Solubility of C5H11OH and C5H11Cl Objectively, what makes the solubility of $\ce{C5H11OH}$ greater than the $\ce{C5H11Cl}$ in water?

In the case of the water solubility what you want to know is how powerful the secondary interaction  of the compound that you are going to dissolve with  water?If the compound that you wish to dissolve have hydrogen bonding then it's definitely dissolve in water .It does not mean others  are  insoluble  in water.

But  the compounds do not have hydrogen bonding are  less soluble in water when compared to    the ones which have hydrogen bonding.Because ones which have hydrogen bonding interact with water molecules easily than others which do not have hydrogen bonding

Now take a look at the image   below,

It shows two water molecules  which posses the hydrogen bonding interaction.

Now give the eye on compounds that you have got,

The  compounds that you going to dissolve  may form  interaction force  with water as above image. If that interaction force is hydrogen bonding ($\ce{C5H11OH}$) then it's solubility is greater than ones which do not have hydrogen bonding($\ce{C5H11Cl}$) (dipole dipole interactions)

So because of that solubility of $\ce{C5H11Cl}$ is less than the solubility of ($\ce{C5H11OH}$) with water.

Then the bottom line is ,the solubility of a compound is vary according the  secondary interaction between the solute and solvent.In the case of water acting as  the solvent solubility is highest in  compounds which having H bonding , than which haaving dipole-dipole forces and  weak dispersion forces.


Why are C6 dispersion forces named C6? I'm reading about water models and their dispersion coefficients, and going back to S. C. Wang's work according to the citation of an equation (Google books here)

As you can see one citation's to a biographical work, and the other to a German work.

Nowhere in the chapter I'm reading (chapter 5, ‘Resolution’, in Rowlinson’s 2002 Cohesion: A Scientific History of Intermolecular Forces) does it say what $C_6$ stands for, and it doesn't seem to be a series of $C_n$, so I'm a little confused.

There's the equation, apparently a Casimir-Polder integral

$$E_\mathrm{DS} = ‒\sum_{n=6,8,10,...} \left(\frac{C_n}{r_n}\right)$$

in something else I'm reading, but it's biochemistry and so they're skipping over actually defining these things (so I can see DFT would be density functional theory but can't actually figure out what either $E_\mathrm{DS}$ is, nor the meaning of a $C_6$ coefficient. Looking for $E_\mathrm{DS}$ is just bringing up a bunch of articles' editors in search results (Eds.).

Can anyone enlighten me as to the latter - the meaning of a $C_6$ coefficient with respect to [London] dispersion forces?

My own answer to this after more reading

  The oscillating electrons in a molecule generate not only instantaneous dipoles but also quadrupoles and higher multipoles. It is to be expected, therefore, that the London dispersion energy is only the first term in a series expansion for the attractive energy

that starts in $$u(r) = −C_6r^{−6}$$

The term comes originally from the wave equation (as Rowlinson's text does eventually describe), the number six arising from a Cartesian coordinate system for each of the two interaction partners

From Calculation of Coefficients in the Power Series Expansion of the Long‐Range Dispersion Force between Atoms J. Chem. Phys. 56, 2801 (1972)

  Highly accurate calculations
  of $C_6$, the coefficient of the $R^{-6}$ term, resulting from
  an induced dipole-induced dipole interaction, have
  been made for many atomic systems, while values of
  $C_8$ (dipole-quadrupole interaction) and $C_{10}$ (dipoleoctopole
  plus quadrupole-quadrupole interactions) are...

In other words, in London dispersion interaction is usually represented as induced dipole - induced dipole interaction, the energy of which is proportional to $R^{-6}$, but there are other components of the interaction.


Hybridisation of cobalt in tris(oxalato)cobaltate(III) Will the hybridisation of of $\ce{Co}$ in $\ce{[Co(ox)3]^3-}$ be $\mathrm{sp^3d^2}$ or $\mathrm{d^2sp^3}$? The question in which I found it has mentioned the answer to be $\mathrm{d^2sp^3}$ while I was thinking of the other one.
According to me, oxalate is a weak field ligand and it should not cause the pairing of electrons in the half-filled $\mathrm d$ orbitals.

As explained in The Chemistry of Iron, Cobalt and Nickel: Comprehensive Inorganic Chemistry, at pages 1104 and 1105:

almost all cobalt(III) complexes are low spin ... only with fluoride ions as ligands are high spin complexes ... found

In table 10, the book specifically lists $\ce{[Co(C2O4)3]}^{3-}$ as low spin and cites to J. Chem. Soc. (A) (1966) 798.
Therefore, according to the historical valence bond theory of transition metal complexes, it would be considered $\mathrm{d^2 sp^3}$ for the following reason:
There are (in the context of the theory) two unoccupied $\ce{3d}$ atomic orbitals in $\ce{Co^{3+}}$.
Six pairs of electrons from the oxalate ions are added to the two $\mathrm{3d}$ orbitals and the $\mathrm{4s}$ and three $\mathrm{4p}$ orbitals.


How many valid resonance forms can be constructed for aniline? Based on the above picture, it seems that there are four. However, I have two questions:

Aren't the top right and bottom left forms simply mirror images of each other, and thus considered the same?
If not, would an alternate version of the top left form be a fifth structure, if the double and single bonds were flipped?

Your four structures are correct and you could also include your suggested fifth structure (your first and fifth structures are comparable to the resonance structures of benzene). While some of the structures are mirror images, we generally think of particular atoms as being fixed when writing out resonance structures. You can think of the carbons as having a number (which they technically would if you named the resonance structures IUPAC style), which leads to the "mirrored" structures not being identical since the double bonds are between different carbons.


Why it is hard to rinse off soap from your body using rain water I have experienced this when every time I take a bath using rain water, I just can't stop wondering why is it really hard to rinse off soap from your body?

Rain water is pretty much devoid of the minerals such as calcium salts that are often found in so-called "hard" tap water taken from underground sources. Calcium ions in tap water react with the fatty acid anions in soap to create insoluble soap scums, which (while they may remain on your skin) do not have the slippery feel of soap. Without the calcium ions present, the soap remains in soluble form (sodium salts of fatty acids) and I would propose that the slippery feel of soap on your skin comes from the traces of it that remain even after considerable rinsing.


Concentration of hypochlorous acid and hypochlorite Given a total chlorine concentration  $c(\ce{Cl})_\text{tot}= \pu{0.1 M}$, with $\mathrm{pH}(\ce{Cl}) = 8$ and $\mathrm{p}K_\mathrm{a}(\ce{HOCl}) = 7.5$, how would you find the concentration of $\ce{OCl-}$ and $\ce{HOCl}$ disinfectants?

So far I think that I should start this way:

$K_\mathrm{a} (\ce{HOCl}) = 10^{-7.5}=3.16\cdot 10^{-8}$

I wrote up the reactions:

$$\ce{H2O + 2e- + OCl <=> Cl- + 2OH-}$$

Where $\ce{OCl-}$ is the conjugate acid and $\ce{OH-}$ the conjugate base.

$$\ce{HOCl<=> OCl- + H+}$$

The conjugated acid is $\ce{HOCl}$, the conjugated base $\ce{OCl-}$.

Now I use
$$\mathrm{pH}=\mathrm{p}K_\mathrm{a} + \log\left(\frac{[\text{conj base}]}{[\text{conj acid}]}\right)$$

and/or for weak acids
$$K_\mathrm{a}= [\text{products}]/[\text{reactants}]$$

But I'm not sure if this is correct and what values to use.

You really overcomplicate things (as I did initially, because I thought you were talking about the dissociation of $\ce{Cl2}$ gas in water and the subsequent reactions).

Here are the basic pieces of information needed:

Reaction equation
Equilibrium equation
Mass balance equation

Reaction Equation

The reaction is a simple acid-base equilibrium reaction:
$$ \ce{HOCl <=>[K_\text{a}] H+ + OCl-} $$

Equilibrium Equation

The equilibrium equation follows easily from the reaction equation:
$$ K_\text{a} = \frac{[\ce{H+}][\ce{OCl-}]}{[\ce{HOCl}]} $$

Mass Balance Equation

The total amount of chlorine is calculated as follows:
$$ [\ce{Cl}]_\text{tot} = [\ce{HOCl}] + [\ce{OCl-}]$$

From the reaction equation also follows
$$ [\ce{OCl-}] = [\ce{H+}] $$

The Solution

Since the concentration of protons is known, we can easily calculate the concentration of the hypochlorite anion:
$$ [\ce{OCl-}] = [\ce{H+}] = 10^{-\text{pH}} = 10^{-8} $$

Since we know the total concentration, simply reforming the mass balance gives us the concentration (strictly speaking, the activity) of the hypochlorous acid:
$$ [\ce{HOCl}] = [\ce{Cl}]_\text{tot} - [\ce{OCl-}] = 0.1 - 10^{-8} = 0.09999999 \approx 0.1 $$

Slap on some units and we're good to go!

$$ \begin{align}
c(\ce{HOCl}) &\approx 0.1~\mathrm{mol\, L^{-1}} \\
c(\ce{OCl-}) &= 10^{-8} ~\mathrm{mol\, L^{-1}}
\end{align}$$


Coordination numbers and radius ratio According to the image above (source), for a radius-ratio between 0.732 and 1, the coordination number is 8. However, for a radius-ratio of 1, the coordination is 12. This is the same for the boundary values between the other coordination numbers, but I’m focusing on the radius-ratio of 1 for now. I don’t understand how the same boundary value could correspond to two different coordination numbers. If you have a monoatomic metal, then all the atoms constituting it have the same radius and the radius-ratio is 1. In order to achieve the closest packing possible, the metal atoms would form a CCP or HCP structure, wouldn’t they? Those two crystal structures have the greatest coordination number. Yet other monoatomic metals, with a radius-ratio of 1, form can BCC structures, with a coordination number of 8. Why would they form such a structure when their radius-ratio can also correspond to a CN of 12? How can one radius-ratio correspond to two CNs?

Coordination number is the number of the nearest neighbours of an atom, and it is related to the Kepler's conjecture. For CN12 (12 is the 'kissing number' of a 3D sphere), there are two possible ways for the spheres to be 'packed' which are HCP and CCP. Space-filling efficiency they two are exactly the same, and it is why one coordination number can correspond to two crystal structures.
The crystal structure of metal may depend on many features, and the space-filling efficiency is just one of them. Even by the thermal expansion, the crystal structure can change. (Tin is the most familiar example.)


Why don't such ions form if half-filled or full-filled configurations are considered stable? We all know atoms want to stabilize by having an octet or duplet configuration. But if you see nitrogen which has an electron configuration of $\ce{2p^3}$. It has half filled p orbital, so it is technically stable, isn't it? Then why don't elements like oxygen form a monovalent oxygen ion?

By similar logic, sodium has $\ce{3s^1}$ configuration which should also be stable because it has a half filed orbital. But sodium is reactive as we know.

Can you please point out the flaw in my argument?

As also pointed out in the comments, there is no such thing as absolute stability. One species is stable only with respect to another species.

You may have come across the fact - that half-filled configurations are "stable" - while studying ionization enthalpy/electron gain enthalpy, or the electronic configuration of $\ce{3d}$ elements. Pay careful attention: the half-filled configuration is "stable", but only with respect to the neighbouring configurations. That is to say, the $\ce{2p^3}$ configuration is much stable than $\ce{2p^4}$ or $\ce{2p^2}$. Yet, $\ce{2p^3}$ is still very less stable than $\ce{2p^6}$. This is why, in nature, nitrogen prefers to exist in diatomic molecules, so as to complete its octet.

Now for the fully-filled $\ce{3s^2}$ configuration (sodium anion, aka "sodide"): again, note that the sodium cation is way more stable than the sodium anion, as the former has a fully filled octet, while the latter doesn't. (Sodide has indeed been formed, but by using special techniques only. See Wikipedia).

Also see: Why are atoms with eight electrons in the outer shell extremely stable? and Why do atoms "want" to have a full outer shell?


How to measure ratio of Na to K in table salt? Some people state, that the salt which we buy at the supermarket contains too much potassium chloride. For me it seems to be a hoax, but I want to have a proof.

Is there any way to measure the percentage of sodium chloride and potassium chloride in salt at home without special equipment?

If it is not possible, what kind of equipment is needed?

To answer your question: no, you cannot do it at home. Some analytical techniques are mentioned in the comments.

However I decided to post because LIKELY people should not worry about the Potassium content of table salt.

Table salt (normally "just" NaCl) containing as much of 28.5 % of Potassium (might be more perhaps, but this is what I have at hand) can be sold free of medical prescription not only in drugstores/pharmacies but also in supermarkets, where it can be found in the special diets section or alongside standard salt.
It is a dietary salt indicated to limit the Sodium intake (for instance in case of high blood pressure).
Its price is much higher that that of common salt. It can be used to replace salt in all kitchen uses. It is less salty and more bitter than the normal salt, and one has to "get the mouth" to it.

I would therefore infer that is highly unlikely that something sold as table salt NaCl could contain so much potassium to worry about. Especially when some of those worrying of it, might be prone to buy Potassium integrators during summer :)

Disclaimer: it is clear that the potassium content of whatever table salt sold in food stores and supermarkets should not be matter of concerns however, as for any diets and integrators (Fe, K, Mg...) there exists health conditions which are incompatible. Before starting a diet or taking dietetic salts one should seek the advice of his/her physician.


Is carbohydrate alcohol inebriant? First of all, though I think that my question is related to this society, if you think reverse, tell me to delete the question:

I would like to know if carbohydrate alcohol is inebriant; Can anyone reference me to some valid source?

I'm assuming you mean "do the alcohol groups in carbohydrates such as sugars cause inebriation".

If that is what you mean then the answer is no. Inebriation is a function of some biological effects that are very specific to small alcohols not just to the OH group in a molecule (I suspect that methanol causes inebriation as well as ethanol, but given that it also kills you this is not something i'm willing to test).

Some of the biological effects are based on specific interactions with proteins in nerves, others are gross physical effects (e.g. dizziness is a product of changes to the density of fluid present in the balance organs in your inner ear when the alcohol concentration in the blood is high). The neuromodulation effects are very specific to the size and shape of the whole molecule not just the OH group and the dizziness is a function of the density of ethanol and its ability to circulate in the blood (larger alcohols and other OH-containing chemicals won't do this).

The body has a whole bunch of mechanisms to handle and manage sugars (as they are a primary way of providing fuel to cells). It would be surprising indeed if animals were inebriated by such compounds as they have evolved sophisticated ways to use them as a key part of their metabolism.


Implication of completely non relativistic Hamiltonian I am studying the
Application of perturbation theory to hydrogenic atoms subject to internal or external electromagnetic fields.

The very first equation it uses is the hamiltonian of hydrogenic atoms
$$H_0=\frac{p^2}{2\mu}-\frac{Ze^2}{r}$$
This equation is said to have no spin and is completely non-relativistic.

What does completely non-relativistic mean?

Your reduced equation is just the normal hydrogen Hamiltonian written in atomic units, that is with $\hbar=\frac{1}{4\pi\epsilon_{0}}=m_{e}=e=1$ (e is the charge of the electron, which they include for some reason despite it being 1 in atomic units). They also make the substitution $p=-i\hbar\nabla$ where, again to be consistent with atomic units, $\hbar=1$.

The form you have normally seen of the Hamiltonian is already spinless and non-relativistic (which are essentially synonymous for quantum mechanical operators). For something as light as Hydrogen, relativity (as in Einstein's theory of Special Relativity) is known to have very limited impact, and so it is neglected in forming the Hamiltonian.

There may be other questions on the Chem or Physics SE related to relativity in QM, but I would suggest unless you are particularly interested in relativity to simply focus on the QM side of things for now until you have that clear and can build on that knowledge.


Why is alcohol boiled with HBr for preparing alkyl bromide? Alkyl halides are best prepared from alcohols ($\ce{R-OH}$, where $\ce{R}$ is alkyl group), which are easily accessible. The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acid ($\ce{HX}$, where $\ce{X}$ is halogen), to get alkyl halides.

Usually we don't need boiling the reactants. But, for the preparation of alkyl bromide requires constant boiling of alcohol with $\ce{HBr}$ ($48\%$). Assume that alcohol is tertiary, so that, we can avoid using catalyst.

What I thought was, alcohol can develop inductive effect, and thus, can get detached with $\ce{OH}$, so that, the bromine from $\ce{HBr}$ can add to $\ce{R}$ group to form alkyl bromide. So, I thought there is no requirement of boiling of reactants, as like we do for preparation of alkyl chloride. Then, why do we need boiling in case of $\ce{HBr}$ and not in case of $\ce{HCl}$?

Assuming i) a primary alcohol $\ce{R-OH}$ to be converted into the corresponding primary alkyl halogenide $\ce{R-X}$ ii) by $\ce{HX}$ acid only, I would expect a decrease of reactivity to be observed in this line: $\ce{HI} > \ce{HBr} > \ce{HCl} > \ce{HF}$, with $\ce{HF}$ of lowest reactivity.  This is based on the decrease of the nucleophilicity of the anions of these acids.
Of course, halogenations with these four acids are reactions of equilibria, and these may be driven forward by increasing the concentration of the reactants, for example.  And I do recall two-step protocols where the alcohol is activated by sulfuric acid before adding hydrobromic acid.  But in both variants, rather $\ce{H3O+}$ then a neutral $\ce{OH}$ would leave the molecule.
(However secondary and tertiary alcohols that are of higher reactivity than primary ones under these acidic conditions, are more prone to isomerization, too.  Simultaneously, as hinted by ssavec, instead of a mechanism of type $\mathrm{S_{N}1}$, the of  $\mathrm{S_{N}2}$ may become more favorable.)
For the preparation of alkyl chlorides, the addition of (dry) $\ce{ZnCl2}$ in presence of $\ce{HCl}$ is sometimes used.  More easy to handle are routes relying on $\ce{PCl3}$ or $\ce{SOCl2}$, for example.


Which elements tend to form the most phases? Some combinations of two elements have very complicated binary phase diagrams across the weight% horizontal axis. Others are rather simple. Is this a function of only one or both of the elements involved? I would think both. However, surely there are some elements that tend to form more phases than others. Has this tendency to form more phases (or less) been quantified for the various elements? If no, is it quantifiable?

Obviously elements form only a given number of known allotropes. However, it is not known if the number of known allotropes is exhaustive. Therefore the number of pseudomorphs in a binary phase diagram seems like a more thorough method of investigating an element's "binding flexibility."

Much empirical data is needed to either produce or validate phase diagrams.  Any phase diagram would likely need to be of practical consequence to justify spending the energy and resources to produce it.

An example of a binary mixture with practical consequence would be a metal oxide formed during corrosion. The many forms of iron and oxide are well studied since iron is commonly used to construct tools.

In 2015 a group claimed to use the computer code CALYPSO (Crystal structure AnaLYsis by Particle Swarm Optimization) to discover a new phase of beryllium.  Beryllium is used in aerospace and the construction of nuclear reactors, so the expenditure of effort was justifiable.

So to answer the  sub-questions:

Is this a function of only one or both of the elements involved?
Practically .. No. In the case of iron oxide in an aqueous solution, multiple iron oxide phases can form at the same temperature and pressure if the pH of the solution or the redox potential value of the solution were changed.

Has this tendency to form more phases (or less) been quantified for the various elements?  - Yes, phases of practical consequence have been quantified at great effort; however, discoveries of new phases are likely to continue.

It is possible to estimate the existence of new phases using computer algorithms; however, these phases must be experimentally validated.  A computer algorithm could be constructed around pseudomorphs; however, if a new phase were discovered it would need practical validation by experimentation.

"Which elements form the most phases?" is an impossible question to give and absolute answer as we are limited to certain pressures and temperatures with which to experiment. The way experimentation is done is likely to bias any verifiable answer toward more practical materials.


Constitutional Isomers Is there a formula to know how many constitutional isomers are there for a compound? For instance in this example, most students would miss the 6th constitutional isomer and will probably only write the first 5. So I would like to kindly ask if there's a formula or pattern for it.

There isn't a good overall formula that I know.
You can look at C/H ratio and easily tell a bit here.
There must be at least one of the following, but not two:

carbon ring
carbon double Bond

(1) 1-Butene

(2) 2-Butene which has two stereoisomers
cis-2-Butene
trans-2-Butene

(3) Isobutylene

(4) cyclobutane

(5) methylcyclopropane


Are there biodegradable pigments that are soluble in oil, but not in water? I'm hoping to find a purple pigment that preferably, but not necessarily, can be used at room temperature. The goal is a pigment that will break down in septic systems.

There are several such pigments which are bio-degradable by some microbial action (mostly bacterial action) and are oil soluble (lyophilic) and water insoluble (hydrophobic).

Some are:

chlorophyll : certain bacteria can decompose chlorophyll.
A paper citing zoo-plankton activity on chlorophyll is here.
Xanthophylls : This is a general category of several pigments whose degradation by marine bacteria has been widely  studied. (You can search for many relevant papers online).
Carotenes : Certain carotenes are also reported to be degraded by fungi and mushrooms. For example here

Anthocyanins (and related polyphenols) which have dark red to purple color are mostly water soluble.


Dissociation energy of benzene dimer It is known that the benzene dimer has two stable structures, T-shaped ($\ce{C_{2v}}$) or parallel-displaced ($\ce{C_{2h}}$) structure.

Now the question is: does the dissociation energy ($\ce{D_0}$) $$\ce{(C_6H_6)_2 -> C_6H_6 + C_6H_6}$$ depends on the structures?
Or it is independent of the structures?

In fact you have three configurations to consider: t-shaped, parallel-displaced and sandwich. The dissociation energy depends on the configurations of the structures since they may interact differently. As the pi-pi interaction is directional, the different structures of the dimer interact differently.


Ionization energy of neon vs its cationic counterpart Which requires more ionization energy: $\ce{Ne}$ or $\ce{Ne+}?$

It seems to me like it should be neon because of noble gas configuration, but the answer given is $\ce{Ne+}.$ Does this anything to do with size of the atom shrinking?

Is this because cationic counterparts always have more ionization energy than their neutral counterparts because it becomes successively more difficult to remove electrons from atoms on successive removals?

Does this anything to do with size of the atom shrinking?

Yes, precisely; this is one way of putting it. Another way is to say that because one electron has been removed, the remaining electrons are less shielded from the nuclear charge: therefore, the effective nuclear charge increases, and the remaining electrons are harder to remove.

  Is this because cationic counterparts always have more ionization energy than their neutral counterparts because it becomes successively more difficult to remove electrons from atoms on successive removals?

Yes. This is true for every element in the Periodic Table, with no exceptions, and it doesn't matter what the electronic configuration is.

If all else were equal, then it would indeed be harder to remove an electron from a noble gas configuration than a non-noble gas configuration. However, that is a very big if! Between $\ce{Ne}$ and $\ce{Ne+}$, the electron configurations are indeed different, but all else is not equal: as discussed previously, the effective nuclear charge in $\ce{Ne+}$ is greater than that in $\ce{Ne}$.


How to modify viscosity of ZnO ink? I have stumbled upon these instructions on how to create $\ce{ZnO}$ nanoparticles.
I am interested in making a DIY ink, for use with a parallel metal plate pen and a piezo-actuated microdrop nozzle.
My question is:
What is a reasonable way of controlling the viscosity of such an ink? Is it possible to replace the water with a mixture of water and isopropanol, for example?

Regular black ink is pretty viscous it has a dynamic viscosity ($\eta$) of 4.177.

In comparison a 50% mixture of methanol and glycerol has a viscosity of 5.8

Some other solvents to consider are

Hexanol which has a viscosity of about 5.4
Heptanol which has as viscosity of
about 7.3

These long-chain alcohols, however, will not dissolve significantly in water.

Some more viscosity data can be found here (it's not an ideal place for viscosities but it is free)

Linseed oil is also not water soluble. In any case here is another link

Ultimately it seems like you'd need to try a few different mixtures to get desired viscosity, solubility and evaporation speed.


Is it possible to make meta-phenol? Normally phenols have the $\ce{OH}$ group at the 1 or 4 position. Can they be made at 2 or 3 positions, where nitro groups and $\ce{-CN}$ can attach? If not, why not?

The hydroxyl groups in phenols are ortho-/para-directing. So, if you want to introduce a group in the meta position with respect to a hydroxyl group you will have to "overwrite" the ortho-/para-preference. This can be done by starting out with meta-directing group which can later be transformed into $\ce{OH}$ in the position where the hydroxyl group shall be. Then you only need to introduce the group you want to be in the meta-position by a simple electrophilic aromatic substitution reaction and subsequently convert the meta-directing group into $\ce{OH}$.

An example for this procedure would be to start out with a carbonyl group at the position of the $\ce{OH}$ group, then nitrate the aromatic ring in the meta position and employ the Baeyer-Villiger oxidation to introduce a $\ce{OH}$ group:

Or you start out with a nitro group at the position of the $\ce{OH}$ group, then introduce chlorine in the meta position of the aromatic ring, reduce the nitro group to an amine, which is then coverted into an diazonium salt which in turn is then transformed into a $\ce{OH}$ group using a Schiemann-type reaction:


XRD: Can stacking faults cause pseudo-centring? Consider a cP lattice with a = 10 Å and an octahedron residing on each of the unit cell's vertices as shown below:

The maximum symmetry will be Pm-3 m.

Now, let's consider there are stacking defects along half of the diagonal of the unit cell's faces. Let's look at displacement by a translation of [0 0.5 0.5]:

If we look onto the bc plane, this will look like an A-centred cell:

If the stacking fault energy is low, random displacement along [0.5 0.5 0], [0.5 0 0.5], and [0 0.5 0.5] is to be expected. Will the situation described cause the powder X-ray diffraction pattern to resemble space group symmetry Fm-3 m, e.g. by integral extinctions of reflections where h + k, h + l or k + l is odd? From my perspective, the average structure the X-ray beam will 'see' is the following:

The outcome depends on the typical length between stacking faults and the coherence length of the X-ray beam.

If the length between stacking faults is large compared to the coherence length, you will get a phenomenon similar to that of twinning (unit cells in same orientation) or of broadened diffraction spots similar to the model of high mosaicity (unit cells in slightly different orientations).

If the length between stacking faults is small compared to the coherence length, you will get a phenomenon similar to partial occupancy of sites in an otherwise perfect crystal (i.e. static disorder - same diffraction pattern with different intensities compared to absence of partially occupied sites).

I found a statement about coherence lengths of synchroton radiation (https://www.researchgate.net/post/What_are_typical_requirements_on_coherence_lengths_for_coherence_experiments_using_3rd_gen_Synchrotron_Light_Sources):

  [Keoki Seu, March 21, 2012] I work at a coherent scattering beamline at the Advanced Light Source. It is an undulator source optimized for coherent flux. At 500 eV, which is usually where we work, coherence lengths are on the order of 6 um vertical and 3-4 um horizontal. We'd love to have larger coherence lengths so we can coherently illuminate larger sample areas, but for the soft x-ray, the values we see here are typical values.

The question was about power patterns rather than single crystal diffraction.

  [OP] Will the situation described cause the powder X-ray reflection pattern to resemble space group symmetry Fm-3 m, e.g. by integral extinctions of reflections where h + k, h + l or k + l is odd?

I don't know what the typical dimensions of the solids in the powder would be, and I don't know what the coherence length of the Xray radiation used in the experiments are, but if you consider all of these factors, you should be able to say in which regime (i.e. twinning-like or static disorder-like) the experiment at hand is in.


Alkanes to alcohols and back? My question: could it be practical, on an industrial scale, to take a mixture of alkanes (mostly ethane,  propane, and butane) convert them to their alcohol forms, send them through a 300-mile pipe, and convert them back to their source alkanes? (They would not have to be handled as a mixture—they could be separated at the source and piped separately.)

I ask this because a company is building a pipeline through my area that is intended to carry alkanes in liquid form (under 1500 psi of pressure, so that they stay liquid). These are known in the industry as “natural gas liquids”.  It looks like this could be a very dangerous pipeline. An engineer friend suggested the alcohol conversion strategy. Even if it were a lot more expensive, it would be far less dangerous and that might justify the cost. Any thoughts?

That's not viable at all.

There's no good reaction to go from alkanes to alcohols. This would be incredible expensive to do, produces a lot of waste and is potentially very dangerous too.
There's no good reaction to convert alcohols back to alkanes. It would be again very expensive. And you would convert more expensive products into less expensive ones, which isn't a very good thing to do.
We know how to handle liquefied petroleum gas very well, it's not that dangerous. There are also hundreds of thousands of km of such pipelines [1] already in use. The safety aspect definitely doesn't justify the cost, danger, waste, ressources and energy needed to convert the product forth and back again.

[1] http://www.j-pipe-eng.com/abstract.cfm?cat_no=2145s


Stability of 4-pyridone vs 4-pyridinol In kryptomerism assignment, there was a question to compare more stable tautomeric form between 4-pyridone (1) and 4-pyridinol (2):

Answer: 4-pyridone.

My friend gave two plausible reasons:

$\ce{C=O}$ is more stable than $\ce{C=N}.$
$\ce{-N=}$ will have greater repulsion between the lone-pair and the double bond than the same in $\ce{-O=}.$

Though, not being convinced by his explanation, I have two questions. First, is the answer correct? And second, what's the reason if it's correct?

I don't think this is an either this or that situation. As with most of the equilibria we know it depends very much on the circumstances we look at it. Many examples of keto-enol-tautomerism are heavily influenced by the solvent. Due to time and resource restrictions I won't be able to go in as much detail as I did in explaining whether enolates get protonated at the carbon or oxygen atom, but I ran a few calculations showing some of the controversy.
The calculations were carried out using Gaussian 16 Rev. B.01 at the DF-B97D3(BJ)/def2-TZVPP level of theory, and thermal corrections were obtained from normal coordinate analysis (in the harmonic approximation) at the optimised structures at the same level of theory using $T=\pu{273.15 K}$ and $p=\pu{1 atm}$ (because that is the G16 standard, sorry for the non-SI). The polarisable continuum model was employed except for the gas phase. All energies are in $\pu{kJ mol^-1}$ for reaction \eqref{tauto}, which means that negative values indicate that 4-pyridinol is more stable than 4-pyridone and vice versa for positive values:*
$$\ce{4-pyridinol <=> 4-pyridone}\tag1\label{tauto}$$
\begin{array}{lrrr}
\text{solvent} & \Delta E_\mathrm{el} & \Delta E_\mathrm{o} & \Delta G \\
\hline
\text{gas}     &  -3.4  &  -4.2  &  -3.7  \\
\text{water}   &  21.0  &  18.7  &  18.9  \\
\text{benzene} &   6.9  &   5.2  &   5.5  \\
\hline
\end{array}
From this we can imply (although there are too few points) that in condensed phase 4-pyridone is the predominant species, while in vacuum it would be 4-pyridinol. Unfortunately, at this time I cannot offer a satisfactory (and easy) explanation for this. The values above have not been calibrated, they are a mere snapshot. I don't want to wander deep into hand-wavy territory and propose a rule of thumb based on bond energies and/or similarities to other molecules.
It is probably true that water is able to stabilise a possible negative charge at the oxygen and at the same time stabilising the positive charge of the proton bonded to nitrogen, but that's about as much of an educated guess I would like to give up.
I want to caution everybody with expressions like the aromatic form when referring to a resonance contributor. Aromaticity is a concept which applies to a (sub-) structure and as such to all resonance contributors. This is why both tautomers are aromatic regardless of the depiction chosen.
In a similar way I want to caution against the argumentation with bond energies. Whenever resonance comes into play, we will have fractional bond orders, which deviate significantly from the conditions under which the tabulated values have been measured (or calculated).
The above notes apply directly to the two reasons your friend offers:

My friend gave two plausible reasons:

$\ce{C=O}$ is more stable than $\ce{C=N}.$

$\ce{-N=}$ will have greater repulsion between the lone-pair and the double bond than the same in $\ce{-O=}.$

There is neither an isolated $\ce{C=O}$, nor a $\ce{C=N}$ bond, as they are both part of the same π-system. The lone pairs cannot (directly) interfere with the π-system as they are orthogonal.
I wish I could offer you a better explanation, but with most chemistry there is nothing easy about simple systems.

Notes:

There is a more extensive theoretical study available, which compares THF, water, and methanol and focuses more on the solvent model. However, obtained values are in a similar region:
Nagy, P. I.; Alagona, G.; Ghio, C. Theoretical Investigation of Tautomeric Equilibria for Isonicotinic Acid, 4-Pyridone, and Acetylacetone in Vacuo and in Solution. J. Chem. Theory Comput. 2007, 3 (4), 1249–1266. DOI: 10.1021/ct6002252.

* For those unfamiliar with the notations: $\Delta E_\mathrm{el}$ refers to the electronic energy in the Born-Oppenheimer approximation; $\Delta E_\mathrm{o}$ refers to the observed energy, i.e. it is the former including the zero-point energy correction; $\Delta G$ refers to the Gibbs energy.


Why does Carbon-13 NMR not have splitting patterns? I read about splitting patterns in proton NMR and found that this is due to spin-spin coupling, where reference to Pascal's triangle can explain the splitting patterns found in doublet, triplet and quartet.

My book mentions this later:

  There are also no splitting patterns caused by the spin-spin coupling, i.e. interaction of neighbouring protons in the 1-H atoms within molecules.

Advanced Chemistry for You, Third Edition (Ryan)

Why do you not find splitting patterns in Carbon-13 NMR?

In 1H NMR, spin-spin coupling leading to multiplet structure most commonly arises due to coupling to other 1H nuclei. This is most often taught in terms of neighbouring 1H nuclei having "up" and "down" spins, both of which lead to the nucleus of interest experiencing a slightly different magnetic field, and hence having a slightly different chemical shift. While this is only a simplification, it is probably sufficient for the purposes of this question.

In 13C NMR, though, proton decoupling is often employed. As the name suggests, this removes all effects of coupling to protons (i.e. 1H). How is this achieved? Essentially, while the signal from the 13C nuclei is being recorded, the NMR spectrometer simultaneously and continually irradiates the 1H nuclei with energy. This means that each 1H nucleus is rapidly undergoing "spin-flips" between up and down spin, and therefore, what the 13C nucleus sees is actually a magnetic field that is very rapidly oscillating between the two possible values. What is actually observed in the NMR spectrum is an average of the two possible values, which is equivalent to what would be seen if there was no coupling to begin with.

Why do we do this, though?

One concern is sensitivity: when peaks are split due to spin-spin coupling, the intensity of the peak is decreased. Since 13C has a naturally very low abundance, it is of interest to prevent the peak intensity from being spread out due to coupling.
Another concern is simplicity: carbon spectra which have C–H coupling can often be extremely difficult to interpret. For an example I refer you to Glenn Facey's blogpost where an inset of a proton-coupled spectrum of toluene is shown.
A subtle point that is often neglected is the fact that the decoupling process also enhances the intensity of 13C peaks via the nuclear Overhauser effect.

All of these factors make proton-decoupled carbon spectra – or in NMR lingo, 13C{1H} spectra – the preferred experiment.

If you're interested in coupling to nuclei other than 1H:

coupling to 13C is all but invisible, considering the low natural abundance of 13C. Recall that in a 13C spectrum you are only observing 1% of all the carbons present. How many of those carbons are coupled to another 13C nucleus? Or in other words, what are the chances of finding two 13C nuclei next to each other? Pretty low is the answer – so much so that an NMR experiment designed to measure C–C couplings was given the acronym INADEQUATE.
coupling to other abundant spin-1/2 nuclei is in fact very often seen. Two simple examples are 19F and 31P, both of which are close to 100% natural abundance – in compounds containing F or P, multiplet structure will be observed in the carbon spectrum. Here's the spectrum of 1-chloro-2-fluoroethane, $\ce{FCH2CH2Cl}$, courtesy of SDBS. The two peaks on the left are from the carbon next to fluorine; coupling to fluorine leads to a 1:1 doublet being observed. Likewise, the two peaks on the right are from the carbon next to chlorine. It is still coupled to fluorine, but because of the greater distance, the coupling constant is smaller.

There is also probably a wealth of examples in organometallic chemistry (for example 103Rh is also spin-1/2), but I don't currently have the energy to find a good example.
coupling to quadrupolar nuclei (nuclei with spin > 1/2) can be seen, depending on the symmetry of the molecule and on some other physical parameters. For example, since deuterium has spin 1, the carbon peak in $\ce{CDCl3}$ is observed as a 1:1:1 triplet (a familiar sight to organic chemists). But this is not always the case: for a more in-depth explanation see, e.g. Symmetry and NMR linewidths of quadrupolar nuclei


Spectrophotometric assay by pNPP Here is a series of related questions that I want to ask.
Background
The activity of acid phosphatase is measured by an enzymatic reaction that converts para-nitrophenyl phosphate (pNPP) to para-nitrophenol (pNP), liberating phosphate. The product, pNP, absorbs light whose wavelength is $\pu{400 nm}$ with an absorption coefficient ($\pu{400 nm}$) of $\pu{19000 M-1 cm-1}$ at extremely alkaline pH. Reaction mixture for an acid phosphatase is slightly acidic. Thus, it must be alkalinized for quantification of pNP.
Two enzyme concentrations are to be examined - They are $\ce{1X}$ and $\ce{0.1X}$ (The $\ce{0.1X}$ enzyme is made by mixing $\pu{1 ml}$ of $\ce{1X}$ enzyme in $\pu{9 ml}$ of $\ce{NaCl}$.) Reaction times for each enzyme is $1$, $10$ and $20$ minutes.
Procedure

Protocol for measurement of acid phosphatase activity:

Mix $\pu{0.12 ml}$ of $\pu{0.5 M}$ $\ce{Na}$ acetate buffer (pH $5.6$) and $\pu{0.24 ml}$ of $\pu{5 mM}$ pNPP in a test tube. Start the reaction by adding $\pu{0.24 ml}$ of an enzyme solution.
After the reaction times of $1$, $10$, and $\pu{20 min}$, respectively, stop the reaction by adding $\pu{0.6 ml}$ of $\pu{0.5 M}$ $\ce{NaOH}$. $\ce{NaOH}$ stops the reaction and converts the pNP produced into a yellow-colored (A400-absorbing) form.
After all reactions are stopped, measure A400 of the samples.

Assay of potato acid phosphatase:
\begin{array}{ll}
\pu{0.5 M} \text{Na acetate buffer (pH 5.6)} & \pu{0.12 ml}\\
\pu{5 mM} \ce{pNPP} & \pu{0.24 ml}\\
\text{Enzyme} & \pu{0.24 ml}\\
\pu{0.m M} \ce{NaOH} & \pu{0.5 ml}\\\hline
\text{Sum} & \pu{1.2 ml}
\end{array}

There are some things I am supposed to calculate after I obtain the results.
First I am going to plot a graph of absorbance versus time and find the slope for each of the lines ($\ce{1X}$ and $\ce{0.1X}$).

I have been asked to find the absorbance change/min/1ml of $\ce{1X}$ enzyme.

My answer: Divide the slope obtained by $0.24$ (the amount of enzyme).

I have to convert the absorbance change to concentration change when
$L$ is $\pu{1 cm}$ and e400 of pNP is $\pu{19000 M-1 cm-1}$.

My answer: This can be found out by $A = eCL$.

The next question is to convert the concentration change to a change in the amount of substance of pNP.

I have no clue how to do this.

Finally, I have to calculate total activity (in moles per minute) in $\pu{4 ml}$ of $\ce{1X}$ enzyme solution.

My answer: Multiply the answer I obtain in the above question by $4$.

For the first subquestion:

Using unit analysis, the answer is very quickly found: The unit of the slope $m$, assuming dimensionless values for the absorbance, is
$$[m] = \frac{[\Delta y]}{[\Delta x]} = \frac{[\Delta A]}{[\Delta t]} = \frac{1}{\text{min}}$$

Now you need a volume in the denominator as well, so dividing by a volume seems like a good idea. Since you need the change in absorption per $\pu{mL}$ of $\ce{1X}$ solution, you divide the slope of the $\ce{1X}$ reaction by $\pu{0.24 mL}$ for the correct value.

For the subquestion with which you seem to have the most trouble: simply apply $c=n/V$.

You will find the amount of substance changed via
$$ \Delta n = \Delta c\times V = \Delta c \times \pu{1.2 mL}$$


Meaning of steady state (kinetics) In kinetics, we use the steady state approximation to relate the concentration of intermediates to the concentration of reactants. For example, given the mechanism
$$\ce{A->[k_1]B}$$
$$\ce{B->[k_2]C}$$
using the steady state approximation for B gives $k_1[A]=k_2[B]$, so $[B]=\frac{k_1}{k_2}[A]$. Since B is at steady state, $[B]$ should be a constant through time. But the equation says that $[B]$ is a function of $[A]$ (and $[A]$ changes with time), so how can $[B]$ be constant? Am I misunderstanding the meaning of steady state here?

Good question.  Let's examine your system in more detail.

$$ \frac{dA}{dt} = - k_1 A $$
$$ \frac{dB}{dt} = k_1 A - k_2 B $$
$$ \frac{dC}{dt} = k_2 B $$

The third equation is redundant because $C$ can always be found from the conservation equation $C = A_0 - A - B$.
This system can be solved exactly without any further assumptions.  So thank you for picking a system that has this nice feature.  An exact solution will let us compare to any approximate solutions obtained via simplifying assumptions.  Assuming that the initial conditions are $A(0) = A_0$ and $B(0) = 0$, then the exact solution is

$$ A(t) = A_0 e^{-k_1 t} $$
$$ B(t) = \frac{k_1}{k_2-k_1} A_0 \left (e^{-k_1 t} - e^{-k_2 t} \right )$$

What about the quasi-steady state approximation (QSSA)?  The QSSA is an assumption that $B$ does not really vary with $t$ as far as $C$ is concerned.  That is, predicting the rate of formation of $C$ can be done by assuming therate of formation of $B$ is zero.  So if $k_1 \gg k_2$, then $A$ is more or less "instantly" and quantitively converted to $B$, after which $B$ converts to $C$ via the usual first order kinetics with rate constant $k_2$. So this case, $k_1 \gg k_2$, is already a clue that the QSSA is invalid for this mechanism.
Why is the QSSA invalid for this case of $k_1 \gg k_2$?  Usually, the QSSA is invoked on catalytic reaction schemes.  In these schemes, it is assumed that the concentration of catalyst is very small relative to the concentration of substrate.  Thus there is an upper bound on the possible concentration of the intermediate, a bimolecular complex of reactant and catalyst: the total amount of catalyst added limits how much bimolecular complex there can be at any time.    This assumption that substrate is in excess relative to catalyst must be satisfied for the QSSA to make sense.
What about the other case?If $k_2 \gg k_1$, then at time scales where $k_2 t \gg 1$, then $B \approx \frac{k_1}{k_2}A_0 e^{-k_1 t}$.  The maximum concentration of $B$  in this case is the ratio of rate constants, a ratio that we've already assumed to be much less than one.  And since the maximum concentration of $C$ is not similarly bounded, this is a case where the QSSA makes sense.  This implies that the conservation equation can be written as $A_0 = A + B + C \approx A + C$.  Proceeding with the QSSA assumption to find the rate of formation of $C$:

$$\frac{dB}{dt} \approx 0$$
$$B = \frac{k_1}{k_2}A$$
$$A_0 = A + B + C \approx A + C$$
$$\frac{dC}{dt} = +k_2 B \approx k_1 A = k_1 (A_0 - C) $$

That last equation is now a single-variable first-order linear ODE and could be solved for $C(t)$ if desired.  But the main thing is that with the QSSA, $B$ was eliminated from the expression for $C$

I can't think of a physical reason why a system where $k_1 = k_2$ exactly couldn't exist.  But someone else will have to come along and explain how to get the math to work out.

If $k_1 = k_2 = k$, then the system of equations is :
$$ \frac{dA}{dt} = - kA$$
$$ \frac{dB}{dt} = k(A - B)$$

The solution is:
$$A = A_0 e^{-k t}$$
$$B = A_0\; k t\; e^{-k t}$$
In this case, there is still no generally valid way to apply the QSSA.  The derivative of $B$ is $B'(t) = k A_0 e^{-k t} \left(1 - kt \right)$ and is equal to zero only when at the precise time where $kt=1$.  It isn't approximately equal to zero in any other time range, except as $t \rightarrow \infty$.  So, the QSSA isn't valid to the case when $k_1 $ is exactly equal to $k_2$, and it isn't valid when $k_1 \neq k_2$ either.
Bottom line: using the QSSA in this problem is equivalent to assuming $k_2 \gg k_1$.  You might not have thought so without going through this whole analysis.  The key fact is that the QSSA requires that variation in the concentration of the intermediate be small relative to the variation in concentration of the products or substrates that you care about.  In mechanisms involving catalysts, this happens any time that catalyst levels are small relative to substrate(s).  However, in other mechanisms, this happens only with arbitrary restrictions on the kinetic parameters that may not be obvious a priori.


Why don't minerals evaporate from seawater along with the water? When water evaporates from the ocean, of course the minerals such as salt remain in the ocean-- that's why seawater is so saline. However, I'm very puzzled: why don't minerals evaporate along with the water? Could you give me a somewhat simple answer for this? (I never studied chemistry, so I really have no clues as to the possible answer.)

In order to get a measure of how much a compound is prone to evaporation one can check boiling points. Boiling point of water is 100 C in atmospheric pressure, for molecular NaCl, sodium sulfate and Magnesium chloride (main minerals in seawater) boiling point is above 1400 C. For ionic salts dissolved in water there is a different story since they are ionized. The ions are surrounded by water molecules in liquid phase, while if vaporized they would not have a chance of having water molecules closeby so they prefer to stay in the liquid phase. All of that being said, one can envision a certain temperature, pressure and concentration (extremely high for three) at which ions would somehow get out to the vapor phase. Since sea level temperature is quite low, mineral or salt evaporation is practically zero.

Side note on why would water evaporate at 4 C instead of 100 C:
Although it take 100 C of kinetic energy (molecule's velocity) to have "all" water molecules to escape from the intermolecular attractive forces, still "some" of them end up having the energy to escape at 4 C (sea temperature average). It is because kinetic energy has distribution over all the molecules, and you will always find some that have the necessary energy. At the same time "some" of the water molecules in the air end up not having enough kinetic energy to remain in the air, and so they end up being attracted back to the liquid. These two numbers (The molecules of water in air that condense back and the water molecules in the ocean that evaporate) are equal so we have thermodynamic equilibrium. The net number of molecules that escape the sea depends on kinetic energy distribution which is exemplified in temperature.


What does it mean in a Ni(II)complex if the HNMR spectrum appears unbroadened and unshifted & what does this tell you about the coordination geometry? I am trying to work out the an unknown Ni(II) complex. I don't known the geometry and I have read that broad lines on the nmr spectrum indicate large compounds with symmetric environments. Does this mean that the complex would have tetrahedral geometry as square planner would be symmetrical and octahedral too large?

If the proton spectrum is unbroadened and not shifted out of the ~0-10ppm range, then I would have to say it is a diamagnetic compound.

It could be low spin square planar.  Tetrahedral and octahedral d8 cannot be diamagnetic.

Image source: http://upload.wikimedia.org/wikipedia/commons/a/a6/Chem507f09sqvstet2.png

Could also be two nickel atoms in the complex that are antiferromagnetically
coupled for a net spin of zero.


Attack at ketone vs carboxylic acid carbonyls Why in this the nonbonding electrons of N group does not attack C2 as I think it is more electropositive ?

Then how pyruvic acid is converted to alanine.

The keto group is more reactive compared to the ester group due to the mesomeric effect (+M effect) caused by the hydroxyl group in the ester functionality, as already pointed out by orthocresol. See also  https://en.wikipedia.org/wiki/Carbonyl_reduction. Generally:
$$
 \mathrm{acylchloride} > \mathrm{anhydride} > \mathrm{aldehyde} > \mathrm{ketone} > \mathrm{ester} \approx \mathrm{carboxyl} > \mathrm{amide}
$$

A very detailed discussion on the subject is given for example in Warren Organic Synthesis: Strategy and Control or in Francis A. Carey Advanced Organic Chemistry - Part A.


Using UV-C (254nm) light to cleave the amino acid off of vyvanse Vyvanse:

Desired product:

Amide bonds absorb in the region of about 190 to 240 nm, so could using UV-C light conceivably cleave the amide bond?

Also I don't think this reaction is as simple as putting vyvanase powder under a UV-C lamp because the reaction intermediates probably need to be stabilized by a proper solvent or something, correct? What would the intermediates likely be, and what solvent would likely help?

Obviously the easiest way to get the desired product is probably simple hydrolysis, as there are not really any other hydrolyzable function groups on the starting material apart from the amide C-N bond. So this is mainly a theoretical question.

Honestly, I wouldn't bet on a cleavage here.

In principle, excited state amides and imides can undergo electron transfer reactions. We did work a bit on that (DOI) and an early example on intramolecular electron transfer between an aromatic amide and a tertiary amine, resulting in amide cleavage, was given by J.D. Coyle (PDF).

In your case, however, the situation is a bit different. You don't have an aromatic amide but an arene and an aliphatic amide. The latter aren't supposed to exhibit a strong absoption at 254 nm, see an older article by Nielsen and Schellman (DOI).

Even if you'd manage to excite the amide, I'd exclude electron transfer from one of the primary amines in your starting material and hence the cleavage of the amide.


Deciding which group twists out of the plane when ortho effect takes place I'm going to take a few stills from 3D conformers of the following compounds available on PubChem in which ortho effect is applicable to show what I'm referring to.
2-Nitrobenzoic acid:

N,N-Dimethyl-2-nitroaniline:

2'-Aminoacetophenone:

In the compounds above due to steric repulsions between the groups present, one group needs to be twisted out of the plane of the benzene ring. My question is, how do we theoretically explain which group shifts out of the plane, and which doesn't?
To my understanding, two things would effect this; one would be the extra stabilisation/destabilisation of a group due to not being in resonance any longer once it moves out of the plane of the benzene ring, and the second would be the "bulkiness" of the group itself, which would energetically be measured as the group's A-Value.
For the above three compounds in the same order, from the perspective of A-Values (the text in brackets mentions what actually happens):

($\ce{COOH}$ twists out of the plane) The A-Values are $\ce{COOH} = 1.2, \ce{NO2} = 1.0$, so it makes sense for $\ce{COOH}$ to stay in the plane of the ring.

($\ce{NMe2}$ twists out of the plane) A-Values $\ce{NMe2} = 2.1, \ce{NO2} = 1.0$ so it makes sense for $\ce{NMe2}$ to stay in the plane.

($\ce{COMe}$ slightly twists out of the plane) A-Values $\ce{NH2} = 1.6, \ce{COMe} = 1.17$ so it makes sense for $\ce{NH2}$ to stay in the plane.

(data from here and here)
I understand that A-Values are not the only thing which we need to consider, since resonance plays a part (maybe even other stuff?), but I haven't been able to figure out convincing reasoning as to why in the above examples the group which shifts out of the plane does so. Does the volume occupied by the groups play a crucial part as well, since so far I am yet to come across a compound in which the smaller group (by volume) shifts out of the plane? Are A-Values even relevant to such a scenario?
(I'm looking for a list of factors which need to be considered, whether this can even be explained/predicted theoretically, and/or explanations for the any of the above compounds. I understand that this could be slightly lengthy, so I'd be more than happy for answers which address any one of these, but I feel talking about the three compounds was necessary to explain the question properly, and hence the somewhat broad question.)

I am not sure the 3D conformation stills as reported on PubChem are all that accurate. Although crystal structures are sometimes referenced for specific molecules, some of the displays are in fact models, some of which do not make sense to me.
Here are some observations regarding the three molecules you noted.
2-Nitrobenzoic acid
The PubChem model shows both oxygen atoms of the nitro- group to be precisely coplanar with the phenyl ring. In the crystal structure, these oxygen atoms have a dihedral angle of $125.5^\circ$ and $129.1^\circ$ with respect to the aryl, i.e. away from coplanarity.
The PubChem model shows both oxygen atoms of the carboxylic acid to be precisely normal to the phenyl ring. In the crystal structure, these oxygen atoms have a dihedral angle of $152.9^\circ$ and $162.7^\circ$ with respect to the aryl, i.e. closer to coplanarity.

Pubchem
Crystal structure entry CCSJWHN

N,N-Dimethyl-2-nitroaniline
I could not find the crystal structure of this molecule. However, I’m not sure it makes sense that the N,N-dimethyl amine adopt an $\ce{sp^2}$ character at the nitrogen center, at least as far as the Pubchem model representation is concerned. In such a pose, there can’t be extended resonance from the phenyl to amine as obtained with unsubstituted aniline. Either the methyls are near coplanar with the aryl to allow the pi orbitals to enter resonance or the amine is in $\ce{sp^3}$.
2’-Aminoacetophenone
I could not find the crystal structure of this molecule. What I found was the 4,5-dimethyl variant as 1-(2-amino-4,5-dimethylphenyl)ethan-1-one.
The 2’-Aminoacetophenone PubChem model shows the acetone oxygen to be precisely $30.0^\circ$ out of the aniline plane. In contrast, the crystal structure of 1-(2-amino-4,5-dimethylphenyl)ethan-1-one indicates that oxygen to be precisely in plane with the aniline. This coplanarity may not be surprising in this case as there is an intramolecular hydrogen bond between the two substituents.

Pubchem
Crystal structure entry 1825318 (BEXTUW)

2'-Aminoacetophenone
1-(2-amino-4,5-dimethylphenyl)ethan-1-one

Original Post:

To my understanding, two things would effect this; one would be the extra stabilisation/destabilisation of a group due to not being in resonance any longer once it moves out of the plane of the benzene ring, and the second would be the "bulkiness" of the group itself, which would energetically be measured as the group's A-Value.

I think steric hindrance, you noted as “bulkiness” and estimated with an A-Values is the primary factor. There is little room for compromise on this one as two atoms can’t occupy the same space. The second factor is resonance as you noted. Another influencing factor is the presence of intramolecular interactions, such as a hydrogen bond. Solvation and pH are other influential factors to the ortho effect.


How can I determine the size of tropolone molecule? How can I determine the size of a tropolone molecule?
It's a seven-membered aromatic ring with =O on the first carbon and -OH on the second carbon.
Is there any software available?

You wanted to measure the distance "across tropolone" but you didn't specify from where to where. The distance from a remote hydrogen to each of the oxygens and to the hydroxyl hydrogen are shown below. You can do the measurements you want yourself here: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ESPotential/ESPmisc2.html


What is the correlation between electronegativity and the sharing of electrons? For example in a C-O bond, the electronegativity difference is 1.2. does this mean that the Oxygen has 1.2 of the shared electrons and Carbon 0.8 (or some other numbers)? Is there a specific equation that can give us how the bonding electron pair is shared?

Is there a specific equation that can give us how the bonding electron pair is shared?

No. The electronegativity scale is based on bond strength, not on partial charges. There are quantum-chemical methods to calculate the electron density in a molecule, and you can derive partial charges from that.

  For example in a C-O bond, the electronegativity difference is 1.2. does this mean that the Oxygen has 1.2 of the shared electrons and Carbon 0.8 (or some other numbers)?

No, the electronegativity difference does not allow you to make a quantitative statement like that. It is useful, however, to predict how a bond would hydrolyze. It also gives you an estimate about how polar a bond is.


Is it true that "Voigt Circuit" would be able to be fitted to various electrochemical impedance spectra? "Voigt Circuit" which is a series combination of Voigt elements (i.e. parallel RC's), is commonly used for Kramers-Kronig relations checking; and as I imagine it is generally believed to be able to be fitted to various electrochemical impedance spectra?
Is this belief true or not?
And if the answer is "Yes", is there any theoretical basis for that or it is only a practical sense (I mean that it is approved according to fitting results of experimental data to Voigt circuit)?

It is known that an electrochemical impedance spectra can be modelized with series-parallel combination of resistors and capacitors elements. What is more, for a given impedance spectra they are an infinity of different combinations of R-C elements which accurately model the same complex impedance.

Similar observations are commonly done in the field of imperfect dielectrics (non-Debye) for the complex permittivity spectra.

For examples: pages 10-11 , page 16 and pp.27-32 in the paper  https://fr.scribd.com/doc/71923015/The-Phasance-Concept . Each example of complex impedance is modeled with several different networks of R-C elements.

On theoretical viewpoint, all this is related to the energy efficiency of the phenomena in the experimental system, that is the presence of dissipative components (R) and storage components (C) distributed on various manner.  Even if the energy storage and energy dissipation are respectively on different physical nature in different experimental devices, possibly in some cases the complex impedance is likely to be similar due to a similar global energy efficiency behaviour. So, an infinity of different R-C models are likely to be derived.

This can be observed not only in experiments but also in purely theoretical studies for very simplified geometrical configuration : https://fr.scribd.com/doc/23155389/Theoretical-Impedance-of-Capacitive-Electrodes

In addition. To go beyond the question raised, the following quotation is extract from the first paper quoted above :

It has long been recognized that many impedance loci result in well-defined semi-circles with center depressed below the real axis. Linear loci are also commonly observed associated to rough electrode surface, porous electrode, diffusion, etc. "This features mean only that the system includes some part(s) transferring energy with constant efficiency in a range of frequencies. Any affirmation of the nature and mechanisms of the phenomena involved would be pure hypothesis if only based on impedance analysis."


Which reagent can I use to remove carbon from oven? I have a practical question: today I have to clean the oven, and it doesn't have self cleaning burn down mode, nor an official scraper I can use...

It's full of charred grease and sugar. I managed to remove 98% of the charred grease with cold grease remover. But it seems that reagent has no effect on the remaining 2%, that's black fully charred carbon or sugar. I also tried vinegar, but it doesn't have any effect on it either.

I don't want to scrape it because I fear that I would scratch the oven.

So which reagent can I use to remove those pesky carbon deposits without damaging the oven, and what will be the chemical reaction?

As a disclaimer, this is for information only.  Anyone using this information assumes all associated risks.  The manufacturer's label should be followed for any cleaning products.

The exact chemical composition of the goo is not precisely known. It is likely either tar or creosote existing as some form of hydrocarbons.  These hydrocarbons are tangled up to the point it is extremely difficult to dissolve.  The temperature required to burn this material is likely too high for anything short of a ultra high temperature industrial furnace.

A good way to clean it is to use oven cleaners which take advantage of the saponification reaction. Saponification will loosen the material; however, some scrubbing and/or scraping will still be involved.

For safety when oven cleaning be sure to have good ventilation.  Wear gloves, eye and face protection, as well as not having skin exposed. Caustic cleaning compounds and vapors from the cleaning are hazardous.

Most oven cleaners have multiple components.

A caustic solution causes a saponification reaction.  The saponification reaction is commonly used to make soap from saponifiable oils.  Because the oils in the material to be cleaned are likely saponifiable,  the saponification reaction makes the material much more likely to dissolve in water.
A foaming component creates a foam. When foam bubbles pop, the popping action works to break up the tangled bonds.
A compound may be included to ensure the entire solution will stick to a non-horizontal surface.
Another compound may be included to seal the solution so as to prevent oxidation.  Oxidation may interfere with the saponification.

The trick to getting oven cleaners to be effective, in my experience, is in the timing.  Timing is important because the contents of tars are not exactly known, so the softening reaction is brief.

A warm surface is easier to clean; however, a hot surface will dry the cleaner before it has a chance to work.
The oven should not be powered when attempting any cleaning.
When any oven cleaner is first applied, it is important to leave the cleaner alone and give it time to work. Ten minutes is usually sufficient.
Once dissolution occurs there is a brief window when the tar or creosote can be easily removed with a nylon brush or sometimes even a paper towel.
When the materials dissolved, vapors which were previously trapped in the tars may be released.  These released vapors are not good to breathe.
Sometimes it is necessary to get deeper layers of tar.  The cleaner can be reapplied up to three times before the solution dries if necessary.  Each time allow it to go undisturbed for ten minutes.  (In my experience, by the fourth application, cleaning the tar is simply not worth the effort.)
If too much time passes and the solution dries, then the tar is more difficult to clean.  The caustic solution can also damage metal components.
Rinse the oven of all cleaner when done.  Caustic residues can damage the oven if left on the surface.


Which substance(s) can solve gelatin (on film stock)? I want to loosen the exposed layer on film stock which is made up of gelatin. Anything which would make gelatin dissolve that's not overly toxic? Thanks

Hot water, about 100 C, should do the trick. Also, this article has some information on the properties of gelatin layers in photographic films.

Properties and Stability of Gelatin Layers in Photographic Materials


Mathematical equivalence of ionic-bond resonance structures and d-orbital participation in hypervalent molecules It is now routine in the scientific community that "main-group hypervalent molecules, under certain conditions, tend to be better represented by resonance structures involving ionic bonds, rather than the utilisation of d-orbitals". For example, PF5 is better represented by a total of 5 "[PF4]+ [F]-" resonance structures, than sp3d hybritals on the phosphorus atom.
However, after reading the Wikipedia article for  "molecular orbital"s carefully, I began to think that they are, strictly speaking, mathematically the same-

The article said that "When the energy difference between the atomic orbitals of two atoms is quite large, one atom's orbitals contribute almost entirely to the bonding orbitals, and the other atom's orbitals contribute almost entirely to the antibonding orbitals. Thus, the situation is effectively that one or more electrons have been transferred from one atom to the other. This is called an (mostly) ionic bond.": maybe the ionic nature of a molecular orbital formed by a 3d orbital of P and a 2p orbital on F can be considered a special case of this statement, as P's 3d orbitals >>>>> F's 2p orbitals in terms of energy levels.

The question is- is my guess true?

You are essentially correct, but the distinction is the degree of bonding.

As you described, in a polarized or partially ionic bond, the atomic orbital of the more electronegative atom is lower in energy. Thus, the bonding molecular orbital is closer in energy to that atomic orbital, and the antibonding molecular orbital is closer in energy to the atomic orbital of the less electronegative atom. Because of this, the relative contribution of the lower energy AO to the bonding orbital is greater, and the contribution from the higher energy AO correspondingly lesser.

The distinction between "covalent" and "ionic" is therefore blurry. There's a continuum of degree of ionic bonding. In the extreme case of a completely ionic bond, there is no stabilization of the "bonding" MO relative to the lower energy AO. Furthermore, the higher energy AO no longer contributes at all. At that point, we no longer have bonding and antibonding MO's, but rather both are nonbonding orbitals comprised entirely of a single AO each.

But there are plenty of cases where there is just a little bit of stabilization, and we still describe the bond as "ionic". That's partly because it's something of a judgement call at what point we conclude that there is no bonding interaction. We can't solve the Schrodinger equation exactly, so we use computational approximations. Our computational methods aren't precise enough to tell the difference between a tiny bit and none at all. For the $d$ orbitals in the compounds in question, most methods predict at least a small participation of the $d$ orbitals in most compounds. But how much above 0 does it have to be before it's no longer correct to say that $d$ orbitals are not involved? We can say with certainty that the participation of the $d$ orbitals is nowhere near the $sp^3d^2$ view that used to be taught.

In your question, you described the compounds as "better represented" by ionic bonding "rather than utilization of $d$ orbitals."

You're correct that this distinction isn't either/or, but the key words are "better represented". Even if the $d$ orbitals contributed as much as 5% to the bonding MO, isn't the ionic representation "better"?


Is this Fischer Projection of (3R, 4S)-3-ethyl-3,4-dimethyl-1-hexene wrong? (3R, 4S)-3-ethyl-3,4-dimethyl-1-hexene

Draw the Fischer Projection of the compound above.

Here is the answer:

Is there a mistake at the upper stereocenter? I think so.

Is there a mistake at the upper stereocenter?

Yes you are thinking correct, the R configuration will be the diastereomeric pair of this compund, the second chiral carbon is in correct configuration. Don't you want to ask why?


What is the IUPAC nomenclature of alkane with large numbers of C atoms (more than 200)? What is the IUPAC nomenclature for alkanes with large numbers of $\ce{C}$ atoms (more than 200)? For example, what is the IUPAC accepted name for $\ce{C_{205}H_{412}}$?

The definitive answer is, of course, in the rules established by the IUPAC’s “Commission on nomenclature in organic chemistry”. The reference you are looking for is:

  “Extension of Rules A-1.1 and A-2.5 Concerning Numerical Terms Used in Organic Chemical Nomenclature”, Pure Appl. Chem., 1986, 58, 1693-1696

which can be found here as a PDF and here as an HTML version. In particular:

  NT-1.1 - The fundamental numerical terms for use in hydrocarbon names or as multiplying prefixes for simple features are given in the following list:

  \begin{array}{|c:c|c:c|c:c|c:c|}\hline
    1&\rm \text{mono- or hen-} &10 &\rm \text{deca-}&100&\rm \text{hecta-}& 1000&\rm \text{kilia-} \\\hline
    2&\rm \text{di- or do-}&20&\rm \text{icosa-}&200&\rm \text{dicta-}&2000&\rm \text{dilia-}\\\hline
    3&\rm \text{tri-}& 30&\rm \text{triaconta-}&300&\rm \text{tricta-}&3000&\rm \text{trilia-} \\\hline
    4&\rm \text{tetra-}&40&\rm \text{tetraconta-}&400&\rm \text{tetracta-}&4000&\rm \text{tetralia-} \\\hline
    5&\rm \text{penta-}&50&\rm \text{pentaconta-}&500&\rm \text{pentacta-}&5000&\rm \text{pentalia-}\\\hline
    6&\rm \text{hexa-}&60&\rm \text{hexaconta-}&600&\rm \text{hexacta-}&6000&\rm \text{hexalia-}\\\hline
    7&\rm \text{hepta-}&70&\rm \text{heptaconta-}&700&\rm \text{heptacta-}&7000&\rm \text{heptalia-}\\\hline
    8&\rm \text{octa-}&80&\rm \text{octaconta-}&800&\rm \text{ octacta-}&8000&\rm \text{octalia-}\\\hline
    9&\rm \text{nona-}&90&\rm \text{nonaconta-}&900&\rm \text{nonacta-}&9000&\rm \text{nonalia-}\\\hline
\end{array}

So, $\ce{C_205H_412}$ is the molecular formula of pentadictane. $\ce{C_7547H_15096}$ is heptatetracontapentactaheptaliane.


Forms of carbon in incomplete combustion When hydrocarbons burn with little oxygen, carbon is produced. What form do these particulates of carbon come in?

Is it a simple molecule like $\ce{S8}$, or a small fragment of a larger structure, like graphite?

Why is this form more stable than others?

It heavily depends on reaction parameters (oxygen/oxidants availability, temperature, pressure, etc.)

In a free-air incomplete combustion at a (relatively) high temperature, you get mainly carbon-based molecules of various kind.

In a internal combustion engine, that usually works at high temperature and high pressure, you can also get nanoparticles of various kind, including graphite.

In a free-air, mixed, relatively cold combustion, the original small molecules, graphite and other carbon-bases compounds can further react with other molecules giving you larger molecules and/or can stick to larger particles forming nano or micro-particles of various kind. Predicting/modelling this kind of reactions can be very complicated.


what is mechanism for reaction phenol + dimethylsulphate -> anisole Dimethyl sulphate breaks the oh bond and forms anisole. How does the reaction proceed?? is it oxygen lone pair attacks on sulphur which has partial positive charge?? the reaction takes place in presence of naoh

The methylation of Phenol using Dimethyl Sulfate is an Electrophilic Methylation and happens through a SN2 nucleophilic attack. Phenol is first deprotonated by the sodium hydroxide as you already thought. The deprotonation of Phenol is favored because the pKa of Phenol is lower than the pKa of water meaning $\ce{OH-}$ is a stronger conjugated base than $\ce{PhO-}$:

The formed Phenolate is a good nucleophile, and will attack the methyl groups of the Dimethyl Sulfate. Those methyl groups are readily attacked by nucleophiles because of the great induction effect caused by the neighbor atoms which are much more electronegative than the carbon atom, which makes those more positively charged. This makes the methyl groups good electrophiles, hence the reaction name "Electrophilic Methylation": If we focus on the methylating reactant, it's performing an Electrophilic Attack.

Below is the reaction mechanism:

EDIT: Thanks @user55119 for pointing out that only one methyl group of the Dimethyl Sulfate reacts under normal circumstances, requiring 1 equivalent of Dimethyl Sulfate per Hydroxyl Group and not 1:2 as I previously thought


Confusion in the line written in my book In my class 11 chemistry NCERT book ,there is a line in chapter P-block.

The line is

  The oxidation state two unit less than the group oxidaton state become progressively more stable for heavier elements in each group . Due to 'inert pair effect'

I could not understand this , what they meant to conclude from it.

Can anybody explain me with an example.

Ok let me explain you expecting you are from a cbse affiliated school.

Group oxidation state = (group number - 10)

This will give you the group oxidation state of that group. Group oxidation state is the maximum oxidation state shown by the elements of the group. For example in 15th group, the group oxidation state = (15-10) =5
That means the maximum oxidation state which the group 15 elements would show is 5.
Generally moving down any p-bock group, the tendency of occurence of the maximum oxidation state (i.e. +5 for group 15 elements, +3 for group 13 elements) decreases. Rather the most occuring oxidation state for heavier elements is the oxidation state 2 units less that the maximum/group oxidation state. The reason behind this is intervening of d and f orbitals for heavier elements i.e. in one word we say the inert pair effect.

Now what is inert pair effect? As we move down the group, gradually for heavier elements d and f orbitals start filling up. The shielding by electrons in d and f orbitals is very poor due to which moving down the group the nuclear charge increases, but there is not much increase in shielding due to which effective nuclear charge becomes effective and hence its not possible for the heavier elements to attain the maximum oxidation state because to attain the maximum oxidation state, they have to loose the more number of electrons. But that is not possible for heavier elements because the nuclear charge is now effective. It wont allow loosing much of the electrons. Hence the maximum oxidation state/ group oxidation state is not achievable for the heavier elements.

What they can achieve is two units less than that maximum oxidation state and that also makes sense. For example in case of group 15 elements, the maximum oxidation/ group oxidation state i.e +5 becomes unstable for heavier elements but +3 state (i.e. two units less than the maximum oxidation state/ group oxidation state) becomes increasingly stable for the heavier elements. Now you might think why only two units less than the maximum oxidation state stable for the heavier elements? Why not 3 units less? Because this is a fact i.e. because we have found the more observing/occuring oxidation state for p block elements is two units less than their corresponding group oxidation state. Now the statement was written because it was observed.

Negative oxidation state for heavier elements is also not achievable because down the group the shell size is large. Though I said effective nuclear charge is bit effective for heavier elements, but that doesn't mean it is stronger enough to gain an electron. It is just strong to get hold of some valence electrons i.e. not allowing loosing much of the electrons. But that strength is not that strong to gain an electron. Hence negative oxidation is not shown for heavier elements.

So basically stable oxidation state for heavier elements is 2 units less that the maximum oxidation state shown by that corresponding group.


Have "molecular clusters" for azeotropes been identified? A different question about azeotropes got me thinking about this point again. Azeotropes have a very specific composition so it seems that the azeotrope ought to have some sort of physical structure. It seems to be a "molecular cluster" of some sort.

The azetrope for water and ethanol is about 95.5% ethanol by weight. A little fiddling and it seems that the ratio is 8 ethanol molecules to 1 water molecule. Is there some particular physical configuration of molecules to which this would correspond?

No, in the absence of extra data, there is no reason to suppose that there is any vapor-phase cluster formation.

Cluster formation in the gas phase would demand very, very strong departures from ideal-gas behavior.  To the contrary, the ideal gas law is an excellent descriptor of gas phase mixtures of ethanol and water.

Check out a Wolfram Demonstration for the ethanol-water system.  It says:

  You can vary the pressure $P$ to any value between 50 kPa and 200 kPa  (i.e., low to moderate pressure so that the ideal gas-phase assumption holds).

If the ideal-gas assumption holds, then there is no significant structure formation in the vapor phase.  The "ideal" gas law describes negligibly small particles that have no attraction or repulsion to each other.  Structure formation means that molecules must be strongly attracted to each other in order for arrangement into a persistent structure to occur.

An "extended" form of Raoult's law that is valid for non-ideal vapor as well as non-ideal liquids, and thus is applicable to azeotropes, is

$y_i \phi_iP = x_i \gamma_i p_{i, \mathrm{sat}^{\star}}$

Here, $\phi_i$ is the fugacity coefficient and takes into account vapor-phase non-idealities (i.e. deviations from the ideal gas law), and $\gamma_i$ is an activity coefficient and  takes into account liquid-phase non-idealities.

For many, many systems of interest, $\gamma_i$ is the driver of non-ideality, including azeotropic behavior.  Fugacity coefficients $\phi_i$ are negligible (except at enormous pressures) a much higher percentage of the time than activity coefficients $\gamma_i$.  This is because liquid phases are often far more dense than vapor phases, meaning that intermolecular forces govern behavior to a much stronger degree than in vapors.


How would you confirm that a colorless liquid given to you is pure water? How would you confirm that a colorless liquid given to you is pure water?
A very easy solution is to take a sample of colourless liquid and put on stove if it starts boiling exactly at 100 ºC then it is pure water. Any other colourless liquid such as vinegar always have different boiling point. Also observe carefully that after some time whole liquid will convert into vapour without leaving any residue.
Now are there any other ways to understand?

Measurement of properties like freezing point and boiling point would give you a quick, qualitative though likely correct, answer.  The simplest to do is the measurement of the liquid's freezing point.

If you can, partially freeze your solution in a kitchen freezer, stir well, then measure the temperature while stirring.  Since the freezing point of water is relatively insensitive to atmospheric pressure, you don't have to worry about correcting for the elevation where you live.  This method is commonly used as a quick way to check the calibration of a thermometer.  Of course a measurement of ${0 ^oC}$ would mean that you likely have water.  If it doesn't freeze, assuming your freezer is cold enough to freeze water, then it's not water.  If this is what you observe, drop a small ice crystal or a grain of salt into the solution to initiate ice formation just to be sure it is not supercooled.  Also, if it doesn't freeze, measure the temperature of the solution to be sure it is at least several degrees below ${0 ^oC}$.

A shortcut, or as a further check, if you have plenty solution to spare you could take a portion and add enough crushed ice that it doesn't all melt upon stirring.  If you get the same value (${0 ^oC}$) as in the first experiment, you can be pretty confident that your solution is water.  If it were some mixture of solutions with a freezing point of ${0 ^oC}$ (unlikely), this test would give you a different measurement than in the first test due to either colligative properties if the amount of ice is small relative to the whole solution or by creating a new, bulk solution with a different melting point than the original solution.

If both of these experiments give you the same result of ${0 ^oC}$, I'll bet my house that it's water.


Why isn't the scandium(2+) cation isoelectronic with potassium? I saw that I got this question wrong on a test, and was very confused.  Why aren't they isoelectronic?  $\ce{Sc^2+}$ with atomic number 21 should have 19 electrons since 21 - 2 = 19, which is the same as the number of electrons as Potassium with 19 protons and electrons.  Does it have something to do with being in different sublevels, one being s and the other d and all?

The neutral $\ce{Sc}$ atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1 4s^2}$, and the $\ce{Sc^2+}$ ion has electron configuration $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 3d^1}$.
The neutral potassium atom has the electron configuration of $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^1}$. While both $\ce{Sc^2+}$ and $\ce{K}$ have the same number of electrons, their electron configurations are different and therefore they are not isoelectronic to one another. (Isoelectronic implies same number of electrons as well as the same structure)


Ignoring the difference of concentrations of reactants. Equilibrium Recently at school, while studying equilibrium we learned that in problems where the initial concentrations of reactants and the equilibrium constant are given, there is a special formula to determine wether the change in reactants can be ignored where C is the concentration of the reactant with the smallest concentration and K is the equilibrium constant: L=C/K. We learned that if L is bigger than 500, we can consider the change in the reactants negligible. However this is where my problem lies: i do not see why the C is included, doesn't the size of K by itself tell us wether a little or a lot of reactants is transformed? Not to mention, some cases turn out flat out wrong. For example:
A+B=(reversible) 2C.

Initial concentrations: A=10^-20M
B=10^-20M.  C=0 M

K=10^-8

By the equation we learned, L=10^-12 and thus the variation in the reactants should not be neglected.
However, after doing the math, x~10^-24 which is 4 orders of magnitude below the beginning concentration of the reactants and completely negligible, is something wrong with my logic? Or did i discover a flaw in my textbook :)

Thanks

Let's look at your example and set up the ICE (intial-change-equilibrium) problem. Then I'll give you more inclusive of thumb.

Let's say reaction equation is

$\ce{A + B <=> 2C}$$

The law of mass action is:

$$K = \dfrac{[\ce{C}]^2}{[\ce{A}][\ce{B}]}=10^{-8}$$

Let's set up the ICE matrix. The change in concentration $x$, will be subtracted from $[\ce{A}]$ and $[\ce{B}]$ and added to $[\ce{C}]$.

$$\begin{array}{|c|c|c|c|}\hline
\ & [\ce{A}]\ (\mathrm{M}) & [\ce{B}]\ (\mathrm{M}) & [\ce{C}]\ (\mathrm{M}) \\ \hline
I & 10^{-20} & 10^{-20} & 0 \\
C & -x & -x & +2x \\
E & 10^{-20} - x & 10^{-20} -x & 2x \\ \hline
\end{array}$$

Now we pug in to the law of mass action:

$$10^{-8} = \dfrac{x^2}{\left(10^{-20} -x \right)^2}$$

Since all numbers have to be positive, let's take the square root of both sides:

$$10^{-4} = \dfrac{x}{10^{-20} -x }$$

Now we have a ratio that says that $x$ is four orders of magnitude smaller than $[\ce{a}]_i -x$. I could have easily used the initial concentration of $\ce{B}$ Put another way:

$$x \approx  K\cdot [\ce{A}]_i$$

If this value is much less than $[\ce{A}]_i$, then we can simplify.


Titration of NaOH with acetic acid During a titration lab where 0.1 N of NaOH is added to 20 mL of acetic acid with the same concentration, the equivalence point occurred after adding about 70ish milliliters of NaOH (different peers acquired different values, ranging from 65 mL and 75 mL of NaOH). This completely puzzled me, because after calculating a theoretical titration curve, my equivalence point occurred after adding about 20 mL of NaOH.

My teacher could not explain why this happened, except that acetic acid can act as a buffer – but how do you calculate this and incorporate it into a theoretical titration curve? I just don't understand how 20 mL of acetic acid would require 70 mL of NaOH to neutralise it (both have the same concentration).

It can't happen.  Neutrality in aqueous solution means equal quantities H+ and OH-; since your solution is unlikely to have a net electric charge, so too must your anion and cation quantities be equal.  Buffering from acetic acid being a weak acid is meaningless if you don't have other anions around to make up that balance.

Solution is electrically neutral:
[H+] + [Na+] = [OH-] + [AcO-]

Solution is pH neutral:
[H+] = [OH-]

Substitute the latter in the former, subtract, you can only get:
[Na+] = [AcO-]

If the concentrations of both solutions were the same, and you're titrating by adding both to the same volume, you can only have equal concentrations if you add equal volumes.

If this was not the case (as you describe) we have to violate one of our assumptions:

Solution actually had a net electric charge (you're doing the titration in space and you're bombarded by beta radiation, i.e. unlikely)
Solution was not truly neutral.  Maybe someone calibrated your pH meters wrong, maybe everyone titrated with a faucet while texting and over-shot by a factor of 3.
Contamination with a strong acid added a variable we don't know about to the ion balance.
Most likely, the solutions were not truly the same concentration.  See if your teacher made her NaOH solution from solid NaOH that she let sit around in open air.  It can pick up a lot of moisture from the air, i.e. it's hygroscopic, and before you know it the weight you're measuring on a balance is half water.


Chirality and number of sterioisomers I have recently read that the number of sterioisomers in a compound having n chiral centres is 2^n. (That is, if no meso form is present)

The below question asks for the number of sterioisomers in the product obtained. And the product obtained has 3chiral centres, so it should be 2^3=8, but the answer given is two.
There is no plane of symmetry.

Do I not consider the carbon attached to the benzene ring when looking for chirality?

Could someone help me out?
Question attached.
!

I believe you are misreading the question. Rather than asking for the number of stereoisomers the product can have (this is where 2^n comes in), it is asking for the number of stereoisomeric products formed. Note that that the two products are diastereomers, which is a type of stereoisomeric relationship. It's not looking for any fancy calculation; just for the related products.

In a Gringard mechanims, the Gringard reagent can attack the carbonyl on either the re or si face, which in turn, can yield an R chiral center or an S center at the location of attack. This lends to the diastereotpic products.


What happens if you autoclave HEPES? Many molecular biology protocols advice against sterilizing HEPES (4-(2-hydroxyethyl)-1-piperazineethanesulfonic acid) solutions by autoclave.

Supposedly, HEPES is heat labile, and the high temperatures used for sterilization somehow impair the function of the buffer, or perhaps generate unwanted chemical species (this is speculation on my part; I've often seen the advice to not autoclave HEPES, but never seen anything about why not). Typically autoclaves heat solutions to 121C and increase pressure to slightly above 2 atm, which is about a degree short of the boiling point of water. This environment is maintained for 20 minutes in order to destroy biological contaminants.

I have seen this claim often, but had trouble locating an actual source or reasoning for why HEPES should not be autoclaved. Has there been a publication that actually tried to heat a HEPES buffer to autoclave temperatures (121C) and see what happens? What exactly is the consequence, especially with regard to the biological uses of HEPES (ie. pH buffer)?

From here:

  Some buffers (e.g., MOPS and HEPES) cannot be autoclaved because they degrade upon heating.

and from here:

  [HEPES] Solutions may be autoclaved under standard conditions.

I guess nobody knows the answer.


How do you calculate enthalpy for a reaction at non-standard conditions? I understand how to calculate the change in enthalpy and entropy for a reaction in standard conditions, but is there a way to calculate these values at non-standard conditions?

Also, is there any way that a reaction with positive change in gibbs free energy will occur?

Do you know how to calculate the change in enthalpy and entropy for a substance as you change the conditions (T, P, ...) in a thermodynamically reversible fashion? Now take the reactants at "standard" conditions, transform them to reactant(s) at the non-standard conditions, react them, and take the product(s) back to the standard condition.  Compare with the reaction occurring at standard conditions. Figure out the enthalpy and entropy differences between reacting at 'standard' and 'non-standard' conditions.

As for your second question, ponder why some reactions do not go to completion.


Coordination dilemma, change of energy related to ligand distance In octahedral field, the baricenter remains conserved and there is lowering of energy of $\mathrm{t_{2g}}$ orbitals.

What is a baricenter and why must it be conserved?
Why is there lowering of energy of $\mathrm{t_{2g}}$ as ligands approach?

Also when ligands are far there an increase in energy but not when they are nearer. What is going on?

I know energy must be conserved but they are many ways to release/utilise energy like sound, heat, light, kinetic energy.

The baricentre should probably be spelt barycentre. It is something analogous to a centre of mass. Simple models such as some flavours of the crystal field theory will assume a destabilisation only of the metal orbitals. However, for an energetic destabilisation to occur, something else must be stabilised — this is the conservation of energy. Thus, the logic says that the raise in energy of the $\mathrm{e_g}$ orbitals must be offset by a lowering of the $\mathrm{t_{2g}}$ orbitals.

When constructing a proper MO scheme of an octahedral coordination complex, this argument is revealed as bogus. As can be seen in this answer, the $\mathrm{e_g}$ orbitals are actually $\mathrm{e_g^*}$ orbitals, i.e. antibonding with respect to the $\ce{M\bond{<-}L}$ bonds which is the reason why they are raised. Disregarding π interactions, the $\mathrm{t_{2g}}$ orbitals are nonbonding and thus their energy is not affected.


Relation of temperature and intensive property I came across a question that inquired about whether there exists any intensive property of a system, that is independent of temperature and then seeks proof for the presence/absence of such properties, from a quasi-theoretical standpoint. The question mentions that the system shall be assumed to be non-ideal (ideal gas laws don't hold true et al).

How to proceed, for the latter? Does there exist any such property, in the first place?

After looking at the list in the Wikipedia article linked in the comment by M.Farooq, I suspect there are few properties that are guaranteed to be independent of temperature, an exception being molality and mole fraction in a closed system and absence of reactions. Other concentrations that are functions of volume are bound to be functions of temperature. Even in the presence of reactions, the atomic mole fraction is bound to be a constant (absent nuclear reactions, that is).

Standard molar quantities and some other properties are defined at fixed temperatures and are therefore strictly temperature independent, but that makes for a circular argument, and presumably this is not the type of property the question is about.


Structures of Thiosulfate, Dithionate & Tetrathionate Ions I am trying to find the structures of thiosulfate, Dithionate & Tetrathionate Ions but I am not getting an exact answer. Please help.

Here you go...

Thiosulfate $\ce{[S2O3]^{-2}}$

There ia a negative charge on the apical sulfur and another negative charge distributed across the 3 oxygens by resonance

Dithionate $\ce{[S2O6]^{-2}}$

There is one negative charge distributed by resonance over each group of 3 oxygens.

Dithionite $\ce{[S2O4]^{-2}}$

There is one negative charge distributed by resonance over each group of 2 oxygens.

Tetrathionate $\ce{[S4O6]^{-2}}$

There is one negative charge distributed by resonance over each group of 3 oxygens.