text:
Predicting pressure inside a container based on temperature
I'm a mathematician and computer scientist, and for this particular problem I would benefit from some chemical expertise.
Suppose I fill a container up with some liquid propane. I believe this is done under very cold conditions so that the propane remains liquid while the container is being filled. It is then sealed off. Suppose now the inside of the container changes temperature. I would like to calculate the pressure that would be exerted on the walls of the container at any given internal temperature $T$.
Intuitively, I would think that the pressure would be a function of temperature and some initial conditions (how full is the tank initially)? If I filled up the tank to 95% capacity with liquid propane, and then heated the tank, I would expect the resulting pressure on the walls of the container to be much higher than if I filled up the tank to only 1% capacity and did the same thing.
However, the only resources I've been able to find so far relate to the vapour pressure, which is a function of only temperature and this seems incomplete.
Can anybody point me in the right direction on how to approach this problem? Some factors that seem to complicate things are:

The containers always seem to contain some level of liquid (due to internal pressure?), the rest is gas. Does the liquid have an effect on the pressure exerted on the walls? Can I just ignore it?
If, rather than fill up the tank with liquid at very cold temperatures, the tank is just directly filled with gas, how does this change things? My guess is that given the initial pressure and temperature you can predict the pressure at a different temperature, but I think the fact that some of the gas becomes liquid at some point is tripping me up and I'm not sure how to approach the problem.

When your tank is 95% full, is the space above the liquid containing air?

As the tank is not full then there is always some vapour in equilibrium with the liquid and so you will need to find the saturated vapour pressure vs temperature. If all the liquid evaporated (say if 1% full) then it will behave as a gas and then pressure is proportional to temperature say via the ideal or van der waals gas laws.

@porphyrin , for the first case, the liquid would also contribute $\rho gh$ pressure on walls.

@William E Ebenezer, yes I has assumed that that was small enough to ignore.

Equations of State - of which the ideal gas law is the simplest, relate pressure, temperature and density to one another

@porphyrin I need to be able to figure this out for specified initial conditions. How does 95% full compare with 85% full, for example

probably the same, you only need a small amount of liquid present at all temps to use vapour pressure. I'm assuming that you are always below the boiling point.

See also the question in the Physics SE https://physics.stackexchange.com/questions/478645/predicting-pressure-of-propane-gas-in-container-based-on-temperature

The problem with answering this question is that propane is not remotely close to being a permanent gas at typical room temperature. This means, in practice, that it is easily liquefied at room temperature with a little pressure and you can't, therefore, use the simple gas law equation to work out the pressure exerted by a given amount of the gas in a vessel of fixed volume.
This means that the tank will fairly certainly contain some liquid (as a butane-filled cigarette lighter will at very modest pressures despite butane being a gas at normal temperatures and pressures).
You can, therefore, estimate the pressure inside the vessel as being the vapour pressure of propane at the given temperature (which ignores any other gas included in the vessel though this may be a good approximation as the filling process is likely to sweep out any other gas). The Wikipedia entry on propane has a convenient chart of the vapour pressure and you can look up the value for a given temperature. The amount of the vessel filled won't make much of a difference to this as long as there is some liquid and some gas in the vessel.
Because simple gas laws don't do a good job of predicting the vapour/gas equilibrium, this isn't a gas-law question and the best answer will always be empirical (see the Wikipedia page).

Well the purpose of solving this problem is to determine initial conditions to keep the pressure of a container safe over forthcoming temperatures, so unfortunately the vapor pressure doesn't help (given that it is only a function of temperature, it provides no information about initial conditions which we can optimize). So I think looking at empirical data may be the best approach. thanks!

@CoffeeDonut Also, find out what the typical fill ratio for commercial propane tanks is and stick close to that. Filling the tank completely with any liquid is likely to be very, very bad as they all expand a bit with temperature. As long as there is sufficient remaining space, then empirical vapour pressure is as bad as it can get.

So in actuality, the real question may be to determine what constitutes "sufficient remaining space"?

@CoffeeDonut exactly, which is why the people who do it have rules about you full the tanks should be (which is usually no more than 80% of the volume). See fill rules.

As long as head space is available the pressure in the tank is the vapor pressure of the liquid; the tank, bottle, flask must be strong enough to hold the weight of the liquid. [Never put mercury in a flat-bottomed Erlenmeyer flask]. For gas delivery to work the VP must be greater than atmospheric pressure at all possible temperatures of use including temperature drop from heat of vaporization. The head space must be large enough to accommodate any expansion of the liquid. This is done with a safety factor because expanding liquid will condense vapor until the head space vanishes. With no head space the heated liquid will expand and distort and eventually rupture the container. In large amounts this can be catastrophic resulting in BLEVE explosions.
The tank must be strong enough to hold the weight and withstand the vapor pressure at possible temperatures. [spherical shape is better]. Head space must be sufficient that liquid can never fill the container. An inert gas head space will not work. The added pressure increases the internal pressure of the liquid increasing the vapor pressure


text:
Chiral carbon in carvone molecules

This is the structure of carvone, with the chiral carbon noted by *.
I don't understand why this is a chiral carbon, to my understanding a chiral carbon is a carbon with 4 different chemical groups attached to it.
In the molecule, the chiral carbon is surrounded by $\ce{H}$, $\ce{C(CH3)=CH2}$ and two $\ce{CH2}$ groups.
Why are both $\ce{CH2}$ groups different and thus making that carbon chiral?

Both CH (which are actually both CH2, but that doesn't matter) are different in what else is attached to them. Imagine a simpler molecule, with the following four substituents: H, Cl, CH3, and C2H5. You'll see right away that they are different. But wait, how are CH3 and C2H5 different if they both start with CH2? Well, just like that.

I edited the CH to CH2 (whoops), I think about benzene every time I see a cyclic molecule.

So one of the CH2 groups have a CO group attached, and then the next carbon is just a C(after the CO), does this matter at all? or what matters is what is immediately next to that CH2 and everything else down the chain is not relevant

Yes it does matter. Everything  down the chain is relevant. You can imagine two chains longer than these, which are very similar and only differ at the very end, and yes, they will be different.

Related: https://chemistry.stackexchange.com/q/39536/7951

If it is a cyclic carbon compound, then the distance, in number of carbon atoms from an electronegative element, will also be relevant right?
If that double bond in the cyclic part wasn't there that would mean that carbon 3 and 4 (counting anticlockwise from the ketone) would be the same, but because carbon 3 is closer to the electronegative element it counts as a different group in terms of chirality, is that right?

Related: Why was the definition of chiral carbon changed? What is the perfect definition for chirality?

If you consider the carbon with as star (C*), it bears a C(CH3)(=CH2) group and an H group (not shown). But it is also part of a ring and in that case, we actually follow the exact same rules!
From C*, if we go to the left or the right, we get a CH2 group. So let's get further: on the left we get a CH group with a double bond to another carbon, while on the right, we get a C atom with a double bond to an oxygen and a single bond to a carbon atom.
Because you get different groups if you follow the ring from C* to the left or the right, it means that C* is chiral!

This is just a general note on handling this type of question.  Many students struggle with line diagrams when they are first introduced.  This is because a lot of important information is implied by the drawing, rather than showing it explicitly.
Take the time to redraw the molecule, clearly showing all of the missing atoms (especially the hydrogen atoms).  This should make things clear for you.
It is well worth your time to redraw the molecule until you are very familiar with line drawings.

I'm afraid that doesn't actually answer the question, I don't think OP has any problem identifying what surrounds the asymmetric carbon (next neighbour), but rather why it is asymmetric in the first place.

I had actually meant to post this as a comment rather than an answer.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Calculating pH of solution

What is the $\mathrm{pH}$ when you dissolve equal amounts of $\ce{NaHCO3}$ and $\ce{Na2CO3}$ in water?

When both of the compounds are dissolved, we will have equal amounts of $\ce{HCO3-}$ and $\ce{CO3^2-}.$ But how do we go from this to actually calculating the $\mathrm{pH}$?
In the solution manual, it is given that $\mathrm{pH} = \mathrm pK_\mathrm{a2} = 10.33$ (It seems like the $\mathrm pK_\mathrm{a2}$ value of $\ce{CO2}$ has been used here — why?), but I don't understand how they can just conclude with that.
My attempt
So, we will start with equal amounts of $\ce{HCO3-}$ and $\ce{CO3^2-}$. We have the following equilibrium:
$$\ce{HCO3- <=> H+ + CO3^2-}$$
We can set up an equilibrium expression (and cancel) until we are left with $K_\mathrm{a2} = [\ce{H+}]$ and $-\log K_\mathrm{a2} = \mathrm pK_\mathrm{a2} = \mathrm{pH} = -\log[\ce{H+}]$.
Now, what I don't understand is why are we using the $\mathrm pK_\mathrm{a2}$ of $\ce{CO2}$? Why that value?

We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
Don't worry, they're not banned. But, we require a minimal effort. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.
See homework

Ask yourself what Ka2, resp pKa2 means.

I have updated my first post with a better attempt.

You may want to look up the Henderson-Hasselbach equation.

@Kdbmvp Ask yourself, how is Ka2 for CO2 defined.

Actually we haven't learned the Henderson-Hasselbach equation, isn't it possible to solve it without it? And is my first attempt wrong?

So Ka,2 for CO2 is defined as Ka,2 = [H+][HCO3(-2)]/[HCO3(-)], so I guess I have it now. What I don't quite understand is why it is written pkA,1 and pkA,2 values for CO2 in tables. Couldn't it have stood H2CO3(carbonic acid) instead of CO2?

In fact, there are 2 versions of pKa1, one for H2CO3, one for CO2. pKa2 is for HCO3-.

So what's the difference between them?

Is not Google or Wikipedia search faster ? Search for the carbonic acid.

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Why do polar aprotic solvents favour SN2 over E2?
I was reading about Substitution and Elimination Reactions and I came across the following on MasterOrganicChemistry:

Polar protic solvents tend to favor elimination (E2) over substitution (SN2). Polar aprotic solvents tend to favor substitution (SN2) relative to elimination (E2).

I understand the first part. But I have trouble understanding the part with the polar aprotic solvents.
Consider the following example (from the same website):

Why is E2 not favoured? I can understand why polar aprotic solvents favour SN2:

A polar aprotic solvent like acetone or dimethylformamide preferentially solvates cations, leaving an almost "bare" nucleophile. This increases its nucleophilicity

But I could find no explanation as to why E2 does not happen because of the solvent. What keeps the nucleophile (cyanide anion here) from abstracting a beta hydrogen?

E2 does happen but product formed is little compared to SN2 in polar aprotic solvent

I have searched incessantly, but no avail.

Competition between substitution and elimination is one of the many points of confusion for introductory organic students. In general, they are taught that there are four factors that will determine the outcome of a reaction where there may be competition between substitution or elimination.
They are:

Type of alkyl halide (methyl, primary, secondary and tertiary);
The strength/basicity of the nucleophile;
The solvent;
The temperature.

I think that many students over-emphasize the impact of any one of these factors.  It is more useful to look at the overall impact of the factors and evaluate the reaction as a whole.
Let's examine the case in question:

Primary alkyl halide generally favours substitution.
The nucleophile is $\ce{^-CN}$ which is a good nucleophile but a weak base.
Polar aprotic solvent.
Low temperature.

The big factor here is not the solvent.  All of the factors point toward substitution, especially the alkyl halide and the nucleophile.  Remember that elimination reactions are essentially acid-base in nature.  If the nucleophile is not a strong base, elimination reactions are less likely.

Actually the polar aprotic solvent favors the nucleophilicity of the base/nucleophile and as a result favors the SN2 to occur over E2.
In the case of protic solvent the E2 is favored as the close approach of nucleophile/base is hindered due to solvolysis.


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Why does my rainwater turn yellow if I increase the pH to 7?
My rainwater is colourless when collected, but if I add a small amount of sodium hydroxide to increase the pH to about 7, after a few days it goes yellow.
The colour change happens more quickly if I increase the temperature.
This change to yellow is accompanied by the disappearance of any dinoflagellates or cryptophytes which were living in the water (determined by 300X microscope.)
The rainwater appears to contain a significant amount of dissolved organic carbon.
The rainwater is collected from the PVC roof of my shed in a covered plastic container which I have recently cleaned.
The pH of the rainwater is between 5.4 and 6 when collected (the best I can do with bromothymol blue indicator solution and narrow-range indicator paper.)
I am located in the south-west of England about one mile from the sea.

Rain water shouldn't contain dinoflagellates and cryptophytes, where did you collect it from? what is the pH when you collect it? Where are you located?

In addition to the comment of @com.prehensible, in my experience adding NaOH to water in order to increase the pH to 7 is very close to impossible unless there is some buffering species present, which again suggests something else is present in the water. At a guess I would say iron, or a chelate of iron (often coloured), is contaminating  the water sample?

com.prehensible, thank you for replying, please see amendments to the text of my question.
I am under the impression that small marine organisms less than 10 microns in size can find their way into the clouds somehow and based on previous experiments mine seem to prefer a pH > 7.

user1136, thank you for replying, surely all rainwater contains acid so that will serve to buffer pH against NaOH won't it, and if the yellow colour is due to the presence of an iron compound, why isn't the water yellow to begin with?

One possibility is that it is only the ionized form of the contaminant that absorbs. One example of such a compound from the world of biochemistry is para-nitrophenol (very popular with the enzymologists). Alkaline solution of pNP are yellow due to aborption by the para-nitrophenolate anion, whereas unionised pNP is colourless (but I am not in any way suggesting that the water is contaminated with pNP!)

tomd, thank you for replying.  I failed to mention that the water doesn't go back to being colourless if the pH is lowered.  I feel increasingly that the yellow colour is due to a water mould growing on sugar in the rainwater.  I am currently doing an experiment and will post results when I'm sure.
I asked this question on biology.stackexchange!


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Why the second and third ionisation enthalpy of Bismuth is more than that of Antimony?
The first ionisation enthalpy values of Group 15 elements follows a decreasing trend due to increasing atomic size and metallic character but why do the second and third ionisation enthalpy values follow such an anomalous trend ?


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Can I determine purity of isopropanol with a hydrometer?
Assuming I start with a known 99.8% purity of isopropanol, after recovery with a distillation apparatus, could I use a hydrometer to test for water content? If so, is there a special one I should use?

Be aware of isopropanol azeotrop with 87.7 w % of isopropanol. https://en.wikipedia.org/wiki/Azeotrope_tables?wprov=sfla1

@Poutnik: Azeotropes of water, you mean?

Rather azeotrop isopropanol - water. But yes, in context of the article, of water.

Theoretically a hydrometer would work. Pragmatically, no. At just 0.2% water there is little density change. The density would also be a function of the temperature. So the lack of sensitivity of the hydrometer and the variation of the temperature would be significant errors in this case.

There is an image of a table provided by Chu and Thompson [1] listing isopropanol water solution densities:

The essential data part is (weight % vs density):

77.87 0.8361
  89.84 0.8069
  99.91 0.7808

There is density gradient -0.0026 / 1 w%, resp. -0.00052 / 0.2 w%, determining purity by density via hydrometers is rather illusory.
Some success could be gained by pycnometers — using special glass flasks with a glass cap with a hollow. You would weight the flask, the flask with water and the flask with isopropanol and you would calculate the density.
In any case, as differences are minimal and as liquids have considerable thermal dilation, one must carefully watch the liquid temperature.
In case of the pycnometry measurements, the glass and both liquids must have the same temperature, the best the one for which density is tabelized.
References

Chu, K.-Y.; Thompson, A. R. Densities and Refractive Indices of Alcohol-Water Solutions of n-Propyl, Isopropyl, and Methyl Alcohols. J. Chem. Eng. Data 1962, 7 (3), 358–360. https://doi.org/10.1021/je60014a011.


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Why doesn't N2O give a positive brown ring test?

A common nitrate test, known as the brown ring test can be
performed by adding iron(II) sulfate to a solution of a nitrate, then
slowly adding concentrated sulfuric acid such that the acid forms a
layer below the aqueous solution. A brown ring will form at the
junction of the two layers, indicating the presence of the nitrate ion.

I know for sure that certain oxides of nitrogen, such as $\ce{NO}$ and $\ce{NO2}$ test positively in the brown ring test for nitrates, but I had never found a reliable source stating the same for the other oxides of nitrogen.
But one of the questions in a class test I took was:

Which one of the following does not give the Brown Ring test?

$\ce{N2O}$
$\ce{NO}$
$\ce{N2O3}$
$\ce{N2O5}$

The answer was given as $\ce{N2O}$ and I am unable to identify the logic behind the answer. Could it be because the nitrogen compound is reduced in the process of forming the brown complex with iron, while $\ce{N2O}$ already has too small an oxidation state to be reduced to $\ce{NO+}$?

You may put it this way, but that's a post factum explanation. Just remember that N2O is relatively inert and doesn't react like that.

The brown ring nitrate test is based on forming ferrous nitrosyl cation, with nitrous oxide formed from nitrates by iron oxidation in acidic solution:
$$\begin{align}
\ce{3 [Fe(H2O)6]^2+ + NO3- + 4 H+ &->3 [Fe(H2O)6]^3+ + NO + 2 H2O} \\
\ce{[Fe(H2O)6]^2+ + NO &->3 [Fe(H2O)(NO)]^2+} + H2O\\
\end{align}$$
Positive results are given also by nitrites and nitrogen oxides able to form nitric or nitrous acid, directly by hydration, or indirectly by disproportionation.
$$\begin{align}
\ce{ [Fe(H2O)6]^2+ + NO2- + 2 H+ &-> [Fe(H2O)6]^3+ + NO +  H2O} \\
\ce{N2O3 + H2O &-> 2 HNO2}\\
\ce{2 NO2 + H2O &-> HNO2 + HNO3}\\
\ce{N2O5 + H2O &-> 2 HNO3}\\
\end{align}$$
Therefore, $\ce{NO}$ and all higher nitrogen oxides oxidizing $\ce{Fe^2+}$, forming $\ce{NO}$, react positively in the brown ring test.
$\ce{N2O}$ cannot do that, because it cannot form $\ce{NO}$ at given conditions.


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Basicity af N containing aromatic species

Hello friends, I got struck in this interesting question on basicity. I am able to get different answers applying different methods.
In general, for these compounds we have to see the lone pair denoting tendency by the nitrogen.
I was able to conclude that IV is most basic. And II is greater than I. Now the problem is between I and III. Whether to consider the -I effect of $\ce{N}$ or $\ce{Ph}$ more effective? And, is -I the correct way to compare among both of these?
I even draw the resonating structures but was unable to conclude anything from there. Please help as to how to proceed.

(This might be cheating but will help any answerers). The pKa' values (pKa of conjugate acid) of pyrimidine (I) and quinoline (III) are 1.10 and 4.85 respectively. (Source: Wikipedia pages). This suggests III is more basic as it attacks $\ce{H+}$ at a lower $[\ce{H+}]$ concentration.

Dupe of https://chemistry.stackexchange.com/questions/50303/why-is-pyrimidine-less-basic-than-pyridine https://chemistry.stackexchange.com/questions/92375/why-is-quinoline-less-basic-than-pyridine

@Mithoron unfortunately, a vs b and a vs c together don’t make duplicates of b vs c. That said, I am of the opinion that there is no way to answer this with beyond stating that, solely based on empirical evidence and experience, an extra nitrogen in the ring is more electron-withdrawing than a fused benzene ring. The proof is in the pKa values.

@orthocresol So I voted for too broad, but q. was kicked from queue because of edit that only corrected spelling...


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Calculating the ph of an unknown solution by measuring the precipitate

You have an unknown metal hydroxide, but no meters or indicators. You have lead (II) nitrate, which is soluble in it, and lead (II) hydroxide that is insoluble in it. You take one liter of the unknown substance and add lead (II) nitrate, causing the formation of a precipitate. You add more lead (II) nitrate until the precipitate stops forming, and then add a little more. After you filter and dry the precipitate, you have 3.81 grams of it. What was the approximate ph of the original solution?

I know that the metal hydroxide would be a base, but I’m having a hard time figuring out anything else about this problem. My first thought was to figure out the molar mass of the precipitate to try and determine how much lead nitrate was added to the unknown substance, but I can’t figure that out if I don’t know what the precipitate is composed of. I thought about figuring out the molarity of the total solution to figure out how much lead nitrate was added, but again I couldn’t see a way to do it without more information.
I also thought, since we’re working with pH, we’re supposed to do something with dilution and changing powers, but I don’t see any way to do it.
This problem has me completely stuck. I just don’t see where to start. Any help would be greatly appreciated.

HINT - Write out the chemical reaction.

@MaxW I tried it, but I’m still not seeing anything. I think I’m misunderstanding the reaction between the unknown substance and the lead nitrate. The only thing we’ve been talking about in class recently is Acid-base reactions forming salt and water or conjugate acids and bases, but i think there’s something I’m missing.

Does lead hydroxide precipitate in an acidic solution or a basic solution? // You need to figure out what the pH is of the solution, not what the unknown cation is.

@MaxW That makes a lot of sense. Thank you very much for your help!

Could you add those equations to the question so we can see how far you got?

You have to assume your unknown hydroxide is completely dissociate in aqueous solution. If so, then you can calculate $\mathrm{pOH}$ of the solution as $\mathrm{pOH} = -\log \ce{[OH-]}$. Then, $\mathrm{pH} = \mathrm{p}K_\mathrm{w} - \mathrm{pOH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}$. Now, you see, since $\mathrm{p}K_\mathrm{w} = 14$, only unknown is $\ce{[OH-]}$.
Now look at the reaction of your unknown hydroxide with lead (II) nitrate:
$$\ce{2 OH- (aq) + Pb(NO3)2 (aq) -> Pb(OH)2 (s) + 2 NO3 (aq)}$$
Since you know the precipitated amount of $\ce{Pb(OH)2}$ isolated from $\pu{1.0 L}$ of unknown hydroxide solution, you can calculate original $\ce{[OH-]}$:
$$\text{Amount of } \ce{OH-}\text{ in original solution} \\
= \pu {3.81 g}\text{ of } \ce{Pb(OH)2} \times \frac{\pu {1 mol}\text{ of } \ce{Pb(OH)2}}{\pu {241.21 g}\text{ of } \ce{Pb(OH)2}} \times \frac{\pu {2 mol}\text{ of } \ce{OH-}}{\pu {1 mol}\text{ of } \ce{Pb(OH)2}} \\
= \pu {0.0316 mol}\text{ of } \ce{OH-} $$
Thus, $ \ce{[OH-]} = \pu{0.0316 M}$, and hence,
$$\mathrm{pH} = \mathrm{p}K_\mathrm{w} +\log \ce{[OH-]}= 14 + \log(0.0316) = 12.5 $$
Asking is approximate $\mathrm{pH}$ of orignal solution. Thus, this answer is acceptable.  However, keep in mind that trace of $\ce{OH-}$ will remain in solution due to the solubility of  $\ce{Pb(OH)2}$ in water $\left(K_\mathrm{sp} = 1.42 \times 10^{-20}\right)$.


text:
pKa and equivalence point
I was doing a titration and using $\ce{NaOH}$ as the titrant. The analyte was a weak acid with a $\mathrm{p}K_\mathrm{a}$ of 4.90. Today I want to use another acid with a $\mathrm{p}K_\mathrm{a}$ of 4.50. Do I need to change the indicator from the last experiment? How close do the $K_\mathrm{a}$ or $\mathrm{p}K_\mathrm{a}$ values have to be in order for me to use the same indicator and still obtain accurate results.

It's not a matter of their closeness; it is a matter of indicator's properties.

$99.9\%$ of the acid $\ce{HA}$ is titrated at  $\mathrm{pH}=7.9$ resp.$ 7.5$.
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log \frac{[\ce{A-}]}{[\ce{HA}]}=\mathrm{p}K_\mathrm{a}+3$$
Considering the final concentration 0.05M,
titrating 0.1M acid by 0.1M hydroxide,
100% of the acid is titrated at:
$$\begin{align}
\mathrm{pH}&=7+\frac12 \cdot ( \mathrm{p}K_\mathrm{a} + \log  c_\mathrm{acid, total}) \\
&=9.45(9.25) + \frac12 \cdot (-1.3) \\
&=8.8(8.6)\\
\end{align}$$
Titration with $100.1\%$ of used hydroxide we can consider
as $0.05 \cdot 0.001=5\cdot 10^{-5}\ \mathrm{M} \ \ce{NaOH}$
with $\mathrm{pH}=-\log ( 5\cdot 10^{-5})=9.7$
The indicator  phenolphthalein with range $8.2-10.0$ works well in both cases.


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Which conductivity law should I use to calculate the conductivity of a solution with high/low concentration?
I have a set of NaCl solutions with concentrations ranging from 0.05 M to 0.1 M.
I have been trying to calculate the conductivity of these solutions and compare the obtained values with the experimental ones obtained with a conductometer.
For this, I used the Kohlrausch's Law. But I observe some deviation on the values. I realized that the law may not apply for high concentrated solutions such as 0.1M. I read that since NaCl is a strong electrolyte, this law should still apply.
Is there a specific law or method to use to calculate the conductivity of concentrated solutions?

You can use the Debye-Hückel-Onsager limiting law. For a strong electrolyte without the formation of ionic pairs, with $z_+ = |z_-|$, just like $\ce{NaCl}$:
$$ \Lambda_m = \Lambda_m^\infty - (az_+^3 + bz_+^3\Lambda_m^\infty)(c/c^\circ)^{1/2}$$
For aqueous solutions at $25^\circ$C and $1$ atm: $a = 60.6 \, \Omega^{-1}\mathrm{cm}^2\,\mathrm{mol}^{-1}$, $b = 0.230$. For $\ce{NaCl}$, $\Lambda_m^\infty = 126.4 \, \Omega^{-1}\mathrm{cm}^2\,\mathrm{mol}^{-1}$.
References
Ira N. Levine, Physical Chemistry, 6th Ed.


text:
Flow of electrons DNMT
In a paper about the modes of action of DNMT, it said there an increased flow of electrons to carbon 5 following the formation of a covalent bond between DNMT and carbon 6 on cystine. Why is this? What is the effect of the flow of electrons?

Why do you think there is?

Said in a research article about the mode of action of DNMT

Then put something about this in your question to show that you have thought about it and are asking the community for more information, not to do your homework for you.


text:
How do you calculate the correct charge on F after F2 dissociates?
When you ionize $\ce{F2}$ to $\ce{F2^5+}$, it quickly dissociates. Experiments show that the products of dissociation are $\ce{F^2+}$ and  $\ce{F^3+}$.
I have tried to model the dissociation of $\ce{F2^5+}$ using Gaussian$09$ (DFT), specifying a 10 Å separation between products.
No matter which functional I use ($\omega$B97X, PBE, M06,etc.), I get a $+2.6$ charge on one $\ce{F}$ ion and $+2.4$ on the other. Why don't these functionals give $\ce{F^2+}$ and $\ce{F^3+}$ ions the corresponding formal charges as shown by experiment? And how do you get the correct charges?

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

@Feodoran It is based on the wavefunctions described in the program, not your wavefunctions. In the calculation packages, the wavefunctions are localised to some extent - they can't be distributed all over the universe. That's why the results showing non whole numbers are wrong.

One should be cautious about attempting to compare partial atomic charges with formal oxidation states (see https://www.nature.com/articles/s41563-018-0165-7). As a trivial example, I just ran an M06/def2-TZVP calculation of the HO2 molecule, which formally should assign charges of H+ and O2-. Based on Mulliken charges, you get a summed partial charge of -0.35 for the O atoms rather than -2 (of course, we then get +0.35 on H). Clearly, this is against the formal oxidation states one would assign. It's not specific to a Mulliken population analysis either.

@Argon But how far apart are the atoms in your calculation?

Fully optimized! I turned this into an "answer" to provide more detail.

If you have two fragments that are very far apart, certain quantum chemistry methods can cause spurious effects from including both fragments in the same calculation.  A well known example of this is dissociation curves for $\ce{H2}$ with Restricted Hartree Fock.
By using a density based method, we are less likely to localise the electrons on any one atoms in integer quantities.  A wavefunction method such as Hartree Fock, which places integer electrons in well defined molecular orbitals is more likely to create this localisation.  If you need higher accuracy, MP2 or coupled cluster may be of more use.
In fact I have achieved Mulliken charges of +2 and +3 on the two different atoms at 10 angstrom separation using Hartree Fock, with a cc-PVTZ basis set in Gaussian 09.
If you insist on using DFT you should use some chemical/physical intuition to partition this system.  From experiment, i.e. time of flight spectrometry, we know that the atoms have integer charge when they strike a detector, and dispersion interactions over that distance are going to very small compared to the coulomb attract of the other atom as a point charge.  So instead you can calculate the energies of the energies of the individual $\ce{F^n+}$ for $n=0,1,2,3,4,5$.  The combination of atoms with total charge $+5$ and the lowest energy, considering the classical coulomb attraction between them, is probably the product of the dissociation.
Gaussian input used:
%chk=FF_HF.chk
%nproc=1
%mem=4GB
# UHF/CC-PVTZ NoSymm

F2 5 plus Ion

5 2
 F      0.00000000    0.00000000   -5.00000000
 F      0.00000000    0.00000000    5.00000000

Thanks mate. UHF does show the correct charges. But in terms of energetics, UHF is horrendous, so I opt not to use it.

By the way, I think cc-pvtz is a bit much for a UHF calculation.

One should be cautious about attempting to directly compare partial atomic charges with formal oxidation states (see this Nature Materials article for a nice discussion on the topic). As a trivial example, I just ran an M06/def2-TZVP calculation of the HO2 molecule (see Gaussian 16 input file below), which formally should assign charges of H+ and O2-. Based on Mulliken charges, you get a summed partial charge of -0.35 for the O atoms rather than -2 (of course, we then get +0.35 on H). Clearly, this is against the formal oxidation states one would assign. It's not specific to a Mulliken population analysis either. Hirshfeld charges provide nearly the same result.
In my experience, partial charges really should not be thought of as quantitative analogues of formal oxidation states. Based on your example, it is clear that the redox states of the two cations are different. If you wanted to gauge whether they could truly be assigned a 2+ and 3+ oxidation state, your best bet is to identify a molecule where conventional wisdom states the fluorine atoms are in the 2+ or 3+ oxidation state, run a population analysis on that molecule, and compare the partial charges with what you get.
#P M06/def2TZVP opt freq Integral=UltraFine

test

0 2
O  9.9430695560218645  9.9578012177096120 11.3238448457162892
O 10.0393101063805297 10.0291419761442047  9.9836293222874293
H 10.6898654975975802 10.5116551761461796 11.6656176919962338

Thanks for your effort mate but I think you got my question wrong. What I'm talking about in my question is when the cations are far far apart from each other - in that case the cations should have the formal charges in the calculations.

Ah! Apologies about that. I now see where the prior answer was going with this.

LOL, maybe I need to rephrase my question...But anyway, if you know the reason why DFT shows me the wrong charges, you are welcome to say something about that.

By the way, I saw you wrote scf=YQC before you edited the answer. Why did you use that? I've never seen anyone use that.


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How do hard drives store information using magnetic layering? (Antiferromagnetic/ferromagnetic films)
I am doing some work with solid state materials in some chemistry courses and I've been doing some research into exactly how a hard drive stores information as 1's and 0's. It seems like a bunch of people know how it functions as an overview but the physics of what's going on is less well known.
So far I can confidently say that a hard drive is composed of many tiny thin films within the sectors on the platter. The film is divided into (at least) two layers. The top layer is the "pinning" layer which is an antiferromagnetic. The bottom layer is a ferromagnetic layer or in some cases a ferrimagnetic layer. The top layer contains magnetic domains which are anti-parallel and therefore contains no overall magnetic moment. The bottom layer being ferromagnetic has all of its magnetic domains aligned with each other, creating a net magnetic moment across the film.
I understand that in order to achieve a 1 or a 0, the magnetic moment of the underlying ferromagnetic layer undergoes a net magnetic moment flip, which is can be read. The pinning layer helps to keep it as a 1 or a 0 by not allowing this moment flip without considerable application of an external force. This force is termed the exchange energy of the film. This however is where I can't figure things out.
How does the "pinning" layer exert any force on the underlying ferromagnetic layer if it has no net magnetic moment? And more importantly how does this exchange energy manifest itself in a "snappy" transition between the 1 and 0 state of the ferromagnetic layer?
Please let me know if there is any more info that I can provide in order to help clarify, I am super interested in the answer to this question and would love to do some more research to figure it out if I can only get pointed in the right direction.


text:
Does acetophenone show haloform reaction?
Although acetophenone is a methyl ketone, as it is aromatic, it is immiscible in H2O so the last step in the haloform reaction should not occur?
I couldn't find any source regarding the haloform reaction of acetophenone.

Who said it will not give: https://www.google.co.in/search?q=does+acetophenone+give+iodoform+test&prmd=inv&source=lnms&tbm=isch&sa=X&ved=2ahUKEwiO8e3GiYviAhWHtI8KHdbIDZ8Q_AUoAXoECA0QAQ&biw=360&bih=560&dpr=2

No creditable research paper shows. My teacher told that it doesn't show.

Your teacher may be wrong.check http://www.webassign.net/sample/ncsumeorgchem2/lab_4/manual.html  which is a lab experiment desinged for the same  acetophenone.

A phase transfer catalyst can be used in last step

Fuson and Bull [Chem. Rev., 15, 275 (1934)] found acetophenone to give a positive iodoform test. They modified the usual procedure, adding dioxane as a cosolvent to better dissolve water-insoluble compounds, since the purely-aqueous method is "uncertain and likely to be misleading" for water-insoluble compounds.
Also cited by those authors as proof of positive haloform test for acetophenone are the following: Lieben, Ann. Spl. Bd., 7, 218 (1870); Kramer, Ber., *13, 1000 (1880); and Cuculescu, Bul. Fac. Stiinte Cernauti, 2, 137 (1928).


text:
What is the IUPAC preferred name of this tetracyclic natural product?
On page 819 of Vollhardt, Organic Chemistry 7th Ed., the following "subunit of the antibacterial platensimycin" appears. I have attempted to apply von Baeyer and skeletal replacement ('a') nomenclature but am stuck deciding between numberings. Here are some vector drawings:
Is von Baeyer the preferred nomenclature for this molecule? and, if so, what is its correct numbering and preferred IUPAC name?

Pubchem has the first bridgehead as the spiro carbon, giving a lower index to one of the secondary bridges: 4-oxo-8-oxatetracyclo[7.2.1.1$^{7,10}$.0$^{1,6}$]tridec-2-ene. I don't have access to the 2013 blue book, so I will refrain from writing an answer. (P-23.2.6.2.4  The superscript locants for the secondary bridges must be as low as possible when considered  as  a  set  in  ascending  numerical  order,  the  decision  being  made  at  the  first  point  of  difference.)

Instead of a von Baeyer name, Chemoffice as well as the usually good ACD/ChemSketch (2017.2.1) prefer a fusion name in this case, i.e. a structure based on 4a,7-methanobenzo[7]annulene. However, I still cannot prove why – when looking into the original IUPAC rules. The resulting name would be 4a,5,6,7,8,9-hexahydro-5,8-epoxy-7,9a-methanobenzo[7]annulen-3(4H)-one.

The pubchem website has given Platensimycin IUPAC name, $3$-$[3$-$[(1S,5S,6R,7S,9S,10S)$-$5,9$-dimethyl-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-en-$5$-yl]propanoylamino]-$2,4$-dihydroxybenzoic acid (see attached picture):

Accordingly, since https://pubchem.ncbi.nlm.nih.gov is usually updating data according to the changes, I believe the name of the cyclic system (without 5,9-dimethyl groups) OP is asking is same as one in the comment by @KarstenTheis (including the correct stereochemistry): $(1S,5S,6R,7S,9S,10S)$-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-ene (see the blue numbering in the picture).
Late correction: After exchanging few comments with Loong, I realized that the given system should be treated as a chemical compound. Accordingly, the name should have been: $(1S,5S,6R,7S,9S,10S)$-$4$-oxo-$8$-oxatetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$tridec-$2$-en-$4$-one (as suggested by Loong).

"since https://pubchem.ncbi.nlm.nih.gov is usually updating data according to the changes" – do we know that they are really doing this?

I have seen changing of data on physical properties section. But not sure for everything, though.

How do we know that the correct name for the moiety as a substituent is the same as its correct name as its own molecule?

Not exactly the same but related. See, the priority changes going from substituent to a molecule. For example, 4-oxo- for substituent becomes 4-one on the molecule.

In the full IUPAC name for Platensimycin, why is it "-4-yl]propylamino" instead of "-5-yl]propylamino"? Isn't the tetracyclic part connected to the propylamino part at carbon 5?

An elder version of Chemdoodle (8.1.0) suggests a name slightly different from the one in the comment by @KarstenTheis, 8-Oxotetracyclo$[7.2.1.1^{7,10}.0^{1,6}]$-tridec-2-3n-4-one.
It may be that I erred while drawing the molecule into the sketcher, hence for a xcheck the screen photo below. The yellowish' spheres indicate Chemdoodle's selection for the current naming -- there are only two atoms of oxygen.


text:
Entropy and total differential
In an exercise, I have derived that
$$\mathrm dS_\mathrm{m} = \frac{3R}{2T}\mathrm dT + \frac{R}{V}\mathrm dV$$
using the 1st and 2nd law of thermodynamics (1 mol ideal gas).
Now, I want to write $\mathrm dS$ as a function of $T$ and $P$ instead of $T$ and $V$. How can I do that?
I have tried to do this:
$$\frac{R}{V} = \frac{P}{T}$$
and
$$\mathrm dV = \frac{\mathrm dV}{\mathrm dP}\mathrm dP = -\frac{RT}{P^2}\mathrm dP$$
This gives me
$$\frac{R}{V}\mathrm dV = -\frac{R}{P}\mathrm dP,$$
which is wrong according to my T.A.
How can I write $\mathrm dS$ as a function of $T$ and $P$?

When differentiating, are you sure $T$ was supposed to be treated as constant? I think this is correct: $$dV=\frac{R.dT}{P}-\frac{RT.dP}{P^2}$$ (Obtained by directly differentiating $V=\frac{RT}{P}$)

The short way is to proceed as noted in the comments. The equivalent long way is to start from the total differential for $S_m$ in terms of $p$ and $T$:
$$dS_m = \left(\frac{\partial S_m}{\partial p}\right)_Tdp + \left(\frac{\partial S_m}{\partial T}\right)_pdT$$
The second term is given by
$$ \left(\frac{\partial S_m}{\partial T}\right)_p = \frac{C_p}{T}=\frac{5R}{2T}$$
The first can be determined using the identity
$$ \left(\frac{dS_m}{dp}\right)_T = -\left(\frac{\partial V_m}{\partial T}\right)_p=-\frac{R}{p}$$
Substituting,
$$dS_m = \left(-\frac{R}{p}\right)dp + \left(\frac{5R}{2T}\right)dT$$
It is not difficult to show that this is equivalent to the result obtained by substituting
$$dV=\frac{R.dT}{P}-\frac{RT.dP}{P^2}$$
into the expansion in terms of $V$ and $T$ shown in the OP.


text:
What is the definition of a bipolaron and a dipolaron?
I wondered about the exact definition of a bipolaron? In particular, if I have an oxygen vacancy in a metal oxide and the 2 excess electrons (when forming a neutral oxygen vacancy) localise on the 2 surrounding metal atoms, is this a bipolaron? And does the term "dipolaron", for having two electrons (holes) localised at the same (metal) centre, exist?

Cross-posted https://physics.stackexchange.com/questions/478328/what-is-the-definition-of-a-bipolaron-and-a-dipolaron

I am afraid that often when physics and chemistry nomenclature and dictionaries met confusion arise. P and BP are encountered in conjugated systems, where they stand for radical-ions and divalent ions, respectively. I would say that bipolaron should refer to bound, rather localized states, at least formally without assuming the real existence of them.

Thanks for your answer! What is yet not clear to me though is if the 2 localised charges in a bipolaron are localised on the same centre or at 2 centres in close proximity (from the examples I saw in textbooks, it seems like the latter one).


text:
Applying Ligand Field Theory, How Can The Color Of Transition Complex Ions Be Explained/Represented?
1.
To my understanding, Crystal Field Theory allows for an explanation for the colors of transition metal complex ions due to the d-d transitions which occur. This transition ∆o, is from the t2g orbitals to the eg.
Now looking at Ligand Field Theory. Is it right to say that this is equivalent to/same as the transition from the t2g orbitals to the eg* or t2g* to eg* orbitals depicted in the diagrams below?

2.
(Assuming octahedral complex with sigma bonding only)
Moreover, I was also initially confused due to the issue that if two of the d-orbitals of the transition metal ion is used for bonding with ligands e.g in a hexaaquachromium(III), [Cr(H20)6]3+, how would the d-d transitions occur if these eg orbitals are involved in sigma bonding with 6 of the ligands.
Is it right to say that the d-d transitions occur from the t2g non-bonding orbitals to the eg* anti-bonding orbitals and that the bonding with the ligands involve eg bonding orbitals which is separate to the anti-bonding eg orbitals?
In summary, I have the impression that CFT provides an explanation for the colors of complex ions and LFT provides an explanation for the bonding of transition metal ions with ligands as well as the colors. Hence, is that fact that there are bonding and anti bonding orbitals which allows for the both the colors and the bonding with ligands?


text:
Does melting HDPE plastic produce toxic fumes? (melting, not burning)
● The question isn't about burning HDPE but melting it at the proper temperature. (At 120 to 180°C depending on it's density, it becomes gooey. According to the source below the "extrusion" temperature range of HDPE is 176 to 260°C ― at 260°C, I guess it becomes a very viscous liquid).
Is it perfectly safe (eg: indoor melting, in a kitchen oven). Or does it produce toxic fumes ― before 180°C? Before 260°C? If so which one?
● And what about melting it canola oil? (This technic seems more energy efficient, and easier to control the temperature).
This experimenter tried with Canola oil, which has a smoke point of over 204 °C. He melted the HDPE in the oil at 176°C (the output needs to form a block and expresses the oil).
Is it safer to melt HDPE in oil ?
○ Bonus: does the Canola oil degrade the HDE? (The experimenter found out that the oil "gets really thick when cool" and he thinks it's because some HDPE dissolves on it. Someone comment "The canola oil probably acts like that because some of the HDPE is converted into paraffin wax which is soluble in canola oil. HDPE isn't lipid soluble".
Any information on the topic is welcome

It's not a duplicated:
This question has been asked about specifically melting empty silicone tubes (made of HDPE). It's off topic since not about HDPE only (it's a complex issue since there are many different types of silicone on the market)

This sounds like a bad idea. Don't.

Silicone tubes are not made of HDPE, they're made of silicone.

Note if you are using HDPE from recycled products you also have to consider the toxicity of any dyes and polymer additives.

This table indicates that polyethylene (HD/LD not specified) "melts at" 135°C, decomposes in the range 335-450°C, and produces vapors that will ignite between 341-357°C.
I'm sure melting/softening temperature is more complex than that, but I'm not sure that you should expect much decomposition into toxic gases at temperatures below 260°C.
However, if you're melting it in an oven, it's possible to overshoot the setpoint temperature. Your oven's thermostat might be slow to kick on and off. Direct thermal radiation from the heating element can heat an object in the oven well beyond the air temperature. Your plastic can even spill/drip onto the heating element.
I wouldn't like the idea of disposing of canola oil contaminated with unknown HDPE breakdown products. I'd be happier using some sort of double-boiler arrangement, where the HDPE is in a container that's heated by a bath of some other fluid. I'd suggest paraffin as a heat-transfer fluid, but its flash point is a bit low (200-240°C?)

thanks! So you also think HDPE could be dissolved in oil? Why do you think paraffin is better than oil?

@JinSnow I'm actually proposing to keep the HDPE away from both oil and paraffin, by putting the HDPE inside a container that's sitting in a bath of paraffin. That way the HDPE melts (although not as quickly as if it were "soaking" in the hot oil or paraffin), and it neither contaminates nor gets contaminated by the oil/paraffin).

I understood, but by curiosity, why would you avoid oil (more than paraffin)? Is it for an ecological reason?

@JinSnow Cooking oils will oxidize with extended heat and exposure to air, but paraffin (and saturated mineral oils) are relatively stable.

sorry, but, it'll oxidize, and so what? What's the problem with it?


text:
"Estrontium" on poster
I was looking at a poster for an international chemistry olympiad. Its background contains some random elements as periodic table tiles.
One of them said "Estrontium." A google search returned nothing but clip art.
Is this an accepted spelling internationally (similar to the way Spanish speakers prefix "e" on words starting with "s" followed by a consonant) or a careless typo on their part? My chemistry teacher was equally confused.

Sounds like Spanish to me. Olympiads may be international, but element names definitely aren't.

That's pretty weird, since in spanish it would be "estroncio". The name strontium comes from that of a scotish village. What was the language of the poster?

Probably multiculturalism at its best. When I studied in a public high school, there was a periodic table in the classroom where element thallium had a handwritten prefix "geni-". "eStrontium" isn't that bad in comparison though:)

The entire poster was in English. The olympiad itself took place in Paris.

Like all good Americans, the only language I know is English. So if it isn't English, it is Greek to me.

Is it just strontium?

Did the tile in question contain more than just the element name (atomic number or weight, for example)? Any chance you took a picture?

@andselisk I'm puzzled how you get "multiculturalism" from a typographical error or a high-school student with a middle-school sense of humor?

@jeffB Episode with Tl is an unrelated story, just a memory to spice things up. As to why I call the precedent multicultural: an incorrectly formed word using Spanish prefix, Scottish name and Latin suffix has somehow managed to pass quality control and show up on a poster of an international olympiad. I thought it was obvious:)

Seems like a typo, as all the spanish periodic tables I could find give the spanish translation of strontium as estroncio. None of the elements seem to use the "-ium" suffix when translated into spanish, so it looks like they just forgot to fully translate the word.

"Estrontium" is not used as an element name in any language. It appears that the error traces to a single user by the name of Alejo Miranda (listed as from Ecuador) who has posted a large collection of periodic tables and individual element clip art to shuttersock, iphoto, etc. in which "estrontium" is incorrectly used as the English name for strontium.  The Spanish periodic table correctly uses estroncio.  The link to the gllery at shutterstock is
https://www.shutterstock.com/g/Alejo+Miranda?searchterm=estrontium&search_source=base_gallery&language=en&sort=popular&measurement=px&safe=true


text:
H-bonding in o-hydroxybenzoic acid

Amongst all hydroxybenzoic acids the ortho isomer is most acidic and the major reason is intramolecular H-bonding in the anion.
My question is that isn't the loss of proton destabilising the molecule more than stabilising it as earlier two H-bonds were possible but after deprotonation only one bond remains??

But how is my question related to ortho effect?

I think the title is misleading.


text:
How to calculate the equivalent weight of KMnO4?
Let us consider the following reaction
$$\ce{2 KMnO4 + 16 HCl → 2 KCl + 2 MnCl2 + 8 H2O + 5 Cl2}$$
Now, in order to calculate the equivalent mass of $\ce{KMnO4}$, first I need to calculate it's $n$-factor which turns out to be $5$ because the oxidation state of $\ce{Mn}$ in $\ce{KMnO4}$ is $+7$ whereas in $\ce{MnCl2}$ it is $+2$. And the $n$-factor of $\ce{HCl}$ is $1$. So, the equivalent weight of $\ce{KMnO4}$ is
$$\frac{\text{molecular weight}}{5}$$
So, ratio of moles of $\ce{KMnO4}$ reacting with $\ce{HCl}$ should be $1:5$.
But from the balanced reaction we can see that this ratio is $1:8$.
Where am I wrong? Is my understanding and hence calculation of $n$-factor and equivalent weight is wrong?
(Please pardon this childish fashion of writing the question. If there is a problem anywhere, please do comment before flagging or downvoting.)

Hint - What are the two redox half-cell reactions?

@MaxW Ummm...I think $$\ce{2 H+ + 2 Cl- -> Cl2 + 2e- + 2H+}$$ and $$\ce{K+ + MnO4- + 5e- -> Mn^2+ + K+ + 4O^2-}$$

You can leave off the $\ce{2H+}$ and $\ce{K=}$ since they are on both sides of the equation.

@MaxW Are not these two the Oxidation and Reduction halves?

Yes, that is right. So per $\ce{Cl-}$ atom how many electrons are exchanged? Per $\ce{Mn}$ atom how many electrons are exchanged?

@MaxW Per Cl atom I think 1 and per Mn I think 5

Sorry, I didn't read the $\ce{MnO4-}$ reaction carefully. $\ce{O^{2-}}$ is not a valid chemical species for the reaction.

1 and 5 are right, so those are your n-values.

Go back a bit. What is the valid half-cell reaction for $\ce{MnO4-}$? // Hint  - It is an acidic reaction.

FYI – the URL for $n$-factor returns 404 error (at least for me, tried with two VPN servers).

Refer to the definition of equivalent (IUPAC Gold Book):

equivalent entity
Entity corresponding to the transfer of a $\ce{H+}$ ion in a neutralization reaction, of an electron in a redox reaction, or to a magnitude of charge number equal to 1 in ions.

In other words, $\pu{1 equiv}$ is the amount of substance reacting with $\pu{1 mol}$ of hydrogen atom.
If hydrogen acts as a reducing or oxidizing agent, then either way $\pu{1 mol}$ of hydrogen atoms liberates or accepts $\pu{1 mol}$ of electrons:
$$
\begin{align}
\ce{0.5 H2 &→ H+ + e-}\\
\ce{0.5 H2 + e- &→ H-}
\end{align}
$$
That's why an equivalent of a redox agent is its amount which liberates or accepts $\pu{1 mol}$ of electrons upon being oxidized or reduced, respectively, assuming electrons don't exist in solution on its own for a significant period of time.
Note that $\ce{KMnO4}$ participates in redox reaction
$$\ce{2K\overset{+7}{Mn}O4 + 16 H\overset{-1}{Cl} = 2 KCl + 2 \overset{+2}{Mn}Cl2 + 8 H2O + 5 \overset{0}{Cl}_2}$$
and the half-reaction for the reduction of manganese is
$$\ce{\overset{+7}{Mn}O4- + 8 H+ + 5 e- → \overset{+2}{Mn}^2+ + 4H2O}$$
Also note that $n$-factor of permangante here is not the number of protons $\ce{H+}$, which are also used up in water formation; rather it's the number of transferred electrons, e.g. $n = 5$ and
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{5} = \pu{31.61 g mol-1}$$
Obviously, the equivalent mass of permanganate isn't a constant and depends on the $\mathrm{pH}$ of the reaction.
For example, in neutral medium half-reaction appears as
$$\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- → \overset{+4}{Mn}O2 + 4 OH-}$$
and there is no explicitly shown protons to count at all!
However, the $n$-factor is $3$, and the equivalent mass of permanganate would be a different value:
$$M_\mathrm{equiv}(\ce{KMnO4}) = \frac{M(\ce{KMnO4})}{n} = \frac{\pu{158.03 g mol-1}}{3} = \pu{52.68 g mol-1}$$


text:
reaction mechanisms of peroxides
Task: Explain the reaction mechanism of 1,4-Dioxane in terms of how you get peroxides. Start here with the autoxidation with one radical X.
My explanation:
in the first step 1,4 Dioxan reacts with Dioxygen, so one gets an initiatorradical and and HOO* (* symbolises here a free electron, a radical). In the second step the initiatorradical reacts with oxygen and one receives a peroxyradical. In the end the peroxyradical reacts with an hydrogen atom and one comes to hydroperoxide. Is that right?

You can't have free hydrogen atoms floating around in solution. The requisite hydrogen atom comes from dioxane itself.

Doesn't the dioxane get the hydrogen atom from another substance? And is the rest correct?

Right, i got it. It comes from the 1,4-Dioxane. Thanks! But what would be the Initation here? the first step, which I draw? and the second line would be the propagation? Is there an termination step, too?

I could not find a definitive mechanism for 1,4-dioxane, but this is the mechanism for the formation of THF peroxide is likely very similar

diagram from here

So, the initiation is always that R* (here an hydrogen-atom) will be separated? Could the molecule do not react with oxygen at first, so that you get an peroxyradical and in the end this radical reacts with a hydrogen-atom to hydroperoxide?

I've edited the question, Would it look like this?

& is it right that 1,4-dioxane prefers to do autoxidation because it is an ether (it contains oxygen-atoms in it, connected to two different alkyl rests.). Or are there any other reasons, why it does like to do autoxidation?

The radical is stablised by the oxygen. This is why all ether-based solvents have a warning for peroxide formation.

But why do cyclic ethers tend more to peroxides than acyclic ethers?

Do they? AFAIK all ethers are prone to peroxide formation, this article may be worth your time, page 5 containing the most useful information http://ccc.chem.pitt.edu/wipf/Web/16340.pdf

Liquid-Phase Photolysis of Dioxane has been studied in several occasions in the past. One of these studies, several neat aliphatic ethers including dioxane were irradiated for $\ce{24 h}$ with a medium-pressure mercury lamp (Ref.1). As a result, the neat liquid-phase dioxane has been reported to yield gaseous, liquid, and solid products after the irradiation. Among the products, the major gaseous products were identified as $\ce{H2, CO, CH4, C2H4,}$ and $\ce{C2H6}$, but the liquid products have not been analyzed. However, the solid products were shown to be the racemic and meso forms of dioxyldioxane (see total reaction illustrated in the diagram to give solid product):

Pfordte rationalized the formation of dioxyldioxane as a simple dimerization of dioxyl radical formed by photoinitiated hydrogen abstraction from dioxane (Ref.1, 2):

Thus, it is safe to conclude that dioxane is susceptible to sun light and/or a foreign radical from any other source. Thus, after a long time exposure to sunlight or a in the presence of a foreign radical, the formation of dioxyl radicals is possible and exposure to environment would supply oxygen (also dissolved $\ce{O2}$) to further react as shown in the propagation step as follows:

References:

Von Klaus Pfordte, “Intermolekulare Dehydrierungen durch UV-Licht, IV – Photoreaktionen der Äther,” Justus Liebigs Ann. Chem. 1959, 625(1), 30–33 (in German) (https://doi.org/10.1002/jlac.19596250104).
P. H. Mazzocchi, M. W. Bowen, “Photolysis of dioxane,” J. Org. Chem. 1975, 40(18), 2689–2690 (DOI: 10.1021/jo00906a029).


text:
How is the potential energy between two atoms measured?
In class, we learned about the interatomic potential graph.

How does one actually measure the potential energy between two atoms experimentally to ensure the graph is correct? What device is used?

We don't. (At least directly, that is.)

@IvanNeretin Then how so indirectly?

Why, quantum chemistry.

Basic vibrational and microwave spectroscopy experiments give you the necessary parameters.

Please consider writing an answer, if you're able and willing to, of course; @Karl...

...and @IvanNeretin.

@orthocresol That'd be a replication of textbook knowledge, a lot of work if done properly. ;-)

@Karl fine to leave out the boring stuff, after all, no point reinventing the wheel, unless you feel particularly inclined to do so... Thanks for heeding the call!

The morse potential (equation see e.g. wikipedia) basically contains the bond dissociation energy, a "force constant" and the bond length at ground state. It does not give any measurable reality, but is just a mathematical model describing (approximating) the same.
The lower part of the curve (usually the ground state and perhaps the first two vibrationally excited states) are approximated by the harmonic oscillator, the energies of which you can get from  IR or Raman spectroscopy. The harmonic oscillator (= an ideal spring) has only one parameter, the force constant.
The bond lengths you can get either directly from microwave rotational spectroscopy, or from the rotational fine structure of you IR spectrum. They give you the reduced mass of the structure, which gives you the bond length if you know the absolute mass of the two atoms.
For the math and actual application, I refer you to any standard physical chemistry textbook.

@CharlieCrown tnx for fixing up my late-night goof-ups

no worries. It is a good answer and was probably the easiest 2 points I have ever gotten :)

@CharlieCrown My pleasure! ;-)

The most common and 'classical' method is to use Infra-red, Raman and Microwave spectroscopy to give the frequencies, and equivalently, the gaps between the vibrational and rotational energy levels. Microwaave spectroscopy also give the average bond length as but only as $\langle 1/r^2\rangle$.
Normally in a text book a model of the potential energy is assumed such as the harmonic oscillator or Morse potential. This is used because the potential exactly solves the Schroedinger equation.  This is a good approximation, but never exact, and is poor at high vibrational energies and close to the molecules dissociation energy. This approach only really works for diatomic molecules.
If you are not satisfied with this approach the next step is through a complicated algorithm (RKR method for example) using the energy levels to define the experimental potential energy which  exists as a set on numbers rather than a formula (as in the Morse case), i.e. it is empirical but correct. This 'data' can then be used to test a quantum model of the molecules bonding to try to reproduce the potential and so try to understand why it has the form it does, such as the fraction contribution from 'normal' bonding plus a fraction from ionic forms etc.
Molecules also have electronic excited states and using femtosecond duration laser pulses it is possible to watch the molecule as the vibrational and rotational wavepackets move on the potential energy. (A wavepacket is a collective motion of oscillators, such as vibrations). Analysing this motion produces the potential, but much more exciting is to see how a molecule dissociates, and in the case of NaI in the gas phase it does so in steps. Each time the bond extends a little dissociation occurs. Look up work by Ahmed Zewail.
Finally it is possible to react, say, a diatomic molecule with an atom, say OH+Cl, in the gas phase and by colliding two molecular beams, and look at the energy and angle at which the products are produced. As the reaction must occur on a potential energy surface the products' properties are defined by this and so can be used to determine the potential.


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Reagents for synthesis of ethyl 2-(2,3-dihydrobenzofuran-2-yl)acetate from naphthalen-1(4H)-one

The following conversion can be performed in a few steps. Show key intermediates and reagents (detailed mechanisms are not required).

So I've been looking at this for a while and I can't seem to figure it out. If someone could please help me by telling what kinds of reactions/reagents I should use in this synthesis. The mechanism is not necesarry to show, since I can do that myself, I just need help with which reaction path to follow.

It is easier to work backwards. How can you make a C–O bond beta to a carbonyl group?

To my knowledge the starting material doesn't exist as drawn. It exists as its tautomer, alpha-naphthol.

Oh, I feel silly for not spotting that now. The original question did indicate that this was the starting material. But of course, it would tautomerise...

So if I use Baeyer Williger to add oxygen next to the ketone, I get a 7 membered ring, but where do I go from there?

If you react the [7] ring lactone under base hydrolysis conditions the phenoxide generated will attack the unsaturated ester 1,4 to give the free acid of the required product. Then just esterify. The big question is whether 1-naphthol undergoes B-V in this manner and I can find no reference for this. Can anyone else?

@Waylander: No to your question. Just a poorly constructed problem.


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What is the chemical equation for the digestion/decomposition of cellulose?
I'm trying to find the chemical equation/pathway from cellulose $\ce{(C6H10O5)_n}$ to methane $\ce{CH4}$. What are the other reactants or products in the reaction?
$$\ce{C6H10O5 + ? -> CH4 + ?}$$
Everything I've found talks about bacteria and enzymes (cellulase, mostly) in terms of the process of enteric fermentation, but I've yet to find a single reaction or series of reactions for this conversion. At least one product would have to have oxygen in it, but that really hasn't narrowed down my search.
I'm mainly interested in how methane is created from the diets of herbivores, so it's very possible that I've been researching the wrong reactants from the beginning.

Not oxygen, water and carbon dioxide. I propose $\ce{ 2 C6H10O5 + 2 H2O-> 6 CO2 + 6 CH4 }$. (note: this is terribly simplyfied)

Biologically, methane is produced by methanogens, which are Archeabacteria. They do not utilize carbohydrates such as cellulose or glucose directly. Instead, they typically consume acetate or a mixture of hydrogen gas and carbon dioxide. These stoichiometry of these two processes is
$\ce{4H2 + CO2 -> CH4 + 2H2O}$ or
$\ce{CH3C(O)OH -> CH4 + CO2}$.
To get overall conversion of cellulose to methane, the cellulose is first fermented by the enteric bacteria you read about, which can produce acetate, hydrogen and carbon dioxide by a variety of pathways. These products are then utilized by the methanogens. One example of an enteric fermentation would be
$\ce{C6H10O5 + 3H2O -> 2CH3C(O)OH + 4H2 + 2CO2}$.
Combining this with the methanogenesis reactions, we can write the net equation of
$\ce{C6H10O5 + H2O -> 3CH4 + 3CO2}$.

Thanks! I knew glucose had to be in there somewhere, but I wasn't sure if it was an intermediate or end product.


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Synthesis of anthranilic acid
Yesterday, I performed the synthesis of anthranilic acid (also known as, 2-aminobenzoic acid. I was instructed to use the common route that is also featured on the Wikipedia page for this compound. That is, using the phthalimide I first synthesised, I dissolved in excess aqueous $\ce {5M}$ $\ce {NaOH}$ and then added an equimolar amount of sodium hypochlorite with heating. To obtain the anthanilic acid, we were asked to add into the large volume of the aqueous solution, glacial acetic acid dropwise with stirring, after adjusting the solution's pH to neutral using conc. $\ce {HCl}$. This was unsuccessful for me and many of my peers. That is to say, anthranilic acid did not precipitate out of solution even when we used a lot of the glacial acetic acid.
My friend and I suspected that the problem could be that there was just too large a volume of solution, such that the concentration of anthranilic acid is just too low. Hence, we decided to boil off some water to obtain a more concentrated solution. He used a hotplate to do so while being more impatient, I used a heating mantle, which was able to reach much higher temperatures much more quickly. Very interestingly, after vigorous boiling and giving off much acetic acid fumes, a white solid was formed as a deposit at the bottom of the flask. I then quickly filtered this greyish solid and performed a recrystallisation using water as a solvent on it. I obtained a white seemingly pure solid as the end product.
Not many of us obtained products yesterday for the practical. Even among the three of us who obtained products, the colour of our products varied with the other two getting products which looked slightly greyish and brown. Likely due to the presence of some impurities as the Wikipedia page does say that the pure anthranilic acid product should be a pure white colour.

The above image shows what I believe to be the most pure product on the right and the purity decreases from right to left.
I would like to ask some questions regarding this procedure I have conducted. Firstly, I believe I saw some benzoic acid crystals in one of my reaction flask, just a small amount at the bottom of the flask after I transferred the contents into a beaker. This was after the hypochlorite addition step. Could it have been one of the possible side products?
Also, how likely is my product actually anthranilic acid? It seems strange that the product crashed out of solution during the violent heating process. Although anthranilic acid is less soluble  in water than benzoic acid, this is still rather unexpected.

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

You have to do the first part of the reaction in an ice bath with vigorous stirring. You added the acids backwards. Start with conc. HCl to about pH 8 then add glacial acetic acid (AA) drop wise. The AA should precipitate at the highest yield just below 7 pH. Dark product is usually a problem, so boil with deactivated charcoal and recrystallize from water twice. I have made it successfully several times. Making N-acetyl AA from this is easy and Chem Player on YouTube shows how. You will need acetic acid anhydride for this.


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Why are the dispersion forces in CS2 stronger than the dipole-dipole forces in COS?
London dispersion forces supposedly have the least strength out of all the intermolecular forces. But $\ce{CS2}$, which has only dispersion forces, has a higher boiling point (and thus stronger intermolecular forces) than $\ce{COS}$, which has dipole-dipole attraction in addition to dispersion forces. Why is this?
I suppose that it has something to do with $\ce{CS2}$ having a thicker/more inducible electron shell, but then a new question arises: how would you know if the dispersion forces in one molecule are stronger that the dipole-dipole forces in another?
(Theoretically, without using boiling points or other experimental data. Also, this is based on question 4a from the 2018 AP chemistry free response.)

COS is not that much of a dipole. To the point of your question: generally speaking, you don't know it until you try it. Chemistry is an experimental science, after all.

Although individual dispersion forces are weak, they are cumulative, and increase with molar mass.  As a general rule, boiling point increases with molar mass.
Polar molecules will have higher boiling points when compared to molecules with similar molar masses.  For example, ethanol($\ce{CH3CH2OH}$) has a higher boiling point than dimethyl ether ($\ce{CH3OCH3}$).
$\ce{CS2}$ is ~16 g/mol heavier than COS.


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Names of NAD+ and NADH
As part of the biochemical process of glycolysis, NAD+ is reduced to NADH. In other words, nicotinamide adenine dinucleotide is reduced to... nicotinamide adenine dinucleotide?
Everywhere I look tells me that NAD+ and NADH refer to the same particle with the same name, just in an oxidized or reduced form. However, NADH has one more hydrogen atom then NAD+, so shouldn't these particles have different names?

The H stands for hydride, but most everyone just shortens nicotinamide adenine dinucleotide hydride to NADH.


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When does ipso attack occur in electrophilic aromatic substitution?
My textbook doesn't mention the conditions required for an ipso attack to occur. It just describes some examples where groups like $\ce{-C(CH3)}$ and $\ce{-SO3H}$ are replaced by $\ce{Br}$ using reagents like $\ce{Br2}$/$\ce{AlCl3}$.
Can someone describe the required conditions in detail?

In ipso substitution first step is usually attack of the nucleophilic species at the ipso carbon of the aromatic
ring (the carbon bearing the leaving group in this case).
Scheme 1

The first step is followed by elimination of the leaving group and regeneration of the aromatic ring.
Scheme 2

This mechanism greatly resembles the tetrahedral mechanism . The attacking species forms a bond
with the substrate, giving an intermediate, such as 1, and then the leaving group
departs. We refer to this mechanism as the SNAr mechanism.The reaction between 2,4,6-trinitrophenetole and methoxide ion involves ipso substitution.Intermediates of this type are stable salts, called Meisenheimer or Meisenheimer–Jackson salts,7 and many more have been isolated .
Scheme 3

In the reaction of dinitro compound  with piperidine,same ipso substitution operates.
Scheme 4
In the reaction between picryl chloride and hydroxyl ions same ipso substitution operates.

Ref
1.MARCH’S ADVANCED ORGANIC CHEMISTRY REACTIONS, MECHANISMS,AND STRUCTURE
SIXTH EDITION  ,Michael B. Smith ,Jerry March
2.Organic Chemistry, Volume 1, 6/E By Finar


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Why is 5-chloro-1-methylcyclopentene not named 3-chloro-2-methylcyclopentene?
Why is the ring numbered the way it is? What is it about the double bond that makes it so?

 source

Acoording to the IUPAC rules for nomenclature of the halo compounds the $\ce{Cl}$ here is treated like $\ce{CH3}$. And also according to rule, the double bond have preference over the groups other than functional groups.

Note that you can significantly improve your answer by quoting the exact IUPAC rule you are referring to, e.g. see how it's done in one of the answers on related topic.


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Which material are piezoelectric buzzers made of?
Which material are these buzzers made of?

Does it depend on the manufacturer or they all are made of the same material? The Wikipedia page does not say anything about the material. I found some datasheet online but it also does not reveal the material. Is there a reference where I could find more about the properties of this material?

Usually, they use the so-called (by non-chemists) piezoceramics. A frequently used material of that kind is Lead-Zirconate Titanate

and used in reverse (force produces voltage) are used to produce a spark as in manual gas lighters.


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What will be the order of acidity in the following compounds?

What will be the order of acidity in the following compounds ? From My point of view it should be 3>1>2>4. But the answer given is 3>2>1>4. Can anyone explain how ?

I think you are correct. The $\mathrm{p}K_\mathrm{a}$ of 4-methylphenylacetic acid is quoted as 4.37 here. The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is quoted as 4.19 by the same source here.

m-nitrobenzoic acid has a pKa of 3.47, so yes the original poster is correct.


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Why are low spin tetrahedral complexes so rare?
As I was going through Concise Inorganic Chemistry by J. D. Lee, I  realised that there are simply no low spin tetrahedral complexes mentioned in the book. Is there any specific condition required for the formation of such a complex?
Usually low-spin complexes are in $\mathrm{dsp^2}$ electronic configuration. But can this kind of orbital form a tetrahedral geometry? Because usually tetrahedron is usually synonymous to $\mathrm{sp^3}$.

Hybridisation won't explain anything in transition metal complexes, so please stop using it, at least to the extent where it is possible to avoid using it. Quite literally everything about transition metal complexes is better rationalised using MO theory, and I am not exaggerating.
The reason why low-spin $T_\mathrm d$ complexes are rare is because the splitting parameter, $\Delta_t$, is significantly smaller than the corresponding octahedral parameter $\Delta_o$. In crystal field theory, there is a complicated derivation which leads to the conclusion that (all things being equal)
$$\Delta_t = \frac{4}{9}\Delta_o$$
For more information, please see: Why do tetrahedral complexes have approximately 4/9 the field split of octahedral complexes? and Why do octahedral metal ligand complexes have greater splitting than tetrahedral complexes?. Of course, this relationship is not exact in the real world, because CFT is a very simplified model; ligands are not point charges. However, it is still true in a qualitative sense.
Since the splitting $\Delta_t$ is smaller, it is usually easier to promote an electron to the higher-energy $\mathrm t_2$ orbitals, rather than to pair the electrons up in the lower-energy $\mathrm e$ orbitals. Consequently, most tetrahedral complexes, especially those of the first-row transition metals, are high-spin. Low-spin ones do exist (e.g. J. Chem. Soc., Chem. Commun. 1986, 1491), but aren't common.

It's just that we just went through the basics of CFT. I can't think without using hybridisation right now..

I can understand that. However, I think it would be a disservice to not at least mention it. It is an absolute travesty that hybridisation for TM complexes is still taught. Also, other people, not just you, will read this too.

CFT contains zero reference to hybridization. It is ridiculous to force into it.

Update: a low spin cobalt(II) complex with a distorted tetrahedral geometry. You can see the complex here. Another example is  $\ce{Cr[N(SiMe3)2]3[NO]}$

There are very rare exceptions to this rule. Only very strong ligands can form low spin complexes keeping the shape still tetrahedral. For example, Ni(CO)4 is diamagnetic in which 2 electrons of 4s are demoted to pair up with the 2 unpaired electrons of 3d.

@Priyansh2003 Not sure what you're getting at here. Ni(0) is $d^{10}$, so the terms high- or low-spin are meaningless. Those terms aren't to do with distribution of electrons in 4s/3d, it's about the distribution of electrons in $e_g$/$t_{2g}$ or $e$/$t_2$ orbitals, and for a $d^{10}$ complex both the $e$ and $t_2$ orbitals are fully filled, so there's no distribution to talk of.

I was saying in the light of Valence Bond Theory. But now I realise that is not the center of the discussion.


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Sodium bicarbonate, varying acetic acid, measuring carbon dioxide volume, understanding results
With my 6yo daughter, I have conducted a simple experiment. We used 1/4 tsp (we don't have sub-gram scales) of bicarbonate of soda for each test. In each test, we used a different amount of vinegar (5% w/w acetic acid), topped to 120ml with water. In each test we left the reaction to (for the purposes of measurement) complete and measured the $CO_2$ volume produced. (the reaction took place in an upturned measuring cylinder).

our first hypothesis was that we would get a linear relationship up to the point that the bicarbonate of soda became the limiting reactant.
we then modified this thinking that a certain amount of $CO_2$ might get dissolved in the water so the graph line might pass below the origin.

The results we got were surprising. With each increase in vinegar, the volume of gas increased by more. See below
vinegar  |  CO2
  (ml)   |  (ml)
--------------------
   5     |  8.5
   10    |  22
   15    |  34
   20    |  110

We have repeated the measurements a few times. We have a fair amount of variability but the same pattern. (I appreciate the woolliness of this statement, more statistic next time).
I guess that the increased rate of reaction which came with increased concentration meant the $CO_2$ had less chance to dissolve. Can someone tell me what the correct explanation is?
As a bonus - is there a way to conduct this experiment in such a way that we will be better able to observe the relationships between the reactants and the product? One thought that I had was saturating the vinegar with NaCl so that it won't allow any carbon dioxide. Could this work?  Is there a better way

Was the upturned measuring cylinder then put into some container of water? I'm having trouble visualizing how your "simple experiment" was done. Please explain the experimental procedure in detail.

I think your vinegar was the limiting reagent in each case. Vinegar is only 5% in acetic acid. You were using pure sodium bicarbonate. Given that more acid you added, you observed more CO2, supports this point.

There is another aspect of creating acetate pH buffer. Acetic acid : sodium acetate 1:1 gives pH 4.75. OTOH, pH of water saturated by CO2 Is 3.85 ( derived as pH=0.5(6.35-log 0.034) from https://chemistry.stackexchange.com/questions/42947/why-does-co2-lowers-the-ph-of-water-below-7). So the acid needs some excess. Another aspect is bicarbonate pH buffer with effective pKa 6.35.

@M.Farooq - yes I'm sure the vinegar was the limiting reagent. The intent was to observe the change in product with the increase in vinegar.

@MaxW the measuring cylinder was upturned on a jam-jar lid providing a small reservoir of water covering the opening of the cylinder, preventing air from entering it. This makes a very simple was to measure the volume of gas produced. The downside is that vinegar will be displaced as gas is produced but this would have the opposite effect to what we observed (reducing the rate of increase in product with reagent)

How did you "isolate" the sodium bicarbonate to the graduate cylinder? Why wouldn't some flow out when you inverted the graduate cylinder?


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What are the enthalpy and entropy contributions to carboxylic acids dissociating in water?
Carboxylic acids such as acetic acid do not dissociate much in the gas phase. In aqueous solution, however, they act as acids (proton donors).
$$\ce{CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$
Are reaction entropy and reaction enthalpy positive or negative, and what are there magnitudes?
In the gas phase, we have clear expectations:
$$\ce{CH3COOH(g) <=> H+(g) + CH3COO-(g)}$$
The reaction enthalpy will be positive (we need to break a bond) while the reaction entropy will be positive (more gas particles). As the reaction does not go forward in the gas phase, the enthalpic contribution must be of larger magnitude.
So another way of stating the question is to ask what the effect of solvation is on the enthalpy and entropy of reaction.

@Withnail The article you cited in a different thread, pubs.acs.org/doi/pdf/10.1021/ed048p338 Journal of Chemical Education, Vol. 48,pg.338, 1971, encouraged me to ask this question even though it is related to this one: Solvent effects on acidity of ethanoic acid

This site lists the heat of solution for acetic acid as -1.5 kJ/mol: http://webserver.dmt.upm.es/~isidoro/dat1/Heat%20of%20solution%20data.pdf

Electrostatic screening associated with solvation is obviously the key factor enabling dissociation in aqueous solution when compared to the gas phase. Without such dielectric screening, dissociation would be too costly ($\approx \pu{1 MJ/mol}$ or $\approx\pu{14 eV}$,  roughly the energy required to ionize a hydrogen atom) to occur to a significant extent at RT. The dielectric constant of water reduces the energy of the gas phase electrostatic interaction by a factor of $\approx 80$. This is an enthalpic effect. This energy is further reduced by water oxygen bonding to $\ce{H^+}$ to form $\ce{H3O+}$ (proton to hydronium).
Differences between dissociation constants of different carboxylic acid groups originate from enthalpy or entropy changes due to anion formation (since the formation of protium is something shared in common by all acids, its solvation cannot explain differences in behaviour of different acids).  The effects on the solvent are often described in terms of "structure making or breaking". As a rule there is an entropic penalty ($\Delta S^\circ<0$) with dissociation that can be attributed to solvent rearrangement around the ensuing charged species. The penalty is seen in thermodynamic data for many weak acids.  Trends are otherwise not quite as clearcut. Beyond the entropic penalty, other factors (enthalpic or entropic) may play a role in encouraging or discouraging dissociation, for instance charge delocalization (inductive, mesomeric effects) or intramolecular rearrangements.

Thanks for the link to the thermodynamic data for a range of weak acids. The lack of entropic driving force for this dissociation is not what I had expected and this has really helped me gain a better understanding of the dissociation process.

@Withnail I'm glad you found it useful. It admittedly presents only a summary of the picture


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Changing of UV-Vis peak absorption after heating Ni(en)2(NO2)2

(c) The trans-isomer of $\ce{Ni(en)2(NO2)2}$ D is red $(λ_\mathrm{max} = \pu{500 nm}).$ It can be heated to give a new trans-isomer E, which is violet $(λ_\mathrm{max} = \pu{560 nm}).$ Assume that, within each of D and E, the two $\ce{NO2-}$ ligands exhibit identical coordination modes.
(i) Rationalize the colours of D and E, explaining the difference in d-orbital splitting. [6]
(ii) Sketch structures for D and E. [2]

I've been stuck on this for some time to the stage that I'm probably just melting brain cells. Any help is appreciated.
My initial thoughts were that it was due to one end of the en ligands dissociating to form a 4-coordinate species in sq planar (owing to $\ce{Ni^2+} = \mathrm{d^8});$ however this would mean that it goes low-spin and then the splitting of d-orbitals becomes even greater, hence would expect a smaller UV-vis peak wavelength as a higher $E$ jump between occupied/unoccupied d-orbitals. (Δ(Sp) > Δ(Oct))
I'm probably going about it the wrong way so yes, any help appreciated! Would an MO approach be more useful here rather than CFT?


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Why do weak acids like ethanol not act as a strong base in water?
In water, ethanol has a $\mathrm{p}K_\mathrm{a} = 15.9$ which means that its $\mathrm{p}K_\mathrm{b}$ is $-1.9$. Which implies that all weak acids in water are in fact strong bases. But this is not true, weak acids in fact do act like weak acids. That would mean that this equation isn't true always.
$$14 = \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b}$$
Considering this is just the ionic product of water it should be valid always?

It's the $\mathrm pK_\mathrm b$ of the conjugate base, not ethanol itself. $$\mathrm pK_\mathrm a(\ce{EtOH}) + \mathrm pK_\mathrm b(\ce{EtO-}) = 14$$

And because pKa > 14, $\ce{EtO-}$ isn't going to exist in aqueous solution to any practical measure, making it a very strong base. $$\ce{EtO- + H2O -> EtOH + OH-}$$

and ethanol is a very weak acid, hence it's conjugate base is strong. A weak acid like EtOOH also has a weak conjugate base.

@Karl did you mean strong conjugate base in the last line of your comment?

@WilliamR.Ebenezer Definitely not.

@Karl Alright. The "very weak" makes sense then.

@WilliamR.Ebenezer ;-) Confuses me again every few years.

There seems to be a misunderstanding: The conjugate of a strong base like EtONa can't be a weak acid, because weak acids, like acetic acid for example, are still quite acidic.

Strong acid (HCl) - very weak conjugate base (chloride anion)
strong base (EtONa) - very weak conj. acid (Ethanol)

A strong acid or base reacts strongly (with water), the resulting conjugate is rather inert, with the equillibrium fully on the side of the latter.

weak acid (acetic acid) - weak conj. base (acetate anion)
weak base (ammonia) - weak conj. base (ammonium)

A weak acid or base just reacts partly towards its conjugate, so does the pure conjugate: They are both rather weak. The equillibrium in an equimolar mixture is rather stable: A buffer solution.
The ideal weak base or acid has a $pKa$ of 7 (in water), that is $1/2$ of the -log ion product of the given solvent. If it's even weaker, the conjugate becomes stronger again.


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What is the mechanism for this reaction?

The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

Moreover, even if I make an amine in this step I cannot proceed further, how will the amine and alcohol react in the next step? Any hints would be appreciated.

First deprotonate and then lose carboxylate. The remaining neutral species reranges to an isocyanate. In methanol the isocyanate forms the methyl ester of a carbamic acid. Compare the similarity of the Lossen and Hofmann rearrangements

My thoughts. I think you have to deprotonate the NH of the hydroxamic acid and lose OH- to give the isocyanate analogous to the mechanism of the Lossen that you have drawn. This is the only option that makes any kind of sense, and it is known that deprotonation of hydroxamic acids proceeds at NH rather than OH.

Seems like methanol/methoxide nucleophilic attack on the isocyanate (instead of water/hydroxide) would lead to the carbamate.  Not sure if that squares with the supposedly two-step reaction scheme though.

Now that a bounty has been placed on this post and a reputable source for the answer is requested, I seriously suspect that one will not be found. Any version of the Lossen rearrangement requires activation of the oxygen into a viable leaving group before base can effect deprotonation and alpha-elimination can occur to form the isocyanate. If the methyl carbamate is the desired product, simply treat the the activated hydroxamic acid with MeONa/MeOH rather than aqueous base.

 This is from March 6th edition page 1610. Does this help? :)

In your question, you have mentioned:

The closest reaction I can find is Lossen rearrangement but in that hydroxyamic acid is first converted to its derivative but how will that be possible here?

You have recognized the reaction correctly but the Lossen rearrangement is the conversion of a hydroxamate ester to an isocyanate (only the first step of given mechanism). Although the original Lossen rearrangement needed hydroxamate ester $\left(\ce{R(C=O)NH-O(C=O)R'}\right)$ to rearrange,  recent research has shown that the rearrangement is possible with free hydroxamic acids $\left(\ce{R(C=O)NH-OH}\right)$ as well (Ref.1):

Without using activating agents, a variety of free aromatichydroxamic acids could be rearranged to aromatic amines in the presence of base alone:

Source of the image: Ref.1
According to $\bf{(a)}$, free aromatichydroxamic acid $\left(\bf{1}; \ce{R(C=O)NH-OH}\right)$ react with a base to give  the corresponding carboxylate $\left(\ce{R(C=O)-O-}\right)$ as intermediate and hydroxyamine $\left(\ce{NH2-OH}\right)$ byproduct. The intermediate $\left(\ce{R(C=O)-O-}\right)$ reacts with another aromatichydroxamic acid $(\bf{1})$ to give necessary hydroxamate ester $\bf{A}$, which undergoes Lossen rearrangement in basic conditions to give the isocyanate $\bf{B}$. Mechanism for aromatic amine from isocyanate is depicted in $\bf{(b)}$.
For instance, the base-mediated rearrangement of free 4-methylphenylhydroxamic acid $\bf{(1a)}$ to  4-methylaniline $\bf{(2a)}$ is given as follows:

A mixture of N-hydroxy-4-methylbenzamide $\bf{(1a)}$ $(\pu{0.363 g}, \  \pu{2.4 mmol}),$ $\ce{K2CO3}$ $(\pu{0.332 g},$ $\pu{2.4 mmol})$, and DMSO $(\pu{2 mL})$ was heated to $\pu{90 ^\circ C}$ and stirred at that temperature for $\pu{2 h}$. The mixture was cooled to rt, and then treated with $\pu{2 M} \ \ce{ HCl}$ $(ca. \pu{3 mL})$. After the mixture became the clear solution, $\pu{2 M} \ \ce{NaOH}$ $(ca. \pu{3 mL})$ was added and extracted with $\ce{Et2O}$ $(\pu{15 mL} \times 3)$. The combined organic layers were dried over anhydrous $\ce{Na2SO4}$, filtered and evaporated under reduced pressure. The residue was purified by silica gel column chromatography $(\ce{hexane/Et2O}, 1:1)$ to yield the pure 4-methylaniline $(\pu{0.253 g}, \ 98\%)$ as a white crystalline solid.

The effects of the amount of base, reaction time, and presence of water are summarized in following diagram (Ref.1):

Thus, as shown in above diagram, presence of water may have direct effect on the conversion of the substrate to corresponding amine (even in DMSO). Further, Ref.1 clearly shows that this reaction requires polar aprotic solvent to proceed through Lossen rearrangement to give thiocyanate intermediate, which eventually converted to corresponding amine in the presence of aqueous base. For instance, if $\bf{(1a)}$ is reacted with $\ce{K2CO3}$ in methanol instead of in DMSO or DMF at for $\pu{90 ^\circ C}$ for $\pu{2 h}$, the yield was found to be only $10\%$ while it was $98\%$ in DMSO and $92\%$ in DMF under the same conditions. The authors reported $86\%$ of the starting compound has been recovered unreacted (none of it recovered in DMSO or DMF).
Note that the first reaction given in the question is in aqueous medium. Thus, according to the results found in Ref.1, the resulting product in the given reaction would be a trace of corresponding amine $\left(\ce{R-NH2}\right)$ and most of substrate would be recovered unreacted. Consequently, there won't be a reactive isocyanate $\left(\ce{R-N=C=O}\right)$, thus the reaction would not proceed to give the anticipated product in methanol as indicated.
Suppose this first reaction has been done in $\ce{OH-/DMSO}$ in place of $\ce{OH-/H2O}$. That would give you the anticipated isocyanate intermediate $\left(\ce{R-N=C=O}\right)$. It is well known that isocyanates reacts very differently in the presence of water, alcohol, and amine. In the presence of water, it would give you corresponding amine with evolution of $\ce{CO2}$. However, in the presence of alcohol or amine, it would give you corresponding carbamate or urea derivative, respectively$:

Since the given second reaction is in methanol (an alcohol), the product would be a methyl carbamate as shown in the scheme above (Ref.2). For example, polyurethane is prepared from diisocyantes:

References:

Yujiro Hoshino, Moriaki Okuno, Eri Kawamura, Kiyoshi Hondab and Seiichi Inoue, "Base-mediated rearrangement of free aromatichydroxamic acids (ArCO–NHOH) to anilines," Chem. Commun. 2009, (17), 2281-2283 (DOI: https://doi.org/10.1039/B822806J).
Greet Raspoet, Minh Tho Nguyen, Michelle McGarraghy, and Anthony Frank Hegarty, "The Alcoholysis Reaction of Isocyanates Giving Urethanes:  Evidence for a Multimolecular Mechanism," J. Org. Chem. 1998, 63(20), 6878–6885 (DOI: https://doi.org/10.1021/jo9806411).

This is like to proceed by a isocyanate intermediate.
The Lossen reaction has been shown to be slow but possible in free base, without the addition of a leaving group or metal catalyst, by Hoshino et. al. in 2009[1] - see Scheme 1(a)
Isocyanates, when attacked by alcohol nucleophiles, form carbamate esters
See also: Org. Biomol. Chem., 2016,14, 9046-9054
[1] Chem. Commun., 2009, 2281-2283

Will try to add diagram etc later.

While this Lossen rearrangement employs base (K2CO3), the reaction is run in DMSO, not hydroxylic solvent. A careful reading of the mechanism illustrates that an effective leaving group is generated in the dimerization. There is still no evidence that the hydroxyl group in hydroxylic base is an effective leaving group.

The paper you mention in your answer contains the answer to the question. However, I can award the bounty to only answer so I am awarding it to Mathew's answer as it is more elaborate even though both answers technically refer to the same paper.

Mechanistically, the N-H is deprotonated to form an amide ion. In basic medium, the hydroxyl group may be slightly deprotonated as well, but the pKa are not drastically different and this is not mechanistically useful. You can envision an isocyanate forming upon rearrangement, which is then subject to nucleophilic attack by methanol. Graphically:

It is also worth nothing that hydroxide ion is the base and the leaving group from the hydrodyxamic acid is also hydroxide. This is likely by design and not accident to prevent unnecessary by-products from existing.
Does this help? :)

Could you please add some reference for the mechanism? This has been unanswered for a long time, so I want to make sure that the answers can be backed up by a good reference.

Did you read my Comment of today? What is the evidence that no activation of the hydroxamic acid is required? If you have hydroxide in methanol, then you also have methoxide. Uncharged methanol is not as good a nucleophile as methanol.

@user55119 Sorry, I missed the comment. The way OP posted the reaction scheme showed two different arrows so I guess it can be assumed that they are two different steps i.e. there is no hydroxide in methanol. And are you saying the reaction OP posted won't happen?

@Shoubhik R Maiti: That is what I  am saying. I would like to have a reference for this reaction as described. I suspect there isn't one and that the question was manufactured ... and quite poorly.

I don't have a reference other than my own knowledge. It follows precedent analogously of Hofmann and Curtius rearrangements to form isocyanates. The isocyanate undergoes nucleophilic addition in a second, separate step to form the carbamate (urethane).
If you consider Hofmann rearrangement, OH group (here) and Br group serve to increase acidity of the N-H hydrogen. Interception of the isocyanate intermediate gives the final product.
Another similar precedent is the Lossen rearrangement, where OR is used instead of OH.
These proposals work mechanistically but may not in practice.

March 6th edition page 1610

@ShoubhikRMaiti this is probably of a case of a poorly formed question. OH- ion is way better at the process than OMe, but we may have to do this anyway because the author ignored the OH- after leaving it out.. This is often a case where I find myself grumpy :(