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Bond dissociation energy of the trihydrogen cation
Recently, in one of the questions in an assignment on the topic of energetics, we were asked to explain why the bond dissociation energy for the trihydrogen cation is much larger than what we would expect it to be. The value given was +849 kJ/mol. To clarify, the bond dissociation process refers to the conversion of the molecular cation into 2 individual hydrogen atoms and a proton.
My proposed explanation is that the additional interaction with the proton lowers the bonding MO of the molecular ion overall. Essentially, I viewed the interaction of the $\sigma$ bonding MO of $\ce {H2}$ with the low-energy 1s atomic orbital of the proton, giving a bonding MO which would be even lower in energy.
The answer given by the teacher is that this molecular ion is particularly stable due to the $\sigma$ aromaticity present in the ion. This results in a very large bond dissociation energy since by breaking up the molecular ion, we destroy the aromaticity of the ion.
I acknowledge that the teacher's explanation is correct but is my explanation also a valid way of looking at the problem?

I wouldn't bother with aromaticity at all and pointed that free proton has so hight charge density, that it attaches itself to noble gases (including helium). Also, yes, the bonding orbital of the cation is very low, but there is no need to invoke any aromaticity concept here, it is enough to say that it has no nodes and so the majority of electron density is located between the protons.

The enhanced stability of the bonding orbital is aromaticity at work, no?

@OscarLanzi Yes I agree. In fact, I feel that my explanation is limited in a sense that I am looking at a localised interaction between H2 molecule and a proton, which is definitely less accurate a representation than an aromatic delocalised molecular ion. But how would MOs look like for H3+, if it were aromatic? Any ideas?

Much like the Frost circle for cyclopropenyl cation, of course.


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Hybridisation in Molecular Orbital Theory
How loosely can the term hybridisation be incorporated while constructing molecular orbitals? Does hybridisation hold any significance when looking at molecules through the Molecular Orbital Theory?

Hybridisation is a convenient tool for chemical interpretation of bonding situations, and thus people like to use it for explaining or even predicting stuff.
From a purely theoretical point of view, you don't really need it. You could construct your basis set by building hybrid orbitals from the atomic orbitals, as some kind of initial guess. Also localization of the resulting molecular orbitals may give something similar to hybrid orbitals. But all this won't change the total energy or the actual electronic wave function of your system. It is just a unitary transformation leaving the final result invariant.

source: Stack Exchange

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What is the chemical equation for extracting biodiesel from cooking oil? (Half complete)
what i got so far:
(k,1,m,n,o is a constant because i dont know the ratio between the chemicals used in the reaction)
kCooking oil(l) + lCH3OH + mNaOH -> nBiodiesel + oC3H8O3
can you tell me the chemical formula for cooking oil and biodiesel, and fill out the constants for me? thanks :D

Neither biodiesel nor cooking oil is a specific chemical substance, so they do not have a certain chemical formula.

Actually you made a slight mistake here in the question. NaOH does not take part in the reaction at all.It acts as a catalyst.
The question should be:-
                             NaOH or KOH
k Cooking oil + l CH3OH ----------------------> n Biodiesel + o C3H8O3.(k,l,n,o be the variables).
Now the cooking oil and the biodiesel are both esters.Converting vegetable oil into biodiesel is called a transesterification reaction.The reaction is given below in form of a diagram.

So the reaction ends up with all gaps filled as:-
C51H98O6(COOKING OIL)+3 CH3OH(METHANOL)->3 C17H34O2(Biodiesel)+ C3H8O3(Glycerol)
SO k=1
l=3
m=invalid as NaOH is catalyst so it does not take part in reaction directly.
n=3
o=1
(GENERAL formula of cooking oil is C51H98O6 and that of biodiesel is C17H34O2.)

oh wow, thank you so much man. that really helped me

You are always Welcome.

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Collision rates between two molecules in a liquid
How would I calculate the collision rate between kaolinite and amylopectin particles in a liquid with a density of ($\pu{1.01g/m^3}$) and a viscosity of ($\pu{0.000434 Pa s}$)? I know this is related to orthokinetic flocculation, however, I do not know the exact equation. If anyone knows the equation, can you please inform me of it?

Your particles are a good deal bigger than molecules.

If you assume that the reaction is diffusion controlled  then this is the fastest that a bimolecular solution phase reaction can be and the second order rate constant is $k=8000RT/(3\eta)$ (dm$^3$/mol/s) where $\eta$ is the solution viscosity and assuming that the molecules are approximately the same size and even if they are not its still a good estimation. In mobile solvents (hexane, water etc)  $k\approx 10^{10} \to 10^{11}$ dm$^3$/mol/s.
If the species a, b are ions (charge $Z_{a,b}$) then the rate constant is multiplied by $\delta/(e^{\delta}-1)$ where  $\displaystyle \delta = \frac{Z_aZ_bq^2}{(4\pi\epsilon_0)\epsilon k_BTd_{ab}}$ where $d_{ab}$ is the centre to centre molecules size, $q$ electronic charge, $k_B$ the Boltzmann constant, $\epsilon$ the solvent dielectric constant and $\epsilon_0$ the permittivity of free space.
[If the species are not the same size radius $r$ then $\displaystyle k= \frac{2000RT}{\eta}\frac{(r_a+r_b)^2}{r_ar_b}$.  ]


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Does presence of alkyl groups on alpha carbon increase yield of alkane in Kolbe's Decarboxylation?
We know that, during the synthesis of alkanes by Kolbe's Decarboxylation reaction,
the reaction proceeds via a free radical $\ce{R.}$ at the anode.
So, an alkyl group, say methyl, when substituted at the alpha carbon will increase the yield of alkane since the radical is stabilized by hyperconjugation. But my book states that the presence of alkyl group on alpha carbon decreases the yield of alkane. Why is it so?

The yield of alkane is not the same as stability of the radical. Self-coupling of these substituted radicals (tertiary?) can yield to disproportionation to an alkane and alkene. E.g.; t-butyl radical to isobutane and isobutylene.

I have done some electrosynthesis work using the Kolbe reaction, what I found was with straight chain carboxylic acids that my platinum anode was quickly fouled.
When I had some branching at the beta site, I had much less electrode fouling and I got a good yield of my alkane.

Beta site? Did you do any work with substitution at the alpha site as the question states?

Sadly no I only worked with things with beta alkyls


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Reaction of primary nitroalkane with nitrous acid
I was reading the reaction of primary nitroalkane with nitrous acid. I read this reaction
Another mechanism which I saw:
I am now confused  whether the first mechanism occur or the second one.I tried to search it on google but didn't get anything which could clear this doubt .Please help me regarding this.This I need to know because the reaction written in my book after 2nd reaction is-
If 1st reaction is right then what will occur after adding $\ce{NaOH}$ in the product. Also when I read Victor Mayer test the 1st reaction was written there. Due to this I am in a dilemma. Please explain.

It could be both.  The two products can tautomerize like a ketone and an enol.

Yes, the two products are tautomers, so the answer is both. And that is not a "mechanism"...

As the comments say, it's both.  The two products can tautomerize in the same way as a ketone and it's tautomeric enol.  In fact, this is basically a keto-enol tautomerization, disguised by occurring around a nitrogen atom instead of carbon.

Could you please tell the mechanism of the reaction?


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Number of geometrical isomers?
How many geometrical isomers will exist for compound of this type which has cycloalkene having carbon atoms greater or equal to 8 in the middle of chain?

I know that due to ring in middle of chain geometrical isomerism exist and also due to cycloalkene having carbon atoms greater or equal to 8. But I am confused in a case when a  cycloalkene exist in middle of the chain. Whether we will consider geometrical isomers due to cycloalkene or due to both, i.e. the ring and also due to cycloalkene case?

Geometrical isomers refers to isomerism in which the connectivity of atoms is preserved, and changes refer only to the spatial arrangement of the atoms. In other words, it only allows transformations around certain features that can be arranged in different ways in three dimensions (and which aren't just conformers that will easily transform into each other due to molecular motion).
I mention this because your reference to a "cycloalkene [...] having [8 or more] carbon atoms" suggest that you are looking for constitutional isomers (i.e. compounds with the same atoms but different connectivity) and not just geometrical isomers (i.e. compounds with the same connectivity but different spatial arrangement).
If we consider geometrical isomers of the compound you gave, 1) the cycloalkene ring, 2) the side chains and 3) the substitution sites in the ring are fixed (as changing them would produce constitutional isomers), so we are left with isomerism due to cis/trans double bonds.
You have three double bonds in the side chains that will produce $2^3=8$ isomers. Cyclooctenes are in principle large enough rings that their trans form can be isolated, so this leads to $16$ isomers; you can figure them out from your posted example (which is cis in the cycloalkene and trans in the side chains) by changing double bonds.
If you want to figure out constitutional isomers, that's a far more complex task; you'll have to account for ring size, number of side chains, ring substitution pattern, branching, position of insaturations and so on; with a hydrocarbon of this size, the number of possibilities will be huge.

could you please explain that- if we just have the ring(without any unsaturation in the ring) then also we will have 16 geometrical isomers i.e. due to the different spatial arrangement of side branches.If we have unsaturation in the ring then won't be there 2× more (i.e 16×2=32) isomers due to the different spatial arrangement of atoms linked with double bonded carbon of the ring?

@Anirudh ,The question is about geometrical isomers. You are also thinking of  Optical isomers (Spatial arrangement of groups around an atom not double bond) which can also be possible. If it is asked to find total no. of stereoisomers, we then also have to think about those. But here in this context, those are not considered.

@SoumikDas thank you for your comment .If we have just a ring in the middle of chain then also there will be 16 geometrical isomers(as disubstituted cycloalkane also show CIS-trans isomerism) and also when a cycloalkene(having  more of equal to 8 carbon) is at the end of  chain then also there will be 16 geometrical isomers .The above case is the combination of two therefore we should consider both(that is my actual doubt)Is my consideration wrong?

As @SoumikDas said, if you consider asymmetric carbons in the cycle, you find four optical isomers for each geometrical isomer (not just cis / trans isomerism, as the two possible cis arrangements are enantiomers, and the same holds for the two possible trans arrangements); so, if we enlarge our scope to consider these, we can find 64 possible stereoisomers, 32 pairs of enantiomers.


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Can a six member ring expand to achieve octet completion to stabilize a carbocation?
(Probably unnecessary background: I came up with this carbocation while predicting the major product of $\ce{NaNO2/HCl}$ with 1-(methylamine)cyclohexan-1-ol)
Will the following carbocation rearrangement occur?

I have opinions both for and against it.
For: The ring expansion leads to octet completion of the empty p orbital of the carbocation. That stabilizes the carbocation.
Against: expansion from six member ring to seven member ring is uncommon. In school, I was taught that - for rings less than size ten - the cyclohexane ring is the most stable ("strain-free"). So, rearrangement shouldn't occur.
A Scholar search gives me no results for the proposed expansion. I know the search probably won't get me anywhere because I don't know the right keywords, and hence I'm stuck. So, can a six member ring expand to achieve octet completion to stabilize a carbocation? Also, does the expansion lead to the major product?

You're basically talking about a pinacol rearrangement or a semi-pinacol rearrangement. It's quite difficult to generate the primary cation, but it can work in an SN2 like fashion.

Would think this is a recipe for epoxide formation

The answer to your question, "can a six member ring expand to achieve octet completion to stabilize a carbocation?" is yes. The most specific and relevant example for that rearrangement is Büchner–Curtius–Schlotterbeck reaction (Wikipedia page).
Your second question is "does the expansion lead to the major product?". The answer is yes, but it is based on the conditions used. If you look at the example you have given, one can determine that the kinetic product is an epoxide (see below). Yet, this reaction is done in acid medium so eventually it proceeds to one ring expansion product (since original ring is symmetrical). Tentative mechanism is given below:

Büchner–Curtius–Schlotterbeck reaction catalyzed by Lewis acids has been used extensively in homologation or ring expansion of aliphatic ketones (addition of diazocarbonyls to ketones would give the exact intermediate as the carbocation structure given in your question) [1, 2]. With unsymmetrical ketones (see above mechanism), questions of regioselectivity in homologation due to different migratory aptitudes of the groups have been studied and concluded that
the migratory aptitudes were uniformly such that the less substituted carbon atom migrates preferentially [3].

References:

Ye, T.; McKervey, M. A. Organic Synthesis with α-Diazo Carbonyl Compounds. Chem. Rev. 1994, 94 (4), 1091-1160. DOI: 10.1021/cr00028a010.
Mock, W. L.; Hartman, M. E. Synthetic scope of the triethyloxonium ion catalyzed homologation of ketones with diazoacetic esters. J. Org. Chem. 1977, 42 (3), 459-465. DOI: 10.1021/jo00423a015.
Liu, H. J.; Majumdar, S. P. On the Regioselectivity of Boron Trifluoride Catalyzed Ring Expansion of Cycloalkanones with Ethyl Diazoacetate. Synthetic Communications 1975, 5 (2), 125-130. DOI: 10.1080/00397917508061442.

The reaction which we are talking about here is called Tiffeneau–Demjanov rearrangement.
According to Wikipedia,

the Tiffeneau–Demjanov rearrangement (TDR) is the chemical reaction of a 1-aminomethyl-cycloalkanol with nitrous acid to form an enlarged cycloketone. The Tiffeneau–Demjanov ring expansion, Tiffeneau–Demjanov rearrangement, or TDR, provides an easy way to increase amino-substituted cycloalkanes and cycloalkanols in size by one carbon. Ring sizes from cyclopropane through cyclooctane are able to undergo Tiffeneau–Demjanov ring expansion with some degree of success. Yields decrease as initial ring size increases, and the ideal use of TDR is for synthesis of five, six, and seven membered rings. A principal synthetic application of Tiffeneau–Demjanov ring expansion is to bicyclic or polycyclic systems.

According to The Encyclopedia of Reagents for Organic Synthesis (DOI: 10.1002/047084289X), the yield of your product will be around 40-42%:

You can check out the Wikipedia page for more details.


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Is food color E-132 indigotine, carcinogenic?
I have seen different opinions on food color E-132 Indigo carmine (indigotine, FD&C Blue 2) if it forders cancer.

E132 was recently reevaluated by the European Food Safety Authority,
here is the abstract of a 2014 publication in the EFSA Journal on the subject:
The ANS Panel provides a scientific opinion re-evaluating the safety of Indigo Carmine (E132). The Panel observed that Indigo Carmine was poorly absorbed and does not raise concern for genotoxicity. No adverse effects in subacute, chronic, reproduction and developmental toxicity studies, and no modifications of haematological and biological parameters in chronic toxicity studies have been identified at doses less than or equal to 500 mg/kg bw/day. The only report of an adverse effect was in testis with a LOAEL of 17 mg/kg bw/day which would give rise to a safety concern if confirmed. The Panel considered that this study has shortcomings since it is not clear to the Panel whether the adverse effects observed were due to the food additive itself or to impurities and/or contaminants present in the material tested and/or to the conduct of the study. The Panel considered that the current ADI of 5 mg/kg bw/day for Indigo Carmine was applicable to a material with the same purity and manufacturing process as material used in studies without adverse effects on testis (93% pure colouring and 7% volatile matter) and concluded that any extension of this ADI to Indigo Carmine of lower purity and/or manufactured using a different process would require new data which would need to address the adverse effects on testis. The Panel noted that at the MPL, exposure estimates of Indigo Carmine would exceed the ADI for toddlers and children at the high level. Exposure estimates using the available usage and analytical data did not show an exceedance of the ADI for any population groups
https://www.efsa.europa.eu/en/efsajournal/pub/3768
As this was published nearly 4 years and E132 remains approved in the EU, I think it safe to say that E132 is not carcinogenic


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How two neutral atoms are together with no force between
How sodium atoms exist together when they all have unstability due to an extra electron present in their outermost shell?
I mean how two neutral atoms Na atoms are existing together in form of white silvery solid referred to as sodium metal

They don't have any instability; also, it is wrong to say there is no force between them. Maybe there is no pure electrostatic force, but that's only a part of the picture.

There's no extra electron. The thing is neutral and with some rearranging of electrons, you can get a more stable system by having more atoms.

It is possible for two lithium atoms or two sodium atoms to form a covalent bond between them. If you consider the s atomic orbitals and the way you can form bonding and antibonding orbitials.
You will be able to form a sigma bonding orbitial and a sigma star antibonding orbitial. You will have sufficent electrons to fully occupy the sigma bonding orbitial. This is the HOMO (Highest Occupied Molecular Orbitial). The antibonding orbitial is the LUMO (lowest unoccupied molecular orbitial).

When there is not any more "stable" option present, sodium or any other element will stay as they are. Sodium will only lose an electron and transform into other compounds if it is thermodynamically more favorable.
From the other way around, if the sodium left with itself, a closed vacuum place with only sodium atoms at STP, the only stable option is having that so-called unstable electron on its outer shell. Stability cannot be thought without considering the surrounding environment.


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Source for medium quantities of salts
Where are good sources for medium quantities (10-1000kg) of chemicals, particularly salts?
For quantities <10kg, I know of lots of options from lab suppliers (Sigma Aldrich, Alfa Aesar, etc). These companies generally charge a premium for quality that make bulk purchases unreasonably expensive.
For quantities >1000kg, I know of industrial sources (Univar, BASF, etc) and it seems they have local resellers in most markets. These companies typically have minimum orders of $1000 or so.
I am in the US.
What keywords should I be searching to find a local or online vendor of in between quantities?

Regarding this hold- wow, for how much time I spend on this site, including reading the rules, I do not understand the rules.

It will depend on the chemical which you want to order, many salts are used in society and industry for a range of purposes. if it is sodium chloride then I suggest you buy lots of packets of table salt.
A word of warning if you try to purchase medium to large amounts of some salts such as ammonium nitrate then you might be investigated as a possible terrorist. I would rather not give out any tips on what people might be on the lookout for, but if you attempt to buy some substances in bulk with no obvious use then people may report you.
I have to ask why you want ton amounts of chemicals such as salts ?

Like @Nuclear Chemist said, if you are working for a ligitimate business, then you can ask big companies such as VWR or Fisher Scientific a quote for your chemical. Sometimes, they'd surprise you by how cheap they can get.

I do indeed want these chemicals for a legitimate business. We are just chemical-industry-adjacent so we don't have relationships with those suppliers. Thanks for the tips.


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Why ethanol doesn't react with NaOH ,while methanol does?
Why ethanol doesn't react with NaOH ,while methanol does?

Who says it doesn't? Who says it does?

my teacher said this, but I don't remember the reason

It would make it a better question if you gave more information in the form of where you obtained the information about the difference between methanol and ethanol from.

Methanol will react a little with NaOH because it is an acid slightly stronger than ethanol. Ethoxide ion is a stronger base than methoxyde because of the induced effect due to C2H5. So ethanol is a weaker acid.

Why is my answer downvoted?

The problem is that methanol, ethanol and water all have similar strucutres. They will have similar pKas. I know that the pKas will increase in the order water, methanol and ethanol. The pKa of ethanol is about 18.
I think an equilibrium will exist between either methanol (or ethanol) and sodium hydroxide. It will be
$$\ce{MeOH + OH- <=> MeO- + H2O}$$
So I think that sodium hydroxide will partly react with both methanol and ethanol. The ethanol will react slightly less as the slightly greater electron releasing nature of the ethyl group will make the ethoxide a stronger base than methoxide.


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Generating geometries for substituted molecules
I would like to automate some of my calculations, and I am looking for a tool to generate reasonable good initial geometries.
The details: I have already optimized molecular geometries, and I would like to perform calculations of several different substituted version of the molecule. Because of the size of the system, it would make sense to keep the already optimized geometry and exchange e.g. selected -H atoms to some R1, R2 groups. Is there any library that can manipulate 3D geometries this way?
I am aware I can do similar things with much graphical software (GausView etc).  However, the point would be automation: a command line tool or a library or something similar that I can call from Python scripts, instead of me clicking things on a screen one by one.
I am also aware that I can generate 3D geometry from 2D or such. However, I would like to keep the already optimized main part of the molecule (it can be pretty big), for more efficient calculations, and insert new groups. I appreaciate any help, thanks.

You might be able to adapt the code described in this blogpost. It's designed for TSs, but can be used for any scaffold http://proteinsandwavefunctions.blogspot.dk/2017/12/tsconfsearch-conformer-search-for.html

You could try cclib in python to parse computational chemistry output files (i.e. your initial optimisation), manipulate coordinates with your own code and then use cclib again to generate new input files for your program of choice.

The key here is keeping the already optimized part, and having the new group insertion not be naive (overlapping atoms). The correct solution would be Avogadro, where you click on an atom to replace it with a new one and it avoids collisions, and then the ability to do a constrained minimization, but entirely from a script. If this is possible from the Open Babel Python interface, I don't know.

@VicLineal Thank you, but my question may not have been clear: I have no problem parsing the output and generating the input files. My question was about the later part, manipulaitng coordinates.

For most of the geometry creation/manipulation I do, I use the Atomic Simulation Environment (ASE) - all in python.
It's pretty straightforward to use and has a gui so you can make sure you're doing the right thing as you go along. In particular, I'd recommend playing with it in a jupyter notebook.
Checkout https://wiki.fysik.dtu.dk/ase/tutorials/manipulating_atoms.html for some examples.
The ASE also has a dataset of existing molecules that you can import and add to your structure in various places.


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Examples for oxidation of fluoride and for fluoride as a base
I have no university degree in chemistry and I’m curious about some examples - explained with the applicable rules (or the exception from the existing rules) - for the following statements (from this answer):

If fluoride is oxidized, it loses an single electron.
If fluoride acts as a base and donates a pair of electrons, it shares a pair of electrons.

Now, what are some examples of

oxidation of fluoride?
fluoride acting as a base?

As far as I think Fluorine ($\ce{F2}$) is the strongest oxidising agent, as the standard reduction potential for the conversion $\ce{F2 + 2e- -> 2F- }$ is very highly positive i.e. $E^0_{F_2/F^-} = + 2.87 V$. If you have any other species which have even more positive reduction potential than this, then only oxidation of $\ce{F-}$  is observed i.e. the reverse reaction occurs. But such species are very rare and thus generally oxidation of fluorides are not observed. You can surely have examples of Fluorides acting as Lewis base. For example, if you react $\ce{KF}$( basically $\ce{K+F-} $ ) with Antimony pentafluoride ($\ce{SbF5}$), $\ce{F- } $  will donate i ts lone pair to $\ce{Sb}$    and then they will form the compound $\ce{K+ SbF6-}$. This is a classic example of fluoride acting as lewis base.


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Non-hydrate forming salts?
Are there any water-soluble salts that don't form hydrates?
Hydration seems very thermodynamically favorable, as evidenced by its abundance. It also makes sense since the solvation and hydration thermodynamics are related. I can't find any evidence of non-hydrate-forming salts by brute force search or keyword googling. Anyone know of any?

https://en.wikipedia.org/wiki/Caesium_chloride

Sodium Chloride.

AlaskaRon Surprisingly, NaCl does indeed form a hydrate: http://www.phasediagram.dk/binary/sodium_chloride.htm


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If i keep adding sulfuric acid into water, will it keep dissolving forever?
If i keep adding sulfuric acid into water, will it keep dissolving  forever?

Sulfuric acid and water are miscible. Does this answer your question? https://en.wikipedia.org/wiki/Miscibility

Yes, it will continue to dissolve the water.

If you remove the last two or so percent of water from concentrated $\ce{H_2SO_4}$, it starts fuming, giving off SO3 until it again contains enough water to be stable.
$\ce{H_2SO_4 <-> SO_3 + H_2O}$
Pure sulfuric acid does not exist, so there is no clear yes/no answer to your question. Also $\ce{H_2SO_4}$ (like all other mineral acids) does not just mix with water, but strongly reacts. So it's clear there can be no miscibility gap like with water/ether or similar pairings.

When you dissolve a small amount of sulfuric acid in water then a chemical reaction occurs. The sulfuric acid dissociates into hydrogen cations (H+) and hydrogen sulfate anions (HSO4-). At very low concentrations of sulfuric acid in water the conductivity of the water is proportional to the sulfuric acid concentration.
When the sulfuric acid concentration is increased then the activity coefficents of the ions will decrease, the ions will be less free to move independently of each other. If we use the Bromley equation then the ion pairing term will become more and more important. One can regard the specific ion interaction of the hydrogen and the hydrogen sulfate ions as a reversal of the dissociation of sulfuric acid.
As the concentration climbs higher and higher, the fraction of the sulfuric acid which is dissociated into seporated ion pairs will decrease. The ions will be in the form of contact ion pairs where the two might be touching. At this point the electrical conductivity of the solution will be less than what would be predicted if we made the assumption that the conductivity is proportional to the concentration (and used low concentration data to make the prediction).
These contact ion pairs will be part way between the truely solvated ions and the intact sulfuric acid. As the concentration of the sulfuric acid goes even higher I think we will get to a situation where the majority of the sulfuric acid is in the form of molecules of sulfuric acid where a covalent bond exists between the hydrogen atoms and the oxygens of the sulfuric acid molecules. The mixture will be very viscous and it will have lots of hydrogen bonds in it.

Yes, there is no chemical reaction between H2S04 and water. it is just a dilution of H2S04. If u want to add 1000 liter water in 1ml H2SO4 , U can add and there will be clear solution without an solid or layer


text:
In the simplest terms, what is exchange current density?
what is exchange current density, and ideally how does it relate to overpotential
I know it's related to the amount of dynamics in a reaction, so it can tell us how simple or complex a reaction mechanism is, but I don't have a clear idea of what exactly it means

https://en.wikipedia.org/wiki/Exchange_current_density

Roughly speaking, it is a measure of how "dynamic" the "dynamic equilibrium" is when ancelectrode is at its equilibrium potential.
Let's say you are dealing with a copper electrode in a slightly acidic copper sulfate solution, a common situation in copper electroplating.  If you apply a high positive potential to this electrode, then of course the copper dissolves.  Upon measuring the rate of this dissolution, as the current density you are producing, you find that this current density increases exponentially as you crank up the potential, as long as you don't get any interference from mass transfer limitations or other electrode reactions.
Now switch to a highly negative potential.  Now copper deposits, and again (in the absence of mass transfer limits or other reactions) the current density varies exponentially with how much you are driving the electrode potential downwards.
What happens at intermediate potentials? The copper dissolution goes along with its exponential rate law and the deposition follows its rate law, but now these reactions are competing against each other so you macroscopically see only the difference between the plating and fissolution current densities.  You might have something like $10 \mu A/cm^2$ for dissolution and $1 \mu A/cm^2$ (these are just numbers I grabbed for this argument, not necessarily true kinetics), in which case you see a difference of $9 \mu A/cm^2$ dissolution.
At the equilibrium potential, the two current density numbers generated by the opposing rate laws are the same so you would not see any net reaction, even though in fact both are occurring together at the atomic level.  That common amount of current density being "exchanged" between the opposing reactions when you macroscopically see equilibrium, is the "exchange current density".


text:
Test for a dipolar compound?
Just surface level here, if I had two compounds and wanted to know which one was dipolar, could I just add them to water? If so, would the dipolar substance dissolve while the other compound is unaffected?
Thanks

I guess this will be helpful: https://chemistry.stackexchange.com/questions/38260/why-dont-polar-and-non-polar-compounds-dissolve-each-other

No generally this won't work; 1-chlorobenzene which has a dipole and 1,4-dichlorobenzene which has no overall dipole are both soluble in water  to differing extents. The possibility to dissolve is determined by the free energy as balance of enthalpy (+/- energy needed) to place victim molecule in solvent vs the entropy gain (positive) on mixing with solvent.  In practice you will need to measure the dielectric constant of your molecules, but values are tabulated for many of these.

In general substances whose molecules have a strong dipole tend to dissolve better in water than compounds which do not. If we consider for example diethyl ether and THF. The two are almost identical the only difference is that THF is held in by ring into one posiition. This makes THF have a dipole which is more permenent. The side groups on diethyl ether can move about thus allowing the molecule to adopt different conformations.

THF is totaly soluble in water, it is infinitely miscible with water. This means any combination of THF and water will mix perfectly. On the otherhand diethyl ether has a limited solubility in water. Only a very small amount of diethyl ether will dissolve in water.
If we look at the solvent accessable surface of THF where the atomic colours represent their partial charges we can see that THF has a strong permenent dipole. Here in these pictures the more blue an atom is the more negative and the more red the more positive it is.

While ether has less of a permennt dipole as the effects of the right hand side partial positive charge is canceled out by the left hand one.

Do be careful as more than dipole strength will determine the solubility of a compound. A better test for liquids would be to measure the relative dielectric constant at different frequencies. If we have a substance whose molecules have a dipole then at a low frequency the relative dielectric constant will be higher than at a high (often microwave) frequency. The reason is that at a low frequency the molecules have time to rotate to line up. At a high frequency this does not occur becuase the molecules have too little time to rotate. Below is shown a random array of dipoles and on the right some dipoles which have been aligned by an electric field.


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Nucleophilicity of halides in solvents
I have studied that as electronegativity increases nucleophilicity decreases. But in the aprotic solvent the order of nucleophilicty is: $$\ce{F- > Cl- > Br- > I-}$$ but how is this possible as $\ce{F-}$ is strongly electronegative as well as a strong nucleophile Isn't this contradictory? There is no effect of the aprotic solvent on anions.
Information given : This causes the most confusion, because it is solvent-dependent. In polar protic solvents (e.g. water and alcohols, any solvent with OH) nucleophilicity increases as you go down the periodic table (F- < Cl- < Br- < I – ). In polar aprotic solvents (e.g. DMSO, acetone) the order is reversed, and the most basic nucleophiles are also the most nucleophilic. (F- > Cl – > Br – > I – ).
On https://www.masterorganicchemistry.com/tips/what-makes-a-good-nucleophile/
I want to know why my concept is wrong.

Just to make sure I understand: did you just claim that iodide is not the best nucleophile in that series?

@Zhe yes I am claiming that in aprotic solvent .

Where did you see the information suggesting this trend? This is counter to my understanding, and I'm guessing it is counter to many others' understanding here based on Zhe's response.

@SendersReagent https://www.masterorganicchemistry.com/tips/what-makes-a-good-nucleophile/

@Zhe http://chemistry.umeche.maine.edu/CHY251/Ch7-Overhead4.html

The short answer to your question is that your question doesn't make sense. Electronegativity is the property of an atom (e.g., fluorine). You can't talk about the electronegativity of your halide. The tendency of the neutral atom to pick up an electron is not the same thing as the anion sharing a pair of electrons.

@Zhe on this site https://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/  they have shown dependence of nucleophile with electronegativity .

Yes, but the terminology is loose. You should not refer to the electronegativity of a halide.

@Zhe Can you please explain me why my concept is wrong ?

Which one? There's a big block of text...

@Zhe how is this possible as F− is strongly electronegative as well as a strong nucleophile Isn't this contradictory?

There a lot of wrong with this statement. Nucleophilicity is a kinetic concept and electronegativity is a thermodynamic concept. It doesn't make sense to compare these two concepts. Also, you're comparing the property of an neutral atom, electronegativity of fluorine, with the property of a anion, nucleophiliticity of fluoride. That also makes no sense.


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Energy of activation vs Bond energy
In the second chapter, Methane Energy of activation and Transition state in the textbook Organic Chemistry by Morrison and Boyd, Energy of activation is given the following definition:

The minimum amount of energy that must be provided by a collision for a reaction to occur is called the energy of activation, Eact.

The Eact for the reaction:
Cl(free radical) + CH3-H ---------> H-Cl + CH3(free radical)
is 4 kcal/mol. Therefore, on an average, it is 4/NA kcal per individual collision (where NA is the Avogadro number).
However, the bond energy of the carbon-hydrogen bond in methane is 104 kcal/mol. Therefore, on an average, it would be 104/ NA kcal for a single molecule of methane. In other words, 104/ NA kcal should be the energy provided by a collision that involves a methane molecule in order to break the carbon-hydrogen bond. Why does this not agree with the statement above which says 4/NA kcal per individual collision is what is required?

The bond dissociation energy is the energy needed for the reaction leading to the rupture the $\ce{C-H}$ bond. I.e. it is the enthalpy of the reaction:
$$\ce{CH_3-H \rightarrow CH_3^{\bullet} + H^{\bullet}}$$
In the case of the reaction in the question, formation of the $\ce{H-Cl}$ bond will counterbalance the energy cost for breaking the $\ce{C-H}$ bond.

So, bond-making and bond-breaking are perfectly synchronized?

The minimum energy pathway (i.e. the one which for which the activation energy is defined in the question) is the pathway for which bond breaking and bond formation balance as best as possible.

I assume that by "balance each  other" you mean take energy from bond making process and use it in bond breaking process. So, if the balance is to be as best as possible then shouldn't activation energy simply be equal to enthalpy of reaction, i.e, 1 kcal/mol (given that HCl bond energy is 103 : so 104 -103 = 1)?

In the general case this is not the case: at some stage in the elongation of the bond (and shortening of the other one) there is some imbalance even on the minimal pathway. At some stage the H atom will be shared between the CH3 group and the Cl atom, and some repulsion between the electrons on the bond orbitals on both sides will rise the energy on the potential energy surface with no other way then go through this minimum. (Think of is as a pass in the moutain: its the lowest path to go but not lower than the valleys on both sides.)

In this reaction the C-H bond is not broken then the Cl reacts with the H to form the products, but rather as the Cl atom approaches its electrons and those in methane interact and a combined transient species called a transition state is formed. This species lasts only a very few picoseconds and has the nature of  long Cl-H and H-C bonds, in a picturesque way something like Cl--H--CH3. As the reaction proceeds the Cl-H bond forms and the H-C bond breaks and the CH3 radical and HCl then depart from one another.  The activation energy is that needed to reach the transition state and involves using some of the kinetic and vibrational energy of the reacting species. The transition state is at a far lower energy than direct CH bond breaking although it is at the highest point on the reaction path, i.e at  the top of the barrier.


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What will be the product of this reaction?

We have an alpha beta unsaturated ether, along with a keto group in the same compound, which is being treated with a grignard reagent shown (assume the grignard reagent is in excess). Will it only attack the carbonyl group, or are things different here?

This reaction is one of a special kind. The one and onlyend product is 3-hydroxy,3- methylcyclohexan-1-one. As the Grignard reagent is in excess, and it is an $\alpha$- $\beta$ unsaturated ketone, there is a possibility of undergoing two different type of reaction mechanisms. But Surprisingly, both will give the same ultimate end product after two steps.   The major mechanism through which the reaction should undergo, is the direct nucleophilic attack of the grignard reagent at the electrophilic carbonyl carbon. As, Grignard reagent is a hard base , it's attack to hard acid centre of carbonyl carbon is more preffered. So, first it will undergo this addition after which it acidic hydrolysis is done, which cleaves the ether and forms an enol which tautomerises to give the final end product ketone. The reaction scheme is shown below,
Now, the minor mechanism is the process of conjugated addition of the methyl group to that $\alpha$-$\beta$ unsaturated ketone. But again after tautomerisation and  hydrolysis it actually forms the same compound, which is 3-hydroxy,3-methyl cyclohexan-1-one. The reaction scheme is shown below,
So, it is seen that however the presence of excess reagent may interfere with the reaction mechanism but the end product is $100 $$\%$ $3$-hydroxy,$3$-methylcyclohexan-$1$-one.

With the minor reaction scheme you would not normally expect cleavage of the ether under the sort of mild conditions that you normally use to workup a Grignard (that do hydrolyse an enol ether). Is this using forcing conditions?

@Waylander, I am assuming that the hydrolysis is done in such acidic conditions which can also protonate the oxygen of the ether and the there will be a $S_N1$ type attack of water molecule at the tertiary carbocation formed after $\ce{MeOH}$ leaves the system. Actually as there is a possibility of tertiary carbocation formation, I think slightly acidic conditions can also facilitate the leaving of $\ce{MeOH}$ and perform the hydrolysis.

This system is essentially tBu-OMe and that usually requires strong acid (HBr or HI) in anhydrous conditions to cleave. I do not think you will get such a hydrolysis in work up. Do you have a reference for this reaction?

The acid is not given, so I assumed that it is a strong enough to cleave the ether.

I do not think that is a good assumption.


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Nucleophiles in protic and aprotic solvent
If the nucleophiles have almost same size (not a large difference) such as $\ce{C- , N- , O-}$ then how their nucleophilicty varies in protic and aprotic solvents?\
Does it depend on types of reaction such as Sn1 or Sn2 ?

I suggest coming up with more descriptive titles for your two questions, right now they look very generic and if I did not click to check I would have thought you posted the same thing twice.

From what I learned in organic I, their nucleophilicity does not change with the solvent because if the solvent is polar Protic, each nucleophile will be hindered in a similar manner (due to the solvent cage). If the solvent is polar aprotic, this hindrance does not matter. In both cases, nucleophilicity matches base strength.
I believe the solvent only affects nucleophile strength going down the periodic table. There are also polarizability effects for larger molecules but I’ve been told in general, a smaller molecule is a stronger nucleophile and larger molecules are weaker in aprotic solvents. This trend reverses for polar Protic solvents.
Also keep in mind, the strength of the nucleophile only matters for SN2 because of the backside attack being the slow step.

This answer is not correct because solvation is infinitely more complicated than what you're claiming here. There will certainly be an effect though determining what that is is certainly not straightforward.

I’m sure it is more complicated. I just wanted to suggest what I learned in my organic I class even though I’m sure it was over simplified as is most material that I have learned.

@Zhe can you please explain it ?


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Reverse electron demand Diels-Alder
Why is the reverse electron demand Diels-Alder reaction much rarer than the normal one? By reverse electron demand, I mean one with an electron-rich dienophile and an electron deficient diene.
In the book by Clayden they have drawn a diagram with the electron deficient dienophile lower in energy and saying that it would prefer to give its LUMO to the reaction.
But, I don't get what's wrong with an electron-rich dienophile lending its HOMO.

I was just reading about this very thing... Hope this helps: https://en.m.wikipedia.org/wiki/Inverse_electron-demand_Diels-Alder_reaction

Thanks. so if i'm not wrong normal and inverse DA are essentially the same thing except that if the energy of the FMO of the diene is greater than the dienophile then theinverse dominates and if it is lower then the normal pathway dominates..

But still i am confused as to why are there remarkably fewer examples of inverse as compared to normal DA or is it just a coincidence..

Maybe because the energy difference between diene HOMO and dienophile LUMO is on average lower than the difference in energy between dienophile HOMO and diene LUMO.

Inverse electron demand reactions are inefficient and slow. Even the prototypical Diels-Alder reaction of 1,3-butadiene and ethylene is quite sluggish. From here you can see the reaction only proceeds to 93.5% yield when ran with ethylene in excess by 50:1 at 200ºC and 350 atm for 4 hours. Yikes!
Now consider that some normal electron demand D-A reactions can occur at room temperature. In fact, 1,3-butadiene spontaneously dimerizes to 4-vinylcyclohexene while in storage, meaning the dimer must be cracked by heating immediately before use. This should give you an idea of the sensitivity of D-A reactions to substituent groups. By lowering the HOMO-LUMO gap, orbital overlap is improved, lowering the activation energy and resulting in a faster reaction.
To supplement and support my answer, here is a paraphrased excerpt from Modern Physical Organic Chemistry by Anslyn and Dougherty (2006):

Electron-withdrawing groups on the dienophile lower its LUMO. This occurs because electronegative substituents have correspondingly lower energy MOs, and lower the energies of all MOs in which they participate. Inversely, electron-donating groups raise the diene MOs.

In support of this analysis, the log($k$) for Diels-Alder reactions correlate quite well with the inverse of the difference in energy between the ionization potential of the diene (related to the energy of the HOMO) and the electron affinity of the dienophile (related to the energy of the LUMO). To further enhance the electrophilicity of the dienophile, it is common to add Lewis acids that can complex the electron withdrawing groups on the dienophile, further lowering the LUMO.

This seems to assume that the rates go in the order normal electron demand > no electronic bias > inverse electron demand, however, I’m curious as to exactly what point the inverse electron demand starts to be faster than reactions with no electronic bias. There has to be some point at which the diene LUMO is so high and the dienophile HOMO so low that that frontier orbits interaction becomes more important than the usual diene HOMO + dienophile LUMO, or am I wrong?

In order to understand why the HOMO of the dienophile is less easily donated, let us again understand the original Diels-Alder properly.
In the original Diels-Alder reaction, there is a close similarity in the energies of the HOMO of the diene and the LUMO of the dienophile. As a result, they form the strongest orbital interactions. The presence of electron releasing groups on the diene and electron withdrawing groups on the dienophile essentially reduce this energy difference even further.
Moving on to the inverse electron-demand Diels-Alder: here, the LUMO of the diene and the HOMO of the dienophile are closest in terms of their energies. Because of this, they form the strongest orbital interactions now, and hence this reaction is able to take place.
As you might have already guessed, the chances of there being a close similarity in the energies of the LUMO of the diene and the HOMO of the dienophile is much less compared to the normal Diels-Alder case. Because of this, we have a greater number of reagents that can undergo the normal Diels-Alder than the inverse one. It's just a question of its rarity than its feasibility, because now you're looking for an electron deficient diene and an electron rich dienophile. In general, due to the resonance stabilisation in dienes, there's a good deal more electron rich dienes than deficient ones, and vice versa for the dienophiles...
I hope that this answers your question.

Also, @Marko 's comment really sums it up: > Maybe because the energy difference between diene HOMO and dienophile LUMO is on average lower than the difference in energy between dienophile HOMO and diene LUMO.

There are perfectly synchronous iedDA reaction as far as we know and there are asynchronous normal demand DA reactions. This is not a special feature of either type as well as the concerted mechanism, there are (ied)DA reactions which (might) proceed by two step mechanisms


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Fast way to isolate 4-nitrophenol from 4-aminophenol
I'm working in the lab with a very old (like 30 years old) sample of 4-nitrophenol, wich of course has been reduced with the time to 4-aminophenol. I need to isolate both reagents and work only with the nitrophenol, is there any quick way to do that?

There are various products of reduction of nitrophenol. Best way would be to cast it aside and buy fresh one.

@Mithoron, I'm screwed, I'm in Cuba, there are no such things in here. I'm trying to get a fresh one, but that's really difficult here :)

One has to ask if the container of 4-nitrophenol was left open during the 30 year period and was it stored in plastic or glassy bottle? Another thinking is that, what was the expiring date for 4-nitrophenol? Like drugs which are also chemicals have expiring date labelled on the container before it goes into the market. Then what taught brought the question raised, however, if you are just assuming, the answers given by Mathew Mahindaratne may be useful. Omar A.Wafudu

I would use solvent extraction. Dissolve the mixture of phenols in an organic solvent. Next shake the organic solution with dilute hydrochloric acid in a separation funnel. Do this several times to wash all of the 4-hydroxyaniline out of the organic phase.
Next shake the organic phase with concentrated sodium chloride solution (brine) and then add anhydrous sodium sulfate. Shake the mixture up in a sealed conical flask. Now decant or filter it to give you a solution of 4-nitrophenol which you can now evapourate down to give you the phenol.
I would like to know how the nitrophenol was reduced into the amine over the years it was in storage. This is a more interesting question to my mind is how did the nitro compound get converted on storage into the amine.

It is a very good question you brought up here. It's usually other way around. Aryl amines air-oxidize. I never heard of nitro to amine reduction by air.

@MathewMahindaratne, thats a good point, I checked the compound by UV-vis, and there is a strong band from 250 nm - 400 nm. I expected only to see the signal around 400 nm. The p-amminophenol absorbs around 250 nm, that's why I conclude that it's the impurity. Of course further analysis is needed, but that's all I can say by now. Again, is a 30 years old reagent stored in very, very bad conditions, anything could have happened to it. Air is the just the tip of the iceberg

@Marange Like everything absorbs around 250. I'm afraid wild guessing isn't good idea.

I would expect 4-nitrophenol to be relatively stable, even when stored for a long time under air, and as Nuclear Chemist says I would certainly not expect any reduction of the nitro group over time.
Have you tried simply recrystallizing the material you have. In Armarego and Chai it has been purified by recrystallization from aq. HCl (0.5N), aq. H2SO4 (1N), EtOH, aqueous MeOH, CHCl3, benzene and Pet Ether. Give any (or all) of the above a go. The expected melting point is 113-114 C so you can check your material for purity without access to any expensive analytical equipment.


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How on the quantum level do the notch filters used in Raman spectroscopy remove only a select wavelength of light?
From what I have gathered regarding the schematics of Raman spectrometers, the notch filter is used to remove a narrow range of light corresponding to that of the laser used to excite the sample. How is this accomplished? I can't find a reference which can explain the physics as to why the filters work. I was thinking that the molecules which make up the filter would absorb at a very selective range. But then why wouldn't one get 'light contamination' from the Raman scattering due to the filter itself? How is this avoided or is the filter in some way immune to the Raman scattering effect?
I assume this can only be explained with QM but if it can be explained more simply I would welcome that response as well.
Edit:
I had this question after watching this video https://www.youtube.com/watch?v=tRrOdKW06sk . If the notch filter is an interference filter as suggested in the comments, the semantics of the question changes to encompass that of interference filters. What is actually occurring at the atomic level so that the angle at which the incident light strikes the filter determines its ability to pass through the filter? Is this due to crystal structure possibly? Why does this angle affect transmittance vs. reflection?  And finally why will the laser still only allow one specific wavelength to transmit?

(guessing) A lens system that focuses only light scattered from the sample position? Also if the filter is always there, substract a background spectrum?

Can you post a schematic (or link to one) that shows this filter?

this reference has a image of a general schematic http://www.chm.bris.ac.uk/pt/diamond/stuthesis/chapter2.htm  Here is another https://www.researchgate.net/figure/Schematic-diagram-of-a-confocal-Raman-system-integrated-with-a-microfluidic-unit_fig1_236459961

that's an interference filter tilted to the exact angle at which it reflects the laser line. No absorption. Plus there are some optical components missing in the schematic imo. Light scattered at the filter would not be focused on the detector.

Polarization helps also. And comb filters can be used.


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Barking dog reaction mechanism
Liebig's blue devil or the "barking dog" reaction is the gas phase oxidation of carbon disulfide with either nitrous oxide or nitric oxide.
What are the gas phase reaction steps in this reaction, I would like to know the nature of the reactive species and the mechanism of the reaction steps. If you can give me literature references to support your answer it would be much better.barking dog reaction

Seems you're rather optimistic about the knowledge about combustion reactions. I doubt we're even close to getting full mechanism. There may be some simple models though.


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Why does the helium hydride ion have such a large dipole moment?
According to Wikipedia, the helium hydride ion $\ce {HeH+}$ has a dipole moment of 2.26 or 2.84. What contributes to this large dipole moment in this molecular ion?

"2.26 or 2.84" That's quite a large variation o.O Did you pick the two values from two different sources?

@GaurangTandon For a particle so rare and obscure, that's OK.

Think of it as a heteronuclear diatomic then a dipole is expected. Either the bond length is large or charge shift great or both. You will have to look into details of bonding to understand further.

https://en.wikipedia.org/wiki/Helium_hydride_ion The bond is 0.77A long (H2 0.74A),

Just a note, the source for those dipole values isn't the best. It seems it was from a self published article using a minimal basis set in a home made program. The reason for the range is that when they compared it to GAMESS calculations, the value varied between those two when using two different STO-3G basis sets. A higher level of theory and better basis set could lead to much different results.

@Tyberius I see. In that case, are you aware of any better sources of this calculated dipole moment data? So you are saying that the dipole moment may not be as large as Wikipedia claims?

@TanYongBoon I would surmise it is pretty large with on the positive charge localizing on the hydrogen side. I just thought it was worth knowing that I wouldn't put much faith in those particular values. For such a small molecule, a much more accurate calculation could be done on a normal computer in a matter of minutes.

@Tyberius I actually just did a calculation using the functional CCST with 6-311G ++ (3df,3dp) basis set and obtained a value of 4 debyes!

@TanYongBoon I guess it isn't too surprising that's it even higher, considering it's ionic

The dipole moment of an array of charges is defined as
$$
\vec{\mu}=\sum_i \vec{r}_iq_i
$$
where $\vec{r}_i$ is the position of charge $q_i$ in the molecule with respect to the center of mass. In a neutral molecule the sum of the charges is zero and therefore the terms in the sum partially cancel. For ions this is of course not the case and the dipole moment is much larger than for the corresponding neutral molecule (provided that the ion is not completely symmetric such as H$_2^+$).


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Understanding the synthesis of twistane

What are the products P and Q?

This was a question in my test series, but I don't have any experience with bridge head compound regarding the cis and trans transformation. I need help understanding this process.

For your information LDA is a common shorthand in organic chemistry for lithium diisopropyl amide. This is a very hindered and very strong base which is used under anhydrous conditions. It is often used in THF or another ether solvent.

The first step is a conversion of the nitrile into a carboxylic acid, in the literature Martin Olbrich, Peter Mayer, and Dirk Trauner (J. Org. Chem., 2015, 80, 2042-2055) used KOH in ethylene glycol at 155 oC for a couple of hours. The next reaction is attack of iodine on the alkene. The cyclic cationic intermidate will be attacked by the carboxylic acid.
I am not sure what SBH is, but it is some reducing agent which functions in the same way as hydrogen with platinium and a base. This combination of reagents removes the iodine by converting the alkyl halide into a sp3 CH. This gives us P.
Another option I would consider is to use nickel chemistry, I would put some hydrated nickel(II) chloride in a flask with ethanol, I would then add a solution of sodium borohydride in water. Do this with care as it emitts a lot of hydrogen gas. It will change the nickel solution from a nice green clear solution to an ugly black slurry. This is not a problem. Add the organic material to this. This black slurry acts like raney nickel. I have used it a few times to remove things like thioether groups from aromatics. For example if you react 2-acetyl pyridine with a base and carbon disulfide and methyl iodide you can make a new compound. If this is then reacted with 2-acetyl pyridine and base, before being boiled up with ammonium acetate in acetic acid you can form terpy with a SMe group on the 4 position of the central pyridine ring. The way to convert this SMe version of terpy to terpy is to boil it with the black slurry made from nickel and sodium borohydride. The problem is that it is hard to do oxidative addition of alkyl halides, many reagents do not work well.
If we consider the Monsanto process for acetic acid then the rate determining step is the reaction of the methyl iodide with the rhodium, if we move to the more modern BP version which uses iridium then the oxidative addition is much more easy. I think that the platinum catylst will work a lot lot better than either the nickel method I mentioned or palladium on carbon.
I worry that sodium borohydride with no modification will be unable to deal with the alkyl iodide in the moelcule.
Another option would be to use tributyl tin hydride and AIBN, this is a selective radical based reducing agent which will convert alkyl iodides into alkyl hydrides (alkanes) without touching alcohols, esters or other carboxylic acid based groups. This could be a method of putting a deuterium label into the product if we very much wanted it.
Keep in mind that if you are ever asked for an example of a free radical chain reaction other than halogenation of an alkane that you can always offer reduction of an alkyl halide by tributyl tin hydride or a free radical polymerisation reaction.

Now a lot of people hate tributyl tin chemistry, there are several disadvantages of it. Firstly it is often hard to separate the tributyl tin side products from the product. Secondly organo tin chemistry is rather toxic, the worst effects are on marine life such as oysters. Tributyl tin compounds can cause female marine organisms to turn into an intersex creature which then dies. But in our case it is not so bad. What we can do is to use a tributyl tin reducing agent and the use the acid base properties of the carboxylic acid to separate it from the neutral tributyl tin iodide.
If shaking a solution of the product in benzene with sodium hydroxide was to fail, I would suggest making a solution of the carboxylic acid in methanol and then passing this through a pad of a weak base ion exchange resin (amino groups on a polymer such as polystyrene). I would then elute the column with more methanol and then with formic acid dissolved in methanol.
The next thing is that we use aqueous caustic potash to break open the lactone ring. The lactone is then oxidised with permanganate. I would have used a chromium oxidant as they tend to be a little more gentle.

Next we react the ketone with ethylene glycol using acid and some dehydration method. One good method is to use toluene and a dean-stark head. Another one which I like is to use triethyl ortho formate with a little acid. This triethyl orthoformate method works under more gentle conditions to remove the water. We might get some ester formation but in the ideal world I would want to do the next step with an ester not a carboxylic acid.
We reduce the carboxylic acid using the cyclic acetal to protect the ketone from the lithium aluminium hydride. Then using acid in water we deprotect the ketone. We then convert the alcohol into a leaving group with tosyl chloride. We then form an enolate with the LDA which then will attack the tosylate to form the ketone.
It is important to keep in mind that under some conditons that enolates can be attacked on the oxygen by hard electrophiles such as tosylates. But I suspect in this case that the tosylate will be unable to reach the oxygen. It would be best to run this reaction at a high dilution. One method would be to slowly add the ketone-tosylate to a large flask containing a large amount of solvent and some LDA. Another method would be to convert the tosylate to the alkyl iodide by reacting the tosylate in acetone with sodium iodide first.

It may be a good idea to read the paper by H. W. Whitlock in Journal of the American Chemical Society 1962, 84 (17), 3412-3413. This is the original synthesis of the compound.

If the sodium borohydride was reacted first with nickel(II) chloride in ethanol to form a black solid of nickel mixed with boron oxide stuff, then this could be better. The black solid is able to act as a hydrogenation cataylst and it already has a lot of hydrogen adsorbed into it. It can be used as an alternative to Raney nickel as a way of removing things like thioethers from organic molecules and replacing them with H atoms.


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The effect of the groups at para position on SN2 reactivity
Compare the SN2 reactivity order for:

The benzyl group stabilises the transition state of SN2 reaction using the π system of the benzene ring for conjugation with the p- orbital of the electrophilic carbon which is electron deficient. (reference: Clayden, page 341)
So in my opinion, groups increasing the electron density of the ring should make the SN2 reaction go faster as compared to groups decreasing it.
Thus the order should be $2 > 3 > 1 > 4$ but the answer given in my worksheet is the exact opposite of this. I would like to know the reason for that and the flaw in my reasoning.
This is what I found in March:

Holtz and Stock showed that electron-withdrawing groups increase the
  rate of SN2 reactions. This can be ascribed to stabilization of the
  transition state by withdrawal of some of the electron density.

I find this contradictory to the content given in Clayden that the TS is "electron deficient". In March its also given that for alkyl groups at para position the Baker Nathan order is followed (methyl > ethyl > t-butyl). (Hyperconjugation is most in methyl so it increases electron density of the ring) (Another  apparent contradiction)
So how exactly do I determine the effect of the p-substituents in SN2.

I can't make answer and not earn bounty because of less time but https://chemistry.stackexchange.com/a/27638/108386 will solve your problem.

This is what I found from your reference to Clayden. You just had to read the next paragraph ( I have highlighted the point ). You can see that March and Clayden have agreed on the same point, that electron withdrawl from the phenyl/allyl group increases the effective conjugation in the transition state.
So I hope You can understand what will be the correct order and why.
P.S. I know this question is old, but it was unanswered, and someone may benefit from an answer.
Reference: Jonathan Clayden, Nick Greeves, S.Warren, P.Wothers -Organic Chemistry (2000) - Pg 422.

Could you explain that why 3 is more reactive than 2 ? Both -CH3 and -NH2 are +R groups but -NH2 shows electron withdrawing inductive effect.

Can you please explain that why there is an effective conjugation due to electron deficient π system?


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Two electron integral evaluation over primitive Gaussians without recurrence relations
I want to evaluate two-electron integrals over s, p, and d primitive Gaussian functions. The procedure is presented for s-type ones in "Modern Quantum Chemistry" by Szabo and Ostlund, but I need to evaluate these integrals for p and d ones. However, most books, such as "Molecular Electronic-Structure Theory" by Helgaker which I read, evaluate these integrals using recurrence relations. Is there any other method which can evaluate two-electron integrals, without using recurrence relations?

You can use any numeric integration technique. It is, however, time-consuming, so people tried to reduce the time by exploiting properties of primitive gaussians.  The relevant data can be found in either in papers on the subjects (see scholar.google.com and look for something like "primitive gaussian integral algorithm)", or in sources of relevant software (libint, gamess, and some other software).

@permeakra Thanks, but I look for an analytical solution.

Sorry for being trivial, but... then use any method of analytical solution. The correct answer doesn't depend on the method, does it?

@Ivan Neretin The problem is that most of  people use the recurrence relations in their codes and I have rarely seen analytical solution, specially for higher angular momentum than $s$ namely $p$ and $d$. Anyway I found a solution in a lecture note (still I'm not sure it's my desire answer) which I would post here, if no one post answer.

Is there a specific reason why you do not want to use the recurrence relations?

@ErikKjellgren I have to use the Recurrence Relations in my codes, but in this step I want to find an analytical solution to use in my proposal, the main reason for this is due to the my special case which I'm studying.

As you mentioned, most implementations such as Obara–Saika (OS), McMurchie–Davidson (MD), Head-Gordon–Pople (HGP), and many others use a series of horizontal and vertical recurrence relations to build up angular momentum and transfer it among each of the four centers. These are analytic solutions, up until the need to solve the Boys function, however we could debate as to whether or not they are closed-form expressions by terminology. They are closed-form in the sense that each part of the recursive expansion does terminate properly with $(ss|ss)$, given in OS equation 44.
In general, any recursive function can be rewritten to be non-recursive using loops of sums and or products, though it may involve considerable effort. In that sense, the recurrence relationships are quite powerful and avoid lots of "boilerplate". If you are interested in a non-recursive expression, there is the earlier Taketa–Huzinaga–O-ohata (THO) paper that, while not very descriptive, includes explicit working expressions for 2-electron and 1-electron (overlap, kinetic energy, and nuclear-electron attraction) integrals. The most relevant expression is
\begin{align*}
\text{ERI} &= \frac{2\pi^{2}}{\gamma_{1}\gamma_{2}} \left( \frac{\pi}{\gamma_{1} + \gamma_{2}} \right)^{1/2} \exp \left\{ - \frac{\alpha_{1}\alpha_{2}\overline{AB}^{2}}{\gamma_{1}} - \frac{\alpha_{3}\alpha_{4}\overline{CD}^{2}}{\gamma_{2}} \right\} \\
&\times \sum_{i_{1}i_{2}r_{1}r_{2}u} B_{i_{1}i_{2}r_{1}r_{2}u}(l_{1}, l_{2}, A_{x}, B_{x}, P_{x}, \gamma_{1} | l_{3}, l_{4}, C_{x}, D_{x}, Q_{x}, \gamma_{2}) \\
&\times \sum_{j_{1}j_{2}s_{1}s_{2}v} B_{j_{1}j_{2}s_{1}s_{2}v}(m_{1}, m_{2}, A_{y}, B_{y}, P_{y}, \gamma_{1} | m_{3}, m_{4}, C_{y}, D_{y}, Q_{y}, \gamma_{2}) \\
&\times \sum_{k_{1}k_{2}t_{1}t_{2}w} B_{k_{1}k_{2}t_{1}t_{2}w}(n_{1}, n_{2}, A_{z}, B_{z}, P_{z}, \gamma_{1} | n_{3}, n_{4}, C_{z}, D_{z}, Q_{z}, \gamma_{2}) F_{\nu}(\overline{PQ}^{2}/4\delta),
\label{tho-2.22-1}\tag{THO eq. 2.22 part 1}
\end{align*}
with
\begin{align*}
&B_{i_{1}i_{2}r_{1}r_{2}u}(l_{1}, l_{2}, A_{x}, B_{x}, P_{x}, \gamma_{1} | l_{3}, l_{4}, C_{x}, D_{x}, Q_{x}, \gamma_{2}) \\
&\quad = (-)^{i_{2}} f_{i_{1}}(l_{1}, l_{2}, \overline{PA_{x}}, \overline{PB_{x}}) f_{i_{2}}(l_{3}, l_{4}, \overline{QC_{x}}, \overline{QD_{x}}) \\
&\qquad \times \frac{i_{1}!i_{2}!}{(4\gamma_{1})^{i_{1}}(4\gamma_{2})^{i_{2}}\delta^{i_{1} + i_{2}}} \cdot \frac{4\gamma_{1})^{r_{1}}(4\gamma_{2})^{r_{2}}\delta^{r_{1} + r_{2}}}{r_{1}!r_{2}!(i_{1}-2r_{1})!(i_{2}-2r_{2})!} \\
&\qquad \times [i_{1} + i_{2} - 2(r_{1} + r_{2})]! \frac{(-)^{u} p_{x}^{i_{1} + i_{2} - 2(r_{1} + r_{2}) - 2u} \delta^{u}}{u! \{ i_{1} + i_{2} - 2(r_{1} + r_{2} - 2u)\}!},
\label{tho-2.22-2}\tag{THO eq. 2.22 part 2}
\end{align*}
and intermediate quantities are
\begin{equation}
\nu = i_{1} + i_{2} + j_{1} + j_{2} + k_{1} + k_{2} - 2 (r_{1} + r_{2} + s_{1} + s_{2} + t_{1} + t_{2}) - u - v - w,
\end{equation}
\begin{align*}
\gamma_{1} &= \alpha_{1} + \alpha_{2},~\gamma_{2}\text{ is similar}\\
\delta &= \frac{1}{4\gamma_{1}} + \frac{1}{4\gamma_{2}} \\
\mathbf{P} &= \frac{\alpha_{1} \mathbf{A} + \alpha_{2} \mathbf{B}}{\alpha_{1} + \alpha_{2}},~\mathbf{Q}\text{ is similar} \\
\mathbf{p} &= \mathbf{Q} - \mathbf{P} \\
\overline{AB} &= A - B
\end{align*}
\begin{equation}
F_{\nu}(t) = \int_{0}^{1} u^{2\nu} \exp(-tu^{2}) \mathop{du},~\text{the Boys function.} \tag{THO eq. 2.11}
\end{equation}
$(l, m, n)$ are the usual polynomial exponents leading to s-, p-, and d-type functions, $(\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D})$ are the usual basis function positions, the $\alpha$ are the basis function exponents, and $f_{j}(l, m, a, b)$ is the coefficient of $x^{j}$ in the expansion of $(x + a)^{l} (x + b)^{m}$.

In \eqref{tho-2.22-1} the summations with respect to the indices $i_{1}, i_{2}, r_{1}, r_{2}, u$ extend from 0 to $l_{1} + l_{2}$, $l_{3} + l_{4}$, $[i_{1}/2]$, $[i_{2}/2]$ and $[(i_{1} + i_{2})/2 - r_{1} - r_{2}]$ respectively. The ranges of $(j_{1}, j_{2}, s_{1}, s_{2}, v)$ or $(k_{1}, k_{2}, t_{1}, t_{2}, w)$ may be found in the same way.

Regarding the indices used in the summations:

If the index $i_{1}$ runs from 0 to $l_{1} + l_{2}$, and $i_{2}$ runs from 0 to $l_{3} + l_{4}$ we can deduce that $i_{1}$ must cover all possible values of the total angular momentum along the x-component for the bra, and $i_{2}$ goes up to the total angular momentum along the x-component for the ket. The indices $j,k$ match with $m,n$, meaning the y- and z-component.
$r,s,t$ are not to be confused with distances themselves, like how we usually define $r_{A} = |\mathbf{r}_{A}| = |\mathbf{r} - \mathbf{A}|$. Both $r,s,t$ and $u,v,w$ are also related to the three Cartesian coordinates, but in a more complicated manner than $i,j,k$; they appear to originate from the closed-form expressions for the Hermite polynomials.

Their implementation is slightly modified, with the necessary expressions in equations 3.3, 3.4, and 3.5, though the above can be implemented directly as well. A quick search revealed this code that implements all the THO integrals in C++. It is worth commenting that in real packages, the Boys function is precomputed from interpolation tables, rather than from the gamma and incomplete gamma functions.

Thanks a lot. If I got it right, there is no way out of the Recurrence Relations, but the THO formula provides an escape! However I didn't understand what is the upper value of $i$ in summation of the expansion?

I added a block about $i,j,k$; essentially they are related to the angular momentum of the x, y, and z components of each integral. Let me know if anything is still unclear.

Thank you so much dear @pentavalentcarbon . In the case which I'm studying, all primitive Gaussian functions have been located at the origin. Can you tell me what happens to the THO formula with such assumption?

Without writing it out on paper, I think $\mathbf{A}, \mathbf{B}, \mathbf{C}, \mathbf{D}$ would all be zero (they are absolute positions, not differences), leading to $\mathbf{P}, \mathbf{Q}, \mathbf{p}$ being zero. That should get you started.


text:
Why are magnesium chloride and calcium chloride more soluble than sodium chloride?
I read that $\ce{MgCl2}$ and $\ce{CaCl2}$ are more soluble than $\ce{NaCl}$ in water. Solubility of $\ce{MgCl2}$ is $\pu{543 g/L}$ and that of $\ce{NaCl}$ is $\pu{360 g/L}$ (both at $20^{\circ} \pu{C}$).
I think that $\ce{NaCl}$ should be more soluble due to its higher ionic nature. $\ce{Mg^{2+}}$ and $\ce{Ca^{2+}}$ are more polarizing and have more covalent nature and thus should be less soluble.
I want to know the reason for why the opposite is happening.

How about a few numbers?

Who said greater ionic character means greater solubility?

there should be balance

@Chemist, It is not advised to add questions apart from what was originally intended by the OP. Hence the reject. Your addition of solubility data was carried forward, duly credited, but again lacked the markup. Please make complete edits and try not to add to old questions (you can ask new ones!)

From my experience by asking new questions it will be immediately marked duplicate without second thought by community members , so I decided to edit this question to make it answerable.

@Chemist Have faith in the SE system :). Even if closed in a hurry, posts can be reopened or the closure discussed on meta. Justice will be done.

If one does not compare solubilities by mass, but by ionic molar amount, NaCl and MgCl2 have near identical solubility.

There are no answers to such why questions, in case anyone is looking for an "answer". Although it is very tempting to rationalize every laboratory observation, there are no definite answers to questions about why certain chemical phenomena occur. While it may be tempting to come up with explanations that allow us to pass exams or understand concepts on a surface level, the truth is that our current understanding of chemistry is not advanced enough to accurately predict the solubility of compounds in water or other solvents. The same goes for melting and boiling points. Tools such as ChemDraw and SciFinder can provide estimates, but there is a significant margin of error in those solubility numbers and the results may be incorrect.
Let us see the complications, e.g.,
Take anhydrous $\ce{MgCl2}$, add it to water and dissolve, the reaction is exothermic, and solution can be shown to be slightly basic.
Take the hydrated version, $\ce{MgCl2.6H2O}$, and add it to water, the reaction is not exothermic in water, and water's pH does not change. So may even ask, which magnesium chloride is being discussed?
As a result, many chemists rely on macroscopic measurements such as thermodynamic properties but such properties are quiet about microscopic details and "whies". A paper titled The solubilities of some inorganic halides, Trans. Faraday Soc., 1958,54, 34-39, is relevant the OP's question with actual experimental data. The abstract reads,
"
They have the thermodynamic data of Group I and Group II halides in Table 1.

"Take anhydrous MgCl2... and solution can be shown to be slightly basic." Not really. Should be slightly acidic, unless the dehydration process left the magnesium chloride impure (e.g. by forming a basic salt).

There is a demo on this topic by the Royal Society of Chemistry. I trust their science. On Youtube, search "Reaction water NaCl, MgCl2". I will not discredit this until and unless there is a solid evidence otherwise.

so, in fact, it does react to produce an imputity. Looking carefully at the video you see there is the basic salt.

Right, the basic salt is formed after hydrolysis. The starting material is not a basic salt of magnesium chloride.

I don't think there is any simple answer.There are some rules of thumb, all of which have exceptions. Most chlorides are soluble. That includes both MgCl2 and CaCl2.
Solubility is the result of the balance between competing interactions: between the ions in the salt and between those ions and the water.
Many factors contribute to the solubility of a compound in water. In this case, the main difference between the compounds is the bonds’ strength.
The longer answer is to do with something called the "enthalpy of solvation".'
you can think it this way that the lattice enthalpy in MgCl2 and CaCl2 will be less and can be easily be overpowered by hydration enthalpy

Why is lattice enthalpy of MgCl2 is less than NaCl

I am afraid solubility is one of the properties that cannot be properly explained yet. For example why is CaF2 absolutely insoluble ? By comparaison CaCl2, CaBr2 and CaI2 are all extremely soluble in water, as they can be dissolved in less than their weight of water.

$\ce{MgCl2}$ and $\ce{CaCl2}$ are more soluble than $\ce{NaCl}$ because In $\ce{MgCl2}$ and $\ce{CaCl2}$ the ionic charge is +2 hence it can interact with more number of $\ce{H2O}$ molecules, as the number of interactions increases more amount of energy is liberated which is a hydration energy which is sufficient enough to overcome the lattice energy of the compound.
Even solubility increases with increase in charge on ion.

If only chemistry were so simple… A brief look at the solubility chart would show this is not really how solubility is predicted/explained.


text:
Why does iron lose exactly 3 electrons in oxidation?
I'm not a chemist, but I have this question. Why does iron (Fe atom) lose three electrons when exposed to oxygen for the first time? Why doesn't it lose less than or more than three electrons?

The electronic configuration of iron is $\ce{[Ar] 3d^6 4s^2 }$.
It can lose two electrons to form the ferrous $(\ce{Fe^2+})$ ion with E.C. $\ce{[Ar] 3d^6}$. This is one stable ion of iron. The second (and more stable) ferric $\ce{(Fe^3+)}$ ion is formed by losing another electron to form $\ce{[Ar] 3d^5}$. This has half-filled d subshell, and hence imparts extra stability.
Though your question is not clear. Iron can exist perfectly in both the oxidation states. Both the oxides $\ce{FeO}$ and $\ce{Fe2O3}$ are well known, although - as expected - the first one is rarer.

Thank you. But why would it become more stable if it loses three electrons instead of two ?

@hellothere Half filled subshells provide more stability to an atom/ion as compared to non-half-filled subshells.


text:
Electronic configuration of vanadium ion in a compound with magnetic moment 1.73 bm
Vanadium has $Z = 23$ and so the electronic configuration of an isolated neutral atom in the gas phase would be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^3}$.
Now, magnetic moment of 1.73 means 1 unpaired electron (calculated using the formula $M^2 = n(n + 2)$). So, vanadium will have to lose 4 electrons.
I worked out the ion's electronic configuration to be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$. But internet and other sources say it's $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1}$. Note the $\ce{3d^1}$ instead of $\ce{4s^1}$.

The linked answer lists what happens but not why, i.e. why electrons are lost preferentially from s rather than d orbitals.


text:
How can an ionic bond have partial covalent in character?
I was going through the topic of 'ionic bond' and read this:

No bond is 100% ionic in character. It has some percentage of covalent character.

I didn't understand how an ionic bond can be partially covalent in character. Please explain along with an example.

One way is Polarization. When a small cation approaches a large  anion it distorts the charge density towards itself thereby forming a type of bond which has directional characteristic i.e from ** (-) ----> (+) ** which is characteristic of a covalent bond.
This polarization has some rules given by **Fajan **

Smaller the cation better distortion/polarization.
More the amount of positive charge more is the polarization.
or more the charge : radius ratio of cation more is the polarization.

So in short every cation has the power of polarization and therefore no bond is 100% ionic.


text:
Reaction of NaOD/D2O with semicarbazone

I managed to solve the question till D but I'm stuck at the last reaction.
I tried to find its solution and the source suggest it as

Upto (D) you are correct From there it's :

I'm not so sure about that four-membered transition state.

@AmanChande Why base don't abstract H+ from -NH2 as in case of Hoffmann bromamide degradation reaction?

In your question NaOD/D2O is given in excess so we have high concentration of OD- which attacks rather than extracting a proton. If it was Dil. NaOD then it will extract proton. THATS THE DIFFERENCE BETWEEN CANNIZARO AND ALDOL CONDENSATION RXN ( conc. NaOH and Dil. NaOH)

@Aman Chande: What do you mean by "attacks rather than extracting a proton"? Are you implying that under strong basic conditions H/D exchange doesn't occur but it does in dilute base? Not from my perspective.

It will be through tuotomerism

Upto D) you are absolutely correct. Now observe carefully. In D) you have imine groups as well as amide groups. Under basic conditions imines don't hydrolyse, but amides can undergo hydrolysis. So, the amide parts will hydrolyse and as the reagents are in excess, you will get the end product as

your answer also makes sense but the book sugest the above one.

But as I know, hydrolysis of imines are possible in acidic medium mainly.

This is what I reffered frm book and conclude d..

nice quote :)...


text:
Extracting metallic magnesium from cholorphyll
My understanding is that when chlorophyll is exposed to an increased concentration of hydrogen ions (or rather the pH is lowered), the magnesium is displaced by a hydrogen ion and chlorophyll is converted to pheophytin.

My question relates to the Mg2+ ion that is released in this process. Is it possible, naturally depending on the acid used to convert the chlorophyll to pheophytin, to isolate the Mg released?
Note my background in chemistry is rather limited. My guess is that the Mg will react to whatever acid (say sulfuric acid) to create a compound like magnesium sulfate which in turn can be processed to extract Mg in metallic form (electrolysis?).
Any clarity would be much appreciated as this has tickled my curiosity for a while now.

See the related question - separating iron from blood

You can make the magnesium salt, but electrolysis of an aqueous solution of this salt will reduce the water, evolving hydrogen, not reduce the magnesium.  You can get metallic magnesium via electrolysis of the molten chloride or high temperature reduction of the oxide with carbon,  neither of these cheap or easy.  See here.


text:
Is there any way of guessing what a compound smells like?
I am aware that we can predict the colours of compounds, on the basis of their electron transition energies, and many other physical properties (on the basis of other theories.)
Smell works in an inherently different manner, as the compound binds with smell receptors in our noses, and we smell a distinctly different smell for many compounds, quite unlike colour, or crystalline/amorphous nature, etc. It's also highly subjective, varying slightly from person to person. But, so does observed colour...
However, I don't see any way of predicting the smell of a random substance with any technique, except when we know what it is (e.g. hydrogen sulphide smells like rotten eggs, ammonia like rotten fish, benzaldehyde like almonds, etc.)
So, I guess my actual question is, is smell an actual physical property affected by the chemistry of a compound, or is it highly random? Are there any theories predicting the smell of compounds on the basis of their structures, functional groups, etc?

https://eic.rsc.org/feature/if-it-smells-its-chemistry/2020168.article

Related: https://chemistry.stackexchange.com/questions/81935/how-does-the-smell-of-a-compound-come-about-and-is-it-possible-to-define-a-smel


text:
Enols and enolates in biosynthetic pathways
I'm having a little trouble unterstanding the way organisms carry out aldol and claisen type reactions. I'm familiar with the laboratory procedures to which I understand there is a correlation.
Inside a cell there are no strong bases, since their existence would lead to the destruction of proteins. However, the reactions mentioned above, need a decent base in order to occur. From what I've read, the presence of enzymes is what obviates the need for a strong base, leading to enolates existing for a minimal ammount of time. On the other hand though, I've read in some notes that these procedures are carried out by the enol form, not the enolate. So my question is what kind of nucleophile is formed in the beggining of these reactions?

I recommend finding a good reference on polyketides. That will get you started.

You mean a generally accpeted book? I use "medicinal natural products" from Dewick. Is doesn't suffice?

That's the one I used. It's decent enough.

authors: Αντώνιος Κελεσίδης


text:
Which has a greater dominance: hydration enthalpy or lattice energy?
When I was reading my notes I came up with a problem. I read that for the s block element salts, when checking the solubility, we consider the hydration enthalpy so we decide their stability in water. But then there is this exception: hydroxides don't follow this. They depend on lattice enthalpy for their stability. Which has a greater dominance and why? Is it the lattice energy or the hydration enthalpy? How can we differentiate them, and where to use these concepts for deciding stability? As in the case of hydroxides lattice energy plays a dominant role, not the hydration energy: why is it so?

The dominant factor depends on situation to situation.Here is an explanation i found helpful while predicting solubilities

The hydration energy is of Form A÷r1 + B÷r2 .

The lattice energy is of form C÷(r1+r2).
Here r1 and r2 are radius of cation and anion while A,B,C are constants.

When Size of anion is large example Iodine,Lattice energy does not decrease with increase in radius but hydration energy decreases drastically since r2>>r1(See formula)
Therefore in such cases,Solubility of Salts decreases with radius of cation.
Order- LiI is more than NaI is more than KI(in terms of solubility)

Now when anion size is comparable(eg Fluorine),The lattice energy decreases much more rapidly than hydration energy(see formula).Hence Solubility increases with increase in cationic size.Thefore order is LiF is less than NaF is less than  KF

lets say i am comparing solubility of sulfate and hydroxides of alkaline earth metals.when comparing solubility two terms come into effect.Lattice energy and the hydration energy both contribute to the solubility but sometimes there is a dominance of one term over the other.In order to account for this dominance a new term was introduced.
The resultant of two effects i.e.
ΔHsolution = ΔHlattice  - ΔHHydration
This term plays the major role in deciding the stability of salts inside water and hence their solubility.
Lets first get to know the definition of lattice energy.
The lattice energy of a crystalline solid is usually defined as the energy of formation of a crystal from infinitely-separated ions and as such is invariably negative. The precise value of the lattice energy may not be determined experimentally, because of the impossibility of preparing an adequate amount of gaseous ions or atoms and measuring the energy released during their condensation to form the solid. However, the value of the lattice energy may either be derived theoretically from electrostatics or from a thermodynamic cycling reaction, the Born–Haber cycle.
By the formula given for the lattice energy we can see that lattice energy is directly proportional to the charge present on ions involved in bonding and inversely proportional to the size of the ions.It is a major contributor in deciding melting points and thermal stability in ionic salts.
Coming to hydration enthalpy,Hydration energy (also hydration enthalpy) is the amount of energy released when one mole of ions undergo hydration which is a special case of solvation. It is a special case of dissolution energy, with the solvent being water.
For example, upon dissolving a salt in water, the outermost ions (those at the edge of the lattice) move away from the lattice and become covered with the neighboring water molecules. If the hydration energy is equal to or greater than the lattice energy, then the salt is water-soluble. In salts for which the hydration energy is higher than the lattice energy, solvation occurs with a release of energy in the form of heat. For instance, CaCl2 (anhydrous calcium chloride) heats the water when dissolving. However, the hexahydrate, CaCl2·6H2O cools the water upon dissolution. The latter happens because the hydration energy does not completely over come the lattice energy, and the remainder has to be taken from the water in order to compensate the energy loss.
Coming to the question now that we have understood the terms involved,we know that sulfate ion is larger in size compared to the hydroxide ion.
Coming to hydroxides of alkaline earth metals
The solubility of the alkaline earth metal hydroxides in water increases with increase in atomic number down the group. This is due to the fact that the lattice energy decreases down the group due to increase in size of the alkaline earth metals cation whereas the hydration energy of the cation remains almost unchanged.Due to its small size the lattice energy plays a dominant role in deciding its solubility whereas hydration energy almost remains constant.
Coming to sulfates,
solubility:The solubility of the sulfates in water decreases down the groups i.e. Be > Mg > Ca > Sr > Ba.
Thus BeSO4 and MgSO4 are highly soluble, CaSO4 is sparingly soluble but the sulphates of Sr, Ba and Ra are virtually insoluble.
Reason
The magnitude of the lattice energy remains almost constant as the sulfate is so big that small increase in the size of the cation from Be to Ba does not make any difference. However the hydration energy decreases from Be+2 to Ba+2 appreciably as the size of the cation increase down the group. Hence, the solubility of sulfates of alkaline earth metals decrease down the group mainly due to the decreasing hydration energies from Be+2 to Ba+2. The high solubility of BeSo4 and MgSO4 is due to high hydration energies due to smaller Be+2 and Mg+2 ions.
This effect can also be explained by the fact that ionic character increases down the group.By fajan's rule more covalent character would mean smaller size of cation hence more polarization effect in solution state.More polarization leads to dissociation of compound in water easily hence the solubility increases from bottom to top for the sulfates.


text:
Comparision of ONO Bond angle in NO2 and N2O4
How can I compare the bond angle between $\ce{O-N-O}$ atoms in $\ce{NO2}$ and $\ce{N2O4}$?
In $\ce{NO2}$ there is lone pair-bond pair repulsion, and in $\ce{N2O4}$ there is bond pair-bond pair repulsion.

I think that NO2 will be a free radiacal with the radical centred on oxygen. Thus it will be a bonded pair-single electron (lone half pair) repulsion.

The bond angle in NO$_2$ is $\approx 134^\mathrm{o}$ and the NO bond length 120 pm, in $\ce{N2O4}$ the N-N bond is very long 175 pm and the angle $\approx 126^\mathrm{o}$.
Nitrogen dioxide is paramagnetic with the odd electron in the $\sigma$ bonding sp$^2$ orbital in the 'third' lobe as it were with the O atoms occupying the other two. Additionally there are 4 electrons in the $\pi$ system, two bonding and two antibonding. NO$_2$ also has several low lying excited states which greatly complicates matters.
Joining up two NO$_2$ naturally leads to $\ce{N2O4}$ which is planar and has a very long single $\sigma$ N-N bond. Why the molecule is planar is not clear as it might be expected to twist to remove oxygen-oxygen repulsion. Intuitively one would expect an MO to extend over the whole molecule and so prevent twisting, but the NN bond is very long and single.
Hopefully someone has more up to date data, otherwise you will have to search the literature for MO based (i.e. numerical quantum/chemistry) approaches to describe the bonding.


text:
Dissolution of Lithium in Methanol
Easy question to kick off my chemexchange membership!
I'm learning chemistry in my spare time, and fittingly for a video art project I'm doing, decided to try a simple inorganic(?) procedure to make some crimson flames.
My research lead me to believe that lithium methoxide works wonders for this, especially in solution with methanol which only has a very faint blue flame on its own. I basically already know it works, based on some demonstrations seen on YouTube, but I'm unable to find much anything in literature about the lithium salt of methanol. So I'd like to verify my understanding of the equation:
To my understanding, dissolving lithium metal in methanol should yield:
$$\ce{Li + CH4O -> CH3LiO + H}$$
...Lithium methoxide in solution with methanol, assuming an excess of methanol, and some released hydrogen gas.
Am I correct? Can someone shortly explain the actual reaction that takes place?
Additionally: I think the solution should be relatively stable and easy to store and handle, like methanol, unless there's a reason to believe otherwise?
Also, if one wanted to produce dry lithium methoxide powder from this, what might be a good precipitant to crash it out of methanol?

You are basically correct. As with the other group I metals lithium reacts with proton sources such as alcohols and water to give hydrogen and the metal salt.  This may be useful to you  http://orgsyn.org/demo.aspx?prep=cv6p0795

You are aware that methanol is rather toxic and easily penetrates your skin? And that this reaction you are thinking of is quite vigorous and can easily lead to an explosion with air of the H2 produced?

@Karl I should probably have clarified that eventually setting fire to some amounts of the solution is obviously an entirely separate thing!
Of course, the dissolving has to be done in a fume hood with proper exhaust and have fire nowhere near it. I will edit my question to rectify the misunderstanding.
As for methanol, yes, I follow good lab practice and generally try not to pour solvents (or anything, really) on myself.
I do appreciate the concern though. If there's something I've yet overlooked, let me know.

@Waylander thanks! I haven't worked with alkali metals before.

And be also ware, methyl- and methoxy- inorganic compounds are notoriously toxic, flammable and explosive. Something as light as lithium is fairly benign, but many aspiring and successful chemists have perished exploring the methyl-metal world.

Why don't you just use a solution of something like lithium chloride in water? If you need it to be more flammable, mix in some ethanol or isopropanol. Obviously, you should be extremely careful if you plan to light it on fire...

@Zhe well, for a visual purpose the methanol is critical, as all the other usual solvents have a way too bright yellow flame of their own, which easily overpowers the lithium's crimson and thus defies the point.
As for lithium chloride, I suppose it would also work, but I don't have any handy so would probably have to produce it anyway.
Is the chloride particularly preferable to the methoxide in this sense?
A 10% LiOMe/Me solution is described as "very stable in room temperature" in the single MSDS I could find.
But yes, any burning will be very careful and small-scale.

I have not done it with lithium but many times with sodium. Lithium is the less reactive, so it should not give more problems. We use sodium to dry methanol, so it is a common operation in the lab. The reaction is:
$\ce{2Li + 2CH3OH -> 2LiOCH3 + H2}$
Of course, hydrogen is dangerous, especially in the presence of oxygen and still more in this exothermic reaction. So, be careful.
The solution is stable, but attracts water and carbon dioxide, so it must be stored well closed.
To get dry lithium methoxide, the easiest is just to evaporate the excess of methanol.
You put it in a round bottomed flask, attach it to the rotary evaporator, open the vacuum and distill it of. With sodium it works perfectly well. A white solid remains. Again, this solid will attract water and carbon dioxide, so it must be stored properly.
Finally, if the only purpose is to have a methanolic solution of lithium, you can better use lithium chloride which is very soluble in methanol, and much safer (no reaction, no hydrogen, no heating).