text:
X-ray structure analysis mirrored coordinates?
How to ensure, that X-ray crystallography structure analysis was done right to provide correct, non-mirrored atomic coordinates?

Example: Following Wikipedia image,

is generated from data taken from the article Redetermination of (+)-methamphetamine hydrochloride at 90 K.
However, the structure presented by the CIF file represents (R)-methamphetamine, known to be levomethamphetamine, (–)-methamphetamine,

the enantiomer of “native” methamphetamine, dextromethamphetamine, (+)-methamphetamine, identified as (S)-methamphetamine.

Also, the article does not claim that the absolute configuration of (dextro)methamphetamine has been for so long mistaken and that they come with revolutionary discovery, rather are just focusing on some inter-atomic distances. (Also the “traditional absolute configuration” has been verified e.g. by synthesis from D-phenylalanine.)

As far as I have known, X-ray structure analysis was unable to provide absolute configuration info. But technology evolves. Can one determine from the experimental data, that it's supposed to be “absolute” or “relative” only (so that the coordinates presented can be mirrored)?
UPDATE: After some comments, I should emphasize that the example is just for an example (of case where the reported coordinates seem to be mirrored; I don’t know how rare or common it is), but can be used as an example in the answer as well, of course (e.g. CIF file analysis, e.g. explaining the Flack parameter value meaning, …)
And a subquestion: When hydrogen positions are provided, does it imply that the configuration is absolute? (I presume that the coordinates can still be mirrored even in such case; I mean “uncertain z-coordinate orientation”.)

X-ray does provide absolute configuration info. It is just that with Mo radiation and light-atom structure it won't be reliable enough.

As Ivan Neretin comments x-ray diffraction is the definitive method to obtain structure. However, x-rays are scattered off the electrons in an atom and this makes it very difficult to detect H atoms, and hence your question. Neutron diffraction could be used instead.

@porphyrin Hydrogen positions are irrelevant to the configuration determination, as the other substituents are carbon and nitrogen based. (Also in the formula there are no hydrogens, yet it’s stereochemically unambiguous ;) )

Ah yes, I mis-read the last two line structures.   In that case as long as you have pure crystalline substances x-ray diffraction should be the technique to use.

In the linked crystal structure all atom positions (including hydrogen) have been refined, there is no doubt that the measured compound is (R)-N-methyl-1-phenylpropan-2-amine.What they were sold for measuring the compound is another question; "The material for this work was purchased from Sigma-Aldrich and was used without any further purification." They did not specify they  how they assigned (+), so I'd guess that was on the label. I don't really understand what the question is; obviously you cannot ensure that the people did their jobs right without checking the raw data.

@Martin: It's just an example. But, I wonder why would they order more expensive enantiomer for the study of a parameter independent of the absolute configuration. However, the price difference is small. And, S.-A. has (+), (-) (sic! typo), or (±) in the names explicitly. (Sometimes both abs. and opt., e.g. S(+)-Methamphetamine solution. I’ve checked the rotation is same for free base and salt.)

I am confused. Where did you get the information that they were not studying the absolute configuration? I read that the only purpose of this measurement were to determine the nuclear positions. Diffraction will give you the unit cell; and depending on the quality of what you obtained/bought, that will give you the absolute configuration(s). After solving the crystal structure with the confidence necessary to publish it, you know the absolute configuration of what you have measured. What I find bothering is that they do not name the absolute configuration at all.

@Martin 1. It seemed to me that they were interested primarily in the $\ce{N-H\bond{...}Cl}$ distance. 2. Naive layman (likely wrong) thought process that can lead to abs. config. doubt: X-ray diffraction is recorded on the plate. If it’s “orthographic projection” (not perspective) the absolute z-coordinate values should be impossible to obtain.

@Martin-マーチン Standard diffraction experiments do not give absolute configuration, see my answer below.

As far as I have known, X-ray structure analysis was unable to provide absolute configuration info.

Routine X-ray structure analysis might not, but there are ways to determine absolute configurations from X-ray data.
The reason that routine X-ray structure analysis does not distinguish between enantiomers is that diffraction intensities show inversion symmetry, or obey Fridel's law. If you rotate the crystal by 180 degrees, you expect the same diffraction image. The phases, however, will be distinct.

Friedel's law breaks down for anomalous diffraction, which occurs when the wavelength of the X-rays is close to the absorption edge of one of the atoms in the structure. This has been used in protein crystallography to find positions of selenium for seleno-methionine derivatized proteins (MAD and SAD phasing methods). However, as Ivan mentions in the comments, the elements present in this molecule would not be near their absorption edges in a standard lab experiment using Mo-radiation.

When hydrogen positions are provided, does it imply that the configuration is absolute?

No, that just indicates that the resolution was sufficiently high to model hydrogen atoms. You can have ultra high resolution without resolving absolute configuration.

How to ensure, that X-ray crystallography structure analysis was done right to provide correct, non-mirrored atomic coordinates?

In a standard experiment, where Friedel mates are averaged, both given coordinates and their mirror image are valid solutions that fit to the diffraction experiment.
If you look at the diffraction data (CIF file), they did not measure anomalous data. The headings are "h,k,l, Fc-squared, Fo-squared, sigma(Fo-squared) and status flag".
Finding the absolute configuration
From classical to cutting edge, here are some strategies:

Derivatize the sample with a chiral substance of known configuration: In this case, you could make an amide with lactic acid purified from a biological sample. You will get a diastereomer, and the lactic acid will tell you the absolute configuration.

Use anomolous diffraction, i.e. choose a wavelength close to an absorption edge of nitrogen or carbon, which would be tricky. Other elements are more suitable because they show a larger effect, and at wavelengths that are easier to handle.

Measure the phases. Using a free electron laser or a triple-beam experiment, measure some or all phases of the diffraction intensities.


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Why do some covalent compounds form lattice while others do not?
Why do some covalent compounds form lattice while others do not? I'm not only talking about amorphous solids but covalent compounds that form molecular lattice in general. For Example: Compounds like SiO2, SiC and AlN form covalent lattices while some covalent compounds like methane and SiF4 form molecular lattices. The question is - Why do only some covalent compounds do this and also, is there a way to identify which compound forms covalent lattice and which forms a molecular lattice? If so, then how?
NOTE: I've looked this up over Concise Inorganic Chemistry by JD Lee and also on the internet but have found nothing.

Off the top of my head, Fluorine has negligible tendency for catenation and multiple bond formation. While on the other hand, C has a high catenation tendency, O can involve itself in an ether kind of linkage, and similarly N can involve itself in multiple bonds on either side. Thus SiF4 would probably interact majorly via Van der Waals forces due to lack of options for polymerization. But of course there may be a better reason here

@YUSUFHASAN Yes. I think so too. But can we treat polymerization the same as lattice formation because if this is true then I've found my answer

No I wouldn't say that polymerization is the same as lattice formation, as lattices tend to have a more ordered, 3-D structure. But still, the idea is to form bonds in both cases, and that will probably depend on the factors I mentioned above

@YUSUFHASAN It actually sounds very convincing. I've trying using the logic on a few compounds and seems like this works. Can you please also tell how to compare the catenation properties of elements and how this sits in the bond formation of AlN and SiO2.

This is not a duplicate of the covalent network question. IT seems to be asking why anything forms a crystalline lattice not just network solids and also why some things don't seem to. That is a much broader question than the dupe.

See, as far as I know, their is no fixed way to compare catenation properties of elements randomly without any context. Some general pointers however, to keep in mind are-:
Carbon has one of the highest catenation powers, as soap molecules like sodium stearate can contain up to 17 carbon atoms.
After carbon, some other elements having significant catenation tendencies are silicon, sulphur, boron etc. (roughly in that order). I will not be elaborating further here as I'm not very sure myself about this except perhaps a few rough pointers such as medium-ish size of atom and the presence of group 13 and 14(which have the maximum number of covalent sigma bonds formed without expansion of octet)
The thing you really need to understand is that the ability to form multiple bonds is more important here to predict the bonding nature of the crystal lattice. Oxygen and nitrogen can form multiple bonds, which would allow them to kind form cross links while satisfying the valency of the central atom. While on the other hand, fluorine will finish it's covalency in satisfying the needs of the central atom (Si) only, so the tetrahedral unit formed will majorly interact via Van der Waals forces (checking it with the facts, this is a pretty good prediction). The exact crystal structure of SiO2 and ALN are quartz and wurtzite respectively, which would be difficult to predict simply by making the covalent structures. However, a fair idea can be taken about the kind of interactions which will prevail in the crystal lattice, provided it is also covalent.

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Why does CH3CH(OH)CH2CH3 give iodoform test?
The iodoform test is given by compounds containing $\ce{-(C=O)CH3}$ group or compounds which can be oxidized to such.
$$\ce{ CH3CH(OH)CH2CH3}$$
Is a 2 degree alchohol and even though it were to be oxidized it would not contain a $\ce{-(C=O)CH3}$ group but according to a question asked by JEE in 1997 it as a matter of fact does give the test.
HOW?

This alcohol can be oxidised to a ketone as $\ce{I2/NaOH}$ is also an oxidising agent (as it produces $\ce{NaOI}$ in which iodine is at higher oxidation number than general).

The substrate you have provided (2-butanol) would indeed give a positive test.
The iodoform test is performed in presence of iodine and potassium or sodium hydroxide, which first oxidizes the aforesaid compound to 2-Butanone.
Mechanism (taken from this Wikipedia page) which shows $\ce{BrO-}$ instead of $\ce{IO-}$, but would be executed in a similar fashion:

Thus 2-Butanone responds to iodoform test successfully.

References:
Ueber Entstehung von Jodoform und Anwendung dieser Reaction in der chemischen Analyse, Lieben, Ann. Spl. Bd. 7, 218 1870

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What chemicals have been named after Star Wars characters?
The molecule below is whimsically named Yoda1 because it "uses the force" to open channels in a class of animal proteins known as piezo ion channels (which appear to be part of their pressure sensing nervous systems). See this Nat. Commun. 2018, 9 (1) for a description and other links.

This raised the question for me of how many and what other interesting chemicals have been named after Star Wars characters?

You should be aware that this is an open-ended question, which is not typically within the scope of stack exchange; typically these are closed as too broad.

@Martin-マーチン I'd agree if there were a vast number of answers but this doesn't seem to be the case.

That does not take away the open-endedness of the question itself. If a huge star wars nerd comes along and starts naming newly designed molecules after SW characters/places/things, then this list would have to be amended... but I am neither for nor against closing this question.

I heard about one other Star Wars compound named after a character: Kinobeon A.[1]

There is also a supramolecular structure named after a Tie fighter.[2]

Apparently there is also a Yoda1 analogue (note the additional sulfur atom) called Dooku1.[3]

References:

Kanehira, T.; Takekoshi, S.; Nagata, H.; Matsuzaki, K.; Kambayashi, Y.; Osamura, R. Y.; Homma, T. A novel and potent biological antioxidant, Kinobeon A, from cell culture of safflower. Life Sci. 2003, 74 (1), 87–97 DOI: 10.1016/j.lfs.2003.06.033. Mirror: www.chm.bris.ac.uk
D'souza, F.; Gadde, S.; El-Khouly, M. E.; Zandler, M. E.; Araki, Y.; Ito, O. A supramolecular Star Wars Tie Fighter Ship: electron transfer in a self-assembled triad composed of two zinc naphthalocyanines and a fullerene. J. Porphyrins Phthalocyanines 2005, 09 (10), 698–705 DOI: 10.1142/S1088424605000812.
Evans, E. L.; Cuthbertson, K.; Endesh, N.; Rode, B.; Blythe, N. M.; Hyman, A. J.; Hall, S. J.; Gaunt, H. J.; Ludlow, M. J.; Foster, R.; Beech, D. J. Yoda1 analogue (Dooku1) which antagonizes Yoda1-evoked activation of Piezo1 and aortic relaxation. Br. J. Pharmacol. 2018, 175 (10), 1744–1759 DOI: 10.1111/bph.14188.

Fascinating. But you should include the structures or some explanation of why the names arose.

Good point. I have added some images of the structures.

Please do not include any update/edit statements into your post, the edit history is available for anyone to check, and those statements are confusing for new readers of your post.

@Martin-マーチン Thank you for your feedback. I saw similar update/edit statements in other Stack Exchange posts, so I thought it was the way to do it. Lesson learned. :-)

I completely understand. Our format is a bit different from the traditional online forum, and probably a bit harder to get used to. Unfortunately it's not always possible to make all posts a better fit so you'll come across these kind of things quite often. These are usually not necessary - just edit the post into what you think is the best version and ignore people who might have already read it. ;)


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Reaction between Salicylaldehyde and 2-Benzoylpyridine
In a paper by Sheng et al (10.1016/j.dyepig.2018.07.036) the authors synthesize the compound MZC as shown in the scheme below.

I'm looking for information about what the reaction mechanism is that underpins this reaction. If anyone can share some insights it is much appreciated.
In the experimental section  it is written

2-Benzoylpyridine (4 mmol, 0.732 g) and ammonium acetate (22 mmol, 1.70 g)
  was dissolved in acetic acid (30 mL) and followed by the addition of salicylaldehyde
  (6 mmol, 0.733 g). The reaction was refluxed for 6 h. The resulting solution mixture
  was cooled to room temperature and cold water was then added to the mixture. The
  solid was collected by filtration and washed with a small amount of cooled water,
  which was purified via recrystallization using acetonitrile to give MZC (0.687 g, 60 %
  yield).

Treat it step by step. For example, step 1: what does a ketone form when mixed with ammonia?

I assume that ammonium acetate $\ce{H3COONH4}$ provides for $\ce{NH3}$ in the reaction mixture. Ammonia will react with the ketone group in 2-Benzoylpyridie and form an imine, and that gives a hint on how an additional nitrogen enters the ring system. So the question then is, does the imine reacts with the aldehyde group in Salicylaldehyde?

The reaction is run in refluxing acetic acid so that is forcing conditions. Imines are not as nucleophilic as amines but they are stiil nucleophiles.

Technically, the reaction is one pot, so the order is not 100% certain. But you could also reasonably form the aldimine first on reactivity grounds (though the reaction is likely under Curtin-Hammett kinetics). Then keep in mind that the pyridine is relatively nucleophilic. I quickly tried drawing this out, and it seems feasible.


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How does an electrolyte conduct electricity at low potential?
My understanding is that electrolytes conduct electricity by:
1) movement of ions, the negative ions migrate to the positive electrode and vice versa. This is however not enough, if these ions just remain ions, the migration and separation of charges will create a potential difference in the solution counteracting the applied potential and at that moment migration of ions will stop. This means no more current will flow.
So therefore:
2) For current to keep flowing the ions must undergo electrolysis at the electrodes to form their neutral forms. In this way their is no build-up of counterpotential, ions keep flowing and the electrons exchanged in the electrolysis can flow from one electrode to the other. (This argument is often given as the reason electrolytes are conductive when I was doing an internet search, also on this site)
Now if we have say a NaCl solution, this would mean that for electrolysis Na+ needs to be converted in to Na, and Cl- into Cl2-gas. The potential needed for the electrolysis of NaCl as calculated from my tables is more than 4V.
So does this mean that if the potential difference is less than 4 V a salt solution does not conduct electricity, ie has very low or zero conductance? If not, how exactly is the current created?
I've read that conductivity is measured in a way that electrolysis is avoided (alternating voltage, wikipedia), so I guess then only the migration of ions is accounted for, but the very definition of conductivity is the ability of a substance to conduct electricity. According to my understanding to maintain a current this requires electrolysis.

Na+ doesn't get reduced in solution.

Ok, so say another reduction takes place, and say H2 is formed, the argument remains the same, you'd need a certain voltage to maintain the current, wouldn't you?

Lower potential makes reaction's K lower then 1 - still non-zero.

Yes, in order to have a continuous current flow there must be both an oxidation and a reduction reaction.

So you're saying that since ln(K) is proportional to the cellpotential there would still be a tiny amount of current? And this is the source of conductivity of salt solutions? But why does wikipedia then say that in measuring conductivity electrolysis is avoided? (by using alternating voltage)

Also, this brings up more questions: even in the absence of an external voltage, and then again even in the  absence of electrodes this would mean that H2 is formed out of a NaCl solution, and since H2 is a gas the equilibrium is constantly shifted so this would mean hydrogen is constantly being produced from salt water?

@MaxW: so if one says electrolytes are conducting it is because of this? Electrolysis of the components in solution? Wouldn't this then mean that the conductivity behaves very nonlinearly with applied voltage, ie almost none at low voltage and very high if the voltage is more than the difference in reduction potential of the compounds? How does this fit in with Ohm's law where resistance (and hence conductivity) is a proportionality constant (so should remain constant!) ?

Here is a good pdf on conductivity http://www.analytical-chemistry.uoc.gr/files/items/6/618/agwgimometria_2.pdf

@MaxW: thanks, this basically says that the conductivity of electrolytes is due to the migration of ions and not due to electrolysis. I still find this very confusing. A metal wire being conductive means that when you apply a DC voltage a current will pass. This does not hold then for a salt solution, at least not like for a metal wire, since the current will be high at first and then drop due to the counteracting field of the migrated ions. Plus at some voltage the conductivity must have a contribution from electrolysis as well, making it nonlinear.

Ok, I should have said: Yes, in order to have a continuous direct current flow there must be both an oxidation and a reduction reaction. // A conductivity meter uses a low voltage and AC current so that it is measuring the "capacitance" of a solution. If a low DC voltage was used the electrodes would charge up and the current flow would stop, just like charging a capacitor.

@MaxW: That makes a lot of sense, but why then do we speak of conductance and not of capacitance of an electrolyte? :-)

While the DC conductivity of a solution is a constant (inversely proportional to the resistance of the solution), during measurement of this conductivity with a DC voltage you polarize the electrodes until they form a capacitative double layer and the DC current drops to zero. In that sense the system will behave like a capacitor. However this is a generally undesirable way to measure a static property in a simple way, so AC current is used, which avoids polarizing the electrodes if the frequency is high enough.

@Mithoron "Lower potential makes reaction's K lower then 1 - still non-zero" I believe Q changes, not K, which is constant (by definition).

Under low potential differences, electrolytes conducts electricity in a similar way that they does at high potentials differences. I think, after reading the question and the comments, that the key point to understand this is: electric potential difference between which spatial points?
The conductivity of a solution is measure of its ability to conduct electricity when the solution is under a difference of electric potential. That is: the solution must be under the influence of an electric field (remember that $\boldsymbol E = - \nabla \boldsymbol \phi$) for having a non zero flux of charges.
You have to look the potential differences inside the solution and not at the electrodes.
You are right in that in order to maintain for long time a continuous current  electrolysis is required (at least for simple experimental devices). If there is not electrolysis you won't see a current. However, this is not because the solution (lets say the bulk, that is where the solution as such is (homogeneous)) does not have the ability of conduct electricity, but because there would not be a net external electric field at all in that region.
In short, the conductivity of the solution is not dependent of your ability for creating an external electric field on it.
EDIT

But then again, what if the potential difference passes a treshold and
  at a certain point allows electrolysis, wouldn't the conductivity
  increase very steeply at that point?

If the potential difference between electrodes is low enough to ignore electrolysis, then there will be a charge accumulation in their surroundings that generates a contrary effect in the potential of the solution "far" from electrodes. So, the mean electric field in the bulk will be null.
According to Ohm's law:
$$ \boldsymbol J = \sigma \, \boldsymbol E$$
For the reasons stated above: $\boldsymbol E = \boldsymbol 0$, and so $\boldsymbol J = \boldsymbol 0$. What is the value of $\sigma$ for the solution? As any value satisfies the equation: We can say nothing from Ohm's law
! (technically, it can be computed even if no electric field is present, but it is not important here).
If the potential allows electrolysis, then a current appears so in principle you can measure the conductivity of the solution using Ohm's law. Notice that it is not an step function the current vs. potential. Due to this, there is not a truly step, but it seems because the growing is almost exponential, so we do not notice the increment till we notice a big increment. You can check out the Butler-Volmer equation for more details.
Once the current due to migration (allowed by electrolysis) is measurable, you could measure the solution conductivity. BUT it is a very bad idea, as normally the solutes concentrations are very small and the process of electrolysis change them! (and of course, many times (but not always) we want to know the conductivity to estimate concentrations)

But then again, what if the potential difference passes a treshold and at a certain point allows electrolysis, wouldn't the conductivity increase very steeply at that point?

No, I am going to edit the answer to clarify that when I finish something that I have to resolve right now.

@Stikke I added more details, I hope them help


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Reading Stereographs
On an old paper, I found an interesting representation of some symmetry operations: stereographs.
Online I found a set of symmetry operations and related stereographs (see: http://newton.ex.ac.uk/research/qsystems/people/goss/symmetry/Stereographs.html), yet it's  not clear to me how the representation is related to the symmetry operations (or: how are the representations obtained?).
Some samples from the website linked above:

C4

C5h

D6

I would like to understand the geometrical (and symbolical) meaning of a stereograph, in order to relate non-mnemonically it to its symmetry operation

Just to be clear: You are asking about the difference between things like $\oplus$ and $+$?

Not only that. Apparently, although I couldn't find a complete description, stereographs are obtained via "mathematical" projection over a sphere of points derived from the symmetry of the object. Then, the location of the points gives the symbol in the stereographs. I think that every strange symbol in a stereograph could be understood in those terms.
I am sorry if the description is terrible, but I couldn't really find anything "complete" about this notation.
If someone could suggest, for lack of a description, a book which explains it, I could make a search in my university's library

I vaguely remember a book with all the space groups (regarding symmetry in solid state chemistry), which had similar pictures. I just don't remember its name. But maybe this can be of help: http://img.chem.ucl.ac.uk/sgp/misc/symbols.htm

The books ' The Basics of Crystallography and Diffraction' by C. Hammond, IUCR texts on Crystallography, publ OUP, and McKie & McKie, 'Essentials of Crystallography, publ Blackwell Scientific, both contain chapters on these projections.

The solid elements are the symmetry operations. The plus and open circle symbols show the positions of atoms and their symmetry mates. The center of the circle is the z-axis. The edge of the circle are directions perpendicular to the z-axis. In the examples in the question, all axes are either along z or perpendicular to z. It gets more complex in cubic symmetry, which also has axes in other directions.

Here, the triangles are three-fold axis along the body diagonals. The symbols are further explained at http://pd.chem.ucl.ac.uk/pdnn/symm2/pntgrp1.htm.


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Implementing the projected atomic orbital (PAO) localization in Gaussian
I am currently attempting to learn how to setup the active space for performing CASSCF calculations.
The textbook I am using as a reference is:
Roos, Björn O. "Multiconfigurational quantum chemistry." Theory and Applications of Computational Chemistry. 2005. 725-764.
My specific problem is discussed in Section 13.1.1. I am working with Gaussian 2016.
I would like to generate a checkpoint file with orbitals localized as atomic orbitals in order to use them as a starting point when selecting my active space.
What I mean is the following...
I am working with BODIPY, with 11 active orbitals (the PI framework).
From a normal SCF I obtain the typical MOs, these would be three of my active orbitals:

If I include the option Pop=SaveNBOs, I obtain localized Natural Bonding Orbitals. These would be three of my NBO active orbitals:

I have found that I can use IOp 4/9 to control the localization method to some extent (for example, "IOp(4/9=20212)" implements the Pipek-Mezey localization). However, after searching for a while, I still don't see how to generate the localized atomic orbitals. Could any of you suggest how to achieve this? I would then be working with the 11 pz orbitals as my active space.
Thanks!

Unfortunately I don't have the mentioned book at hand an I'm not a Gaussian user, but I want to point out that atomic orbitals are local by definition. PAOs are not localized AOs. In PAOs the contribution of your occupied molecular orbitals are projected out. They are an alternative (local) representation for the virtual orbital space. Maybe this helps you a bit in searching...


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Electroplating an alloy from separate metal solutions
Similarly to a question asked here, would mixing two solutions containing metallic ions in different proportions allow to electroplate something with the alloy of the metals? For example if I take a $\ce{CuSO4}$ solution and mix it with a $\ce{ZnSO4}$ solution, will I be able to electroplate brass? If I want to make 70/30 brass, would mixing a 0.7 mol/L and 0.3 mol/L solution work?
I know a "brass" solution can be made by reverse electroplating (is that the term?) a brass anode in an acid solution so mixing two solutions should have the same result, copper and zinc ions in a solution.

In short, yes, it is possible to electroplate alloys, but it isn't as simple as just mixing the ions you want in the ratio you want.
In the simplest approximation, the alloy deposition would behave as two simultaneous, separate depositions at the same voltage and the composition could be controlled by changing the voltage, assuming each component has a different current-voltage response.
In the real world, unfortunately, there are numerous interactions between the ions in solution, the metals on the electrode, etc. and the outcome becomes quite complex to predict.
Here is a decent open-access review of the chemistry and applications of this type of process.


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H-N-H bond angle in ammonia boron trifluoride adduct
In the reaction:
$\ce{NH3 + BF3 -> NH3-BF3}$ does the HNH bond angle increase or decrease?
Surely, the FBF bond angle decreases because boron changes from $\ce{sp^2 -> sp^3}$. But how can we comment upon the NHN bond angle? In my first attempt, I expected it to increase because though N still remains in $\ce{sp^3}$ hybridisation  lp-bp repulsions are greater than bp-bp repulsions.
But the answer key states that the NHN bond angle will decrease. Please let me know how to determine this.
I couldn't find the bond angle data on net or in JD Lee.

I think bond angles should increase.

Yes, that's what I think too


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Which is more acidic - 3,5-dimethylphenol or 3,5-dimethyl-4-nitrophenol?
Does steric hinderance stops the benzene  ring resonance?

Not for phenols to my knowledge

I agree, the 4-nitrophenol will be the more acidic

But I think due to steric crowding created by two methyl groups, nitro group will go out of the plane and shouldn't take part in resonance. However it decreases the acidity by $-I$ effect , but not by $-R$ effect probably.

The only difference between both compound is the presence of nitro group.
Acidic strength is the ease of giving away H+ and become a stable anion. So let's make the anion in both case.
The phenoxide group, which is present after deprotonation of phenol is highly stable in the first case along with it's +I tendency.
In the second case we have more resonating structure along with already discussed effects of phenoxide we also see that when the nitro group takes of the -ve charge via it's -M effects we see that all atoms have complete octet, making it more stable than first one and because of the higher stability of conjugate base it's acid strength would be more.
So second (nitro, one) is more acidic than the first compound.


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Is there a way to oxidize salicyl alcohol to salicylic acid without using chromium trioxide?
I'm doing a project for school and I need to convert salicyl alcohol to salicylic acid, but I'm not allowed to use $\ce{CrO3}$ because of its toxicity. I know it is also possible to oxidize a primary alcohol to a carboxylic acid with potassium permanganate, but I read it might be a too strong oxidizer. Is there any good alternative?

@IvoFilot Please don't use something like CrO$_3$ as it may break in some browsers (especially with the responsive design); instead use the mhchem macro $\ce{CrO3}$. More detail on meta: here and here. We also usually do not correct Brittish spellings (if used consistently), because that has lead to edit wars in the past. In any case: Thanks for your effort.

Hmm, I think there are too many ways...

@Waylander: A general Procedure given in J. Org. Chem. 1999, 64(7), 2564–2566 as follows:

Typical Procedure. A mixture of alcohol 1 ($\pu{40 mmol}$), TEMPO ($\pu{436 mg}$, $\pu{2.8 mmol}$), $\ce{MeCN}$ ($\pu{200 mL}$), and sodium phosphate buffer ($\pu{150 mL}$, $\pu{0.67 M}$, $\pu{pH = 6.7}$) is heated to $\pu{35 °C}$. Then mixtures of $\pu{9.14 g}$ of 80% sodium chlorite ($\ce{NaClO2}$, $\pu{80.0 mmol}$) in $\pu{40 mL}$ of water and 2.0 mol% dilute bleach ($\pu{1.06 mL}$ of 5.25% $\ce{NaOCl}$ diluted into $\pu{20 mL}$ with water) are added simultaneously over $\pu{2 h}$ (Caution! Do not mix bleach and $\ce{NaClO2}$ before being added to the reaction mixture). The mixture is stirred at $\pu{35 °C}$ until the reaction is complete (<2% starting material, $\pu{2−5 h}$ reaction time), then cooled to room temperature. Water ($\pu{300 mL}$) is added, and the $\mathrm{pH}$ is adjusted to $8.0$ with $\pu{2.0 N}$ $\ce{NaOH}$ (about $\pu{48 mL}$). The reaction is quenched by pouring into cold ($\pu{0 °C}$) $\ce{Na2SO3}$ solution ($\pu{12.2 g}$ in $\pu{200 mL}$ of water) maintained at $\pu{<20 °C}$. The $\mathrm{pH}$ of the aqueous layer should be $8.5$−$9.0$. After stirring for $\pu{0.5 h}$ at room temperature, methyl tert-butyl ether (MTBE, $\pu{200 mL}$) is added. The organic layer is separated and discarded. More MTBE ($\pu{300 mL}$) is added, and the aqueous layer is acidified with $\pu{2.0 N}$ $\ce{HCl}$ (~$\pu{100 mL}$) to $\pu{pH = 3−4}$. The organic layer is separated, washed with water ($2 \times \pu{100 mL}$) and brine ($\pu{150 mL}$), and then concentrated to give the crude carboxylic acid 2 in 85−100% yield. The products may be purified by silica gel column chromatography or crystallization.

Converting alcohol to carboxylic acid requires strong oxidizing agent and those reagents are not commonly in school laboratory. In that case, mild oxidizing agent should be used to first convert salicyl alcohol to salicylaldehyde and then convert to salicylic acid.
You should try $\ce{MnO2}$ or (Conc. $\ce{HCl}$, ethyl acetate, sodium Hypo (8 - 12%)) and then use $\ce{KMnO4}$ to convert it to salicylic acid.


text:
STO 3G basis expansion of p orbitals
I am expanding $2p_{x},2p_{y},2p_{z}$ (slater type )orbitals in gaussian STO-3G basis as follows,
$$|2p_{x}\rangle=C_{1}x\exp(-\alpha_{1}r^{2})+C_{2}x\exp(-\alpha_{2}r^{2})+C_{3}x\exp(-\alpha_{3}r^{2})$$
$$|2p_{y}\rangle=C_{1}y\exp(-\alpha_{1}r^{2})+C_{2}y\exp(-\alpha_{2}r^{2})+C_{3}y\exp(-\alpha_{3}r^{2})$$
$$|2p_{z}\rangle=C_{1}z\exp(-\alpha_{1}r^{2})+C_{2}z\exp(-\alpha_{2}r^{2})+C_{3}z\exp(-\alpha_{3}r^{2})$$
I thought the value of $C_{1},C_{2},C_{3},\alpha_{1},\alpha_{2},\alpha_{3}$ are the same for all the $p$ orbitals. I tried to evaluate the following two integral $$\langle2p_{x}2p_{y}||2p_{y}2p_{z}\rangle=\langle2p_{x}2p_{y}|2p_{y}2p_{z}\rangle-\langle2p_{x}2p_{y}|2p_{z}2p_{y}\rangle$$
where,
$$\langle2p_{x}2p_{y}|2p_{y}2p_{z}\rangle=\sum_{i,j,k,l}C_{p_{x}i}C_{p_{y}l}C_{p_{y}j} C_{p_{x}m}\int d^{3}\vec{r_{1}} \int d^{3}\vec{r_{2}}  x_{1} e^{-\alpha_{i}r_{1}^{2}}y_{1}e^{-\alpha_{l}r_{1}^{2}}\frac{1}{|\vec{r_{1}}-\vec{r_{2}}|}y_{2}e^{-\alpha_{j}r_{2}^{2}}z_{2}e^{-\alpha_{m}r_{2}^{2}}$$
The above integral vanishes completely. I think something went wrong in the basis I have chosen or does it really vanish ? I guess my STO-3G expansion is wrong. Can someone help me fixing this please ?

I really don't get the point. Can you please elaborate ?


text:
What should be the reaction between lead oxide and dilute sulfuric acid?
I know for a fact that dilute $\ce{H2SO4}$ is a non-oxidizing acid which makes it a reducing agent. I also know for a fact that $\ce{PbO2}$ is a strong oxidizing agent. So, I thought Redox reaction should have occurred between them, but my teacher says that no reaction occurs between the said reactants because $\ce{PbO2}$ does not react with low concentration of acid. If it were so, why does $\ce{PbO2}$ react with dilute $\ce{HCl}$. Since, both dilute $\ce{HCl}$ and dilute H2SO4 are reducing acids of low concentrations, both should be eligible for the Redox reaction.
I have tried this on the internet but all I get are reaction balancing links .JD Lee's Concise Inorganic Chemistry has nothing about this also.

YOur logic is wrong.

I know for a fact that dilute $\ce{H2SO4}$ is a non-oxidizing acid...

sort of, but so far so good.

...which makes it a reducing agent.

This is where you are wrong, just because it is non-oxidizing does not make it reducing. Concentrated $\ce{H2SO4}$ is an oxidizer, diluting it does not make it less of an oxidizer, but limits the strength/mechanism.
A similar example is mixing nitroglycerin and diatomaceous earth doesn't make nitroglycerin less explosive, it just makes it harder for it to explode (dynamite).

So, what you mean is dilute sulfuric acid is still an oxidizing acid but its capacity has been diluted to the point that it cannot oxidize appreciably anymore.

@Lucifer - Yes.


text:
How does mesh size of Barium Chloride affect sulfate results via turbidimetric method?
I am analyzing sulfate using Standard Methods 4500 $\ce{SO4 ^2-}$ E (turbidimetric method) using $\ce{BaCl2}$ and measuring %T at 420 nm. In the past, we used an appropriate 20-30 mesh size for the $\ce{BaCl2}$; however, it appears that it only is sold in mesh size 10 now.
This smaller mesh size is causing higher %T in the standards (less turbidity), but a check standard will still give accurate results ($r^2$ = 0.98). The main issue is sample results being anywhere from 3-10 mg/L too high  (e.g., routine samples of 7-8 mg/L now give 10-11 mg/L, and 5-6 mg/L now is 15 mg/L). How does the size of the crystals affect the sample results? Is there any way to correct the results or in some way account for the smaller mesh size?


text:
Can bridged and bridgehead positions be chlorinated using free radical mechanism?

When the above compound is treated with $\ce{Cl2}$ in the presence of sunlight ($h\nu$), what is the number of monochloro structural isomers formed?

I expected only 1 product becuase I don't believe that monochlorination can take place at the bridgehead or the bridge positions. And rest of the 4 positions, by symmetry, are equivalent. This is because chlorination takes place through free radical mechanism in which a $\ce{sp^2}$ hybridised carbon free radical is formed.
But the answer given in the test's answer key is $3$. I think they have considered the products of bridgehead and bridged positions too but I don't understand how that is possible.

You're getting 3 products as one derivative is formed on chlorination of the 5 membered ring and two enentiomers for the 6 membered ring. I'll draw them outand post them in jee chem by tomorrow

The question in the test(this is an AITS problem) asked for number of monochloro 'structural' isomers.

The "bridged position" is just another ring, when viewed differently.
Suppose you look at the molecule a little differently (rotate it around a bit), then you could have a five-membered main ring, with a two membered "bridge", as opposed to what we have now (6 membered main ring with 1 membered bridge).
It's definitely possible to have a free radical halogen attack on that position, and with two stereoisomers at the positions which you have identified, it becomes total 3 monochlorinated derivatives.

(in case you're wondering how to check if the two stereoisomers are not identical, you could try to superimpose them, or go from one form to the other without breaking any bonds. In that case, you'll find that they're indeed different. You could even check for a plane/center of symmetry)
--EDIT--
My answer is wrong, as @Abcd and @AkshatJoshi pointed out, because I counted stereoisomers along with the required structural isomers.
I read about attacks at Bridgehead positions and there's NO place that mentions the existence of Bridgehead free radicals. Thus, there will only be 2 structural isomers, but 3 isomers overall.
So, I believe the answer ought to be 2.

I don't understand your 3rd product's mechanism.

See it's like a 60 degree angle at the bridge head position. How on earth can it become 120 degrees?

@Abcd that's not a bridge head

@AvnishKabaj Sorry I meant Bridged not bridge head

It's not a 60° angle either it's a five membered ring

@AvnishKabaj I know still 72 degree is way less than 120 degree requirement!

But monochlorination does take place on 5 membered rings and I don't think it's 72° it prolly exists in a conformer like cyclohexane

Abhigyan C question asks for structural isomers

@Abcd Yeah. I didn't notice. Did the solutions come out?


text:
Is there such a thing as a "minimal soap" molecule?
Wikipedia's Soap gives sodium stearate as an example of soap, and apparently I've been eating it:

Sodium stearate is the sodium salt of stearic acid. This white solid is the most common soap. It is found in many types of solid deodorants, rubbers, latex paints, and inks. It is also a component of some food additives and food flavorings.

What would be the smallest or simplest molecule that we could reasonably call a soap? Perhaps a functional definition would be that it could perform some of the functions of soap in the same way that soap does.
In Why does bleach feel slippery? and its follow-up Is it known for sure that bases feel slippery because of the production of soap/surfactant? the saponification of other existing molecules is discussed, and I'm not looking for that here.

Source

It boils down to the definition of soap. Wikipedia defines a soap as the salt of a fatty acid.
IUPAC claims the smallest fatty acid can be considered to have 4 carbons.
Therefore the simplest soap molecule would be a (generally sodium) salt of butyric (butanoic) acid, i.e. sodium butyrate.
Now apart from the chemical definition, a soap must adhere to its function in order to be defined as such. Therefore, before calling something a soap, we would have to know if the substance does in fact act as a surfactant in a oil-water surface (reducing the interfacial tension) - but that will depend on the nature of the oil in question and the purity of the water. Only for a well defined system we can then conclude that such molecular salt is acting as a soap in it.

I've use the everyday-chemistry tag to highlight my personal interest on the practical aspect from the point of view of a non-chemist. I don't have access to sodium butyrate. I wonder if you would be willing to venture a guess how it would feel if I touched a solution? Would it tend to remove a fingerprint from a window, or some oily residue from a dish?

There is another dimension to Vinicius' answer. I doubt there is a real yes or no answer to if a compound is a surfactant. It is more a matter of how good of a surfactant a particular compound is. Since vinegar is often used as a cleaning solution, I'd guess that sodium butyrate would remove a finger print off glass very well.

@MaxW I think I see what you are getting at, will give it further thought.

@uhoh: For practical purposes, I'd say the biggest issue with using sodium butyrate as a cleaning agent would be its rather nasty rancid smell. Basically, imagine something like the concentrated essence of stinky cheese, gym socks and vomit.

@IlmariKaronen hmm... that's what people say about durian and that doesn't stop durian-lovers from eating as much as they can in season! The smell evokes reactions from deep appreciation to intense disgust, and has been described variously as rotten onions, turpentine, and raw sewage But don't worry, while I love to eat durian, I won't be crossing paths with sodium butyrate any time soon. Thanks!

@MaxW But is it the surfactant qualities of vinegar which are being used in cleaning, or is it some other property? (e.g. the acidity) -- I wonder if sodium acetate would work as well.

@ViníciusGodim You say the IUPAC link claims the smallest fatty acid can be considered to have 4 carbons. It doesn't. It says  Natural fatty acids commonly have a chain of 4 to 28 carbons (usually unbranched and even-numbered), which may be saturated or unsaturated. By extension, the term is sometimes used to embrace all acyclic aliphatic carboxylic acids. So that means that 4 is an approximate lower bound, not a hard and fast rule. The definition leads immediately to the question of  what does aliphatic mean? The same site defines it so broadly that ethanoic acid is definitely included.

@DavidRobinson Thanks for your comment. I wonder if that might be a great new question post. Doing so would provide more space for responses as answers.

@MaxW surfactants lower the surface tension (or interfacial tension) between two liquids which can be determined in a quantitative and objective matter. I think it is soap that has the subjective and ambiguous definition.

@uhoh When I started to write the new question, of what is the shortest fatty-acid molecule, I found that this question had been asked as part of another question on this site. Unfortunately the only answer doesn't actually go further than my comment above. Perhaps the answer is that there is no actual bound other than the obvious 1-carbon limit, but that it is rather arbitrary what counts as a 'proper' fatty acid. I suggest answering or commenting on that question if anyone has any views.

@DavidRobinson okay I see what you mean. Thank you for the speedy follow-up!

@A.K. - Understand but a "large" change in interface surface would indicate a surfactant, but a "small" change wouldn't. The problem is that the difference between "large" and "small" is fuzzy.

While I agree that the definition matters I disagree with the definition of soap as "as the salt of a fatty acid". For one SDBS, AOT, SDS, Cocamide DEA (not even a salt) and CTAB are all popular soaps that are not fatty acid salts. I also do not think that soap and surfactant are interchangeable words. Surfactants lower surface tension, soaps create micelles either by them selves or with oils. Essentially all soaps are surfactants but not all surfacants are soaps. For example, ethanol may be considered a surfactant as it does reduce the surface tension of water, but it does not help form micelles in water appreciably and thus should not be considered a soap.

What would be the smallest or simplest molecule that we could reasonably call a soap?

This is somewhat a matter of estimate, but given that propionic acid is not micible with water and acetic acid is, I would assert that sodium (or other alkali metal)  propionate is the simplest soap since it has an aliphatic group to interact with an oil phase, yet has carboxylate salt "head" to retain interaction with the water phase. It is possible sodium acetate may do so as well but empirical data is needed. I suppose technically lithium salts would be the smallest since the lithium ion  is smaller than the sodium ion, but that is perhaps too technical and may not interact with water as strongly as sodium salts.

Ýou can distinguish between detergents and soaps, the latter being of natural origin, although many if not most people will use the words as synonyms.

As always, there is no sharp border and no minimal molecule. You can make the head piece more charged, which means the molecule becomes more hydrophillic and water soluble and you can make the tail piece shorter, so it will not bind fats very well anymore and fails to form micelles at some point.
Consequently, the question makes no sense, because the issue is a task-specific trade-of and the term soap is too broad. It is not even clear, whether you want a detergent for water or a e.g. fluorinated oil.

"the question makes no sense" Apparently it makes some sense to several people.

No it does not make any sense, because of a too drastic oversimplification. The correct question would be, "what influences the property of a "soap" in a certain solvent".


text:
molar solubility of CaF2 at pH3 given molar solubility at pH7 and pKa of HF

Here was my attempt #$1$ at a solution.
$K_{sp}=[Ca^{2+}][F^-]^2$
Using the information at pH=7, I estimated that $K_{sp}\approx3.7\cdot10^{-11}$, which means that barely any HF formed.
I assumed that only $HF$ forms and not $CaOH$ because it's insoluble.
I got $\displaystyle \frac{[H^+][F^-]}{HF}=10^{-3.17}$
We have $  \displaystyle \sqrt{\frac{3.7\cdot10^{-11}}{[Ca^{2+}]}}$ fluoride ions.
We rewrite $K_a=10^{-3.17}=\displaystyle\frac{10^{-3}[F^-]}{2[Ca^{2+}]-F^-}$ and plug in the concentration of fluoride ions in terms of calcium ions.
This yields $[Ca^{2+}]=3.7*10^{-4}$, which gives a multiplier of 1.76.
The correct answer is 1.83, which I guessed based on my answer. How do I arrive at that?

The $K_{sp}$ of $\ce{CaF2}$ which you found out as $3.7 \times 10^{-11}$ is indeed correct, and after that essentially you have to solve two equations.
Now, suppose at $pH =3$, (i.e. $\ce{[H+] = 10^{-3}}$),  $\ce{CaF2}$ dissociates as $\ce{Ca^2+}$ and $\ce{F-}$ and their concentration in the solution are as follows, $$\ce{CaF2 <=> Ca^2+ + 2F-}$$ $\ce{[Ca^2+ ] = S }$ mol/L. and $\ce{[F-] = 2S -}$ $x$ mol/L. ,as, this $x$ mol/L. of $\ce{F-}$ is in equilibrium with $\ce{HF}$, as, $$\ce{H+ + F- <=> HF}$$ This equilibrium satisfies the equation $$\ce{\frac{[H^+][F^-]}{[HF]}= 10^{-3.17 }   => \frac{10^{-3}(2S - x)}{x} = 10^{-3.17}  ...(i)}$$.
And another equation which you have to solve is just regarding $K_{sp}$, i.e $$S(2S-x)^2 = 3.7 \times 10^{-11} .....(ii)$$ From first equation, you will get $\frac{2S}{x} = 1.676$ ; if you substitute for $x$ in the second equation by this, you can now solve for $S$ and the answer will be $S= 3.845 \times 10^{-4}$.
Thus , the factor will be $\frac{3.845 \times 10^{-4}}{2.1 \times 10^{-4}} = 1.83$

Here was my attempt #$1$ at a solution.
$K_{sp}=[Ca^{2+}][F^-]^2$
Using the information at pH=7, I estimated that $K_{sp}\approx3.7\cdot10^{-11}$, which means that barely any HF formed.

This is sort of backwards. At pH=7 there is barely any HF formed, so $\ce{[F-] \approx 2[Ca^{2+}]}$, and thus $K_{sp}\approx3.7\cdot10^{-11}$

I assumed that only $HF$ forms and not $CaOH$ because it's insoluble.

Nice! But to quibble it should be $\ce{Ca(OH)2}$.

I got $\displaystyle \frac{[H^+][F^-]}{HF}=10^{-3.17}$
We have $  \displaystyle \sqrt{\frac{3.7\cdot10^{-11}}{[Ca^{2+}]}}$ fluoride ions.
We rewrite $K_a=10^{-3.17}=\displaystyle\frac{10^{-3}[F^-]}{2[Ca^{2+}]-F^-}$ and plug in the concentration of fluoride ions in terms of calcium ions.

Again a bit sloppy but all correct to this point. Id write out that $\ce{ 2[Ca^{2+}] =  [F^-] +[HF]}$and thus $\ce{[HF] = 2[Ca^{2+}] - [F^-]}$

$$K_a = 10^{-3.17} =\dfrac{\ce{[H+][F-]}}{\ce{[HF]}} =\dfrac{(10^{-3})\ce{[F^-]}}{\ce{2[Ca^{2+}]-[F^-]}}\tag{1}$$
$$ 6.761\cdot10^{-4} =\dfrac{10^{-3}\ce{[F^-]}}{\ce{2[Ca^{2+}]-[F^-]}}\tag{2}$$
$$ 0.6761 =\dfrac{\ce{[F^-]}}{\ce{2[Ca^{2+}]-[F^-]}}\tag{3}$$
$$ \ce{2[Ca^{2+}]-[F^-]} =\dfrac{\ce{[F^-]}}{0.6761} = 1.4791\ce{[F^-]} \tag{4}$$
$$ \ce{[Ca^{2+}]} = 1.240\ce{[F^-]} \tag{5}$$
$$ \ce{[Ca^{2+}]} = 1.240\sqrt{\dfrac{3.7\cdot10^{-11}}{\ce{[Ca^{2+}]}}} \tag{6}$$
$$ \ce{[Ca^{2+}]^3} = (1.240)^2 \times 3.7\cdot10^{-11}\tag{7}$$
$$ \ce{[Ca^{2+}]} = \sqrt[3]{(1.240)^2 \times 3.7\cdot10^{-11}} = 3.846\cdot10^{-4} \tag{8}$$

So you had the right equations, you just flubbed the math.

Soumik got the right answer, but the method seems weird. Let's do this a different way.
We have three chemical equations to use.
$$\ce{CaF2 <=> Ca^2+ + 2F-}\tag{1}$$
$$\ce{HF <=> H+ + F-}\tag{2}$$
$$\ce{[Ca^{2+}] = \frac{1}{2}([F-] + [HF])}\tag{3}$$
From equation (2) we know that
$$K_a = 10^{-3.17 } = 6.761\cdot10^{-4} = \ce{\frac{[H^+][F^-]}{[HF]}}\tag{5} $$
rearranging we get
$$\ce{[HF]} = \dfrac{\ce{[H^+][F^-]}}{K_a}\tag{6}$$
Substituting this expression for $\ce{[HF]}$ into equation (3) gives
$$\ce{[Ca^{2+}]} = \frac{1}{2}\left(\ce{[F-]}+ \dfrac{\ce{[H^+][F^-]}}{K_a}\right)= \frac{1}{2}\ce{[F-]\left(\dfrac{K_a+\ce{[H^+]}}{K_a} \right)}\tag{7}$$
rearranging we get
$$\ce{[F-]} = \ce{2[Ca^{2+}]}\left( \dfrac{K_a}{K_a+\ce{[H^+]}} \right)\tag{8} $$
From equation (1) we get
$$K_{sp} = \ce{[Ca^{2+}][F-]^2}\tag{9}$$
or substituting (8) and simplfying
$$\ce{[Ca^{2+}]} = \sqrt[3]{\dfrac{K_{sp}}{4\left( \dfrac{K_a}{K_a+\ce{[H^+]}} \right)}}\tag{10}$$
Now for pH=7 and pH=3 let's evaluate the term $4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2$.
$$\text{@pH=7}\quad4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2 = 4\left(\dfrac{6.761\cdot10^{-4}}{6.761\cdot10^{-4}+1\cdot10^{-7}}\right)^2 = 3.999\tag{11}$$
$$\text{@pH=3}\quad4\left(\dfrac{K_a}{K_a+\ce{[H^+]}}\right)^2 = 4\left(\dfrac{6.761\cdot10^{-4}}{6.761\cdot10^{-4}+1\cdot10^{-3}}\right)^2 = 0.6509\tag{12}$$
So
$$\text{@pH=7}\quad K_{sp}= 3.999\ce{[Ca^{2+}]^3} = 3.999(2.1\cdot10^{-4}) = 3.703\cdot10^{-11}\tag{13}$$
$$\text{@pH=3}\quad \ce{[Ca^{2+}]} = \sqrt[3]{\dfrac{K_{sp}}{0.6509}}=3.846\cdot10^{-4}\tag{14}$$
and  finally
$$\text{ratio} = \dfrac{\ce{[Ca^{2+}]}_{\text{pH=3}}}{\ce{[Ca^{2+}]}_{\text{pH=7}}} = \dfrac{3.846\cdot10^{-4}}{2.1\cdot10^{-4}}=1.831 \ce{->[Rounding]} 1.8$$
NOTE - There are only two significant figures in all the data given, so the answer should be 1.8, not 1.83.

i understand both of the methods, and they make perfect sense. so where did I make a mistake in my method?


text:
Organic reduction of alkene
Can Wolff Kishner and Clemmensen Method be used to reduce alkyne or alkene. If so do they produce syn addition or anti addition product. ?

Zinc plus a proton source is known to reduce propiolates in THF Tet. Letts. 46 5613
Rieke Zinc has been reported to reduce alkynes to *cis-*alkenes Tet. Letts. 32 299
If you add catalytic Ni then Zn will reduce alkynes selectively to cis alkenes or trans alkenes depending on conditions 3 though this probably operates through Ni(O).
Zn/H+ does not reduce unactivated alkenes. Alkenes conjugated with an EWG may be reduced.
W-K reduction relies on the hydrazine adding to a ketone to form a hydrazone in the 1st step. Nothing comparable can occur with an alkene. However hydrazine can reduce alkenes under different conditions by Di-imide reduction review article.


text:
structure information form IR and Raman spectra?
Is it possible to gain the idea of full molecular structure from IR or Raman spectroscopy?

Practically speaking: no. While these are very useful methods to show the presence of certain chemical groups, there are many molecular features (eg groups with less characteristic signals, the difference between isomers) that not well represented in the spectrum.

For a pure "simple" molecules you can pattern match an unknown with spectra from a library of known compounds. But pragmatically it is hard to get "pure" compounds. In fact it often words the other way. You know you have a reasonably pure compound when there are no stray peaks when compared to a known pure sample of the compound.

No, you cannot get bond lengths from IR/Raman you need to do microwave spectroscopy and this is only a productive approach for simple molecules. NMR is the only spectroscopic technique, this generally gives topology, i.e. this grouping is next to that, e.g. CH2 next to CH3  etc. Xray diffraction is the only definitive structural method .


text:
If Mg(OH)2 is insoluble, why does the reaction of MgO and water produce a slightly alkali solution?
The equation of the reaction is:
Mg(OH)2 + H2O -> Mg(OH)2.
Why is it that this reaction produces a solution of around pH 9? There are no OH- ions produced.

That is not the definition of "insoluble."

There are lots of hydroxide ions produced.  Most of them are reprecipitated as solid magnesium hydroxide.  Magnesia is not only a base in water, but a moderately strong one.

As you noted
$$\ce{MgO + H2O -> Mg(OH)2}$$
but the Magnesium Hydroxide is not totally insoluble.
$$\ce{Mg(OH)2 <=> Mg^2+ + 2 OH-}\tag{1}$$
$$\mathrm{K}_{\mathrm{sp}} = 5.61\times10^{-12}= \ce{[Mg^2+][OH-]^2}\tag{2}$$
The dissolution of $\ce{OH-}$ from the $\ce{Mg(OH)2}$ is in equilibrium with the autodissociation of water:
$$\ce{H+ + OH- <=> H2O}\tag{3}$$
with
$$\mathrm{K}_\mathrm{w} = 1.00\cdot10^{-14} = \ce{[H+][OH-]}\tag{4}$$

We will ignore $\ce{CO2}$ from the atmosphere making $\ce{CO3^{2-}}$...

Let's assume that a "considerable" amount of $\ce{Mg(OH)2}$ dissolves, but not all, so that we can ignore the initial amount of $\ce{H^+ \text{ and }OH-}$ from the autoionization of the pure water. Then using $\ce{[OH-] = 2[Mg^{2+}]}$ we can substitute into equation (1) for $\ce{[OH-]}$ and get
$$\ce{[Mg^{2+}]} = \sqrt[3]{\dfrac{5.61\times10^{-12}}{4}} = 1.1194\cdot10^{-4}\tag{5}$$
which means that
$$\ce{[OH-]} = 2\times1.1194\cdot10^{-4} = 2.2388\cdot10^{-4}\tag{6}$$
This is basic enough that the autodissociation of water can be ignored, so our assumption was valid. We can calculate $\ce{[H+]}$ from equation (4):
$$\ce{[H+]} = \dfrac{1.00\cdot10^{-14}}{\ce{[OH-]}} = \dfrac{1.00\cdot10^{-14}}{2.2388\cdot10^{-4}} = 4.4667\cdot10^{-11}$$
and
$$\mathrm{pH} = -\log{(4.4667\cdot10^{-11})} = 10.35002 \ce{->[Rounding]} 10.350$$

Because Ksp of Mg(OH)2 is not zero !
First of all understand that solubility (diffusion of ions from solid to liquid phase) has some equilibrium constant. And that K (eq.constant ) of any reaction can't equal zero. So, for this small but finite K, some ions Mg and OH diffuse into the water and make the solution Alkaline !

An insoluble base means that the compound $\ce{XOH}$ remains as its own phase (usually solid) in the medium, while some of it dissolves in the water as $\ce{X+}$ and $\ce{OH-}$. In this case, the $pH$ of the medium is dictated by the solubility product ($K_s = [\ce{X+}][\ce{OH-}]$), instead of the acid/base constant as for something like $\ce{NH3}$ - where both itself and its conjugated acid $\ce{NH4+}$ exist in solution and in equilibrium and the $pH$ will be defined by the relation $K_b = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]}$. Note that because the base itself is insoluble, it does not appear in the equilibrium constant because it has no chemical activity in the aqueous medium.
In a complete counter scenario, something like $\ce{NaOH}$ is considered a soluble base, but in this case the conversion into ionic species in solution is total, hence the base is also considered strong. For those, the pH is controlled by the concentration of the base itself ($pH = 14 - pOH$ with $pOH = -\log([\ce{NaOH}]$).
All the considerations above of "pH determination" are valid for systems where the base is the only solute in water.


text:
What is the purpose of tetrazole in oligonucleotide synthesis?
The ATDBio book on nucleic acids describes tetrazole as an activator required to couple nucleotides.

The diisopropylamino group of the nucleoside phosphoramidite is protonated by the activator, and is thereby converted to a good leaving group. It is rapidly displaced by attack of the 5′-hydroxyl group of the support-bound nucleoside on its neighbouring phosphorus atom.

The first molecule in the scheme they provide is clearly already protonated; a second step in which the diisopropyl amino group is replaced by a tetrazole molecule is displayed.
What is the purpose of this step; why doesn't the support bound nucleotide attack the phosphorous, directly? Are the arguments purely kinetic (i.e. sterics and availability)?

Er no, the first molecule in the scheme is not protonated - it is protonated by the tetrazole. The alcohol will not, on its own, displace the di-isopropylamino group from the phosphorus centre.

In the image in the question, the leftmost amine is already drawn in its protonated form as $\ce{HNR3+}$ and is nucleophilically substituted for tetrazole, no?

The language here is not quite clear. That is a substitution and there is a nucleophile involved, but the arrow-pushing shown for step 1 (or step 2, for that matter) are almost certainly not what mechanistically happens in this process. Phosphorus compound do not under $S_{\mathrm{N}}2$ reactions normally; they undergo addition-elimination-type substitutions.

@jacob No, it is shown as protonated by the tetrazole

@Waylander, it may well have been protonated by tetrazole but what I’m trying to say is that this is not shown as an individual step in the diagram. The non-protonated form is never represented diagrammatically. Regardless, that’s really not the point of the question.

Each step of catalysed reaction is much easier then uncatalysed, because activation energy is lower. Your tetrazole is both better nucleophile then alcohol and leaving group then amine.


text:
Do molecules containing only carbon exist?
Why doesn't carbon form bonds with itself to form a molecule? Carbon shows the property of catenation, so why doesn't it form a cyclic molecule such as:
 C = C
 ‖   ‖
 C = C

graphene, diamond, fullerenes ...

also relevant : https://chemistry.stackexchange.com/questions/43887/how-small-is-the-smallest-known-carbon-ring-containing-only-double-bonds

buckyball also...

Diamond and graphene if not engineered down do not qualify as molecules. Buckyballs is a nik for Buckminsterfullerenes. See Oscar Lanzi below for an answer

That specific molecule would be monstrously unstable due to ring strain. But plenty of other molecules with just carbon would not be and many exist.

They do form.  Besides fullerenes, there are carbon molecules with varying numbers of atoms around some stars[1]. Vaporized graphite in such envirinments produce chains of 3 to 30 carbobs and larger spheroidal clusters of 40 to as many as 100 carbins. These may combine with hydrogen and nitrogen to form circumstellar polyynes. Such molecular forms are, of course, less stable than the macromolecular structures we more commonly see under ambient conditions.
Reference

H. W. Kroto, J. R. Heatth, S. C. O'Brien, R. F. Curl. R. E. Smalley (1987). "Long Carbon Chain Molecules in Circumstellar Shells". The Astrophysical Journal 314, 352-355.


text:
How do you call the opposite of an ion?
According to IUPAC, an ion is defined as:

An atomic or molecular particle having a net electric charge. (source)

But how do you call an atomic or molecular particle that does not have a net electric charge? Is there a noun or an adjective that one can use to refer to the opposite of an ion?

"Atom" or "molecule" sounds good enough for me. If you want an umbrella term, go with "uncharged particle". I don't think there is a single word for that.

I checked: according to IUPAC, an "atom" consists of Z protons and Z electrons (source). That is, by definition, an "atom" has a net electric charge of zero. I did not expect this, because using this definition, ions do not qualify to be called atoms. That doesn't seem to reflect the common usage of these terms.

In my field, one could call it a neutral.

The common usage of many terms is just wrong.

@Ivan Well, the idea behind having definitions is to reduce ambiguity, but obviously in this case the definition actually introduces ambiguity itself, because IUPAC's idea of an atom is very different from how the word is actually used. Therefore I think talking about "atoms" would be misleading. I appreciate the other suggestions though.

@FeliGoldstein Atom and molecule definitely neutral. "Common usage" depends on context. My adviser, for one, insisted that terms be correctly used, so the common usage was the correct usage. For example, you abstract a hydrogen atom in a radical reaction but deprotonate a hydrogen ion.

@FeliGoldstein What ambiguity are you talking about? How else is the word 'atom' actually used?

atom is occasionally used when the term nuclei would be more appropriate.

When or where an ion is called atom? Never saw/eared that.


text:
Fischer Projection

i'm confused of how to get the Fischer projection of the molecule on the top. i can see that the hydrogen and OH group are sticking out.
However i don't understand how the CH2OH group is pointing away. From my perspective the bond it is pointing straight down. i would appreciate some help.

Forget the table.

Stand up higher so that that bond is pointing more away from you.

You drew it correctly. Fischer projections don't give you the option of drawing a bond in the plane of the paper. You have to choose (a) coming out of the page, or (b) going behind the page. In your example, the hydrogen and hydroxide are definitely coming out of the page. The other two bonds have to be represented as going back into the page.
It might make more sense if you imagine rotating the molecule a little so the CH2OH is actually pointing back a little. You would still have the hydrogen and hydroxide coming forward and the aldehyde going back.

Here is a 3D tutorial: http://proteopedia.org/wiki/index.php/Fischer_projection


text:
Homemade spectrophotometer to exemplify Beer-Lamberts law
I have built a 12-band spectrophotometer, and i'm currently keen on proving Beer-Lamberts law through a small practical example.
The solution that I'm going to test with my spectrophotometer, is orange juice that I thin with water. I naturally know I0 and I1, I.e. the absorption, and the path length that the light has to travel. Meaning that I have 2 unknowns, respectively the concentration and the molar absorptivity.
I am aware that I can figure out the molar absorptivity by doing multiple assays at different concentrations, plot them into a coordinate system and find the slope between the coordinates, which will be equal to epsilon*pathlength.
Unfortunately, I don't know how I find the molarity concentration of the orange juice that I intend investigate. I do however have the weight and the amount. Will that be sufficient to calculate the concentration molarity?

The molar absorptivity is a proportionality constant. You do not know what is the compound responsible by the color. So, this constant will not be in L/mol.cm but will be in terms of the concentration units. For example, if you use the mass to express the concentration (g/L) the constant will be express in this unit L/g.cm. With this system, you can estimate concentrations and so on. To known the molar absorptivity you must know the concentration in mol/L and it is not possible with this experiment.

Thank you very much for your answer. What would you suggest that I do instead? I only have access to "home-chemicals", and my spectrophotometer measures intensities from 450 nm to 860nm with roughly 50nm intervals between each of the 12 bands.

I know that carotenoid is responsible for the yellow color, but i don't know the molecule structure and exactly what kind of carotenoid that is in oranges.

Why you need the value in mol?

To know the molar absorptivity you need the pure compound to produce a solution with a known concentration. Why are you using orange juice? What is your objective?

Just to prove that there is a coherence between beer lamberts law and what i can observe with my spectrophotometer. However, i guess i can still prove it, since, as you say, epsilon and c are proportional. epsilon doesn't necessarily need to depict the M^-1 cm ^-1 right?
Meaning, i can do a lot of observations at a certain wavelength, at different concentrations, and then based on my regression (which obviously is linear) be capable of plotting my OBSERVED constants into beer lamberts law?

Exactly, if you plot the absorbance versus concentration (g/ml) you prove the linearity of your equipment. In this case, your epsilon will be in L.g-1.cm-1 instead L.mol-1.cm-1 but it doesn't matter. If you want you can buy some copper sulphate used in flower shops, purify by crystallization and obtain a constant in molar units for this case...


text:
The mechanism of the reaction between 2-chloroethanol and hydroxyl ions in aqueous solution
The mechanism of the reaction between 2-chloroethanol ($\ce{CH2ClCH2OH}$) and hydroxyl ions in aqueous solution forming ethylene oxide ($\ce{CH2CH2)O}$ probably contemplates two steps:

$\ce{CH2ClCH2OH + OH- <=> CH2ClCH2O- + H2O}$
$\ce{CH2ClCH2O- <=> (CH2CH2)O + H2O + Cl-}$

Is the mechanism valid for the following experimental reaction rate?
$\frac{-d[\ce{CH2ClCH2OH}]}{dt} = K_{exp}[\ce{CH2ClCH2OH}][\ce{OH-}]$

According to Ballinger and Long (1959), the reaction mechanism is as such:
$\ce { (1) CH2ClCH2OH (aq) + OH^- (aq)  -> CH2ClCH2O^- (aq) + H2O (aq)}$ (equilibrium)
$\ce { (2) CH2ClCH2O^- (aq) -> (CH2CH2)O (aq) + Cl^- (aq)}$ (slow)
This is not surprising since acid-base reactions, such as these proton transfers, take place very quickly. Although the second step may appear to be intramolecular, it may be slower since we are forming a strained epoxide ring and the $\ce {C-O^-}$ and the $\ce {C-Cl}$ bonds also need to be in a trans orientation for most effective orbital interaction of the relevant orbitals. Based on the above mechanism, your rate law seems to be correct, based on the assumption that the first step is much slower than the subsequent step and has come to establish an equilibrium.
Reference
Ballinger, P.; Long, F. A. The Reaction of Chlorohydrins and Hydroxide Ion in the Solvents H2O and D2O. J. Am. Chem. Soc., 1959, 81 (10), 2347–2352.

Yes it is. Because alcohol is a weak acid and it is reasonable to consider first step is irreversible because of the second step almost occuring simultaneously. And also because that the second step is much faster than the first.


text:
Multiplication symbol in SMARTS
There is a SMARTS pattern that was published which looks like the following[1]
[#6 × 4]

and there are actually spaces around the × symbols between the 6 and 4.
Question is, should I change this multiplication notation to a lower case "x", or really use some sort of ASCII symbol?  I didn't see a multiplication symbol anywhere in the Daylight SMARTS description.
References:

Hou, T. J.; Xia, K.; Zhang, W.; Xu, X. J. ADME Evaluation in Drug Discovery. 4. Prediction of Aqueous Solubility Based on Atom Contribution Approach. J. Chem. Inf. Comput. Sci. 2004, 44 (1), 266–275. PDF

The question was answered on the Open Babel mailing list
Suffice to say, there's no multiplication symbol in SMILES or SMARTS - only standard ASCII characters. It's likely a lower-case 'x' that's been mangled by a journal.


text:
Is an isobar the same as an isotope?
I am a little bit confused about what an isobar is. Its online definition is that it's an element with the same number of neutrons but a different number of protons from an element $\ce{X}$.

To me, it doesn't make sense from the get-go, because once you change the number of protons the element changes as well so why exactly is it defined as the same element $\ce{X}$ with the same number of neutrons and a different number of protons.
Definition of an isotope: An isotope is an element $\ce{X}$ with the same number of protons and a different number of neutrons.

So to the actual question now. Isn't an isobar just an isotope? Here is an example to clarify what I mean.
If we take for example carbon $\ce{^12C(p:6, n:6)}$ and turn it into an isotope it will be $\ce{^13C(p:6, n:7)}$, and that makes sense, but if we turn it into an isobar it would be $\ce{^13C(p:7, n:6)}$, which doesn't make sense, because it looks exactly like an isotope of nitrogen $\ce{^13N(p:7, n:6)}$.
If the atomic number changes than the element changes as well. So isn't an isobar just an isotope of the following element with a smaller neutron number?

Related: What are compounds with the same mass called? and What are isodiaphers?

To me an isobar is a line on a weather chart connecting points of the same atmospheric pressure.

It is easier to think of isotope and isobar as a relation between two or more nuclei. Any given nuclide, such as $\ce{^12C}$ is both an isotope and an isobar. The isotopes to $\ce{^12C}$ are $\ce{^13C}, \ce{^14C}, \ce{^15C}, \ldots$ and $\ce{^11C}, \ce{^10C}, \ce{^9C}, \ldots$. And the isobars to $\ce{^12C}$ are $\ce{^12N}, \ce{^12O}, \ce{^12F}, \ldots$ and $\ce{^12B}, \ce{^12Be}, \ce{^12Li}, \ldots$.

Not quite, an isotope has same number of protons ($ A- N = Z = \mathrm{constant}$), but a different number of neutrons ($\mathrm N$ varies; e.g. $\ce{^3_\color{red}{1}H}$ and $\ce{^2_\color{red}{1}H}$, or $\ce{^235_\color{red}{92}U}$ and $\ce{^238_\color{red}{92}U}$ are isotopes).
An isobar has a fixed number of total nucleons ($Z + N = A = \mathrm{constant}$; e.g. $\ce{^\color{red}{40}_19K}$ and $\ce{^\color{red}{40}_20Ca}$, or $\ce{^\color{red}{3}_2He}$ and $\ce{^\color{red}{3}_1H}$ are isobars). Not nearly as mainstream as isotopes, but isobars are important to consider when doing mass spectroscopy.
Extra fact: For nuclei of the same number of neutrons ($A - Z = N = \mathrm{constant}$), the term is isotones.

People who don't know as much as A.K. often use "isotope" to mean "you know, like, carbon-14 or something", i.e. all nuclei with a given pair of values for $Z,,N$. In case anyone's curious, the term you want for that is nuclide.

I believe the definition you found may have been a little bit misleading. Here is another definition of isobar I found: each of two or more isotopes of different elements, with the same atomic weight. An isobar is referring to completely different elements. The prefix iso- means only one component must be the same between the different elements, and in the case of an isobar: mass.
I saw in your question the example that an isobar of carbon-13 would look like so $\ce{^13C(p:7, n:6)}$; however, this is not proper notation, as an isobar cannot be of the same element. Once the number of protons changes, your element is no longer the same.
Isobars are simply two different elements with the same mass while isotopes are two of the same elements with different masses.

Isobar is of more interest to physics than chemistry.
As others have explained, your definition is confusing.  The one here may be clearer: isobar (Wikipedia).
In chemistry, the number of protons is most significant since it determines the number of electrons and hence the chemical behaviour.  The number of neutrons is relatively unimportant: variants may be useful for labeling and may have slightly different behaviour (most noticeable for $\ce{^1H}$ and $\ce{^2H}$).  So, isotope is used fairly frequently to discuss these variants of the elements.  Isobars may have very different chemical behaviour and are unlikely to be an interesting grouping.
In nuclear physics, the number of neutrons and protons have a similar significance hence the term nuclide (Wikipedia) for atoms with a specific number of protons and neutrons is useful.  Isobars have a closer connection with each other than isotopes since isobars can interconvert via beta processes relatively easily.  It is rare for an atom to decay into an isotope.  For example, $\ce{^{14}C}$ decays into $\ce{^{14}N}$  rather than $\ce{^{13}C}$ .
Isobars will be relevant in some specific situations in chemistry such mass spectroscopy.  So, it is still useful to understand the concept.

Thank you for the explanation. Yes, your right, this question is more of a physical topic than chemical topic, but since the first time I heard about it was in chemistry I mixed it up. Thank you for pointing that out. I will look into the right stackexchange next time.

@BeatriceHurbean I'm sure isobar is discussed more in physics than chemistry, but as Loong's related post suggestion shows isobars are important to consider in mass spectroscopy, so very much a chemistry topic too.

In a nuclear physics context, one considers neutrons and protons as two different states of one object called a nucleon. This means that nuclei which are isobars are treated as a nucleus with some number, A of nucleons but having different charges (given by number of neutrons - number of protons, since N-P differs for different nuclei with N+P=A). This is useful for studying nuclear structure, since the assumption about the strong nuclear force binding neutrons and protons is that it does not depend on the electric charge. The coulomb energy is a separate contribution.
That implies that the mass difference for a given set of isobars, say, for A=53, 53Co, 53Fe, 53,Ni, 53Mn, should depend only on the difference in the coulomb energy since the strong force doesn't distinguish between neutrons and protons. It can be shown that this mass difference should fit a parabola. If you have a multiplet with more than three members, then you can test the assumption about the charge independence of the nuclear force and perhaps learn additional information about nuclear structure just by measuring the masses of those isobars.  For more detail, you can look up Isobaric Mass Multiplet Equation and/or nuclear isospin.


text:
How to distinguish between an insoluble ionic compound and an insoluble metallic compound?
Insoluble ionic compounds and insoluble metallic compounds could both have high melting points, insolubility, and no conductivity in water. Is there a way to distinguish between the two in a laboratory setting?

How well does solid salt conduct electricity?  Now how about solid copper?

Welcome to Stack Exchange! Could you edit your question? It doesn't make sense right now: "Insoluble ionic compounds and insoluble ionic compounds..." I'd edit, but I'm not sure if you mean insoluble/soluble ionic compounds, or insoluble ionic compounds and metallic compounds.

@miltonaut Ok, thank you! I just edited the question.

Whack it with a hammer. Metals are malleable and ionic salts are not.

@MaxW The compound we used were in tiny pieces. Is there another method I could have used?

As solids, metals...

conduct heat and electricity well (ionic solids do not)
are malleable and ductile (ionic solids are not)
have lower melting points and boiling points than ionic solids

Test these properties by...

applying heat and a thermometer; using a conductivity apparatus (larger/chunk samples)
hit the sample with a hammer as MaxW suggested (larger/chunk samples)
test for relative melting/boiling points by applying equal amounts of heat (granular samples)

What if the unknown substance was in little pieces (like iron filings) and there were only conductivity testers that worked on solutions? I'm just curious if there is another method.

Test #3 - relative melting points. The "points" of a substance are intensive properties. As long as you have similar amounts, and equipment to carry out the tests, it should work.


text:
Collison of two alpha particlesre
I was reading on nuclear chemistry - in particular, the bombardment of nuclei with alpha-particles - and was wondering what the effect of
$\ce{^4_2He + ^4_2He -> ^8_4Be}$
would be. My instinct is to recognize this as a "fusion"-like reaction, but I am not comfortable enough with the subject to fully calculate it on my own.

The Be-8 nucleus lies almost 1MeV above a ground state of 2 alpha particles. The nuclear reaction Li7 (p,$\alpha$) $\alpha$ was the first human-controlled reaction by Cocroft and Walton. So, no, you can’t fuse two alphas.

Just to be clear, it isn't that you can't collide two alpha particles, the problem is that the $\ce{^8_4Be}$ nucleus is highly unstable and decays to yield two alpha particles. To get a stable product 3 helium nuclei have to collide within a very short time to yield $\ce{^12_6C}$ .

My memory is clearly failing - the $\alpha$+$\alpha$ state is only 98keV below Be-8. Still, the Be-8 lifetime is 6E-17 sec, so it really likes falling apart.

This is a very important nuclear reaction that happens in red giant stars that have burned all their hydrogen so they start burning He instead. As stated by Jon Custer, the $^8$Be nucleus is unstable so its lifetime is short. At the temperature in a star, a lot of collisions between two alpha particles have the same energy as the Be nucleus with respect to two free alpha particles resulting in an efficient production due to resonance (the kinetic energy lies very close to the Gamow peak).
The density of He in the star is so high that the Be nucleus can collide with another alpha particle before decaying into two alpha's again. This process forms a nuclear excited state of carbon, which by accident, also has roughly the same energy with respect to the free particles as the collision energy of the particles in the star so that there is another resonance phenomenon. There is a small change that the Carbon atom decays to its ground state before falling apart in 3 alpha particles. This process is slow, but since it happens over cosmological time scales it is nevertheless very important and responsible for the formation of all elements heavier than lithium.
See also this explanation on the hyperphysics page.

An easy place to find this (correct) info is at TUNL (http://www.tunl.duke.edu/nucldata/index.shtml) - in particular, find the 'Energy Level Diagram' page for the appropriate A (total protons and neutrons) and pull up the isobar diagram.


text:
Resonance in benzene: creation of current
I am a student. I recently studied resonance in Benzene. I know that π electrons in Benzene are involved in resonance this means the electrons somewhat move in the process of resonance.
I have a question :
If electrons are moving in Benzene there  should be a current?
However I have not seen any such fact in the books i am currently studying. Please tell me if this thing happens in Benzene or not.

The current is within the molecule, the more interesting question, is shouldn't there be a magnetic field?

See https://en.wikipedia.org/wiki/Aromatic_ring_current

I can't comment, but benzene has a ring current and that creates a magnetic field perpendicular to the current. Conversely, a magnetic field will create an induced current (Ampere's Law) in the ring (delocalized electrons from the pi bonds).


text:
Can you make imines from Formaldehyde or Acetaldehyde?
Title says it all. Wondering if imine formation would work with an aldehyde, even better would be formaldehyde.

Yes it does. They are reactive species, methylene imines particularly are easily hydrolysed back to amine + aldehyde.
They are the basis of reductive methylation/alkylation of primary amines where reaction with alkyl halides, especially reactive ones such as MeI risks dialkylation. More details here reductive alkylation

Thank you, I figured it would but couldn't wanted to make sure before I used it in a presentation. Also, I was wondering about alkylation of ammines for a while now!

@ronnie   Please accept the answer if you are happy with it


text:
Most reactive site for electrophilic aromatic substitution
I was given a problem in which I had to identify the most reactive site for electrophilic substitution out of the positions 1,2,3,4 (shown in the figure).

My Approach:  Upon drawing the resonating structures of the molecule, negative charge appears on position 3 and 2.
In the resonating structure in which the negative charge is under 3, there is a positive charge under -NO2 group making that structure less stable.
This does not happen with the structure in which the negative charge is under 2.
So I concluded that 2 would be the most reactive site.
But the answer given is '4'.
What is wrong in my reasoning?

@AvnishKabaj I think that should happen only if bulky groups are attached to the phenyl ring which would align them in mutually perpendicular planes because of free rotation. Can you give a source which mentions this?

@AvnishKabaj Also, phenyl ring is said to be ortho para directing because of its resonance effect, (while it deactivates the ring via inductive effect).

consider only the deactivating effect of the ring.. 4 is the best option because that ring is comparitively less deactivated

@AvnishKabaj Look at the substitution on that specific biphenyl in the linked question...

Waaiiiit the answer below just reiterated what I said

As you know, in deciding which position undergoes aromatic electrophilic substitution, three elements must be taken in account: steric effect, mesomeric effect and inductive effect.
Let's call the ring containing atoms 1 and 2 "A", and the one containing atoms 3 and 4 "B".
Before deciding the importance of any of those effects, some considerations about the geometry of the molecule should be taken in account. It could be guessed, by thinking about the spatial arrangement of such molecule, that the high steric effect between rings A and B prevents their rotation, thus keeping them in a constantly "skewed" position in respect to the nitrated ring.
A representation of the molecule, with the Van Der Waals radius of the elements, follows:

Steric Effects: Because of the presence of  a nearby phenyl (the nitrated one) to any of the ortho positions of rings A and B, and because of the presence of the other phenyl in the middle of the possible electrophile's path, all the ortho positions might be considered as non favourable.
Mesomeric effect: One might guess that the resonance structures of the nitrated ring could extend to ring B, while leaving ring A untouched.
Actually, such a resonance structure requires a "double bond behaviour" for the aromatic carbons between ring B and the nitrated ring. But the molecule is extremely unlikely to behave that way, since the copresence of two aromatic rings prevents their own rotation, and locks them in a skewed position. This conformation, in other words, renders the pi system "less conjugated", and the mesomeric effect much less relevant.
Inductive effect: This is the final and, in this specific molecule, most important consideration. Simply, everything boils down to the proximity of the nitro EWG: this makes ring A more electron-poor, and ring B more electron-rich.
The answer is, hence: the meta position of ring B is attacked, or the "4" position in your image.

's if rings are skewed and mesomeric effect is very very less then it stands to reason that meta substitution is not possible. During eas intermediate formed could be more stable if para attack takes place.Inductive effect from NO2 inductive effect decreases with distance and could be negligible .


text:
Equation for the equilibrium constant for the hydration of carbonyls using UV spectroscopy
I was looking through the literature when I found a paper by Greenzaid et al. [1] which states that the equilibrium constant $K_\mathrm{hyd}$ is:

$$K_\mathrm{hyd} = [\text{hydrate}]/[\text{carbonyl}] = (\varepsilon_0^\mathrm{w} - \varepsilon^\mathrm{w})/\varepsilon^\mathrm{w}, \label{eqn:1}\tag{1}$$
where $\varepsilon^\mathrm{w}$ is the molar absorption coefficient measured under conditions of hydration, and $\varepsilon_0^\mathrm{w}$  is the molar absorption coefficient of the carbonyl compound in water in the absence of hydration.

(In water solvent, $\varepsilon_0^\mathrm{w}$ usually taken as the molar extinction coefficient in cyclohexane.)
How did the researchers get to the above equation in terms of the molar absorption coefficients?
As $[\text{hydrate}]$ is equal to $[\text{carbonyl}]_0 - [\text{carbonyl}],$ the equation \eqref{eqn:1} turns to:
$$K_\mathrm{hyd} = \frac{[\text{carbonyl}]_0 - [\text{carbonyl}]}{[\text{carbonyl}]}$$
However, the Beer–Lambert law states that
$$A = c\varepsilon l,$$
where $A$ is the optical density, $c$ is the concentration, $\varepsilon$ is the extinction coefficient and $l$ is the path length.
Denoting the corresponding parameters for carbonyl without hydration as $A_0$ and $\varepsilon_0$, and the parameters for carbonyl with hydration as $A$ and $\varepsilon$, the following can be written:
$$
\begin{align}
K_\mathrm{hyd} &= \frac{[\text{carbonyl}]_0 - [\text{carbonyl}]}{[\text{carbonyl}]} \\
  &= \frac{\frac{A_0}{\varepsilon_0l} - \frac{A}{\varepsilon l}}{\frac{A}{\varepsilon l}} \\
  &= \frac{\frac{A_0}{\varepsilon_0} - \frac{A}{\varepsilon}}{\frac{A}{\varepsilon}} \\
\therefore  &= \frac{\varepsilon}{A}\left(\frac{A_0}{\varepsilon_0} - \frac{A}{\varepsilon}\right) \\
  &= \frac{\varepsilon}{A}\frac{A_0\varepsilon - A\varepsilon_0}{\varepsilon\varepsilon_0} \\
  &= \frac{A_0\varepsilon - A\varepsilon_0}{A\varepsilon_0} \\
\end{align}
$$
But we require
$$K_\mathrm{hyd} = \frac{\varepsilon_0 - \varepsilon}{\varepsilon}$$
Shouldn't this mean that $K_\mathrm{hyd}$ is actually proportional to a fraction containing the $1/\varepsilon$ values?
References

Greenzaid, P.; Rappoport, Z.; Samuel, D. Limitations of Ultra-Violet Spectroscopy for the Study of the Reversible Hydration of Carbonyl Compounds. Trans. Faraday Soc. 1967, 63 (0), 2131–2139. DOI: 10.1039/TF9676302131.

The assumption discussed by the paper (which they argue should be discarded because it is not a good one) is that the absorption of carbonyl species is independent of solvent.  That is the same as saying the concentration of carbonyl is uniquely determined by absorbance.  The other assumption is that the only thing that affects absorbance of a compound is the extent to which its hydrated: hydrated carbonyls are assumed not to absorb at all, and apparent per-molecule absorbance of carbonyl compounds in water is thus lowered by an amount proportional to the equilibrium constant.

Under those assumptions, there is no real difference between $A$ and $\epsilon$, they are both direct measures of the concentration of carbonyl in a solution.  They key is that the units of $\epsilon$ are L/mol/cm, where mol is mole of total carbonyl containing compound.

The law is not as you state although it often gets reported as this. The '$A$' you quote is the optical density. The Beer-Lambert law is $I_{tr}=I_0e^{-\epsilon_\lambda [C]\ell}$ where $I_{tr}$ is the intensity of transmitted light for a molecule at concentration $[C]$ at wavelength $\lambda$ and cell path length $\ell$, and $\epsilon_\lambda$ is the extinction coefficient at wavelength $\lambda$.
From the definition you can see where the expression you ask about comes from even with their different notation; it looks as if $w=-{\epsilon_\lambda [C]\ell}$ where ${\epsilon_\lambda [C]\ell}$ is the optical density.

I'm afraid I still don't follow. In your last equation, what is w, and why is it the negtative of the optical density? Additionally, doesn't it still imply that the extinction coefficient at a given wavelength is still inversely proportional to the concentration of the species? Working through the maths I get the relation:
(e(water, measured) - e0) / e0, where e0 is the extinction coefficient for the initial concentration and e(water, measured) is equilivent to eW in my post.

The equation you quote has $\epsilon^{+w}$ and from Beers law it should be $e^{-\epsilon [C]L}$ so $-w$, it may be my misunderstanding from what you wrote I has assume that it was a typo and $\epsilon^{w}$ should actually be $e^w$.

Ah that’s my fault - I didn’t define the terms in my equation because they were defined in the link. I’ve now edited it to make it more clear - the terms in the equation are all extinction coefficients; the W superscript says that they are the extinction coefficients in water solvent. This means when I wrote e in my first reply to you I meant epsilon, the extinction coefficients.

I realized that in my calculations I erroneously cancelled Ao and A with each other. Now working through the mathematics I can't seem to get the desired equation at all.