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Matchmol for substructure determination
I'm using Checkmol/Matchmol for substructure determination and retrieval, and I'm having problems determining whether a given molecule is the substructure of another molecule when they have the same connectivity map.
I work in drug design so charges, isotopes, radicals and geometry are a very important concern.
Given the following molecules, for example:
+---------+-----------------+-------------------------------------+
| mol_id  | formula         | smiles                              |
+---------+-----------------+-------------------------------------+
|    7959 | C7H15Cl2N2O2P   | ClCCN(P1(=O)NCCCO1)CCCl             |
|   41041 | C7H15Cl2N2O2P   | ClCCN([P+]1([O-])NCCCO1)CCCl        |
|  572776 | C7H15Cl2N2O2P   | ClCCN([P@]1(=O)NCCCO1)CCCl          |
|  572791 | C7H15Cl2N2O2P   | ClCCN([P@@]1(=O)NCCCO1)CCCl         |
| 1742146 | C7H13Cl2D2N2O2P | ClCCN(P1(=O)NCC(CO1)([2H])[2H])CCCl |
+---------+-----------------+-------------------------------------+

You can see that connectivity map is identical in all cases, though some variations in stereochemistry, charges and isotopes can be found:

First question: "subestrcture" is a very wide term but I think about it as to whether is molecule A contained in molecule B, or not. Would you say all these are substructures of each other given they share the same connectivity map?
In a practical scenario, ignoring charges, isotopes and stereochemistry, every molecule in the example will be a substructure of eachother:
matchmol someMol.sdf someOtherMols.sdf

According to matchmol help (which is the only documentation on matchmol usage I could find), adding the isotopes constraint (-i) would exclude 1742146; adding charges constraint (-a) would exclude 41041; and adding geometry constraints (-gG) would exclude both 572776 and 572791... or that is what I would expect.
However, running the program for each molecule using all constraints results in the following:
matchmol -sGgaid someMol.sdf someOtherMols.sdf

7959  7959    T
7959  41041   F
7959  572776  T
7959  572791  T
7959  1742146 F
41041 7959    T
41041 41041   F
41041 572776  T
41041 572791  T
41041 1742146 F
572776    7959    T
572776    41041   F
572776    572776  T
572776    572791  T
572776    1742146 F
572791    7959    T
572791    41041   F
572791    572776  T
572791    572791  T
572791    1742146 F
1742146   7959    F
1742146   41041   F
1742146   572776  F
1742146   572791  F
1742146   1742146 T

Notice that, as expected, for the plain molecule (7959), neither 41041 (charged) nor 1742146 (with D isotopes) are a substructure; but both stereoisomers (572776 and 572791) are.
Interestingly, for the the charged molecule (41041), the same molecule is a substructure (bug?), but plain and stereo are.
Stereoisomers 572776 and 572791 should not be substructure of each other when geometrical constraints are in place, yet they are...
So... second question: Am I doing something wrong? Am I misunderstanding the results?


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Predicting products of chemical equations?
Consider the equation $\ce{2 NaOH (aq) +Cl2 (g) —> NaOCl (aq) + NaCl (aq) + H2O (l)}$. How would one know the products of this reaction? Is there a general way of doing this? I would have no earthly knowledge of predicting this. Is it that we would look at each molecule and decide which goes with which on the basis of octet? Also, why does it just so happen that 2 moles of $\ce{NaOH}$ react with 1 mole of chlorine? What would happen if it was a 1:1 ratio?

Welcome to Chemistry.SE. We’d like you to take the Tour to get familiarized with the site and to understand our policy towards homework questions. Meantime, let us know what you have done so far to solve this specific problem as starting point.

Yes there is a way of doing this. It is called chemistry. It does not boil down to one or two simple rules, otherwise it would never be called a science in its own right. As to the ratio, that's quite simple: take any other ratio, and one of the reagents will be in excess.

I agree with Ivan Neretin. In addition to that, in general, knowing molecular orbital diagrams of individual reactants and possible product candidates will help quite much to guess the outcome. I did not memorize any reaction in inorganic chemistry for instance, but you can attain this kind of comprehension mostly at the end of the undergraduate program.

One would not "know" what the products of the reaction are, a priori. As mentioned by @IvanNerenten, the way you know is by studying chemistry. In fact, the reaction is different if you use cold NaOH vs hot NaOH. The reaction as written is with cold NaOH and is a disproportionation reaction.This is where the chlorine is both oxidized and reduced in the same reaction.

How would one know the products of this reaction? Is there a general way of doing this?
It's not easy at all, in chemistry, there are patterns that are valid in many cases (e.g. you know that you are adding a strong acid with a strong base you can predict that you will have a slat and water as products).
For more complex reactions it's harder to find valid patterns. There are some techniques that use ab-initio calculations and some softwares but in many cases, they are not so accurate.
I find useful these rules of thumbs not always valid:

Get a good knowledge of electrochemistry.
Write the equation highlighting the ionic compounds (e.g. $\ce{Na+ + Cl-}$ ).
Check the possible products in agreement with the stoichiometry.
Stable compounds (e.g. water) are more likely to occur compared to unstable compounds.

Why does it just so happen that 2 moles of NaOH react with 1 mole of chlorine? What would happen if it was a 1:1 ratio?
A chemical formula it's a model a representation of the expected behaviour.
If you had 1:1 ratio you would have a limiting reagent. So  you will end with 1 mole of NaOH reacting with half mole of $\ce{Cl2}$.

source: Stack Exchange

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Identifying an unknown gas using the Ideal Gas Law
So my chemistry teacher gave me this packet having to do with The Ideal Gas Law. One of the homework questions says, "An unknown gas at $\pu{20.0 ^\circ C}$ and $\pu{1900 torr}$ has a density of $\pu{3.95 g/L}$. Identify the compound." I have tried this problem many times and have derived many formulas to no avail. Can someone please help me?

Have you clarified what you have tried or where specifically you seem to be getting stuck. One thing you could do is suppose you have one mole of the compound. Using the data you have, you can then calculate the volume. Can you see a way forward from there? @ThomasMcCarthy

Suppose the mass of the gas is $m$ and its molar mass and the volume under the given conditions are $M$ and $V$, respectively. Thus, the number of moles ($n$) and the density ($d$) of the gas under the given conditions is:
$$n=\frac{m}{M} \space \text {and}$$
$$d=\frac{m}{V} \space \Longrightarrow V=\frac{m}{d}$$
Substitute these values in the Ideal Gas Law, $PV=nRT$.
$$P\frac{m}{d}=\frac{m}{M}RT$$
Thus,
$$M=\frac{d}{P}RT$$
Since you know $P$, $T$, and $d$, numerically, you can calculate $M$ in $\mathrm{gmol^{-1}}$, using appropriate $R$ value.

Ok so I did the math for that and my answer was 38.0268 g/mol. The final part to the question is to identify the compound and I was unable to find a gas with the molar mass of 38.0268 g/mol.

Well it could be C-13 trideutero methyl flouride, that will have an approximate formula mass of 38

@ThomasMcCarthy - 38.0268 has far too many digits. The density was only given to three significant figures so no more should be used. Thus the answer should be 38.0. So I'd guess $\ce{F_2}$

@MaxW: Actually it is 30.00 so MaxW assumption is correct: the gas is $\ce{F2}$.

PV = nRT
Assume you have 1 liter volume so V = 1
P = 1900 torr
n = moles = ?
R = 62.63 L-torr/mol-K
T = 20 + 273 = 293 K
n = PV/RT = (1900)(1)/(62.63)(293)
n = 0.104 moles
Molar mass = g/mole = 3.95 g/0.104 mole = 37.98 = 38.0 g/mol
Unknown could be F2 gas

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Boiling points of alkanes, alkynes and alkenes
I know that when comparing between these three families, alkynes have the highest MP and BP, followed by alkanes and, lastly alkenes (when comparing the same number of carbons).
I know alkynes "win" because of the triple bond (pulling the carbons tightly together, etc.  I also know the alkynes have the least number of dispersion forces.
So, how come when comparing alkanes to alkenes, the number of dispersion forces "wins out" in determining MP and BP and not the double bond (which also pulls tighter than a single C-C bond)...but with alkynes, the triple bond can "beat out" the decreased number of dispersion forces?


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why would an alpha particle ejects neutron when bombarded on berreylium?
when an alpha particle hit berrelyium it emits neutron. Does the alpha particle causes unstability to the nucleus or any other reason?

Alpha particle is itself a nucleus, just a relatively little one. Then again, beryllium is not awfully large either. Imagine a head-on collision of Earth and Mars; I have a vague feeling you wouldn't want to describe it in terms like "Mars causes instability to Earth".

So then it is due to the kinetic energy of the alpha particles. Is this true?

Yes, you may put it this way.

I disagree with @IvanNeretin's comment. This is a quite well known nuclear reaction, 9Be($\alpha$,n)12C (amongst other things well known as the neutron spark plug for the first atomic bomb). The 13C intermediate nucleus that is formed is some 10MeV above the ground state. One relaxation path is through releasing a neutron to leave 12C. (Remember that you need to conserve both energy and momentum in the center-of-mass frame of the interaction.) The overall Q value for this reaction is about 5MeV.

The reaction can be regarded as a fusion reaction, the fast moving helium nucleous has sufficent energy to overcome the electrostatic repulsion of the positvely charged beyllium nuclous. What happens is that the alpha particle will enter the berylium nucleous to form the product.
The product will be carbon 13 in an excited state, the carbon 13 could relax down to the ground state by the emission of gamma rays (photons) but by the ejection of a neutron it is able to convert itself into the very stable carbon-12.
In the same way as in chemistry we have the chemcially very stable noble gases (full electron shells) in nuclear science we have some nuclides which are blessed with additional stability.
Carbon-12 has both an even number of protons and neutrons which will increase its stability more than a odd,odd, even,odd or odd,even nuclide. Also there are some magic numbers which correspond to complete shells of nucleons (protons and neutrons). Carbon-12 is not one of these magic nuclides (full shells of protons and/or neutrons) but it is more stable than carbon-13.


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Turning Soda Glass into Lead Glass and Fused Silica

What would be the method of turning Soda Lime Glass into:
A. Lead Glass
B. Fused Silica

Soda Lime Glass is $\ce{SiO2}$ ~70%. Lead Glass is $\ce{SiO2}$ ~ 62.9%. Fused Silica is $\ce{SiO2}$ ~ 99.9%. Can one start with a batch of Soda Glass, chemically separate the $\ce{SiO2}$, and then use that to make Lead Glass/Fused Silica? I'm not expecting it to be cost effective, but I doubt it's physically impossible. Is it?

Two completely different chemical compositions. So the only way to make a conversion is to sell the soda glass and use the money to buy lead glass.

Cannot be done.

@Johan88 I added your comment to the question. As your question previously stood it was violating the homework policy and would have been closed. Hope that's fine with you.

@AvnishKabaj much appreciated. Thank you. Really don't understand the difference between a hw and a non-hw question but I'm sure you do. Grazie

The policy is very badly titled. It's not limited to questions which are literally homeworkA "homework question" is any question whose value lies in helping you understand the method by which the question can be solved, rather than getting the answer itself. This includes not just questions from actual homework assignments, but also self-study problems, puzzles, etc.

@AvnishKabaj yes, I read, but I still don't understand. Thanks again.

It was just a copy of textbook question, or at least looked like. You can't just copy/paste a textbook and expect someone do it for you, or any other assignment. Is it clear now?

@Mithoron Copy/paste? A textbook? I haven't been a student or part of a learning institution for over half of my life. THANKS.

@Mithoron And no, it's not clear, and it is misleading, just like a Moderator comments on the homework question post itself.

@Mithoron And, in case you haven't read the policy yourself: "As a general rule, we do not discourage homework questions, as long as they are related to chemistry." None the less, mine was not a homework question, and I came to Chemistry Stack Exchange because: "it's a place to get specific conceptual chemistry questions answered." THANKS

Everything you suggest, can be done. It is a bit impractical though, the best is to start from the right raw materials.
Soda glass can be purified by high temperature. The sodium oxide will evaporate as the temperature becomes high, but we are talking on the upside of 1600 $^o$C perhaps even up to 2000 $^o$C. All the other alkali and many rare earth oxides will evaporate as well, so if you want any of those back, you'll have to add them. In addition, high temperature silica is a mess to work with when it is pure, viscous and prone to "boiling over" when releasing gases - so YMMV. So there you have pure silica. It'll never be "fused" silica though, the name derives from the process it was taken from - and it is usually in the form of micro-sized globes or spheres. You'll end up with a "lump" or "pile" of silica.
To make lead glass, you have forgotten an important ingredient - lead oxide. So, solidify your silica glass and crush it. Take a batch of molten lead oxide dissolve silica the in it (there is something with "molten lead oxide" that is depressing and discouraging but don't let that keep you from your plan). One should respect lead and its oxide, especially when there is an appreciable vapour pressure of them - you'll be permanently heavy metal poisoned if exposed. So, as the viscosity of this mix will become higher as you add silica in it, you are going to have to apply generous amounts of elbow grease to get this mix. I take no responsibility for your elbows though, take care when sloshing around hot and sticky liquid-ish substances.
Now, with all that being said - use the right raw materials and it is much easier. Still not easy, that's why money can be made making glass - but easier.

Thanks for the answer. So the sodium oxide and all the other alkali and earth oxides evaporate off, and the sodium dioxide silica alone will remain as a molten heap?? And: that addresses part A. of the question, but can that pile of purr silica then be made into fused silica? No obstacles there right - just need enormous heat?

@Johan88, silica is not "sodium dioxide silica", but SiO2.

@Johan88 Well, fused silica is called fused silica because you take an amorphous source of silica, usually in the form of a fume, microscopic and condensed powder. If you use anything else they prefer to call it fused quartz. The properties are similar. Now, can you get to that purity by the above method alone? Possibly not, purity is so paramount for fused silica. Can it be done? yes. with more purification. It will be hideously more expensive.


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Organic Chemistry Help Please!

I'm not sure how to even start these two problems. How do you predict the structure of something when its molecular formula isn't provided to you? Also how do you calculate percent purity from this given data?

Look first at the C13 spectrum which is labelled up to show you have one methyl group, two CH2 groups in very different environments and a quaternary C. Then look at the proton spectrum which shows a methyl group with very small coupling with a shift suggesting it is attached to a double bond, 2 protons with a shift suggestive of being on a double bond, and a singlet CH2 with a shift suggesting that the carbon it is on has a heteroatom attached. The MS shows that the molecule contains Cl. This is not compatible with 3-Cl-1-butene.
I would suggest that the CH2 at 4ppm is due to a -CH2Cl group.
The two protons at 5ppm are due to =CH2 which is why they show little coupling. The very small coupling shown by the Me group is suggestive of allylic coupling.
If you draw out the possible structures of C4H7Cl the one that best fits is 3-Cl-2-methyl-propene.
The purity of the sample is most easily measured by comparing the integration of the Me group with that of the small doublet beside it which is the material named on the bottle, 3-Cl-1-butene.

Ahhhh that's what I missed, I didn't see that there was a Cl from the mass spec! Thank you!

One other question actually, why do we look at the integration of the methyl group and compare it to the small doublet beside it when it comes to analyzing purity? This was a "challenge" problem assigned by the professor and I don't think he went over this in our ochem2 class...

The integration is a quantitative measure of how much material is present in the mixture. If the value of the integral of the small doublet is 5% of the main Me peak then the sample is 95% pure.


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Monobromination of napthalene derivative

How does this reaction take place? Why does the bromine attach to that one particular location?
I do know that

since is $\ce{FeBr3}$ is present only 1 $\ce{Br}$ will attach
non-substituted areas of benzene will get $\ce{Br}$

Observe the carbonyl group. It is important

Does it do resonance? i mean the carbonyl group

It does with the NH, but it also deactivates the right hand right

oh now i understand. On right hand side there is more negative charge so bromine prefers left side.

But why it not attach to place adjacent to ch3? Is it because of steric hindrance?

Look up the the directive properties of acetanilide, which this substrate closely resembles.

This may be useful to you  http://websites.rcc.edu/grey/files/2012/02/Bromination-of-Acetanilide.pdf

You have electronically asymmetric rings. The right ring is deactivated by the carbonyl group that acts like an acyl group (remember that acyl groups are strong meta-deactivators). The left ring is activated by both $\ce{-CH3}$ and the nitrogen substituent acting as an amine (activating ortho/para directing), so bromination will occur on the left ring (since it is more nucleophilic).
Monobromination is not guaranteed experimentally. Although the right ring is deactivated under the right condition (at higher temperatures) you can end up with more than one $\ce{Br}$ being added. Maybe this question is just showing that a bromination will occur probably in that particular position.


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pH value of HF, hydrogen fluoride or hydrofluoric acid?
I've been looking for a source but I find different results when I try looking for an answer to this.
I'm looking for weight percentages of 1 to 10% HF (~0.5 to ~5.1M) and what their pH values are.
Now Honeywell says that around 5 to 10% it's pH is above 4.4
However if you calculate this using the $K_a$ of $6.8\times 10^{-4}$, you would get about 1.5 pH, which is quite the difference.
http://www3.imperial.ac.uk/pls/portallive/docs/1/7276108.PDF (Honeywell / imperial university)
This is just an example, because looking for these values online, for the same molar concentration, different answers are given of around pH 1 to 4.5.
So my question is can somebody explain how to properly get the correct pH values, or confirm what they are.
Also, secondary I would like to create a buffer solution with HF having a pH within the range of 4-7 (specific etching chemistry, $\ce{SiO2}$ vs $\ce{HfO2}$, not etching $\ce{HfO2}$). So other options to achieve this from stock $\ce{HF}$ (49%) with standard chemicals is also an option, however $\ce{NaF}$ is not readily available for me (mostly only access to solutions, such as $\ce{NH4OH}$).

Not directly pertinent to the question proposed, however...Simply would like to point out that if you do not know what you are doing/are not comfortable handling, HF can be VERY dangerous so please ensure you have the proper protective equipment [for example, simply inhaling a small amount can destroy your vocal cords...]

I very well know how to work with HF, we have all the facilities, there's just a bunch of different values regarding pH to be found (a value usually not necessary to know when working with HF). I need a confirmation

Sure, I didn't mean any disrespect, just wanted to make sure.

@H.Khan - Never apologize for bring up safety. It isn't discussed enough here. HF is a very very nasty chemical.

@H. Khan, Don't worry I understand that you meant well. I just wouldn't be working with it if I was not safe. I talked with the supervisor to get some pH indicators, because we're not allowed to bring in stuff that can tear. He had some special ones left so I was able to use those.

I was able to get some rough pH indicators that are allowed in our lab (cleanroom environment). No clue why a supplier like Honeywell has completely wrong data.
Apparently some sources are completely off, doing the calculation based -log(1/H+) gives the correct values in the end, tested for 50, 10, 5, 1 and 0.5%.
for a quick calculation see:
http://www.endmemo.com/chem/phcal.php

All the calculations based on a value for the $K_a$ of HF acid assume that the concentrations equal the activity of the species. For dilute solutions that is fine. For 5 molar solutions the assumption isn't so good.

How were the indicators validated?

These are special double indicator strips that do not leave particles, but it's still a visual inspection, so it's an approximate answer.
http://www.mn-net.com/Testpapers/pHTests/DUOTEST/tabid/10472/language/en-US/Default.aspx
for pH 1.0 to 4.3, however due to the HF, the range is more like 0.6 or 0.9 even when checking (held above and dipped with teflon tweezers).

Why don't you get a proper pH meter that is compatible with HF? More likely to give you reliable results than indicator strips that are often not very precise. Also, in a clean room enviroment easier to handle. Just make sure you use a pH probe that is suitable, otherwise it might get attacked and give false values. And always keep proper safety routines as well as medications close by (as mentioned by others).


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Distinguishing Brown Ring Test for Nitrates and Nitrites
My question is how will one detect the presence of nitrates by brown ring test when nitrites are also present in the solution?
Brown Ring Test for Nitrite
$\ce{NO_2^- + H+ -> HNO2}$
$\ce{3HNO2 -> H2O + HNO3 + 2NO}$
$\ce{FeSO4 + 6H2O -> [Fe(H2O)6]SO4}$
$\ce{[Fe(H2O)6]SO4 + NO -> [Fe(H2O)5NO]SO4 + H2O}$
Brown Ring Test for Nitrate
$\ce{2NO_3^- + 6Fe^2+ +4H2SO4 -> 6Fe^3+ + 4SO4^2- + 2NO + 4H2O}$
$\ce{FeSO4 + H2O -> [Fe(H2O)6]SO4}$
$\ce{[Fe(H2O)6]SO4 + NO -> [Fe(H2O)5NO]SO4 + H2O}$
Note: I personally think that it may have something to do with preventing oxidation of nitrites to nitrates and the release of NO (nitric oxide). But I am not quite sure how to go about that.

Yes, you are quite right. It has something to do with prevention of oxidation of nitrites. By controlling the pH, the nitrite present is reduced to nitric oxide ($\ce{NO}$) which reacts with ferrous sulfate to form a brown-colored complex ion, $\ce{FeSO4.NO}$. Nitrates do not react in such way and thus not registered in the test. It uses a special agent to prevent color formation or precipitation of common
interfering ions. Here is the relevant information[1]:

The high range nitrite test is a modification of the classical brown
  ring test for nitrate using ferrous sulfate. By controlling the sample
  pH, the nitrite present is reduced to nitrous oxide, which reacts with
  the indicator to form a greenish-brown color. Nitrates are not
  registered in the test. [...]
In an acidic medium ferrous sulfate reduces nitrogen in nitrite
  ($\ce{NO2–}$) to form nitrous oxide ($\ce{NO}$). Ferrous ions combine
  with the nitrous oxide to form a brown-colored complex ion, the color
  intensity of which is in direct proportion to the nitrite present in
  the water sample. Color development follows Beer’s Law.
$$\ce{2Fe^2+ + 4H+ + 2NO2‾ → 2Fe^3+ + 2NO + 2H2O}$$
$$\ce{NO + FeSO4 → FeSO4·NO}$$

Reference

https://www.hach.com/asset-get.download.jsa?id=7639984196


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Is Avogadro's number an integer?
I have heard that Avogadro's number, $N_\mathrm A=6.022 \times 10^{23}$, is the number of atoms contained in $12$ grams of $\ce{^{12}C}$. I think it should be an integer, but I couldn't find the exact number. Is it an integer?

There is no exact value. The value with something like a dozen decimal digits you might have seen in a textbook is as exact as it gets.

It is nonetheless an integer, for sure. You can't have half an atom.

@FelipeS.S.Schneider You're assuming that 12 g is such that an integral number of carbon-12 atoms go into it?

The old (1971 I guess) definition of mole that you stated surely leads to floating point numbers (although only if you measure weights with more than 23 significant digits). The new definition (2018) fixes one mole to be exactly $6.02214076 \times 10^{23}$, which is an integer.

I stand corrected. As @Felipe says, since very recently there is an exact value, and (not that it is of any slightest importance) it is an integer.

@orthocresol Yes, since I wasn’t absolutely sure whether the question is really about Avogadro’s number or actually about the Avogadro constant

@Felipe to reduce confusion could you turn your observation into an amswer?

Ok, I'll do it in the other question (sorry for being late!).


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Is Sodium Hydroxide a Bronsted-Lowry Base?
I know that sodium hydroxide dissociates into sodium cation and hydroxide, so it must be an Arrhenius base. I am also aware of the rule that all Arrhenius acids and bases are considered to be Brønsted-Lowry acids and bases. However, when it comes to sodium hydroxide itself, it seems as if it does not accept proton, but hydroxide does. And someone on this StackExchange stated that it is not a Brønsted-Lowry base, but an Arrhenius base. I am confused now. Is this an exception?

OH- + H2O = H2O + OH- acceptor and donator based on Bronsted and Lowry concept

All Arrhenius acids and bases are also Bronsted-Lowry acids and bases.
According to this website,

The Bronsted-Lowry theory doesn't go against the Arrhenius theory in any way - it just adds to it.

Remember, when we write $\ce{NaOH (aq)}$, what we really mean is $\ce{Na+ (aq)}$ and $\ce{OH- (aq)}$. Sodium hydroxide is a Bronsted-Lowry base because the hydroxide ions that are a part of sodium hydroxide accept protons.

Well, let's take very simple Brønsted-Lowry base $\ce{NH3}$, as it accepts a $\ce{H+}$ in acidic medium or in aqueous medium, so we call ammonia as a Brønsted-Lowry base, whereas $\ce{NaOH}$ (sodium hydroxide) is an Arrhenius base, as $\ce{OH-}$ from $\ce{NaOH}$ will accept $\ce{H+}$, but according to Brønsted-Lowry theory, in this case hydroxide ion should be the Brønsted-Lowry base not sodium hydroxide.

Sodium hydroxide dissociates into Na+ and OH- in solution - so they are effectively the same thing for aqueous medium

Well, it is a kind of wordplay, nothing much, if you follow the definition of Brønsted Lowery and Arrhenius than the answer will really make sense .


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Pinacole Rearrangement through epoxide

This reaction supposedly passes through an epoxide intermediate. Can someone please explain why this is necessary?
Source

Well, these types of questions are always answered with: run a mechanistic study and check. If it's the way it is, then it's necessary because that's what nature has chosen. To be fair, the mechanism isn't clearly defined here either because the reaction conditions are not clearly defined.

@Zhe This sites tries to show a possible expoxide mechanism-https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/rearrang.htm Would you like to say anything about it?

@Zhe I read the mechanism but couldn't understand it fully...

I don't have anything to say about it. It's conceivable, and we don't tell nature what to do because I think it's better. It does what it does, and we try to reconcile what happens with our simplistic rules.

@Zhe Anyways... I couldn't understand the necessity of epoxide formation... couldn't the group just migrate without forming eopxide as it did in the previous example?

Yes, there are several simple mechanistic possibilities, and all of them may operate to some extent depending on conditions. Only way to test would be to run experiments. Though, presumably, if you assert that the mechanism is such and such, you have run the experiments to rule out the others.

Epoxide formation is suggested to explain the scrambled labelled carbon (50:50) obtained during mechanistic study (as @Zhe told, only nature can explain why, but we can only speculate).

A phenyl shift is also mentioned. This can give a stable intermediate

As I mentioned above, during an isotope labelled mechanistic study, the formation of epoxide intermediate is suggested to explain the scramble isotopic carbon (between carbonyl carbon and neighboring carbon). The detailed arrow pushing mechanism is given bellow:

These steps are subjected to change according to many factors including the reaction conditions and nature of the diol used as correctly pointed out by Zhe and aforementioned MSU website in your comments, which states:

Many factors must be considered when analyzing the course of a pinacol rearrangement. These include:
• Which hydroxyl group is lost as water? or Which intermediate carbocation is more stable?
• What is the inherent shifting tendency (migratory aptitude) of different substituent groups?
• What is the influence of steric hindrance and other strain factors on the rearrangement?
• Are epoxides formed as intermediates in the pinacol rearrangement?
• Does product stability govern the outcome of competing rearrangements?
• Do the reaction conditions (i.e. type of acid, concentration, solvent and temperature) influence the course of rearrangement?

A reference for the rearrangement through an epoxy intermediate:
C. A. Bunton,  T. Hadwick,  D. R. Llewellyn,  Y. Pocker, Tracer studies in alcohols. Part III. Intermediates in the pinacol–pinacone rearrangement, J. Chem. Soc., 1958, 403-408 [DOI: 10.1039/JR9580000403] (http://pubs.rsc.org/en/Content/ArticleLanding/1958/JR/JR9580000403#!divAbstract).


text:
How many stereoisomers can exist for 4,5-bis(3-hydroxyoct-1-en-1-yl)cyclopentane-1,3-diol?

Q.10 For the given compound $\ce{X}$, the total number of optically active stereoisomers is ____.

This is from JEE Advanced 2018 second question paper.
I thought the answer to be at least 32 because of exactly 5 stereocentres present.  One of the double bond cannot show geometrical isomerism possibly.
But the answer given in the answer key is 7:

I did not find any plane of symmetry because of the different kind of double bonds.

Two issues. (1) The image is not clear enough; it's hard to tell exactly which bonds are wavy versus wedge. (2) You're misreading the question. The wedge bonds are fixed. You can change configuration only for the wavy bonds.

I'm so sorry for the picture quality. I can clarify it, The long chain on the bottom of the cyclopentane system contain the 2 wavy bond S,  all others are in dark bond.

Well, then there are 3 configurations to worry about, for a maximum of $2^{3}=8$. It can't be 32, but 7 is certainly reasonable. I'll let you figure out why it's 7 and not 8.

Okay,  but where did you get the third stereogenic centre there?  Those two in wavy and then?  Does it show synanti about that sigma bond in cyclopentane. If it is 8 then it will be easy because there must be a plane of symmetry.

There's no third stereogenic center. But if the double bond isomerism that generates a separate isomer, that isomer can be chiral. If it is chiral, it is optically active...

I think the answer should be $6$, if the trans configurations around the double bond is fixed. As there are two wavy bonds, so there are all total $4$ possible arrangements of them in space. Among them, if both are dark bonds, that is meso compound and optically inactive. and other three stereoarrangements will be optically active and each will give two enantiomers each. Note that the other chiral centres are fixed in absolute configuration, so you cannot include them also for counting enantiomers.

@Zhe Oh, I have misread your comment, in a sense "I'll let you know, when I figure out why it's 7 and not 8", and made an answer.

The question is tricky, because

the two ends of the wavy bond are both stereogenic units (chiral cyclopentane ring carbon atom, and the double bond), so there is some additional unintentional ambuguity, but it's not a big deal, because the legend at right explains the situation somewhat;
it in fact asks “How many … stereoisomers does this structure represent?” (because of some stereogenic units with unspecified configuration) instead of “how many … stereoisomers does it have?”.

The unspecified-configuration stereogenic units are two chiral carbons and one double bond, which give total 8 stereoisomers.

The first one (top left) has a plane of symmetry, the other 7 stereoisomers are optically active.

(Just note that the question has been changed afterwards, original one did not include the answer key link/quotation :-| )


text:
Why aqueous KI is used when collecting Iodine?

In room temperature solid di-iodine pentoxide($I_2O_5$) reacts with carbon monoside to form carbon dioxide & Iodine. This can be used to measure the amount of carbon monoxide in an air sample. $5dm^3$ air sample is passed through a tube containing solid $I_2O_5$. $\bbox[yellow]{Discharged \space I_2\space is \space sent\space into\space a\space container\space which\space has\space excess\space aqueous\space KI\space solution.} $ This solution is titrated with $0.005mol/dm^3$$Na_2S_2O_3$ solution( Starch indicator was used). If$10cm^3 of Na_2S_2O_3 $ was used, find the concentration of carbon monoxide in ppm. (density of air sample =$1.4\times10^{-3}gcm^{-3}$

Question: Why have they used KI solution when collecting $I_2$?
My View: Is it because $I_2$ is readily dissolving in KI ?
P.S. I already did the calculation. Just interested in the following phenomenon.
Thanks

Hi Emil, note we have mhchem for chemical equations. If you could use that, that would be great! (Use \ce{formulae} inside a MathJax expression)

Your guess is right. Pure water does not dissolve iodine very well, but a solution of $\ce{KI}$ does, thanks to the formation of $\ce{I3-}$ and other polyiodide anions. And of course we would prefer to have our iodine in a dissolved form when we are going to titrate it.
When it comes to the starch reaction or redox reaction with thiosulfate, $\ce{I3-}$ is as good as $\ce{I2}$.
So it goes.


text:
Why does blood turn reddish brown?
Why does blood take on such a colour after a couple of hours?

Is this the iron in my blood being oxidised to iron (III) oxide or is this something else?
I have checked this article, and no I do not find a definite answer to my question. I want to learn about the chemistry behind this change in colour of blood from bright red to brown. thanks

"As red blood cells degrade within a bruise, haemoglobin breaks down into bilirubin and biliverdin, and it is these pigments that pass through a series of colour changes. " https://www.quora.com/Why-does-a-blood-clot-change-colour


text:
Mechanism for the intra-molecular exchange of an ester and a nitrile group

When the isolated product 4 is treated with sodium ethoxide in ethanol it rearranges to 5 as shown:

My TA says that you need to start with a reverse Claisen at the ketone but I am not quite sure where to go from there. I do not know how to move the nitrile alpha to the carbonyl furthest from the methyl group.

Have the anion next to the CN attack the ketone. When the tetrahedral intermediate breaks down the anion of the ester leaves giving on workup 5

A good exercise to solving this type of problem is trying to map the carbons of the reactant onto the carbons of the product.

Following is a reasonable suggestion for the rearrangement:


text:
Book about biochemistry for physicist
I want to understand the basics of biochemistry, from a physicist point of view. I am looking for a textbook that discusses things like the covalent bond and the oxidation number, starting from first principles (quantum mechanics), and builds up to explain things like why ATP is an energetic currency of metabolism, the organization of glycolysis, and oxidative phosphorylation. Of course I do not expect to have detailed mathematical derivations of all of these. But I do want to end up with a firm grasp of the physical principles involved.
Maybe there is single textbook covering these areas, so I would also appreciate suggestions to build a "reading-list", which can include classic papers.
Thanks.

This question doesn't quite make sense. There is no physicist point of view in biochemistry. It's a chemical view of biology. There is no physical model capable of capturing the complexities here in a useful and familiar way. There's a reason chemistry is its own discipline, and not just physics.

@Zhe I disagree. There are many books tackling even more complex problems from a physical perspective. By this I mean that the discussion always carries the relevant physical principles  to the forefront and involves quantiative models. An example is the book Phillips, R., et al. Physical Biology of the Cell, which deals with very complex topics like signalling, protein folding, etc., but always maintaining the "physicist perspective". Do you see what I mean?

There are many possible books those that come to mind are  'Protein Physics' by  Finkenstein & Ptitsyn (publ AP), 'Applied Biophysics' by Waigh (publ Wiley), 'Molecular & Cellular Biophysics' by Jackson (publ Cambridge) & 'Molecular Biophysics' by Daune, (publ Oxford).

There is no way for you, for example, to capture the concept of nucleophilicity from a physical perspective. At some point, you've left physics behind to discuss chemistry. Physics is good at describing simple systems and aggregate systems (frankly, just like chemistry). But chemistry ignores the smallest scales to build abstractions of intermediate size systems that physicists don't care about. You asked about biochemistry. To ignore the chemistry and rely so heavily on physics means that you're not looking at biochem. It's either physics or biophysics.

@Zhe Why do you say that nucleophilicity cannot be captured from a physical perspective? There should be at least a simple example or model system where a discussion can be given in terms of the quantum mechanical principles involved.

Because it's inherently a chemical concept. If you feel that everything can be captured from a physical perspective, then I challenge you to explain an elephant from it. Even if you were able to, I think the insights you derive would not be the ones you were most interested in. There's nothing wrong with the physical perspective. It's just not the appropriate (or even feasible) lens to approach every problem.

@Zhe There are papers that calculate the nucleophilicity from first principles and numerical computations (ex. https://pubs.acs.org/doi/abs/10.1021/jp105914c). This means that the quantum mechanics principles involved can be stated precisely enough. I'd suspect that there are textbooks explaining the principles involved in an accessible manner. Probably some simple example, like an hydrogen atom or an approximate model of a larger atom, can even be solved analytically in some form. I am only trying to describe the intent of what I am looking for.

If you focus on the physics too much, you're going to miss the important features of the other discipline, in this case, chemistry or biology or their interplay. Those features don't have a physicist point of view. As someone who has studied both chemistry and physics, my warning to you is this: don't fall into that trap.

@Zhe Thanks. I don't want to miss those features. I am already studying books on biochemistry (such as Lehninger) which I think offer this perspective you mention. It's the physical aspect that I am lacking (specially since I am a physicist by training) and would like to read more about it

Physicists normally do not understand that chemistry in general and organic chemistry in particular is very empirical and has rules that are based loosely on ideas, for example, about charges being here and there in a molecule and reacting in this way or that. These rules  are based primarily on a vast amount of experimental data as the effects are subtle and cannot be easily computed with current technology. In proteins the situation is somewhat easier, only 20 amino acids, limited geometry of reaction site, seen via xray  diffraction,  and so on, so these are  more easily understood.

Your question will probably get closed because any answer is going to be an opinion. Instead of recommending a book-I'll shed a little light on what your journey will be like.
The Subject(s)
Your pursuit to describe high level biological functionalities using first principals covers many fields. I don't think you'll find a single book that explains it all. Instead, you're going to have to learn from a number of different sources.

Physical Chemistry - This will give an overview of electronic structure using quantum mechanics
Physical Organic Chemistry - Learn how to apply physical chemistry to organic molecules.
Bio-Chemistry - Apply organic chemistry to biological processes.
Physical Bio-Chemistry Combines physical chemistry with bio chemistry.

You're really after point 4, but it will need to learn a fair amount before getting there.
The Reality
The reality of things is that we don't need to dig down to first principals to describe many biological processes. Chemistry is able to abstract enough of the nitty-gritty away to come up with easier & faster high level concepts.
Take programming for example, web developers don't need to write websites in assembly language because someone has gone ahead and abstracted it all into a nice user friendly language. You could spend a lifetime coding a site like stackexchange in assembly-or you could use a higher level language and do it in a few months.
Just how you won't find a book on website design using assembly language (first principals), you probably won't be finding one describing biology using pure quantum mechanics.
Further Steps
Assess your current knowledge of physical and (general) chemistry. Once you know which level you're at, go to different university sites and check which books the professors are using to teach. If you find a course that is covering the topics that you're interested in-contact the TA or professor and ask for recommendations; I'm sure they'd be happy to help.

Thanks. If you can suggest some specific books that would be great, or at least University Departments websites that offer these kind of courses.


text:
What are the differences between D and L glucopyranose?
I have just studied the concept of cyclotron in carbohydrates. My reference book used Glucose as an example and began by saying we can find D-glucose or L-glucose:

Depending on what enantiomer we have, we will get opposite chair conformations at cyclisation:

After saying this, my book states that significant differences appear at cyclisation of these enantiomers, and there I just do not follow. The only difference I see is that one is the mirror image of the other. But visually, I see both structures identically equal.
Could someone explain to me what differences I should be observing in both chair conformations, and also, what consequences these have on the molecules' behaviors?

They're different. Therefore, they have different chemical properties.

They're as visually identical as a left hand and a right hand are visually identical.

They are mirror images of one another and therefore their chemical and physical properties are the same, except for the deflection of polarized light, which will be in opposite directions.
Mirror image compounds that are not the same are called enantiomers and it does not matter if they are in the open or cyclic form.
They will, however, have different properties if they are in a chiral medium.
For instance, if placed in a chiral solvent, they will have different solubilities.
Chiral means that the compound has a mirror image that can not be superimposed on the original structure.

When you state "their chemical and physical properties are the same", are you suggesting that they behave the same biologically, and that one can substitute L-glucose for D-glucose in one's diet?

No. Biological environments are chiral, so they can not be replaced. Many biochemical reactions occur only on one of the enantiomers.

Ah ha, makes sense.

Thank you, I knew about the polarizing of light and completely forgot about it until I read your answer. Is this the only difference when in a non-chiral medium? And when you say that the chiral means the compound's mirror image cannot be superimposed with the original structure -- could you give an example diagram or something?

@Bee  The two structures you have drawn in your question are like that. No matter how you turn them, they will never be the same. Also if you look at your hands, there is now way to get the same disposition, so a left hand will never fit into a right hand glove. In fact, chiral comes from the Greek kheír wich means hand.

But, both with my hands as well as the molecules I included, by just flipping one of the molecules over I get the other one. Is this not enough?

If you rotate one of the chair conformations by 180 degrees along the axis that goes through the center of the ring, everything will be at opposite sides. If you rotate it 180 degrees along the axis that goes through the screen, you will have to twist the molecule to the other chair conformation to have the oxygen in the same position. But then the substituents will be axial instead of equatorial. The Fisher projections can not be used for this because the third dimension has been eliminated.


text:
Radioactive Decay Process
Please bear with my long winded description.
It is classical to model radioactive decay of some particles $A \rightarrow A^\ast$ by the differential equation
 \begin{align}
 \frac{\mathrm dN_A}{\mathrm dt}=-\lambda N_A
 \end{align}
where $N_A$ refers to the number of particles in state $A$ and $\lambda$ is the rate of decay. If we initially start with
$N_0$ number of particles in state $A$ then by solving the above differential equation we see that
\begin{align}
N_A(t) = N_0\mathrm e^{-\lambda t}
\end{align}
gives us an explicit number of particles in state $A$ as a function of time $t$. In particular, we also see that
\begin{align}
\lambda = \frac{\ln 2}{t_{1/2}}
\end{align}
where $t_{1/2}$ is the half-life. However, the model is completely deterministic, meaning no probability required.
However, quoting Wikipedia

Radioactive decay is a stochastic (i.e. random) process at the level of single atoms, in that, according to quantum theory, it is impossible to predict when a particular atom will decay, regardless of how long the atom has existed.

But collectively, the mean number of particles in state $A$ is well modeled by the above deterministic model. However, the
difficulty here is to articulate the connection between the deterministic model and the stochastic model in a rigorous manner.
Let $T$ be a continuous positive random variable (RV) that denotes the lifetime of a single (microscopic) particle in state $A$. It is natural to make the assumption that in the stochastic model the decay of a particle in state $A$ is independent of the time that the particle spends in state $A$. With this assumption and a little bit of math (which is not important for the discussion), one can show that $T$ has an exponential distribution with a parameter $\lambda$, i.e. the
  PDF of $T$ is given by
  \begin{align}
  p_\lambda (t) =
  \begin{cases}
  \lambda\mathrm e^{-\lambda t} & \text{ if } t \geq 0,\\
  0 & \text{ if } t<0.
  \end{cases}
  \end{align}
  and the cumulative distribution function (CDF) is given by
  \begin{align}
  P(t\geq T) = 1-\mathrm e^{-\lambda t}
  \end{align}
  for $t \geq 0$ and 0 otherwise. In particular, we see that the probability of the particle in the decayed state $A^\ast$ at time $t$
  is precisely $1-\mathrm e^{-\lambda t}$ and the probability in of the particle in the state $A$ is $\mathrm e^{-\lambda t}$. Hence at each fixed time
  $t$ we have a Bernoulli distribution. If we make another assumption that a collection of the radioactive particles decay independently, then we see that given $N_0$ particles at time $t$ the probability that $N$ particles survives and $N_0-N$ particles decay is given by
  the binomial distribution
  \begin{align}
  P(N_t=N) = \frac{N_0!}{N!(N_0-N)!}(1-\mathrm e^{-\lambda t})^{N_0-N}\mathrm e^{- \lambda N t}
  \end{align}
  By standard computations, we also see that the mean of the binomial distribution is $\operatorname{E}[N_t]= N_0\mathrm e^{-\lambda t}$ and the variance is $\operatorname{Var}[N_t]= N_0\mathrm e^{-\lambda t}(1-\mathrm e^{-\lambda t})$ which leads to the relative standard deviation
  \begin{align}
 \frac{\sqrt{\operatorname{Var}[N_t]}}{\operatorname{E}[N_t]} = \frac{1}{\sqrt{N_0}}\left(\mathrm e^{\lambda t}- 1\right)^{1/2}.
  \end{align}
I have two questions.

Question 1: Looking at the relative standard deviation, it seems to me that for very large time $t$ the deterministic model seems to be a poor approximation of the stochastic model since the relative standard deviation grows exponentially in time. Is my assertion valid?

I actually disagree with the above calculation but have no valid argument against it.

Question 2: My interest in the above analysis comes from my desired to give a rigorous justification for using Monte-Carlo simulation of radioactive decay to arrive at the above deterministic model. However, I'm stuck after the above analysis. Could someone provide me with some reference or explanation to rigorous justify the connection?

For the Monte Carlo approach look at the Gillespie method (J. Phys. Chem. v 81, p2340, 1977) and his book 'Markov Processes' publ. Academic Press.

The average disintegration rate is given by $\bar n=\sum_{n=0}^{N_0} nP(n)$ where $P$ is the binomial distribution. If $M$ is the average number of events in time $t$ then $\displaystyle M=N_0(1-e^{-\lambda t})$. For times smaller than the lifetime $1/\lambda$  the $\displaystyle e^{-\lambda t} \to 1-\lambda t$ then $M=N_0\lambda t$ and as the disintegration rate is $M/t$ this leads to $\displaystyle \frac{dN}{dt}=-\lambda N$. Thus provided the measuring time-bins are small compared to the lifetime this rate law should hold.
As the decay proceeds the number of counts becomes small, perhaps $10^6$ at short times to just 1 or 2 at long times then of course the signal to noise becomes huge, zero counts in some time bins for example and 1 or 2 counts in others. (This becomes an issue in analysing data, see Anal. Chem., 2003, 75 (16), pp 4182–4187).
The expected variance is, as you quote, $V=\sigma^2=N_0e^{-\lambda t}(1-e^{-\lambda t})$ and can be rewritten as $V=Me^{-\lambda t}$ where $M$ is the mean number of events recorded in time $t$. As the time bins are normally short so that $\lambda t <<1$ then $\sigma=\sqrt{M}$ so that $\displaystyle \frac{\sigma}{M}=\frac{1}{\sqrt{M}}$

Does not look like a good method to me, the method is not a general method that can be done quickly with a pocket calculator. When the half life is very high when compared with the time of observation such as decays from 1 mg of radium-226 during one hour. The count rate is given by lambda N. But for dealing with a system where the observation time is similar or longer than one half life then the Bateman equations are normally the most common method used in the nuclear / radioactivity world. The method of using time bins is normally only used for things like multicompartment animal models.

This answer has nothing to do with the Bateman equations. By 'time bins' I mean the total time divided by the number of data points that are measured in an experiment or number calculated if that is what you are doing. Photon counting the emission from molecules is entirely similar to radioactive decay and a few thousand points is normally sufficient. Of course it does depends on how many orders of magnitude in intensity you need to follow the decays over, 4 decades would be typical. The matrix method I commented on (see below) is equivalent to the Bateman but based on linear algebra.

Have you read the masters thesis from the USAF, I am not very interested in the case of the single radionuclide with a moderate half life or hours or days. It is rather simple to measure its halflife. The interesting cases are mixtures of two or more radionuclides or complex chains of radionuclides. The time bin method is a poor first choice as it is often so computationally demanding, I see it as a last resort method. I would always try with the Bateman equations first, if they fail I would then try a "time bins" method

I had a quick look at the thesis, it describes standard known methods, nothing new; the moments method is known to be very poor, for example, and the matrix method effectively the same as in my comment. I think that you misunderstand what I mean by time-bins: you have to calculate data at discrete time points these are 'time bins' as they are discrete separated by $\Delta t$

Having run excel based simulations of radioactive decay many times, the problem I see is that anything other than the Bateman equation method requires a lot of cells to calculate a value. Can I ask you are you doing the simple system of a radionuclide decaying to a stable product or are you doing decay chains, it is not clear to me which of the two are you trying to deal with. Out of interest have you ever done radioactivity work ?

I normally now use python and formerly IDL,  MathCad,  or Maple to do simulations of up to 100 linked chemical equations. Others who do this seriously, such as for atmospheric reactions, have many  thousands of reactions , but all can be performed using matrix methods. I would never use a spreadsheet do to other than summing list of numbers.

There are several methods of dealing with the problem of radioactive decay calculations. I rather like the Bateman equations as a method. They can be used to calculate the amount of a radionuclide at time t, given the amounts at zero time and the half lives of the radionuclides in question.
This includes decay chains such as A --> B --> C --> D --> E --> F etc etc
The weakness of the Bateman equation method is that the dynamic range of a digital computer causes errors to appear. The core of the problem (or nucleous of the problem) is that in a Bateman equation we have terms such as (Lambda1-Lambda2).
When the two lambda values are very close to each other then error due to the way in which excel (or even more specialist maths software) subtracts one number from another causes some errors to appear.
I have on occasion tried to use Bateman equations to model things like the conversion of nuclides into each other in a reactor core. Such as
U238 --> U239 --> Np239 --> Pu239 --> Pu240 --> Pu241 --> Am241  --> Am242
But what happens is that the further we get along the chain the worse the errors become, after a while a spreadsheet starts to give an unreasonable output such as predicting an almost infinite amount of a later radionuclide. For the early radionuclides in the chain the spreadsheet works perfectly.
In general the greater the difference between the halflives the better an excel sheet using Bateman equations works. For example if we were to try to consider the system.
Pu241 --> Am241 --> Np237
As the half lives of the three nuclides are very different the lambda values are very different. In case anyone does not know what a lambda value is, it is the decay constant used in the equation.
A = Ao exp (-lambda t)
We get a lambda value by dividing ln(2) by the half life.
Here in our Pu241, Am241, Np237 system it would work very nicely.
If on the otherhand we were to attempt to consider
Y-95 --> Zr-95 --> Nb-95
As the half lives of the Zr and Nb nuclides are rather similar then excel is less able to make a good prediction.
There is a thesis written on this subject which was done by a USAF officer (Captain Logan J. Harr) who considered the evolution of mixtures of fission products in the seconds after a short duration fission event (Nuclear bomb detonation). This is an interesting but rather complex problem as the decay chains are typically very long.
The Monte Carlo method has been considered by others both on Stack exchange and elsewhere. I am sure that with plenty of computational power that the Monte Carlo method will work well, the disadvantage of it I think is that it is just a brute force attack on a problem. It is a rather clever bute force attack but it is still a brute force attack.
Another method would be to copy the method used to deal with multicompartment models for things like plutonium and lead in a human. Here if we treat a human as being four compartments, we can assume a series of first order reactions take plutonium (or lead) from one compartment to another. If we set up a spreadsheet such that we calculate the amount of metal in each compartment by assuming that the flow is constant over a series of short time slices, then it is possible with short time slices to get a good prediction.
For example here is one paper on the subject.

For a sequence of species such as you give as we usually are not interested in what  the exact equations actually are, it is easier to use a matrix method to solve the equations. If $K$ is the matrix of rate constants and $C_0$ the initial values then $C=Xe^{-t\lambda}X^{-1}C_0$ where  $X$ are eigenvectors, $\lambda$ are the eigenvalues of $K$ ,  and $e^{-t\lambda}$ a diagonal matrix of these values with zeros elsewhere.


text:
why does ring flip of L-fructose occur?

when drawing the chair conformation for beta-L-fructose, I got the structure above (in the picture). The answer said that the conformation above had a ring flip. Why does a ring flip occur and how can we tell when it occurs?

For starters, this isn't fructose in the picture ^

ummm, what do you mean? I'm pretty sure it is beta-L-fructose

You are missing an OH on the terminal (exocyclic) methyl group.

These ring-flips are called 1C4-4C1 transitions and these happen to all of the pyranose sugars in solution when the conditions allow. However, the abundance of each forms depends on their energy. There is a so called 1,3-diaxial interaction between the axial substituents of a cyclohexane analogue. Two axial substituents on the same side are much closer than two equatorial substituens. If these substituents are relatively big, it will be energetically unfavorable to bring them close together due to steric effects. Therefore the molecule will "choose" the conformation that has less energy, so less "large" substituents on the same side. In your case the OH and Me groups are considered the "large" substituents, while the H atoms (undrawn) are the "small" ones. Therefore the conformation you drawn has an unfavorable Me-OH interaction, which will drive the conformation change from 4C1 to 1C4.

(source: prenhall.com)

Molecules vibrate and twist all the time, the energy they have is more than just zero point energy because they in a 'heat bath' of other molecules at, say, room temperature. In the case of (non-aromatic) rings such as fructose or cyclohexane the activation energy needed to move from the chair to the boat form is not much greater than normal thermal energy so both these forms co-exist in equilibrium with one another but with the chair being favoured do to lower steric interaction as described in the answer by @fasekaszs. The ring strain is particularly small in 6 membered rings.


text:
Why is acetone less acidic than cyclohexanol?
The $\mathrm{p}K_\mathrm{a}$ of acetone is 19, whereas the $\mathrm{p}K_\mathrm{a}$ of cyclohexanol is 16.
However, acetone is resonance stabilized and its major resonance structure displays the negative charge on an oxygen atom, like cyclohexanol. Is there any qualitative reason to explain why acetone is less acidic than cyclohexanol?
Thank you very much.

Acetone is a CH-acid, and cyclohexanol is an alcohol. If anything, one should be surprised that the difference in their acidities is not much greater than it actually is.

@IvanNeretin I suppose what you're saying is that in acetone it is a C-H bond that is broken, which is much less polar than an O-H bond?

Yes, exactly. $!$

Not only less polar, but stronger.

The average C-H bond energy is $\approx 410 $ and the O-H $\approx 460$ kJ/mol so the result is surprising; a 1000 times difference in equilibrium constant but it seems it is the wrong way round?

Bond dissociation energies are usually for homolytic bond cleavage, which might explain the difference.

if you consider the structure of acetone, it has 2 $\ce{CH3}$ groups and a $\ce{C}$ double bond $\ce{O}$ ($\ce{C=O}$). Therefore, for the $\ce{H}$ to disassociate, it must be removed from the $\ce{CH3}$ group which leaves now a $\ce{C=O}$, $\ce{CH2}$, and a $\ce{CH3}$ group where the negative charge lies on $\ce{C}$ a part of the $\ce{CH2}$ group. Conversely, the cyclohexanol would have the negative charge lying on the $\ce{O}$ because the $\ce{H}$ a part of the $\ce{OH}$ group would be removed.
Just by simple electron negativity rules, $\ce{O}$ is much more favored to possess a (partial) negative charge than $\ce{C}$. Thus, cyclohexanol is a much more stable acid than acetone which is reflected in its $\mathrm{p}K_\mathrm{a}$ value.

While that is true, my point was that with acetone, the negative charge can be on an oxygen as well (in the major resonance structure). Therefore, it's not just the carbon atom that bears the charge.

Acetone has no resonance structures, it's enolate has. However the second one, with the negative charge on the carbon atom, is just very, very unfavourable. So there is no real stabilisation.
So both of your deprotonated species have a negative charge  on the oxygen atom, but coming from acetone, you also have to create an expensive double bond!


text:
Which alkene on heating with alkaline KMnO4 solution gives acetone and CO2?

Which alkene on heating with alkaline $\ce{KMnO4}$ solution gives acetone and a gas that turns lime water milky ?

2-methyl-2-butene
isobutylene
1-butene
2-butene

I know that $\ce{CO2}$ turns lime water milky but, I'm not able to find the alkene in the question. Also, I want the mechanism involved.

Permanganate chops the double bond, leaving oxygens on both ends, and then finishes off any aldehyde groups it can find. See where this gets you.

What diagrams? Draw a structural formula of any alkene on a sheet of paper. Cut it in half with scissors across the double bond. Glue each half on a separate blank sheet. Attach $\ce{=O}$ to the severed bonds. That would be it.

@Oscar Yeah, that's what I said in the first comment.

The best approach, and good practice, is to draw out the reaction products for each case.  You should know that alkaline potassium permanganate replaces a carbon-carbon double bond with two carbon-oxygen bonds, one to each of the original double-bonded carbons.  For instance, with 2-butene we have:
$\ce{CH3-CH=CH-CH3 -> CH3-CH=O + CH3-CH=O}$
Here the product molecules are the same because the alkene is, of course, symmetric.
But wait, there's more.  The carbonyl compounds pictured in the above example are stable in the alkaline permanganate medium if they are ketones, with only carbon atoms attached to the carbonyl function.  If there are aldehydes, the carbon-hydrogen bond(s) to the carbonyl group will be oxidized further.  Thus the "acetaldehyde" identified above for 2-butene is further oxidized to break the carbon-hydrigen bond at the carbonyl group:
$\ce{CH3-CH=CH-CH3 -> 2 CH3-CH=O -> 2 CH3-C(O)-OH}$
where in the alkaline medium the acetic acid will appear as acetate ion.
As you work through your choices you find that there is one aldehyde where the carbonyl carbon is bonded only to hydrogen atoms allowing that function to be oxidized all the way to carbon dioxide (or, again accounting for the alkali, carbonate or bicarbonate ion).  Find it, then see which choice gives that along with the other given product acetone.  Bonus question:  Would the acetone hold up against further oxidation?  How would you tell?

Can you give me the mechanism of what you stated in your first paragraph?

https://www.organic-chemistry.org/namedreactions/ozonolysis-criegee-mechanism.shtm.

In ozonolysis, $\ce{O3}$ is used as reagent, here $\ce{KMnO4}$ is being used.

Attempt 2, then -- https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.youtube.com/watch%3Fv%3D_lB3HcdPNGM&ved=0ahUKEwj0pNWMvdPbAhXr44MKHUrxDIIQwaICCEEwDA&usg=AOvVaw1OHhpJ0glNJcUQO_Tsz3tm.


text:
positive rays ?? (What and why)
Question :
We were recently studying the chapter on Atomic Structure and we had a topic about cathode rays  and the single particle of a cathode ray is called cathode ray particle.
The next topic was about +ve rays / canal rays and there was a discharge tube diagram showing the creation of positive rays and the diagram showed positve rays emitting from anode which is the expected thing (since cathode rays emit from cathode).
Our class teacher told us that the diagram was wrong and that positive rays don’t emit from anode and also they aren’t called anode rays.
Now I am confused whether my teacher is right or wrong. If he is right then what is the reasoning behind this
Teacher’s answer :
This is what our teacher told us :

positive rays are not called anode rays because they unlike cathode rays (which originate from the cathode) these are not produced by the anode.  Origin :
  These are produced by ionization of atoms or molecules of residual gas when bombarded with cathode rays.
Equation :  $$ \ce {e}^{-} + M → \ce {M}^{+} + \ce {e}^{-} + \ce {e}^{-} $$

Cathode rays and anode rays are quite different things.

@IvanNeretin : OOPs. That was a mistake on my part, updated

Your teacher was absolutely right, because anode rays are not generated from anode.
They are the result of fast-moving cathode rays that strike the gas molecules. They knock out electrons from gas molecules and thus become positive ions which start moving towards the cathode.
They are called positive rays due to their positive charge and canal rays as they were passed through a perforated cathode or canal.

Your teacher was wrong (or you misunderstood).
Anode rays = positive rays = canal rays, and they come out of anodes.
See the Wikipedia article on anode rays.
The whole business is tricky to memorize because long before you were born, people defined the charge of electrons as negative. This is how I remember my definitions:
(1) Electrons are negatively charged.
(2) Cations are positively charged. Remember that cats are cute and furry and cuddly, which is a positive thing.
(3) Cations are attracted to cathodes. Anions are attracted to anodes. That's why they are called that. So the electrodes have charges opposite to the ions they attract: anodes are positive (not enough electrons), cathodes are negative (too many electrons).
(4) Particles that leave an electrode do so because they have the same charge and are repelled. Anodes make positive particle beams, cathodes make negative particle beams.
I can't tell if the diagram you were given is wrong without seeing the diagram...

I can’t show the diagram it is on my book

"Anion = A Negative Ion" works better for me, even though I do think cats are a positive thing (amber vs cat fur included).

"Cat-ions are purrsitive" is how I remember it. Although I probably shouldn't.

Positive rays don't come out of anodes, it just looks that way. They are generated in the gas from gas molecules not from the anode.


text:
Is the ionic solution of a half-cell part of the electrode in a electrochemical cell?
A simple electrochemical cell consists of two half-cells, that is: a metal solid submerged in a solution. Additionally, there is salt bridge connecting the half-cells.
Ex: $\ce{Cu(s)}$ in $\ce{CuSO4(aq)}$ and $\ce{Zn(s)}$ in $\ce{ZnSO4(aq)}$
According to my text-book (Chemistry: Collins 3rd Edition, pg.506), the 'electrode' in a electrochemical cell is the whole half-cell.
"In work on electrochemical cells, however, the term electrode is extended to include what is known as a half-cell. So it refers not only to the conductor, but also to the conducting solution in which it is placed."
From other sources, I get the idea that the electrode is the metal solid (ex. $\ce{Zn}$), while the $\ce{ZnSO4}$ is an electrolyte (a solvent that seperates the cations and anions). Even when they are talking in the context of an electrochemical cell.
However, would not the saltbridge be considered the electrolyte as that is the compound that actually neutralizes the half-cells, through a seperation of cations and anions that travel to respective half-cells. While the redox reactions is actually happening between the $\ce{Zn(s)}$ and the $\ce{Cu^2+(aq)}$, which includes the metal solid and the ionic solution. As they are both conducting electrons.
Or are both the ionic solution and the saltbridge made of an electrolyte. Both consists of polar solvents, that seperate cations and anions. Therefore the electrode would be a metal solid in an electrolyte.
I guess the essence of my question, comes to down how to define electrode and electrolyte in the context of an electrochemical cell.

I think you should clarify this a bit. It seems to me this is about cases like glass electrode where you have it kinda "all in one".

Your question is about definitions.
I haven't read the Collins book, but, in general, the electrode is the metallic conductor that connects a circuit to the "non-metallic" part of the circuit. The electrolyte is this non-metallic part - dissociated ions that conduct electricity. Note electrolyte has broader uses than electrochemistry.
The term electrode can be used in reference to the whole half-cell since it's the site at which the reaction is taking place. At other parts, all we have is storage of reactant and product for the reversible reaction.
Consider the Standard Hydrogen Electrode(SHE), the whole half cell is referred to as an "electrode", though this is more convention than technical.
-
I suspect the Collins book is for UK A-levels; they should use the clear general definitions;
the electrode is the solid metal connecting circuit to non-metallic part, & the electrolyte is the solution of ions:
Here are some additional brief notes on contextual half cell setups:

SolidMetal/MetalSolution half-cell: Metal Electrode in metal ion solution.
All AqueousIons half-cell: Platinum electrode in solution.
Gaseous/AqueousIons half-cell: Glass/Platinum Deliver tube (for gas) bubble into solution.

The salt bridge is an electrolyte; A-level questions on electrode, electrolyte and salt bridge definitions aren't asked too often, consult your spec & exam papers. Importantly, solutions need to be $1dm^3$,  at $298K$ and $100kPa$, & salt bridge is typically $KNO_3$


text:
Modeling Belousov-Zhabotinsky reaction using ordinary differential equations
Is it possible to model the BZ reaction using first order ODEs? If so what would these equations be?
Or, is there something about the reaction that needs higher orders?

Yes, I did a project on this once - look up the FKN model, and for more complex schemes, the MBM and GTF models. Along with an understanding of kinetics it's not too difficult. The true mechanism isn't know, these are approximations, and are already quite complicated.

Maybe related: I had the chance to perform this very reaction about 3 years ago, for an undergraduate lab course, and there were many things which confounded me back then too. I have a question on the topic here. It contains an experimental procedure as well, in case you are interested.

Is it possible to model the BZ reaction using first order ODEs? If so what would these equations be?

TL;DR: Yes.
Many variants of the reaction exist. The only key chemical is the bromate oxidiser. The catalyst ion is most often cerium, but it can be also manganese, or complexes of iron, ruthenium, cobalt, copper, chromium, silver, nickel and osmium. Many different reductants can be used.
Similarly, there exist many models to describe this reaction. One that is the simplest, and I particular like is called the "Oregonator" (discussed herein). The precise mechanism/model, as far as I know, is still an open question.
We characterise the reaction using the following equations:
$$  \ce{A + Y ->[k_1]  X } \tag{1}$$
$$  \ce{X + Y ->[k_2]  P} \tag{2}$$
$$  \ce{B + X ->[k_{34}]  2 X + Z} \tag{3} $$
$$  \ce{2 X ->[k_5] Q} \tag{4}$$
$$  \ce{Z ->[k_6] fY} \tag{5}$$
This model is a simplified version of the FKN model, by applying the steady state approximation, and partial equilibrium approximation to the FKN model.
Overall, $$\ce{fA + 2B -> fP+Q }$$ wherein $f$ is a suitable stoichiometry co-efficient. For any, $f$
$$\begin{cases}
     (f-1)A + 2B \rightarrow (f+1)P & f > 1 \\
      2B \rightarrow 2fP+(1-f)Q &f < 1 \\
      B \rightarrow P &f=1
   \end{cases}
$$
Now, one can simply write down the rate equations
$$ \frac{\mathrm{d}X}{\mathrm{d}t} = k_1 AY  - k_2XY + k_{34}BX -2k_5X^2$$
$$\frac{\mathrm{d}Y}{\mathrm{d}t} = -k_1 AY - k_2XY + k_6Z$$
$$ \frac{\mathrm{d}Z}{\mathrm{d}t} = k_{34}BX - k_6Z$$
The system evolves with 2 different time scales, and we can expect it to show relaxation oscillations. We can make this more apparent via some mathematical shenanigans, which you may perform as an exercise. I might revisit this later in the day, and write a bit more about them.


text:
Systematic Approach to Finding Number of Isomers
I have this compound

I want to know how many isomers (including stereoisomers) may have this constitution. Is there a systematic approach towards solving such types of problems related to the total number of isomers?
Note: The above compound has 64 stereoisomers (from textbook).

Yes there is a systematic approach: there are 6 things which may be one way or another, hence $2^6$ isomers.

@IvanNeretin Do you mean the chiral centers and the double bonds?

I mean the double bonds and the atoms in the ring with attached side chains (the latter are also chiral centers, but that's irrelevant).

@IvanNeretin Ok Thankyou. But what about total number of isomers including the structural isomers also?

That number is a great deal greater and can't be estimated easily.

To my knowledge, there is no direct method or formula to find out how many isomers for given framework of an organic compound. Interestingly, this could be a good question for a mathematician. The only formula of this kind is the one to find theoretical optically active stereoisomers of a compound if the number of stereocenters (chiral centers) are given.
For instance, the given framework has 4 stereo centers, hence theoretically, you have $2^4 = 16$ stereoisomers for that compound alone. If your book says it has 64 stereoisomers, it probably has counted geometric isomers of conjugated double bonds (E/Z isomers) as well. All these 64 isomers are illustrated as compound 1 in the scheme below where each red asterisk represent a chiral center. Note that, since the ring has three different substitution groups, none would lead to symmetry in the molecule. Thus, all theoretical amount is equal to the actual number of isomers. Now, question is whether you need to count the stereoisomers for this framework only or not. If it is only for this framework, then the number is 64. But, if you want to change the framework by moving the substitution around, that would be a different story.

Now, for example, if you just move the carbonyl group around three other available carbons (say, move to the carbon between two hydroxy groups as shown in the compound 2 in the diagram), then you have additional 64 stereoisomers by that single move (compound 2). There are other 4 movable groups in the ring and all of them create multiple combinations such as 2,3,4,5-substituted cycloocta-1-one. Also note that chloro group and one of the hydroxyl groups can switch their positions to give another option. Or, the chloro group can just occupy the ring as well, creating another chiral center (e.g., 6-alkenyl-3-chloro-1,4-dihydroxy-8-methylcycloocta-1-one) or move to side chain. Upon all, there are changeable positions for the two double bond on heptyl chain as well (see compound 3). Double bond(s) in the side chain and an alkyl radical(s) can also move to ring (as methy, ethyl, etc.) are also possibilities. Thus, you are going to have tons of stereo isomer as any moving to the ring creating more stereo centers. Good luck on that, if you are not a mathematician.

If, for example,  you want to move substituents around the ring then this is the same as the number of ways of choosing 5 objects out of 8 which is $8!/(5!3!) = 56$,  then you have the stereo-centres to consider also, $2^4$ for each of these 56 and then multiply by the number of  E,Z configurations. Quite a few thousand.


text:
What is the purpose of the electrolyte in the half-cell where oxidation is taking place?
In an electrochemical cell that consists of two half-cells, there is an electrolyte and an electrode in each half-cell.
Example:
$$
\begin{align}
\ce{CuSO4(aq) + 2e- &-> Cu(s) + SO4^2-(aq)} \\
\ce{Zn(s) + SO4^2-(aq) &-> ZnSO4(aq) + 2e-}
\end{align}
$$
In the half-cell where the reduction takes place, $\ce{Cu^2+}$ gains two electrons. Here the electrode and the electrolyte serve a purpose, as it is the $\ce{Cu^2+}$ ion from $\ce{CuSO4}$ that goes through a reduction reaction.
However, in the half-cell where the oxidation takes place, it is $\ce{Zn(s)}$ that loses two electrons and forms $\ce{Zn^2+(aq)}$. Thus, the $\ce{ZnSO4}$ does not seem to play any role in the oxidation. Why is it included in the electrolyte in the zinc oxidation half-cell

The primary role of the anode chamber electrolyte (the anolyte) in this case is to provide sufficient conductivity in that chamber.
If deionized water were used in the anode chamber, the conductivity of that liquid would be extremely low. Thus, it's important for there to be something dissolved there in order for any appreciable amount of current to pass through the overall electrochemical cell. The overall electrochemical circuit involves both electronic conduction (in the wires and electrodes) and ionic conduction (in the two electrolytes), and it will not function efficiently/effectively if any element of that circuit is too poorly conductive.
That said: It's not crucial that it be specifically $\ce{ZnSO4}$, however.  $\ce{NaCl}$, $\ce{H2SO4}$, $\ce{NH4NO3}$, or a wide variety of other chemicals might work just as well (or even better!).  However, a sulfate salt makes sense here because (at least):

It provides relatively high conductivity
Its salts with metal cations, including $\ce{Zn^{2+}}$, are relatively water-soluble.
In this specific cell configuration, using sulfate makes sense because it's the same anion as in the $\ce{CuSO4}$ used for the catholyte, and thus you don't have to worry about counter-diffusion of different anions across the salt bridge.

It further makes sense to use $\ce{ZnSO4}$ specifically as the anolyte salt because it provides the same cation as you'll be forming from your anode reaction. This will result in a more stable electrochemical cell, because having a different, potentially electrochemically-active cation in the anolyte would most likely change both the thermodynamic and the practical operating characteristics of the cell.

If the anode and the cathode are connected through a metal wire, the electrons would go through the metal wire. In the anode chamber, the electrons (I assume) never interact with the electrolyte. Does then electrolyte serve a function?
Or is the anode's ability to oxidize and move electrons through the metal wire dependent on the electrolyte?
It seems in this case that not even the zinc ions, after oxidation, are dependent on the electrolyte, as it rather the solvent (water) that makes it able to neutralize with the ions from the salt bridge? (Not sure)

@BernardBengtsson See my edit; does that help clarify at all?

To clarify, as ZnSO4 has a high ionic conductivity, the negative ion in the salt bridge is better able at neutralizing Zn 2+ ions in the solution?

@BernardBengtsson Mmmm, no, that really doesn't have anything to do with it. If the anode-side electrolyte had minimal dissolved ions in it, then it would allow very little current to pass, and the electrochemical cell would not function well.


text:
Entropy change of evaporating NaCl
What is $\Delta S$ for this process?
$$\ce{NaCl(s) -> NaCl(g)}$$
(A reference that I could quote would be ideal.)

The heat of sublimation at 900 K is 211.3 J/mole

Under what conditions?  And do you want a numerical answer or an equation for its calculation?  If the former, a starting point is to google "standard molar entropy NaCl" and you'll find references to the standard entropies of both NaCl(s) and NaCl(g).  I would recommend the NIST Chemistry WebBook, which should be in the first few hits, as an authoritative source.  That will get you the answer under standard conditions.
If you want an equation, you can google "Third Law entropy calculation" and you'll find equations for integrating the heat capacity of the various phases up from 0 K to find the entropy at some other temperature.  In your case, since you only want dS, you just have to integrate between your two state points (where your NaCl is solid, and where it's a gas).
You'll also need those equations if you want a numerical answer at some other temperature than the standard entropy, because you'll need to integrate the heat capacities of NaCl(s) and NaCl(g) from 298K to the temperature of interest.  I would guess you can just use Cp of NaCl(s) at 298K, but for NaCl(g) Cp may change a fair amount.  Maybe the ideal gas approximation is good enough.  Anyway, fortunately the bulk of dS is going to be in the phase change anyway, and the standard molar entropy difference will get you that.

Thank you Christopher!*
I was looking for a number.
Background: I'm teaching a class (~undergraduate college level) that includes some thermodynamics. We discussed the lattice energy of NaCl (Madelung etc.) and how it describes the reaction NaCl(s) --> Na+(g) + Cl-(g); and how dissolving NH4NO3 in water results in a cold solution. So I meant to discuss the reason why NH4NO3 dissolves, but NaCl does not evaporate, even though both processes are "bad" energy-wise.
So I'd say standard conditions.
__________
*I'm told not to say that but I can't help myself.

Given all that, would you say it's reasonable to take standard molar entropies for gas (229.79 J/(mol·K)) and solid (72 J/(mol·K)) from the Wikipedia data page for NaCl, and say that ΔS = 229 - 72 = 157 J/mol/K?

Now I'm reading that NaCl(g) is actually mostly NaCl molecules, with some Na2Cl2, and no ions. Maybe I'm better off using a different example...


text:
What, if any, hydration states does sodium bicarbonate have?
Sodium carbonate has a few hydrates in addition to the anhydrous form. I haven't found any references online that assert any hydrates of sodium bicarbonate though. Do they simply not exist?

To my knowledge sodium hydrogen carbonate hydrates do not exist.
However, there exists a mixed hydrate of sodium carbonate and sodium hydrogen carbonate with the formula $\ce{Na2CO3 \cdot NaHCO3 \cdot 2H2O}$ (dihydrate of trisodium hydrogendicarbonate, $\ce{Na3H(CO3)2}$) that occurs naturally as the mineral trona.
The region labeled trona in the phase diagram below shows the conditions under which the hydrate can be crystallized from an aqueous solution of $\ce{Na2CO3}$ and $\ce{NaHCO3}$.

The picture is taken from US patent US7638109B2.


text:
KCl + CuCl2 in solution, is there a reaction?
I'm wondering if the reaction of $\ce{KCl + CuCl2}$ in solution (aq) will produce any products (have a net ionic equation/solid precipitate)?
Using the metal activity series, I see that potassium is the most reactive, so it should displace the copper and form KCl again, but wouldn't that mean there was no reaction taking place?
I've came up with this $\ce{KCl + CuCl2 -> KCl + CuCl2}$, but not sure if it is correct.

There will be $\ce{[CuCl_x]^{n-}}$ in solution where x=3,4,5 and n=1,2,3 depending on concentration. They are called chlorocuprate ions.

For a dilute solution of copper(II) chloride in dilute aqueous potassium chloride there will be no reaction.
There will not be a redox reaction even if the potassium chloride solution is more concentrated, the only reaction which will occur when the chloride solution is high is the complexation of chloride anions to copper cations to form chlorocomplexes such as [CuCl4]2-.
In the solution as there is no reducing agent it will be impossible for a redox reaction to occur.
There will be exchange of chlorides between the two compounds, this could be shown by a radioisotope experiment using chlorine-36. I am sure if one was to get chlorine-35 enriched potassium chloride and chlorine-37 enriched copper chloride it could also be done if we had a mass spectrometer suitable for the measurement. But the radioactive experiment would be more able to give a better result.
For the dilute solution as the copper(II) and potassium chlorides fully dissociate and become seporated (fully solvated) ion pairs the chlorides will be interchanged between the two metals. If we were to start the experiment with copper(II) chloride and radioactive potassium chloride (how about a Cl-36 radiotracer in it). Then if we were to combine the two in water, then to evapourate the mixtrue down to a mixture of solid copper(II) chloride and potassium chloride. Then if we were to heat the mixture and evapourate off Cu2Cl4 vapour (this is the most likely gas state form of copper(II) chloride) then the radioisotope would be distributed between the copper and potassium chlorides.
If we were to start with a chloride complex of a metal which very slowly exchanges chloride (and other ligands) with what is in solution such as RhCl3(H2O)3 and the radioactive chloride in the form of KCl. Then if I was to dissolve the two together and then to seprate again the two chlorides then I would expect the majority of the chloride radioactivity to remain in the potassium chloride.

I agree with the theoretical renditions of my colleagues.
I also suspect that there may have been a reaction performed by Glace using his prepared cupric chloride (which likely had a high chloride concentration and was in contact with copper metal at some point), which eventually strangely appeared to produce a precipitate.
I explain my presumed formation of what I surmise is copper oxychloride as a result of a comproportionation reaction from the action of copper metal acting on cupric chloride (see https://en.wikipedia.org/wiki/Copper(I)_chloride ) in the presence of a high chloride concentration creating a soluble CuCl2- complex:
Cu(ll)L + Cu  <--> 2 Cu(l)L  (L here refers to a chloride ligand)
So Glace’s starting solution could have been a mix of complexed cuprous plus cupric chloride.
Of note is that there also exists an equilibrium reaction between water and the aqua cupric complex providing an acid source:
[Cu(H2O)6]2+ (aq) + H2O (l) = [Cu(H2O)5(OH)]+ (aq) + H3O+ (aq)
The final piece of the puzzle is the electrochemical reaction consuming the created H+  (moving the above equilibrium to the right) activated by the oxygen in air in the presence of cuprous:
4 Cu+ + O2 (from air) + 2 H+ → 4 Cu2+ + 2 OH-   (source: Wikipedia reference above and also Eq (7) below)
A related reaction also occurs with ferrous, oxygen and a souce of H+ (see https://wwwbrr.cr.usgs.gov/projects/GWC_coupled/phreeqc/html/final-78.html  ).
Here is an extract from Wikipedia (see https://en.wikipedia.org/wiki/Dicopper_chloride_trihydroxide ) on the reactions associated with the preparation of copper oxychloride:
“ CuCl2 + Cu + 2 NaCl → 2 NaCuCl2 (eq.6)
6 NaCuCl2 + 3/2 O2 + H2O → 2 Cu2(OH)3Cl + 2 CuCl2 + 6 NaCl (eq.7) “
where Eq (6) produces a soluble cuprous complex which feeds Eq  (7), an electrochemical reaction.
Copper chemistry can be a bit surprising at times.


text:
Reading CGTF basis from Basis Set Exchange site
I'm making a simple electronic structure code as an exercise, and I want to confirm that my understanding about basis set data from BSE is correct.
The CGTF of H in aug-cc-pVTZ basis is given as:
H    S
 33.8700000              0.0060680
  5.0950000              0.0453080
  1.1590000              0.2028220

H    S
  0.3258000              1.0000000

H    S
  0.1027000              1.0000000

H    S
  0.0252600              1.0000000

H    P
  1.4070000              1.0000000

H    P
  0.3880000              1.0000000

H    P
  0.1020000              1.0000000

H    D
  1.0570000              1.0000000

H    D
  0.2470000              1.0000000

END
S-type GTF is simple, but I have questions about the number of other high-l GTFs.
P-type GTFs
I think the values
H    P
  1.4070000              1.0000000

mean that $\chi = (1.00000) g_{p}$. Then Does that means there are three p-type function ($x$, $y$, $z$)?
D, F-type GTFs
If that so, in the upper case. How D, and F-type GTFs? Specifically, if we consider D-type GTFs, are there six gaussians ($x^{2}$, $y^{2}$, $z^{2}$, $xy$, $yz$, $xz$) or five gaussians following the atomic convention ($xy$, $xz$, $yz$, $x^{2}-y^{2}$, $3z^{2}-r^{2}$)?

Yes it is correct that for $p$-type basis functions that you have all three $p$-functions, i.e. $p_x$, $p_y$, $p_z$.
For basis functions of larger angular momentum, you are also correct that there is two "different" conventions. The cartisean coordinate and the spherical coordinate convention.
In many electronic-structure programs both conventions are used, since the integrals are calculated in cartisean coordinates and then afterwards transformed to spherical coordinates, i.e. as you stated for $d$-type functions going from 6 basis functions to 5 basis functions.
In regard to make you own electronic-structure code, I would recommend keeping everything in cartisean coordinates, because this will make your life simpler. (Keep in mind that if you compare you energy with from a Hartree-Fock calculation with other programs you will have a slightly lower energy, because you have 1 more $d$-type basis function).
For a simple electronic-structure program a very nice guide for molecular integrals can be found here, http://joshuagoings.com/2017/04/28/integrals/


text:
Formulation of a suspension of Magnesium Hydroxide
For a school project of mine, I have to formulate a suspension of Magnesium Hydroxide which contains at least 50% of solid matter.
This suspension is prepared from Magnesium Oxide 98% with a granulometry of around 75 micrometers. Unfortunately, I don't have a way to crush it more than it is until it is the ultimate solution.
The hydratation process will be made prior to the addition of additives to maintain the whole thing in suspension.
As my particles are relatively large and heavy, I know I have to fight against gravity. For that, I need additives that have dispersive properties and allow the mix to have a decent viscosity while avoiding sedimentation.
But, as I'm more specialized in enviromental chemistry, I'm not very good at proposing which surfactant/dispersant would be the most suitable for this project even after in-depth studies of papers talking about this.
From what I know, I think a polymer could seal the deal ? Maybe Sodium Polyacrylate as I saw it works fine with Calcium. Some Polyelectrolyte ? Or even multiple additives ? And in what quantities ?
I don't even know if the project is viable without crushing the particles of MgO so if you have any idea, I thank you in advance for your answer.
S.Berthels


text:
How can we say whether or not an organic compound is soluble?
In the exam recently conducted by my school, they asked whether 2-hydroxyacetophenone is water soluble or not.

I thought that since there is intramolecular H-bonding in this compund, it won't be soluble. But my teacher marked it wrong. Also, when I checked it on internet, I saw in some websites it was mentioned to be water-soluble. My question is how can we say whether or not an organic compound is soluble?

Well, the hydroxyl group and (to a lesser extent) ketone oxygen are good for solubility in water, and it just turns out they are good enough to pull the unwilling hydrocarbon part into the solution. To be able to deduce that, think of acetone and phenol. As for chelation, this is not a thing at all unless you have a metal, which isn't the case.

@IvanNeretin, sorry I edited the q, by chelation I meant intramolecular H-bonding..

Well, intramolecular H-bonding might be slightly bad for solubility, but no more than that.

What's your threshold for calling something soluble?

The traditional rule of thumb has always been that 1 oxygen can dissolve 4 carbons.
In this case, the compound is $\ce{C8H8O2}$ which makes an oxygen:carbon ratio of 1:4, so it should be soluble in water.
The source of this rule of thumb is Dr. Pedro J. Victory, SJ, who taught us Organic Chemistry. That was a long time back but I recall this rule of him.
The context was liquid-liquid extraction.
I do not remember if he added a corollary for the case of esters, where you should count only one of the oxygens of every ester group.
We perform extractions with ethyl acetate all the time and there the ratio is 1:2.  Very probably, he did add that.
To support this rule I have searched for some solubilities at PubChem. On basis of this data, the threshold for soluble (as asked in a comment by @Zhe) would be around 10% (100 mg/mL).
                    mg/mL T (C)
acetone             completely soluble
methyl ethyl ketone 223    25
3-pentanone          43    25
2-pentanone          46    25
isopropanol         completely soluble
1-butanol            63    25
2-butanol           181    25
1-pentanol           22    25
2-pentanol           45    25
3-pentanol           55    25
methyl ethyl ether  230    25
diethyl ether        60    25
methylacetate       243    25
ethylacetate         80    25
n-propylacetate      19    20

I've never heard this before, but my experience doesn't go past undergraduate organic chemistry. Is it something from medicinal chemistry or the pharmaceutical industry?

@pentavalentcarbon I have updated


text:
If acyl chlorides are reacted with aqueous ammonia, will carboxylic acids also be formed?
As part of my A-level chemistry course, I have been taught that an acyl chloride will react violently with water to produce a carboxylic acid and hydrochloric acid. I have also been taught that an acyl chloride will react with an aqueous solution of ammonia to produce an amide and ammonium chloride. My question is, would the acyl chloride also react with the water in the aqueous ammonia solution? If so, this could decrease the yield of the amide, because some of the acyl chloride will react to produce the carboxylic acid.
I would appreciate any additional relevant details about the reaction, such as answers to the following questions (assuming that the acyl chloride does react with the water):

Is there an easy method to separate the amide from the
carboxylic acid?
Would it be best to use a concentrated ammonia solution, so that
there is plenty of ammonia for the acyl chloride to react with
instead of reacting with the water?
Is the reaction with water negligible compared to the reaction with ammonia?

Thanks in advance!

Acyl halides react with water but usually not violently. The acyl halides are normally not soluble in water so the reaction is slow. Then, when the reaction proceeds, acid is formed and the pH decreases, reducing the reactivity of water.
Ammonia and amines are much more nucleophilic and react much faster with the acyl halide. This reaction is so much faster that it can be carried out in water. In this context it is called the Schotten–Baumann reaction, which normally proceeds with high yields.
An excess of ammonia is required at least to absorb the HCl formed, but normally a huge excess is used. The acyl halide is more expensive than ammonia so we want to have it completely transformed to the amide. The excess of ammonia will also help to dissolve the acid, if it is formed in the reaction.
The other points:

To separate acids and amides, extract with alkali. The acid will go to the aqueous phase and the amide to the organic phase.  Or, if they are small, recrystallize it. The amides are less soluble than acids.
Huge excess of ammonia or simple amines is normally used.
The reaction of water is negligible under these conditions.

Thank you for your helpful answer, it all seems to make sense if the statement that "Acyl halides react with water but usually not violently. The acyl halides are normally not soluble in water so the reaction is slow." is correct, however I have seen several sources which appear to offer contradictory information e.g. "Ethanoyl chloride reacts instantly with cold water. There is a very exothermic reaction in which a steamy acidic gas is given off (hydrogen chloride) and ethanoic acid is formed." - https://www.chemguide.co.uk/mechanisms/addelim/water.html

Yes, but only because it is a very small acyl halide. When the size increases the solubility decreases and the reaction is slow. Then, once the reaction sets on, heat is generated, reaction rate increases, the solubility increases and you end up with a violent reaction. Therefore, if you have to react an acyl halide with water, you must use hot water and add the acyl halide slowly to it.

Most acyl halides react slowly with cold water, acetyl chloride is one of the exceptions. Acyl halides derived from aromatic acids e.g. benzoyl chloride are really quite unreactive. Ammonia can be bought in solution in non-nucleophilic solvents (e.g. THF) to ensure no carboxylic acid is formed in the amide forming reaction.

We could, of course, still get the carboxylic acid.  Given the amide formed from ammonia, hydrolysis with hot sulfuric acid solution.