text:
How to balance this redox reaction
How to balance this redox reaction (using half reaction method):
$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2 + H^2O}$$
My attempt:
$$\ce{Cu -> Cu(NO3)2 + 2e-}$$
$$\ce{3e- + HNO3 -> NO}$$
$$\ce{e- + HNO3 -> NO2}$$
So we multiple Cu half reaction by 2 to balance electrons and we get:-
$$\ce{2Cu + 2HNO3 -> 2Cu(NO3)2 + NO + NO2}$$
I would have solved further but the correct answer happens to be different.
Correct answer: $\ce{2Cu + 6HNO3 -> Cu(NO3)2 + NO + NO2 + H2O}$

Please avoid using Latex in titles due to searching issues. Also, you can get better formatting using the \ce command. Thirdly, what is your understanding of the half reaction method? Your equations don't balance so clearly something is wrong.

Your very first half-reaction looks somewhat suspicious: multiple N and O appear out of nowhere.

Your "correct answer" can't be correct. You have two copper atoms on the left and one on the right. You have 6 H on left and two on the right. And so on...

There is no way to "balance" this equation as written.  It has hydrogen on the left, but none on the right.

The point is that in a half cell reaction all the atoms have to balance. You just to add electrons as a "reactant" or a "product". (e.g. $\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^-}$)

How to balance this redox reaction (using half reaction method):$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2}$$

You really can't balance this equation for two reasons.
(1) There is hydrogen on the left hand side, but none on the right.
(2) There are two competing redox reactions. One produces $\ce{NO}$ and the other produces $\ce{NO_2}$. The overall reactions are:$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  ->  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$ $$\ce{3Cu_{(s)}  +  8HNO3_{(aq)}  ->  3Cu(NO3)2_{(aq)}  +  2NO_{(g)}  +  2H2O_{(l)}}$$
Thus the ration of $\ce{NO2_{(g)}}$ to $\ce{NO_{(g)}}$ is a kinetics problem. In concentrated acid the top reaction predominates and you get mostly the first reaction. In dilute acid the second reaction predominates and you get mostly the second reaction.
Now as an example let's look at the redox reaction of the first reaction.
Looking at the copper first we have: $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$
Just looking at $\ce{NO3^{-} -> NO2 + e^{-} }$ the charge balances, and the $\ce{N}$ atoms balance, but the $\ce{O}$ atoms do not balance and hence this is not a valid half cell reaction. The valid reaction is: $$\ce{e^{-} + 2HNO3_{(aq)}  -> NO^{2-}_{3} + NO2_{(g)}  +  H2O_{(l)}}$$
Using "redox" I can simply balance the electrons. So I need the Cu reaction as is, but I need to multiple the nitrite acid reaction by two. $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$ $$\ce{2e^{-} + 4HNO3_{(aq)}  -> 2NO^{2-}_{3} + 2NO2_{(g)}  +  2H2O_{(l)}}$$
Now that the electrons match, I can add the two equations:
$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  ->  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$

If I added the two competing reactions and divided them by common factor 2 I get $2Cu+6HNO_3 \longrightarrow 2Cu(NO_3)_2 + NO + NO_2 + 3H_2O$. Do you think this is the balanced equation.

No. Kinetics adds a whole different level of consideration. That is not a "general solution." It would be true for some very specific reaction conditions. (1) The concentration of acid couldn't change, and (2) the geometry of the Cu would have to sat the same throughout the reaction. So for a copper sphere the geometry of the Cu changes (surface area) as the reaction proceeds.  Thus to get your balanced equation the copper would have to be like a film on a glass side so the surface area stays the same.

The "wrinkle" with the Cu is caused by the fact that the Cu is a solid and not a liquid. Thus there are two "phases" in the reaction. If the reaction were $\ce{Cu^{+}{(aq)} -> Cu^{2+}{(aq)}}$ and then it would be a lot easier to control the Cu. The point is that we want to products to stay the same from the time the reaction starts till it finishes.

If you used just enough $\ce{HNO3}$ to dissolve the copper then it would be sort of like 90% of the reaction in strong acid and 10% of the reaction in weak acid. That would really be an interesting case to solve exactly. But again, you'd have to make some assumptions about the Cu geometry.

First, identify the oxidation and the reduction (which you successfully have done). Metallic copper is oxidized to Cu(II), in the form of a copper dinitrate. Then we have N(V) being reduced to both N(II) and N(IV). Assuming this is carried out in an aqueous solution, we can just work with the nitrate ion, and not the nitric acid molecule, and just work with the $\ce{Cu^2+}$ ion, and not the denitrate.
\begin{align}
\textrm{Oxidation: } \ce{Cu^0 &-> Cu^{II}} \\
\textrm{Reduction: } \ce{N^{V}O3^- &-> N^{II}O} \\
\textrm{Reduction: } \ce{N^{V}O3^- &-> N^{IV}O2}
\end{align}
Step 1: Now you need to find out how many electrons are needed for each of these reactions, which you have correctly done. Step 2: Then you balance each reaction with respect to both mass and charge. To balance for oxygen, you add water on the oxygen-deficit side. Then you can add protons to balance for hydrogen. If done correctly, your charges will be the same on both sides. Step 3: The last step is to multiply each reaction with a factor so that all three reactions involve the same amounts of electrons. The only way to you can directly add them together, is if each reaction involve the same number of electrons. In other words, find the smallest common multiplum of electrons, and multiply each reaction with the appropriate factor to get to that multiplum.

To balance a reaction, it's not just the electrons which need to balance, it's all entities - both electrons and atoms.
As Ivan mentions in the comments, your first big problem is that your half reactions aren't balanced with respect to atoms. You need to re-write them such that the number of oxygens/nitrogens/hydrogens on the right match the number of oxygens/nitrogens/hydrogens on the left. This may involve adding additional species (such as $\ce{H2O}$ and $\ce{H+}$ or $\ce{OH-}$ if you're in an aqueous solution). A hint here is that the $\ce{Cu(NO3)2}$ is really a salt, and can be thought of as "equal to" $\ce{Cu^{2+}}$ and two $\ce{NO3-}$
Once you have your half reactions balanced, then you need to balance their combination. This may involve playing around with proportions.
A final note, you may want to double check the "correct" answer. The one you give currently fails that key property of a balanced reaction - that is, having an equal number of each element on both sides. For example, there are two $\ce{Cu}$ on the left, but only one on the right. Likewise, there are 6 $\ce{N}$ on the left, but only 4 total on the right, etc.


text:
What is the electron configuration for Og with a charge of 2-?
I was writing a cool program that determines the electron configuration of elements for me, and I wondered what would happen if you went over 118 — the "limit" of my program.  So, what is the electron configuration after you go over 118 electrons?

I'm curious how do you handle Cu for instance. As a free atom in outer space, or as the metal in a solid? Or can you program do both?

@MaxW I'm only in 9th grade, so I haven't really learned that much about electron configurations. Do the configurations really differ for Cu based on its environment? My program takes the input of however many electrons an atom has, and spits out the electron notation along with the noble gas notation. So if the Cu in outer space has x electrons and the Cu "as the metal in a solid" has y electrons, I can input both x and y into my program, and it will give me  the correct (to my knowledge) notation for both x and y.

Yes. Cu in a metal is more like a Cu+2 ion. A gas of Cu atoms in free space doesn't conduct electricity. A solid piece of copper will.

@10Replies Not that it's necessarily relevant, but how does your program calculate the lowest energy configuration? Does it take account of 'anomalous' electron configurations such as chromium and copper?

@10-Replies - Great you're thinking about chemistry. The Periodic Table is the ultimate lego kit.

@bon I'm not sure if I understand what you are saying... But it uses the  Aufbau principle, the Pauli exclusion principle. And it doesn't know about special cases where the rules don't apply.

According to SCF Relativistic Hartree–Fock Calculations on the Superheavy Elements 118–131 J. Chem. Phys. 53, 2397 (1970), for neutral atoms:
119 $8s$
120 $8s^2$
121 $8p8s^2$
122 $7d8p8s^2$
123 $6f7d8p8s^2$
124 $6f^38p8s^2$
125 $5g6f^38p8s^2$
126 $5g^26f^28p^28s^2$
127 $5g^36f^28p^28s^2$
128 $5g^46f^28p^28s^2$
129 $5g^56f^28p^28s^2$
130 $5g^66f^28p^28s^2$
131 $5g^76f^28p^28s^2$
Later calculations reported in Electronic Configurations of Superheavy Elements J. Phys. Soc. Jpn. 65, pp. 3175-3179 (1996) are different starting with element 122:
122 $8p^28s^2$
123 $6f7d8p8s^2$
124 $6f^28p^28s^2$
125 $6f^48p8s^2$
126 $5g6f^48p8s^2$
127 $5g^26f^38p^28s^2$
128 $5g^36f^38p^28s^2$
129 $5g^46f^38p^28s^2$
130 $5g^56f^38p^28s^2$
131 $5g^66f^38p^28s^2$
However, Electronic structure of eka-thorium (element 122)
compared with thorium J. Phys. B: At. Mol. Opt. Phys. 35 (2002) 1693–1700 finds:
122 $7d8p8s^2$
specifically agreeing with the 1970 paper and disagreeing with the 1996 paper.
See A suggested periodic table up to Z[=]172, based on Dirac–Fock
calculations on atoms and ions Phys. Chem. Chem. Phys., 2011, 13, 161–168 for additional information.

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text:
Getting structural formula of complex molecules?
How is it possible to precisely map huge molecules? Somehow, scientists easily know the structure of a compound of any complexity.
What techniques are used to know the sturcture of molecules like this one, or complex proteins that can be constructed of thousands of atoms, like the ones used in intracellular processes? It is so common to see them drawn as structural formulaes or pictured in a video, with a fine per-atom precision... but how?
UPDATE
This update addresses the comments about the question being too broad.
Firstly, I want to say that for me this question is rather exact - here and there, in lots of places over the internet, I see structural formulas and 3D models of countless chemical compouds that claim to bear a 100%-precise depiction of the actual molecule structure (I have linked an example). And I am interested in how all this data is gathered and why is it 100%-correct? I am so used to the pictures of all these complex compounds that I take them for granted. But recently I became to wonder, how do we even get these, how this knowledge is received?
Secondly, although there's certainly a difference between a structural formula and a dynamic 3D model like in this video, it is not so important in this question. The fact that we can precisely depict any compound or molecule is not even in question, it looks like that there's no problem in doing so for any chemical.
The very exact question is: what is(are) the method(s) of getting such precise formulas and pictures for objects (molecules) that are not even visible to the naked eye or microscope?
Of course, there can be differences in how the mapping is done for different molecules, or some data, like the 3D model can be complemented by some computation conducted with help of a computer, but all in all, what is the basis of our confidence of being able to exactly know a structure of any molecule?
Though, especially, after reading the comments, I think that a proper answer could be too large and include historical data..

Your question is rather too broad in it's current form imo. Also are you asking about constitutional data (sequence) or conformational (3-D structure)?

I agree with @mith it's too broad now. There are many methods that can be used for structural elucidation, such as IR spectroscopy, NMR spectroscopy, X-ray crystallography, and mass spectrometry, not to mention a whole host of microscopy techniques. Often, it is a combination of these that are used to arrive at the correct structural formula.

Historically, before the technologies mentioned above were developed, chemists applied various degradation and derivatization reactions to produce relatively simple fragments that could be positively identified by their properties. By working backwards from these fragments to the original compound, one could infer its structure. Some of these historical structure elucidations remain among the classics of the chemical literature and illustrate the importance of logical thinking in chemistry.

We didn't really mean to say that the question was too unclear, or unspecific. It is actually a very good question (I mean it), but it is probably not too suitable for SE because any proper answer would be too long. There have been so many methods, used in the past and present, in order for chemists to actually find out what they were doing - as you see, the two comments directly above already listed multiple chemical and physical methods, but without even saying anything about what they are. So... it might be good to narrow the scope a little bit, or perhaps do some reading on your own.

Agreeing with @orthocresol there. There are a plethora of methods that one can use and historically has used. Some are still used today, some are not; some can do all the work by themself, some need assistance by others. Often, structures are drawn even though they are not truely proven just because it makes more chemical sense than anything else. It is nigh impossible to answer this properly unless one is very bored and has lots of time.

Okay, I really appreciate the explanations and insights. I think, the full set of these comments actually gives a very good answer to the question. Since a complete formal answer is unfeasible, the overview of that answer that you guys have inscribed, is fairly enough. I gather enough from your words as to how that happens. It would be good if someone could write a quintessence of all that matter as a short meta-answer, so I could mark it, but that is not necessary. Still, there are questions in this for me of course, but I see that they are not easy to tackle with.

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text:
Why is logP quite a bit lower in cyclohexane than in hexane?
http://www.rsc.org/suppdata/nj/b3/b303016d/b303016d.pdf
If anyone doesn't know the definition, logP is the the log of the ratio of a molecule in dioxane to water. So a high logP would mean the molecule is mostly in the organic dioxane layer rather than water. E.g a molecule with a logP of 2 would prefer the organic layer 100 times more than water, and would be considered hydrophobic/lipophilic.
Hexane logP=4.15, very hydrophobic and almost all of it will stay in the organic layer.
Cyclohexane logP=3.66, still very hydrophobic, but hexane is almost 4 times more hydrophobic than cyclohexane. Can anyone please try to explain why the cyclic molecule appears to like water slightly more than its straight chain counterpart?

https://en.wikipedia.org/wiki/Partition_coefficient - octan-1-ol not dioxan

Think of the hole in the water that the molecules occupies as a bubble. The bigger the bubble the greater the area over which surface tension is acting.
The highest volume to surface ratio is for a sphere. So hexane is like a rod and cyclohexane is like hexagonal pyramid. Guess which has a higher volume to surface ratio?

edit more information...
Not that it is exactly the same, but look at the densities of the liquids. The cyclohexane molecule occupies a small volume in the liquid as compared to hexane, hence it is denser, even though it has 2 hydrogen atoms less per molecule.
Now water has 18 ml/mole so we can figure out roughly how many water molecules are displaced by a molecule of each hydrocarbon.
                                                     # water
                                                     molecules
             density         molecular mass  ml/mole displaced
Cyclohexane  0.7781 g/mL     84.16 g/mol     108.2    6.0
Hexane       0.6548 g/mL     86.18 g/mol     131.6    7.3

I'm not fully convinced/haven't grasped this fully so if a bit more depth could be possible that would be brilliant!

@Aced125 - I added some additional information which might be helpful.

Ok, I don't seem to understand how a greater amount of water molecules displaced correlates to a higher logP though.

Neither hexane nor cyclohexane will participate in much hydrogen bonding. So essentially the molecule has to make a "hole" in the hydrogen bonds between the water molecules that would be there otherwise. So it is like the molecules need a "bubble" devoid of anything into which that can slip. Hexane requires a bigger bubble, thus takes more energy and is hence less favorable.


text:
Describing evaporation without entropy -- what about Helmholtz?
I have in mind a rigid, impervious, heat-insulating vessel containing a monatomic liquid and vacuum. The vessel has an insulating forcefield inside, exactly at the surface of the liquid. It can be toggled.
Step 1: The liquid is in thermal equilibrium. The field is switched off for a very short, predetermined time. Some portion of the molecules at the surface exceed some threshold of kinetic energy, enough to escape into the vacuum.
Step 2: Now the liquid is no longer in thermal equilibrium because some of its fastest molecules have escaped. The same can be said for the vapor, except it lacks slow ones. We wait until both regain equilibrium -- they will do so separately because of the forcefield -- and then turn the field back on.
We return to step 1. This time there is some condensation as well as evaporation.
We repeat steps 1 and 2 again and again. At each iteration we can always predict the proportion of liquid that will evaporate, and the proportion of vapor that will condense, based purely on the temperatures of the liquid and vapor, the volume of the liquid, the pressure of the vapor, and the length of time the field is off. We can do this without considering changes in entropy.
But the Helmholtz energy is what determines the direction of this process, and its calculation requires knowledge of the change in entropy.
So is there a problem with my model?

This is very similar to Maxwell's Demon.

What is the entropy change introduced by the switched force-field?

@airhuff - Hm. Very good question. Hm.


text:
TLC plate and Rf values

I have calculated the Rf values for 6 samples. How do I determine which is really Sriracha ?

I was tempted to vote close this as a homework question, but the problem posed is flawed and unanswerable as was given. That's worth pointing out to the OP.

I don't see how this is unanswerable. My interpretation of the diagram was that all spots were run at the spotting line, so the real compound is the one that moved up 2cm (the third one).

Tl;DR - You do not have enough information.

I cannot help but think that this question is not answerable. You have four samples: Huy Fong Sriracha and three "knock-offs". There are six lanes on the plate. All six samples contain potassium sorbate.
If the question was something more like: You have five mysterious hot sauces and one of them is Sriracha... Then the TLC plate makes sense. One of those poorly labeled lanes is sriracha and the others are the mystery sauces. You just need to pick the one that gives the same pattern of spots with the same Rf values.
You don't have that. You have a mismatch between number of lanes and number of samples. You don't know which one is the authentic standard. All of the sauces contain the potassium sorbate (and they should - its a common preservative). You've determined the Rf values and that is all you can do. Well, maybe not all.
I can tell that lane 1 and lane 4 probably contain the same sauce and likewise lanes 2 and 5. Other than that, Jan's comment is correct. The potassium sorbate should always be at the same Rf (and since it is ionic, that likely means at the bottom of the TLC plate).

I’m not sure where you took the certainty that all samples contain potassium sorbate. If I looked at the TLC in an unbiased manner, I would say that exactly one contained sorbate.

This is sort of odd so I'm just going to propose a solution. You really have to assume some quirky chemistry and a bit of logic puzzle analysis to "solve" this problem.
I think this is the sort of problem where if you really know chemistry that you're easily overcome with analysis paralysis. This problem has a number of chemical uncertainties which render the problem rather meaningless for a real chemistry problem. First the extract conditions are unspecified so it is very unclear what sort of species would be extracted from the sauce. Second the development solution is unspecified. Third the reagent used to develop the TLC plate is unspecified. Fourth sorbate is a preservative, not really a "key" flavor component, so making the analysis about sorbate is sort of weird.
So in order to "solve" the chemical part of the problem you have to assume ( ass out of u and me) that (1) sorbate is in all the sauces, (2) that sorbate is the only thing that the developer will reveal, and (3) the various sorbate species extracted, and/or sorbate interactions cause different Rf values in the sauces.
(1) isn't hard to believe for commercial sauces.
(2) is a really big leap since the sauces probably contain hundreds of chemicals.
(3) Isn't too hard to believe since even through potassium sorbate may be added to the hot sauce, it may be varioous sorbate species such as sorbic acid, or potassium sorbate, or sodium sorbate that might be extracted. Also the sorbate could be effected by pH changes in the development.
So all in all any "real" chemistry is ignored in this problem.
The last aspect is the logic problem within the chemical problem. The problem states that you ran 3 other sauces besides the "best" Sriracha. There are 6 samples on the plate, and you should have run the "best" Sriracha 2 times so that it is alongside each of the other three sauces.
Let's number the samples 1 to 6 from left to right. Sample 1 is a pure potassium sorbate reference. Both samples 2 and 5 must be the "best" Sriracha since these are the only Rf vales within 2-6 that line up, and the blob shapes are the same.
So:

Sample 1 is next to 2. (Potassium sorbate reference)
Sample 3 is next to 2. (i.e. 1st of 3 test sauces)
Sample 4 is next to 5. (i.e. 2nd of 3 test sauces)
Sample 6 is next to 5. (i.e. 3rd of 3 test sauces)

So Sample 3, test sauce "A", doesn't seem to have as much of the whatever makes Rf value of the Huy Fong Sriracha sorbate so high.
Sample 4, Test sauce "B", seems to be potassium sorbate.
Sample 5, Test sauce "C" seems to have something which inhibits sorbate traveling up the TLC plate.
Thus the whole problem has a decided slant towards a logic puzzle than a real chemistry problem.

I cleared the comment train here because some had been deleted already.  Let me know if you still need any of yours to incorporate into your answer.


text:
Are `dTDP-4-dehydro-6-deoxy-alpha-D-galactose` and `dTDP-4-keto-6-deoxy-D-glucose` the same thing?
Are dTDP-4-dehydro-6-deoxy-alpha-D-galactose and dTDP-4-keto-6-deoxy-D-glucose the same thing?
I am looking for the function of this enzyme in uniprot. It says

Catalyzes the synthesis of dTDP-4-amino-4,6-dideoxy-D-galactose (dTDP-Fuc4N) from dTDP-4-keto-6-deoxy-D-glucose (dTDP-D-Glc4O) and L-glutamate.
dTDP-4-amino-4,6-dideoxy-alpha-D-galactose + 2-oxoglutarate = dTDP-4-dehydro-6-deoxy-alpha-D-galactose + L-glutamate.

Note that glucose in the first line and galactose in the second line.
Or it is a typo there?

There are four stereocentres in hexoaldoses giving rise to a set of $2^4 = 16$ different molecules with the same connectivity; you can group these to eight pairs of enantiomers (perfect mirror images of each other). The enantiomers are labelled D and L depending on the orientation of the stereocentre which is farthest from the aldehyde group. In linear orientation, these are the following molecules:

(All images taken from Wikipedia; all D-isomers.)
In the order of reading these are allose, altrose, glucose, mannose, gulose, idose, galactose and talose. If you compare glucose (3rd in the first row) with galactose (3rd in the second row) you will realise that they differ only in the orientation of the hydroxy group in 4-position. The posh terminology is epimerism — D-glucose is the C4 epimer of D-galactose and vice-versa.
You asked about two labels where the sugar is preceded either by 4-dehydro-6-deoxy or 4-keto-6-deoxy. The agree in the 6-deoxy part which means that the oxygen on C6 is missing; you will find a methyl group $\ce{CH3}$ there instead. The labels 4-keto and 4-dehydro may also be confusing. The first says that a keto group is there instead of the hydroxy group, while the second says that two hydrogens have been removed from that position. The result is the same, instead of having $\ce{CHOH}$ as fourth carbon we have $\ce{C=O}$. This also deprives us of the stereocentre (a carbonyl group is never stereogenic).
Putting all of this together: Yes, the two are identical. If you start from the C4-epimers and remove the stereocentre at C4 it does not matter which you started from. Both galactose and glucose are common sugars so both choices are understandable (even though glucose is probably known even a tad more). I would, however, strongly object agains 4-keto-D-talose in the place of 4-keto-D-mannose (mannose and talose being C4-epimers) since mannose is much more known than talose.

Glucose and galactose are two different carbohydrates and these molecules are actually different. Both actually are isomers of each other, both with the formula of C6H12O6.
So the two biomolecules are different, not the same.

But this page suggests they are the same.

TanMath, please don’t answer if you really don’t know.


text:
Terminology for "proton stripping"?
This is going to probably be an easy question for an organic chemist, but I am looking for the proper terminology to describe the following reaction mechanism.
In particular, what is a more precise way than "hydrogen stripping" to describe the process where a photogenerated hole on an oxygen can pick off the hydrogen and its associated electron to create a radical that reforms to make formaldehyde?
Further, would I say something like "an HCO radical attacks an adsorbed methoxy to create methyl formate" or is there a more precise way to talk about it? The proper reaction names would be ideal.  Many thanks from a physical chemist who took orgo way too long ago!
EDIT: Thanks Orthocresol for the hint about hydrogen abstraction. Upon looking into things more, hydrogen abstraction aka hydrogen atom transfer (HAT), can be viewed as a subset of proton coupled electron transfer reactions (PCET). The difference is that HAT occurs with minimal solvent reorganization and between similar donors whereas PCET is more general where there can be heat transfer (i.e. it is diabatic versus adiabatic.) So I believe that PCET may be the proper name compared to "hydrogen/proton stripping".
One last question: does anyone know what the reaction between the HCO radical and the methoxy would be called?  Thanks again!

"Abstraction", perhaps.

Photolysis, maybe?

You're welcome for the "hint", but sadly that's all I can suggest, I have absolutely no experience with photochemistry. I hope you find what you are looking for.

Photocatalysis - reactions caused by light. The term seems to be predominately related to free radicals.
This is obviously a misnomer. The photon is consumed which isn't true for a "real catalyst."
I know that lasers have been used to create excited molecular states which then react. I investigated this for reactions in gas phase of different atomic isotopes of a chemical.
I think if you get any more specific than that then it degenerates into "commonly accepted names" for certain reactions. Sort of like you just have to know what acetic acid is. There is no way to "figure out" what the name means if you don't just know.

Right, I know that this is a photocatalytic process but that term is so generic that I want to find the proper terms. I could say something like "photo-oxidative homolytic cleavage" or something but that isn't as precise as "McMurry Coupling" which would say describe the coupling of 2 carbonyl molecules through loss of oxygen into a C=C bond, and so I was looking for the equivalent specific names here.


text:
How to know properties of ion Aluminum -27?
How do I know the charge/properties of this Ion Aluminum -27? All the information given is that Al has an atomic number of 13. I want to know how to know that it has an electric charge of +3 from the given information.

Your question doesn't make sense. If you mean aluminum-27 not aluminum -27 then it's not relevant to its electric charge. $\ce{^27Al}$ can have +1, +2, +3 etc. positive charges depending on how much energy you provide.

The $\ce{Al^2^7-}$ ion will be extremely reducing. You probably need to keep it in vacuum fixed to the negative pole of your stabilised DC wiring. However, it should make for very impressive lightning. Was that not the question you asked?

Given that a -2 ion will auto-ionize extremely quickly back to -1, he must mean the isotope, not the charge state.

Aluminum-27 refers to an isotope of aluminum. For almost all chemical reactions the isotope makes no difference in the chemistry. You need "exotic" (meaning very unusual) conditions for the isotope to matter.
Unusual conditions would mostly happen in the gas state. For example a lighter gas would diffuse faster. This was the basis for the separation of uranium isotopes at Oak Ridge during WWII to make the first atomic bomb.

The main isotope separation mechanism at Oak Ridge during the war were the Calutrons doing magnetic separation on an ion beam. Liquid diffusion, developed at the Naval Research Lab started to be used late in the war early in the process to make slightly enriched feedstock for the Calutrons. Gas diffusion was developed post-war.

For this you must know about noble elements like Helium, Neon,Argon,etc.
When elements except those I have mentioned above want to become stable they tend to form ions.
In order to solve your question, you must know the nearest noble gas to aluminum, which is neon with atomic number 10.
Aluminium atom naturally 13 protons and electrons each and it has to donate its 3 electrons to become like neon.
When it donates 3 electrons, the number of protons become more and hence positive 3 charge.

Did you understand the question, since I didn't. But if I understand correctly what you're getting at, Al can form ions of whatever charge it wants. It's just that its most common oxidation state in Al compounds is +3. Careful what your choice of words are.


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Does anyone know the particle size of carbamate and protonated MEA?
Does anyone know the particle size of carbamate and protonated MEA ?  that's the product when CO2 is absorbed into monoethanolamine (MEA)

There are no particles: everything stays in the solution.


text:
Why does fructose reduce Tollen's reagent and Fehling's solution?
Even though fructose is a ketohexose (ketone-containing hexose, a six-carbon monosaccharide), it reduces Tollen's reagent and Fehling's solution. Generally, a ketone does not reduces Tollen's reagent and Fehling's solution. So, why is this possible in the case of fructose?

The presence of the hydrogen atom in the aldehyde group gives reducing properties to the aldehyde molecule. Aldehydes are easily oxidized by mild oxidizing agents such as Tollens’ reagent ($\ce{Ag+}$ in aqueous ammonia), Fehling’s reagent ($\ce{Cu^2+}$ complexed with tartrate), or Benedict’s reagent ($\ce{Cu^2+}$ complexed with citrate). Ketones do not have such reducing properties and are not oxidized under similar conditions. Therefore, Tollens’ reagent, Fehling’s reagent, or Benedict’s reagent can be used to distinguish aldehydes from ketones.
However, fructose reduces such reagents even though it contains no aldehyde group. Reduction occurs because the reagents are basic solutions and fructose is readily isomerized to a mixture of aldoses (glucose and mannose) under basic conditions.

Under basic conditions, the proton alpha to the carbonyl group (aldehyde or ketone) is reversibly removed. If the resulting enolate ion reprotonates on the oxygen, an enediol intermediate results. Thus, the base-catalyzed enediol rearrangemet can move the carbonyl group up and down the chain.
Furthermore, in the enolate ion, the affected carbon atom is no longer asymmetric. Reprotonation can occur on either face of the enolate, giving either configuration. Thus, the base-catalyzed epimerization results in a mixture of stereoisomers.
Therefore, under strongly basic conditions, the combination of enediol rearrangements and epimerization leads to a complex mixture of sugars (see also: Lobry de Bruyn–van Ekenstein transformation).

However, everywhere, it is written only 'alpha-hydroxy ketones'. Not all alpha-hydroxy ketones are fructose, that is, not all alpha hydroxy ketones can double tautomerization to form an aldehyde.

This answer nicely elaborates the mechanism stated in words in this answer.

In the presence of very dilute alkalis, glucose, for example, is converted into a mixture of glucose, mannose and fructose. It is explained by the presence of an intermediate enodiol, and is also called epimerization, since a mixture of epimers is obtained. All monosaccharides are oxidized with Benedict, Fehling and Tollens reagents, also giving aldonic acids. These reactions do not serve to distinguish aldoses from ketoses, as occurred with aldehydes and ketones, since the latter also reduce the previous reagents, since, under the conditions of the tests, the ketoses equilibrate with the aldoses through an ene diol, as just seen. The carbohydrates that test positive for these tests are
called reducing sugars.

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text:
How can you boil milk without causing it to foam and overflow?
I have looked at this Why does milk overflow...
But I've not found a solution to prevent the milk overflowing.
I remember reading (cannot recall the source) that in industries where milk is boiled a that some sort of silicone treatment on the boiler prevents the milk from overflowing. Can someone elaborate on this?
What other methods can be used to prevent milk from overflowing when boiled?
This is a huge problem for housewives in India, who have to boil milk everyday.
[Unlike in the US, in India you can never be sure that the milk you buy is uncontaminated or even pasteurized. Also, milk is boiled to sterilize it, rather than pasteurized, this probably kills off all pathogens. Boiling it beyond sterilizing, boils off some smelly volatile substances that makes better yogurt]

A double boiler would help, but that would take longer and use more fuel. But without the bubbling due to the direct heat you're less likely to drive off the "volatile substances" to which you refer.

The classic "rest a cooking spoon on the surface" method seems to keep liquids from boiling over.
Also non-toxic, porous stones at the bottom of the pot ("boiling stones") prevent it.

I found this:
Wikipedia - Milk Watcher

The milk watcher was invented by Vincent Hartley in 1938... ...A milk
  watcher disrupts this process by collecting small bubbles of steam
  into one large bubble and releasing it in a manner which may puncture
  the surface film. The device also rattles when boiling occurs,
  alerting the cook who may then lower the heat setting of the stove.

Have to make one and see if it'll help. Reviews on Amazon indicate this device works.

Some tips, after experimenting with a DIY milk watcher made out of a small plate. 1) Make sure the plate has some uneven edges and holes to allow steam to escape. (2) Put a weight on the lid, like a heavy spoon to keep the plate horizontal. (3) Keep the heat source on low flame / temp; low enough that it does not overpower the ability of the punctured foam to allow steam to escape without the milk overflowing. This device works, but you need patience to find the right combination of plate size, heat intensity and vessel size. Happy experimenting!


text:
Chemical reaction for the flow color change from blue to red and again to blue
I am trying to demonstrate color change in the liquid as it passes through a channel to demonstrate blood getting purified after passing through certain organs of the body.
I want the initial color of the liquid to be blue in color and as it passes and comes out of a small pipe, it should change into red color and again after going through a small pipe, I want it to change back into blue color.
In short, blue to red to blue.
I have tried to search things online and I found the following.  There is a reaction which involves sodium hydroxide and sucrose. Under stirring upon addition of $\ce{KMnO4}$ the color change happens. But I am not quite sure if I will be able to handle stirring.
So, I am looking for something simpler than that. I am really not that sure if such a reaction exists.

Would passing through an ion exchange column be ok? $\ce{H^{+}}$ solution exchanged to $\ce{Na^{+}}$ solution in column 1, then exchange $\ce{Na^{+}}$ for $\ce{H^{+}}$ in column 2. So you need an indicator that is blue in acid and red in neutral solution. The ion exchange columns would work for a while. Then you'd have to take the system apart and "reload" columns using a flush with HCl acid and NaCl solutions to reload the columns again.

Instead of coloring and recoloring the liquid it is probably easier to color the channel the liquid flows in.

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text:
What is the mechanism for this nucleophilic conjugate addition?

What is the mechanism for this reaction-

My working so far:

But my final product doesn't match with the one given. Where did I go wrong (if at all)? How should I have proceeded?

Firstly, there is no way the -SH proton will survive in ethoxide. You have to take it off first. Apart from that, your starting is probably correct. However, after the initial addition, you probably wouldn't attack the carbonyl with the sulfur as that would form a disgusting four-membered ring with a positive charge on the sulfur that you can't get rid of. At this stage, you have to remember that you still have EtO- in your solution, and there are two ester groups in the molecule. What is likely to happen under these conditions?

@orthocresol will it deprotonate to form a conjugated carbonyl for the S to attack?

If you deprotonate a carbonyl group, you are forming a nucleophilic enolate. The sulfur is also nucleophilic. You need an electrophile.... somewhere..... is there one in the molecule?

Please see my edit- does this happen? @orthocresol

I can only see the deprotonation, did something happen with the upload? Anyway, it is a deprotonation that happens, but there's a subtle difference between the two esters. The other ester has a sulfur next to it, whereas the one that you deprotonated doesn't. Would a sulfur stabilise, or destabilise, a negative charge?

@orthocresol it can stabilise negative charge

@orthocresol could you please incorporate some diagrams because i am getting confused...

I'm sorry, I know it is pretty confusing. This is partly why I am not a fan of mechanism questions on SE.. or the Internet, for that matter. However, if you give me 30 mins to eat lunch, I'll write up a complete answer with some step by step instructions.

To figure out what's going on in an organic reaction, there are a few things you really should look out for:

what functional group(s) are present on the molecule(s);
what conditions the reaction is being run under;
which of the functional group(s) are most reactive under the given reaction conditions.

If you apply these steps to every stage of the reaction, then deducing the mechanism will be easier, and you also save yourself a lot of time by not having to memorise a bunch of mechanisms that you have been previously taught. If those mechanisms were taught by a competent lecturer, you should be able to apply the three steps above to see exactly why those reactions happen.

The solution (courtesy of @Jan)
So, without further ado:

What functional groups are present on the molecule?

That's easy.
$\hspace{30 mm}$

What are the conditions?

That's easy too. $\ce{EtO-}$ is strongly basic, but it can also act as a decent nucleophile.

Which of the functional groups are most reactive in base?

Well, to begin with, the thiol is acidic, so it will be deprotonated easily. No question about that.
Now you have a couple of electrophiles - the two esters - as well as a couple of nucleophiles: the thiolate anion, as well as the ethoxide anion. A hard base such as ethoxide is not likely to react in a conjugate addition. Maybe it might react directly with the esters at the carbonyl carbon? But wait! Both esters are ethyl esters $\ce{RCO2Et}$. So, you can use ethoxide to attack those esters all you like, but you won't get anywhere doing it - you'll just get back the same product. In fact, that's exactly why we use ethoxide instead of a different base like methoxide, $\ce{MeO-}$: to prevent transesterification.
So, let's talk about the thiolate anion instead. It's pretty nucleophilic, and you also have an electrophilic α,β-unsaturated carbonyl, so these can react:
$\hspace{15 mm}$
Now, you have a nucleophilic enolate, as well as an electrophilic ester on the other side. They can immediately react with each other to form the product that you are given:
$\hspace{25 mm}$
This reaction has a name - it is called a Claisen condensation, but that's not important! The important part is that you realise that it is just the reaction of a nucleophile with an electrophile.

An alternative (my original answer)
There might actually be another possible product, though. The enolate that you formed above could pick up a proton from somewhere (either ethanol, or perhaps another thiol molecule) to give the conjugate addition product, which is exactly the same as the one you drew. However, your initial addition product had a positive charge, which is never a good idea in a basic medium. Under basic conditions, you should always deprotonate first and reprotonate at the end of any reaction.
If this were to be the case, you should probably go through the three steps again:

Functional groups?

That's still pretty easy. You have a sulfide and two esters.
$\hspace{45 mm}$

Conditions?

Still basic.

Most reactive functional group?

Well, the sulfide is indeed nucleophilic and you could use it to attack either one of the electrophilic carbonyl groups. But there are a few things that don't really favour this:

Sulfur is a pretty soft nucleophile. It might love to react with $\ce{CH3I}$, but probably wouldn't like a carbonyl so much. In fact, this is the reason why the initial step consisted of conjugate addition instead of direct addition.
There's no more hydrogen on sulfur that can be taken off, so if sulfur does attack the carbonyl, it will be stuck with a positive charge. This makes the sulfur a pretty good leaving group, which means that any addition to the carbonyl group will be highly reversible.
If sulfur attacks the carbonyl group further away from it, it will form a four-membered ring. This is not particularly favourable. If it attacks the carbonyl group nearer to it, it will form a three-membered ring, which is even worse!

So, the sulfur is stuck where it is. You are better off thinking about the reactivity of the carbonyl group in base. You should know that they can be deprotonated to form enolates. Does it matter which carbonyl group is deprotonated? Of course it does. This molecule is not symmetric, so the two carbonyl groups are not equivalent. This obviously leads to the question of which carbonyl group should be deprotonated. Let's look at the possible anions formed from the deprotonation:
$\hspace{10 mm}$
The first of these two is more stable than the other, because the electronegative sulfur atom is capable of withdrawing electron density from the negatively charged carbon. (Actually, there are several theories as to why sulfur stabilises the carbanion, but I won't discuss that here. See this paper, for example.) So, it is that first conjugate base that will be preferentially formed.
At this stage, you have a good nucleophile (the anion) as well as a good electrophile (the carbonyl on the other side). So, what's left is to react them in the same Claisen condensation as above:
$\hspace{25 mm}$
There we go, side product.


text:
Are primary, secondary, and tertiary amines considered functional group or chain isomers?
Consider $\ce{C3H7NH2}$.  Can one make $3$ different isomers with this molecular formula? My question is that are they functional group isomers or chain isomers?
I have been told that they are functional group isomers but I intuitively think that they are chain isomers.

I can make four different isomers of this.

Yes I know we can , but my question is are they functional group or chain ?

When we talk about chain isomers we mean isomers with different carbon skeleton.
If the carbon skeleton is changing as well as the functional group we report it as functional group isomerism.
For the given compound in your question, primary, secondary and tertiary amines will show functional group isomerism, as they are forming different types of amines itself.

That's incorrect - for amines it can be considered position isomerism, definitely not functional.


text:
What are most common condition for reduction of hematite to magnetite?
I found that hydrogen can reduce $\ce{Fe2O3}$ to $\ce{Fe3O4}$ under anaerobic conditions, but since I dont have $\ce{H2}$ in my lab what should I use to convert hematite into magnetite? and what conditions are used - temperature, pressure, partial pressure of oxygen and other important factors.

There is an old saw "If you have to ask how much it costs, you can't afford it." This is an analogous situation. Such chemistry would only happen under exotic conditions. If you have to ask what such conditions are, then you don't have sufficient knowledge to undertake such an experiment SAFELY.

We only have Argon and CO2 I am not sure why you think Argon is dangerous? It is inert gas so I dont understand your opint?


text:
How to identify if a compound is organic or inorganic?
What is the actual basis of distinction of organic compounds? The presence of carbon is not the only basis.

This isn't trivial nor standardised. Check out my answer to a different question. And you might also be interested in this question

There's also http://chemistry.stackexchange.com/questions/6690/what-is-the-definition-of-organic-compounds

In short, if you read of it in an organic chemistry textbook, it is organic.

The organic compound consists of C-C or C-H bonds.


text:
Why can't Pd/C and H2 reduce both the alkene and carbonyl portions of α,β-unsaturated carbonyls?
Why is it that the major product of the reduction of chalcones the ketone and not the monoalcohol? In other words, Why isn't the major product a benzyl alcohol?
From what I understand, catalytic hydrogenation can be used to reduce carbonyls as well as alkenes.
My TA told me that nucleophilic hydrides are preferred for reducing carbonyls. Why wouldn't hydrogenation work as well? Does it have to do with resonance involving the carbonyl since in a chalcone, the carbonyl is adjacent to an aromatic ring and is also conjugated with the alkene? Could the reason that catalytic hydrogenation can't effectively touch the α,β-unsaturated carbonyl be the same reason catalytic hydrogenation can't effectively reduce carboxylic acids, esters, and amides — all of which are also resonance stabilized?
My first thought had to do with heats of hydrogenation and how resonance-stabilization found in carboxylic acids and its derivatives reduce the heat of hydrogenation. However, later I found a resource online implying that all pi bonds - even the delocalized, resonance-stabilized ones found in benzene — could be reduced through catalytic hydrogenation with enough time.

In addition, it was noted elsewhere that chalcones could be completely reduced to the benzylic alcohol, although not with ease.
This leads me to believe that there is an activation energy barrier that's impeding the hydrogenation of certain substrates within the confines of a 3 hour undergraduate lab period … am I on the right path? Sterics, perhaps? It was noted that the more highly substituted an alkene is, the slower it is reduced because of the difficulty in getting a highly substituted alkene to approach the catalyst's surface with the $\ce{M-H}$ bonds in an appropriate manner.

In your question,  do you mean reduce down  to the allylic alcohol or the saturated one?

@Beerhunter - the allylic alcohol.

you are considering the chalcones to break down into two parts making allylic alcohol as one product or something else? enlighten me

I am considering the "complete" reduction of the chalcone. In other words why isn't the product an allyl alcohol with no double bond?

It is the vacation now and I do not have access to my university library - however, March's Organic Chemistry 6th ed has a table listing the same reactivity order you have given above towards catalytic hydrogenation (except the aldehyde and alkene are swapped). The reference given is House, Modern Synthetic Reactions, 2nd ed - so there might be info in there - and if nobody gets around to doing it, I will try to do some research when I am back in the UK. Hudlicky, M., J. Chem. Educ., 1977, 54, 100 also contains many literature links.

What about a symmetry argument? A C=C bond has $D_{ooh}$ symmetry, and a C=O bond has $C_{oov}$ symmetry, whereas an H-H bond has $D_{ooh}$ symmetry.
So, wouldn't the group orbitals of H-H vs. C=O transform under different IRREPs? We can assume that $D_{ooh}$ and $C_{oov}$ "reduce" to $D_{2h}$ and $C_{2v}$ symmetry respectively.

You hit it right on the nose. The real key piece of information is that given enough time, all the unsaturated bonds will be reduced. This tells you that though the reduction is thermodynamically favorable, it is the difference in the energy barriers ($\ce{\Delta \mathrm{G^{‡}}}$) that prevents the carbonyl reduction from occurring at the same rate as the alkene reduction. This means to produce the alcohol faster, we must manipulate the kinetics of the reaction.
To understand why the carbonyl-reduction transition state is higher in energy, we should consider the differences between the carbonyl and alkene bonds. The bonds are much more polarized, and as such the carbonyl $\pi$ bond is considerably stronger than alkene ($93\ \mathrm{kcal\ mol^{-1}}$ vs. $63\ \mathrm{kcal\ mol^{-1}}$)$^{\mathrm{[1]}}$. This means that at low temperatures, only a small fraction of the molecules have enough energy for the carbonyls to associate onto the $\ce{Pd/C}$ surface. Catalytic hydrogenation of carbonyl compounds (aldehydes, ketones, and especially esters) requires high temperatures and pressures to increase the presence of both the substrate and hydrogen on the catalyst.

$^{\mathrm{[1]}}$ Fox, M. A.; Whitesell, J. K. Organic chemistry; Jones and Bartlett: Sudbury, MA, 1997.

Implication being, adsorption requires the breaking of the pi bond?

@orthocresol I believe so. At any rate, the bond must be broken at some point in the reaction.

This issue is discussed in the following references:
Hydrogenation of chalcones using hydrogen permeating through a Pd and palladized Pd electrodes Electrochimica Acta vol. 55, pages 5831–5839.
and
On the role of promoters in hydrogenations on metals; α,β-unsaturated aldehydes and ketones Applied Catalysis A: General 149 (1997) 27-48
the latter of which says:

Palladium is a very good catalyst for the C=C or $\ce{C#C}$ bond
  hydrogenations, but a very bad catalyst for hydrogenation of carbonyl groups. It is known that this is due to a too weak adsorption of carbonyls under reaction
  conditions [reference 28]. The mentioned weakness of adsorption through the carbonyl
  group could be in its turn caused by the change in the electronic structure of the
  Pd surface atoms ($4d^{9.7}$ $5s^{0.3}$ $\ce{->}$ $4d^{10}$ $5s^0$) induced by the hydrogen atoms in the interstitial positions [reference 29].

Reference 28 is: Concentration Dependence of Ketone Hydrogenation Catalyzed by Ru, Pd, and Pt. Evidence for Weak Ketone Adsorption on Pd Surface  Bull. Chem. Soc. Japan, 55 (1982) 2275.
Reference 29 is: V. Ponec and G.C. Bond, Catalysis by Metals and Alloys, Series: Studies in Surface Science and Catalysis, Vol. 95, Elsevier, Ansterdam, 1995. (google books link)


text:
Why the backbone of polylactic acid's repeating unit contains only one oxygen?
The monomer unit of polylactic acid is $\ce{CH_3-CH(OH)-CO_2H}$:

So why the backbone of the repeating unit is $\ce{-C-C-O -}$:

and not $\ce{-O-C-C-O -}$?

Can you draw a mechanism for the condensation of two molecules of lactic acid?

@orthocresol what do you mean?

How is the polymer formed from its monomers? Can you draw a mechanism for this?

@orthocresol both ways seem fine. in other words, why the polymer doesn't have O-O bonds all over? why water is released and not H2?

I believe there is only one way that "seems fine". You need to be able to draw a mechanism. I do not mean "draw the products", because if all you look at is the product, it will be hard to tell what makes sense and what does not make sense. So, I mean "draw a mechanism", with all the curly arrows and whatnot. If you can do so, then you will see why you do not form O-O bonds all over the place.

@Sparkler If you want a rule of thumb, O-O bonds are a big red warning sign, so is releasing $\ce{H2}$ (unless you have a metal around, which is not the case), and combining both of them together is not just impossible, it is ten times impossible.

a hint: PET (but here, you have single monomer)


text:
Why is the bond angle of AB3E1 approximately 109.5?

Since there are 3 B atoms and the angle between each atom is equal, shouldn't the bond angle be equal to 120?

Your drawing shows you why.

How does your second statement follow from the first. You seem to be ignoring something else around the central atom.

Aren't all 3 B atoms on the same plane?  And if so, doesn't that mean, to maximize distance from each other, that they will occupy as much of 360 degrees as possible?  To occupy the maximum amount of space, they have to be at 120 degrees.

How is it possible that you put picture with huge lone pair, while being ignorant about her influence?

I understand that a lone pair will cause greater electron repulsion and subsequently make the angles smaller.  However, would the angle between the B with straight line and the B with the dashed line also be 109.5?

Yes all three B-A-B angles are equal to 109.5

But if all three B-A-B angles = 109.5 and all three B-A-B angles are on a plane, then their total sum has to equal 360 degrees.  109.5 degrees would not make this add up.  And if the lone pair squished the three angles together, then angle B (with the solid straight line) - A - B (dashed line) should be greater than 109.5.  But since all the angles are 109.5, this cannot be the case.  Clearly I'm missing something obvious here.

True. Then consider an idea that all three B-A-B angles are not on a plane.


text:
Are the lone pair electrons on an carbonyl oxygen in sp2 orbitals?

Carbonyl Group Double Bonds
[...]
One pair of the oxygen lone pairs are located in 2s while the other pair are in 2p orbital where its axis is directed perpendicular to the direction of the pi orbitals.

http://chemwiki.ucdavis.edu/?title=Organic_Chemistry/Aldehydes_and_Ketones/Properties_of_Aldehydes_%26_Ketones/The_Carbonyl_Group
How can a lone pair on the carbonyl oxygen be in a p orbital? Neither of the electrons in a carbonyl oxygen can be delocalized.
This seems to be more along the lines of what I'm thinking:

It's may be possible 'cause hybridization is lame ;) Also I heard that's just the situation in water molecule. That simply means that sp2 hybr. would be more appropriate for oxygen.

@Mithoron It may be lame but hybridization represents the upper bound of some professors' knowledges.

I can write up an answer later, but you could actually look at the MO diagram for formaldehyde and see which hybridisation scheme best reproduces the MO energy levels qualitatively. Sorry that was a bit of a mouthful.

Short answer: Both descriptions are wrong and both descriptions are somewhat correct.


text:
Why are overcoats used on aluminum if aluminum oxide forms on surface anyway?
If a thin layer of aluminum oxide is formed on the surface of aluminum to prevent further oxidization, why are optical components using extra "overcoat" materials to prevent dulling of the metal over time?

Sapphire is an aluminum oxide which is transparent when in crystalline form. However when aluminum metal corrodes the aluminum oxide takes up more volume that the aluminum metal from which it was formed. So the aluminum oxide can't form the same crystalline structure as aluminum metal. Thus instead of one nice nice uniform crystalline sheet of aluminum oxide the oxide forms small crystals. Such an oxide coating reduces the reflectivity of the mirror. (Thinking of it like dropping table salt on a mirror.)
The overcoating materials keep oxygen away from the aluminum metal so that the oxide coating doesn't form.

I think you should read about allotropes of Al2O3

"Such an oxide coating reduces the reflectivity of the mirror." How much and any source?  Theory is fine but the reduction might be so low it wouldn't even matter. I googled tests claiming 92% reflectance in the visible spectrum.

Well I tried giving a half-answer that I thought would be satisfactory. If there is 92% reflectivity, where does the other 8% go? Some is randomly scattered and some is absorbed. If you think of taking a picture with the telescope, then some of the scattered light from the surface of the mirror would fog the film. So it isn't just reflectivity. // Adding a coating is an expense. If a coating wasn't necessary for long term stability then it wouldn't be done.

Adding silicone monoxide coating increases the reflectivity, because it's less detrimental than a layer of aluminum oxide to the reflectance of aluminum. I think it's around 96%. Yes, it also protects from corrosion, but in what conditions and how long? For outdoor telescopes exposed to humidity, extra $ wasted on silicone coating might make sense. For indoor projects, who knows, maybe the natural aluminum oxide layer is sufficient? That's why I want more info.

There is humidity and oxygen indoors too. In winter the humidity is usually higher indoors for comfort. // I'm not prepared to debate the economics of silicone monoxide coatings.

nobody wants you to debate that though


text:
Does copper(III) nitrate exist?
I was balancing this reaction when I found an unusual compound had been formed.
$$\ce{Cu(s) + HNO3(aq)-> Cu^{3+}(aq) + NO(g)}$$
My attempt:
Balancing oxidation number:
$$\ce{Cu -> Cu^{3+} +3e^-}$$
$$\ce{3e^- + NO_3- -> NO}$$
Balancing hydrogen and oxygen atoms:
$$\ce{3e^- + 4H^+ + NO3- -> NO + 2H2O}$$
Adding $\ce{3NO3-}$ on both sides to form compounds:
$$\ce{3e^- + 4H+ + 4NO3- -> NO + 2H2O + 3NO3-}$$
Adding both half reactions
$$\ce{Cu + 4H+ + 4NO3- -> NO + 2H2O + Cu^{3+} + 3NO3-}$$
$$\ce{Cu + 4HNO3 -> NO + 2H_2O + Cu(NO3)3}$$
So does this compound $\ce{Cu(NO3)3}$ actually exist? or have I done a mistake.

Could you clarify where the first equation came from. I'm guessing that you were trying to determine what the products would be for copper reacting with nitric acid and you developed the equation rather than trying to "balance it".

I had a test yesterday and it was in the paper.

Did some digging with Google. Evidently some Cu+3 compounds exist as an anion complex where copper has organic ligands. http://pubs.acs.org/doi/abs/10.1021/ja00244a020 The equation does balance. However $\ce{Cu(NO3)3}$ does not exist.

That reaction would not happen in water.

Copper (iii) salts need a very basic anion such as fluoride to form , for this reason it mostly exists as an oxide or fluoride or something like that. I'm not sure whether the nitric acid would form the copper (iii) nitrate salt.

I'll ask my teacher, perhaps it should be $Cu^{+2}$ instead of $Cu^{+3}$

With nitric acid , under normal conditions, I'd think it would be copper (ii) but copper (iii) nitrate does exist on @chemspider so there must be a way to do it.

It's mentioned in a patent for haircare treatment!

Inorganic chemists have cooked up some weird chemicals. I'd be curious as to how copper (iii) nitrate could be prepared.

Copper(III) nitrate cannot be obtained from aqueous nitric acid, and likely doesn't exist.
Reaction carried under strongly oxidative conditions between $\ce{Cu(NO3)2}$ and fuming $\ce{HNO3}$ yields in nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]},$ [1, 2] sometimes written as adduct $\ce{Cu(NO3)2 · N2O4},$ which is contradictory to the crystal structure consisting of copper(II) centers octahedrally coordinated by nitrate groups and the voids in cross-linking mesh of octahedra are occupied by nitrosyl ions [3]:

Figure 1. Crystal structure of nitrosyl copper(II) trinitrate $\ce{[NO+][Cu(NO3)3−]}$ (ICSD-422033). Color code: $\color{#3050F8}{\Large\bullet}~\ce{N}$; $\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.
According to Wiberg's Anorganische Chemie [4, p. 1707], oxidation of $\ce{Cu^2+}$ to $\ce{Cu^3+}$ is possible only with strong oxidation agents $(E^\circ(\ce{Cu^2+/CuO+}) = \pu{+1.8 V}),$ and copper(III) is highly unstable in water and has a tendency to oxidize the medium $(E^\circ(\ce{H2O/O2}) = \pu{+1.23 V}).$
The reduction potential or $\ce{Cu^3+}$ can, however, be reduced down to $\pu{+0.45 V}$ by coordinating it with appropriate strongly donating ligands (which nitrate really isn't).
Examples of such complexes are:

$\ce{K3CuF6}$ (oxidant: fluorine $\ce{F2}$; $\ce{CuF3}$ is unknown);
$\ce{K5[Cu(IO6H)2]},$ $\ce{K5[Cu(TeO6H2)2]},$ $\ce{Na5[Cu(TeO6H2)2]}$ (oxidant: peroxodisulfate $\ce{S2O8^2-}$ in strongly alkaline medium);
$\ce{Na[Cu(OH)4]},$ $\ce{KCuO2},$ $\ce{Ba(CuO2)2}$ (oxidant: hypobromite $\ce{BrO-}$/oxygen $\ce{O2}$; $\ce{Cu2O3}$ is unknown).
Cu(III) complex of 2-ethoxy-5,10,15,20-tetrapentafluorophenyl-3,7-diaza-21,22-dicarbaporphyrin (Cu(III)-$\ce{N2CP}$, a doubly N-confused porphyrin) [5]
A variety of Cu(III) peptide complexes [6, pp. 213–223]

References

Znamenkov, K. O.; Morozov, I. V.; Troyanov, S. I. Synthesis and Crystal Structure of Copper(II) Nitrato Complexes $\ce{NO[Cu(NO3)3]},$ $\ce{Na2[Cu(NO3)4]},$ and $\ce{Ag2[Cu(NO3)4]}.$ Russian Journal of Inorganic Chemistry 2004, 49 (2), 172–179. (in Russian)
Gagelmann, S.; Rieß, K.; Wickleder, M. S. Metal Oxidation with $\ce{N2O5}$: The Nitrosylium Nitrates $\ce{(NO)Cu(NO3)3},$ $\ce{(NO)2[Zn(NO3)4]}$ and $\ce{(NO)6[Ni4(NO3)12](NO3)2·(HNO3)}.$ Eur. J. Inorg. Chem. 2011, 2011 (33), 5160–5166. https://doi.org/10.1002/ejic.201100565.
Blackwell, L. J.; King, T. J.; Morris, A. Crystal Structure of the Copper Nitrate–Dinitrogen Tetroxide Adduct $\ce{Cu(NO3)2 · N2O4}.$ J. Chem. Soc., Chem. Commun. 1973, No. 17, 644–644. https://doi.org/10.1039/C39730000644.
Wiberg, E.; Wiberg, N.; Holleman, A. F. Anorganische Chemie. Nebengruppenelemente, Lanthanoide, Actinoide, Transactinoide, 103. Auflage.; De Gruyter: Berlin; Boston, 2017; Vol. 2. ISBN 978-3-11-049573-7. (in German)
Furuta, H.; Maeda, H.; Osuka, A. Doubly N-Confused Porphyrin:  A New Complexing Agent Capable of Stabilizing Higher Oxidation States. J. Am. Chem. Soc. 2000, 122 (5), 803–807. https://doi.org/10.1021/ja992679g.
Karlin, K. D.; Tyeklár, Z. Bioinorganic Chemistry of Copper; Springer Netherlands: Dordrecht, 1993.

Copper(III) compounds indeed do exist, but are very strong oxidizers. It is safe to assume that they are not formed in water solutions unless some special ligand is present, such as periiodate.
I highly doubt that Cu(III) nitrate was ever obtained in water solutions.

Yes, copper (III) nitrate does exist according to ChemSpider. Although I think it is quite rarely used. Copper (III) salts require a very basic anion such as oxide or fluoride to be produced and are usually found in these forms.
Reacting nitric acid with solid copper, under normal conditions, would most likely form the copper (II) nitrate salt as the nitrate anion may not be basic enough for copper (III) ions to form. Fuming nitric acid, possibly under specific conditions, or another nitrating agent must produce the copper (III) nitrate.


text:
Meaning of negative entropy value
I was looking through the thermodynamic data chart at the back of my textbook  and found that $\ce{S^{2-}_{(aq)}, F^{-}_{(aq)}, OH^{-}_{(aq)}}$ all have negative entropy values. But according to the Boltzmann equation, $S=k\ln W$, so if $S<0$, $W<1$. But since $W$ is the number of microstates, how can $W$ be less than 1?
I read in some other places than this is based on the entropy of $\ce{H^+_{(aq)}}$ being set to $0$, but why do we set this equal to $0$ and how does it follow Boltzmann's equation?

Boltzmann equation is applicable to compound as a whole. Entropy of particular ion is somewhat arbitrary anyway.

Googling "entropy of aqueous ions" will bring up some references. "Real" entropy would be measured relative to absolute zero. But measuring relative to say $\ce{H^+}$ in a 1 N solution at 20 degrees C would give you a stake in the ground against which you could measure other entropy values. My PChem is rusty at best. I'd also guess that choosing  $\ce{H^+}$ as the stake in the ground has something to do with the standard electrode being the hydrogen electrode for electrochemical cells. The gist is that the tables are relative values - when the whole reaction is considered the bias drops out.

I assume were talking about the Entropy of Hydration.

And for many solid-state tables (free energy of elements and alloys), the reference state is the stable state at 298K, and as for @MaxW's comments the values are relative to that. It is hard to perform measurements at absolute zero (tongue firmly in cheek).

It is especially hard to preform experiments in liquid water at absolute zero.

The numbers in the table are relative to some standard state and they will tell you what that standard state is.

The reason that we can arbitrarily assign any entropy value to a chosen standard ion (in this case, $\ce{H3O+} \rm{(aq)}$, and the value assigned is $0$.) is that ions never appear on their own in chemistry equations due to charge conservation. If this were not the case we would not be able to use such a "standard". That's why we cannot apply the same to $\ce{H2}\rm{(g)}$ for example to "simplify" our calculation.


text:
Is it possible to express Gibbs free energy as following?
We know $$\Delta G =\Delta H -T\Delta S$$
$$\Delta S= \frac{Q_{rev}}{T} $$
Hence $$\Delta G=\Delta H-Q_{rev} $$

Yes... It is possible. But a better question for me to ask is: why would you not think it's possible?

Yes, but only if the temperature T in the initial and final equilibrium states of the system is the same and you are able to identify a reversible path between the initial and final equilibrium states over which all the heat transfer takes place at the temperature T.


text:
Supersaturated conversion of sodium acetate utilizing electricity
Is it possible to use electricity to take crystallized sodium acetate back to a supersaturated solution?

Well if you have a solution which has crystallized sodium acetate, then putting it on a hot plate will heat the solution and dissolve more sodium acetate. You can't keep the same temperature of solution and "persuade" more solid sodium acetate to dissolve into a saturated solution by passing an electric current through the solution.

You could use a microwave oven. which does pass electricity through the sodium acetate, but that's just a variation on @MaxW's answer.

A microwave oven wouldn't pass electricity in a sodium acetate solution. The microwaves would cause the water molecules to rotate and hence heat the liquid. The solid would be heated some, but not much. Ice doesn't absorb microwaves as well as water.

The microwaves can create a current in metal. So if you have a plate with a silver or gold band around the rim, then using it in the microwave will create a current in the metal band. In fact it will heat such a metal band so much that it will destroy the band.

What about using ultrasound to excite the crystal structure? Or another type of radiant heat?

@Johnny-V - Since you seemed determined to find an interaction I'll mention liquid crystals. In the presence of an electric field (like a capacitor, not actually current passing through crystals) the liquid crystal can "dissolve." Sodium acetate isn't such a molecule.  https://en.wikipedia.org/wiki/Liquid_crystal


text:
Nicotine from tobacco?
I'm reading a couple of methods of extracting nicotine from tobacco. I have read that many people trying to commit suicide with nicotine. The method is simple:

Boil some tobacco with water
Leave it for a few hours
filter the solution
boil it till you get a thick black syrup thing

That's pure nicotine and they claim that 2-3 drops of that can kill you. Is that true? What do you think?
I am not thinking to commit suicide. I just asking because it is really  weird that so little a dose can be fatal.

For any chemical the $\ce{LD_{50}}$ is the lethal dose. It is usually per kg of body weight for humans. Wikipedia https://en.wikipedia.org/wiki/Nicotine#Overdose  lists that the fatal dose is 0.5–1.0 mg/kg for humans.

A non-polar solvent alkaloid extraction would produce a much purer nicotine product. The water extraction will contain a lot of impurities . Dichloromethane would be a choice non-polar solvent for this.

Nicotine solutions are sold over the counter for vaping with "electronic" cigarettes. Seems that intake matters. Ingesting it isn't the same as inhaling it.

@MaxW Do the solutions sold over the counter have as much nicotine as eating pure stuff would?

What is weird about this level of toxicity (2-3 drops/person)?

MaxW's comment linked to the relevant Wikipedia page:

Nicotine is regarded as a potentially lethal poison.[17] The LD50 of nicotine is 50 mg/kg for rats and 3 mg/kg for mice. 30–60 mg (0.5–1.0 mg/kg) can be a lethal dosage for adult humans.[11][56] However the widely used human LD50 estimate of 0.5–1.0 mg/kg was questioned in a 2013 review, in light of several documented cases of humans surviving much higher doses; the 2013 review suggests that the lower limit causing fatal outcomes is 500–1000 mg of ingested nicotine, corresponding to 6.5–13 mg/kg orally.[13] Nevertheless, nicotine has a relatively high toxicity in comparison to many other alkaloids such as caffeine, which has an LD50of 127 mg/kg when administered to mice.[57]

You ask:

That's pure nicotine and they claim that 2-3 drops of that can kill you. Is that true?

"Drops" is an imprecise non-quantitative term, but let's suppose that a drop is about 20 μL in volume.  Then 3 drops is 60 μL or 0.06 mL.  The density of the nicotine "oil", if pure, is about the same as water's.  So 60 μL should equal about 60 mg of nicotine.  According to the paragraph from Wikipedia, this corresponds to a widely used figure for a lethal dose.
However, note the text later in the paragraph that a lethal dose is closer to 500-1000 mg.  That would require ten to twenty drops, not just two or three.
Undoubtedly what is a "lethal" dose varies widely from person to person and I have no doubt that consuming 40 or 60 mg of nicotine at a time is very dangerous.  That's the same amount of nicotine as smoking 50-70 cigarettes or chewing ten to fifteen pieces of nicotine gum at the same time.  That can't be good for you.

Boil some tobacco with water

The word you haven't thought about is "some".  According to a different Wikipedia page:

It constitutes approximately 0.6–3.0% of the dry weight of tobacco[6]

Since plants are probably 70% water, that means that nicotine is roughly 0.2% to 1% of the wet weight of tobacco leaves.  So to obtain three drops of pure nicotine (approximately 60 mg), we would need at a minimum 6 g of tobacco.  That isn't very much! But keep in mind this calculation doesn't allow for any losses during extraction, and also I'm not sure if the Wikipedia figure for the nicotine content of tobacco refers to "fermented" leaves or to freshly harvested leaf tissue.  So for a variety of reasons this calculation could be way off.

You missed the point that ingesting nicotine is very different than inhaling it like vaping with an "electronic" cigarette. Vaping would be close to injecting it since the lungs don't have digestive juices to break down "food."

Feel free to add your own answer MaxW.  I'd appreciate it if you could provide links to back up your assertions.  Thanks!


text:
Symmetrize nearly symmetric molecule
In the overall good paper of Beruski et. al.,[1] an algorithm of how to symmetrize an "almost symmetric" molecule by symmetrizing the distance matrix is given for the example of Methane.
The main idea is that symmetric equivalent atoms have the same permuted list of distances in their respective column and row of the distance matrix. Thus by averaging and clustering these distances one can get the distance matrix of the closest symmetric molecule. Depending on the "amount of averaging" the molecule is forced into higher and higher symmetries.
However when I try to apply this algorithm to a more complex molecule (say Hexanitrocuprate(II)) the new distance matrix averaged over it's Symmetric Equivalent Atoms (SEA) with the tolerance of 25.0 pm, gets non-Euclidean whereas the original distance matrix is Euclidean. To get the distance matrices the molecule positions have been converted from Angstrom to pm and resorted into the order of the elements as groups of O, N and Cu.
For an explanation of this problem which is called the Euclidean distance problem, see e.g. the dissertation of Al-Homidan[2]. Digging a bit into the theory one can apply a projection method which transforms the non-euclidean distance matrix back into a Euclidean one. The algorithm mentioned in Beruski et. al. reconstructs the atom positions in the molecule directly through a quasi-Newton optimization of a score function on the distance matrix.
However both approaches "de-symmetrize" the distance matrix back to a state which is close to the distance matrix of the original molecule (at least for the example Hexanitrocuprate(II)). I assume the main problem is the constraint on the distances which returns the molecule back to its original shape.
Of course there are algorithms which determine the symmetry (the point group) of the molecule first (see e.g. the answer to this and this questions from computational chemistry), and then symmetrize the molecule with this information. However I would like to just use the local symmetry of the molecule without the use of a rather complex point group recognition algorithm.
Any idea how to solve this problem?
References

O. Beruski, L. N. Vidal. J. Comput. Chem. 2014, 35, 290–299.
Suliman Saleh Al–Homidan. Hybrid Methods for Optimization Problems with Positive Semi–Definite Matrix Constraints. PhD Thesis, University of Dundee, Dundee, December 2003.

Referenced File contents
Hexanitrocuprate(II) (mol file)
 OpenBabel10241500433D

 19 18  0  0  0  0  0  0  0  0999 V2000
    6.2190   -0.6777   -1.0333 O   0  5  0  0  0  0  0  0  0  0  0  0
    5.5672   -1.4099   -0.2754 N   0  3  0  0  0  0  0  0  0  0  0  0
    6.0808   -2.3487    0.3568 O   0  0  0  0  0  0  0  0  0  0  0  0
    3.7257   -1.2558    0.0053 Cu  0  7  0  0  0  0  0  0  0  0  0  0
    3.5134   -2.5214    1.3633 N   0  3  0  0  0  0  0  0  0  0  0  0
    4.2665   -2.2644    2.3200 O   0  5  0  0  0  0  0  0  0  0  0  0
    2.7186   -3.4711    1.3724 O   0  0  0  0  0  0  0  0  0  0  0  0
    2.1847   -0.7314    0.9234 N   0  3  0  0  0  0  0  0  0  0  0  0
    2.1383   -0.1356    2.0089 O   0  5  0  0  0  0  0  0  0  0  0  0
    1.1629   -1.0996    0.3148 O   0  0  0  0  0  0  0  0  0  0  0  0
    4.4060    0.3341    0.7046 N   0  3  0  0  0  0  0  0  0  0  0  0
    4.2381    1.3976    0.0898 O   0  5  0  0  0  0  0  0  0  0  0  0
    4.9585    0.2682    1.8160 O   0  0  0  0  0  0  0  0  0  0  0  0
    3.4004   -0.4833   -1.6643 N   0  3  0  0  0  0  0  0  0  0  0  0
    2.3776    0.2229   -1.6723 O   0  5  0  0  0  0  0  0  0  0  0  0
    4.1301   -0.6158   -2.6565 O   0  0  0  0  0  0  0  0  0  0  0  0
    3.3488   -2.7126   -1.0984 N   0  3  0  0  0  0  0  0  0  0  0  0
    2.1566   -2.9077   -1.3889 O   0  5  0  0  0  0  0  0  0  0  0  0
    4.2641   -3.4679   -1.4571 O   0  0  0  0  0  0  0  0  0  0  0  0
  1  2  1  0  0  0  0
  2  3  2  0  0  0  0
  2  4  1  0  0  0  0
  4  5  1  0  0  0  0
  4  8  1  0  0  0  0
  4 11  1  0  0  0  0
  4 14  1  0  0  0  0
  4 17  1  0  0  0  0
  5  6  1  0  0  0  0
  5  7  2  0  0  0  0
  8  9  1  0  0  0  0
  8 10  2  0  0  0  0
 11 12  1  0  0  0  0
 11 13  2  0  0  0  0
 14 15  1  0  0  0  0
 14 16  2  0  0  0  0
 17 18  1  0  0  0  0
 17 19  2  0  0  0  0
M  CHG  8   1  -1   2   1   4  -3   5   1   6  -1   8   1   9  -1  11   1
M  CHG  5  12  -1  14   1  15  -1  17   1  18  -1
M  END

Original distance matrix (csv file)
0.,285.286949929365,365.2230816911768,231.30761855157303,273.544384149995,270.1046991446095,123.91014849478631,124.29027476033676,302.6233946673654,388.3080251552883,431.26815544391866,445.4489939375776,312.76083674270984,274.96343866048807,384.58956837127033,289.23154098403586,388.7863095583486,270.7369295090716,186.91339037104862
285.286949929365,0.,227.223583723169,306.3413243426358,365.1435658751226,247.45944091911304,405.7485996278977,276.3043665959697,124.43807697003355,123.8434471419461,282.8381340979324,294.0360137806251,418.39638311056177,317.41328674143426,424.68919800249216,449.11493284013613,309.2691185682787,306.8235470103297,186.84160805345257
365.2230816911768,227.223583723169,0.,247.33251666531837,286.97859484637524,306.0711925353316,448.4098021899164,425.63235379843957,293.8447174954826,282.7119788760285,123.91290691449379,124.5006200787771,307.3106168032598,308.05683501587816,277.2283785978629,407.5981909920602,317.55003684458933,418.0803374950801,186.88456249781575
231.30761855157303,306.3413243426358,247.33251666531837,0.,270.01965354395963,369.4707259039612,270.7582805751285,318.1718988220048,306.2871146489842,421.57825655505525,265.7871304258353,356.7231046624258,123.9839384759171,124.29034636688402,312.6727620052632,350.3630121745159,446.674488302164,437.58755123974913,186.53714348622367
273.544384149995,365.1435658751226,286.97859484637524,270.01965354395963,0.,230.05833738423826,289.4902126152109,384.045655228646,444.93924203198793,431.431883036013,389.95115001753743,305.0093236279835,270.4883160138345,388.6504335517973,124.29409800951929,123.87409656582768,273.8722139976964,311.3959969235315,186.81917728113459
270.1046991446095,247.45944091911304,306.0711925353316,369.4707259039612,230.05833738423826,0.,351.89634922800775,311.670567266144,356.77044566499615,266.0324591097861,421.4445697835007,306.2198891319765,436.99448245944706,446.8783960989835,314.4797433858022,272.66601438389785,124.24955130703692,123.9724917068299,186.61386711603188
123.91014849478631,405.7485996278977,448.4098021899164,270.7582805751285,289.4902126152109,351.89634922800775,0.,217.80078627038978,419.21944170088295,508.3628252537749,511.8687892419306,524.9716223759146,308.0939549877602,325.6086231966224,399.6968266073675,264.61506117377365,464.78430610337955,343.31409073907815,276.2143091876306
124.29027476033676,276.3043665959697,425.63235379843957,318.1718988220048,384.045655228646,311.670567266144,217.80078627038978,0.,267.449905963715,368.7223068923279,481.3579984793023,507.42245929797,418.3490884416984,319.9529268501853,494.40938108009243,398.5994203457903,433.1200195096044,280.05786437806034,262.00150133157643
302.6233946673654,124.43807697003355,293.8447174954826,306.2871146489842,444.93924203198793,356.77044566499615,419.21944170088295,267.449905963715,0.,217.94643975068732,302.61870216495214,387.42636771391795,428.77549603492974,267.33766588342917,506.8832586503523,524.4238664477429,431.52580583320855,396.42026209567035,258.21647294469807
388.3080251552883,123.8434471419461,282.7119788760285,421.57825655505525,431.431883036013,266.0324591097861,508.3628252537749,368.7223068923279,217.94643975068732,0.,344.49151789267614,302.7000578130107,525.9104743204873,438.13579470296645,479.9178584924716,513.5864635093102,287.36807842904193,322.40736250898493,279.11945310207244
431.26815544391866,282.8381340979324,123.91290691449379,265.7871304258353,389.95115001753743,421.4445697835007,511.8687892419306,481.3579984793023,302.61870216495214,344.49151789267614,0.,217.96045444070813,323.15332413577306,285.548470316337,370.6348631739869,509.54701883143224,438.51930561835024,525.7033852848962,279.08958705046666
445.4489939375776,294.0360137806251,124.5006200787771,356.7231046624258,305.0093236279835,306.2198891319765,524.9716223759146,507.42245929797,387.42636771391795,302.7000578130107,217.96045444070813,0.,396.7806431770582,430.47714968392916,267.8258529343275,422.69329270760846,267.5679164249705,428.5547332605253,258.61424032717144
312.76083674270984,418.39638311056177,307.3106168032598,123.9839384759171,270.4883160138345,436.99448245944706,308.0939549877602,418.3490884416984,428.77549603492974,525.9104743204873,323.15332413577306,396.7806431770582,0.,218.50141784437005,281.8931845575554,340.6993872903208,500.54774028058506,510.5552453946586,270.3742881636491
274.96343866048807,317.41328674143426,308.05683501587816,124.29034636688402,388.6504335517973,446.8783960989835,325.6086231966224,319.9529268501853,267.33766588342917,438.13579470296645,285.548470316337,430.47714968392916,218.50141784437005,0.,433.95084733181466,463.36765974763495,531.1155093762561,501.53582304756657,266.85213752188685
384.58956837127033,424.68919800249216,277.2283785978629,312.6727620052632,124.29409800951929,314.4797433858022,399.6968266073675,494.40938108009243,506.8832586503523,479.9178584924716,370.6348631739869,267.8258529343275,281.8931845575554,433.95084733181466,0.,217.79172573814643,315.11605671561716,415.056597706867,261.1181544818361
289.23154098403586,449.11493284013613,407.5981909920602,350.3630121745159,123.87409656582768,272.66601438389785,264.61506117377365,398.5994203457903,524.4238664477429,513.5864635093102,509.54701883143224,422.69329270760846,340.6993872903208,463.36765974763495,217.79172573814643,0.,327.93776269286224,309.6932477468632,276.7366726691639
388.7863095583486,309.2691185682787,317.55003684458933,446.674488302164,273.8722139976964,124.24955130703692,464.78430610337955,433.1200195096044,431.52580583320855,287.36807842904193,438.51930561835024,267.5679164249705,500.54774028058506,531.1155093762561,315.11605671561716,327.93776269286224,0.,218.174964879108,267.10750756951785
270.7369295090716,306.8235470103297,418.0803374950801,437.58755123974913,311.3959969235315,123.9724917068299,343.31409073907815,280.05786437806034,396.42026209567035,322.40736250898493,525.7033852848962,428.5547332605253,510.5552453946586,501.53582304756657,415.056597706867,309.6932477468632,218.174964879108,0.,270.58962895868723
186.91339037104862,186.84160805345257,186.88456249781575,186.53714348622367,186.81917728113459,186.61386711603188,276.2143091876306,262.00150133157643,258.21647294469807,279.11945310207244,279.08958705046666,258.61424032717144,270.3742881636491,266.85213752188685,261.1181544818361,276.7366726691639,267.10750756951785,270.58962895868723,0.

new distance matrix (csv file)
0.,284.90364177745516,365.1833237831497,239.03947838006067,277.35510979538634,269.22803953053614,123.88514977926349,124.29218638492802,301.66740166629756,404.11562988556216,434.849076029974,431.45426939883123,312.7048608574015,273.827178600421,376.9980984165578,277.0514027606322,388.71837155507296,274.0233685968097,186.76829146761784
284.90364177745516,0.,237.30978125769235,310.23931151726777,381.83722368114843,239.03947838006067,404.11562988556216,271.73490110511295,124.46934852440532,123.88514977926349,277.0514027606322,301.66740166629756,410.5138853378226,317.61308082818346,429.3481046605877,434.849076029974,317.61308082818346,312.7048608574015,186.76829146761784
365.1833237831497,237.30978125769235,0.,239.03947838006067,277.35510979538634,310.23931151726777,434.849076029974,429.3481046605877,301.66740166629756,277.0514027606322,123.88514977926349,124.46934852440532,312.7048608574015,317.61308082818346,271.73490110511295,404.11562988556216,317.61308082818346,410.5138853378226,186.76829146761784
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Without digging too deep into the theory, Cartesian coordinates are the least suitable for working with molecular systems. I found internal coordinates much better for this purpose. It might be also usable for you...

@ssavec: I'm not clearly understanding how internal coordinates would help here. Since the symmetrization works on the distances between the atoms, using internal coordinates would not change this, or am I missing something?

@Rainer: I had time to crunch through your question and the article. So, I had hard time reading the "mol" file, but I succeeded in converting it manually into .xyz file. It is not cuprate, there is cobalt inside, but it does not matter. What is worse, your original matrix seems quite wrong, in what units do you have them? The "new" matrix is even more wrong, the last column has the same distance to all other atoms, which is super-weird.

@ssavec: Sorry for the confusion. The mol file positions are in units of Angstrom and the distance matrix is in units of pm. Furthermore the distance matrix is indexed with the clustered order of the atoms in the molecules in groups of O, N and Cu. I also corrected Co to Cu as you rightly pointed out. The new distance matrix is heavily averaged, since the tolerance for the averaging is 95pm. However if one choses a smaller tolerance the result is similiar with respect to breaking the symmetry.

Consider this option: each distance can be associated with 'interaction'. Let each 'interaction' to produce a force. If the distance is below average in the cluster, than the force should be a repulsion, and if it is above average it is attraction. Shift atoms according to summary forces scaled by some value so the shif was below, say, 0.1 A. Run several iterations. || that said, I highly recommend to divide atoms into classes based on their neighborhood before the procedure, otherwise you'll have troubles in some cases.

I've done some work in both symmetry detection and in distance matrix methods. I think it's a great idea in concept, but the devil will be in the details for large, more complex molecules.
The first problem is that distance geometry methods are over-determined. For each atom, there are range constraints to the other $N-1$ atoms (a lower bounds and an upper bounds) so that's a total of $N(N-1)/2$ constraints but only $3N-6$ degrees of freedom for the molecule. For a small molecule like methane, this isn't a huge gap:

5 atoms = 10 constraints
3N-6 = 9 degrees of freedom

In your case, you have 19 atoms, so:

19x18/2 = 171 constraints
3N-6 = 51 degrees of freedom

That's a gap of 120 constraints. My experience with distance geometry methods is that you must be very careful to smooth out all the constraints. This is algorithmically challenging and numerically unstable. For methane, it's trivial - it's a $5\times5$ matrix and you only need to consider bonding pairs and bond-angle pairs (i.e. H-H interactions).
Consider for your molecule.. you make sure all the Cu-N constraints are averaged out to be symmetric. Ok, if the Cu-O or the O-N or O-O are slightly off, there will be a "pull" back to the asymmetric geometry you mention.
I would actually take the advice in a comment - convert first to a z-matrix representation and average out all symmetric bond lengths and bond angles. Then I'd generate the distance matrix and start your procedure. The starting point should be closer than what you're using right now.
The second problem is in the optimization after you create the initial atomic coordinates from the distance matrix. As you said, it's easy for the distance matrix to become non-Euclidian.
I'd first treat this as a minimization problem on the distance matrix. You want the Euclidian matrix closest to the symmetric form. By definition, there must be a Euclidian, symmetric distance matrix (i.e., the perfectly symmetric molecule). Finding that is tricky.
So you might to iteratively smooth the matrix:

Perform the Euclidian projection
Re-symmetrize the new projected matrix
Repeat, checking the trace of the difference between the Euclidian and symmetrize matrix - hopefully the magnitude will decrease

Ok, so now you find a distance matrix that's nearly Euclidian and symmetric. Once you generate the coordinates, you'll need to ensure optimization in Cartesian space retains symmetry (e.g., for any perturbation on atom X, its symmetric partner X' must also move).

By the way, for anyone finding this old question, for the general question of “I just want a symmetric molecule,” I’d suggest using Avogadro (avogadro.cc). For v1.x on Linux or Mac, there’s View => Properties => Symmetry. For 2.x, there’s an even better symmetry tool that will clean up a molecule and visualize the symmetry elements.


text:
Mechanism for synthesising azetidine

Problem
   A synthesis of azetidine is shown below. Provide a mechanism for all the steps. Why does the formation of azetidine in the last step require high temperature?

My working is thus:

I am stuck on the last step and my version seems wrong. Can someone please guide me in the right direction?

You lost a carbon atom on the last step of your first image (but no harm done, because you regenerated it from scratch at the beginning of the second).

There is not much to say other than: your version is more or less correct. It is debateable whether the ester hydrolysis happens before or after the cleavage of azetidine (I would suspect before due to the lower overall charge). Remember that you always have to consider aquaeous work-up at the end of a reaction, which will give you the three protons you need to regenerate acrylic acid, ethanol and azetidine.
You also have another question written on your paper after the halogenation of the terminal alcohol, namely:

What is the point of this step?

It’s not really a point. It’s just that the reaction conditions contain $\ce{HCl}$ which will protonate any strong enough base around and the secondary amine is a strong enough base. It’s more of a consequence of the reaction conditions than an actual pointful step.


text:
Cleavage of type 1 collagen by pepsin.
Where does pepsin cleave a human type 1 collagen molecule? Does it occur at multiple locations or only one?

There are multiple types of collagen variants, so the answer here is somewhat non-specific.
Pepsin is an endo type of serine protease and it has low specificity. Proteolysis depends on residues upstream (P1’) and downstream (P1) of the recognition site. Hydrophobic residues at P1’ and P1 increase the likelihood of successful proteolysis. An aromatic residue such as Phe, Trp and Tyr at P1’ and a hydrophobic residue Phe, Leu and Met at P1 are associated with highest likelihood of successful proteolysis. Positively charged residues such as His, Lys and Arg at position P1 are detrimental to proteolysis.
A collagen fiber contains multiple fibrils that are made up of tropocollagen molecules, strengthened by cross-links. Tropocollagen is in fact comprised of three left-handed procollagen molecules, forming a right-handed triple helical structure.
Treatment of acid-soluble tropocollagen with pepsin releases multiple peptides whose composition mainly includes Glycine, Tyrosine, Leucine and Phenylalanine amino acids among others.


text:
Why don't halogens form cations like H+?
The first ionization energy of hydrogen is $1312\;\mathrm{kJ\cdot mol}^{-1}$, which is larger than that of halogens such as chlorine, bromine, and iodine.
Given that observation, why don't halogens form $\ce{X+}$ ions like $\ce{H+}$?

It is a subtle matter. The halogens do exist in the +1 oxidation state, for example, in $\ce{OCl-}$. However, you aren't likely to see a naked $\ce{X+}$ cation. It is electron-deficient and would greatly prefer to form a bond to something else. "Well, why does hydrogen form $\ce{H+}$ then?" In fact, the naked $\ce{H+}$ ion does not exist either. In water, any naked $\ce{H+}$ ion will immediately form $\ce{H3O+}$. We only write $\ce{H+}$ as shorthand notation for "a Bronsted acid"; $\ce{H+}$ itself does not actually exist. Not in solution, at least.

@orthocresol What if we consider a case other than solution?

Even in superacidic media people have problems formulating true $\ce{H+}$. Superacids will protonate just about anything, showing how very unhappy the hydrogen would be to form a $\ce{H+}$ ion. $\ce{CH5+}$, anyone?

One context where $\ce{X+}$ form fairly readily is in ionizing plasma. But, just about everything forms cations in ionizing plasma, so ...

Iodine does form some compounds of $\ce{I+}$ in which it acts kinda like a metal.

https://chemistry.stackexchange.com/questions/40643/why-does-iodine-form-cations-sometimes .... https://chemistry.stackexchange.com/questions/136529/what-is-the-correct-formula-for-iodonium-ion