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Shi Epoxidation Issues - Any help would be appreciated
As part of my PhD project, I'm currently looking into running a Shi Epoxidation reaction. The first step is to reproduce the current literature on this reaction by using trans-stilbene as my substrate. However, from the multitude of reaction conditions I have tried, none of them seem to work as described in the papers (very low to no conversion observed). I have tried the reaction conditions from the following papers:

"An Efficient Asymmetric Epoxidation Method for trans-Olefins Mediated by a Fructose-Derived Ketone" - the original Shi paper - using oxone as oxidant
"An Efficient Catalytic Asymmetric Epoxidation Method" - Shi - optimized conditions - using oxone as oxidant
"Asymmetric epoxidation using hydrogen peroxide ($\ce{H2O2}$) as primary oxidant" - Shi - using hydrogen peroxide as oxidant

The catalyst I have used in my reactions is the Shi Diketal Catalyst purchased from Sigma (https://www.sigmaaldrich.com/catalog/product/aldrich/520160?lang=en&region=CA). What I've tried so far:

testing different batches of oxone, including titrating it and running test reactions ($\ce{aldehyde -> COOH}$ oxidation) - > 90% conversion observed
trans-stilbene + m-CPBA -> oxidized product is quantitative
I've left reactions run for > $\pu{6-24 h}$ with low conversion observed (< 30%).
$\mathrm{^{13}C}$ and $\mathrm{^{1}H}$ NMR of the catalyst show that it's clean

I am slowly running out of things to try out. It's weird because it's a published and highly used reaction ... If you have any experience with this reaction I would really appreciate your help!

What pH are you running your reactions at?

Please add proper citations to the papers so that they are easier to find.


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Can you give examples for a halide ion reacting with interhalogen compound to give polyhalides?

Interhalogen compounds are formed when two halogens $(\ce{F},$ $\ce{Cl},$ $\ce{Br},$ $\ce{I})$ react together. They are represented by $\ce{XX'_n},$ where $\ce{X}$ is halogen of larger size and $\ce{X'}$ of smaller size. The total number of halogens in $\ce{XX'_n}$ depends on the radii of $\ce{X}$ and $\ce{X'}.$ The correct statement(s) regarding $\ce{XX'_n}$ is/are
❌ 1. $\ce{FCl3}$ is one of the stable interhalogen compounds.
✅ 2. Iodine can form an interhalogen compound containing maximum numbers
  of halogen atoms.
✅ 3. Interhalogen compounds are covalent in nature and more reactive than
  halogens, in general.
✅ 4. Halide ions reacts with interhalogen compounds to give polyhalides.

Regarding the 4th option: will it be like the following:
$$\ce{X'^- + XX'_n -> XX'_{n + 1}}?$$
Something like this? ($\ce{X}$ and $\ce{X'}$ are halogen compounds), so the polyhalide ion here would be $\ce{X'_{n + 1}}.$
What would be you views on this?

$\ce{Cl- + ICl3}$ will do.

TLDR: There is no exact generalized reaction. The reaction depends on type of interhalogen compound(value of n) and the nature of metal halides(ionic or covalent)

$\ce{XX^{'}}$ form

Let us take iodine monochloride($\ce{ICl}$). If it is reacted with ionic chloride like $\ce{KCl}$, it will form $\ce{ICl2-}$ ion.
$$\ce{MCl + ICl -> M+ICl2-, (M=K, Rb)}$$
If a different counter-ion is taken like bromide, it will form $\ce{ClBrI-}$ ion
$$\ce{ICl + KBr -> K+[ClBrI]-}$$
But if we add pure diatomic halogen, it will form a additive product
$$\ce{Cl2(excess) + ICl -> ICl3}$$
Covalent halides like $\ce{AlCl3}$ however gives different product:
$$\ce{ICl + AlCl3 -> I+ + AlCl4-}$$

$\ce{XX^{'}3}$ form

Let us take iodine trichloride. It will form $\ce{ICl4-}$ ions on reacting with ionic chloride:
$$\ce{MCl + ICl3  -> M+ICl4-}$$
If we take any other counter ion like fluorine, it will form $\ce{MICl3F}$ which is again kind of an additive product.

$\ce{XX^{'}5}$ form

Taking chlorine pentafluoride as examples, reacting with covalent fluorides like $\ce{AsF5}$ and $\ce{SbF5}$ will give adducts: $\ce{ClF5.AsF5}$ and $\ce{ClF5.SbF5}$
Iodine pentafluoride gives iodine heptafluoride on reacting with fluorine at 250 C
So, there is exact no generalized reaction. It all depend upon type of inter-halogen compound, reaction conditions and nature of metal halide. The close generalization I can make is that interhalogen compound of form $\ce{XX^{'}_n}$ if reacted with ionic metal halide will form metal polyhalide of form $\ce{MXX^{'}_{n+1}}$ or $\ce{MXX^{'}_{n}X^{''}}$(if a different counter-ion is taken). Additive product is formed when inter-halogen compound is reacted with pure diatomic halogens provided the second halogen is big enough to occupy electrons like bromine or iodine. In some case, covalent halides forms adducts with interhalogen compounds.
Thus the generalized reactions are(of course, there are certain exceptions):
$$\ce{X^{'-} + XX^{'}_n -> XX^{'-}_{n+1} (ionic)}$$
$$\ce{X^{'}_2 + XX^{'}_n -> XX^{'}_{n+2}}$$
Reference

http://www.vpscience.org/materials/Study%20Material%20on%20d%20Inter-halogen%20%20Compounds%20By%20Dr.%20D.%20M.%20Patel.pdf (reaction sources)

All interhalogen compounds are prepared by direct combination of elements, and where more than one product is possible, the outcome of the reaction is controlled by temperature and relative proportions of the halogens. For example, reactions of $\ce{F2}$ with the later halogens at ambient temperature and pressure give $\ce{ClF, BrF3}$, or $\ce{IF5}$, but increased temperatures give $\ce{ClF3, ClF5, BrF5,}$ and $\ce{IF7}$. For the formation of $\ce{IF3}$, the reaction between $\ce{I2}$ and $\ce{F2}$ is carried out at $\pu{228 K}$. Same applies for halogen and interhalogen compounds hence these reactions are uncommon to find being pretty complex.
For this:
$\ce{X'^- + XX'_n -> XX'_{n + 1}}$
I can write reactions like:
$\ce{I2 + IF3 -> 3IF}$ at $\pu{-78 °C}$ in $\ce{CCl3F}$
$\ce{Br2 + BrF3 -> 3BrF}$
$\ce{2I2 + IF5 -> 5IF}$ (*not a point to mention that $\ce{I-}$ and $\ce{Br-}$ ions exist as $\ce{I2}$ and $\ce{Br2}$, respectively)
Keeping in mind that the generalization of above stated reactions seems incorrect.

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Why doesn't anyone manufacture 4.5 V cells?
Every 4.5 V lantern batteries are a series of three 1.5 V cells. It would have more capacity if it consisted of single 4.5 V cell. But I don't see any 4.5 V cell.
According to Wikipedia's Standard Electrode Potential data page, there are vast possibilities of a 4.5 V cell. The one I think most plausible is:
$$
\begin{align}
\text{Anode:} &\quad &\ce{Ca + 2 H2O &→ Ca(OH)2 + 2 H+ + 2 e-} \quad|\cdot 5\\
\text{Cathode:} &\quad &\ce{2 BrO3- + 12 H+ + 10 e- &→ Br2 + 6 H2O}\\
\hline
\text{Overall:} &\quad &\ce{5 Ca + 2 HBrO3 + 4 H2O &→ 5 Ca(OH)2 + Br2}
\end{align}
$$
Have anyone tried this or anything else 4.5 V?

To put it mildly, I'd say that your perspective on what is and what isn't plausible in chemistry is rather skewed.

@IvanNeretin Well, to be honest, I thought it's plausible because it's a calcium battery.

Calcium is not that good in a battery, for it will react with water by itself. (Sure, some workaround might be possible.) Bromine is not good anywhere near you. HBrO3 is even worse, and moreover, it doesn't even exist in a pure form.

@IvanNeretin HBrO3 is in aqueous solution. (The cathode reaction was edited accordingly.)

Would it really 'have more capacity' if it were a single 4.5V cell? That depends on many factors that have nothing to do with the electrode potentials. Engineering a useful product is way harder than you think.

Perhaps you consider the volume allotted to the batteries as equivalent to capacity. A better measure would be ampere-hours at 4.5v, or watt-hours. And weight might be a factor to consider.
An electrochemical cell with a high voltage (4.5v) that requires complicated or dilute ingredients could better fill the volume allotted, but if it runs out of power in a shorter time, it has less capacity. You could consider 6 cells in a 9v battery to be a complicated device, but it seems so simple. Three 1.5v cells in a package would give you a simple overall structure, and could give you high capacity due to the simple internal composition of each cell.
One reason that lithium ion batteries are so highly regarded is their high electrochemical capacity.
"The more energy a [rechargeable] battery can hold when charged, the less often you need to charge it. No one wants to be replacing the batteries in their wireless keyboard every week or to have their kid’s toy conk out after just 15 minutes. Whereas a common disposable AA battery’s capacity is anywhere from 800 mAh for budget brands up to 2,000 mAh for premium versions, most rechargeable batteries start at around 1,800 mAh and go up to nearly 3,000 mAh. Most trustworthy companies right now make AA batteries around 2,000 mAh. We don’t consider batteries with stated capacities lower than 1,700 mAh, which is too underpowered, or higher than 2,500 mAh, which makes us suspicious that a company is overpromising." https://thewirecutter.com/reviews/best-rechargeable-batteries/

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Bracketing Standards and dilutions
You are running an ICP-MS (Inductively coupled plasma mass spectrometry), with bracketing standards of 10 ppm(mg/L) K+ and 100 ppm K+ and sample of 500 ppm, prepared 5 mL diluted to 100 mL. Shouldn’t you adjust your upper limit bracketing standard, instead of just diluting your sample? Your original sample concentration of K+ (500 ppm) should not change adjusted for the dilution factor. What am I not understanding mathematically?

You have standards at 10 and 100 and your diluted sample will be about 25, ignoring units. That seems fine, so what is troubling you about it?

Thanks for the feedback. I was trying to parse out the mathematics. I think I am confused as to how the software is reporting the data. I get that 5/100 would generate 500 ppm and 5/1000 would generate 50 ppm.  However, I enter my dilutions into the software and I thought it is reporting the K+ in my original, undiluted sample.  That, should not change.  I should stick to calculating it myself instead of the software, ha.

As Ed said you are following the correct protocol by diluting the sample. The key reason is for doing so is to avoid a non-linear region in the calibration curve. This is why one would dilute the sample rather than make more concentrated standards. Several interesting effects happen with alkali metals at high concentrations in the flame or plasma. One of them is the concept of self-absorption.
Imagine you are monitoring the resonance line of K emission. Once you introduce lot of potassium atoms in the plasma (by using a concentrated standard) what may happen is that the ground state K atoms may absorb the light emitted by the excited potassium atoms? What is the result? You get less intensity than expected.
Try watching a sodium street lamp with a diffraction grating, or take a CD outside at night. You will see a prominent dark line in the emission spectrum of the lamp like this at 589 nm. This is the same self absorption phenomenon. There are so many Na atoms in the ground state that that they are self absorbing the light emitted by excited sodium atoms in the electrical discharge.

This is very useful information. Thanks for taking the time.  I did neglect the linearity of my standards and it make more sense to dilute the sample. I think I need a better understanding of how the software is reporting my data.

I always tell analytical chemists to take software output with a grain of salt. Modern softwares have become black-boxes. Check numbers manually in the good old Excel and see if they make sense.


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How to derive this temperature-pressure-specific volume relationship?
(for isentropic, adiabatic, ideal gas flow)
 $$\frac{T_x}{T_y} = \left(\frac{p_x}{p_y}\right)^{\frac{(\gamma-1)}{\gamma}} = \left(\frac{V_y}{V_x}\right)^{\gamma-1}$$ where $V$ is the specific volume, $x$ and $y$ are any 2 rocket nozzle axial sections (though I presume this relation holds true for many other things as well), and $\gamma$ is the specific heat capacity ratio $c_p/c_v$, numerically equivalent to the molar heat capacity ratio $(\bar{C}_p/\bar{C}_v)$.
Thanks!
p.s. I may have just learnt the hard way that it is probably not the best idea to write a markdown-dense blurb on a phone ;)

For an adiabatic system like a piston where $\delta Q = 0$, using the first law of thermodynamics gives you the following expression:
$$\mathrm dU = \delta Q + \delta W$$
$$\mathrm dU = - p\,\mathrm dV$$
This expression is pretty much useless however, in that you can't integrate it, since $T$, $V$, and $p$ are all constantly changing interdependently in the system. However if you assume the gas is ideal and making a few substitutions...
$$C_V\,\mathrm dT = \frac{-nRT}{V}\,\mathrm dV$$
$$C_V \frac{1}{T}\,\mathrm dT = -nR\frac{1}{V}\,\mathrm dV$$
$$C_V \int_{T_1}^{T_2} \frac{1}{T}\,\mathrm dT = -nR\int_{V_1}^{V_2} \frac{1}{V}\,\mathrm dV$$
$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -nR \cdot \ln\left(\frac{V_2}{V_1}\right)$$
$$C_V \cdot \ln\left(\frac{T_2}{T_1}\right) = -(C_p - C_V) \cdot \ln\left(\frac{V_2}{V_1}\right) = (C_p - C_V) \cdot \ln\left(\frac{V_1}{V_2}\right)$$
$$\ln\left(\frac{T_2}{T_1}\right) = \left(\frac{C_p}{C_V}-1\right) \cdot \ln\left(\frac{V_1}{V_2}\right)$$
$$\ln\left(\frac{T_2}{T_1}\right) = \ln\left(\frac{V_1}{V_2}\right)^{\gamma-1} ; \gamma = \frac{C_p}{C_V}$$
$$\frac{T_2}{T_1} = \frac{V_1}{V_2}^{\gamma-1}$$
From here it's just more cycling through variables; see if you can work the mathematics a bit so to get your second relationship.

What happened to the n that is supposed to be in front of the Cv?

as in why am I starting with $C_vdT$ instead of $nC_vdT$?

Yes, that's what I'm asking.

Because the constant volume heat capacity is defined as $C_v = (\frac{\partial U}{\partial T})V$, whereas $C{v_m} = \frac{1}{n}  (\frac{\partial U}{\partial T})_V$ when utilizing the molar volume

You can look at the units in that expression to see that $C_v$ is the correct equality as well, as everything cancels out to units of energy if $C_v$ is used in units of $\frac{J}{K}$ instead of $C_{v_m}$ with units of $\frac{J}{mol \cdot K}$

It was my understanding that Cv is a fundamental physical property of the gas and is always used to designate the specific (molar or mass) heat capacity.

I guess we were just taught different conventions; I usually distinguish the two in my work as I did herein. Though, I'm not entirely sure how one would refer to $(\frac{\partial U}{\partial T})_V$ and $\frac{1}{n}(\frac{\partial U}{\partial T})_V$ separately if the latter expression is assigned $C_v$. Maybe the non-molar heat capacity is just typically unused in your field so it's not an issue?

@ChetMiller: Among the physical chemistry textbooks I've used, $C_V$
refers to the total constant-V heat capacity of the substance or system (this is also the IUPAC convention: https://goldbook.iupac.org/terms/view/H02753 ) and a tilde, subscript-m, or some other designation  (e.g., $\tilde{C_V}$, $C_{V,m}$) is used to indicate that the value is a molar constant-V heat capacity.

Why is this relationship valid only for reversible adiabatic processes and not irreversible adiabatic processes?

why is this relation only valid for a reversible adiabatic process and not an irreversible adiabatic process?

From the open system (control volume) version of the first law of thermodynamics, between cross sections x and y, $$\dot{Q}-\dot{W}_s-\dot{m}\Delta h=0$$where $\dot{Q}$ is the rate of heat addition to the control volume, $\dot{W}_s$ is the rate of doing shaft work, $\dot{m}$ is the mass flow rate, and $\Delta h$ is the change in specific enthalpy between cross sections x and y.  If the process is adiabatic and reversible, no shaft work is being done, and the gas is ideal, this equation reduces to $$dh=C_pdT=VdP=\frac{RT}{P}dP$$ along the path between the two cross sections, where V is the molar volume.  This equation integrates to $$\frac{T_x}{T_y}=\left(\frac{P_x}{P_y}\right)^{(\gamma-1)/\gamma}$$


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Role of hcl gas in organic reactions
I encountered many questions in the competitive exam field of high school organic chemistry where we add some compounds and treat it in excess alcohol solutions in the presence of dry hcl gas, it is obvious how the mechanisms might go if the hcl is in liquid solution but when i see dry hcl gas i have no idea how it might influence the reactions.
From what i can tell it is used to signify that the expected mechanism we are targeting is not common as there is this gas component present but how exactly does the presence of this acidic gas influence the mechanisms of the reactions?

And what you'd expect when you wouldn't expect that to be a catch?

Well the hplus would activate the carbonyl carbon even more amd the alcohol will attack and deprotonate to give product in the option a but this wont happen in dry hcl as hcl needs moisture to ionize

That's not true. While in vapor phase HCl won't protonate anything, but it's only a way to deliver it to the reaction mixture. Anhydrous mixture lowers acidity but not much.

The important thing about HCl gas in these reactions is that it is dry, there is no water present ruling out any hydrolytic pathways. This means that the first step to consider is a protonation by HCl as HCl can ionise in alcoholic solvent, followed by nucleophilic attack by the alcohol solvent. In the example shown the oxygen of the carbonyl is protonated and MeOH attacks. The transient product is the hemiacetal A, this is then protonated, water leaves stabilised by electron donation from the OMe group, a second MeOH adds to give B as the final product. Further reading and a diagram of the mechanism is here

But how that protonation happen without moist hcl in solution and if there is no role of hcl why is it even present we can very well remove it from our catalyst

HCl can ionise in protic solvents. You need something to protonate the carbonyl group. TFA or Methanesulfonic acid will do as well. Dry HCl in MeOH/EtOH can be generated by adding Acetyl chloride or Me3SiCl to the anhydrous alcohol.

While we appreciate your contribution to the site, it would be preferable to not answer questions, that do not comply with our homework policy.


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Term symbols for nitrogen: determining J
I'm trying to figure out what terms are possible for nitrogen with the electron configuration $\ce{[He] 2s^2 2p^3}$. There is an old question on StackExchange and the corresponding answer was a great help already but doesn't mention $J$ or specify what terms exactly are possible. I'll try to explain how I would determine the possible terms:
The s-electrons needn't be considered for they complete the s subshell, this means I'm left with three p-electrons. I do get the same results for the possible values of $L$ and $S$ as user orthocresol provided with his answer namely:
$L=3,2,1,0$
$S=3/2, 1/2$
Now the possible values of J range from $L+S$ to $|L-S|$:
For $L=0$ there are two S-Terms*: $^4\!S$ and $^2\!S$ giving
$$^4\!S_{3/2} \quad ^2\!S_{1/2}$$
$L=1$ gives two terms*: $^4\!P$ and $^2\!P$, the former term has the following possible J values: $J = 5/2, 3/2, 1/2$ and the latter: $J=3/2, 1/2$. The P-Terms of this electron configuration are:
$$^4\!P_{5/2} \quad ^4\!P_{3/2} \quad ^4\!P_{1/2} \quad ^2\!P_{3/2}\quad ^2\!P_{1/2}$$
Doing the same for $L=2$ there are two D-Terms*: $^4\!D$ and $^2\!D$, possible J values for the former: $J = 7/2, 5/2, 3/2, 1/2$ and for the latter: $J = 5/2, 3/2$. This gives six different terms.
I'm stopping at this point because I think I'm not doing it right. Using the NIST ASD I only see five terms with this electron configuration in the Grotrian diagram:
$$^4\!S_{3/2}, ^2\!D_{5/2}, ^2\!D_{3/2}, ^2\!P_{1/2}, ^2\!P_{3/2}$$
and according to the book "Physical Chemistry: A Molecular Approach" Table 8.4 for three p-electrons only $^2\!P,^2\!D,^4\!S$ Terms are possible. What's wrong with my approach that I'm that much off?
*Question on the side: I don't think it's correct to use the term "term" here because what I'm talking about is split into actual terms. Is there a way to refer to, for example, the $^4\!D$-Terms in general, maybe term system?

Firstly, regarding the extra question: Atkins' Molecular Quantum Mechanics (5th ed.) uses "term" for $^2\!S$, and "level" for $^2\!S_{1/2}$.
Back to the main question. It's been a long time since I did term symbols, so I am happy to be corrected, but if I am not wrong, your $^4\!P$, $^4\!D$, and $^2\!S$ terms are not allowed because of the Pauli exclusion principle. The argument is pretty similar to the one in the comments on my answer, which you're already aware of.
A term with $S = 3/2$ must have a set of states with $M_S = +3/2, +1/2, -1/2, -3/2$. Likewise, a $P$ term with $L = 1$ must also have a set of states with $M_L = +1, 0, -1$. Consequently, the total number of states associated with a single term $^{2S+1}\!L$ is $(2S+1)(2L+1)$. (The total number of states in a level, $^{2S+1}\!L_J$, is $2J+1$, and you can also show that the sum of this over the allowed values of $J$ is equal to $(2S+1)(2L+1)$.)
One of those twelve states in the purported $^4\!P$ term must have $(M_S, M_L) = (+3/2, +1)$. However, $M_S = +3/2$ can only be achieved if all three electrons have the same spin (spin up in this case), which in turn necessitates that the three electrons are all in different p-orbitals, such that $M_L = \sum m_l = +1 + 0 + -1 = 0$.
Ergo, a state with $(M_S, M_L) = (+3/2, +1)$ cannot exist, and the entire term $^4\!P$ cannot be possible. The same argument also applies to the $^4\!D$ term.
I suspect that showing that the $^2\!S$ term is forbidden is somewhat trickier. I had a similar issue before with carbon, which is an easier system to understand (only two electrons to consider instead of three), so maybe you will find this useful: Pauli-forbidden term symbols for atomic carbon.
Out of the three surviving terms ($^2\!P$, $^2\!D$, $^4\!S$) you already see from NIST that all the possible $J$ values (from $|L-S|$ to $L+S$) are allowed, so there is no issue with your calculation of $J$.

So for each term there is a set of microstates and if one of these microstates can't be fulfilled without disregarding the Pauli principle than the whole term needs to be discarded?

Yes, pretty much.


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Supplement weight by element or compound
When a vitamin or supplement claims, for example, that it contains

600 mg of calcium (as calcium carbonate)

what does that mean? Does each tablet contain 600 milligrams of the element calcium?
Or does a tablet contain 600 milligrams of the compound calcium carbonate? In that case, using the molar masses of Ca and CaCO3, each tablet would contain only
600 * (40.1 / 100.1) = 240.4 milligrams of the element calcium.

It means it has calcium carbonate sufficient to provide 600 mg of calcium. Since chalk is 40% calcium, the mass of chalk is 2.5 times 600 mg = 1.5 g.

Otherwise, there would be written  "600 mg of calcium carbonate".

Similarly, 600 Euros ( in dollars ) does not mean 600 dollars, but 600 Euros exchanged into the equivalent number of dollars.


text:
Diketone intramolecular reaction question

I realize that one of the methyl groups on either carbonyl compound will become deprotonated and then it will attack the other carbonyl to form a pentane ring. However, with the methyl group in the picture below there should be an OH group. There is no heat in the reaction at all that would make this reaction an aldol condensation, thus how does the double bond form?

Why do you think you need heat? The initial ring formation, an aldol REACTION, forms the beta-hydroxycyclopentanone. Base again deprotonates the ketone and elimination of hydroxide occurs with net loss of water---an aldol CONDENSATION. Some reactions do occur at ambient temperature.


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what is the bonding energy of I2 and I3-?
I need some information of bonding energy.
I found that the bonding energy of $\ce{I2}$ is $\pu{151 kJ/mol}$ (https://www.chegg.com/homework-help/dimethyl-ether-ch3och3-ethanol-ch3ch2oh-constitutional-isome-chapter-9-problem-90p-solution-9781259631757-exc),
but I have some difficulty in searching the bonding energy of $\ce{I3-}$.
Would you inform me the bonding energy of $\ce{I3-}$? ...or is there any theory related how to guess the bonding energy?

What is 'bonding energy'? Does that refer to the strength of one I-I bond, or something else?


text:
How to decide whether to pick 4-bromo-2-methoxyphenol or 4-bromo-6-methoxyphenol as the name for the compound?
Why is the following compound named as 4-bromo-2-methoxyphenol and not 4-bromo-6-methoxyphenol?
Shouldn't the substituents be numbered according to their alphabetical order?

When choosing which name to use when the only difference is the numbering, we choose the combination that gives the lowest possible numbers, 2 and 4 in this case.
So the name will be 4-bromo-2-methoxyphenol. The substituents are given in alphabetical order.

A more explicit last line would be "The substituents are written down in alphabetical order".

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic groups, which are expressed as a suffix or implied by the name of the functional parent compound
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The compound given in the question has the functional parent compound phenol. The name ‘phenol’ implies the presence of the characteristic group $\ce{-OH}$. Since there are no other characteristic groups and low locants are assigned first to the principal characteristic group, the locant ‘1’ is assigned to the substituent group $\ce{-OH}$.
Finally, low locants are assigned to the remaining substituents as a set according to Rule (f); i.e. low locants are assigned to the prefixes ‘bromo’ and ‘methoxy’, all considered together as a set in a series of increasing numerical order. Thus, the systematic name for this compound is 4-bromo-2-methoxyphenol and not ‘4-bromo-6-methoxyphenol’ since the locant set ‘2,4’ is lower than ‘4,6’.

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.

Only if no decision can be attained by application of Rule P-14.4(f), the lowest set of locants is selected according to P-14.4(g) in the order of the citation in the name (which indeed is based on the alphabetical order).


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Liquid solutions and vapour pressure
"The vapour pressure of water at 80° C is 355 torr. A 100 mL vessel contained water-saturated oxygen at 80° C, the total gas pressure being 760 torr. The contents of the vessel were pumped into a 50 mL vessel at the same temperature. What were the partial pressures of oxygen and of water vapour, what was the total pressure in the final equilibrated state? Neglect the volume of any water which might condense."
This question is given in my module and the solution given to the same is quite vague. Could somebody please explain it to me? Thank you very much. What I could particularly not understand is how the water vapour's pressure in the 50 mL vessel remains unchanged.

That's precisely because it is saturated vapor. We see it pretty much every day, and still it feels weird. If you put it under the piston and press, its pressure wouldn't grow at all.

The key to understanding this type of question is that the equilibrium (saturation) vapor pressure of a pure liquid is to a first approximation (that is, assuming ideal behavior) independent of the presence of other gases, such as (in this case) oxygen. Therefore, you can think of $\pu{355 Torr}$ as being always the pressure of the vapor when in equilibrium with the pure liquid at $T=\pu{80 ^\circ C}$. For equilibrium to be established requires that there is (more than) enough substance to fill the volume with vapor at that pressure. The minimum amount of substance which will fill the volume with vapor at that pressure is, under the assumed ideal conditions, given by the ideal gas law: $$n=\frac{pV}{RT_b}$$where in the present problem $T_b=\pu{80 ^\circ C}$, $p=\pu{355 Torr}$, and V is the volume occupied by the vapor, which changes from $V<\pu{100 mL}$ to $V<\pu{50 mL}$. In the present problem you don't know how much of the water is initially a liquid, only that the inequality $V<\pu{100 mL}$ is satisfied by the vapor. During the volume reduction some of the vapor condenses into additional liquid in order to retain the equilibrium pressure. The oxygen is treated here as a spectator, although it contributes to the total pressure in the vessel. In practice it also alters the vapor pressure of the liquid, although here by a negligible amount.

Sir, this is a different question, but related to vapour pressure (I couldn't find any other related question): If we are given a solution of some solutes whose concentrations and temperatures are given, can we compare their vapour pressures? I know that higher temperature means higher VP, but is there an empirical relationship b/w concentration and temp? For example, out of 0.02 M $\ce{NaCl}$ at $50^{\circ}$C , 0.005 M $\ce{CaCl_2}$ at $50^{\circ}$C , or say, 0.03 M sucrose at $15^{\circ}$C. They also have different Van't hoff factors.. so can you explain that in brief?

I am not sure what you mean by "relationship between concentration and temp". I recommend you read about colligative properties and Raoult's law if you haven't already. Raoult's law says that vapour pressure depends on concentration on solute, or rather on mole fraction of solvent: $p=\chi p^\circ$. This will result in an effect on the boiling point T at a particular pressure, known as boiling point elevation. Chem SE has many Q&As on the subjects, and the web is full of resources. Maybe you should take a look at this, for instance: http://users.stlcc.edu/gkrishnan/solution2.html

Can you have a look at this question I asked? I am not looking for a relation b/w concentration and temperature, but VP and conc. or temp, so that it easy to compare the VP's. I have read colligative properties and Raoult's law, but in this question, since temperatures and concentrations are not same, can we compare VP's? For example, one thing that I am looking for is this.But concentration is still not taken

And congrats for becoming a moderator! I hope, you also answer my question if I have made myself clear enough :).


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Which substitutent should be placed on axial position in SCl2(OCH3)2?
According to Bent's rule, in $\mathrm{sp^3d}$ more electronegative element is placed on the axial position, so between $\ce{Cl}$ and $\ce{OCH3}$ in $\ce{SCl2(OCH3)2}$ which should be placed on axial as oxygen is more electronegative, but wouldn't $\ce{CH3}$ reduce its overall electronegativity due to its positive inductive effect?The electronegativity of Oxygen is more than that of chlorine but the negative inductive effect of OCH3 is less than that of Chlorine
But my teacher said that $\ce{OCH3}$ should still be placed on axial position considering electronegativity of oxygen.
Can someone please explain this? Because we are comparing a group's electronegativity (which cannot be found on the Pauling scale) to that of an atom.

What can be done to improve the quality of the question?

Related: Geometries of SCl2(OCH3)2 and SF2(OCH3)2

This post does not explain why OCH3 is placed at axial...I have seen this post before and found it to be giving a  conceptual understanding of Bent's rule at a basic level but my question is slightly different

EN of $\ce{O > Cl}$, but -I effect of $\ce{OCH_3^- < Cl^-}$. May I add this point in question?

Thanks for the suggestion have added that in question...am hoping for a proper answer to this now


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Question on identification of a compound
Question:

Multiple correct answers are possible
My Approach:
I just found structure Q for all options A, B, C and D for which I have uploaded the structures below with their corresponding options.

Of course, option B cannot be correct since the molecular formula of the compound Q doesn't match with the one obtained after the reductive ozonolysis.
What I know about Molisch test is it gives the violet ring when an alcoholic solution of 1-naphthol is added to the carbohydrate solution. But I don't know how to proceed after this step as I don't know how to identify which of the compounds react with Molisch reagent.

Generally Q is dehydrated with H2SO4 to form a furfuraldehyde before the Molisch test. In which case Q must be at least a C5 aldehyde. Try this: https://chemistry.stackexchange.com/questions/106734/clarification-in-the-mechanism-for-molischs-test-for-glucose/106758#106758


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Which of the canonical structures contributes more to resonance hybrid?
I am quoting a rule on resonance from Organic Chemistry by T.W. Graham Solomons, Craig B. Fryhle, 12th edition, page 25.

Structures in which all the atoms have a complete valence shell of electrons
  (i.e., the noble gas structure) are more stable.

I tried to apply to the following.

According to this rule structure 4 must be most stable. But structure 4 is not aromatic, while structure 1 is aromatic. Since aromatic species are more stable then nonaromatic species, structure 1 must be more stable.
My question is which of the structures is more stable?

One resonance structure simply can't be (or not be) aromatic, much like an island inhabited only by Tom Hanks can't be democratic. It is a collective feature.

There is no more (or less) stable resonance structure. Only the complete set describes the molecule, they simply do not exist on their own. See: What is resonance, and are resonance structures real? The question you pose is flawed and cannot be answered. You might be able to find the contribution of the different conformations though.

Better question would be to ask, "Which cannonical structure contributes more to the resonance hybrid? "

From my studies, I would have chosen number 4 because it has the greastest number of bonds (covalent), which carries more weight than the precense and placement of charge. My text (Organic Chemistry 7th ed. by Francis Carey) doesn't cite aromaticity as a main rule for determining resonance contributions.

There are few things you should keep in mind while deciding comparative stability of resonance structures:

Non-polar structures are more stable than dipolar.

Resonance structures with a greater number of covalent bonds are more stable than those with lesser number. Thus, nonpolar structure of buta-1,3-diene is more stable than any other resonating structures.

Resonance structures with similar charges on adjacent atoms are insignificant due to electrostatic repulsion and thus are unstable.

Structures with positive charge on electropositive element and negative charge  on electronegative element are more stable.

Structures in which charge is delocalised are more stable than those in which there is charge separation.

Structures with complete octet are more stable.

All the atoms in a molecule taking part in resonance should be co-planar. because this enables effective overlap of p-orbitals and delocalization of electrons.

But when you have to decide among these properties itself, you can use the following observational facts:

Although during comaprison, there may be a positive charge on the more electronegative atom(O in this case) with complete octet, but it will be preferred over a structure with an electropositive atom ( C in this case ) having positive charge and incomplete octet.

But when number of covalent bonds are same, positive charge on C atom with incomplete octet is more preferred than positive charge on more electronegative O atom.

A stable neutral structure is a major contributor of a resonance hybrid.

An important note: when you are comparing the stability of the resonance structures , you are actually trying to figure out the major contributor to the resonance hybrid (that which best describes it). Technically, all the resonance structures are contributors to the same true structure of the ion. Since all of your mentioned structures represent the same species, they cannot have have different energies, and therefore they cannot have different stabilities.
Your question is actually asking "which contributor is more important in describing the structure and behavior of the hybrid?"
So fact number 1 and 2 makes it very clear that structure 4 in this case will be the most stable.

I can only reiterate: There is no such thing as a more (or less) stable resonance structure. They simply do not exist on their own and are only a tool for approximating the electronic bonding situation; only a complete set would actually do that. This answer is unfortunately wrong.

what's a complete set ?.. It must be the pack of all the possible resonance structures of a single compound..in this case it looks complete..

A complete set is all possible resonance structures. The one in the question is by far not complete, it is missing the charge separated structures. The only part in which it could be considered complete is the pi system.

but don't you think it's not possible to make charge separated structures here .. I mean the "O" atom should be pi bonded in such a case ... how can a negative charge develop over the "O" atom..??

With charge separated structures I mean those representing ionic bonds, i.e. in the way like $\ce{[H-CH3 <-> H+\quad^-CH3]}$. These do have non-negligible contributions and are necessary for complete description. What you describe in your answer is approximated beyond the useful, even if you disregard that resonance structures do not exist.

My dear friend @Martin-マーチン, there are more or less significant resonance structures. Only when we know the more significant resonance structures can we begin to predict the properties of a molecule. Sure, resonance structures could not be isolated but that does not mean they do not "exist" since the real structure is essentially a superposition of all possible resonance structure but adding a different linear factor to each structure.

Ionic resonance structures, I agree, they do exist. It is with ionic resonance structures that we could explain things like HF2- and B2H6. However, in this case, we should ask ourselves, is such an ionic resonance structure stable? Well, no, because a CH3- the structure is inherently unstable and the C-H bond has hardly any ionic character.

@user85426 Resonance structures are here mathematical tricks just as orbitals are; that is not only my point of view. They do in none of the accepted definitions of stable exist. I respect that you gave a different view and you can disagree with me, but there won't be convincing me. Please don't call me dear friend, that is a term of endearment that has to be earned otherwise it sounds condescending.

Sorry if whatever I said offended you, it is just how I usually talk. Thank you for responding. Of course, I agree that resonance structures are in fact, as you said, "mathematical tricks" (or I would like to call them tools). But when we are talking about the term "existing", there is really no scientific definition for the term. Of course
they could not be isolated but that does not mean that they are not a representation of the true structure. In that sense, they very much exist.

From a fundamentally, philosophical level, what really does existence mean anyway? How do we really know that the true structure exist even when we ourselves are not sure of our own existence? Is this a phone I am typing on? Is there really a forum? Am I really talking to someone else from the other end of the globe? Well, I will not truly know. Do resonance structures exist? Well, if my existence could be doubted, does anything actually exist?


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Favourability of leaving groups in nucleophilic aromatic substitution
It is known that the order of leaving group favourability of the halide ions changes when we go from nucleophilic substitution, in the form of $
\ce {S_N1}$ and $\ce {S_N2}$ reactions, to nucleophilic aromatic substitution reactions ($\ce {S_NAr}$). In the former, the primary consideration is the strength of the $\ce {C-X}$ bond since the rate-determining step would always involve $\ce {C-X}$ bond cleavage. Thus, the order starts with $\ce {I^-}$  as the most favourable and $\ce {F^-}$ as the least. However, in the latter, the primary consideration is the ability of the halogen atom to withdraw electron density from the carbon atom it is attached to. This is because the rate-determining step is the initial addition to form the meisenheimer intermediate, which does not involve $\ce {C-X}$ bond cleavage. Thus, the order starts with $\ce {F-}$ as the most favourable and $\ce {I^-}$ as the least.
Apart from analysing the favourability of the halides as leaving groups in $\ce {S_NAR}$ reactions, I hardly see this concept being used to analyse the favourability of other leaving groups in $\ce {S_NAr}$ reactions. How would conventionally "bad" nucleophiles, such as $\ce {CH3O^-}$ or $\ce {OH^-}$, fare in such reactions? Would they be as superior as leaving groups like $\ce {F^-}$ or $\ce {Cl^-}$ in $\ce {S_NAr}$ reactions?
Some information is provided in Carey & Sundberg (2007) on this:

Groups other than halogen can serve as leaving groups. Alkoxy groups are very poor leaving groups in $\ce {S_N2}$ reactions but can act as leaving groups in aromatic substitution. The reason is the same as for the inverted order of reactivity for the halogens. The rate-determining step is the addition, and the alkoxide can be eliminated in the energetically favorable rearomatization.

Reference
Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer, 2007.

That is a pretty good question. Do you have any example where alcoky groups act as leaving groups in SNAr?


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Why draw "up" arrows first when filling orbital diagrams?
I am a 11th/12th grade student studying orbital configurations. I have gone over this lesson and it says to always draw spin up first. However, it never says why. Is drawing down spins first incorrect?

There is definitely no physical reason behind this. Or, for that matter, behind the very idea of drawing spins up/down, rather than left/right.

By doing it consistently, you are sort-of automatically following Hund's Rule. But it makes more sense to learn the rule itself and follow it consciously.

Consistency is just a good habit to get into so that you don't forget to follow Hund's Rule. I would gather that up is typically drawn first because it's easier to draw an up arrow vs a down arrow. If a down arrow is faster/easier for you, go for it.

And there was at least 2 more already - one of them also in "related" , just right from my comments...

There is no real reason about that and drawing them "down" would be exactly the same as long as you draw them always in the same direction (it needs to be consistent).
The same applies to orbital lobes which can be hollow (white) or plain black depending on their relative sign. When two molecules' orbitals interact, you just consider they have the right sign for an interaction.


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Computing enthalpy of formation of dissolved gas from its enthalpy of dissolution
Provided that I know the enthalpy of formation $\Delta H_f^0(X_{g})$ of a substance $X$ in gas phase and its enthalpy of dissolution $\Delta H^0_{sol}(X)$ in the same conditions, considering that the dissolution reaction is $X_g \leftrightharpoons X_{aq}$
 is it legit to consider that the enthalpy of formation of that substance $X$ in aqueous phase would be $\Delta H^0_f(X_{aq}) = \Delta H^0_{sol}(X) - \Delta H_f^0(X_{g})$?
As a concrete exemple, considering that the enthalpy of formation of $H_2$ in gas phase is 0 kJ/mol and that its enthalpy of dissolution is approximately -4.15 kJ/mol (using the data provided here), can we conclude that the enthalpy of formation of $H_2$ in aqueous phase is approximately -4.15 kJ/mol?

That seems correct.


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Boiling point elevation and liquid solutions

$\pu{1 L}$ of an aqueous solution of urea having density $\pu{1.06 g mL-1}$ is found to have elevation in boiling point $\Delta T_\mathrm{b} = \pu{0.5 °C}.$ If the temperature of this solution is increased to $\pu{101.5 °C},$ then calculate the amount of water which must have vaporized by doing so. Ebullioscopic constant of water is $K_\mathrm{b} = \pu{0.5 K kg mol-1}.$

I tried solving this question using the Clausius-Clapeyron equation and the basic ideal gas equation. I calculated the change in the number of (gaseous) moles of water that would occur by raising the solution's temperature to $\pu{101.5 °C},$ and then multiplied it by its molar mass, hoping to get the mass of water thus evaporated.
Am I doing something incorrectly? My answer does not seem to match the answer given in my book at all. If yes, please help me out here.

Use the ebullioscopic equation (the first equation) in this Wikipedia article,
$$\Delta T = K_\mathrm{b} m$$
First solve for $m_\mathrm{init}$, the initial molality of urea. Second figure out at what molality $m_\mathrm{fin}$ the boiling point is elevated by $\Delta T = \pu{1.5 °C}.$ Since the amount of urea is constant, the % change in the mass of water is then given by
$$\% \left(\frac{\Delta w_{H2O}}{w_{H2O,\mathrm{init}}}\right)= \frac{m_\mathrm{init}^{-1}-m_\mathrm{fin}^{-1}}{m_\mathrm{init}^{-1}}$$
where $m$ is the molality of urea, $w$ mass solvent.


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Potential energy of electron in excited hydrogen atom

The angular momentum of electron in an excited H atom is $\frac{h}{\pi}$. The potential energy (PE) of electron is?

Let $\frac{h}{\pi}=\frac{nh}{2\pi}$, therefore $n=2$.
So, $$E=\frac{-13.6}{4}=\pu{-3.4eV}$$
The answer is $\pu{-6.8eV}$. What’s wrong with my solution? (I think the issue is with the formula, yet I would like to confirm it)

The Bohr model is fundamentally flawed and incorrect. The (orbital) angular momentum of an electron in an orbital with quantum number $l$ is given by $\sqrt{l(l+1)}\hbar$, not $n\hbar$.

This answer owes a big debt to @orthocresol, who got me back on track after my over-simplified answer to the OP’s question.
In the long obsolete Bohr theory of the hydrogen atom, the total energy of the $n$-th energy level, $E_n$, is $-2.179\times10^{-18}/n^2\ \mathrm J$, which is approximately $-2.18\times10^{-18}/n^2\ \mathrm J$. This is depicted in the figure below, with original figure reference therein.

For the $n$-th energy level, $E_n$ equals the sum of the kinetic energy, $T_n$, and the potential energy, $V_n$. Using the virial theorem, as per this link, the average kinetic energy, $T_\mathrm{ave}$, is minus one half of the average potential energy, $V_\mathrm{ave}$. Thus $$-V_\mathrm{ave}=2 T_\mathrm{ave}$$
Let $T_n$ be estimated by $T_\mathrm{ave}$ and $V_n$ be estimated by $V_\mathrm{ave}$. For integer $n\ge1$, $E_n=-2.18\times10^{-18}/n^2\ \mathrm J$. With $n = 2$, $E_2=-5.45\times10^{-19}\ \mathrm J$. Since $1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm J$, $E_2 = -3.40\ \mathrm{eV}$. Therefore, $-3.40\ \mathrm{eV}=V_\mathrm{ave}+T_\mathrm{ave}=V_\mathrm{ave}-V_\mathrm{ave}/2=V_\mathrm{ave}/2$, so $V_\mathrm{ave}=-6.80\ \mathrm{eV}$. Then $T_\mathrm{ave}=3.40\ \mathrm{eV}$. In summary: average potential energy = $-6.80\ \mathrm{eV}$, average kinetic energy = $3.40\ \mathrm{eV}$, and total energy = $3.40\ \mathrm{eV}$.
This link is recommended for further information.

Actually, the -6.8 makes complete sense in the context of quantum mechanics: it is a known result (the virial theorem) that for a hydrogen atom, $\langle V\rangle = -2\langle T\rangle$, where $V$ and $T$ are potential and kinetic energies respectively. Now, because -3.4 is the total energy $E = T + V$ (the formula for $E$ is derived from a Hamiltonian which includes both kinetic & potential energies), this leads to the result that $T = +3.4$ and $V = -6.8$. What is baffling is the simultaneous and incoherent use of quantum mechanics with the old Bohr model for angular momentum.


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Pncn space group diagram
I'm trying to make a diagram of the Pncn space group.
Firstly, I know that this space group belongs to the orthorhombic system and derives from the point group (2/m 2/m 2/m).
Therefore, although the abbreviated notation does not appear, this spatial group also has binary axes (which may be normal or helical).
We also know that the multiplicity of the space group is 8. That is, there are 8 positions equivalent to (x,y,z). Next, we try to determine them:

By the translation of the primitive lattice: (x,y,z)
As the point group 2/m 2/m 2/m/m is centrosymmetric: (-x,-y,-z)
For the plane type "n" perpendicular to the axis "a": (-x, 1/2+y, 1/2+z)
By the plane type "c" perpendicular to the axis "b": (x, -y, 1/2+z)
By the plane type "n" perpendicular to the "c" axis: (1/2+x, 1/2+y, -z)

If they were correct, I'd still be missing three positions. These three positions are given by the binary axes. Here I have a serious problem because I don't know if this binary axis is normal ($2$) or helical ($2_1$).
After making these deductions, I check to see if I've done well:

After that, I realise that some of the equivalent positions do not coincide with those I have deduced. For example, the two of the sliding planes of type "n".
I also have mistakes that I can't explain or understand. These are the following:
a) Why is one of the binaries at 1/4? I've searched the Internet and I can't find any explanation that helps me much.
b) Why is the plane type "n" displaced? I would have placed it on the line of axis "b".

I thought of adding a link to a very similar recent question... then I realized it was your question. What happened? Are you moving back in time? How can we help?

Indeed, a month ago I asked that question and you answered it very cordially. The need for the question is that there are still aspects that I do not understand in the realization of these diagrams. It is not a time travel.

Well, my former answer applies to this problem just as well. How is it different, really?

The problem is obviously similar. But, I went to solve it to the answer you gave me in your day and I didn't get the right diagram. The seven points you told me to follow are not very well understood and, therefore, I asked this new question. I hope that all of you can help me. @IvanNeretin

Go by the steps. Draw all the planes. Then deduce all axes and screw axes. Then the center. At each step compare your picture to the known result. Stop when you encounter the first differences. Where are we now?

But, do I plane when I know it's on the axis or at 1/4? For example, in this case, the plane "n" is 1/4 of the axis and the plane "c" is not. How do I know that?

When you start, you don't know anything of that. Just put the planes in place (at 0, not 1/4). If the final picture shows some of them at 1/4, this must have been a result of origin shift sometime later. Take a note of that and move on.

I just drew them on the line of axis a and axis b. I have also drawn them both to 1/2, that is, both planes are cut in the center of the diagram. What do I do next?

Go for the axes. See if they are positioned relative to the planes the same way they are on the picture.

How can I see that?

Each axis would be along the intersection of the corresponding planes, unless there is a shift, in which case go for the corresponding step in the instruction and see what the shift does to the axis.

What means "in which case go for the corresponding step in the instruction"?

I vaguely remember it tells you what to do when one or both combined elements have shift in them.

I still don't understand when to set the symmetry elements to 1/4. I would like to know which is the algorithm, set of steps, that you perform to know the diagram.

You set the axes to 1/4 when they are at 1/4. This may result from a few causes: (1) the plane that gave birth to this axis was itself at 1/4, or (2) the plane was at 0, but the other plane contained a shift that moved the axis, or (3) someone shifted the origin afterwards.

The second reason means that if we have an ordinary plane of symmetry perpendicular to a sliding plane of "tipio b", does the binary axis move to 1/4?

If the shift is perpendicular to the axis, yes.

And could you indicate some kind of algorithm that you can use to solve this kind of problems in a correct and efficient way?

My old answer is this algorithm.

In that algorithm, I find some shortcomings. For example, the 1/4, that we have commented before, and know if the binary is normal or helical.

The binary is helical if one of the parent planes had a shift along it. Otherwise it is normal. This is not exactly a shortcoming; the answer did actually contain a phrase to that effect ("If the shift is parallel to the resulting element..."), but you had to decipher it yourself.


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Does ordered mesoporous carbon (for example, CMK-3) contain aromatic rings?
I'm studying the article Synthesis of ordered mesoporous phenanthrenequinone-carbon via π-π interaction-dependent vapor pressure for rechargeable batteries, in which authors explain some properties of the resultant composite of phenanthrenequinone (PQ) and ordered mesoporous carbon CMK-3 by the $\pi-\pi$ stacking between these moieties.
While PQ obviously contains aromatic rings, does the ordered mesoporous carbon do? As far as I know, $\pi-\pi$ stacking is possible only between aromatic compounds.
UPD1: So I did several things to elaborate on this question.
First, I tried to figure out, how ordered mesoporous carbon is produced. In this article authors report on several ways therefor; most of those include pyrolysis of hydrocarbons. Well, in my non-professional opinion, this means, that the product will contain hydrogen in some form. Thus, there will be hydrocarbons, including, probably, cyclic and aromatic (for example, pyrene is often a product of burning, as far as I know). Is this right?
Second, I tried to figure out whether it is in general necessary for $\pi$-stacking to have both compounds aromatic. This answer suggests, that not, but that p-orbitals are needed. Can anyone confirm, please? Also, in this article (still to be read) there should be some insights.
Third, I decided to check on the question what is $\pi-\pi$ stacking in general, by reviewing this article. While I'm still in process of reading it, I've already understood, that there's a lot of confusion in this field since there is a variety of this interaction types; possibilities depend strongly on the molecules themselves and their surroundings. Am I wrong?
Fourth, I tried to track back one of the statements regarding multi-walled carbon nanotubes $\pi-\pi$-interacting with PQ from this article (nanotubes just seem to me somewhat more clear than CMK-3, the chemical formula of which I even don't know). This led to surprisingly branched and sad journey to several articles aged past to '70s, not in any of which the $\pi$-stacking was ever proven, but always assumed (I have a long report for myself on this journey, here it is, however, not the correct place to post it). Anyways, it's an ongoing literature research and I decided to make a pause when the amount of articles to check reached six.


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Carbon-Carbon Chain stabilities
While learning about Carbon and its Compounds in a class 10 coaching, I suggested a structure for some isomers.
It contained a connection of two consecutive carbons connected by double bonds. The teacher said that it can be counted as one, but is not a stable compound.
A recent research on C18 carbon carbon chain allotrope involved alternating triple bonds.
My question is, why is consecutive connection of carbon atoms via double bonds not stable, whereas an alternating triple Bond chain is found to be existing? Also explain some physics behind it

Sequential double bonds are less different from alternating triple bonds than you seem to think. Also, both are not very stable.

related https://chemistry.stackexchange.com/questions/41888/why-does-treatment-of-a-geminal-dihalide-with-sodium-amide-lead-to-an-alkyne-and https://chemistry.stackexchange.com/questions/47568/do-allenes-form-rings


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Vaporized silver
Silver vaporized in a propane oxygen flame condensed onto borosilicate glass is highly reactive to the flame atmosphere.
The silver coating seems to disappear in an oxidizing flame yet turns yellows and browns in a reducing flame.
The level of reduction and even the pressure in the gas line will dramatically change the appearance of the silver in the finished glass work.
What is causing this to happen?  Is the vaporization process in the flame forming something other than elemental silver? Why does it change color/disappear in different flame environments? How is this effecting the outcome in final product?
Any help is greatly appreciated!
I use this glass coloring technique extensively, I teach other glassworkers how to use it but have very little understanding as to how or what is happening. An array of colors can be made by controlling the gas pressure and flame atmosphere.  There is no data I could find that explains this behavior of silver.
I want to add that if the glass and silver is not heated beyond the softening point of the borosilicate glass (about 1500f) the silver can be made to turn yellow/brown to clear over and over again by simply adjusting the amount of oxygen in the flame. It is a repeatable process. The amount of reduction in the flame when the silver is vaporized also has an effect on the outcome. Less reduction will give more transparent colors ranging from blues to greens reds and yellows while more reduction is associated with more yellows, white and more opaque or translucent colors.

Thin films of pretty much anything can have surprisingly different colors, for reasons having nothing to do with chemistry.

Yes, this is true. I’d like to know those reasons!

In one word: diffraction.

How would this be so greatly effected by flame atmosphere ?

Thickness of the layer is one of many obvious ways. Then again, besides all that thin film diffraction stuff, there may be some genuine chemistry going on, after all.

I was going to suggest thin film interference: https://en.m.wikipedia.org/wiki/Thin-film_interference . Given the circumstances, the silver layer (and whatever may be reversibly on it) can change thickness, thereby changing color or appearance. By any chance, is this behavior observed with any other metal layer you might use?

I should add that once the silver is worked into the glass it is no longer a thin film, it leeches out into the glass like water color might spread on paper. This is greatly effected by reduction. Too much reduction prevents this from happening.

The only other metal I would use is gold which doesn’t seem to be effected in the way that silver is by flame atmosphere.

Curious. I remember reading that molten silver can “absorb” a considerable quantity of oxygen. If this is true, then perhaps it contributes to the film thickness and color effect. There is no silver oxide at the temperatures you are using, so I have no further hunches, unfortunately. But, if you could post a couple of small photos, maybe someone here could provide an answer or a clue.

Gold nanoparticles have been used to make ruby glass for hundreds of years. The color is due primarily to surface plasmons. In a similar way, the color of metallic silver varies with particle size. If the particles are large, there is a large conduction band and the silver is highly reflective across the spectrum. Nanoparticle silver, though, exhibits various colors due to surface plasmons. According to that reference, 10 nm particles would absorb best at ~400 nm wavelength... i.e., absorb blue, leaving yellow reflected.
That said, there is certainly oxidation and reduction of silver, which is much more reactive than gold, in different parts of the flame, as you surmise. Perhaps various ionic compounds of  silver form , such as $\ce{Ag2SiO3}$, silver silicate. It is even suspected that gold might react in glass!
It's hard to know which effect predominate, colored compounds or surface plasmons. Perhaps could add $\ce{Ag2SiO3}$ to a glass frit and heat it by radiation or conduction in vacuo to avoid oxidizing or reducing the compound, and check its color? That would be interesting to report here!

Very nice answer! And the Lycurgus cup goes back to the 4-th century. +1!


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Calculate the degree of polymerisation. It's molecular weight and its repeat structure are given
I understand how to calculate the degree of polymerisation normally, but in this question, the extra H and OH outside the bracketed monomer have thrown me off. Do I include their atomic masses in the overall Atomic mass for the degree of polymerisation or are they not part of the monomer chain and so not part of the calculation?

It's not like can have no terminal groups (unless oligomer is cyclic)

The degree of polymerisation is $n$. The molar mass of the molecule of course includes the end grous. You need to use the correct terms to not get confused!


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Calculation of equilibrium constant in a gaseous system A ⇌ B + C
I am supposed to determine whether the following two statements are true or false:

For the reaction $\ce{A (g) <=> B (g) + C(g)}$, $K_p = \pu{1 atm}$. If we start with equal moles of all gases at $\pu{9 atm}$ of initial pressure, then at equilibrium the partial pressure of $\ce{A}$ increases.

The reaction quotient $Q_p > K_p$, hence the equilibrium shifts in the backward direction.

I discussed it with my friend, who gave me a weird explanation abut how to find the reaction quotient $Q$ at the start. He claimed that it would be equal to the total pressure at the starting point, $\pu{9 atm}$. But I believe that my friend is wrong, according to the following calculation:
Let us assume that $x~\mathrm{mol}$ of each gas is present. Hence the mole fraction of each gas is $x/(3x)$ = $1/3$. The partial pressure of $\ce{A}$ is thus $p_\ce{A} = (1/3) \times 9 = 3$ . Similarly we get $p_\ce{B} = 3$ and $p_\ce{C} = 3$, which gives
$$Q_p = (p_\ce{C} \times p_\ce{B})/p_\ce{A} = (3 \times 3)/3 = 3$$
Regardless of whether $Q$ is 3 or 9, it is still larger than $K$, so the eventual answer is not changed. However, I wanted to know what the correct value of $Q$ was.

Your analysis is correct, except that you don't account for how the the units cancel out in your calculation.
The initial reaction quotient, $Q_P$, is indeed 3, as you have calculated.  And since $K_P$ is 1, the reaction will shift to the left, increasing $P_A$.  Hence Statement-1 is nearly correct.  I say "nearly" because $K_P$ is dimensionless, so it should be 1, not 1 atm (each partial pressure should be divided by whatever you define as the standard pressure, 1 atm or 1 bar -- that's what you've left out of your calculation; you should try this yourself as an exercise).
Statement-2 is not quite correct, because it says the "equilibrium shifts in the backward direction."  Assuming you are not changing the temperature, the equilibrium does not shift (it's a constant: $K_P$ = 1).  Rather, the reaction  (and thus $Q_P$) shifts in the backwards direction until equilibrium is attained.

How you set up an equilibrium constant is your choice; there is no reason for it to be dimensionless. The worst part about this statement is the use of 'moles' as a synonym for 'amounts of substance'.

@Martin-マーチン Yes, equilibrium constants, depending on how they're expressed and the convention used, can sometimes have dimension. However, I've never seen a modern physical chemistry textbook that expressed $ K_P$
as anything other than dimensionless. And the statements here concerned $K_P$.  Also, I don't understand your objection to the use of "moles".  "Amounts of substance" is ambiguous—it could mean the number of molecules or the masses of the molecules.  So they weren't using "moles" as a synonym for the ambiguous "amounts of substance", they were using "moles" to mean "moles".

I have seen in plenty of textbooks that $K_p$ (not a capital p) is defined in partial pressures, therefore it had a unit. All I wanted to convey was that your boiler plate statement is not correct, and why I down voted your answer attempt. The term 'amount of substance' is not ambiguous; it is clearly defined in the gold book. They were using a unit to refer to a quantity, which is wrong. You don't have to understand my objection, but I think it is necessary to point out, so that users who come here to learn get the chance to learn from correct information.

@Martin-マーチン  Is your objection that the statement said "equal moles" instead of "an equal number of moles"?  If so, I think you are confusing an English typographical error with a nomenclature error.  Both statements are clearly written with awkward English, so I implicitly read "equal moles" to be a bad English version of (not a bad nomenclature version of) "equal numbers of moles".

That usage is wrong. The quantity is called 'amount of substance'. The unit for this quantity is 'mole'. Don't use the one as a synonym for the other. I cannot explain it in any more simpler terms. It's been an issue, which had been brought up on this platform multiple times, I'm on mobile and I don't have the patience to look up the references. We're also beyond the point where any more commenting serves a purpose to improve your answer attempt. If there were more questions which need to be resolved, maybe chat is more appropriate.

@Martin-マーチン I read the answer at: https://chemistry.stackexchange.com/questions/75288/is-amount-of-substance-the-same-thing-as-number-of-moles, which said: " 'Number of moles of [substance X]' " rigorously means the numerical value (thus dimensionless)". Since the statements didn't specify a dimension (just that their quantities are equal), I don't see the problem. So please move this discussion to chat.

@Martin-マーチン Also, regarding , note the answer given by porphyrin at: https://chemistry.stackexchange.com/questions/107374/what-are-the-units-of-kc-and-kp/107414#107414 : "The equilibrium constant  is dimensionless as it is defined, ultimately, in terms of the ratio of activities, which are themselves dimensionless."


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Why is cis proton coupling larger than trans proton coupling in cyclopropanes?
In Clayden, Greeves, Warren Organic Chemistry Second Edition Page 815, it states:

In four- and three-membered rings, cis couplings are larger than trans ...

but I did not find an explanation for the statement. That being said, why is it true that cis coupling is larger than trans coupling in cyclopropanes and cyclobutanes? Previously, the textbook stated that coupling is a through-bond effect, not a through-space effect, which led me to reason that the trans coupling in cyclopropane should be larger than the cis coupling, as in the case of cyclohexane.

For one there is the normal (Karplus equation-like) coupling across bonds for $^3J_{\mathrm{H-C-H}}$.  Yet, building a model of cyclopropane or cyclobutane shows that hydrogens «on one side» of these plate-like molecules are this much squeezed into proximity that likely there is additional coupling across space for those sharing the same hemisphere of the cycle which in sum increases the interaction of cisoide H nuclei over their trans interaction.  Higher cycloalkanes are not this much pushed into one conformation, so the through-bond-only coupling is more significant for them.

@Buttonwood Thank you for the reply. So are you saying that coupling can also occur through space?

Coupling may occur both across bonds (as seen in multiplet splitting pattern of the typical monodimensional $^1H$-NMR spectrum), and coupling may occur across space; the later is the foundation for H,H-NOESY and H,H-ROESY two-dimensional NMR spectroscopy. The intensity of the corresponding coupling signals through space however decay rapidly over distance r by about 1/(r**6), which is why you need specialized techniques to register them.  Today however they are commonly used in structure elucidation of organic compounds, typically acessed within less than 5 to 10 minutes per sample.


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How can a neutral atom attract electrons when it's supposed to have zero charge?
A neutral atom is an atom with an equal number of protons and electrons
We know the force of attraction between the proton and electron is equal
and assuming that:

one proton's positive charge attracts one electron.
The "neutral" in a neutral atom means electrically neutral.

How can a neutral atom attract electrons when it's supposed to have zero charge ?

Comments are not for extended discussion; this conversation has been moved to chat.

Possible duplicate of Fundamental forces behind covalent bonding

The answer lies in electronegativity. When a proton attracts an electron, the electron doesn't magically suck out the charge of the proton. The proton's charge is still distributed in all directions. The reason why 1 proton on average can attract only 1 electron is because electrons push each other out.
Now let's first take H - it has 1 proton which attracts 1 electron. If another electron jumps in, only 1 electron stays in the end. Then He - it has 2 protons, so it attracts electrons even more. So even though electrons are fighting for the place, the nucleus charge is enough to hold them.
It gets interesting with Li. It should have 3 electrons, right? But 1st shell can take only 2 electrons, so the 3d electron must go to the 2nd shell which is further away. In such case the inner shell of electrons has a much greater effect on the outer electron, this is called electron screening, not to mention that the further you are from nucleus - the weaker the attraction. So even though Li has more protons than He, it's too weak to hold electrons on the 2nd shell, so some other atom will take the electron away and Li will be ionized and become Li$^+$ with only 2 electrons.
How strongly an atom wants new electrons is called electronegativity. It increases to the right of the periodic table because nucleus gets larger and larger and can hold on more and more electrons. In the last columns atoms want electrons so much that they can mug other atoms with weaker electronegativity.
But then the row of the table finishes and new row starts. At this point previous shell is completely filled and a new shell starts, and the electron screening kicks in again.
You can see these trends here. Electronegativity is the reason why Na & Cl can't form a molecule (covalent bond) - Cl (strong electronegativity) simply takes Na's (weak electronegativity) electron and both become ions: Cl$^-$ and Na$^+$. In the end they form an ionic bond instead.

This explanation is not quite complete. There is e.g. hydrogen anion, which is a bound system, so even $\ce{H-}$ can exist, despite being on the left of the periodic table. Your explanation here seems to imply that this situation is impossible.

My explanation is simply not complete because you can't put everything into 1 post. If you follow the link in the answer you'll see that it's consistent with what you're saying - Hydrogen's electronegativity is not that low. But the fact that H$^-$ exists in ionosphere or in stars' atmosphere doesn't make it a common case anyway.

It’s certainly possible to give incomplete answers that avoid being wrong or giving a wrong impression along the way.

Almost all answers are incomplete. You (maybe) get complete answers when you read several books and spend several years in the field. Completeness is not the only criteria. Answer should be useful to the person who asked the question. Here we're helping each other to learn more, get deeper insights - not simply document known facts for people who already know the answers.

An atom is only neutral when viewed as a single object from large enough distance. But as an electron comes closer to the atom, it "notices" the electron cloud first. This cloud also "notices" the electron and deforms—the atom polarizes—so as to keep the atomic electrons farther on average from the extra electron, since like charges repel. But this deformation leads to less shielding of the nucleus than in isolated atom, so the extra electron begins to feel a non-negligible attracting electric field.

A neutral atom is an atom with an equal number of protons and electrons [...] The "neutral" in a neutral atom means electrically neutral.

Neutral in this case means net neutral. It does not mean that the opposite charges "destroy" each other. After all, the electrons are still attracted to the nucleus even though the overall system is net neutral.

We know the force of attraction between the proton and electron is equal [...]

I am unsure what that statement is supposed to mean. Maybe it means a proton and an electron attract each other, i.e. both experience a force? That is certainly true.

one proton's positive charge attracts one electron

That is not quite true. The charge of the proton attracts any number of particles with negative charge simultaneously.

How can a neutral atom attract electrons when it's supposed to have zero charge?

It has a net zero charge. Irrespective of the net charge of an atom or ion, when an electron approaches, it is attracted by the nucleus and repulsed by the electrons that are already there. Depending on which force is larger, it will stay or go.
This so far is a non-quantum argument, but I think it is sufficient to address some of the misconceptions. You have to consider quantum effects to explain why the electrons don't fall into the nucleus, and why all the electrons are in distinct states.
What else can we learn from this?
Neutral particles such as helium atoms do attract each other (dispersion forces). Anions (such as fluoride) do form and are stable in the gas phase. Neutral molecules can be polar, and bind to anions.

The electronegativity of atoms/elements is indeed an indication of the relative attractiveness of that atom/element for an electron.
But it seems to me that the question asks what force or energy is involved. In general, atoms and molecules are neutral. It takes energy to produce a free electron and hang it in space. If it detects a positive charge, it will be attracted and become stabilized by emitting energy as photons or heat. A less stabilizing destination for an electron would be a "non-bonding" or slightly anti-bonding orbital - this is electronegativity. it's just a better (more stable) place for an electron than plain old free space. And the force of attraction can be measured by the energy required to remove the electron from thin new place (whether it is a really good place or just an OK place - but still better than free space).


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Is it good practise to mix double and triple zeta basis sets?
I need to compute extremely accurate single point energies using the coupled cluster method. The methods to be employed scale largely with the number of basis functions. A few basis functions may have a huge impact in the computational cost.
I planned assigning a 3ζ basis set to a central atom, and 2ζ for the surrounding atoms. I cannot compute everything at 3ζ.
Clearly, these small basis sets are intended for estimate the contribution of high excitation.
Are there any reason which prevent this mixture, or any argument supporting the usage of 2ζ for every atom?

It is perfectly fine, and actually quite common, to use big basis sets for the most important atoms (perhaps at the active site of the chemical reaction) and a smaller basis set for the surrounding hydrogen atoms (for example).
The only problem you might face would be, that this complicates the basis set extrapolation to get closer to the complete basis set limit. Such an extrapolation works best if you are, for example, extrapolating from 2Z & 3Z to infinity (not a mixed extrapolation such as 2Z/3Z & 3Z/4Z to infinity). In fact you won't be able to extrapolate because you likely won't be doing calculations with a 1Z basis set, and you won't be able to do a 3Z/4Z mix (because if you could do that, then you wouldn't need to mix 2Z into your 3Z calculation!).
I do want to point out that if you need to compute "extremely accurate single point energies" then even if you could use 3Z on the whole molecule, it would not likely even get you to 1 kcal/mol precision for energy differences. So as long as "extremely accurate" means in your context "+/- 10 kcal/mol for energy differences", you are doing everything just fine (and don't need a basis set extrapolation anyway). If you are aiming for better accuracy than that, you'll probably need a basis set extrapolation (meaning you want all atoms represented at 3Z level).

In the following reference a "mixed" basis set was used;
in this paper TZP (3Z) is used for the atoms Fe, S, and Mo while SVP (2Z) is used for C, H, O, and N:
Li, Z.; Li, J.; Dattani, N. S.; Umrigar, C. J.; Chan, G. K. The electronic complexity of the ground-state of the FeMo cofactor of nitrogenase as relevant to quantum simulations. J. Chem. Phys. 2019, 150 (2), 024302. DOI: 10.1063/1.5063376.
Since the molecule is FeMoco (an iron-sulfur complex) it is likely that the Fe and S are very important (so they have a big basis set). It is also believed that the molybdenum (Mo) is important, because why else would a biological molecule have Mo in it, so it has a big basis set too. The rest have a small basis set.

Thank you! I can compute the energy of these molecules up to something like CCSD(T)/aug-cc-pV5Z. The question is related to the tiny contribution of even higher excitations. Could you add a reference in which such mixing is done?

No, excuse me for the delay, I worked up to 2am tonight and from 7 to 18 today. I was completely out of time. Thank you very much for the reference.

I've corrected for def2-SVP, which is a double zeta basis set (i.e. split valence with polarisation functions). If 'extremely accurate' means +/- 10 kcal/mol for energy differences, then that is not accurate at all, and there wouldn't be any reason to run CC calculations. You'd get qualitatively better results with DFA.

@user1420303: How many spin orbitals (2x the number of spatial orbitals) do you have for the aug-cc-pV5Z calculation? If you can do CCSD(T)/aug-cc-pV5Z then with my program you might be able to do CCSDT(Q)/aug-cc-pVTZ (not with MRCC or other programs, but we are currently developing a powerful code that treats the matrices differently in RAM).

@Martin-マーチン , Thank you for comment. As the contributions of higher excitation are very small, the usage of small basis sets for estimating them does not induce large errors.

@user1271772 , I am currently demanding to the fullest to my new hard drive due to a CC calculation. I'll give you that info as soon it finish. Although of course, the feasibility depends of the used code (and approximations involved). I am really curios about your software. Do you have any reference or link to read about it? I am currently computing CCSDT(2)_Q using the Nwchem TCE engine.

@user1420303: our software is currently being developed and is in the testing phase. We don't have any website yet unfortunately, we're just a small team but we are focused on HPC. If you can do CCSDT(2)_Q on NWChem, we can probably do CCSDT(Q) for you. This was already true for a system where we did CCSDTQ for 55 electrons. If we can do CCSDTQ for 55 electrons, we can do CCSDT(Q) for much more !!


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Does the natural atomic orbital analysis result in natural electron configurations of molecules?
In my understanding, the natural atomic orbital (NAO) analysis is a method of calculation to get occupancy of electrons in center of each atoms. In other words, the result of NAO-occupancy indicates the electron configuration about a non-bonded molecule. More specifically, electrons are rearranged in each atom according to the surroundings of such atoms within a molecular configuration.
In this way of understanding, is it correct to say the data in table is result of the NAO analysis?
And, have I no misconception about NAO method?

NAO means "Natural Atomic Orbital"? Seems a rather obscure method. From what book do you have the table?

Yes, it is the "Natural atomic orbital" method. It might be more correct to say that it is useful to calculate local populations, rather than configurations. The method is not widely used per se (as far as I know) because it has a strictly local character, which doesn't describe correctly bonding and delocalization, but it is involved, for instance, in the first steps of the NBO algorithm

Please add a citation for the table you are using. Not quoting the source material properly is plagiarism.


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Choosing LC-MS modifier - why only acids?
During method selection our chromatographists have options like NH4, TFA, Formic Acid. What's the point of choosing only between acid? Why not use the strongest that guarantees ionization? The sample contains small organic molecules.
And if the analyte is an acid, don't we want to have a base as a modifier? Otherwise how can we ionize it using ESI?

What are your analytes? And LC-MS is no method to "purify" anything.

Just guessing: Any free electron pair can accept a proton, finding one in your analyte that could be abstracted is often not so easy.

Added info about the analyte, thanks. We do use LCMS for purification. Sure, not all the sample is passed through MS if you allude to the fact that it would destroy the sample.

I am more alluding to the fact that typical LCMS systems are not set up for preparative work. Especially your modifiers would get separated in the LC too. I assume you actually do MS on the fractions you collect from your LC system, right?

I never thought about this.. I know that during drydown the modifiers also participate in salt creation, so somehow they get into the fraction. But yeah - we run MS for QC after purification.

Stanislav, that is a Nobel Prize winning question. In short, nobody knows nitty-gritty details of how the electrospray ionization imparts a charge to the analyte. When Fenn (reinvented) the electrospray phenomenon and used for ionizing large biomolecules he was shocked and gave an analogy that it is like making the elephants fly. He got a Nobel Prize as well and a lot of patent issues in the university. There are three electrospray ionization models, you can read on Wikipedia, https://en.wikipedia.org/wiki/Electrospray_ionization#Ionization_mechanism
The key point in electrospray is that you need an electrolyte (forget about acids or bases). All you need is a conducting liquid between the electrode to generate the electrospray. In that process, the ion is created. There are studies, I am sure, on the effect of pH on the efficiency of ionization. I can tell what to avoid in electrospray: TFA (because it forms ion pairs), hexanes, all other pure non-conducting organic solvents. So you you can get the point that ionization in the solution has little to do with ionization in the electrospray.
Now coming to your chromatography part, which is purely empirical. The reason for avoiding bases in mobile phases is that it they will eat (=dissolve) the silica columns. Silica is not pH stable above 8. Silica based HPLC columns can tolerate low pH like 2.5, and below that the bonded starts to peel off, e.g. if you use a C-18 column and use pH 1, it will become a bare silica in a few days.

Huh.. Then what's the main purpose of the modifier if it doesn't impact ESP much?

Modifiers are used for chromatographic purposes and you need some electrolytes in electrospray. Buffers do have an impact on the efficiency but solution ionization has little to do with electrospray ionization. I mean there is no simple correlation that can be generalized.

It's worth noting that newer column derivatization chemistries are good enough that basic modifiers can be used, even with silica columns.


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I was told that the more exothermic the solution, the more soluble the salt. Is this statement wrong?
I was also told that $$\Delta H(\text{solution}) = \Delta H(\text{hydration}) - \Delta H(\text{lattice E})$$
However, for $\Delta H(\text{solution})$ to be exothermic, it has to have a negative value <0. Which means lattice energy is going to have to be greater than $\Delta H(\text{solution})$ to produce a negative value, based on the equation above.
Yet I am also told that hydration energy released needs to be enough to compensate for the lattice energy required to break down the solid. So it can't be lesser than $\Delta H(\text{lattice E})$.
Isn't this contradictory? I am confused.
I have read a similar question at   Enthalpy of solution
but the answer given is that dissolving is dependent on Gibb's free energy instead.
Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."?

See also: https://en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle

Does that mean that I should disregard the statement "the more exothermic the solution, the more soluble the salt."?

No, based on the Gibbs free energy equation $$\Delta G = \Delta H -T \Delta S$$ and the condition that $\Delta G<0$ for spontaneous dissolution, if you assume $\Delta S$ is constant then the statement is true.
The point of that other post is that as a general rule you should not simply ignore the $\Delta S$ term since sometimes it is the decisive factor determining solubility (such as in the case of endothermic dissolution reaction).

I saw your previous comments on the magnitudes, rather than the values, of ΔH(hydration) and ΔH(lattice E) being compared as well as the fact that both are usually negative values since energy is evolved. I can't believe I missed that. Thanks a lot for the clarification! And thanks for the explanation on Gibs free energy too.

@TetsuDau Yea, I rewrote the comment because I was using the wrong sign convention for the lattice enthalpy - it is the enthalpy of formation of the lattice from ions in the gas phase.


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What effect does the dye color have on the kinetics of a commercial lightstick?
According to thoughtco:
If fluorescent dyes weren't put in glow sticks, you wouldn't see any light at all. This is because the energy produced is usually invisible ultraviolet light. Fluorescent dyes may be added to light sticks to release colored light.
The dyes all have there own unique chemical structure, but how do these chemicals affect the amount of energy released in the chemical reaction? I am planning to do an investigation on the activation of energy of a lightstick, and I am wondering if the colors have anything to do with the reaction rate.

The chemiluminescence of glow sticks has been investigated in great detail, so you can easily find lots of information on it. Definitely search on wikipedia and do a search.

Check this one out: https://chemistry.stackexchange.com/a/115635/79678

One last thing: the activating reaction does not produce UV light. It produces an excited species that then transfers energy to the recipient dye.

@EdV Some sources say that there is a two-step process, and that first, the hydrogen peroxide and the phenyl oxalate ester combine to produce UV light, which is then absorbed by the dye and is excited, releasing visible light when dropping down to the ground state. Is this wrong? Is there an existence of UV light as an intermediate energy transferer between the chemical reaction and the excitation of the dye in a glowstick reaction?

I think it is wrong and I would be very surprised if UV was produced. Certainly the dyes can be excited by UV and I have used 405 nm laser light to excite the dyes in 3 different unbroken color dye capsules from glow sticks. Of course, the UV emission idea can be tested experimentally, and maybe I am simply wrong: that would be neat! What I think happens is just collisional transfer from whatever the excited species is to the dye molecule. As I say, I would be happy to be wrong! If UV is actually generated, it must be in the reported literature and can be detected easily.

@EdV According to Wikipedia, "Chemiluminescence differs from fluorescence or phosphorescence in that the electronic excited state is the product of a chemical reaction rather than of the absorption of a photon (electromagnetic radiation)" Doesn't this mean that the article I was reading was wrong? At least for glow sticks of course

Yes. In fact, you can easily demonstrate chemiluminescence using 30% hydrogen peroxide and ordinary 5.25% by weight NaClO bleach. I used to do this as a lecture demo in a dark lecture hall. Either add the peroxide to the bleach or use a syringe to shoot bleach into the peroxide. Then faint red light is emitted from the singlet delta oxygen produced. No UV is produced in this reaction. This does not say that the glow sticks cannot involve UV, but I seriously doubt it. In any event, detecting UV at the surface of the light-emitting liquid would be an easy experiment.

What you cited about difference between chemiluminescence and fluorescence does not preclude UV emission by cl followed by fl of the dye. However I have the same opinion of @Ed V. No UV is involved in commercial lightsticks. Finally I don't see a real correspondence between the title and the body of the Q.


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Why is the dipole moment of XeF$_6$ zero?
In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 73 under the section 3.6 (VSEPR Theory chapter Chemical Bonding):

...according to this theory, the position of the lone pair in $\ce{XeF6}$ molecule should be in axial position and the shape should be pentagonal pyramidal. But this is not observed and the actual shape of molecule is octahedral. This implies that the lone pair is placed at equatorial position, and this explains the structure of $\ce{XeF6}$ shown in figures 3.7 a,b,c.
According to Figure 3.7, $\mu_\text{expected}$ for $\ce{XeF6}$ is not equal to zero. But $\mu_\text{real}(\ce{XeF6})\sim0$ . To explain the above facts, it is considered that the actual molecule is in dynamic equilibrium of all the three structures shown in Figure 3.7. Here it is considered that the lone pair is present in the stereochemically active $s$ orbital and similar structure is observed as $\ce{IF6-}$. But anions like $[\ce{SbX6}]^{3-}$, $[\ce{TeX6}]^{2-}$ ($\ce{X}=\ce{Cl, Br, I}$), $[\ce{BrF6}]^-$,$[\ce{ICl6}]^-$ have been assigned perfectly octahedral structure on the basis of X-ray crystallography and the lone pair is present in stereochemically inactive $s$ orbital.

It is explained the three structures given in the above diagram are in dynamic equilibrium, so dipole moment is zero. But I feel it must be towards the right since the lone pair of electrons stays mainly towards right since the vertical component cancel (structure c and d)
Could you please explain how according to the author, the dipole moment of XeF$_6$ be zero.
Kindly note: I have read this question related to the same passage but a different concept. My doubt is not about stereochemically active or inactive s orbitals but about the non-existence of dipole moment in the given compound.

Carbon dioxide is not always symmetrical if you consider its vibrational modes, but the time average dipole moment is still zero.

@Zhe, I understand your case, but here only the vertical components of the dipole moment cancel and I think the net dipole moment must be towards the right since the lone pair stays mainly to the right side of the molecule as given in the figure. This is contrary to what is being said in the book.

By symmetry, that lone pair does not stay on the right side. I would imagine that they did not draw out all 12 possible transitions.

@Zhe, If my calculations are right, there must be [ 12 (lone pair of electrons on the edge of the octahedron) + 8 (lone pair on the face of the octahedron) = 20 transitions instead of 12 ]. Am I right?

You might want to look at https://en.wikipedia.org/wiki/Pseudorotation

I only counted the transitions as drawn, which must happen over an edge. There are 12, though accounting for symmetry, just 1. Sorry, I don't have time to write an answer right now.

https://chemistry.stackexchange.com/questions/50187/what-is-hybridisation-of-xef6-in-solid-state


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Dipole moment in NF3 versus NH3
The usual explanation for the molecular dipole moment of $\ce{NF3}$ being smaller than that of $\ce{NH3}$, despite the $\ce{N-F}$ dipole being stronger than the $\ce{N-H}$ dipole, is that influence of the lone electron pair on nitrogen is to oppose the net $\ce{N-F3}$ dipole, while enhancing that of $\ce{N-H3}$. See, for example, this good example.
What bothers me about this explanation is that when the $\ce{N-F}$ dipole is calculated (experimentally, I think), the effect of the electron pair on $\ce{N}$ is already factored in. Likewise for the calculation for $\ce{N-H}$.
Putting the question in terms of the language of the linked article:

[T]he lone pair's negative concentration will augment the polarity contribution from the polar bonds while for the $\ce{NF3}$, the lone pair's dipole would subtract from the dipole contributed by the NF bonds.

If the lone pair on $\ce{N}$ subtracts from the $\ce{N-F}$ bond in the molecule it must have had the same effect in the original diatomic dipole calculation. Likewise for $\ce{N-H}$.
If this is true, assuming identical $\ce{N-F}$ and $\ce{N-H}$ bond angles in $\ce{NF3}$ and $\ce{NH3}$, it seems that the vector calculation should still give a stronger dipole moment for $\ce{NF3}$. [Please regard this as a question!]
One thing that would lessen the molecular $\ce{NF3}$ dipole moment is a smaller $\ce{NF}$ bond angle (compared to $\ce{NH}$), which increases the lateral component of the dipole at the expense of the axial, making the (axial) resultant smaller. I think fluorine is larger than hydrogen and maybe this would tend to push the fluorine atoms out laterally, decreasing the resultant axial dipole. However I do not see this as an explanation in texts so I am guessing it is wrong.
My confusion is that the vector algebra doesn't seem to support the argument but there is probably something I am not taking into consideration.

Do you have a table of bond dipoles as a reference?

You should probably be cautious in general applying the concept of bond dipoles to predict molecular dipole moments, and it does seem ad hoc if not downright flimflam to invoke a dipole to the nitrogen lone pair to predict dipole moments in the examples you present (particularly in the case of nitrogen trifluoride).
Now if you read the wikipedia page on nitrogen trifluoride, it explains that

$\ce{NF3}$ is slightly soluble in water without undergoing chemical reaction. It is nonbasic with a low dipole moment of $\pu{0.2340 D}$. By contrast, ammonia is basic and highly polar ($\pu{1.47 D}$).[9] This difference arises from the fluorine atoms acting as electron withdrawing groups, attracting essentially all of the lone pair electrons on the nitrogen atom. $\ce{NF3}$ is a potent yet sluggish oxidizer.

The low basicity of $\ce{NF3}$ supports the argument that electron density associated with unpaired electrons on N is withdrawn by the fluorine atoms. So it would appear that in the case of $\ce{NF3}$ there might not be a significant dipole moment between the $\ce{N}$ nucleus and an electron pair (or the charge opposite the three fluorines). Instead the electron density (relative to ammonia as a reference) is polarized toward the fluorines. This makes the application of a method of linear summation of bond dipoles somewhat sketchy.
Continuing along similar sketchy lines, the website to which you provided a link argues that the difference in molecular dipoles cannot be explained simply in terms of the order of electronegativity ($\ce{F > N > H}$). But note that electronegativities are computed from dissociation energies, electron affinities, etc, and not from partial charges or dipole moments, and the relation between charges or dipole moments and energies is bound to be highly non-linear, making quantitative prediction of dipole moments based on electronegativities difficult.
What seems necessary is to perform some ab initio electronic structure calculations.

OK but if F withdraws all the electron density there would still be a large net dipole moment. Yet it is smaller than that of $NH_3.$ Maybe it is true that calculations are missing and if so that's a good answer. +1 Flimflam?

@daniel flimflam is a good word :-) even if my usage is not necessarily accurate as the original user perhaps did not intend to deceive. I'll modify my answer, to make a little more clear what I mean, when I get a chance.