoryl transfer" reaction, that is, the transfer of $\ce{PO3^-}$ from one compound to another. From a chemistry point of view, this is a nucleophilic substitution at the electrophilic phosphorus atom.
In the uncatalyzed reaction, water is the nucleophile and tyrosinate (or tyrosine if the O has been protonated) is the leaving group. In the enzyme active site, water is replaced by cysteine as the nucleophile as you described.
Examples of reactions in which phosphate is truly a leaving group are glycosyl transferases that use glucose 1-phosphate as a substrate and the hydrolysis of pyrophosphate, in which one phosphorus atom is the electrophilic site and the other is part of the phosphate leaving group.
With respect to your main question, since the leaving group in both the catalyzed and uncatalyzed reaction is tyrosinate or tyrosine, a key question is the protonation state of the tyrosine oxygen. Tyrosine (protonated) is a much better leaving group than tyrosinate (unprotonated).
On the other side, a sulfide ($\ce{-S^-}$) group is a much better nucleophile than a sulfhydryl ($\ce{-SH}$) or water, but not as good as hydroxide, so that will also make a difference.
Lastly, the protonation state of the oxygens on the phosphoryl group will have an effect on how electrophilic it is, an effect which is independent of the identity of the nucleophile.
All of that is to say that you need to be very specific about what two reactions exactly you are comparing.


text:
Faradic and non-Faradic current in Voltammetry
Below is a schematic drawing for a three-electrode cell used for cyclic voltammetry:

Image modified from Wikipedia Voltammetry
This is what I understand with the voltammetry working principle from self-study, please correct me if I am wrong.

At WE, a half-reaction, $\ce{Ox + e- <=> Red}$ is modulated by the potential applied on the WE. Electrochemical equilibrium can be assumed at the surface of WE. When the potential at the WE is increased, WE draws electrons out from the analyte, equilibrium position shifted towards the Ox. When potential decreased, WE 'injects' electrons into the analyte and the equilibrium position shifted towards the Red. In voltammetry, the potential varied continuously, and it is this continuous shifting of equilibrium position (that involves giving or accepting electrons) leads to motion of electrons and hence gives rise to Faradic current.

Using the quoted understanding above (which might be wrong), I couldn't answer the following questions:

When you increase the potential, electrons got drawn into the WE then go towards the CE via external circuit. What will happen when the drawn electrons reach the CE? Will they get injected back into the solution? If yes, what species will accept the electrons?

A typical reversible CV will have an onset potential, beyond which the current increases sharply. If the argument above is correct, even a slight increase of potential should already lead to shifting of equilibrium and hence Faradic current, and so there should be no onset but rather a direct increase in current as soon as potential is applied.

Say you keep changing the potential applied, the equilibrium position will be shifted continuously and hence Faradic current observed. What if you stop changing the potential at one point, but hold the overpotential at a non-zero constant? What will be observed? 0 current?

When there is no redox-active species in the electrolyte, current will increase very slightly when potential increased, and that is due to non-Faradic current (capacitive current), due to ions accumulating at WE surface. I don't see how ions accumulating at WE surface will lead to current flowing between WE and CE and hence gives rise to non-zero current.

Consider for a moment an electrochemical reaction where we can (for small overvoltages) us the Butler–Volmer equation to predict the current.
I = Io {(exp ka E)-(exp -kc E)}
Where E is the overvoltage and ka and kc are two constants (in real life they are the combination of many constants for the electrode). As long as the current is not limited by mass transport being too slow we can use this equation. This current is the Faradic current which flows as a result of the chemical change.
The surface of the electrode is constantly bombarded with either the reduced or oxidized form of the electroactive "thing", the probaility of the thing reacting when it hits the surface is a function of the potential of the electrode vs the bulk of the solution it is in. This current is a function of the potential of the electrode.
As we change the potential of the electrode relative to the bulk liquid which it is immersed in then we will have a current (Non - faradic) which flows due to the change of the potnetial difference between the two terminals of the capacitor which is present. The electrode is one of the two terminals of the capacitor while the liquid it is immersed in should be regarded as the other one.
Here for this current we have
I = C (dE / dt)
This current is a function of the rate of change of the electrode's electrical potential.
So for a real electrode with a small current flowing we can write the following equation for the total current.
I = (C (dE/dt)) + Io {(exp ka E)-(exp -kc E)}
Now lets deal with sub questions 1 to 4

Imagine we have a solution of [Fe(CN)6]4- (1 mM) in  1 M NaNO3, now imagine that at the working electrode one molecule of [Fe(CN)6]4- is oxidized to [Fe(CN)6]3-. What will happen is that one electron from the counter electrode is needed to perform some electrochemical event there. If by luck some [Fe(CN)6]3- has diffused there then this could be reduced. As an alternative a proton could be converted into a hydrogen atom. Depending on the nature of the counter electrode this could be adsorbed onto the surface (if it is Pt) or it could combine with another hydrogen atom to form a hydrogen molecule. There are other things which could be reduced at the counter electrode such as dissolved oxygen if the chemist has failed to deoxygenate the cell.

As I wrote above the Butler-Volmer equation predicts that the current will change suddenly as we pass through the point where E is zero.

If we maintain the potential at a fixed value then the non faradic current will become zero, however the faradic current will continue to be dictated by a combination of mass transport and the Butler-Volmer equation. When a reaction is occuring at the electrode then the small volume of liquid around it will have different concentrations of the reduced and oxidized forms as the bulk. What will happen is that diffusion will move molecules of both forms between the bulk of the liquid and the special zone around the electrode. The more the cell is stirred the thinner the stagnant zone is around the electrode.

If there is no electroactive species in the cell and as we increase the potential of the workeing electrode then ions are repelled or attracted to / from it then at the counter electrode a similar but opposite number of ions will be repelled / attracted thus allowing charge to flow through the charging effect.


text:
Finding final volume of the diluted solution with defined mass concentration given initial molar concentration

To what volume should you dilute $\pu{126 mL}$ of an $\pu{8.15 M}$ $\ce{CuCl2}$ solution so that $\pu{49.5 mL}$ of the diluted solution contains $\pu{4.56 g}$ $\ce{CuCl2}?$

I know you need to use $c_1V_1=c_2V_2,$ but do not know how to incorporate the $\pu{4.56 g}$ of $\ce{CuCl2}$ into it.

First figure out the concentration of the resultant, 49.5 mL contains 4.56 g CuCl2, then starting with 8.15 molar solution, how much must it be diluted.

I figured I need to use resulting molar concentration:
$$c_2 = \frac{m_i}{M(\ce{CuCl2})\cdot V_i} = \frac{\pu{4.56 g}}{(\pu{134.45 g mol-1})(\pu{0.0495 L})} = \pu{0.685 M}$$
$$V_2 = \frac{c_1V_1}{c_2} = \frac{(\pu{8.15 M})(\pu{126 mL})}{\pu{0.685 M}} \approx \pu{1.50 L}$$

I took a liberty to brush up formatting, notations and significant figures. Keep in mind it's advantageous to first solve the problem using symbolic algebra, and plug in the values at the very last step. Also, please visit this page, this page and this one on how to format your future posts better with MathJax and Markdown. Finally, refrain from asking in the answering section — if you post an answer, it shouldn't be a proofreading request.


text:
Which is more acidic alcohol or ketone?
I have read about it and i found that the alcohol is more acidic, but isn't the conjugate base produced from losing the alpha hydrogen of the ketone more stable ?!
The negative charge in this conjugate base is spread over the oxygen atom and the alpha carbon atom by resonance ( charge is delocalized ) as shown in the attached image.

while in the conjugate base produced from the alcohol we see that the negative charge is carried only by the oxygen atom and no resonance structures which means that ketone should be more acidic since its conjugate base is more stable and it will easy be formed.

Resonance is only one factor when it comes to anion stability. For a more extreme example, consider HCl and acetic acid...

No, I want this example what makes a ketone less acidic than alcohol although the conjugate base of a ketone is more stable than the conjugate base of the alcohol by resonance .

It isn’t more stable. If the pKa is larger, then it’s a weaker acid, and the anion is less stable. No exceptions.

What makes an apple taste different from an orange?

You are considering two different Hs. I mean: one is a $\ce{O-H}$ bond; the other is an $\ce{C-H}$ bond.
You can't compare these two bonds.
Yes, $pK_a$ of the alcohol is around 16/17 and $pK_a$ of the ketone is around 25/26, but are two bonds completely different.

no it is not about bonds at all H2O is more acidic than NH3 and here we are comparing OH bond with NH bond .

@AhmedHossam If not for enol / ketone tautomerism, and hydration equilibrium, you could perhaps simply compare them with alcohols, but their properties aren't nearly as simple as both you and Andrea seem to think.


text:
Commercially available conductive polymers
I am looking for intrinsically conductive plastics (not plastics that are made conductive by filling them with carbon black for example). I have found that there are conductive polymers like polyanilin, polythiophene and polypyrrole that are sometimes used as conductive coatings. But are any of these materials commercially availabe as sheet material? As it is the case with conductive PE UHMW for example? I am looking for sheets with a thickness of about 5 mm that I can machine into different shapes.
Additional question: Does anyone know if there are any commercially available metal / polymer composites?

There are many plastic sheet stock suppliers that provide 'conductive' material. Often they mean it does not build up static charge. How conductive do you mean by conductive?

Yes, i know, but as I understood these materials are made conductive by mixing them with carbon black which is not what I am looking for. A surface resistance somewhere below 10^6 ohms is sufficient.

So, something like "Semitron ESd 410C
Standard Sheet Size: 12" x 12"
The Semitron ESd products are inherently dissipative and do not rely on atmospheric phenomena (e.g. humidity) to activate, nor are surface treatments used to achieve dissipation.
Does not contain carbon or graphite powder to provide electrostatic dissipation, sloughing is not a problem."

Yes exactily, thank you! Can you send me a link where you found this text? I find the material but not this exact description of it.

Googling 'Semitron ESd' will bring up the links of Alro and their many suppliers. Multiple grades, you will need to select what might be best.


text:
Calcium Hydroxide, 98%, extra pure + DI resulted in a bright pink solution
When combined with DI, my 98% calcium hydroxide resulted in a pink solution. I am trying to make limewater for a class demonstration and have only ever seen limewater as a colourless solution. Anyone else had this issue?
Thanks!

Why [tag:organic-chemistry] tag? Is there an organic substance or indicator present in the solution? DI usually stands for "deionized". Deionized what, water? How clean was the glassware? Could you upload a photo and specify lighting conditions to eliminate possible light scattering effects?

Try different water ( even tap one ) and different glassware/vessel. Still pink ? If still pink then my educated guess says there must be a pink dye in the calcium hydroxide lot. If not there may be dye contamination of the vessel, possibly some phenolphthaleinum from prior activities.

Thank you for your responses, it seems a phenolphthaleinum contamination was the issue and subsequent attempt to make limewater have resulted in a clear solution. Thanks!

It's almost as if there were phenolphthalein in the solution.


text:
What is the mechanism for the reaction of (Ph)2CO and succinic acid with NaH and quenching with acid?

I'm guessing the C=O will be attacked first on the PhCOPh, but how do you get rid of the O?
Just out of interest, how else could you convert C=O to C=C?

I presume this is multiple equivalents of NaH. How strong a base do you think NaH is?

Reminiscent of the Stobbe Condensation with succinate esters. The reaction is conceivable but questionable as described. Do you have a legitimate reference for this reaction?

Interesting. The Stobbe condensation apparently goes via a cyclic intermediate that ultimately results in the 'hydrolysis' of one of the two esters https://pubs.acs.org/doi/10.1021/ja01157a133 , https://fr.wikipedia.org/wiki/Condensation_de_Stobbe . NaH has also been used apparently https://pubs.acs.org/doi/pdf/10.1021/ja01157a130 . But yeah, starting from the dicarboxylic acid... hmm...

Thank you, Stobbe condensation does seem to provide a viable mechanism. The source is a past-paper question from a first-year undergrad chemistry exam.

Starting from the anion of succinate dicarboxylate, the first step in the reaction will be similar to the Stobbe condensation BUT it is highly unlikely that the resultant alkoxide will add to a resonance stabilized carboxylate to form a lactone. I suspect that this question is made up.

Welcome to Chemistry! On Chemistry mathematical and chemical expressions can be formatted using MathJax (and LaTeX Syntax).  If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details. cc @user55119


text:
Reaction of Grignard reagent with hydrogen cyanide
Grignard acts as a base with compounds containing acidic hydrogen:
$$\ce{RMgX + R'-OH -> R-H + R'-O-Mg-X}$$
and it acts as a nucleophile with others with an electrophilic site:
$$\ce{RMgX + R'-CHO -> R-COH-R' + HO-Mg-X}$$
Now the compound of interest is hydrogen cyanide:
$$\ce{H-CN}$$
This compound has both an acidic hydrogen and an electrophilic site(carbon). Also its worth mentioning that the alkyl counterparts of hydrogen cyanide go through the nucleophilic action of Grignard on the carbon atom.
So how does the reaction proceed here?

Compare the pKa's of R-H and HCN to get an answer.

An old name for hydrogen cyanide is prussic acid. Does that gives you a clue?

A Grignard reagent will certainly react with water to form a hydrocarbon, and based on $pK_a$ values $\ce{HCN}$ is a much stronger B-L acid than water. While a nucleophilic attack on the cyanide function is theoretically possible, it would be kinetically less favored than extraction of the sufficiently available protic hydrogen forming the hydrocarbon instead of an eventual aldehyde.
There are various other ways to make aldehydes, but this answer reports the existence of $\ce{HMgX}$ compounds which can serve as hydride-ion sources. If that hydride transfer works with nitriles, the reaction would yield a product that could be worked up to an aldehyde.

This question is best answered by research in Chemical Abstracts and actual research in a properly equipped Lab if nothing is found. I can think of addition schemes to remove the acidic H then possibly react with the with the CN moiety.


text:
Is intensity in NMR spectra acquired at regular time intervals?
I'm writing a parser that reads fid binary files and ASCII procpar files from FT-NMR experiments, most of them acquired using Varian instruments. This instruments do not keep the time values at which each intensity is acquired. So there is no obvious way you'd plot Intensity vs Time.
How do the standard visual programs get the time values?
What I found though, is that the acquisition time (time they measure intensity) is stored  in an ASCII file, probably in seconds. procpar files store every detail (spectrometer frequency, pulse width (time), pulse width to rotate the magnetisation by 90° etc.)
I could assume intensity is measured at regular intervals, and divide total-time/data-points, but there may be something else I am missing.
I could also test Fourier transforming the data but it would be time consuming.
The 1D FIDs seems to store real data, also some 2Ds that I've found elsewhere. But I am aware that they may store complex data, in which case the data-points are divided by 2, and a different treatment altogether may be needed.
The FIDs may be 1D, 2D etc, only limited by the instrument capabilities. 2D FIDs seem to hold a large set of fid results, all would be linked to the same time axis, or some similar rearrangement afaiu. There can also be '1D arrayed' data, that I am not yet sure how to interpret. May just be a set of 1D FIDs.

In order to apply discrete Fourier transform, the fundamental requirement is to have uniform sampling rate.

@BuckThorn updated, we can chat for more details if you are keen to. I am writing a parser in Typescript for those files. There are parsers in other langs, but I like to do it by myself from scratch. Plots show well w/o time axis bc I plot using array indexes, which are also equidistant.

By intensity, do you mean individual data points in the FIDs?

Yes. It is an electric current induced in a wire by the variable bulk magnetization field, I believe. I have to say that I included the required details but it is going to confuse a reader. The answer is "yes" if I assume the $\Delta t$ are equidistant. Which seems to be correct. I was expecting some further insights maybe. Thanks for the edit @orthocresol.

1D data
Generally, yes, data points in the FID are acquired uniformly. So, the sampling interval will simply be the acquisition time divided by the number of points.
As Buck Thorn pointed out in some (now deleted) comments, however, there is a slight subtlety in that FIDs in NMR are inherently complex-valued, whereas computers only store real numbers (i.e. complex numbers must be stored as a pair of real numbers). I'm not sure what Varian machines do, but on Bruker machines, the data array actually alternates between real and imaginary points. So, data point 1 is actually the real part of the first complex data point, data point 2 is the imaginary component, etc. So instead of $N$ real data points you actually get $N/2$ complex data points, with an effective sampling interval that is twice of what was earlier mentioned.
This is not as big a problem as it might seem. Using $N/2$ complex data points sampled at half the rate for a complex FFT, or $N$ real data points for a real FFT, will actually give you the same spectral window by the Nyquist theorem. The resulting spectrum will also be the same, although you may need to use the fftshift() function to correct the frequencies.
There is more in-depth discussion of this issue in this article: Turner & Hill, J. Magn. Reson. 1986, 66, 410.
I'm aware of examples where 1D data is acquired in a non-uniform manner and the missing points are reconstructed. See e.g. Ndukwe et al., ChemPhysChem 2016, 17, 2799. These are pretty niche, though, and I'd say that virtually all 1D data you will encounter is uniformly sampled.
2D data
2D data is a lot more complicated. Generally, these are functions of two time variables: $S = S(t_1, t_2)$, so they are stored as matrices, built up of many FIDs.
Within each FID, $t_1$ is a constant, and $t_2$ varies. Between each FID, the value of $t_1$ changes. So, it's often the case that each row of the matrix is one FID, and so the columns represent different values of $t_2$, and the rows represent different values of $t_1$. However, you should consult the manufacturer's manuals as to the exact format in which this data is stored (row-major or column-major).
The $t_2$ dimension is entirely analogous to 1D spectra (essentially, 2D NMR is about collecting many 1D spectra with different values of $t_1$). Hence, it's still almost always the case that data is uniformly sampled in the $t_2$ dimension.
However, the $t_1$ dimension is way more complicated, for three reasons:

There is a similar issue to the real/complex difference mentioned previously for 1D data. In the case of 2D data, it's very often the case that the first and second FIDs actually have the same value of $t_1$, but are actually different experiments designed to yield the real and imaginary part of a complex number (the States method). Alternatively, the first and second FIDs have the same value of $t_1$, and are meant to allow you to regain the real and imaginary parts of a complex number through linear combination (echo-antiecho method). Alternatively, the first and second FIDs actually have different values of $t_1$ (TPPI method, or magnitude-mode spectra).
In general, it is not sufficient to directly perform Fourier transformation on a 2D data matrix read from a file without any extra processing.
The entire reason for this is so that positive and negative frequencies can be discriminated in the $t_1$ dimension after Fourier transformation, and is technically known as quadrature detection. This is an extremely important topic in 2D NMR and there is no way I can explain this fully; I instead refer you to a textbook such as Keeler's Understanding NMR Spectroscopy.

The 'complex' signals in $t_2$ are in fact orthogonal to the 'complex' signals in $t_1$, in the sense that the imaginary units used in both dimensions are to be considered separately. That means that, instead of a 2D matrix of complex numbers $a + bi$, you actually have a 2D matrix of hypercomplex numbers (or quaternions) $a + bi + cj + dij$, where $i$ and $j$ represent the imaginary units in the two dimensions. The number $a$ is real in both dimensions, $b$ and $c$ are imaginary in one of the two dimensions each, and $d$ is imaginary in both dimensions. On Bruker machines, each of these four components are stored in separate files. I don't know what Varian data looks like, but I expect it to be broadly similar.

In recent years non-uniform sampling (NUS) in the $t_1$ dimension is becoming more and more common, and there is a non-negligible chance that any random 2D dataset you obtain from someone in fact uses non-uniform sampling. In fact, when people talk about non-uniform sampling in NMR they are almost always referring to the $t_1$ dimension of 2D NMR (or higher-dimensional NMR). A simple Google search for NUS in NMR will likely bring up several reviews.

It is not necessary to acquire NMR data at a uniform sampling rate (regular time intervals), but it is the standard practice in a large majority of cases. Most experiments acquire data with a uniform sampling rate in order to take advantage of numerical methods such as the Fast Fourier Transform (FFT), which is designed for application to regularly sampled data and whose development was a key to widespread adoption of NMR. There are likely other advantages with regular sampling, including simpler electronics, scheduling of acquisition, data processing, and ease of troubleshooting (although tools to process irregularly sampled data can come to the rescue as a troubleshooting tool when regularly acquired data is partly corrupted). Some modern NMR acquisition methods and complementary processing techniques attempt to circumvent shortcomings of regular sampling by applying irregular sampling.
One shortcoming of regular sampling, particularly of a complete FID, is that it requires a lot of time (and memory, but that is secondary), particularly in the case in multidimensional NMR, where experimental acquisition times can become rather lengthy (days, possibly more) and therefore also costly.
One way to shorten acquisition times is to truncate FIDs in either the direct or the indirect dimensions. In the direct observe dimension it is also common to truncate acquisition before completely sampling an FID to avoid problems with application of high-power pulses when suppressing particular NMR interactions during acquisition. Truncation effects can be ameliorated using prediction and apodization algorithms that are included as standard tools in current NMR instrument software.
Irregular sampling is another more complex way to shorten the required number of points acquired in the indirect dimension, with potentially large rewards in saved time. There are a number of techniques available as standard tools on modern NMR instrumentation to schedule irregular sampling and process the data. Older data (maybe more than 15 years old) was most likely not acquired with irregular sampling techniques.
On Varian (now Agilent) instruments the number of points is stored in the procpar file. To determine the acquisition time, sweep width (half the sampling rate), or number of points for your data set I suggest looking at parameters np,sw, and at, which are described in the VnmrJ users manual:

np Number of data points
Description: Sets number of data points
to be acquired. Generally, np is a dependent parameter and is
calculated automatically when sw or at is changed. If a particular
number of data points is desired, np can be entered, in which case at
becomes the dependent parameter and is calculated based on sw and np.
Values: np is constrained to be a multiple of 2 (Acquisition
Controller or Pulse Sequence Controller board) or a multiple of 64
(Output board). (See the acquire statement in the manual User
Programming for a description of these boards.
at Acquisition time
Description: Length of time during which each
FID is acquired. Since the sampling rate is determined by the spectral
width sw, the total number of data points to be acquired (2swat) is
automatically determined and displayed as the parameter np. at can be
entered indirectly by using the parameter np. Values: Number, in
seconds. A value that gives a number of data points that is not a
multiple of 2 is readjusted automatically to be a multiple of 2
sw Spectral width in directly detected dimension
Description: Sets
the total width of the spectrum to be acquired, from one end to the
other. All spectra are acquired using quadrature detection. The
spectral width determines the sampling rate for data, which occurs at
a rate of 2*sw points per second (actually sw pairs of complex points
per second). Note that the sampling rate itself is not entered, either
directly or as its inverse (known on some systems as the dwell time).
If a value of sw is entered whose inverse is not an even multiple of
the time base listed above, sw is automatically adjusted to a slightly
different value to give an acceptable sampling rate. To enter a value
in ppm, append the character p (e.g., sw=200p). If a DSP facility is
present in the system (i.e., dsp='i' or dsp='r') and oversampling in
the experiment has not been turned off by setting oversamp='n', then
the oversampling factor will be recalculated.

Somewhere in the metadata of your measurement, there is a value called "dwell time". That is the length of the interval which is represented by a single point in your FID, or, if you look at an oldschool AD converter, the time interval between triggering the readout of one value and the next.
In the old days (of FTNMR, not the very old days of CW NMR), the real and imaginary FID intensities were sampled alternatingly, using a single ADC.
Later (probably since the eighties?), one would always have two ADCs running at the same acquisition rate, each connected to one output of the quadrature receiver.
Today, the transients are sampled at extremely high rates (oversampling at 100s of MHz) and digitally filtered/downsampled to a reasonable bandwidth/data rate before they are accumulated and stored like a normal FID from the eighties. ;)
That's regular 1D data, and the first dimenson in nD. For all dimension >1, as was explained in the other answers, everything depends on the exact experiment that was carried out. The exact, absolute time in which each single point was measured could be reconstructed, but it is rarely of any importance, unless you're doing reaction monitoring, i.e. measuring just one spectrum after the other.

I guess it's kinda off topic, but do you know where I could read more about the oversampling / digital filter techniques? I've always known they've existed (and the more practical consequences, e.g. the lack of spectral aliasing in direct dimensions), but never found any resource which got into the details. (+1'd it yesterday, naturally.)

@orthocresol There is nothing NMR-specific about it afaik, standard DSP techniques.  Oversampling is simply replacing e.g. 4 consecutive datapoints by a single point having their average value. The simplest filter uses a running average, or you can use filter functions that are not just a boxcar, but e.g. a gaussian. Mathematically that is done via convolution/deconvolution. I've long wanted to have a comprehensive book explaining that in more details. If you find one, let me know.

What you describe as oversampling is actually called downsampling. Oversampling simply means that we are way above the Nyqyuist limit.


text:
Calculate pH of imidazole in solution
I am having trouble understanding how to know how a molecule will reaction in solution. For example, if I have $\pu{1 M}$ of imidazole in solution, how can I calculate the pH of the solution by only knowing that the $\mathrm{p}K_\mathrm{b} = 7.1$?
I assume that imidazole reacts with water (we only work with water solutions) but would the reaction look like this:
$$\ce{C3N2H4 + H2O <=> C3N2H5+ + OH-}$$
So then:
\begin{array}{ccccccc}
\ce{C3N2H4 & + & H2O & <=> & C3N2H5+ & + & OH-}\\
  \pu{1 M} &   &  -  &     & 0       &   & 0   \\
       1-x &   & -   &     & x       &   & x \\
\end{array}
\begin{align}
  K_\mathrm{b} &= 10^{-7.1} = \frac{x^2}{1-x} &
  \implies x &= 2.82 \times 10^{-4}
\end{align}
So then,
$$\mathrm{pOH} = -\log(2.84 \times 10^{-4}) = 3.55,$$
and
$$\mathrm{pH} = 14 - 3.55 = 10.45.$$
Is that the correct way to think when solving problems like this?

seems correct,
confirmed by PubChem: "pH 10.5 for 38 g/L at 20 °C
sourced from: OECD; Screening Information Data Set (SIDS) Initial Assessment Report for SIDS Initial Assessment Meeting (SIAM) 17 Imidazole (CAS number 288-32-4) (2003). Available from, as of July 30, 2018: https://www.inchem.org/pages/sids.html"


text:
Machine learning-based drug design: how to determine the desired properties given a target?
I am working on a machine learning-based drug design project. The machine learning part was easily understood for me being from a computer science background. However, the biochemistry part is what I have troubles understanding.
In machine learning-based drug design such as "https://www.science.org/doi/10.1126/sciadv.aap7885", a generator model is trained to output valid molecules (in smiles string), and then, this generator is biased/fine-tuned to output only the molecules with certain chemical properties.
My questions are:

how do we determine these desired chemical properties? For example, if I want to train my generator model to output molecules that can serve as drugs for Covid-19 disease, what should the chemical properties be, by chemical properties I mostly refer to (logP, hydrogen bond donors, hydrogen bond acceptors, polar surface area, molecular weight, etc.)?
is there a standard/well-established approach to determine these desired properties given a certain target (in this case, the target is the Sars-Cov-2)?

I guess some heavy quantum chemistry computations must be involved, like the one prof. Zahradník of the "Prague school" of quantum chemistry developed.

The answer is you cannot, unless you have a large team of pharmacological chemistry geniusses. Firstly you have to define what kind of drug you want. Research the way they work. Find a suitable target region at the virus, or at the human body. And THEN you can try to do whatever machine-learning magic you think you're capable of. https://en.wikipedia.org/wiki/Clarke%27s_three_laws

Oh, and the list of chemical properties you show is not even nearly sufficient. The spatial arrangement of anything on the drug that can interact is the key.