Is 1,3 alkyl shift allowed in Cyclobutyldicyclopropylmethyl Cation? While reading a post , i came across the following mechanism .(source : Which ring expansion in cyclobutyl(cyclopropyl)methanol is favourable?)

The rearrangement is a series of shifts.

The first ring expansion is 1,2 alkyl shift.
The second ring expansion is 1,3 aalkyl shift.
Third step is again 1,2 Hydride shift.

I am aware of 1,2 shift but not not 1,3 shift.So I researched from various sources which are tabulated below .

ORGANIC CHEMISTRY ( SECOND EDITION ) by Jonathan Clayden ,Nick Greeves, Stuart Warren .Quoting from page no. 942 "The methyl group can slide smoothly from one orbital to another—there are bonding interactions
all the way. The next step, migration of H, is just the same—except that the HOMO is now a C–H σ bond.

in all the above , 1,2 shifts are seen.

Advanced Organic Chemistry , FIFTH EDITION ,Part A: Structure and Mechanisms
by Francis A. Carey and Richard J. Sundberg.

Quoting from page no.446 it says

"The preferred alignment of orbitals for a 1,2-hydride or alkyl shift
involves  coplanarity of the p-orbital at the carbocation ion center and the
orbital   of the migrating group "

1,2 shift pathway is seen.

https://www.masterorganicchemistry.com/2012/08/15/rearrangement-reactions-1-hydride-shifts/

Quoting from the above link.

"Here are some examples of “allowed” rearrangement reactions. Notice how we’re always going from a less substituted carbocation to a more substituted carbocation. One exception is at the very bottom; the rearrangement is favorable because the new carbocation is resonance stabilized."

A rearrangement can occur to give the more stable tertiary carbocation.(involves 1,2 shift)

https://www.masterorganicchemistry.com/2012/08/22/rearrangement-reactions-2-alkyl-shifts/

Quoting from the above link.The Mechanism Of Alkyl Migration Reaction is a 1,2 shift in two examples below.

Ring-Expansion Reactions

From https://en.wikipedia.org/wiki/Sigmatropic_reaction .This talks about Thermal alkyl [1,3] shifts.This type of mechanism is not possible for the rearrangement mechanism i am researching .The structure involved is totally different.

Therefor ,the above sources have all talked about 1, 2 shifts only .

So my question is , is 1,3 alkyl shift as given in the post (Which ring expansion in cyclobutyl(cyclopropyl)methanol is favourable?)

  allowed ?

There may be 1,3-alkyl shifts but they are not needed here. Cation 1 should expand the more strained cyclopropane ring first to form cyclobutyl cation 2. After a 1,2-hydride shift to form a tertiary cation 3, the cyclobutane ring in 3 expands to form spirocyclopentyl cation 4. Another cyclobutane ring expansion of 4 leads to cation 5 and eventually to alkene 6.


When simulating spectral line broadening, which convolution is preferred? Many computational chemistry packages permit the calculation of vibrational and electronic spectra. These spectra are obtained as a set of discrete eigenvalues however they are often convolved with some distribution to give a continuous spectrum that is realistic for a finite temperature, rather than a sequence of impulse functions.

The ADF package allows the user to convolve the spectrum with either a Gaussian or Cauchy (Lorentzian) function. My understanding is that the latter affords a more realistic line broadening, however the Gaussian must be there for a reason.

Which spectral broadening scheme is preferred? Why the choice?

I'm not familiar with the computational packages -- I'm an experimentalist, not a theoretician.

As an example for why both might be present, we can turn to gaseous infrared spectroscopy:
one will quickly find that the line-widths in vibration-rotation spectra depend on pressure. At lower pressures, say less than a torr, Doppler-broadening is the main mechanism. This is Gaussian in nature. While at higher pressures the broadening is instead due to collisions, which are Lorentzian. You'll find combinations of the two and so forth depending on the specifics of your case.


Stability of carbocations with -M groups in beta position First part
The original question was to arrange the following compounds in the order of their rate of dehydration with conc. $\ce{H2SO4}$:-

(A) $\ce{CH3-CH(OH)-CH2-CH=O}$
(B) $\ce{CH3-CH(OH)-CH2-C(=O)-CH3}$
(C) $\ce{CH3-CH(OH)-CH2-C#N}$
(D) $\ce{CH3-CH(OH)-CH2-NO2}$

In each of the cases, a carbocation will be formed (+ve charge on the $2$nd carbon via E1 mechanism). Since all the groups(not talking about $\ce{-OH}$, because that’s the leaving group) are electron-withdrawing, the weaker the electron-withdrawing group, the higher the rate.
So, the order according to me should be B > A > C > D. But the answer given is the exact opposite.
"Edit- My teacher who gave me this question now told me that my answer is correct when the reaction takes place in acidic medium and reverses in basic medium."
Second part
Here is a similar question,

Arrange the following compounds in the order of their rate of dehydration with conc. $\ce{H2SO4}$:-
(A) $\ce{CH3-CH2-CH(OH)-NO2}$
(B) $\ce{CH3-CH(NO2)-CH2-OH}$
(C) $\ce{NO2-CH2-CH2-CH2-OH}$

Here, the answer given is C > B > A which I agree with because the further the $\ce{-OH}$ group is from the nitro group, the more stable will be the carbocation. But I feel it contradicts the first question.
Is my answer for the first question correct? If not, what is the correct reasoning?

These would be the arguments I propose under the assumption that the RDS approximation is valid within the scope of the question:
The order (B > A > C > D)  was said to be correct for the conc. $\ce{H2SO4}$/ $\Delta$ case since there would be no effect of $\ce{CH-}$ acidity since the protonation of the $\ce{-OH}$ group takes place and so the carbocation stability takes precedence meaning that the stronger the -I group, the slower the rate since the transition barrier increases.
However, for the scenario where only heat is used in dehydration (I assume basic conditions here), the reaction proceeds via E1cB  since the two main requirements for E1cB is that there is a relatively acidic hydrogen and a poor leaving group. Here, the RDS of the reaction would be the $\ce{C-H}$ bond cleavage. This would be stabilised by the presence of a strong -I group since a carbanion is being generated here which is stabilised by the delocalisation of the electrons via the inductive effect.
So, for such a scenario the rate order would be D > C > A > B.
The answers depend upon the medium as you can see.
This entire argument seems hand wavy with the assumption taken but otherwise as Martin pointed out, this would not be trivial and would require quantum modelling:

Protonation and deprotonation reactions in a catalytic cycle are notoriously hard to calculate. The RDS approximation will break down.


Lipinski's Rule of 5 Oxygen Rules I would like clarification on the role of oxygen as a hydrogen bond acceptor in Lipinski's Rule of 5 (for drug absorption). I have learned in my medicinal chemistry class that both electron pairs from an oxygen atoms are able to individually form hydrogen bonds.

I was wondering if every sp3 hybridized oxygen would account for 2 possible hydrogen bond acceptors in this rule for drug absorption? If not, what other mechanisms or in vivo findings are not causing oxygen to not count as double?

In Advanced Drug Delivery Reviews, 2001, 46, 3-26, (DOI) Lipinski and coworkers write:

  Too many hydrogen bond acceptor groups also hinder permeability across a membrane bi-layer. The sum of $\ce{N}$s and $\ce{O}$s is a rough measure of H bond accepting ability.

So, for the empirical rule of five it's simply the number, nothing more.

In the cited article, the situation for amides is addressed on page 19 in the context of log P calculations and correction factors that increase lipophilicity. Here, the authors write:

  PRX, a proximity correction factor for nitrogen
  and oxygen atoms that are close to one another
  topologically. [...] In addition, for each
  carboxamide group, we add an extra 1.0 and for
  each sulfonamide group, we add 2.0


Quality of water for drinking purposes As I’m reading about water cations and anions, I read that the benefits we gain from water are of course lots, but the most important of them is the neutralisation of acids ($\ce{H+}$) from the anions present in water such as:  $\ce{HCO3-}$ , $\ce{CO3^2-}$ and $\ce{OH-}$.

Does this mean that the higher the concentration of bicarbonates, the better the quality for drinking?

Should we choose the bottle of water with the highest concentration of these anions?

And also, when boiling the water we loose bicarbonates, so is boiled water unhealthier?

Does this mean that the higher the concentration of bicarbonates, the better the quality for drinking?

Not necessarily. This is more a subjective question involving what tastes good to you.

  Should we choose the bottle of water with the highest concentration of these anions?

Assuming the water is actually monitored for safety/quality, the upper limit on caustic (hydroxide) alkalinity is likely too low to do any damage. And carbonate alkalinity will be limited by pH and equilibrium with atmospheric carbon dioxide. So: If it tastes good to you, then drink it.

  And also, when boiling the water we loose bicarbonates, so is boiled water unhealthier?

When you boil water, you are increasing the concentration of dissolved species that do not boil off at that temperature. You can, for example, increase levels of nitrate and nitrite in water by boiling it, which is a bad idea if your water is contaminated by large amounts of nitrate/nitrite. Conversely, you can kill some if not all harmful bacteria in water by boiling it. And I don't think that boiling water results in the loss of bicarbonate in any measurable way, at least not on timescales we're interested in (like cooking).


Is there any selectivity between the C-2 and C-4 positions of pyridine in SNAr reactions? In nucleophilic aromatic substitution (SNAr) reactions on pyridines, displacement of leaving groups at C-2 and C-4 is favoured because the negative charge in the intermediate can be delocalised onto the nitrogen atom.

However, is there any intrinsic difference in the rates of substitution at C-2 and C-4 (assuming the leaving groups at both positions are the same)? If so, why?

True, the para (C4) and ortho (C2) positions give similar resonance structures but consider the fact that:

For an $\ce{S_{N}Ar}$ reaction the electron rich nucleophile must attack. Nitrogen, being highly electronegative, does draw electrons away from the ortho and para positions but because the electron clouds would be larger around nitrogen, it is harder for the nucleophile to approach the ortho than the para. You could think of it as the lone pair on Nitrogen being a steric shield of some sort.
An attack on the para position gives a symmetric intermediate. This would be lower in energy compared to the intermediate produced by an attack on the ortho position.

As such, the para product would be the kinetic product of the reaction whereas the ortho product would be the thermodynamic product. Statistically ortho should be attacked more often as well, as there are two positions whereas there is only one para position.


Using Roman Numerals to indicate a compound I don't completely understand it's use. Here's an example I don't understand:

Shortly after discussing $\ce{[CuCl4]^2-}$, an exam paper asked a question about copper(II) chloride. I understand that copper(II) chloride could refer to $\ce{CuCl2}$, calling it copper(II) chloride, but couldn't it also refer to $\ce{[CuCl4]^2-}$? I would understand the question in reference to $\ce{CuCl2}$, but it got confusing when I realised it could equally be referring to $\ce{[CuCl4]^2-}$, especially after it just mentioned it.

No, those are two distinctive species with different names:

$\ce{CuCl2}$: copper(II) chloride;
$\ce{[CuCl4]^2−}$: tetrachloridocuprate(II).

Oxidation number denoted with the Roman numerals in parentheses refers to the element it's placed after and doesn't carry any additional information besides the fact that copper exists as $\ce{Cu^2+}$ in both compounds.

Nomenclature rules were designed that way so that a name would unambiguously define a single compound only (I'd suggest the Red Book's Nomenclature by IUPAC [1] as a reference).

References

IUPAC “Red Book” Nomenclature of Inorganic Chemistry, 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; IUPAC Recommendations; Royal Society of Chemistry: Cambridge, UK, 2005. ISBN 978-0-85404-438-2. (PDF)


Reason for same frequency of spectral line in infrared and visible region I encountered this question while trying to understand the basics of spectroscopy and atomic structure.

Which electronic transition in Balmer series of hydrogen atom has same frequency as that of n=6 to n=4 transition in Helium positive ion?

I am aware that lines of Balmer series correspond to visible region of electromagnetic spectrum and that to Brackett series correspond to infrared region.
How is it that the frequency of a spectral line belonging to infrared region be same as that of line belonging to visible region knowing that frequency is characteristic property unique to an electromagnetic radiation?

According to Bohr's theory of the structure of the hydrogen atom, the atom is defined as a system with a positively charged nucleus that has electrons revolving around it in discrete orbits of fixed size and energy and stated that the orbits were held stationary using electrostatic forces.
Upon calculation of the various parameters of the model, we get the energy of the $\mathrm{n}^{\text{th}}$ orbital in the hydrogen atom to be:
$$E=\frac{\mathrm{-13.6}}{\mathrm{n^2}} \mathrm{eV} \tag{1}$$
If we state that the atom to be considered were to be made mono-electronic and had an atomic number of Z, satisfying the conditions for the Bohr's model to work, we get that the energy of the $\mathrm{n}^{\text{th}}$ orbital to be:
$$E= -13.6 \cdot\frac{\mathrm{Z^2}}{\mathrm{n}^2} \mathrm{eV} \tag{2}$$
A spectral line is observed when the electron de-excites itself from a higher energy state to a lower energy state that is more stable. This is done by emitting a photon whose energy is the same as that of the difference of energy between the two levels. Applying $\frac{\mathrm{hc}}{\lambda} =E$, and $\frac{\mathrm{c}}{\lambda} = f$ we get
$$f = \frac{\mathrm{c}}{\lambda} = \mathrm{c} R_H\big[\mathrm{\frac{1}{n_1^2}-\frac{1}{n_2^2}\big]Z^2} \tag{3}$$
Here $R_H$ is the Rydberg constant whose value is $\approx\pu{109,678 cm^-1}$
In the question, you have been asked to figure out which line in the Balmer series($\mathrm{n_1} = 2$) of hydrogen would have the same frequency as that of the $3^\text{rd}$ line in the Brackett series($\mathrm{n_1} = 4$). I believe you would know how to solve the question and if not would now be able to via equating the frequencies of the two spectral lines.
Moving on to your doubt which is based on a misconception.

How is it that the frequency of a spectral line belonging to infrared region be same as that of line belonging to visible region knowing that frequency is characteristic property unique to an electromagnetic radiation?

The issue with this statement here is that you are assuming that the Balmer series is always visible for all atoms. This is only true for hydrogen as the the wavelengths provided for the different series are based on measurements taken on the hydrogen atom. That is, the Paschen series is infrared for hydrogen and this may change if the atom changes. As the atomic number increases, the frequency becomes greater as you can see from equation ($3$). This implies that at higher atomic numbers, there is a greater shift towards the UV region.
TL;DR: What you say is correct, frequency is a characteristic property unique for a certain electromagnetic radiation. However, the Balmer series is defined to be visible specifically for the hydrogen atom and not for any other mono-electronic atom/ion. As the atomic number becomes greater, a shift towards the UV region takes place.


How can I prevent the effervescence of a solvent during evaporation? How can I prevent the effervescence of a solvent like petroleum ether when I evaporate it on a water bath?

Break a ceramic utensil (mug, plate, any porous item will do) and add a few chips in the solution. This should lead to a smoother boil rather than the liquid jumping around.

This is a cheap alternative of anti bumping granules and works in a similar way by allowing the gas to accumulate on the rough surface and escape the solvent in a controlled fashion.


How to make 0.3% ammonia water from 28-30% ammonia hydroxide Can someone tell me how to make 0.3% ammonia water from 28-30% Ammonia hydroxide?
I am trying to make a gallon or 3.78 Liters.

Hint : Ammonia water is actually diluted Ammonium Hydroxide ruled by the following equilibrium :

$$\ce{NH3 + H2O <=> NH4+ + HO-}$$


Are all degenerate d-orbitals identical? This discussion spun off from the comments to another question. The basic idea of that question was that electrons don't have a preferred order of filling in $p$ orbitals, i.e. the first electron will fill in the $p_x$ orbital just as well as it will fill in the $p_y$ or $p_z$ oribtal. This is because in the absence of electromagnetic fields, all p-orbitals are degenerate, and the coordinate axes are things we arbitrarily assign, so, the degenerate $p$ orbitals are all identical.

I wonder if the same can be said about $d$-orbitals. Four of them ($d_{xy}$, $d_{xz}$, $d_{yz}$, $d_{x^2-y^2}$) feel identical to me, as they all have four perpendicular and coplanar lobes (but I could be wrong). The fifth one - $d_{z^2}$ - doesn't look identical to me. Its positioning of the lobes is wildly different from the other four.

This makes me wonder,

  Are all degenerate $d$-orbitals identical?

By identical, I mean that the electron filling order has no preference. (but this might not be the best definition of identical orbitals please mention if you have a better one)

Remarks:

I didn't mention $f$-orbitals because they have even ridiculous orbital structures (4 types have six lobes, 2 have 8 lobes, and 1 seems to be a advanced version of $d_{z^2}$ with two rings. Crazy!), but in case their answer is remotely similar, please consider mentioning them in your answer for completeness.
I have talked about atomic orbitals and not molecular orbitals (just in case that made the answer different).

Yes, they are identical.
One thing that we don't really teach well with orbitals is thinking about the symmetry of the orbital with respect to the name of the orbital.
$p_{x}$ has the same symmetry as the function $f(x,y,z)=x$.
Likewise, $d_{x^{2}-y^{2}}$ has the same symmetry as $f(x,y,z)=x^{2}-y^{2}$.
What about $d_{z^{2}}$? You should note that $d_{z^{2}}$ is really $d_{2z^{2}-x^{2}-y^{2}}$.
Imagine taking $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$, summing them up, and renormalizing. So the symmetry is like: $z^{2}-x^{2} + z^{2} - y^{2}=2z^{2}-x^{2}-y^{2}$.
What you get is a big lobe of the same phase along the $z$-axis. And around it in the $x$-$y$-plane (but smaller), you see a smear of the opposite phase that's shaped kind of like a torus. That's why it looks different. It's really two of the other looking orbitals put together.
Why don't we just use $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ instead? Because they're not linearly independent with the 4 other "normal"-looking orbitals. $d_{x^{2}-y^{2}}$ summed with $d_{z^{2}-x^{2}}$ is $d_{z^{2}-y^{2}}$. The 5 conventional orbitals are just the nice linear combinations of the 5 spherical harmonic solutions to the angular part of the Schrodinger equation. See here.
You don't seem to have a problem with $d_{xy}$, $d_{zx}$, $d_{yz}$, and $d_{x^{2}-y^{2}}$ all being degenerate. So hopefully, you can see that $d_{z^{2}}$ is not really different and therefore also of the same energy as the other 4.
EDIT:
Based on comments, it's also important to point out that the pictures of orbitals with smooth surfaces are not real pictures of orbitals. These are boundary surface diagrams where we've drawn the surface that represents a constant probability of finding the electron for that orbital. This basically means that electron densities for $d_{z^{2}-x^{2}}$ and $d_{z^{2}-y^{2}}$ will smear together in the $x,y$-plane (the orbitals are the same phase) to create something that is symmetrical and shaped like a torus.
This analysis extends fully to $f$ orbitals. Just identify the full functional name for each orbital and apply the same analysis here.


Can I synthesize an Amide simply by heating an amine and a carboxylic acid (with a vacuum pump)? I am interested in creating an amide of stearic acid. I thought that simply mixing both of them in some solvent, heating the solution, and using a vacuum pump to distill the liberated water (to drive the reaction forward) would be enough.
But a bit of research showed that this synthesis is normally done with acyl chlorides derivatives or methyl ester derivatives or other more elaborate methods.
Without access to those tools, is it still possible for me to produce a good yield of amide, using only a fatty acid, a primary or secondary amine, a solvent, and heating & vacuum?
(Secondary optional question: If I use ethanolamine, what will be formed preferentially, the amino-ester or the hydroxy-amide?)

It can be done but it is not a good reaction. From Lundberg et al. [1]:

Formally, the amide bond is formed through the condensation of a carboxylic acid and an amine with the release of one equivalent of water. This reaction has been considered challenging due to the competing acid–base reaction, which occurs when the amine and the carboxylic acid are mixed. Although the amide bond can be formed from the corresponding ammonium carboxylate salt upon heating, this reaction has generally been considered to be of limited preparative value. Furthermore, the high activation barrier for the direct coupling of a carboxylic acid and an amine can only be overcome using forcing reaction conditions.

Use of borate ester catalysis under Dean–Stark conditions has recently been reported to give good yields without resorting to acyl activation groups [2].
References

Lundberg, H.; Tinnis, F.; Selander, N.; Adolfsson, H. Catalytic Amide Formation from Non-Activated Carboxylic Acids and Amines. Chem. Soc. Rev. 2014, 43 (8), 2714–2742. DOI: 10.1039/C3CS60345H. (PDF)
Sabatini, M. T.; Boulton, L. T.; Sheppard, T. D. Borate Esters: Simple Catalysts for the Sustainable Synthesis of Complex Amides. Sci. Adv. 2017, 3 (9), e1701028. DOI: 10.1126/sciadv.1701028.


Factors affecting Kp (pressure constant) Initially, I found that $K_\mathrm{p}$ depends only on temperature; but later I found that it depends on pressure of $\ce{CO2}$ gas in case of $\ce{CaCO3 (s, aragonite) → CaO (s) + CO2}$.

How do we know whether it depends upon partial pressure of reactants and products or not? Or, this is a special scenario?

If there is a high gas pressure in the environment the reaction is taking place in, $K_\mathrm{p}$ will shift towards the side of the reaction that has fewer gas molecules, as it is "fighting against" the high pressure, in a way. So in your case, since only one side has gas (the product side) if you increase the pressure of $\ce{CO2}$ or any other gas, the equilibrium will shift towards the reactant side.


How to convert Sodium Carbonate to Sodium Bicarbonate at home (CO2 scrubber) Is it possible for me at home to react atmospheric $\ce{CO2}$ with sodium carbonate to make sodium bicarbonate.

I am seeing if I can take $\ce{CO2}$ out of the air.

A process commonly used to teach $\ce{CO2}$ scrubbing to chemical engineering students involves not sodium bicarbonate, but sodium hydroxide ($\ce{NaOH}$).

The reaction is exothermic and is as follows:

$$\ce{CO2 (g) + 2 NaOH (aq) -> Na2CO3 (aq) + H2O (l)}$$

The aqueous sodium carbonate is subsequently treated with calcium hydroxide to precipitate out calcium carbonate:

$$\ce{Na2CO3 (aq) + Ca(OH)2 -> 2NaOH (aq) + CaCO3 (s)}$$

Disclaimer:

This is usually done with much higher concentrations of $\ce{CO2}$ (e.g. >10%) than you would find in the atmosphere (~400 ppm). The point of scrubbing in a chemical plant is to reduce emissions to a level that is still much higher than atmospheric $\ce{CO2}$ levels but lower than governmental standards. I suspect that you would not see a noticeable change in ambient $\ce{CO2}$ levels from putting something like this in a place with such (relatively) low $\ce{CO2}$ to begin with. There is a very low driving force to push gaseous $\ce{CO2}$ into the liquid and solvate it when the partial pressure of $\ce{CO2}$ is low.

There exists a unique relationship between the partial pressure of $\ce{CO2}$ above a solution, the dissolved carbon in the solution, and the pH of the solution, shown here (own work):

Note that the equations on the right-hand side are in solution.

So if you wanted to see what was going on in a controlled environment with no other species in the solution / that could enter the solution, you could just use a pH probe to see how much carbon was pushed into the solution!


Cathodic protection of active metals Cathodic protection prevents the removal of positive form the metal surface by applying a negative charge to the metal.

Is is possible to prevent the reaction of an active metal (i.e. sodium) with water by aggressively forcing current into the metal using a DC source? We allow the "water" to be a dilute solution (~1%) of the metal hydroxide. There is also an anode that completes the circuit but its OK if the anode disintegrates.

Edit: I will let the answer stand but it begs one question: If the power supply forces the sodium metal to be at a strongly positive potential, it may end up coated in layer(s) of hydroxide (or even oxide?), which may prevent the water from reaching it. But if the metal is liquid (NaK) then it may cause the metal to have "negative surface tension" making it impossible to maintain a high potential as the capacitance runs away to "infinity". So it's not completely settled.

It is unlikely that a potential applied to a sodium electrode in water would prevent the spontaneous oxidation of sodium by water.  Consider the following reduction potentials:

$$\ce{2Na+ + 2e- -> 2Na(s)}\ E^0=-2.71\ \mathrm V\tag{1}$$
$$\ce{2 H2O + 2e- ->H2 + 2 OH-}\ E^0=-0.83\ \mathrm V\tag{2}$$

The reduction of water will always occur before the reduction of sodium ions.  If we reverse equation (1) so we can combine the two and get the full redox equation for your system, we get:

$$\ce{2 Na(s) + 2 H2O <=> 2Na+(aq) + H2 + 2 OH-}\ E^0= 1.88\ \mathrm V\tag{3} $$

A positive potential indicates a spontaneous reaction under the given conditions so thermodynamically, it will be near impossible to prevent the reaction from being spontaneous.

How impossible?  We can do a thought experiment and ask the question at what pH would this reaction be non spontaneous? In other words, when would the Nernst Equation for this system evaluate to zero:

$$E=E^0-\frac{RT}{nF}\ln\frac{\text{Products}}{\text{Reactants}}\tag{4}$$

Entering our values, converting Ln to Log and substituting $\mathrm{pH} = -\log[\ce{H}]$ (and assuming all other concentrations are fixed at 1 M or 1 atm):

$$E=1.88+0.059(14-\mathrm{pH})\tag{5}$$

The reaction would not be spontaneous until the pH reached 45 units.  Something that can't be done in water.


Electrons jumping from 1 water molecule to another Electrons can be anywhere in the universe and where they are is random.

But they are still discrete particles and thus it takes energy for electrons to move.

Is it possible for an electron to jump from 1 water molecule to another causing $\ce{H2O^+}$ and $\ce{H2O^-}$ to form?

Electrons can be anywhere in the universe

Kind of: There is a non-zero probability that an electron can be at any distance from some point of origin (for instance, the nucleus of an atom), but there is some distance which corresponds to the place where we're most likely to find it (the place where the probability amplitude is highest) most of the time.

  and where they are is random.

Again, sort of: They are most likely located in certain regions of space for most of the time, but since it's a probability, it's not a certainty. That said, there is a common sense element that can be applied here. Yes, an electron from a hydrogen atom in a flask in your laboratory can be out near Jupiter, but realistically, it's not. And if it's on Jupiter, it's not really that hydrogen atom's electron anymore.

  Is it possible for an electron to jump from 1 water molecule to another causing $\ce{H2O}^+$ and $\ce{H2O}^−$ to form?

Yes, it is possible (but probable? I don't know, but certainly there are conditions under which such a reaction will occur).


How do restricted open shell calculations mathematically proceed in the context of the self consistent field iterations? I'm specifically concerned with HF theory. I understand restricted closed shell calculations use the Roothaan equations. If I understand correctly these equations assume each filled molecule orbital is doubly occupied. Each of the electrons in a pair is described with the same spatial wave function.

I also think I understand the Pople-Nesbet equations which allow the wavefunction for each electron to be described with their own spatial wavefunctions furthermore their spins are not constrained hence spin contamination can result.

I however, have not been able to figure out how restricted open shell calculations work. All I understand is that the spin is constrained. I do not understand if the wavefunctions of the paired electrons are described with the same wave function (similar to the Roothaan restricted closed shell approach) while the unpaired electrons are simply allowed to have their own spatial wave function. Coming back to the title question is there a bunch of equations with a famous person’s name (Roothaan, Pople-Nesbet) that are used to describe the scf procedure for restricted open shell calculations or does it use some combination of Roothaan and Pople-Nesbet?

The paper by Tsuchimochi and Scuseria describes it quite well.

Briefly, Roothan-Hall equation is the same but the ROHF matrix is constructed from the $\alpha$ and $\beta$ Fock matrices from UHF.

$$\mathbf{F}_\mathrm{ROHF} =
\begin{bmatrix}
    \mathbf{R}_{cc}       & \mathbf{F}_{co}^\beta & \mathbf{F}_{cv}^{cs} \\
    \mathbf{F}_{oc}^\beta & \mathbf{R}_{oo} & \mathbf{F}_{ov}^\alpha \\
    \mathbf{F}_{vc}^{cs}  & \mathbf{F}_{vo}^\alpha & \mathbf{R}_{cc} \\
\end{bmatrix}$$

where $c$ is doubly occupied, $o$ is singly occupied, $v$ is virtual and

$$\begin{align}
\mathbf{R}_{cc} &= A_{cc}\mathbf{F}_{cc}^\alpha + B_{cc}\mathbf{F}_{cc}^\beta\\
\mathbf{R}_{oo} &= A_{oo}\mathbf{F}_{oo}^\alpha + B_{oo}\mathbf{F}_{oo}^\beta\\
\mathbf{R}_{vv} &= A_{vv}\mathbf{F}_{vv}^\alpha + B_{vv}\mathbf{F}_{vv}^\beta
\end{align}$$

  Different values for these parameters have been suggested in the literature. Although they do not affect the ROHF wave function and energy, they affect orbital energies whose physical meaning is obscured because of this dependence.

Takashi Tsuchimochi and Gustavo E. Scuseria, J. Chem. Phys. 2010, 133, 141102. DOI: 10.1063/1.3503173; available as ePrint: arXiv:1008.1607 [cond-mat.str-el].


How to remove CH3 labels from a molecule's image? [OpenBabel in Python] I would like to remove CH3 labels from svg images generated in openbabel in Python.
I am thinking I should use obConversion.SetOptions("w",openbabel.OBConversion.OUTOPTIONS) and replace "w" with another character. This does not work. Do you have any ideas? The image is below, for reference.

obConversion.AddOption("C",openbabel.OBConversion.OUTOPTIONS)

Referenced openbabel documentation. Setting it to "C" will remove terminal carbons.


Bridged ring system reduction through Wolf Kisher: Is it possible? Is it possible for  a ketone like this to undergo the Wolf Kishner mechanism? If so, what would be the product? It seems unlikely to me as disrupting the stable equitorial carbonyl seems unfavorable...

If you are thinking of reducing the ketone with hydrazine under alkaline conditions then I see a problem. The hydrazine will also act as a nucelophile on the alkyl bromide in the molecule. I think that you will have a SN2 reaction on the alkyl bromide to form an alkyl hydrazine at the same time as the hydrazine is reacting with the ketone. If you then heat it up with base you will remove the ketone but assuming that you use a large excess of hydrazine you will end up with an alkyl hydrazine in the top left of the molecule rather than an alkyl bromide.

Why not react the original compound with benzyl alcohol using a mild base to form a benzyl ether at the start. Then do the reduction with hydrazine, then hydrogenate with hydrogen gas and palladium on carbon. This hydrogenation will remove the benzyl group and give you a alcohol. Then you can react the alcohol with tosyl chloride/pyridine and use the tosylate as an electrophile. If you very much want an alkyl bromide then react the tosylate with sodium bromide in DMF.


Why does ionisation energy to remove electrons from the same subshell increases? Electrons in the same subshell possesses the same energy.
So why does ionisation energy to remove electrons from the same subshell increases?

For example, $\mathrm{2s^2}$ has two electrons. Since both of them are in the same subshell shouldn't ionisation energy to remove each of them stay the same?

The ionisation energies would be same if there was no shielding effect but it exists. The electrons due to mutual repulsion  prevent the outer electrons from the nuclear pull. Thus the force with which the nucleus attracts the electrons decreases as we move away from the nucleus both because of the increasing distance and also due to the increasing shielding effect of the inner electrons . Now in the same shell when one electron is removed the shielding effect decreases . The shielding effect due to the electrons of the same shell is the greatest. Thus the successive ionisation energies of any atom increases due to decreasing shielding effect.


Taste same but why not LiCl a substitute for NaCl (common salt)? A clinical paper from 1949 [1] observes the toxicity of LiCl, at very low concentrations. That just went over my head... Is toxicity the only reason to avoid LiCl as a substitute for NaCl (specially for those who were on a salt restricted diet)? Why did they even consider it to be a substitute in the first place?

What physical and chemical property of Li behaves differently than Na (e.g. being toxic)?

P. S.: not sure what tag to put in.

Bibliography

Hanlon, L. W. Journal of the American Medical Association 1949, 139 (11), 688. DOI 10.1001/jama.1949.02900280004002.

The reason why anything is ever considered a substitute for normal salt (NaCl) is that some people take too much salt with their diet and this is widely thought to be bad for your blood pressure. This, however, is disputed by some mainly on the grounds that only a small proportion of people are sensitive to hight sodium in the diet and the rest of us can cope with widely differing amounts with no discernible harm.

However, this doesn't mean that anything that tastes like salt is necessarily a good substitute. Not having enough salt is also a problem as the Na/K balance in cells is important and a serious lack of sodium can be an acute danger. For most people with an adequate diet, using a salt substitute (for table salt) can reduce their sodium intake but, since there is already a lot of sodium in the diet, won't cause a problem. But not for all people.

The most common table salt substitute is KCl, which, again, won't be harmful to people with a normal diet. The body is good at processing potassium and sodium and can regulate their levels within reasonable bounds. So KCl is safe as the body is used to coping with it.

This isn't true for lithium. There is no major biological role for lithium in the body, and, while lithium is regarded as an essential trace element, nobody really knows why. Whatever the normal role, the body is not designed to cope with significant amounts of it (unlike K and Na which are major components of many biological systems). So this might suggest that using large amounts a salt substitute has an unknown risk for harm far greater than the use of, say, potassium. Moreover, lithium does has significant known effects on the nervous system, though the mechanisms are poorly understood. This is illustrated by the use of controlled amounts of lithium salts as a significant treatment for bipolar disorder. This fact alone (regardless of any other toxic effects in high doses) would suggest it isn't a safe alternative when used as a substitute for table salt in uncontrolled doses.

The use of lithium as a treatment for bipolar disorder is relatively recent (started in the early 1950s) so it might have looked like a harmless alternative to sodium chloride in earlier times.


Steam Distillation - Condensate not going through condenser Basically what the title says.  I set my equipment up exactly like pretty much every diagram I could find.  There's condensation through the claisen adapter and into the distillation adapter, but for some reason, the condensation won't fall slide down into the condenser tube.

The boiler flask has been at a roaring boil for several hours, the water going through the condenser tube is hooked up properly and ice cold, and the receiving flask is sitting in ice.  Everything is identical to the diagram.  I've tried it with and without the thermometer - and with and without the separatory funnel but the condensate won't get in the consdenser tube.

Any ideas?

First of all, make absolutely sure that there is nothing blocking the path between the distillation flask and the final vent, and that all the joints are tight.

When beginning the distillation, you should be able to view the condensation line move up within the apparatus after the mixture begins to boil. If it moves up toward the thermometer, but then stops and never makes it up to the condenser, that might indicate that you need to insulate the apparatus above the distillation flask and up to the thermometer. Try wrapping it with glass wool and foil and see if that helps.


Ullmann's Reaction (similar to Wurtz) - aryl chloride with deactivating groups - stability of free radical Ullmann's Reaction:

Mechanism:

Question:

My textbook says with $\ce{X} = \ce{Cl}$, the reaction gives less yield compared to using $\ce{X} = \ce{I}$ since $\ce{Cu}$ is not as reactive as $\ce{Na}$ ($\ce{C - I}$ bond is weak).

But if aryl halides with substituents such as $\ce{NO_2}$ are used, they can help in obtaining good yield.

What I observe from both the mechanisms is that adding deactivating groups in the ortho or para position relative to the halogen should actually give lesser yield since the deactivating group would cause a partial positive charge to exist on the $\alpha$-carbon hence destabilizing the free radical intermediate.

Can anyone clarify my doubt? I suspect my last argument to be wrong regarding the $\alpha$-carbon. The $\alpha$-carbon is already electron-deficient and the partial positive charge should make it worse?

Courtesy: http://www.name-reaction.com/ullmann-reaction

Your assumption that electron-withdrawing groups destabilize radicals is wrong. Radicals are in fact stabilized by both electron-withdrawing and electron-donating groups and even more so if both types are attached to the same carbon atom (captodative effect).

I modified an image from Ian Fleming's Molecular Orbitals where you have the proof. In the image you can see that the combination of the carbonyl group's (which is electron-withdrawing) pi bonding/antibonding pair of orbitals with the p orbital of the radical results in an overall decrease in energy:


Does Combining Solid/Liquid Concentration with $K_c$ Matter when Comparing $K_c$ of Different Reactions? I know it is convention to take liquid/solid concentrations as constants and lump it with the equilibrium constant. However, if we were trying to compare the equilibrium constants of two different reactions and if one of the reactions had a solid reactant, doesn't lumping the solid reactant concentration with the equilibrium constant make one reaction seem more product favoured than it actually is?

The equilibrium constant of a reaction containing solids or liquids doesn't contain hidden values for their concentrations.

The issue is that we use the concentration (or pressure) of a substance in the equilibrium expression because it is a shortcut.  What we are actually putting into the expression is the activity of that substance, a more complicated idea.  It very often reduces to the ratio of the concentration of a solution to the concentration at standard state.  Since the concentration of a solution at standard state is 1M, there's no point in dividing through by 1 for each term.

For solids and liquids, the concentration at standard state is its molar density. ($C = \frac{mol}{L}$)  This value doesn't depend on the quantity of substance available, so the ratio between the density at standard state and the density at any point in a reaction is 1.  The values ARE in the equilibrium expression; they're just all 1's.

As to your concern about comparing equilibrium constants, we can only do that directly for reactions with the same number of reactants and products.  Adding (or removing) any term to either side interferes with the comparison.

Compare:
$\ce{A(aq) +B(aq)->AB(aq)}$
$\ce{2A(aq) +B(aq)->A2B(aq)}$

We have two similar reactions but the equilibrium constants can vary by a huge amount due to the presence of a second A atom in the reactants.  Similarly, with a constant solid/liquid concentration (activity = 1) the equilibrium constant effectively ignores that product or reactant.

It's just one of the limitations of the equilibrium constant.

Check out the wiki on Thermodynamic Activity for more information.


Casimir effect and van der Waals forces The Casimir effect makes two close metal plates in a vacuum attract.
I was pondering and figured out that two close molecules should experience the same effect, as they are close and between molecules there is nothing. Maybe even stronger as they are usually way closer than metal plates can be.
So, is there any theoretical work on this? Is the Casimir "force" considered one of the van der Waals forces? Am I talking nonsense?
Also, could the attraction between the plates result from van der Waals interactions and not from the quantum field stuff? How is it known that this is not the case?

You are correct in your ideas. The effect is important in the van-der-waals interaction and appears as an additional  $1/r^7$ term compared to the usual $1/r^6$ term. (This is sometimes called a retardation effect). It is present only at larger distances (a few nm) and arises only in induced-dipole interactions (dispersion interaction) because of the time it takes for the interaction to propagate between molecules. During this finite time (governed by the speed of light in the medium) the first dipole's direction has changed during the time it takes to cover distance $2r$ so that the inducing and induced dipoles become less correlated the farther the two dipoles are apart. The equation describing this effect is called the Casimir-Polder eqn. $W(r)=-23\alpha_0^3hc/(8\pi^2(4\pi\epsilon_0)^2r^7)$ where $\alpha_0 $ is the polarisability and $\epsilon_0$ the permittivity of free space.


acid-catalyzed ring opening of epoxide I have some questions about the mechanism above:

why does ethanol attack the tertiary C before O-C bond breaks?  Shouldn't the mechanism proceed via an $S_{N}1$ mechanism?
Why is ethanol, as a weak nucleophile, able to attack tertiary carbon?
In a lecture it's said that a partial tertiary carbocation is involved, but left secondary carbon is also losing its electrons to oxygen, therefore becoming a partial carbocation; and if two carbon are both partial carbocation, then the tertiary carbon would be more stable and ethanol should attack the secondary carbon.

Ring-opening reactions can proceed by either SN2 or SN1 mechanisms, depending on the nature of the epoxide and on the reaction conditions.

If the epoxide is asymmetric, the structure of the product will vary according to which mechanism dominates.

When an asymmetric epoxide undergoes solvolysis in basic methanol, ring-opening occurs by an SN2 mechanism, and the less substituted carbon is the site of nucleophilic attack.

In the case of acid catalysed ring opening, the mechanism is not exactly SN1 or SN2. It is a hybrid of both.

In the first step, the oxygen gets protonated.

Second step the C-O bond starts to break ( the C with more alkyl substituents is preferred for cleavage of C-O bond. In your case the tertiary carbon atom is preferred to the secondary C atom)

Third step the nucleophile ( or alcohol in this case) attacks the electrophilic carbon (the C with which C-O bond is still breaking). Unlike in SN1 mechanism the reaction does not wait for the carbocation to be formed completely. The incoming nucleophile attacks the carbon from a direction opposite to that of direction of C-O bond because C and O are still bonded to some extent.

So hence the mechanism proceeds similar to what you have shown.
In acid catalysed ring opening the nucleophile attacks the more substituted carbon because this carbon has more fractional positive charge then the secondary one.


Why is the alkene a more stable species in this case? How can we account for $\mathrm{S_N1}$ substitution in this case versus elimination in others?
When we dehydrate alcohols, we usually carry out the reaction in concentrated sulfuric acid and at high temperature. The hydrogen sulfate ($\ce{HSO4-}$) present after protonation of the acohol is a weak nucleophile, and at high temperature the highly reactive carbocation forms a more stable species by losing a proton and becoming an alkene. (Emphasis mine) Furthermore, the alkene is usually volatile and distills from the reaction mixture as it is formed, thus drawing the equilibrium toward alkene formation. The net result is an $\mathrm{E1}$ reaction.
 (Solomons, 11th edition page 511)

The author seems to imply that the formation of an alkene from a carbocation is energetically more favorable than the formation of an $\mathrm{S_N1}$ product. Why?
Once the carbocation is formed, the step leading to the $\mathrm{S_N1}$ product requires the formation of a sigma bond whereas the step leading to the $\mathrm{E1}$ product requires the formation of one pi bond.
It seems to me that the $\mathrm{S_N1}$ product should be more favorable (not by a lot though, since the rate limiting step for both reactions is the same).

I don’t think that your interpretation is what the author meant. I think that they are comparing the reactivity (or inverse stability) of the alkene to that of the carbocation. I think we can both agree that a carbocation is a much more reactive species than an alkene.

When it comes to comparing the products of a nucleophilic substitution to those of an elimination, I don’t think there’s an easy call that can be made. It depends far too much on variable factors such as the steric congestion around the reactivity site, the substitution pattern of the double bond, the size and shape of the nucleophile etc.

In the case of this specific reaction, however, we can make that call. The only mildly nucleophilic species is the hydrogen sulphate anion $\ce{HSO4-}$ deriving from deprotonated sulphuric acid. Hydrogen sulphate has three equivalent oxygens that each carry one third of the negative charge, and they are also practically symmetry-equivalent (not ideally, though, since the proton will angle off). The high oxidation state of the central sulphur atom reduces the electron density on the oxygens. Each of these oxygens is therefore a very bad nucleophile — compare to typical sulphonic acid residues such as tosylate or mesylate which are equally terrible nucleophiles. This should make a nucleophilic attack highly unlikely.

A second thing to remember is that sigma bond does not equal sigma bond. Usually when discussing sigma bonds, $\ce{C-C}$ or $\ce{C-H}$ bonds are meant which are very unpolar, have a bond dissociation energy of something around $\pu{400 kJ/mol}$ and are thus very hard to dissociate. Here, we would be making a $\ce{C-O}$ bond. But not any kind of $\ce{C-O}$ bond: the oxygen belongs to a sulphate residue. This makes the $\ce{C-O}$ bond very weak and very easily cleaved, especially since sulphate is a very good leaving group.

Therefore, in this specific case we can actually make the call and say that the elimination product is more stable than a product of a nucleophilic substitution.

One more thing: you do write cleavage of a sigma bond. But what is actually happening is akin to a deprotonation: a proton leaves leaving its electrons behind in a type of acid-base reaction. As you may know, these are typically very rapid as protons have an easy time ‘jumping’ from one molecule to another if the overall conditions for acid-base reactions are favourable.


Ionizable hydrogen atoms for acids Given that hydrogen atoms that are attached to a compound must be ionizable for the compound to be an acid, what prevents a hydrogen from being ionized? Is it solubility. If the bond is nonviolent, water could not dissolve the compound, and so hydrogen atoms are not ionizable. Is this correct?

Not all hydrogen atoms in a compound are equally acidic. The acidity of a hydrogen at a certain position in a molecule depends on its chemical environment, i.e. the solvent (if the compound is dissolved), which atoms and functional groups it is attached to, and how well the negative charge of the corresponding conjugated base is stabilized. A quantitative measure for the acidity of a proton is its $pK_a$ value which is defined as follows (reference):

$$pK_a=-\mathrm{log}_{10}K_a$$

with $K_a$ being the equilibrium constant of the dissociation reaction in water $\ce{HA + H2O \rightleftharpoons H3O+ + A-}$, in which the hydrogen is abstracted from the compound by heterolytic bond cleavage. $[\ce{H2O}]$ can be assumed to be constant (approx. 55 mol/l) and is therefore included in $K_a$:

$$K_{a}=\frac{[A^{-}][H^{+}]}{[HA]}$$

Let's take formic acid ($\ce{HCOOH}$) as an example. As the name of the compound suggests, it has at least one acidic proton, and if all hydrogens would be equally acidic and dissociate from the molecule, the reaction would look like this:

$$\ce{HCOOH + 2H2O \rightleftharpoons 2H3O+ + COO^{2-}}$$

However, this does not happen in aqueous solution, only one of the protons is acidic:

$$\ce{HCOOH + H2O \rightleftharpoons H3O+ + HCOO-}$$

The reason for this is that both hydrogens are situated at different positions in the structure:

With a $pK_a$ of 3.77, the hydrogen of the carboxyl group is significantly more acidic than the hydrogen bonded to the carbon atom. The typical $pK_a$ of an aldehyde hydrogen (H bonded to the carbon of a carbonyl group) is in the range of 17 (source), so such a position cannot be deprotonated in water.


Depression of the Freezing Point Recently we learnt that there is a depression in the freezing point when a non-volatile solute is added to the solvent. Our teacher explained this phenomenon with a graph where the vapour pressure of the solution and pure solvent intersects with the vapour pressure of the solid solvent. my question is: wouldn't the vapour pressure of the solid solution be lower than the vapour pressure of the solid solvent(that would make another curve in the graph)? Because, the non-volatile solute is still blocking the surface of the solvent, thus reducing the vapour pressure.
The solution is of a solid non-volatile solute in a liquid solvent

In chemistry we assume a mixture of two solids as two different phases, which do not interfere with each other and thus, the solid solute in the solid mixture will not interfere with the solid solvent. So, the vapour pressure of the solid solution and solid solvent will always be same. In a similar way, when two immiscible liquids are mixed together, the total vapour pressure is counted as the sum of Vapour pressures of the liquids if kept separately.
Hope this helps!!


Effective nuclear charge and Ionization energy A common reason given on why 3rd ionization energy > 2nd > 1st is because of increasing effective nuclear charge.

As per my book $Z_\mathrm{eff}$ = Atomic number $-$ Number of inner electrons.

Now let us consider a Carbon atom. C has an Electronic configuration of $\ce{[He] 2s^2 2p^2}$ or $2,4$

Right now the effective nuclear charge is $6 - 2 = 4$

Now if we remove the first electron the first ionization energy will be $=\pu{1086.45 kJ/mol}$

After doing this effective nuclear chrarge is still $6 - 2 = 4$

Now if we remove the second electron the second ionization energy will be $=\pu{2352.62 kJ/mol}$

If the effective nuclear charge in removal of 1st and 2nd electron is same then why is 2nd ionization energy > 1st.

note : these values are taken from here

Effective nuclear charge $Z_{\text{eff}}$ is given by Slater's Rules.

According to these rules, for $\ce{C}$, effective nuclear charge experienced by an outer shell electron is:
$$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 3 = 3.25$$

For $\ce{C+}$, effective nuclear charge is
$$Z_{\text{eff}} = 6 - 0.85\times 2 - 0.35\times 2=3.6$$

Since the effective charge for $\ce{C+}$ is more than that on $\ce{C}$, the second ionisation energy is more than that of the first.


Definition of Electronic energy Using Born-Oppenheimer approximation, we separate the nuclear and electronic parts of the wavefunction, and treat them separately. We include the terms that have direct dependence on the electronic positions in the electronic Hamiltonian,

$$\hat{H}_{e} = \hat{T}_e + \hat{V}_{eN} + \hat{V}_{ee}$$

Thus, the total Hamiltonian is given by:

$$ \hat{H} = \hat{H}_{e} + \hat{V}_{NN} $$

So, what is the definition of electronic energy? According to the source mentioned below, electronic energy does not include $\hat{V}_{NN}$, which makes sense. But, all the electronic structure packages quote the Total energy as the electronic energy. The following is quoted from Gaussian website:

Zero-point correction=                  .023261 (Hartree/Particle)
Thermal correction to Energy=           .026094
Thermal correction to Enthalpy=         .027038
Thermal correction to Gibbs Free Energy= .052698
Sum of electronic and zero-point Energies=-527.492585
Sum of electronic and thermal Energies= -527.489751
Sum of electronic and thermal Enthalpies=-527.488807
Sum of electronic and thermal Free Energies=-527.463147

Why is this discrepancy? What is the actual definition of electronic energy?

Source for my argument: http://vergil.chemistry.gatech.edu/notes/quantrev/node31.html

Electronic energy is indeed an eigenvalue of electronic Hamiltonian, and, as you correctly pointed out, it doesn't include the contribution from repulsions of the fixed nuclei (which has nothing to do with electrons). But sometimes the total energy for a particular nuclear configuration (one that includes the above mentioned interactions contribution) is still (incorrectly) termed "electronic energy" while just an eigenvalue of electronic Hamiltonian is called "pure electronic energy". That is essentially wrong, but it is sort of a custom.

And I would not say that "all the electronic structure packages quote the Total energy as the electronic energy". Did you check all of them? For instance, I quickly checked one of my DALTON calculations, and found this:

@    Final DFT energy:           -210.248305383291
@    Nuclear repulsion:           158.997839473485
@    Electronic energy:          -369.246144856776

So the first one is the (total) energy, then come its two components with the right names.

Same for ORCA:

Total Energy       :         -210.25104538 Eh           -5721.22181 eV

Components:
Nuclear Repulsion  :          158.99783872 Eh            4326.55115 eV
Electronic Energy  :         -369.24888411 Eh          -10047.77296 eV

GAMESS-US (here first two terms = electronic energy)

        ONE ELECTRON ENERGY =    -596.6619030866
        TWO ELECTRON ENERGY =     228.7817683952
   NUCLEAR REPULSION ENERGY =     158.9978510234
                              ------------------
               TOTAL ENERGY =    -208.8822836680

NWChem (middle three terms = electronic energy)

         Total DFT energy =     -210.248032406037
      One electron energy =     -597.377034042581
           Coulomb energy =      258.596296569204
    Exchange-Corr. energy =      -30.465145956066
 Nuclear repulsion energy =      158.997851023405

And I don't want to continue, since I'm pretty sure that most of the programs use the right terms.


Number of stereoisomers of 1-(2-methylcyclohexyl)ethan-1-ol How many stereoisomers are possible for the compound in the picture? I tried to draw a rough picture and indicated the chiral carbons. I'm getting 8 stereoisomers are possible but my book gives the answer as 2. Can someone explain where I'm going wrong?

The comments on the question are incorrect [edit: the offending comments seem to be gone now], as is the answer given in the book. In general, answers printed at the back of a book should never be taken as the absolute truth. There are typos in all books. Even well-known and well-written textbooks have many errata (usually corrected in subsequent printings).

As you said, there are three chiral centres and therefore 8 stereoisomers.

There are no other elements of symmetry present in this molecule which render any of those 8 stereoisomers equivalent.


Difference between Excess property and property changes of mixing I have some confusion in my mind about the difference between an excess property and a property change of mixing. I know that the Excess property is defined as the difference between the actual value and those of an ideal mixture:
$$\ M^E(T,P,x_i)=M^{actual}-M^{id}(T,P,x_i)$$
On the other hand the property changes of a mixture is:
$$\ M^{mix}=M{actual}-\sum n_i*M_i$$
So the latter seems to me that the whole change of a mixing process is with respect to the pure, ideal gas, species.

Excess and mixing are the same for enthalpy and volume but are different for entropy and Gibbs free energy.

Now come the problem: property changes of mixing and excess properties are related each other by the residuals properties (deviations from ideal gas)?

Otherwise I do not understand why they are different.

Thanks a lot

The difference is subtle, however they are different as you suspect.

You can look at the property of mixing as the change in property resulting from mixing the pure components (with everything at the same temperature and pressure) together to form the mixture.

The excess property is deviation between the actual property and the property in an ideal mixture at the same temperature and pressure.

They are the same for volume and enthalpy because the volume of mixing and the enthalpy of mixing in an ideal solution are both 0. This is not the case for G or S.

In other words the difference arises because $M^{id}=\Sigma x_i M_i$ is not true for all properties. It is only true for enthalpy and volume. So the second term in your expression for the excess property above, cannot in general be written as $M^{id}=\Sigma x_i M_i$ (if it could, the excess property and mixture property would be the same). It should instead be $M^{id}=\Sigma x_i M_i + M_{mix}^{id}$.

The extra term for mixing the components would be be given by $$
\frac{G^{id}_{mix}}{RT} = \Sigma x_iln(x_i)
$$
and
$$
\frac{S^{id}_{mix}}{R} = -\Sigma x_iln(x_i)
$$
for Gibbs free energy and entropy, respectively.


How do I create a transparent liquid for ferrofluid display? Hello~ I am a media art artists working in Korea.
Recently, create a kinetic artwork using ferrofluid.
I need to create an artificial pond with a diameter of 3 meters. So I need a lot of suspension liquid.

Prototyping
https://youtu.be/3o33upltO34

simulation videos
https://youtu.be/NE19B_2byGA

Unfortunately, I tried my solution failed.
- case 1: Deionized water 40% / isopropyl alcohol 40% / Propylene Glycol 20% (Stains occur on the glass )
- case 2: Deionized water 60% / Coco betaine surfactants 40% ( no stain but, This seems strange movements)
- case 3: Deionized water 35% / isopropyl alcohol 35% / Coco betaine surfactants 30% (Stains the glass )
- case 4: Deionized water 60% / LES surfactants 40% ( no stain but, This seems strange movements.)
- case 5: Deionized water 60% / APG surfactants 40% ( no stain but, Decomposed into many particles)
- case 6: Windex(Glass Cleaner) 100%  ( no stain but, Decomposed into many particles)

I used glass types
- pyrex glass (Small stains occur)
- tempered glass (Small stains occur)
- soda-lime glass (Big stains occur)

I was always cleaning solution into a container before adding the alcohol.

And most it had hardened over time, the ferrofluid.

There's a really good video on YouTube by Nilered/Nileblue on ferrofluids. In it he suspends magnitite with oleic acid (main component of olive oil) as a surfactant and uses Kerosene (Paraffin Oil) as a carrier.
There's a PTFE 'dry lubricant' sold under the WD-40 brand name that might be able to prevent the glass from staining as it creates a thin coat of PTFE on the glass which is highly inert. Just spray the piece of glass with the PTFE Spray and wait for it to dry (~7mins). [The PTFE spray has a myriad of other uses too, you can put it on any surface you want to have low friction, to be lubricated, an unreactive surface, dust repellant surface, etc].
This might be a way around the staining issue you are getting. Or at least point you in the right direction. Good luck!


What is the difference between galvanization and sacrificial protection? If we use zinc in both methods, then the same process occurs: zinc corrodes in preference to iron, and eventually zinc has to be replaced. So what's the difference?

They are used in different media. Galvanization is used to protect iron exposed to non-corroding medium (air), while sacrificial electrodes are used to protect iron in electroconductive and corrosive medium (sea water or moist soil).

Protective zinc oxide layer on galvanized surface is desired to stay undisolved,  while the purpose of sacrificial electrode is to keep being oxidized to assure electron flow to other metal surface. In latter case medium electrconductivity is a must for ion transfer process, while corrosivness is needed for oxidized zinc surface layer disolution.


Open data in chemistry? I am a student in computer science, which is a field where sharing research data and source code is very very much recommended. I am looking for sites where data from chemistry experiments could be openly distributed. Do such exist? I would like to test some algorithms for analyzing certain kinds of experiments, without having to do them myself (which I easily can't). The motivation is that I think (.. or maybe hope) that I would have some insights to offer.

I did not find many suitable tags, sorry.

There are a variety of "open data" sources in chemistry. In your case, you probably want "open notebook science.".

There are a few practitioners in chemistry, although Jean-Claude Bradley sadly passed this last year.

JC Bradley
Andrew Lang
Open Source Malaria

I believe there will be a collection of articles published in the open access Journal of Cheminformatics as a memorial to Jean-Claude, but I don't think these have cleared review yet.


Why do bubbles form and stay when I shake a bottle of tea? I have some bottled tea and I noticed that shaking it creates some foamy bubbles that stay for quite a while (at least half an hour if not more). I originally thought it has something to do with my saliva, but then I realized that a brand new bottle of tea also does the same thing. The listed ingredients are water, oolong tea and vitamin C. What could be the reason?

Tea is may be considered as aqueous solution of many different substances extracted from solid tea leaves.  (It is not so much about «tea» as in tea, or a tisane.)
In this solution may be ingredients altering the surface tension of water sufficiently enough that air bubbles are trapped for a prolonged time.  It could be a component added to act as a preservative, which equally happens to act as a surfactant, or an ingredient intentionally added to ease formation and preserve these bubbles; like the tapioca pearls in  bubble tea.
Compounds acting as surfactants may be artificial (like CTAB) made for cleaning, or processes from natural sources, e.g., polysaccharides (like starch), or plain natural.  Washing clothes with leaves of climbing ivy, for example  works because they contain a small amount of saponins.  That a compound acts as a surfactant does not say if it is safe in food, or not.


How is solid titanium(II) oxide an electrical conductor? Most salts in the solid state are electrical insulators, since the ions are not mobile (e.g. sodium chloride). However, solid titanium(II) oxide, $\ce{TiO (s)}$, is a conductor. How is this so?

The electron configuration of $\ce{Ti^{2+}}$ is $\mathrm{[Ar]~3d^2~4s^0}$.

$\ce{TiO}$, like e.g. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$, adopts the rock salt ($\ce{NaCl}$) structure.

Due to the lower nuclear charge of $\ce{Ti}$ its 3d orbitals are less contracted than that of $\ce{Fe}$, $\ce{Co}$ and $\ce{Ni}$ which allows for overlapping of the 3d orbitals and the formation of a metal d band in $\ce{TiO}$. The d electrons of $\ce{Ti^{2+}}$ partially fill this band and lead to electrical conductivity. $\ce{FeO}$, $\ce{CoO}$ and $\ce{NiO}$ on the other hand are electrical insulators.

Note that in this case it is electrons that act as the mobile charge carriers, not the ions.


Assignment of the 13C NMR of 2-heptanone The 13C NMR of 2-heptanone is given below:

I understand the assignments of the carbons labelled 7, 6, 4 and 1.

My question is: Why is the shift of environment 5 higher than that of 3, despite being further away from the electron withdrawing carbonyl group?

For open chain alkanes, chemical shifts in 13C are dependent on substituents at the α, β and γ positions. α substituent effects are largely correlated to electronegativity. β substituents cause a shift of similar magnitude, but are not related to electronegativity. In fact a number of studies have shown an inverse relationship to electronegativity, but this is not consistent. The origins of the β effect are unclear. The downfield shift is larger for primary/secondary carbons than for tertiary/quaternary (hence the larger downfield shift for C-5 which has 2 β carbons). The γ effect is small and has a dependence on sterics of the bonding pathway.

The chemical shifts for straight chain alkanes can be calculated quite accurately using additive tables (such as in Pretsch, or used by Chemdraw etc). As a β substituent, an alkane -CH2- has a shift increment of +9.4, versus -C(O)- having an increment of +0.5. Hence your difference of about 8ppm for carbons 3 and 5.

From Chemdraw

Node     Shift    Base + Inc.   Comment (ppm rel. to TMS)
CH2     22.9         -2.3      aliphatic
                     18.2      2 alpha -C
                      0.5      1 beta -C(=O)-C
                      9.4      1 beta -C
                     -2.5      1 gamma -C
                     -0.4      general corrections
CH2     31.3         -2.3      aliphatic
                     18.2      2 alpha -C
                     18.8      2 beta -C
                     -2.7      1 gamma -C(=O)-C
                     -0.7      general corrections


Grignards and anhydrous ethanol I remember using both 95% ethanol and absolute ethanol quite a few times. You'd need absolute ethanol for a reaction that would fail if water was present, such as a Grignard reaction. I can't remember whether the absolute ethanol was in an ordinary bottle or the type with a rubber septum (where you keep the liquid under nitrogen to avoid contact with air). If the ethanol was to be used as a reagent, a septum bottle would be more appropriate.

http://boards.straightdope.com/sdmb/showpost.php?p=5314683&postcount=10

First, why would one use ethanol for a Grignard reaction? If a Grignard is destroyed by water (as suggested by the poster) then wouldn't a Grignard also be destroyed by ethanol?

I'm guessing that the poster just forgot all his chemistry. Ironic for someone posting on a message board that proclaims to be "fighting ignorance since 1973."

Second (assuming that one could somehow run a Grignard reaction in ethanol), do you really need anhydrous ethanol for Grignard reactions to work? Why not just use a slight excess of Grignard to account for the water present in 95% ethanol? Seems to me that anhydrous ethanol would be pretty expensive given that the usual distillation methods wouldn't get you from 95% ethanol to 100% ethanol.

Finally, is there a reaction in which using excess Grignard to account for acidic protons wouldn't chemically work in some way?

First, why would one use ethanol for a Grignard reaction? If a Grignard is destroyed by water (as suggested by the poster) then wouldn't a Grignard also be destroyed by ethanol?

As well as being excellent nucleophiles, Grignard reagents are also very strong bases because they essentially function as alkyl anions. Therefore they will react with ethanol to form magnesium ethoxide, the alkane, and magnesium bromide.

Grignard reactions are usually carried out in ether solvents such as diethylether and tetrahydrofuran because they are unreactive towards the reagent and they actually stabilize it by coordinating to the magnesium ion to form a tetrahedral complex.

(image source)


Ring Strain and C=O Stretching Frequency Why does $\ce{C=O}$ stretching frequency increase with ring strain?

Why does conjugation decrease $\ce{C=O}$ ring strain?

I am told that cyclopropanone exhibits a higher $\ce{C=O}$ stretching frequency than does cyclohexanone. I know that higher stretching frequencies are correlated with a shorter, stiffer, and stronger bond. What I can't see is why would the $\ce{C=O}$ bond have more of these characteristics in a strained ring system than in a less strained ring system.

Are there hyperconjugative effects going on?

Are there other rationalizations that we can make? I.e. would smaller $\ce{C-C-C}$ angles suggest higher $p$-character in the $\ce{C-C}$ bonds and therefore higher $s$-character in the $\ce{C=O}$ bond?

Why does C=O stretching frequency increase with ring strain?

The carbonyl carbon prefers to be (more or less) $\ce{sp^2}$ hybridized.  Imagine taking the two other bonds attached to the carbonyl carbon and bending them towards each other so that the R-C-R angle is forced from 120 to 90 degrees (the same thing as incorporating these bonds into a smaller and smaller ring).  As we do this, the hybridization of the orbitals the carbonyl carbon uses to form these bonds changes from $\ce{sp^2}$ to $\ce{p}$.  As these bonds use more p-character, more s-character remains for use in the C-O sigma bond.  This would strengthen the CO bond and raise the energy required to stretch it (increase its ir stretching frequency).

An alternate explanation can be made based on resonance structures.  Look back at the acylium ion resonance structure used in our earlier acid chloride discussion (resonance structure IV).  As we strain the carbonyl by placing it in a small ring, the bonds between the carbonyl carbon and the adjacent carbons become weaker.  This means that resonance structure IV, with a strong $\ce{C#O}$ triple bond, becomes more important.  Here, in the present case, we would say that resonance structure IV involves carbon-carbon hyperconjugation.

  Why does conjugation decrease C=O ring strain?

Do you mean, "why does conjugation to the carbonyl decrease the ir stretch frequency"?  Look at the resonance structure on the right side below for the conjugated ketone, cyclohexenone.  It's inclusion leads to a weakening of the carbonyl pi bond.


Would ABS plastic degrade in a dishwasher? I printed an object with a 3D printer and I used a roll of light green ABS plastic filament.

I would like to use the dishwasher to clean it. The temperature would be kept below 70°C, well below the glass transition temperature of about 110°C.

The dishwasher uses standard tabs, not one of the phosphate-free tabs.

What would be the effect of said cleaning on objects made of ABS plastic?

According to several application guides/compatibility charts for ABS plastic (like this one), ABS is fairly resistant to the conditions present in a dishwasher (mild to strong alkaline, salts, no organic solvents, mild temperatures).

Please note that while the ABS structural backbone may be attacked by nucleophilic agents at the nitrile carbon, and that the nitrile group may be hydrolised in acidic or basic environments, the conditions required are much harsher that a common dishwasher.

For example (p.8), glacial acetic acid was found to cause significant swelling, while 25% w/w sodium hydroxide and 25% w/v HCl barely caused changes.

So, while the concern for release of toxic additives present in the formulation of the original ABS filament may or not be significant, from the chemical point of view it is almost sure it won't degrade noticeably.

Also, anecdotally, I used to wash old LEGO pieces in a dishwasher (kids are dirty), and not degradation was observed at all.


Definition of a cluster Wikipedia defines a cluster as an ensemble of bound atoms or molecules that is intermediate in size between a molecule and a bulk solid. What is meant by a bulk solid? I can't find a definition anywhere? Further, is this a good enough definition?

The properties of a cluster depend on the exact number of constituents, and the geometry of the arrangement.

The properties of a bulk solid otoh are just indifferent to the exact size of the specimen (e.g. crystallite).

(In between you have nanocrystalline substances, where properties change with size, but only gradually.)

I would say that clusters are much like molecular compounds in that they have a distinct structure and absolute composition, however they are held together not by covalent bonds but by polar or dispersive forces (also metallic), and they usually consist of a number of identical particles.


What is the meaning of parentheses in a condensed molecular formula? In the formula $\ce{(CH3)3-C-CH(CH3)-CH2-CH3}$, what do the parenthesized portions represent? What is the molecular structure, and how do I name the compound?

The parentheses just have a grouping function: $\ce{(CH3)3-C -}$ represents a carbon atom to which 3 methyl groups are attached, $\ce{-CH(CH3)}-$ is a carbon atom in the chain to which one hydrogen atom and a methyl group are attached.

Draw the molecule, find the longest chain for the base name, determine the position of the substituents and you're done.


Effective rate constants I have the following reversible bimolecular reactions

$$\begin{align}
\ce{A_0 +B &<=>[$3k_+$][$k_-$] A_1} \\[5pt]
\ce{A_1 +B &<=>[$2k_+$][$2k_-$] A_2} \\
\ce{A_2 +B &<=>[$k_+$][$3k_-$] A_3} \\
\end{align}$$

that can be written as follows

$$\begin{align}
\ce{A_0 +3B &<=>[$k_1$][$k_2$] A_3} \\[5pt]
\end{align}$$

The initial conditions are $[\ce{A_1}]_0 = [\ce{A_2}]_0 = [\ce{A_3}]_0 = 0$. I know rate constants $k_+$ and $k_-$ and would like to write rate constants $k_1$ and $k_2$ in terms of $k_+$ and $k_-$.

It would be great if you have any ideas. I am quite new to the topic. So far, I can only think of using some fitting algorithms to find $k_1$ and $k_2$.
Thanks!

Finding the equilibrium constant of your overall reaction is straightforward, as you've done in the comments, and we have $$K_\text{tot} = K_1K_2K_3 = \left(\frac{3k_+}{k_-}\right)\left(\frac{2k_+}{2k_-}\right)\left(\frac{k_+}{3k_-}\right) = \frac{k_+^3}{k_-^3}.$$ But this is insufficient to conclude that $k_1 = k_+^3$ or that $k_2 = k_-^3$. Indeed, let us consider what is stated when we have, as in your last reaction, rate  constants for the overall reaction. Such a scenario presumes an overall rate law of the form $$\frac{\text{d}\ce{[A_3]}}{\text{d}t} = k_1\ce{[A_0][B]^3} - k_2\ce{[A_3]},$$ and there is no a priori reason to suggest that this is likely. (Often this is possible in certain limiting regimes, but not in general.) To investigate $k_1$ and $k_2$, we will have to solve our system fully. We can write out the rate equations for our system as
\begin{align*}
\frac{\text{d}\ce{[A_1]}}{\text{d}t} &= 3k_+\ce{[A_0][B]}-k_-\ce{[A_1]}-2k_+\ce{[A_1][B]}+2k_-\ce{[A_2]}\\
\frac{\text{d}\ce{[A_2]}}{\text{d}t} &= 2k_+\ce{[A_1][B]}-2k_-\ce{[A_2]}-k_+\ce{[A_2][B]}+3k_-\ce{[A_3]}\\
\frac{\text{d}\ce{[A_3]}}{\text{d}t} &= k_+\ce{[A_2][B]}-3k_-\ce{[A_3]},
\end{align*}
a coupled system of differential equations, which can be solved using matrix methods---diagonalize the coupling matrix, find the eigenvectors and eigenvalues, and solve---but these calculations are often quite messy, so I will leave them to you if you're interested.

Once $\ce{[A_2]}$ and $\ce{[A_3]}$ are known, we can plug them into the last differential equation and compare it with the presumed form of the overall rate law to determine $k_1$ and $k_2$.


Are all concerted electrophilic addition reactions of alkenes syn? The Organic Chemistry book I am using (8th edition Carey, et al.) doesn't say it straight out, but I am wondering if it's a universal rule that all concerted electrophilic additions of alkenes are syn.

In every concerted example I have seen so far in the textbook, the bonds that are being made wind up on the same side, but I don't know if it's a rule. For example: epoxidation, transition metal catalyzed hydrogenation, and hydroboration are all concerted and all syn.

If they are concerted, there is a very, very large probability of them being syn-specific. For explantion, let’s look at the π orbitals of ethene (scheme 1). (Remember that double bonds are planar and have two distinct sides when viewed across an entire molecule.)

Scheme 1: π and π* orbitals of ethene, viewed from the side.

Reactants have one of two possibilities, typically. They are either electrophilic and will thus attack the highest occupied molecular orbital (HOMO). Or they are nucleophilic and will attack the lowest unoccupied molecular orbital (LUMO). It just so happens that the π bond is the weakest bond in a typical alkene and thus the bonding molecular orbital is typically the HOMO and the antibonding one is the LUMO.

To attack this orbital, you somehow need to interact with both lobes. If you do so, you will either attack syn-stereospecificly, i.e. arrive from one side due to simplicity. Or you are a very large attacking molecule and are able to grab around the double bond, interacting with one atom on one side and with the other one on the other side.

Most reactants are not that large. Typically, there will be a single attacking atom (e.g. halogens, meta-chloroperbenzoic acid), two atoms bonded together directly (e.g. borane) or two atoms that are still in very close proximity (e.g. $\ce{OsO4}$ — two bonds — or a diene for a Diels-Alder reaction — three bonds). None of these even have the chance to reach across, especially if you consider that the largest of these, the diene, is in fact a large π system and inherently restricted to being planar (within very small possible deviations).

Of course, that does not make it impossible. If you use a large enough π system in a Diels-Alder like reaction, that may be able to wind itself downwards like a spiral staircase. I haven’t seen that type of reaction yet, though.


Mole Concept: how it applies to alloys I was recently asked to find the total molar mass of a gold and palladium alloy, given the total number of atoms in the compound.
The method was to simply convert the total number of atoms to the total number of moles by using Avogadro's number as the conversion factor.

But as I understand it,  Avogadro's number and his concept only works if all the atoms are the same.  If multiple types of atoms are present in a group of $6.022\times10^{23}$ of them,  how could the different masses add up to a mole?

One mole is a number, not a mass.

See the following figurative example:

Let's say we are dealing with a sample of pure chemistry books.

A book's weight is 1 kg.
One mol of books would be 6,022 × 10²³ books.
The mass of one mol of books (the molar mass of books) would be as follows:

1 kg (mass of one book) × 6,022 × 10²³ (# of books)

Now let's say we're dealing with a mixture (alloy) of 60% chemistry and 40% history books. What is the mass of 1 mol of this mixture?

A history book's weight is 2 kg

So, our entire collection of books, sum up to Avogadro's number = 6,022 × 10²³ units. And we know that it is 60% Chem. and 40% Hist. So the mass of the entire collection would be:

m = N × 0,6(1kg) × 0,4(2kg)

where N is the Avogadro's number.

Enough of analogies. You have the mass of the gold and palladium atoms from the periodic table. I am assuming you also have the proportion Au/Pd.

You also know that there's 1 mol of atoms. The exact same equation would apply:

m = N × P.M(Au) × P.M(Pd)

where N is the Avogadro's number, P and M are the percentages and atomic mass, respectively, of gold and palladium.


How to determine the bond order using the Lewis structure? What are the steps to follow in order to determine which molecule has a higher bond order by drawing the Lewis structure? For example, with $\ce{CO}$ and $\ce{CO3^{2-}}$?

Determining the bond order starting from a Lewis structure is a task that can range from very easy to rather difficult. Luckily, most cases you will encounter are the easy ones. The first step should always be to draw out your molecules. For carbon monoxide and carbonate, this is what you should initially arrive at:

The second step is to check if you are missing any resonance structures. This is most obvious in carbonate whose three oxygens are all equal and whose bonds can be pushed around as shown below to give these additional two resonance structures:

However, also carbon monoxide can be drawn in a different resonance structure. These two resonance structures are obviously not equal since one equips all atoms with an octet while the second leaves carbon with a sextet:

Therefore, in a third step, you must weight the resonance structures. As stated above, the three of carbonate are all equal, we must consider them contributing the same to the overall structure. Those of carbon monoxide, however, are not equal; the triple-bonded structure will contribute far more than the double-bonded structure.
Depending on the level of your exam, you may be expected to completely ignore the lesser resonance structure (introductory level) or to assume weightings for both structures (advanced level) — or, anything in-between. Here, let us assume that the double-bonded resonance structure contributes almost nothing to the final result so we may ignore it (which is close enough to the truth).
Having done all this preparative work, we are now able to take a look at the actual bond whose bond order we want to determine. For each resonance structure:

Count the number of electron pairs in one specific bond

Multiply that with a weighting fraction

Sum up the values obtained in this way
$\displaystyle \text{B. O.} = \sum_i \frac{n_i(\ce{e-})}2 \times x_i$

For carbon monoxide, of which we have a single non-negligible resonance structure, we have 3 electron pairs in the resonance structure which contribute 1 (or the entirety) to the final structure so our bond order is 3.
For carbonate, we have three resonance structures that we must consider separately and we must multiply the individual structure’s electron pair count with $\frac13$ because each contributes a third to the overall picture. If we consider the upwards-pointing $\ce{C\bond{...}O}$ bond of the second figure, from the first resonance structure we have a contribution of $\frac23$, from the second we have $\frac13$ and from the third we also have $\frac13$. Adding these three values we have an overall bond order of $\mathbf{\frac43}$.

Finally, although this is already well into the advanced level I wish to point out why the weighting step is important. Consider the structure of a carboxylic ester as shown below. At first sight, this might look like ‘half a carbonate’, which would mean that both shown resonance structures are equal, contribute $0.5$ to the overall picture and therefore both bond orders would be $1.5$. However, this is only the case if we are talking about the carboxylate anion, not about the ester. In the ester’s case, as can be seen below, one of the resonance structures includes charge separation. Therefore, we must weight the non-charge separated resonance structure stronger than the charge separated one. We might choose weighting factors of $0.75$ and $0.25$ (without any additional data the choice of factors is entirely arbitrary!). This leads us to two different bond orders for the two $\ce{C-O}$ bonds, one being something like $1.75$ the other corresponding to $1.25$.

It is important to note that in the ester’s case we cannot ignore the contribution of the second, minor resonance structure since comparing the reactivity of an ester to that of a ketone shows a clear difference (the ester being less reactive, i.e. the double bond not being a full double bond). Likewise, replacing an ester e.g. with a thioester or selenoester further reduces the contribution of the second resonance structure closer to where it may safely be ignored again.


Find the partial pressure in a container I have an air mass which consists of 78.1% $\ce{N_2}$, 20.9% $\ce{O_2}$, 0.964% $\ce{Ar}$, and 0.0378% $\ce{CO_2}$ in a container which volume $\mathrm{V=40~L}$ at a temperature $\mathrm {T = 298.15~K}$.

The total pressure is $\mathrm{p_{total} = 1.5~atm}$.

Now I am removing all the $\ce{O_2}$ in the container. What will the partial pressure of $\ce{N_2}$ be?

I know the formula $\mathrm{pV = nRT}$.

Before I remove all the $\ce{O_2}$ I can calculate the amount of substance $$\mathrm{{n_{total} = \frac{pV}{RT} = \frac{1.5 \cdot 40}{0.082 \cdot 298.15}~mol = 2.45~mol}}$$.

The molar mass of $\ce{N_2}$ and $\ce{O_2}$ is $\ce{M(N_2)} = \mathrm{28~g\ mol^{-1}}$ and $\ce{M(O_2)} = \mathrm{32~g\ mol^{-1}}$.

If I remove all the $O_2$ the new mass/volume percentage of $\ce{N_2}$ will be

$$\mathrm{m_\%(N_2) = \frac{78.1}{78.1+0.964+0.0378} = 98.7 \%}$$.
So I thought I could find the amount of substance of $\ce{N_2}$  by saying
$$\ce{n(N_2)} = \mathrm{\frac{98.7~g}{28~g/mol} = 3.525~mol}$$ but this will exceed the total amount of substance before I removed $\ce{O_2}$ so this makes no sense.

When I have found the amount of substance of $\ce{N_2}$ I think I can just replace $\ce{n}$ in $\mathrm{p = \frac{nRT}{V}}$ and get the partial pressure of $\ce{N_2}$.

Maybe I have the right percentage of $\ce{N_2}$ but I am calculating the amount of substance wrong when it's a gas? Could the correct answer be $\mathrm{98.7 \% \cdot n_{total} = 2.422~mol}$ and then inserting this in $\mathrm{p = \frac{nRT}{V}}$ which gives me around $\ce{p(N_2)} = \mathrm{1.5~atm}$?

Since we're assuming these are all ideal gases, we can use Dalton's Law of Partial Pressures to solve this fairly easily; you never need to find moles of anything.  Unless something is strange, the percentages given are by volume or by pressure rather than by mass.  If they are intended to be by mass, what follows would need to be modified.  I suspect that the use of the words "air mass" in the problem are what caused the confusion.

The total pressure, 1.5atm, is equal to the sum of the partial pressures, each of which is proportional to its mole fraction in the container.  Because these are gases that have the same volume, 40L, their mole fractions are their percent of the total amount of gas.

The nitrogen, which is 78.1% of the total, has a partial pressure of:

$ 1.5 \cdot 0.781 = 1.1715\ atm$

That's true whether the oxygen is in the container or not.  To see that this is the case, we can remove the oxygen and do the same calculation.

If we remove 20.9% of the gas (in the form of oxygen) we are left with 81.1% of the original 1.5 atm remaining.

$1.5\cdot 0.811 = 1.1865\ atm$

We can determine the mole fraction of nitrogen in this new mixture of gases by adding up the remaining percentages and finding nitrogen's contribution to it:

$78.1 + 0.964 + 0.0378 =  79.1018\ units\ of\ gas\ total$

$\frac{78.1}{79.1018} \cdot 100\% = 98.73\%\ N_2\ after\ oxygen\ is\ removed$

As before, Dalton's Law of Partial Pressures tells us that the pressure of nitrogen will be its mole fraction times the total pressure, or 98.73% of 1.1865 atm.

$1.1865 \cdot 0.9873 = 1.1714\ atm$


Why are powder XRD axes sometimes plotted in reverse? Just plotting some powder XRD data and I wanted to know why the axes in some of the literature that I'm looking at is plotted in reverse. Is there a reason it starts at 60 Degrees (2 theta) and then goes down toward zero?

Best reason I can think of is that you're being given data in momentum space rather than position space.

You can convert from "position-time" space, which is what we're comfortable with as humans, and describe the system in "momentum-energy" space, which is simpler from a physics point of view. Position is the reciprocal of momentum and energy is the reciprocal of time. If you convert the diffraction equations to momentum ($k$), which is proportional to $\frac{1}{sin(\theta)}$, then the diffraction equation $\lambda = 2dsin(\theta)$ is transposed to $k' = k + K$ where $K$ describes the peak in momentum space, $k$ describes the momentum of the incoming photon, and $k'$ describes the diffracted photon. This equation is linear and much easier to manipulate than the "real-space" form of $\lambda = 2dsin(\theta)$.

The equation in "momentum" space (k-space) is difficult to understand but is nice and linear from a physics standpoint.

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


How do you explain the thread sticking to an ice-cube when salt is sprinkled over it? So this experiment goes like this :

You take an ice-cube
You put a thread on top face of it
You sprinkle some common salt on the ice-cube wherever its touching the thread
You wait for sometime (approx 1 min)
Voila! now you can lift the thread and ice-cube will stick to the end of the thread so you can lift the whole thing altogether

How do you explain this in chemistry?

I conjecture that there are several things happening.

Step 3:  Addition of salt will melt the ice at its immediate point of contact. The string will naturally be surrounded by the highly concentrated brine (salt water).
Step 4:  This is enough time for the brine to get diluted, noting that there is an infinite reservoir of potential water by way of the ice cube.
Step 5:  Once the salt gets sufficiently diluted, then the comparatively large ice cube provides the necessary cold for the (now) very dilute salt water to reform as ice.

Recalling that in Step 3 the salt water surrounded the string, then ice naturally forms around the string in Step 5, hence you can pick up the ice cube with the string.


What is the oxidation state of copper and iron in chalcopyrite? What is the oxidation state of Cu and Fe in chalcopyrite ($\ce{CuFeS2}$)?
Some websites tell that it is $\ce{Cu(I)Fe(III)S2}$, and others tell that it is $\ce{Cu(II)Fe(II)S2}$. Which is right?

A number of papers report that the formal valency states of chalcopyrite are best considered as $\ce{Cu+Fe^3+S2}$, based on various computational and spectroscopic evidence.

References:

Klekovkina, V. V.; Gainov, R. R.; Vagizov, F. G.; Dooglav, A. V.; Golovanevskiy, V. A.; Pen’kov, I. N. Oxidation and magnetic states of chalcopyrite CuFeS2: A first principles calculation. Opt. Spectrosc. 2014, 116 (6), 885–888. DOI: 10.1134/S0030400X14060149.
Pearce, C.; Pattrick, R.; Vaughan, D.; Henderson, C.; van der laan, G. Copper oxidation state in chalcopyrite: Mixed Cu d9 and d10 characteristics. Geochim. Cosmochim. Acta 2006, 70 (18), 4635–4642. DOI: 10.1016/j.gca.2006.05.017.
Goh, S. W.; Buckley, A. N.; Lamb, R. N.; Rosenberg, R. A.; Moran, D. The oxidation states of copper and iron in mineral sulfides, and the oxides formed on initial exposure of chalcopyrite and bornite to air. Geochim. Cosmochim. Acta 2006, 70 (9), 2210–2228. DOI: 10.1016/j.gca.2006.02.007.


Relation between acidity and reactivity? Up till high school, I've been taught that reactivity and acidity of a substance in a relation that if a substance is acidic and if it's acidity is higher (comparatively to some other substance) than its reactivity will also be higher. Examples were mostly given of $\ce{HCl}$ and $\ce{H2SO4}$ stating that since $\ce{H2SO4}$ provides more hydrogen ions when ionized hence its reactivity is much higher than $\ce{HCl}$.

However today I stumbled onto a contradiction of this theory. It was stated in a website that the acidity of phenol is much greater than an average alcohol due to a greater stability of conjugate base.
Some links ahead it was also stated that the reaction of acyl chloride is less vigorous with phenol as compared to other alcohols due to reactivity of $\ce{-OH}$ group being stabilized by delocalization.

My question is, why does the theory "Acidity is proportional to reactivity" not applies for reaction of acyl chlorides with phenols and alcohols?

The issue is that reactivity is not an inherent property of a compound. It can be reactive towards specific types of reactions and completely inert to others. Acyl chloride reacting with phenol doesn't rely on phenol behaving as an acid, the type of reaction for which you would describe it as more reactive.


The intuition behind carbon's covalent bonds As an introduction to hybridization, my textbook discusses that "The formation of covalent bonds often starts with the excitation of the atoms." Taking carbon as an example, the author says:

  Carbon  forms four covalent bonds. Yet if we consider the elctron configuration in the carbon atom, we would not predict this as it has only two singly occupied orbitals available for bonding. ($1s^2\, 2s^2\, 2p_x^1\, 2p_y^1$)

I don't understand why carbon's bonding is counterintuitive. Why don't the 4 added electrons just go into the $p_x$ and $p_y$ orbitals then into some newly formed $p_z$ orbital? Why does the existence of "only two singly occupied orbitals available for bonding" complicate things?

As an extension, why is the excitation of the s-orbital electron to $p_z$ needed?

Thank you for your answers and please try to respond keeping in mind that I am a high schooler (more specifically, part of the IB Diploma program).

The most naive answer would guess that since every orbital wants two electrons and there are two orbitals with only 1 electron in each, then 2 more electrons are needed.

However carbon typically combines the one $2s$ orbital and three $2p$ orbitals to form four equivalent $2sp^3$ orbitals each with one electron. The four hybrid orbitals have a tetrahedral shape. So after hybridization, the carbon atom wants to create four bonds to fill the four electron shell vacancies.


How real are bond dipoles? In general chemistry, it is common to teach students to determine a molecular dipole by having them first determine "bond dipoles" which are just based on electronegativity. Then, by adding up these vectors, one determines the direction of the overall dipole.

In quantum mechanics, however, the molecular dipole is found by calculating the expectaion value $\langle\psi|\hat{\mu}|\psi\rangle$. This can be decomposed into $x$, $y$, and $z$ components if one so desires. Regardless, the actually quite useful idea of a bond dipole is nowhere to be found.

I assume that there is not a rigorous way of defining a bond dipole so that a sum of these bond dipoles equals the expectation value above. This is because there is not a well-defined way to partition electron density which is exact. I suppose one could do this using the theory of atoms in molecules.

Nonetheless, other ideas like local mode vibrations are approximate in the sense that a local mode is not an eigenfunction of the polyatomic vibrational Hamiltonian, yet local modes are still very useful and even more accurate in some contexts.

The same is sort of true of bond dipoles in that if one imagines a very long molecule which is polar at one end, a nearby molecule which moves along this large molecule will experience a changing electric field because the dipole of the molecule is only a true point-dipole when the molecule is very far away. So, the field experienced by nearby molecules is more akin to the sum of bond dipoles in the region nearby.

So, is there a rigorous way to determine bond dipoles from first principles? Is it as simple as projecting the total dipole onto a bond axis? What is the form of this projector? Also, in cases where the partitioning of atoms is well-defined, such as the theory of atoms in molecules, are bond dipoles well-defined and do they relate to the total dipole as we expect?

Answers to any of these questions would be welcome.

To be more clear, I am talking about the bond dipoles described on this wikipedia page. They do not provide any formal way of actually calculating the bond dipoles. They mention one can get these dipoles after calculating the total dipole, but I am not sure of the uniqueness of this under unitary transformations.

If by "bond dipole moment" you mean "electric dipole moment", then the electric dipole moment is the first term in a multipole expansion of the molecular charge distribution. In that respect, an electric dipole moment is an approximation.

For a static charge distribution, we are interested in the resulting electric fields or potentials. For a general charge distribution, this does not have a closed form expression, but can be increasingly approximated by higher order terms in a multipole expansion. The dipole approximation is good enough for many cases. Depending on the experimental setup, the only surviving term might be the dipole term. So in that sense it maps to an experimental value and can be considered "real".