Is this mechanism for the formation of a tosylate correct? I've been given this mechanism by my lecturer for the formation of a tosylate. It claims that NEt3 is needed to remove the hydrogen from the O+ in the second step. However surely the Cl- from TsCl would remove the H and no NEt3 is needed?

No, $\ce{Cl-}$ will not remove the $\ce{H+}$ because it is not basic enough. The $\mathrm{p}K_\mathrm{a}$ of $\ce{Cl-}$'s conjugate acid ($\ce{HCl}$) is $-7$. The $\mathrm{p}K_\mathrm{a}$ of $\ce{NEt3}$'s conjugate acid ($\ce{HNEt3+}$) is $10.75$ (source). This means that $\ce{HCl}$ almost completely dissociates in solution, and is very unlikely to form $\ce{HCl}$ molecules. $\ce{NEt3}$, on the other hand, is much more basic and will easily pick up the $\ce{H+}$.

Also the mechanism is incorrect. A lot of the arrows are wrong. See here (pyridine serves the same function as $\ce{NEt3}$):


How does intramolecular hydrogen bonding cause the molecules to be separated from each other? I learnt about intramolecular hydrogen bonding today, which occurs between molecules such as ortho-nitrophenol. What I was told is that in case of intramolecular bonding, the molecules separate from each other, opposite of what happens during intermolecular hydrogen bonding.

I don't understand why is this. Why would attraction within a molecule cause separation of the molecules? There is obviously some intermolecular hydrogen bonding present in this case, but I understand that this might be negligible. Still, why should the molecules spread away from each other?

Molecules with inter-molecular hydrogen bonds tend to associate with one another, while molecules with intra-molecular hydrogen bonds tend to associate with themselves.

Ortho- and para-nitrophenol are often used as examples of molecules with intra- and inter-molecular hydrogen bonds respectively (see picture).

Because of their molecular structures the ortho isomer can form an intramolecular hydrogen bond with itself, whereas the para isomer can only hydrogen bond with another molecule of p-nitophenol.  One would say that the para isomer associates with other p-nitrophenol molecules through hydrogen bonding, whereas the ortho isomer associates with itself through hydrogen bonding.  Because the para isomer associates with other molecules of p-nitrophenol through hydrogen bonding, it takes more energy to break up this association and cause the molecule to melt or boil.  On the other hand, the ortho isomer melts and boils at lower temperatures because there are fewer intermolecular forces (such as the hydrogen bonding) holding the ortho molecules together.


Open flame combustion in bizarro world Humans typically achieve combustion by starting a chemical reaction of a fuel (often a carbon $\ce{C}$ containing fluid like methane $\ce{CH4}$), which is is brought into an an oxidizer (a gas containing $\ce{O}$, like air). For example, when you turn on a lighter.

What would happen if the situation were turned around?
We have an open space entirely filled with fuel and we let in some oxygen-containing fluid. How would the combustion look?

I know the (adiabatic?) temperature maximum of a flame will be achieved at conditions which are slightly on the fat side. But I have no idea how, for example, the production of unburned $\ce{C}$'s, soot production and so on, would look.

tl;dr: You are likely to see more soot, $\ce{CO}$, $\ce{H2}$, less of $\ce{CO2}$ and same amount of $\ce{H2O}$.

There are different type of combustion that you can talk about. For example when you mix fuel and oxidizer before burning them, it is referred to as premixed system. For example a gas cooking range, gas turbine engines, IC engines, etc. On the other hand if you do mix the fuel and oxidizer you get what's called a diffusion system. For example a candle flame, flame thrower, apartment fires, etc. Typically premixed systems will produce a blue flame, whereas diffusion systems will produce yellow-orangish flame.

It is important to distinguish between a premixed and diffusion flame system because they will behave differently. A lighter produces a diffusion flame. For this answer we will focus on diffusion flame.

An important characteristic of a diffusion flame is that it burns where there is just the right amount of fuel and oxidizer. That means irrespective of whether there is fuel entering oxidizer or oxidizer entering fuel, the flame will only be present where there is just the right amount of fuel and oxidizer. So for example for a methane-oxygen system at the flame we get, $$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$

So in both the cases we would expect to see same amount of $\ce{CO2}$ and $\ce{H2O}$ begin produced when same amount of fuel is burnt. That doesn't sound interesting at all. Plus our ideal situation ignored the effect of heat produced and its effect on the surrounding.
What happens when we take this in account?

Soot

Well in most methane systems it is very difficult to produce soot, so let us consider something like propane $(\ce{C3H8})$ or higher order hydrocarbon (anything with more carbon atoms), which is more likely to soot. Soot is produced when unburned $\ce{C}$-atoms combine to form larger molecules, which combine to form particles. How soot is produced is still an active research area. But we do know that most soot is produced at relatively low temperatures (compared to the flame temperature) in fuel rich regions.

In a candle/lighter flame the soot is produced near the wick, but most of it is oxidized as it reaches the flame. Only some close to the center escapes from the tip because there is not enough oxidizer there to burn it all. This is the main reason diffusion flame appear yellow.

Now imagine your lighter which starts off with oxidizer in the center, there won't be any soot particle inside the flame. So the flame will appear bluish. But the heat produced from the flame may be enough to break (pyrolyse) the fuel in the surrounding region. This will cause soot to be produced. Also because there is no reason for the soot to approach the flame it will not get oxidized easily and you will be left with a lot more soot.

Unburned fuel

Next consider other unburned combustion products. When you burn methane in oxygen you produce more than just carbon dioxide and water. The unbalanced reaction can written as:
$$\ce{CH4 + O2 -> CO2 + H2O + CO + O2 + H2 + C + ...}$$

The unburned $\ce{C}$ forms the soot. A side effect is production of $\ce{H2}$. So now not only will you see more soot, there will be excess hydrogen too. Also in case of higher hydrocarbon fuels you might see some pyrolyzed fuel which is yet to form soot or get oxidized fully.

In a premixed flame system when you have lot of fuel lying around you are more likely to produce $\ce{CO}$ rather than $\ce{CO2}$. A simple reason being you first need to produce $\ce{CO}$ before you get $\ce{CO2}$. A side effect of this is that the temperature will be lower. This is because most heat is released when $\ce{CO}$ is oxidized to $\ce{CO2}$.
So what happens in a diffusion flame system?

As mentioned earlier in diffusion flame systems, the flame is around where there is just the right amount of fuel and oxidizer. The products formed at the flame diffuse out (into the fuel and oxidizer) and since there is scarcity of oxygen, $\ce{CO2}$ will get reduced to $\ce{CO}$ i.e. you will see increase in $\ce{CO}$ production and decrease in amount of $\ce{CO2}$.

Side note: Saturn's moon Titan has an atmosphere which primarily consists of nitrogen (98.4%). The next abundant gas is methane (1.4%) followed by hydrogen (0.1-0.2%). So it's not a bizarre world after all.


Dehydration of methylcyclobutanol In one of the exams I saw it was written that dehydration of 1-methyl-cyclobutanol yields cyclobutylmethylene (double bond outside the ring) which is the anti Zaitzev product, instead of methylcyclobutene.
Why?
Thanks!

tl;dr
The thermodynamic product of the dehydration reaction of 1-methylcyclobutan-1-ol is 1-methylcyclobut-1-ene, not methylenecyclobutane. This is supported by [1],[3] and my own calculations.

Some details
I was unable to find any reference to why the dehydration of 1-methylcyclobutan-1-ol should yield methylenecyclobutane (3) as the main product. From what I could gather up until now, the major product is 1-methylcyclobut-1-ene (1). I also tried finding suitable mechanisms and/or protonation (hydration) energies for these two molecules according to the following scheme, to compare their relative thermodynamic stability to each other.

Wiberg and Hao report the enthalpy difference between 3 and 1 to be only $\Delta H = -0.54~\mathrm{kcal/mol}$. They obtain this value from equilibrating 1 and 3 over sodium on alumina at various temperatures.[1]  This would translate to a ratio of 1:3 of about 1.0 : 0.4.[2]

Lajunen and Himottu quite similarly report, that 1 is more stable than 3 by about $4.3~\mathrm{kJ/mol} \approx 1~\mathrm{kcal/mol}$. They obtained this value via equilibrium constants measurements in perchloric acid.[3]  This would in return translate to a ratio of 1 : 3 of about 1.0 : 0.2. It is also reported, that the transition states of protonation are quite similar, but lower for 1.

I also conducted some quantum chemical calculations, using DF-BP86/def2-SVP and the more sophisticated MP2/def2-TZVPP. These are in reasonable agreement with the previous findings. The enthalpy differences, approximated at $T=298.15~\mathrm{K}$ and $p=1~\mathrm{atm}$, are $1.4~\mathrm{kcal/mol}$ and $1.3~\mathrm{kcal/mol}$ respectively. This gives a 1:3 ratio of about 1.0 : 0.1. (More values, as well as the proton affinities (PA),[4] yielding 2, are given in the table at the end.)

It is not surprising, that these structures are quite similar in energy. They both contain a four membered ring. In 1 two $\ce{sp^2}$ centres are present in the ring, forcing planarity. As a result of the shorter double bond, two angles are slightly larger than $90^\circ$, while two are forced smaller.

In 3, only one $\ce{sp^2}$ centre is present. However, the dihedral angle is a mere $18.4^\circ$, the out-of-plane angle is $26.8^\circ$, respectively. This is apparently not enough to significantly reduce ring strain. Because of symmetry constraints, i.e. $C_\mathrm{s}$, this arrangement forces three inner ring angles to be larger than $90^\circ$.

The concept of ring-strain, or strain in general, is not well defined and it cannot be measured directly. There are relative methods, to obtain this virtual quantity, sometimes involving hypothetical molecules. There are different reference systems and as far as I am aware none of them are considered the ultimative choice. In this sense, any arguments can be made. It is therefore also almost impossible to derive ring strain from just a pen-and-paper exercise, especially when the molecules are that similar.

References

K. B. Wiberg and S. Hao, J. Org. Chem. 1991, 56, 5108-5110.
How the ratio is determined
M. Lajunen and M. Himottu, Acta Chem. Scand. 1987, 41A, 252-258.
How proton affinity is defined

Appendix

\begin{array}{llrr}\hline
  & & \mathsf{\pmb1} & \mathsf{\pmb3} \\\hline
\text{DF-BP86/def2-SVP} & \Delta E_\mathrm{el}    & 0.0   & 1.3   \\
\text{DF-BP86/def2-SVP} & \Delta E_\mathrm{o}     & 0.0   & 1.6   \\
\text{DF-BP86/def2-SVP} & \Delta H                & 0.0   & 1.4   \\
\text{DF-BP86/def2-SVP} & \Delta G                & 0.0   & 1.6   \\
\text{DF-BP86/def2-SVP} & \text{Ratio} (\Delta H) & 1.0   & 0.1   \\
\text{DF-BP86/def2-SVP} & \text{PA}               & 200.2 & 198.8 \\\hline
\text{MP2/def2-TZVPP}   & \Delta E_\mathrm{el}    & 0.0   & 1.1   \\
\text{MP2/def2-TZVPP}   & \Delta E_\mathrm{o}     & 0.0   & 1.5   \\
\text{MP2/def2-TZVPP}   & \Delta H                & 0.0   & 1.3   \\
\text{MP2/def2-TZVPP}   & \Delta G                & 0.0   & 1.7   \\
\text{MP2/def2-TZVPP}   & \text{Ratio} (\Delta H) & 1.0   & 0.1   \\
\text{MP2/def2-TZVPP}   & \text{PA}               & 399.3 & 398.0 \\\hline
\end{array}


When temperature is decreased, why do reactions occur at all? I admit that my knowledge of collision theory may be lacking, but, as I understand it, when particles collide, a reaction will not occur without overcoming the activation energy.

That being said, as the temperature of the environment in which the collision takes place is decreased, I believe it is logical that the kinetic energy of these particles will also decrease. Hence, to me, it would make sense if none of these particles had the energy required to react.

So, my question is: Why do reactions still occur when the environment's temperature around a collision decreases? That is, shouldn't there be a point (such as the temperature in a freezer, perhaps?) in which the activation energy cannot be overcome at all?

The activation energy varies a lot for various reactions. As long as T > 0 K, there is still disordered kinetic energy in the molecules. But the reason gasoline doesn't spontaneously combust, is that the temperature isn't high enough to overcome the activation energy - a spark is needed. So you are onto something, just remember that the activation energy varies with each reaction.


Why isn't ΔH = TΔS in ΔG = ΔH - TΔS? Applying a Legendre transformation on $U = q - w$ we get the familiar $G = H - TS$.
Making an innocent approximation delivers $\Delta G = \Delta H - T\,\Delta S$.

When one wants to predict the spontaneity of a chemical reaction, one gets the $\Delta H$ and the $\Delta S$ for the reaction in question from tables, and both refer to the system undergoing the transformation. Therefore we can write:

$\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - T\,\Delta S_\mathrm{sys}$

In which:

$\Delta G$: Variation of the system's Gibbs energy

$\Delta H$: Variation of the system's enthalpy

$\Delta S$: Variation of the system's entropy

However: $\Delta S_\mathrm{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

So: $\Delta G_\mathrm{sys} = \Delta H_\mathrm{sys} - q$. If we are making predictions based on Gibbs energy, we are under constant pressure and so $\Delta H_\mathrm{sys}=q$.

Therefore: $\Delta G_\mathrm{sys} = q - q$, $\Delta G_\mathrm{sys} = 0$ and we are always on equilibria.

Where am I mistaken? I've already seen some say that some stuff refers to the surroundings, but that makes no sense to me. If both are at thermal equilibrium, $q$ is going to have the same effect on the entropy of the system and that of the surroundings, from $\Delta S = \frac{q}{T}$.

I'm a chemistry undergratuate and have seen no mention to anything related to this on common p-chem textbooks (Atkins, McQuarrie, Ball).

However: $\Delta S_{sys} = \frac{q}{T}$, assuming the process is reversible (as usual).

No. $\Delta S_{env} = \frac{q}{T}$, assuming the process is reversible (as usual). So you get $\Delta S_{universe} = -\frac{\Delta G}{T}$ for a process where the heat exchange with the environment is reversible (and there is no non-PV work).

For more background, see https://chemistry.stackexchange.com/a/114323


Why can unsymmetrical tert-butyl ethers be prepared by dehydration of the constituent alcohols? Practice problem 11.11 what stops R-O-R from forming at the same rate as the desired product?

In order to understand why this is OK, you have to understand the mechanism by which this occurs. It is essentially an $\mathrm{S_N1}$ reaction that is most favoured by the conditions (acidic + polar protic solvent).

So, for the formation of an ether $\ce{^tBu-O-R}$ where $\ce{R}$ is a primary alkyl group:

competing formation of $\ce{R-O-R}$ is slow because primary carbocations are very unstable, and
competing formation of $\ce{^tBu-O-^tBu}$ is slow because $\ce{^tBuOH}$ is a poor nucleophile (tertiary alcohols are very bulky).


Why are transition metals said to have different valencies? I understand that the term valency refers to the number of bonds an element would be able to form with either chlorine or hydrogen. If this is directly related to the number of valence electrons an atom has, why do transition metals have variable valencies?

I think it's kind of a poor use of terms here. When talking about transition metals, the valency generally refers to the oxidation state of the metal atom. E.g: hexavalent chromium is $\ce{Cr^6+}$, not necessarily a chromium bonded to six other things.


Why is ionization potential considered a periodic property? Why is ionization potential considered a periodic property???
A periodic property is a one which appears at regular property
But we see that every element has some kind of ionization potential

The ionization energy is described as the energy required to remove an electron from an atom in the gaseous phase. Apart from some exceptions, it is possible to identify a general trend in ionization energies within the periodic table: moving from left to right in a period the ionization energy increases, whereas it decreases from top to bottom in a group.


What happens when an alkyne reacts with ammoniacal Cu2Cl2? My text book provided me with 2 reactions concerning alkynes (Specifically Terminal Alkynes).
The first one was the dimerisation of an alkyne on passing it through ammoniacal $\ce{Cu2Cl2}$.
All was fine until I encountered the next reaction.
It was a test to distinguish between terminal alkynes and other alkene and alkynes.
It involved the fact that treating a terminal alkyne with  ammoniacal $\ce{Cu2Cl2}$ would result in the alkyne forming a red precipitate as copper replaces hydrogen of one of the terminal hydrogen.
However now I was confused. Which of these reactions actually happens? Or are there specific conditions for each of these reactions that my text missed out?
Please help me in this aspect.

Terminal alkynes (including acetylene) form a $\color{red}{\text{red precipitate}}$ of copper acetylides when it is passed through ammoniacal cuprous chloride solution:
$$\ce{R-C#C-H + CuCl + NH3 -> R-C#C-Cu + NH4Cl}$$
However, in the presence of oxygen, this reaction proceeds further to dimerize the corresponding alkyne. This reaction, the oldest homo-coupling of terminal alkyne reaction,  was first discovered by Carl Andreas Glaser in 1869 (Ref.1, 2), hence the reaction is named Glaser-coupling after him:
$$\ce{2R-C#C-Cu ->[O2][NH3/EtOH] R-C#C-C#C-R }$$
The actual Glaser's reaction was (ref.2):
$$\ce{2Ph-C#C-H ->[CuCl/O2][NH3/EtOH] Ph-C#C-C#C-Ph}$$
References:

Carl Glaser, "Beiträge zur Kenntniss des Acetenylbenzols," Berichte der deutschen chemischen Gesellschaft 1869, 2(1), 422–424 (https://doi.org/10.1002/cber.186900201183).
Chao Liu, "Chapter 1: Introduction," In Lecture Notes in Chemistry 102: Transition Metal Catalyzed Oxidative Cross-Coupling Reactions;  Aiwen Lei, Editor; Springer-Verlag GmbH DE (Part of Springer Nature): Berlin, Germany, 2019, pp. 1-10 (ISBN: 978-3-662-58102-5).


How large is the energy difference between keto- and enol-form of guanine and uracil? The nucleobases in DNA and RNA are generally present in the keto-form, and not in the enol-form. As an interesting historical note, James Watson and Francis Crick did initially believe them to adopt the enol-form, which isn't compatible with the correct model of the DNA they later proposed.

But there are some reports indicating that in certain RNA structures guanosine and uridine form essentially a Watson-Crick basepair where one of the two bases is present in the enol-form, instead of the usual G-U wobble base pair.

Is there anything known on how large the energy difference between the keto- and the enol-form of those bases actually is?

I am not an expert in the field, but a quick literature check turns up a good amount of papers on the topic. In particular, I found this paper [1] which nicely answers the question for guanine:

So, in the gas phase, the most stable enol tautomer has a 0.9 kcal/mol free energy difference with the most stable keto tautomer. That free energy difference is very exacerbated in aqueous solution, with now a ∆G° = 8.7 kcal/mol. We also notice that the most stable keto tautomer is not the same in the gas phase and in solution, and that both keto and enol have many tautomers close in free energy, showing the limits of the simple keto vs. enol line of thought.

Regarding uracil, the first reference that comes up in a bibliographic search is this paper [2]. I only quote the abstract:

  The effects of the solvent on the tautomeric equilibria of cytosine and uracil are studied using Onsager's reaction field model in the framework of density functional theory. […] Our results are in good agreement with available experimental results and confirm that the polarization of the solute by the continuum has important effects on the absolute and relative solvation energies.

pKa Values of Guanine in Water: Density Functional Theory Calculations Combined with Poisson-Boltzmann Continuum-Solvation Model, Y. H. Jang et al., J. Phys. Chem. B 2003, DOI: 10.1021/jp020774x
Solvent effects in density functional calculations of uracil and cytosine tautomerism, L. Paglieri et al, Quantum Chem., 2004, DOI: 10.1002/qua.560560517


How to interpret this formula about hydrogen bond energy I want to evaluate some chemistry related formulas, which I don't understand.
In proteins, hydrogen bonding often occurs between the oxo group = O oxygen of one amino acid and the α-amino group (N − H) of another amino acid.
The bond energy is determined by the distance between oxygen and hydrogen and by the angle enclosed by the N-H and H-Bonds.
Let's assume the $E = (\frac{1}{d_{HO}^{10}} - \frac{1}{d_{HO}^{12}}) * cos(\alpha)$ model describes the energy of the hydrogen bond.
What is the meaning of the number 10 and 12 in the formula?

I know the angle and the distances between one O, one H and one N atom (that's all the input we got).
maybe related:
https://www.sciencedirect.com/topics/biochemistry-genetics-and-molecular-biology/hydrogen-bond

These exponents $10$ and $12$ are a little bit arbitrary. They have been chosen so as to obtain an energy curve corresponding to the measured H-bond energies. Other attempts have been published, like the famous Morse curve, which is, for covalent bonds versus HO distance $d$ :  $$E(d) = D[1 - e^{-k(d - d_o)}]^2$$ where $D$ is the dissociation energy of the covalent H-O bond. Apparently the curve with the two exponents $10$ and $12$ gives a better fit.


What would be the enthalpy change for a isothermal expansion? I was doing a question on thermodynamics and came across this one stating

During isothermal expansion of gas of an ideal gas :
a) Internal energy decreases
b) Enthalpy increases
c) Enthalpy reduces to zero
d) Enthalpy remains unchanged

I cannot agree to any of these as $\Delta H = \Delta U + W$, $\Delta U = 0$ as $\Delta T = 0$ but $W \neq 0$ so $\Delta H = W$ which is $\neq 0$ and as $W \lt 0$ the enthalpy should decrease but how much how do I know and to add to the confusion the answer given is enthalpy remains unchanged in the light of the equation $\Delta H = \Delta U +\Delta n_\mathrm gRT$ and $\Delta U$ and $\Delta n = 0$  so $\Delta H = 0$ is it right.

You have already mentioned that the change of internal energy is zero $(\Delta U=0)$ since, for an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$, and  in a closed system $n$ is constant $(\Delta n=0)$ and during an isothermal process also $T$ remains constant $(\Delta T=0)$.

Enthalpy $H$ is defined as

$$H=U+pV$$

and the ideal gas law states that

$$pV=nRT$$

Thus

$$H=U+nRT$$

Since $n$ (closed system) and $T$ (isothermal process) are constant, the product $nRT$ is constant, and therefore, according to the ideal gas law, also the product $pV$ is constant.

Furthermore, since $U$ is constant during the given process, the sum $H=U+nRT$ as well as  $H=U+pV$ has to remain unchanged (answer d).

Note that your assumption

$$\Delta H = \Delta U + W$$

(where $W=p\Delta V$ is the pressure-volume work) only applies to processes under constant pressure $p$, since enthalpy is defined as

$$H=U+pV$$

and thus

$$\begin{align}
\mathrm dH&=\mathrm dU+\mathrm d(pV)\\
&=\mathrm dU+V\mathrm dp+p\mathrm dV
\end{align}$$

which simplifies to

$$\mathrm dH=\mathrm dU+p\mathrm dV$$

at constant pressure $(\mathrm dp=0)$.

However, the pressure does not remain constant during the process given in the question.


What is the chemical composition of the grignard reagent? What is the actual chemical composition of the grignard reagent and what is the method of preparation? Can it be prepared in college laboratories?

A Grignard reagent is an organomagnesium halide of the generic form $\ce{RMgX}$, where $\ce{R}$ is a hydrocarbon group and $\ce{X}$ is a halide, usually $\ce{Cl}$, $\ce{Br}$, or $\ce{I}$.

These reagents are produced by the reaction of magnesium metal with the corresponding organic halide in a non-protic (but coordinating) solvent, like diethyl ether.

For example, you would make phenylmagnesium bromide from bromobenzene and magnesium:

$$\ce{PhBr(C6H5Br) + Mg ->[][\ce{Et2O}]PhMgBr}$$

They are moisture and air sensitive, but not so much that you need to get crazy with inert atmosphere. These reagents are not stable outside of solution.

Grignard reagents can be made in most laboratories set up for organic chemistry. In fact, they are very common reactions to include in undergraduate organic chemistry lab courses because they are very well behaved.


Copper plating, is sulphuric acid necessary? Alternative to H2SO4? Many links and tutorials on copper plating use a solution that contains copper sulphate / sulphuric acid and hydrocloric acid.

In Sweden, Sulphuric acid is almost impossible to get a hold off. You need a license to buy it, and it costs 100 USD to get one...

What does the H2SO4 do in the electrolyte? Is there an alternative to the acid?

I want to plate PCB copper clads.

Copper acetate can be used, at least for a thin layer, and it can be made with chemicals found at home.

To get a bright copper coat, extra acid (vinegar, 5% acetic acid, in this case) or gelatin can be added.

That said, if the plating is for through-hole continuity, it might be simpler and cheaper to use a solderable conductive ink.


What would be SMILES notation for a compound with delocalized bonding? For example, the cyclopentadienyl rings in ferrocene $\ce{Fe(}\eta^5 \ce{-C5H5)2}$ are often drawn with delocalized electron circles in them instead of explicit pi bonds.

The Wikipedia article for ferrocene provides the following SMILES string: [cH-]1cccc1.[cH-]1cccc1.[Fe+2]. In this SMILES string, the charge on the cyclopentadienyl ligands are explicitly attached to carbon atoms in the ring.

Can explicit or implicit delocalization be stored in a SMILES string? For example, would c1cccc1.c1cccc1.Fe also be an acceptable SMILES string for ferrocene that can be parsed and understood?
Does the localized version of the SMILES string provide enough information for the delocalization to be reconstructed when the string is parsed back into a structure? If not, how can this be done?

There are general problems representing delocalized bonding and organometallic compounds in cheminformatics (generally) and SMILES in specific.

To quote the Open SMILES standard:

  This simple mental model [of a connection table between atoms using valence bonds] has little resemblance to the underlying quantum-mechanical reality of electrons, protons and neutrons, yet it has proved to be a remarkably useful approximation of how atoms behave in close proximity to one another. However, the valence model is an imperfect representation of molecular structure, and the SMILES language inherits these imperfections. Chemical bonds are often tautomeric, aromatic or otherwise fractional rather than neat integer multiples. Delocalized bonds, bond-centered bonds, hydrogen bonds and various other inter-atom forces that are well characterized by a quantum-mechanics description simply don’t fit into the valence model.

Other interesting cases are agostic interactions, 3-center 2-electron bonds (e.g., diborane).

There are several suggestions for advancing beyond the typical connection-table valence bond view, including:

"Yet Another Representation of Molecular Structure" J. Chem. Inf. Comput. Sci., 1995, 35 (5), pp 787–802
"Overcoming the Limitations of a Connection Table Description:  A Universal Representation of Chemical Species" J. Chem. Inf. Comput. Sci., 1997, 37 (4), pp 705–714

That both papers were published 17+ years ago should be some indication of:

Chemists like the valence bond view of molecules, despite the limitations
How complicated bonding can be in non-organic chemistry
How far cheminformatics needs to go to properly handle organometallic and non-organic compounds

By the way... anyone interested in implementing such efforts should let me know.. Many people would be happy to help. :-)

UPDATE

I should point out that there are some practical solutions to representing such things. For example Alex Clark suggested recording zero-order bonds (and sometimes dummy atoms):

"Accurate Specification of Molecular Structures: The Case for Zero-Order Bonds and Explicit Hydrogen Counting" J. Chem. Inf. Model., 2011, 51 (12), pp 3149–3157


Interpreting X-ray diffraction data So I've got my data for some x-ray crystals of small molecules of Ag, Cu,... I was curios to know what the best way to determine what type of lattice each of these atoms may have?

Your question has no sense – sorry.

You talk about a few small molecules (with coinage metals) but ask about lattice type for atoms. Atoms in molecular crystal have no their own lattice. Lattice describes how your molecule are aligned in 3D-space.

Assuming that you have a single crystal data then in the beginning of experiment your X-minion (that's the person that took X-ray data for you) has already found the cell parameters and then usually also Bravais cell. To be sure that selected lattice was right whole data set need to be collected, the data have to be integrated and then using Lattice exceptions (see International Tables for Crystalography Volume A,  2.2.13. Reflection conditions) you can tell what lattice type you have. If you do the same for glide planes and screw axis exceptions then eventually you will find space group.


Does mixing two different foaming soap make it ineffective? To save soap during this pandemic, I mixed foaming soap of two different brands and then the solution became cloudy (both originally transparent) . It seems foaming okay. But does this make the soap less efficient in washing away germs/virus? Does the foaming show that the mixture is still working as a regular soap?
I know that soap is both hydrophilic and hydrophobia which makes it capable of washing away bacterias. Is it possible for the ingredients of two soap make it non polar?
Thanks.

According to the answers, more details are added. The ingredients of the two foaming soaps are given below. According to the ingredients, can we tell if the effectiveness of soap is reduced by mixing them?

About 1/3 is method foaming hand soap with ingredients--WATER (AQUA), SODIUM LAURYL SULFATE*, COCAMIDOPROPYL BETAINE*, GLYCERIN*, ALOE BARBADENSIS EXTRACT*, TOCOPHERYL ACETATE (VITAMIN E), CITRIC ACID, SODIUM CHLORIDE*, SODIUM CITRATE*, METHYLISOTHIAZOLINONE, METHYLCHLOROISOTHIAZOLINONE, FRAGRANCE (PARFUM: LINALOOL, LYRAL), EXTERNAL VIOLET 2 (CI 60730), BLUE 1 (CI 42090).

2/3 is equate anti-bacteria foaming soap with ingredients in the picture.

The soap solutions were originally transparent. An interesting experiment, if you still have some of the original soaps, is to shine a light thru the solutions and look for a Tyndall effect:

The light source is on the far right, out of the picture. The beaker on the right shows light scattering from particles comparable in size to the wavelength of light due to the Tyndall effect; the beaker on the left shows no scattering, even tho the light beam is passing thru it. Its contents are unknown, but there seem to be no particles of sufficiently different index of refraction and size to scatter light.
The tiny particles which scatter the light (just like in a smoky room) are micelles, aggregates of several hundred or thousand soap molecules. In the original soap solutions, this scattering is small, due to selection of manufacturing conditions (exact formulation, concentration, pH, other ingredients, etc., etc.), so the solutions appear transparent. What is optimal for each soap separately may not be optimal for the combination: perhaps the mixing produced larger micelles which scattered more, so the mixture scattered more light and became cloudy to the eye. I'm going to suggest that both soap solutions show a Tyndall effect, and the mixed solution shows a much larger Tyndall effect. It may change with time.
Now the question becomes: is the combination as effective as either solution alone. Well, it is different (cloudy). It could be worse; it could be better! It might not be observably different.
The foaming test could cast some light on the effectiveness: If you can test the foaming (height of foam and time stability) of the original solutions and of the mixture, and if they are all the same (+/- ~10-20%), the combination is probably OK.
There is a case where this foaming test will indicate immediately that the combination is not good. The molecules in soaps and detergents have a hydrophilic end and a hydrophobic end. Soaps are salts of fatty acids, and have a negative (hydrophilic) carboxyl group at one end and a (hydrophobic) hydrocarbon at the other end, and a soluble alkali metal cation. There are nonionic detergents which have no charge; they are compatible with anionic or cationic soaps and detergents. Cationic detergents (usually amine salts) have a hydrophilic positively charged nitrogen at one end and a hydrophobic hydrocarbon at the other, and some anion, usually chloride. If you should mix solutions of a cationic detergent with an anionic soap, the positive and negative charges associate to produce a neutralized ion-pair with two hydrophobic hydrocarbon groups - together this looks like a big fatty molecule with a little polarity in the middle. Overall, this molecule is not very soluble in water and precipitates out (but it floats like scum). And since it is not soluble in water, it won't foam.


Making carbon monoxide gas from formic acid and sulfuric acid? Can I create carbon monoxide gas by just mixing formic acid and sulphuric acid together? Or do they have to be heated?

The formation of carbon monoxide already takes place when mixing formic acid and sulfuric acid at room temperature, but the reaction is relatively slow. Since the temperature of the mixture may significantly increase during the course of the reaction, the reaction rate may increase, which makes the reaction difficult to control.

For laboratory applications, concentrated sulfuric acid (or phosphoric acid) is usually heated to about 80 °C. Formic acid is added slowly, drop by drop. The formation of gas should start upon addition of the first drop, otherwise the temperature should be increased.

The generated gas contains various impurities, which can be removed using gas washing bottles. Usually, the gas is passed first through $\ce{KOH}$ solution; however, you should take care to prevent any backflow of $\ce{KOH}$ solution to the sulfuric acid.


What is the role of size in formation of complex compounds? In my textbook, these reasons are given as of why transition metals form a large number of complex compounds:-

Smaller size of metal ions
High ionic charges
Availability of d orbitals for bond formation

What is the use of the first reason, i.e small size? Shouldn't it be opposite? In interhalogen compounds too, in XYn, X should be of greater size than Y. So why opposite in case of complex compounds?

The 3d electrons are attracted strongly to the nucleus (positive charge) and this overalls 'pulls' the electron cloud closer to nucleus making the atomic radii much smaller, which is why there is a general trend of decreasing atomic radii across the period.

Since in transition elements series the 4s orbitals are filled first before 3d orbit, this is turn makes the transition element very reactive and readily form compound as the 4s electrons are released easily, (in addition 3d orbitals provide a screening effect) and excited to a higher energy level during typical reaction to form  complex salts.


can sodium persulfate etch or damage a titanium heater? I use Sodium Persulfate to etch copper on a PCB process. I have been using glass heater on the etching tank but the glass is too fragile; it breaks too easily.

I read that titanium heater is used commercially (in aquariums as well). I don't know much about chemistry, could anyone tell me if it is okay to use titanium with Sodium Persulfate?

Sodium Persulfate to water ratio is 1:10 and the maximum temperature is about 50 C.

If the surface of the immersed element is all titanium, the answer appears to be that it won't get corroded unless the temperature is much higher than 50$^\circ$C; see http://www.azom.com/article.aspx?ArticleID=1336

The case is changed if there are joints or parts made of something else, as in this illustration: http://www.saltwatereddies.com/content-popup_image/pID-2199/popup_image.html

I would think that would not be so safe.

Of course, if you've been breaking glass heating elements right and left, you have little to lose by trying the titanium one...


Is orbital hybridisation real? Is the concept of orbital hybridisation an accurate description of a physical process that actually occurs in bonding atoms, or is it, like VSEPR, a heuristic tool to reason about the characteristics of classically infeasible chemical bonds? Forgive the naivete of my question: I ask this because I am not actually acquainted with the bulk of the underlying theory of chemical bonds, but only with a somewhat flimsy and superficial description that seems inadequate and overly contorted to me.

Thanks in advance for any answers!

Orbital hybridisation is, like most other topics bordering quantum mechanics, just a model of reality (in fact, I would go as far as to say like all topics ever thought of, but lets not get too philosophical). As such, there are limits to its usefulness, so models should never be thought of as a "true picture" of anything.

Models are always simplified and in the end "wrong", but sometimes they are useful. Through models we can gain insight and make new predictions that can be falsified or not and as such we advance our understanding (and the most learned even advance science as a whole). Without models I would not understand anything, but I considert it counter productive for myself to think of them as representing some "truth".

$$\text{Observation} \rightarrow \text{interpolation} / \text{modelling} \rightarrow \text{prediction} \rightarrow \text{experiment} \rightarrow \text{observation}$$


Can redox reactions be reversed? Suppose I did this reaction by dipping a $\ce{Zn}$ rod in $\ce{CuSO_4}$ solution.

$\ce {CuSO4 + Zn -> ZnSO4 + Cu}$

Is there a way to reverse this reaction to

$\ce{Cu + ZnSO4 -> Zn + CuSO_4}$?

If so, how?

Short answer: The reverse reaction can occur using an electrolytic cell.

Long answer: There are two types of cell-electrochemical cell and electrolytic cell.

Electrochemical cell uses a spontaneous reaction to convert chemical energy to electrical energy.

$\ce{CuSO4 + Zn -> ZnSO4 + Cu}$

This reaction is spontaneous and the cell setup for this reaction is an electrochemical cell.

Electrolytic cell is just the opposite of electrochemical cell. Non spontaneous reactions can be made to happen using an electrolytic cell.

$\ce{Cu + ZnSO4 -> Zn + CuSO4}$

This reaction is non spontaneous. Energy from an external source will be used to make this reaction happen. If you want to know how these cells work, refer to this site.


Chemical reaction notation in differential equations It's common to see chemical reactions in this notation:

$A \overset{k}{\longrightarrow} \emptyset$

which means $A$ is degraded with rate $k$. What's the proper way to convert this to an ordinary differential equation? does the notation suggest that $A$ is degraded in linear proportion to its concentration:

$\frac{dA}{dt} = -k[A]$

or does it mean that the degradation rate is constant irrespective of $[A]$?

$\frac{dA}{dt} = -k$

or is the original arrow notation perhaps ambiguous? references would be helpful.

In general, there is no way to know exactly how a net reaction should be turned into a differential equation. Or, perhaps it's more accurate to say that you can't know which differential equation should describe the reaction.

In your case, since you're talking about a decay process, this will always be first-order with respect to concentration and should thus be written as,

$$\frac{dA}{dt} = -k[A].$$

Some other examples are:

$\textbf{Elementary Reactions:}$

If a reaction consists of only a single step, or the rate is known to be affected by only a single step, then the rate can be written down similar to how you guess above. One example are decay processes, which are always first-order. Consider, $$\ce{H2O2(l) ->[k]H2O +\frac{1}{2}O2}$$
Because we know this is a decay process, we can write down,$$\frac{-d[\ce{H2O2}]}{dt}=k[\ce{H2O2}]$$
$\textbf{Reactions with Simple Physical Models:}$

I'm thinking specifically of the derivation of the Michaelis-Menten Equation. I'll provide a sketch of this.

The goal is to describe the kinetics of enzyme catalysis. Because the reactant is relatively localized (i.e. it's trapped in an enzyme), there can't be that many different states. At most, we can have this sequence of steps:
$$\ce{E +S<=>[k_1][k_{-1}]ES<=>[k_2][k_{-2}]EP<=>[k_3][k_{-3}]E +P}$$where $\ce{E}$ is the enzyme, $\ce{ES}$ is the enzyme-substrate complex, and $\ce{EP}$ is the enzyme-product complex.

On nothing but physical intuition, we make two assumptions:

The rate of change of $\ce{ES}$ is zero (i.e. constant concentration)
The reverse step involving $\ce{EP}$ can be ignored, assuming $k_3\gg k_{-3}$

From these assumptions, and equilibrium reactions, we can write down a number of differential equations which give us a very useful and quite accurate equation.

The differential equations one can write down abide by the law of mass-action, which basically just says if we write down all the places some mass can go, then we can know the rate of change for a particular step.

For instance,
$$\frac{d[\ce{ES}]}{dt}=k_1[\ce{E}]+k_1[\ce{S}]+k_{-2}[\ce{EP}]-k_{-1}[\ce{ES}]-k_2[\ce{ES}]$$

As you can see, you basically just pick a step and tally up all the ways things are reacting to form a molecule at this step and all the ways things are leaving this step.

$\textbf{Takeaway:}$

Pretty much, in order to write a differential equation describing rate of changes of reactions, we must have some physical intuition about what is going on. Once we have this intuition, we can almost always get somewhere meaningful.


Why is NH3 a nucleophile in NH3+CH3 -> H3NCH3 In the image below:

$\ce{NH3}$  has no charge as it has a full octet.

$\ce{CH3}$ needs 2 more electrons to form a full octet, so has a positive charge.

Why does $\ce{N}$ of the $\ce{NH3}$ want to bond with the $\ce{C}$, even though $\ce{N}$ already has a full octet and neutral charge, and bonding would make it have a negative charge?

Is it because $\ce{N}$ is more electronegative than $\ce{C}$ ?

When working with nucleophiles and electrophiles, it is easier to think of them as "positive-ophiles" and "negative-ophiles" respectively.

With that said, NH3 (ammonia) has a lone pair of electrons that is attracted to positive charges.  The carbon atom with three bonds is electron deficient and positively charged (which your image does not show).  Thus, as a "positive-ophile"/nucleophile, the ammonia through its lone pair is attracted to and forms a bond with the positively charged CH3+ ion.

Yes, nitrogen does have a full octet, but its lone pair of electrons is nonetheless still attracted to the electron deficient and positively charged CH3+ ion.  This is similar to how water molecules (also fully satisfied octets) can bind with positively charged sodium ions in solvation process of salt.


Does aluminium oxide ionize after melting point? My chem is very basic but I would like to know is there a temperature after melting point where aluminium oxide can ionize enough to pass electrical current through? Is there a temperature where the resistance of aluminium oxide deteriorates?

I'm going to try and answer a bit of confusion in the original question "Does aluminium oxide ionize after melting point?". Aluminium Oxide is already ionised in the solid, or more pedantically the ionic model fits Aluminium Oxide sufficiently well that we can think of it comprised of $\ce{Al^{3+}}$ and $\ce{O^{2-}}$ ions and make reasonable predictions. The point is that in most solids, ionic or not, the constituent atoms and molecules are energetically constrained to vibrate about their "normal", so called equilibrium, position, and so never move very far, whereas in liquids they can move about reasonably freely. Now as conductivity involves the nett movement of charges most ionic solids are not conducting while the liquids are, as in the liquid the charged ions are free to move toward the appropriate electrode while in the solid they are not.

However it's not that quite that simple, but the above is probably good enough to University level. But for completeness I'll add that there are things called fast ionic conductors. These are "ionic" solids that conduct electricity well above a certain temperature. A well known example is Silver Iodide, which above 419.15 K start to conduct electricity pretty well. A simple model here is that at that temperature the Iodide sub-lattice stay rigid, so providing a solid like framework, but the Silver sub-lattice melts, and so the silver ions can move, and thus nett charge migration is possible, and so the solid conducts. For a bit more of an introduction, see Wikipedia article of fast ion conductor. Interestingly one form of Aluminium Oxide can act as a fast ion conductor, but in this case the charges that are moving are "impurities".


Quantum mechanical model of atom and the quantum numbers I studied Bohr's model of atom and then the drawbacks of it and then quantum mechanical model of atom. Now quantum model is according to uncertainty principal and dual nature of matter and it says we can not talk about fixed orbits (uncertainty principle) and can only talk of  an area of maximum probability of finding an electron known as "orbital". Now when I'm studying principal quantum number, it says it is the "shell number". So how come we are talking about fixed shells again and returning to what Bohr said?

So how come we are talking about fixed shells again and returning to
  what Bohr said?

Nope, we are not returning to the old quantum theory. We do not talk about orbits anymore, we rather have orbitals. Note that it is important to bear in mind that orbitals also aren't real, i.e. this description is still approximate, but it is a much better way of thinking about atoms (and molecules) than Bohr model.

Now, to understand the concept of orbitals you have to learn the basics of quantum chemistry first, namely, the Hartree-Fock method. Until then just realize that:

We do use hydrogen-like one-electron wave functions (orbitals, or, spin-orbitals, to be more precise) to describe the state of each and every electron in any atom.
These hydrogen-like orbitals are described by a set of quantum numbers: the principal quantum number, the azimuthal quantum number, the magnetic quantum number, and the spin quantum number.
The set of orbitals belonging to a given principal quantum number $n$  constitutes what we call a shell in the orbital model.


How does solutions of two liquids work? When two liquids mix, and they are not hindered by a polarity mismatch (like oil and water), why don't they always form a homogenous mixture? Due to entropy, a liquid will disperse itself evenly. This even distribution will lead to homogeneity, no? According to an answer to this question (one given when this question looked different), said that they don't always distribute themselves evenly. So, let me lay done some ground rules:

Both the liquids are non-polar or polar
The liquids do not react in a way that creates products

Now, what hinders these liquids from dispersing evenly in each other?
Also, I'm going to clarify a certain conception I have, in case it is wrong:

A homogenous mixture is a mixture where the parts are evenly distributed. A solution is a homogenous mixture, where the solute has been deconstructed into its ionic components.

It has already been pointed out that your text is incorrect. Saccharose (commonly known as sugar) is not an ionic compound yet it dissolves very well in water. It is not alone although I am too lazy to think up a non-carbohydrate example.
I’ll also point out that ionic compounds exist that are insoluble or poorly soluble in water, although some may rightly argue that their bonds tend to be more covalent than ionic.
As for two different liquids mixed together, the answer is: You can’t say a priori. (Or, in a highly logical reading of your question: no, the result will not always be a solution.) Many polar liquids will form a solution with many other polar liquids but in some cases the pairing will not. Conversely, a polar liquid such as ethyl acetate or acetone can form a solution with a practically unpolar liquid such as hexane.


Is oxygen above the critical point always supercritical fluid? Would it still appear to roughly follow the ideal gas law? In this answer I've asserted (without a "Chemist's license") that as long as oxygen is above it's critical point in both temperature (154.5 K) and pressure (50.4 bar) it's going to be a supercritical fluid.

This comment by a user who I believe is an experienced diver says:

  Scuba divers use the idea(l) gas law for air and oxygen bottles to calculate the amount of remaining gas for 200 down to 10 bar. This would not be possible if there is supercritical oxygen in the tank.

In other words, the ideal gas law works pretty well for oxygen between 10 and 200 bar; for a given temperature the amount of oxygen (number of atoms or mass in kg) is close to linearly proportional to pressure.

The comment proposes that this is evidence that the oxygen couldn't be a supercritical fluid.

Question(s):

When oxygen is above both 50.4 bar and 154.5 K is it always a supercritical fluid, end of story?
If it is, would it nonetheless still tend to follow fairly closely the ideal gas law?

Yes.  Any fluid with a temperature is above critical temperature and the pressure above the critical pressure is by defintion a supercritical fluid.  Don't be mislead by all the claims that supercritical fluids are special and wonky with all sorts of amazing, bizarre properties.  This is true of some supercritical fluids near the critical point, but the farther you go away from the critical point, the more like a regular gas supercritical fluids become.  (The only exception here is that at gargantuan extremes of P and/or T, eventually fluids will become plasmas, metallic solids, black holes, etc.  But as long as it is a fluid of the same substance [no chemical reactions], then it will be supercritical.)
Yes, supercritical fluids can often be modeled by the ideal gas law.  A good look into why would take into the theory of corresponding states.  This theory says that most gases will most often have similar compressibilities $Z = \frac{PV}{nRT}$ when they are in "corresponding" states, which means that at similar temperature and pressure relative to each gas's critical point.  There are charts that you can use (approximately) for many, many real gases.  Things to realize:

The compressibility factor $Z$ is a measure of non-ideality.  Ideal gases always have a compressibility factor of 1.0.
You can use the chart to find approximate compressibility factors for many, many real gases if you know that gas's critical temperature $T_c$ and critical pressure $P_c$.  The "reduced temperature" is just $\frac{T}{T_c}$ and likewise for pressure.
If you check out the generalized compressibility factor chart, look at the dependence on reduced temperature!  At a reduced temperature of 2.0, the line is nearly flat at $Z = 1.0$.  This is an indication that the gas behaves nearly ideally.  It takes pretty close approaches of $\frac{T}{T_c}$ to 1.0 to get strong non-ideality.
Oxygen's critical temperature of 154 K means that a reduced temperature of 2.0 just requires 309 K, about ambient temperature.  At these temperatures, oxygen will be fairly ideal, regardless of temperature, up to reduced pressures of 5 or 8, which correspond to 250 to 400 bar or so.


Why does the alkyl group anti to the hydroxyl migrate in Beckmann rearrangement? Background: Many organic reactions involve the migration of an alkyl group from one position to the adjacent one. The migration of the alkyl group is decided by its migratory aptitute i.e. electron-richness. It generally follows the priority order of hydride > phenyl > higher alkyl > methyl.

Main question: The Beckmann rearrangement also involves an alkyl migration. However, this migration is not governed by migratory aptitude. In fact, the $\ce{-R}$ which is in anti-position to the hydroxyl group in the oxime migrates, irrespective of its migratory aptitude! My question is why is this so? And are there other organic reactions which have such a rule for alkyl migration?

MasterOrganicChemistry  and Wikipedia don't even mention the word "trans" or "anti" anywhere. My textbook obviously doesn't mention anything either, hence I am driven to ask this question.

Oximes can undergo the Beckmann rearrangement.  Oximes also exist as stable syn and anti isomers.  In the figure below, $\ce{R_1}$ is anti to the hydroxyl group; another isomer exists where it is syn the the hydroxyl group.

In the first step of the Beckmann rearrangement the protonated hydroxyl group makes for a good leaving group.  As the hydroxyl group begins to depart, the $\ce{R}$ group which is anti to the leaving group begins to migrate and form a bond with nitrogen.

(Wikipedia)

This step is said to be concerted (rather than stepwise) with the $\ce{N-O^{+}H2}$ bond being broken and the new $\ce{R-N}$ bond being formed at (more or less) the same time.  Whether you think of the $\ce{R}$ group as participating in something similar to an SN2 reaction or whether you think of the $\ce{R}$ group as starting to bond to the available $\sigma^{*}$ orbital of the elongating $\ce{N-O}$ bond (equivalent descriptions), you see that having the $\ce{R}$ group approach the nitrogen from a 180° angle from the breaking $\ce{N-O}$ bond would be preferred. Hence, the $\ce{R}$ group anti to the $\ce{N-O}$ bond migrates preferentially.

There are many reactions showing this geometric preference, the Baeyer-Villiger oxidation being an example.


What is the difference between the Bohr model and the quantum model? Specifically, Bohrs atom model gives the radius of an atom as 0.53Å, while a quantum description offers a similar value. What is the difference between these?

The Bohr model is a comparatively simple model, whereas the quantum description is part of the huge theory of quantum mechanics, so to list all the differences including implications would need to explain quantum mechanics itself.

But I think quite a good starting point would be to say, the Bohr model has electrons travel on certain, specific, classical trajectories, picked out by certain rules. The quantum model does not have electron trajectories. Instead electrons are quite different, they are spread out a bit like clouds of gas, in areas called atomic orbitals. I would recommend reading about atomic orbitals, the Wikipedia page is quite interesting. You will see very quickly they are quite different from the Bohr model.


Why is FORTRAN so commonly used in computational chemistry? I've been using Ruby to write scripts for research, but I want to get into some heavier stuff that Ruby is just too slow for. I noticed there are a few things written in C and C++, but there is an oddly large proportion of software used in computational chemistry that is written in FORTRAN (in which I have zero experience.)

Why is FORTRAN used in computational chemistry? From what I understand, FORTRAN is kind of on the ancient side (“punchcard” old.) I was a bit shocked to find fairly recently written tutorials for FORTRAN.

Is it a sort of, "this is how we've always done it," thing or is there an efficiency aspect to FORTRAN I'm overlooking?

Note: I may have FORTRAN confused with later programming languages with similar names.

I don't think that's really true anymore.

Some Fortran use is historical (i.e., early codes were developed in FORTRAN because that was the best programming language for number crunching in the 70s and 80s). Heck, the name stands for "formula translation."

Some Fortran use is because of performance. The language was designed to be:

  especially suited to numeric computation and scientific computing.

Many times, I find chemistry coders sticking to Fortran because they know it and have existing highly optimized numeric code-bases.

I think the performance side isn't necessarily true anymore when using modern, highly optimizing C and C++ compilers.

I write a lot of code in C and C++ for performance and glue a lot of things with Python. I know some quantum programs are written exclusively or primarily in C++. Here are a few open source examples:

Psi4 - Written in C++ and Python
MPQC - Written in C++
LibInt - Written in C++ for efficient quantum integrals.
LibXC - Written in C with Fortran "bindings" for DFT exchange-correlation functionals

This is my opinion, but my recommendation for faster performance in chemistry would be Python with some C or C++ mixed in.

I find I'm more efficient coding in Python, partly because of the language, partly because of the many packages, partly since I don't have to compile, and that's all important.

Also, you can run Python scripts and functions in parallel, on the GPU, and even compile them, e.g. with Numba. As I said, if I think performance is crucial, I'll write pieces in C or usually C++ and link to Python as needed.


What effect on boiling and melting points does intramolecular hydrogen bonding have? I know that intermolecular forces increase the boiling and melting point of a compound, but what is the effect of intramolecular forces on boiling and melting points?

I found contradicting answers while searching on the web.

The effect of intramolecular forces versus intermolecular forces on boiling and melting points is sometimes seen when we compare the boiling points of ortho- and para-nitrophenol (or similar compounds).
Para-nitrophenol shows intermolecular hydrogen bonding, which causes an increase in boiling point as different molecules bond better with each other, as compared to ortho-nitrophenol, where there is intramolecular hydrogen bonding, which does not help in increasing the boiling point, hence p-nitrophenol has a higher boiling point than o-nitrophenol.
However, while comparing intramolecular bonding with no hydrogen bonding, no difference can be expected unless the intramolecular bonding affects the intermolecular bonding in some way, for instance by changing size and thus packing.


Reason behind markovnikov's rule in hydration of alkenes via mercury catalysis Recently I came across an interesting reaction regarding the hydration of alkynes using mercury catalysis (HgSO4) in the presence of sulphuric acid. I've read that after the mercury complex acts like a protecting group by preventing any carbocation rearrangement, water attacks the more substituted site to open up the 3-membered ring formed as an intermediate. But this is a nucleophilic attack (SN2 type) and the more substituted site is more sterically hindered and carries less partial positive charge on it( due to influence of other groups). So why does this occur via Markovnikov's rule itself?

If I could postulate, the reaction is like an epoxide ring opening during acidic conditions, where the oxygen is protonated, making it electron-deficient and have a +1 formal charge. This creates a good leaving group (an electron-deficient O) and in your case, an electron-deficient Hg.

Credits: Clayden Organic Chemistry Vol 2

Since it is a good leaving group, it wants to leave and as predicted by the Sn1 mechanism, the C-Hg bond bonded to the more substituted carbon prefers to break in order to take advantage of the more stable carbocation intermediate. This makes that particular bond weaker. The nucleophile thus prefers to attack the more substituted end in order to take advantage of the weaker bond. As the bond is weaker, the net enthalpy change would be lower, making the attack at the more substituted end more thermodynamically favourable. Additionally, notice that in the transition state, the developing positive charge is more resonance stabilised if the nucleophile attacks the more substituted end, making the reaction more kinetically favourable. From both perspectives, kinetics and thermodynamics, an attack at the more substituted end is favoured.


Find the final temperature and volume of the gas $\pu{10 L}$ of a monoatomic ideal gas at $\pu{0 ^\circ C}$ and $\pu{10 atm}$ pressure is suddenly released to $\pu{1 atm}$ pressure and the gas expands adiabatically against this constant pressure. What are the final temperature and volume of the gas? (Answer: $\pu{174.9 K}$, $\pu{64.04 L}$)

There are some points I don't get:

First of all, it is not given whether it is reversible of irreversible expansion.
If it is reversible, I am at wits end deciding whether to use one of the following (without success):
$$ pV^\gamma = \mathrm{constant} \quad \mathrm{or} \quad T^\gamma p^{1 - \gamma} = \mathrm{constant}$$
If it is irreversible, I tried using
$$W_\mathrm{irreversible} = \Delta E$$
i.e.
$$-P_{ext} \left(\frac{nrT_2}{p_2} - \frac{nrT_1}{p_1}\right) = nC_\mathrm{v}\Delta T$$
I am having trouble using $r = \pu{0.0821 L atm mol-1 K-1}$ and $R = \pu{8.314 J mol-1 K-1}$.

The process should indeed be considered irreversible, as the problem statement gives a rather explicit hint stating that the pressure was released suddenly, so that your 3rd suggestion is correct. With an irreversible expansion against a constant external pressure, the adiabatic equation $pV^\gamma = \text{const}$ is inapplicable, so one must use the first law of thermodynamics. Work is done at the expense of the loss of internal energy:

$$W = - \Delta U = nC_\mathrm{v}(T_1 - T_2) \tag{1}$$

For ideal monoatomic gas the following is applied:

$$n = \frac{p_1V_1}{RT_1} \tag{1.1}$$
$$C_\mathrm{v} = \frac{3R}{2} \tag{1.2}$$

so that (1) can be rewritten as such:

$$W = nC_\mathrm{v}(T_1 - T_2) = \frac{3p_1V_1R}{2RT_1}(T_1 - T_2) = \frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} \tag{1.3}$$

At the same time work by gas expansion is

$$W = p_2(V_2 - V_1) = p_2V_2 - p_2V_1 \tag{2}$$

and from the ideal gas law the reduced product is

$$p_2V_2 = nRT_2 \tag{2.1}$$

so that (2) can also be rewritten:

$$W = p_2(V_2 - V_1) = nRT_2 - p_2V_1 \tag{2.2}$$

I would leave for OP the task to to equate (1.3) and (2.2) and express the final temperature $T_2$ from there:

$$\frac{3p_1V_1}{2} - \frac{3p_1V_1T_2}{2T_1} = nRT_2 - p_2V_1$$

Here I bring up the final result:

$$T_2 = \frac{3p_1V_1 + 2p_2V_1}{5p_1V_1} T_1 = \frac{3 \cdot \pu{10 atm} \cdot \pu{10 L} + 2 \cdot \pu{1 atm} \cdot \pu{10 L}}{5 \cdot \pu{10 atm} \cdot \pu{10 L}} \cdot \pu{273.15 K} = \pu{174.7 K} \tag{3}$$

Volume after expansion is also to be found using ideal gas law and (1.1):

$$V_2 = \frac{nRT_2}{p_2} = \frac{p_1V_1RT_2}{RT_1p_2} = \frac{p_1T_2}{p_2T_1}V_1 = \frac{\pu{10 atm} \cdot \pu{174.7 K}}{\pu{1 atm} \cdot \pu{273.15 K}} \cdot \pu{10 L} = \pu{64.0 L} \tag{4}$$


Working with wave number equations, use trial-and-error? I have a wave number question which I have partially solved:

  What transition in $\ce{He+}$ ion shall have the same wavenumber as the first line in Balmer series of $\ce{H}$ atom?
  (a) $7\to5$ (b) $5\to3$ (c) $6\to4$ (d) $4\to2$

Here's how I solved it:

Note: symbol $Z$ is the atomic number
and $1/\lambda$ is wave number (I have written frequency by mistake)

At this point my teacher told me to continue using a trial and error method, is there any other way to solve this, perhaps using some other equation.

[…] my teacher told me to continue using a trial and error method […]

But nobody told you to do it manually once you have understood how to adjust the Rydberg constant for $\ce{He+}$ and use the formula to get the energies for the given  transistions, in order to compare them with the $\ce{H_{\alpha}}$ line (656.3 nm):

import scipy.constants

transitions = ((7,5), (5,3), (6,4), (4,2))

def get_he_line_nm(n_upper, n_lower):
    '''calculate the energy for a line in a He+ spectrum and return
       it in nanometer'''

    energy = 4 * scipy.constants.Rydberg * (pow(n_lower, -2) - pow(n_upper, -2))

    return 1.0/(scipy.constants.nano * energy)
    # energy in nm

for upper, lower  in transitions:
    wavelength = get_he_line_nm(upper, lower)
    print('{0} -> {1} : {2:7.2f} nm'.format(upper, lower, wavelength))

will give the output

7 -> 5 : 1162.81 nm
5 -> 3 :  320.37 nm
6 -> 4 :  656.11 nm
4 -> 2 :  121.50 nm


Bleach turned towel bright pink before fading to white We have some relatively new white towels that were getting dirty, so we decided to clean them with bleach (standard Clorox). When we sprayed the diluted bleach on the towels they immediately turned bright purplish pink, but the color faded back to white within seconds. We initially thought it was somehow transferring color from the pink nozzle on the spray bottle, but we noticed the same effect when the diluted bleach was poured directly onto the towels. No harm appears to be done, but what would cause the towel to turn pink, especially since it fades almost immediately?

Although Dr. Laundry says it only happens to the incoming brighteners within a detergent and not to those fixed to the fabric, I think that your new towels should have enough brighteners - deposited into and/or onto them - to react as well.

Optical brightners and bleach reaction:  https://www.clorox.com/dr-laundry/detergentbleach-reaction/

Additional note : brighteners absorb light in the near UV and emit in the blu tail of the Vis spectrum, so making the fabric "extrawhite". Search for Optical brighteners in Google or Wikipedia.

Perahps someone here could know or propose the ongoing reaction and is transient behaviour. As the OP said in a comment, the towels were washed with a detergent containing a stilbenedisulfonic triazine derivative(s).


Prediction/computation of mass spectral ionisation intensity There are quite a few approaches (ranging from empirical additive rules based on presence of functional groups to full blown in silico QM calculations) to predicting molecular properties such as UVmax absorption or IR vibrational frequencies. While not perfect, they give a general ballpark range of what to look for.
I’ve recently been doing some quantitative MS work which has involved a lot of dilutions and calibrations.
It got me curious  if there are approaches to predict peak intensity as it’s not really something I’ve ever read about (or even included in software similar to basic IR/NMR prediction). It obviously wouldn’t be useful for quantification, but could give a vague sense of how strongly something will be observed (if at all).
To give an example, imagine mixing 1 mmol of paracetamol and 1 mmol of aspirin in DMSO and recording an LCMS of this—although you have the same amount of both the intensity in the MS TIC will vary significantly.
It seems to me that the parameters are quite well controlled (voltage, temperature, $\mathrm{p}K_\mathrm{a}$ etc.)—is it possible?

The absolute response coefficient (i.e. peak intensity in counts per mM or similar units) will be way too strong a function of instrument-specific parameters like solvent composition, pH, one or two or three temperature variables specific to the ion source, one or or two or three more voltage parameters particular to the ion source, whether a given emitter is dirty or fresh, how big it is, etc.
However, it may be possible to predict relative response coefficients across a series of compounds.  Most approaches I've seen rely on prediction of gas-phase basicity values, or alternately proton affinity of gas-phase neutral species (the assumption being that you are predicting the ionization efficiency of singly-charged ions in positive ionization mode arising from neutral molecules).
Folks have been working on "machine learning" approaches in this area.  Here is one example...


Between 3-chlorophenol and 3-fluorophenol, which one is more acidic? Between 3-chlorophenol and 3-fluorophenol, which one is more acidic?

In this case, I know that both F and Cl are electron-donating groups. Furthermore, I know that F is more electron-donating than Cl. But I can't figure out whether that makes 3-chlorophenol more acidic, or 3-fluorophenol.

Experimentally, the $\mathrm pK_\mathrm a$ values for 3-chlorophenol and 3-fluorophenol appear to be (perhaps unexpectedly) equivalent.

The table below1 shows the measured $\mathrm p K_\mathrm a$ values of substituted phenols in dimethyl sulfoxide (DMSO):

$$\begin{array}{ccc} \hline
\text{substituent} & \sigma & \mathrm p K_\mathrm a \\ \hline
\ce{H} & 0 & 18.03 \pm 0.04, \color{silver}{16.4} \\
\ce{p-CH3} & -0.170 & 18.86 \pm 0.05 \\
\ce{m-CH3} & -0.069 & 18.23 \pm 0.01 \\
\ce{p-Cl} & 0.227 & 16.74 \pm 0.03 \\
\color{red}{\ce{m-F}} & \color{red}{0.337} & \color{red}{15.88 \pm 0.07} \\
\color{red}{\ce{m-Cl}} & \color{red}{0.373} & \color{red}{15.83 \pm 0.05} \\
\ce{m-CN} & 0.56 & 14.76 \pm 0.05 \\
\ce{m-NO2} & 0.710 & 14.39 \pm 0.12, \color{silver}{13.75} \\
\hline
\end{array} \\
\small\text{(Greyed out values are taken from ref 2.)}$$

In general, halides have slightly unusual effects on aromatic rings. Despite being highly electronegative (and deactivating), they actually donate into the pi system via their lone pair.

This donation means that the phenoxide generated upon deprotonation (i.e. the dissociation being measured by the $\mathrm pK_\mathrm a$ value) isn't as stabilised as it would be with a purely withdrawing group (such as the nitrile or the nitro groups).

References

Bordwell, F. G.; McCallum, R. J.; Olmstead, W. N. Acidities and hydrogen bonding of phenols in dimethyl sulfoxide. J. Org. Chem. 1984, 49 (8), 1424–1427. DOI: 10.1021/jo00182a020.
Chantooni, M. K., Jr.; Kolthoff, I. M. Comparison of substituent effects on dissociation and conjugation of phenols with those of carboxylic acids in acetonitrile, N,N-dimethylformamide, and dimethyl sulfoxide. J. Phys. Chem. 1976, 80 (12), 1306–1310. DOI: 10.1021/j100553a009.


Formation of Bands in Semi-Conductors At school, we are learning about semi-conductors and their applications in modern electronics. One of the features of semi-conductors is that there is a small energy gap between the valence and conduction band. My teacher explains that he bands occur because in the compound, since there are millions of atoms close together, due to the Pauli Exclusion Principle the energy levels of the atoms slightly change in value, forming band. When electrons are excite form the valence to conduction band, the substance is able to conduct an electric current.

However I fairly certain this is a superficial explanation as it doesn't really explain what the conduction and valence band is and doesn't explain why electrons in the conduction band are able to conduct electricity while electrons in the valence band can't.

After some researching on the web this is my rough understanding of semi-conductors now:

The s and p orbitals of the metals combine together to form MOs. As the AOs of millions of atoms combine together, these MOs become a continuum, forming two bands. One band is the valence band which consists of 2s sigma bonding and anti-bonding orbitals while the other band is the conduction band which consists of 2p pi bonding and anti-bonding orbitals. This explains why electrons in the conduction band are able to conduct an electric current since they are conjugated (not sure if this is the correct term) across the entire substance and hence explains why electrons in metals are described as a 'sea of de-localised electrons'.

This is just my understanding which isn't that solid. Could someone please offer a more in-depth explanation into how these bands form and correct any wrong things that I said. Also I still have some questions.

One website stated that 'it is only when these bands become filled with 2p electrons that the elements lose their metallic character'. Which I interpret as that when the conduction band is completely filled with electrons, the substance isn't able to conduct electricity any more. However why is this so? Aren't there still electrons in the pi MOs which are 'de-localised' along the substance?

Also the above explanation using MOs seems to explain why these bands form. Does this mean that the explanation that my teacher gave me about the Pauli Exclusion Principle being responsible for the formation of bands is wrong and not needed?

doesn't explain why electrons in the conduction band are able to conduct electricity while electrons in the valence band can't

Electrons in the valence band can conduct electricity, as long as the valence band is not completely filled. This occurs in semiconductors, when you thermally promote electrons from the valence to the conduction band, you are leaving behind some holes in the valence band. The electrons in the valence band can then conduct electricity as well by "jumping" through the holes. The conductivity of semiconductors arises therefore not only from electrons in the conduction band, but also holes in the valence band.

The origin of conductivity lies in a partially-filled band, and it doesn't matter which band it is. For a more rigorous treatment google for "free electron model". The idea is that each crystal orbital (i.e. MO of the crystal) has an associated value of electron momentum. For each orbital with a positive electron momentum, there is an orbital with an equal but negative momentum. When the band is fully filled, all of the crystal orbitals are filled, and there is no net electron momentum because the positives and negatives cancel out exactly. When the band is incompletely filled, the application of a potential difference in a certain direction across the metal causes the electrons to preferentially populate orbitals with positive momentum in that direction. This leads to a net positive electron momentum in that direction, meaning that electrons "flow".

  The s and p orbitals of the metals combine together to form MOs. [...]

Yes. Bands are simply a collection of crystal orbitals that have such small energy gaps between them such that they are almost continuous. I don't see how the Pauli exclusion principle comes into play except for when you are discussing the population of the orbitals (obviously any one orbital can have a maximum of two electrons). The two ways that I know of to obtain the bands (nearly free electron model, and the LCAO/tight-binding approach) both do not invoke the PEP. However I will add a disclaimer that physicists approach solid state chemistry from a very different direction and there could very well be some explanation that I do not know of.

However, yes, the electrons populate orbitals that are "delocalised" across the entire solid and this is the origin of the "sea of delocalised electrons" description of metals.

  'it is only when these bands become filled with 2p electrons that the elements lose their metallic character'

I honestly don't see how that is relevant or even correct at all. Let's use the example of sodium versus chlorine. The band structure of Na metal is vastly different from that of chlorine since the molecular structures are different to begin with. Sodium is a metal and chlorine is a molecular gas, you wouldn't use a band structure to describe chlorine. You cannot just take a diagram for $\ce{Na}$ and say that it applies to $\ce{Cl2}$ and try to make statements such as "filling of the 3p bands leads to decrease in metallic character" because those bands don't even exist in chlorine. Band structures are just large MO diagrams, it is almost like trying to compare the physical properties of sodium and chlorine using MO diagrams for $\ce{Na2}$ and $\ce{Cl2}$, which is plain absurd.

However, if chlorine were to adopt a structure that has close-packed atoms like in sodium, then you would be right that it would still be metallic due to the partially-filled 3p bands.

If you want to read further I would recommend finding a solid state chemistry textbook, all of them would cover basic band theory. In particular I find this book by Moore and Smart clear and accessible.


Oxidation state of tungsten in W3O8 Recently I came across this equation:

$$\ce{WO_3 +SnCl_2 +HCl\rightarrow W_3O_8 + H_2SnCl_6 +H_2O}$$

The compound $\ce{W_3O_8}$ that appears in the right hand side of the equation seems to be mixed oxide. Normal calculation of the oxidation state of W in the compound yields $\ce{\frac{16}{3}}$ which is a fraction. A google search on the compound does not give any good results too.

So, is $\ce{W_3O_8}$ a mixed oxide? What are the oxidation states of the different tungsten atoms?

Note that $\ce{W3O8}$ is a non-stoichiometric compound and it is not that simple compound you may think. It has a very complex structure as such its oxidation state is very difficult to calculate. I am not going in details about the structure of the compound and I will only mention it a non-stoichiometric compound actually a oxygen-deficient cluster.

Please give a good read to Xin Huang et.al., J. Phys. Chem. A 2006, 110, 85-92, the link that @JM97 has posted. It says that the $\ce{O-O}$ is much longer than normal oxide bond and so it is assumed to be a peroxo bond $\ce{O2^2-}$. Now, peroxide has oxidation state of -1. So, assuming $\ce{W3O8}$ to be a simple compound (no stoichiometric defect or any kind of defect), the oxidation state is:

$$3x + 8(-1) = 0$$
$$x = +{8/3}$$


Do electrons go to even lower ground states after exothermic reactions? I am a bit confused about what happens to the energy levels during an exothermic reaction.

I understand that during an exothermic reaction, the energy of the electrons decreases.

Is the potential energy of electrons related to the energy level diagram, shown below?

So if the potential energy decreases, does that mean the ground state for the new molecules formed will be different from the old molecules?

Yes, electron energy levels are like that, especially for single atoms.

Yes, forming molecular bonds creates different electron energy levels, with exothermic reactions leading to chemical bonds with lower ground levels then the reagents had.

Note that by an electron ground energy level is meant the minimal energy of  energy levels unocuppied by other electrons.

As an electron is forbidden by quantum rules to be in the same quantum state as any other electron within an atom or molecule.

Thermal, photon or collision excitations may keep part of electrons in excited state.

Much more info you can get reading about

molecular orbitals
molecular orbitals diagrams
molecular orbital theory

on Wikipedia.


What is the Si–N–C bond angle in silyl isothiocyanate? In silyl isothiocyanate, supposedly, the value of the bond angle $x$ is $180^\circ$.

I don't know how to figure this out. There is a lone pair on nitrogen and two sigma bonds. Shouldn't it have sp2- or sp3-hybrid orbitals and thus $x$ be smaller than $180^\circ$?

The answer that you are supposed to give is presumably something along the lines of "dπ–pπ" bonding between silicon and nitrogen. This essentially means that the lone pair of nitrogen is involved with backbonding into a silicon 3d orbital, leading to a linear geometry at nitrogen (i.e. the bond angle is 180°). This can be represented with the following resonance structures:

To be very clear, this is very likely incorrect. The extent of d-orbital involvement in main group compounds is usually tiny. If there is any backbonding, the acceptor orbital is probably a Si–H σ* orbital rather than a d-orbital (cf. backbonding from transition metal to phosphine).

Electron diffraction studies of H3SiNCS show that the Si–N–C bond angle is actually 163.8°.1 A computational study (TZVP + MP2 level of theory) found 165.91°.2. In fact, the linear structure above is not an energy minimum, but rather a transition state for the interconversion of the two energy minima:3

I don't actually know how one can predict this a priori without access to a computer, but clearly this system is not amenable to bond angle prediction with VSEPR theory, and invoking the so-called "Si–N dπ–pπ bonding" doesn't give the correct answer either.

References

Glidewell, C.; Robiette, A. G.; Sheldrick, G. M. Electron diffraction study of large amplitude vibrations in H3SiNCO and H3SiNCS. Chem. Phys. Lett. 1972, 16 (3), 526–529. DOI: 10.1016/0009-2614(72)80416-0.
Palmer, M. H.; Nelson, A. D. The structures of the azido-, isocyanato- and isothiocyanato-derivatives of methane and silane and their derivatives. A comparison of ab initio with experimental results. J. Mol. Struct. 2004, 689 (1–2), 161–173. DOI: 10.1016/j.molstruc.2003.11.008.
Palmer, M. H.; Guest, M. F. The methyl, silyl and germyl esters of the pseudo-halogen acids. A comparison of structural data from experimental and ab initio studies. Chem. Phys. Lett. 1992, 196 (1–2), 183–190. DOI: 10.1016/0009-2614(92)85951-6.


Reactivity of different alcohols with NaNH2 and CH3I Identify the final product (B) in the following reaction:

I'm not sure how $\ce{NaNH2}$ and $\ce{CH3I}$ will react with the different types of alcohol present here. According to my solution,ether was formed in place of $\ce{CH2OH}$ in the end, whereas the answer says that $\ce{CH2OH}$ group will remain unaffected. What am I doing wrong? Can anyone explain to me the full mechanism?

I suspect, possibly just to a minor extent, that you have also managed to produce an expected photolysis product here (see below).
Interestingly, the photo sensitive methyl iodide, in the presence of a suitable catalyst ($\ce{CO2, Cu,}$..) with lab light (rich in UV) may have commenced in a radical reaction chain in the presence of ethanol:
$$\ce{CH3I ->[$h \nu$] ^.CH3 + I^.}$$
$$\ce{C2H5OH + I^. -> HI + C2H5O^.}$$
$$\ce{C2H5O^. + C2H5O^. -> (C2H5)2O2}$$
$$\ce{(C2H5)2O2 + ^.CH3 -> (C2H5)2O + CH3O^.}$$
....(the reaction chain continues)
where, I have stopped at the formation of Ether, albeit possibly created in very small amounts but detectable by its smell, nevertheless.
Repeat the experiment controlling for light/catalyst and see if you can still smell the Ether.
My advice, any time working with a photoreactive iodides, be mindful of light exposure.
[EDIT] The implication of my analysis, check your product yields, reagents and procedure, as a minor photo-induced side product may not (but still could) be connected to your results. In particular, I would start with the $\ce{NaNH2}$ as per Wikipedia:

$\ce{NaNH2 + H2O → NH3 + NaOH}$
$\ce{4 NaNH2 + 7 O2 → 2 Na2O + 4 NO2 + 4 H2O}$

which suggests that either exposure to moisture/water/air can degrade the compound.


Is these a pair of diastereomer or identical compound? I attempted on this question (e) and found out that they are diastereomers, but the answer from the book is that they are identical. I worked out the configuration at each of the centres for the molecules then compare the results. If you find that one is RS and the other is SS then you'll know that they differ at only one stereo-centre and hence I think the book is wrong. I've got this question from Solomons: Organic Chemistry, 11th edition..

Maybe the answer of the book is wrong? For the fisher projection, it is Diastereomer definitely.


Why does paper turn yellow over time? Yesterday I saw a bill of some book which I bought 6 months back. Then this question struck my mind: "Why does paper turn yellow over time"?

I remember my teacher saying some organic compound was used to make paper white. And this effect is reversed over long periods of time. Can anybody explain?

from "Stuff Matters" by Mark Miodownik:

Paper yellows with age for two reasons. If it is made from cheap low-grade mechanical pulp, it will still contain some lignin. Lignin reacts with oxygen in the presence of light to create chromophores ( meaning, literally, 'color carriers'), which turn the paper yellow as they increase in concentration. This type of paper is used for cheap and disposable paper products, and is why newspapers yellow quickly in light.
It used to be fairly common to increase the textural quality of paper by coating it with aluminum sulfate, a chemical compound that is used primarily to purify water, but what wasn't appreciated at the time was that this treatment creates acidic conditions. This causes the cellulose fibers to react with hydrogen ions, which results in another form of yellowing. It also decreases  the strength of the paper. Large number of books from the 19th and 20th centuries were printed on this so-called acid paper and can now be easily identified in book shops and libraries by their bright yellow appearance. Even non-acid paper is susceptible to this aging, just at a slower rate.


Remove stain from synthetic leather I have a valuable soccer ball. It is not real leather, but a synthetic leather.

I used this cleaning agent to remove a few marks:

https://www.amazon.co.uk/Motsenbockers-Lift-Off-Graffiti-Remover-Trigger/dp/B00N3B25LU

However, I stupidly forgot to wait for the ball to dry and then placed it within an orange plastic carrier bag. Now the cleaning agent has reacted with the dye and there's a yellow-tint stain.

I have tried using acetone and these:

https://www.amazon.co.uk/Dr-Beckmann-Stain-Devils-Survival/dp/B07TKJPH7D/ref=sr_1_14?keywords=dr+beckmann&qid=1575488301&sr=8-14

but I cannot remove the stain.

Does anyone with chemistry knowledge know what else I could try?

UPDATE:

Got it out in the end. I used Dr Beckmann Stain Devils tea & wine, cut some cloth in to the shape of the area I wanted to clean, poured water, then the powder, then water on top and left for hours. Took multiple attempts.

The dye has probably migrated a bit into your ball`s surface, so whatever (solvent) you try to get it out with again, it might take a while.

Remember the classic school (chromatography) experiment: You draw a line with a felt pen on a piece of paper, 2 cm from the edge, and stick the paper into a glass filled only 1cm high with acetone.

You can try to transport the dye so far into the ball you dont see it any more, or try to "fill the glass up to 3cm with acetone". ;)

And remember: Its hard to tell how resistant your ball and its original painting is to acetone or hair bleach or soda.


What is an "attack" in Organic Reactions? I began to study organic reactions and found the term "attack" everywhere, what this "attack" precisely mean?

You might have heard the term "attack" in statements like "the nucleophile attacks the electrophile", and terms like frontside and backside attack, and what this basically means is that the nucleophile approaches the electrophile, and begins to share its electrons with the electrophile to form a bond between them (and possibly break a different bond in the molecule too). Frontside and backside attack mean the nucleophile approaches the front or the back of the molecule and forms a bond.

If you've learnt about arrow pushing, you'll know that an arrow has its tail starting at the nucleophile (the electron rich species) and it's head points to the electrophile (the electron poor species) - you can also think of "attack" as the nucleophile "attacking" the electrophile with electrons to react/form a bond, in the direction the arrow is pointing. I suppose we call it attack because it makes it sound more action-packed :-)


Getting an “index” of 7 for a Set of Powder Diffraction Data I know a value of $h^2 + k^2 + \ell^2 = 7$ is not allowed but I cannot understand where I am going wrong with the indexing here. Any insight would be greatly appreciated as I am wondering if this is made up data that has a typo? I calculated d using $d= \lambda{} /(2\sin{\theta{}})$ where $\theta$ (theta) is half the value of $2\,\theta$ (obviously) given

I think:
$$\frac{1}{d^{2}} =\frac{\frac{h^{2}\sin^{2} \alpha }{a^{2}} +\frac{k^{2}\sin^{2} \beta }{b^{2}} +\frac{l^{2}\sin^{2} \gamma }{c^{2}} +\frac{2hk}{ab}(\cos \alpha \cos \beta -\cos \gamma ) +\frac{2kl}{bc}(\cos \beta \cos \gamma -\cos \alpha ) +\frac{2lh}{ac}(\cos \gamma \cos \alpha -\cos \beta )}{1+2\cos \alpha \cos \beta \cos \gamma -\cos^{2} \alpha -\cos^{2} \beta -\cos^{2} \gamma }$$
Have you explored the case where $a\neq c$ such as tetragonal?
$$\frac{1}{d^{2}} =\frac{h^{2} +k^{2}}{a^{2}} +\frac{l^{2}}{c^{2}}$$


Alkaline Water Electrolysis with RbOH In industry, it appears the alkaline water electrolysis is largely done with solutions of NaOH or KOH - presumably due to low cost.

I presume that alkaline water electrolysis is possible under RbOH, however, I can not seem to find any evidence in the literature showing the use of RbOH for water electrolysis. Can water electrolysis be done with RbOH to give oxygen and hydrogen gas evolution at the working (TiO2) and counter (Pt) electrode respectively?

Thanks!

Of course! Rb is more electropositive than hydrogen, so hydrogen is released.

As you mention, $\ce{RbOH}$ is a bit more expensive than $\ce{NaOH}$. At present, $\ce{RbOH}$ is ~US\$500/kg and $\ce{NaOH}$ is ~US\$500/tonne, three orders of magnitude cheaper.

You might use $\ce{CsOH}$ or $\ce{FrOH}$, too. The Fr will even illuminate the reaction with Cherenkov radiation.


Miller Indices of Goldnanoparticles from Turkevich and Frens method I need the miller indices of gold nanoparticles that were synthesized after the route of Turkevich and Frens. I during my research I found a lot of publications, but it looks like no one measured it before.
Has someone an idea of what the miller indices can be? Or better a source or paper where I can found them?

Assuming you refer to a record like this:

which I found in Kimling et al. in «Turkevich Method for Gold Nanoparticle Synthesis Revisited» (2006JPhysChemB15700) when reporting X-ray diffraction.


What does the colon in µ-η2:η2-O2 mean? In the formula $\ce{[nBu4N]2[(UO2)2(\mu-\eta^2:\eta^2-O2)(NO3)2(\mu-Au(CN)2)2]}$, what does the colon stand for? Is this an official notation?

This notation is mentioned in the IUPAC Nomenclature of Inorganic chemistry, Recommendations 2005 (‘Red Book’):

  IR-10.2.5.1 The eta (η) convention

  The special nature of the bonding of unsaturated hydrocarbons to metals via their π-electrons
  has led to the development of the ‘hapto’ nomenclature to designate unambiguously the
  unique bonding modes of the compounds so formed. (…) The Greek
  symbol η (eta) provides a topological description by indicating the connectivity between the
  ligand and the central atom. The number of contiguous atoms in the ligand coordinated to the
  metal is indicated by a right superscript numeral, e.g. η3 (‘eta three’ or ‘trihapto’), η4 (‘eta
  four’ or ‘tetrahapto’), η5 (‘eta five’ or ‘pentahapto’), etc.

  (…)

  If an unsaturated hydrocarbon serves as a bridging ligand, the prefix μ (see Sections
  IR-10.2.3.1 and IR-10.2.3.4) is combined with both η and κ, where necessary. The colon is
  used to separate the locants of the bridging ligand which indicate binding to different metal atoms.

  (…)

  Examples:

  21.

  (μ-η2:η2-but-2-yne)bis[(η5-cyclopentadienyl)nickel](Ni—Ni)

(This convention is not used for unsaturated hydrocarbons only, there is also a more complicated example (25) with carbonyl (CO) ligand.)

Use of this notation for formulae does not seem to be explicitly mentioned or “allowed”. If it is not an omission in the rules, I think that such construct could be called a name-formula hybrid, which might useful for detailed description the discussed structure detail in a shortest way possible.

The $\ce{[^{n}Bu4N]2[(UO2)2(\mu-\eta^2:\eta^2-O2)(NO3)2(\mu-Au(CN)2)2]}$ polymeric complex is described in [1], which talks about $\ce{U-O2-U}$ linkage with bidentate peroxo ($\ce{O2^{2-}}$) unit.

References:

Brown, M. L.; Ovens, J. S.; Leznoff, D. B. Dicyanoaurate-Based Heterobimetallic Uranyl Coordination Polymers. Dalton Transactions 2017, 46 (22), 7169–7180.


If I mix carbon dioxide and sodium hydroxide in the right amounts in water, do I get baking soda? Suppose I need a sodium bicarbonate solution (for non-food related reasons). Can I get something resembling it by mixing $\ce{NaOH}$ and $\ce{CO2}$ in water?

Of course one should not attempt to make food in a chemistry lab, so lets rule out the food safety requirements for food-grade baking soda.

This is certainly one method by which sodium bicarbonate can be produced.  According to this Wikipedia article:

$\ce{NaHCO3}$ may be obtained by the reaction of carbon dioxide with an aqueous solution of sodium hydroxide. The initial reaction produces sodium carbonate:
$$\ce{CO2 + 2 NaOH → Na2CO3 + H2O}$$
Further addition of carbon dioxide produces sodium bicarbonate, which at sufficiently high concentration will precipitate out of solution:
$$\ce{Na2CO3 + CO2 + H2O → 2 NaHCO3}$$

The net reaction for this two-step process is then simply:
$$\ce{CO2 + NaOH -> NaHCO3}$$
Note that this is not the common method by which sodium bicarbonate is produced commercially, but it is a perfectly valid method of it's synthesis.


Acid catalysed dehydration of 5-methylidenecyclopent-2-en-1-ol I think the following reaction is NOT possible.

I could not find a mechanism for ring expansion. Is ring expansion possible in any way?
Is the following mechanism correct?

I'd say your mechanism is acceptable. However, the final product, pentafulvene (PIN: 5-Methylidenecyclopenta-1,3-diene) is said to be not stable at room temperature. Generally, fulvenes are thermally unstable, sensitive to oxygen and photosensitive (Ref.1). They are also prone to acid- and cation-catalysed polymerisations. For example, there have been numerous reports of pentafulvenes undergoing dimerization via Diels–Alder cycloadditions at room temperature similar to that of cyclopentadiene (e.g., Ref.2):

Based on these facts, I'd like to suggest an alternative path for the reaction giving much stable cyclopentaenone derivative:

References:

Ellen Swan, Kirsten Platts, Anton Blencowe, "An overview of the cycloaddition chemistry of fulvenes and emerging applications," Beilstein J. Org. Chem. 2019, 15, 2113–2132 (doi:10.3762/bjoc.15.209).
Beat Uebersax, Markus Neuenschwander, Hans‐Peter Kellerhals, "Fulven‐Dimere: Synthese, Strukturbeweis and thermisches Verhalten," Helvetica Chimica Acta 1982, 65(1), 74-88 (https://doi.org/10.1002/hlca.19820650109).


I am having problem with this question including topics of redox reactions and charges What is the oxidation number of $\ce{Fe}$ in $\ce{K4[Fe(CN)6]}$?
This seems to be a pretty easy problem. Just input the oxidation numbers of everything and equate for 'x'. But the problem is that I don't know the charge on the Cyanide ion right there. Now, I knew from doing many questions that $\ce{Fe}$ usually has +2 or +3 so i guessed +2. But I don't want to be guessing on the actual day.
I essentially need to know the best way to find charges on polyatomic ions because there's just loads of them and I don't think I can remember over 300 oxidation states. I know the charge can be calculated by adding up Oxidation states but most elements have variable Oxidation states too so I am really confused.
Even if you know a certain website that may be useful, please link it. Any help is appreciated.

As the formula of the substance is known, you should first state the formula of the ions produced when the substance is dissolved into water. Here, $\ce{K_4Fe(CN)_6}$ gets dissolved in water and produced $4$ ions $\ce{K^+}$ so that the $4$ corresponding negative charges  must be fixed on the remaining anion, which has the formula $\ce{[Fe(CN)_6]^{4-}}$ with $4$ negative charges. Now you have to think how to synthesize this complex anion from $\ce{1 Fe^x+}$ ion and $\ce{6 CN^-}$ ions. The equation is $$\ce{Fe^{x+} + 6 CN^-  -> [Fe(CN)_6]^{4-}}$$ Counting the charges gives x positive charges from $\ce{Fe^{x+}}$, 6 negative charges form the six cyanide ions, and this makes a total of $\ce{4-}$. So the initial $\ce{Fe}$ ion must have been charged $+2$
The nature of the ion $\ce{CN^-}$ remains to be explained : Well ! The carbon atom has 4 independent valence electrons. Nitrogen has 5 valence electrons, but 2 of them are taken in a doublet. So there is only 3 available electrons in N atoms for bonding. When 1 C is attached to 1 N, this makes 3 covalent bonds between C and N. As a consequence one electron is still available on the carbon atom. This is not a stable compound. The carbon atom needs one more electron to get 8 electrons in its outer layer. This electron can be offered by one H atom, which would make a covalent molecule HCN, with one covalence between H and C, and three covalences between C and N. But this last electron can also be given by a sodium Na or a potassium atom K. This would produce the compound NaCN or KCN, where  Na (or K) and the carbon atom are bound in a ionic bond between $\ce{Na^+}$ (or $\ce{K^+}$) and the ion $\ce{CN^-}$. This is the origin of the ion cyanide $\ce{CN^-}$.