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Formating chemical equations for proteins binding in hugely complex configurations
In a previous question I asked about a simple case, I'm now trying to write equations for proteins that bind multivalently. In the simplest case proteins $A_1$ and $A_2$ can bind to up to two $L$ proteins. This gives a massive chemical equation:
`
where $(AL_1AL_2)_r$ is a tetramer where $A_1$ and $A_2$ are both bound to both of the $A$ proteins in a ringlike structure. But this diagram misses out one possible state, $A_1LA_2$ which is connected in this way:
.
I can't see a way of joining this extra state to the above chemical equation without arrows crossing each other.
What is the best way of formatting these kinds of complex protein reaction equations?

Anyway, this is a question for TeX.SE. Or maybe even for math.SE (these things look mighty reminiscent of so-called commutative diagrams), just remove everything chemical before asking it there.

Agreed that it is a maths problem first, then a TeX one. Frankly, if the diagram is going to be too complex to layout, then it will be too complex to read. Several different diagrams (as you show here) will be more understandable,

How would you divide it in to several different diagrams?

@IvanNeretin i've corrected it

This isn't a Tex problem, I can use xypics fine to generate the above, what I'm interested in is how protein chemists format these kinds of equations (even if it means splitting them up).  My external examiner has insisted on chemical equations for every system I investigate in my thesis, so  I have to include no matter how ugly it is.
There are well studied systems dramatically more complex than this (like microtubules ), so clearly there must be a standard notation used by protein chemists.

Generally speaking, for these sorts of complex reactions, you want to go to more than two dimensions. (At least for the initial bit.)
For each potential binding reaction you have a different dimension. That is, for $L$ binding to $A_1$ on site 1, you go horizontally. For $L$ binding to $A_2$ on site 1, you go vertically. For $L$ binding to $A_1$ on site 2, you go in the Z dimension, and then for $L$ binding to $A_2$ on site 2, you go in the 4th dimension.
Obviously, you can't put an actual 4D representation on paper. In fact, you can't put a 3D representation on paper. But if you're lucky, you can do a bit of topological rearrangement of the network and then do a projection of your network onto a 2D plane.
For example, if you have three potential binding interactions, each independent, then your system can be represented as the corners and edges of a cube. You can then represent this system by any of the standard projections of a 3D cube into 2D space.
For four potential binding interactions, it's the the corners and edges of a tesseract. Projecting a tesseract onto 2D space is much harder, in that there's no way to do it without lines crossing. But depending on what you want to represent, you might be able to adequately mark the crossings, or add "jumps" between two sub-graphs. (That is, have two more sub-graphs and have a systematic way of going from nodes on one graph to the node on the other graph with implicit reaction edges.)
For an arbitrary reaction graph, there's no guarantee that the crossing number of the graph is going to be zero -- that is, there's no guarantee that you'll be able to represent the network on a 2D page without crossings or jumps. Often the best you can do is attempt to be systematic about things, and then add crossings/jumps as appropriate.


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Why is the oxidation state of Carbon (-4)?
Based on my understanding, atoms try to achieve the electronic configuration of the nearest noble gas, by losing or gaining electrons. For example, Sodium is closest to Neon, so loses an electron so the outer shell is full (2nd shell is full). This doesn't include the transition metals or the lanthanides or the actinides, which I realize follow something different because of their orbitals.
In the case of Carbon, it is equally distant from Neon as it is from Helium. There's the octet rule, but why exactly does it prefer Neon's configuration over Helium?

Valency don't have a sign. Do you mean oxidation state?

And -4 Is far from the only possible oxidation state for carbon.  We can get any whole number from -4 to +4 inclusive, with some fractions thrown in on the side (e.g. cyclopropenyl ion =+(1/3)).

Basically, Carbon forms covalent bonds in majority of compounds. It doesn't mean that it has lost its electrons to the atom to which it is bonded to - rather it shares its electrons so as to get noble gas configuration.
Oxidation state is a hypothetical number assigned to each atom in a molecular structure (refer this) which assumes all bonds to be purely ionic. So it doesn't mean that that the carbon atoms in $\ce{C_2H_2}$ having an oxidation state of $-1$ have gained $1$ electron each to result in 5 electrons in the valence shell.
If you meant the $\ce{C^{4-}}$ (carbide) ion, carbon actually prefers this because removing 4 electrons is difficult as it is quite endoergic as it requires nearly 14000 kJ/mol (refer this) to form a $\ce{C^{4+}}$ basically due to its small size, which means that electrons are closer to the nucleus and we can remove it only at the expense of high energy. Thus, $\ce{C^{4-}}$ is comparatively easy to form as it is easier to add electrons into the valence shell.

$\ce{C^{4-}}$  is impossible to form.  No atomic ion with 2 or more negative charges is bound.  Only when imagining a compound as purely ionic does  $\ce{C^{4-}}$  come to mind.

Yes. I was just comparing the ability to remove or add four electrons to a carbon atom

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Froth floatation process - Use of stabilizers in froth floatation method
How exactly do the stabilizers affect the process of Froth floatation method? I mean how do they stabilize the froth in the process?

Bubbles tend to collapse when the liquid that makes up the walls drains away and the walls weaken. The walls can be strengthened by choice of surfactant (and addition rate) and by adding thickeners like water-soluble gums (e.g., xanthan) or polymers or glycerol (also for children's bubbles).
The small particles being floated may act as imperfections that break the walls more easily, so coating the particles with appropriate polymeric dispersants can make the (coated) particles more compatible with the (thickened) wall.
On the other hand, many industrial operations sometimes develop undesirable bubbles/foam which be eliminated by spraying a small amount of low surface tension silicone or non-viscous solvent like isopropanol on the top of the foam. Avoid these if you need to stabilize the foam.

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Permanganate oxidation of side chains in Benzene Derivatives
What is the exact mechanism of permanganate oxidation of side chain derivatives of benzene? Is it same for all functional groups? I want to know that in any situation I mean when side chain has only one benzylic hydrogen gives the carboxylic acid. Why does it occur?

I have changed the question. @Mithoron I wanted to know why exactly this is occurring. Even with one benzylic hydrogen, the product is same. Is there any definitive answer to my question?

Answer to original describes mechanism in general way. And does answer this.

But I want more exact mechanism for my question.


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How large an energy difference in Hartree can be used to distinguish two different structures?
For example, I use two different geometries of the same molecule to do optimisation in Gaussian, and the two resultant geometries look roughly the same. The energy difference between the two optimised geometries is only 0.0000001 a.u.. So can I view them as the same geometry? Or is the energy difference large enough to distinguish them? How large an energy difference is large enough for you to see two geometries as different geometries?

When you do a geometry optimization, one of the convergence criteria is the change in energy between iterations.
I think  0.0000001 a.u., is below the threshold for convergence in energy (if you use standard settings).
This suggest that the two structures are the "same".

I wouldn't use energy as a general criteria to discriminate conformations; a root mean squared deviation (RMSD) of superimposed structures would be a better metric. This is because two very different conformations may have very similar energy. For example, if you perform a relaxed energy scan of one of the dihedrals of cyclopentane, you get nearly the same energy for most of the dihedral angles, bu very different ring conformations.
If you found an energy difference of 0.0000001 a.u. and the superimposition of the conformations is nearly perfect, I'd say they are in the same minimum of the energy surface.

Your talking about confirmation is a good example. But in my case, the two geometries are not conformers. They are just slightly different in bond lengths (around 0.0002 angstrom). I'm thinking whether I can see them as the same point on the potential energy surface, or they are actually two different points...

I consider them to be in the same point in the energy surface - 0.0002 angstrom and 0.0000001 a.u. is pretty much nothing! Gaussian stops the minimization because it considers that both structures have optimized. If the minimization was continued, they would certainly land in the same exact conformation.

By the way, what are the details of your criteria for superimposed structures? How small a deviation is acceptable to see two geometries as the same minimum on PES?

I doubt there is a need for a cutoff. If a molecule is fully optimized from multiple starting structures, there is either a very small deviation (e.g. 0.0002) or something much bigger (e.g. 0.5 angstroms). Did you ever find something between - like 0.01 angstroms?

That's a pretty incredible threshold. To put things in perspective, if $1 E_{\mathrm{h}} \approx \pu{627 kcal/mol}$, then $\pu{0.0000001 a.u.} = \pu{6.27e-5 kcal/mol}$. Energy differences between conformers are often only a few kcal/mol, but this is a full 5 orders of magnitude smaller. Even $k_{\mathrm{B}}T$, which is $\pu{207 cm^{-1}}$ at room temperature, is $\pu{0.59 kcal/mol}$. If two molecules are truly the same conformer, then small geometry differences (bond lengths, angles) will 1. computationally converge to identical structures by slightly tightening convergence criteria, and 2. exist experimentally as identical vibrationally-averaged structures. That is, any vibrational mode has enough energy to cover this difference. Using a similar argument to the above, $\pu{1 cm^{-1} = 0.0029 kcal/mol}$, which is 2 orders of magnitude larger than the criterion you gave.
Consider the default energy convergence criteria for a few popular packages:
$$
\begin{array}{lc}
\hline
\text{Program} & \text{Max energy change}\,(E_{\mathrm{h}}) \\
\hline
\text{Gaussian 16} & \pu{1.0e-7} \\
\text{Molpro 2015.1} & \pu{1.0e-6} \\
\text{ORCA 4.0} & \pu{5.0e-6} \\
\text{Psi4 1.1} & \pu{1.0e-6} \\
\text{Q-Chem 5.0} & \pu{1.0e-6} \\
\hline
\end{array}
$$

So can I view them as the same geometry? Or is the energy difference large enough to distinguish them?

For two molecules you know to be the same conformer, $\pu{0.0000001 a.u.}$ is indistinguishable and not far from numerical noise due to loss of precision.

How large an energy difference is large enough for you to see two geometries as different geometries?

I am not aware of an official reference. However, for the case of a typical organic molecule that doesn't have shallow basins in its potential energy surface, I would consider the default geometry optimization convergence criteria enough.
Useful links

Gaussian manual (see the "Options" tab)
Psi4 manual
Q-Chem 5.0 manual


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Do all decomposition reactions involve heat?
When water decomposes into Hydrogen and Oxygen as in 2H2O—-> 2H2+O2 does this involve heat? Is it always implied that heat is involved in a decomposition reaction or does this just spontaneously occur? Chemistry 1st semester so excuse the fundamental nature of the question. Thanks

Water: yes. Always: no.

Plainly put: no, decomposition reaction does not always have to entail providing heat. A decomposition reaction (in simple language) is when one compound breaks into multiple other compounds, like so:
$$\ce{A -> B + C}$$
As a general rule, if the product side is relatively more stable, it will decompose without you having the to provide extra heat. In the case of the reaction you gave, $\ce{H2O}$ is more stable than $\ce{H2}$ and $\ce{O2}$, and thus you will have to provide heat for the breaking of bonds.

Check out the iron oxygen phase diagram.  Wustite ($\ce{FeO}$) decomposes upon ... cooling, below 570°C, to magnetite ($\ce{Fe_3O_4}$) and metallic iron.  Steel strip manufacturers deal with this reaction because it impacts the acid pickling of hot mill scale.

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Arrhenius equation in exothermic reactions
According to the Arrhenius equation, the rate of reaction increases when the temperature increases. But according to Le Chatelier's  principle, the reaction rate decreases when the temperature increases.
So is Arrhenius equation applicable on exothermic reactions?

It would then be the case that although both the forward and backward rates increase, the rate of the backward reaction increases more than the rate of the forward reaction for the exothermic reaction. Thus, both physical postulates are applicable.

If you keep the temperature of the reaction mixture constant, the rate of the reaction (whether endothermic or exothermic) is higher at a constant higher temperature than at a constant lower temperature.

An important distinction is that Le Chatelier's principle indicates thermodynamic behavior while the Arrenius equation is a matter of kinetics. If you run a hydrogenation reaction for example, increasing the temperature has an undesireable effect on maximum yield following Le Chatelier's principle, but you are also getting that yield much quicker, so it often pays off.

The Arrhenius equation is absolutely applicable to exothermic reactions and it will give results in accordance to Le Chatelier's principle.
Applying the Arrhenius equation for the forward reaction yields:
$$k_\mathrm{f} = A \exp\left(\frac{-E_\mathrm{f}}{RT}\right),$$
applying the Arrhenius equation to backward reaction yields
$$k_\mathrm{b} = A \exp\left(\frac{-E_\mathrm{b}}{RT}\right).$$
We know that $E_\mathrm{f} -E_\mathrm{b} = \Delta H$. Dividing $k_\mathrm{f}$ by $k_\mathrm{b}$ we obtain:
 $$\frac{k_\mathrm{f}}{k_b} = A \exp\left(\frac{-\Delta H}{RT}\right).$$
For an exothermic reaction $\Delta H < 0$. So $-\Delta H/(RT)$ is a positive number. As $T$ increases, the exponent decreases and thus $k_\mathrm{f}/k_\mathrm{b}$ decreases, which is nothing but the decrement of the equilibrium constant.
Thus the result is in perfect accordance with Le Chatelier's principle.
Also, you can apply the van't Hoff's equation which states:
$$\ln\left(\frac{k_2}{k_1}\right) = \frac{\Delta H}{R}\cdot\left(\frac 1{T_1} - \frac 1{T_2}\right),$$
which also gives the decrease of rate of reaction with increase in temperature as $\Delta H < 0$.

You've assumed that the frequency factor is the same ($=A$) for both the forward and backward reactions. What is the justification for doing so? Afaik, it depicts the rate of collision of molecules with each other, and since the reactant side and the product side have different molecules, they should have different $A$ values. Thanks!


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Shouldn't enthalpy be constant measured at constant volume?
Shouldn't enthalpy be measured at constant volume? Then the work done on and by the gas is 0 meaning that the temperature change is caused solely by heat released or absorbed by the system. You'd be able to calculate
$$\mathrm dQ = \mathrm dT \cdot C_\mathrm s$$
(specific heat).
However, since enthalpy is at constant pressure the substance will do or have work done on it and thus there is no easy way to relate the enthalpy change to the change in temperature. Am I mistaken somehow?

If you want to define a new function then go ahead , but it wouldn't be enthalpy.

Enthalpy is heat flow associated with a constant pressure process.  Here is how:
Start with the familiar first law of thermodynamics for energy change:
$\Delta E=q+w$
$E$=energy
$q$=heat flow into the system
$w$=work done on the system
The work done is rendered thusly:
$w=-\int_\text{start}^\text{end}P\,\mathrm dV$, $P$=pressure, $V$=volume
Pressure is deemed constant so:
$w=-\int_\text{start}^\text{end}P\,\mathrm dV=-P\Delta V=-\Delta(PV)$
And then our first law balance equation becomes:
$\Delta E=q-\Delta(PV)$
$q=\Delta(E+PV)$
So heat flow is the change in a combination of state variables involving energy, pressure and volume.  This combination is what we define as enthalpy and usually label $H$.  Thereby, this enthalpy variable is designed to automatically include (mechanical) work inherent to a constant pressure process.

Trying to figure out the answer to my question I've read and tried to understand this equation a million times. I see how the terms cancel but if we picture the situation things start falling apart. React a gas exothermically so that it expands and it will do work. The enthalpy change is no longer equal to the change in heat.

Why not, if the reaction was run at constant pressure?  We see above that enthalpy change must equal heat flow in a constant pressure process.  Your example just has a big work term included in the enthalpy change.

Enthalpy was just defined for constant pressure.
It's not like that there was a need of new quantity, and it was taken at constant pressure, but there was a need of a quantity defined for constant pressure. This is because in real life most reactions are not carried out at constant volume, but in test tubes or flasks under constant pressure, that is, atmospheric pressure.
Also, there is already a quantity defined for constant volume, Qv. It's not necessary that a new symbol should be given to a quantity, as you said, it will be equal to heat absorbed or released, therefore it's symbol is related directly to heat.


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Why doesn't oxygen replace chlorine?
If oxygen is more electronegative why doesn't it replace chlorine in compounds?
Example:
$\ce{2NaBr + Cl2 -> 2NaCl + Br2}$
works, while
$\ce{4NaCl + O2 -> 2Na2O + 2Cl2}$
doesn't.

You can displace chlorine from hydrogen chloride.
With sodium chloride, if you did form sodium oxide it would be a strong base.  As such it extracts a $\ce{Cl^+}$ moiety from any potential chlorine molecules.  Sodium chloride is oxidized by oxygen, but the oxidized chlorine ends up as chlorine oxyanions such as hypochlorite (i.e., oxide coordinated to chlorine).

It's redox not acid base.

Why would it not be both?

I'm still not sure what you even mean in this. Proper answer is that O2 has to low redox potential.

Then say so in your own answer.

Re: "chlorine gets oxidized to oxyanions such hypochlorite" So, the products formed from $\ce{Na2O+Cl2}$ are $\ce{NaCl+NaOCl}$? Thanks!

In order to know if a reaction is spontaneous or not (if it will happen towards the direction you have drawn it) you must look at its Gibbs energy, ΔG. Electronegativity of one element in a reaction is one of the many factors that contribute to it. In this case you are dealing with a redox reaction (because the oxidation states of the elements involved change) and therefore you need to use the relationship: ΔG=-nFE$^0$cell with E$^0$cell calculated from the relevant half reactions. For more details read an introduction to electrochemistry from some inorganic chemistry book.


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How do energy changes in reactions avoid violating the conservation of momentum?
As mentioned in this question, the idea of chemical potential energy explains how the thermal/kinetic energy of particles can change in a reaction; energy is taken in / released by the bonds broken / formed.
However, consider the reaction A -> B + C occurring in a closed system of just these single particles.
A reaction such as this is often endothermic. Therefore, the kinetic energy of the products must decrease. However, it is impossible to remove kinetic energy from the system without changing momentum (unless B and C are travelling in opposing directions to begin with, which is not the case. If they split in opposite directions, that will have already resulted in a KE gain).
How can this happen without violating the principle of conservation of momentum?
EDIT: Here is an explanation showing that it does not work if B and C travel in opposite directions:

Another point is that the average KE relates to T for a statistical ensemble.  Your standing still A molecule would be in your picture at 0 K. It is not clear what you mean, at least to me. The point I want to make is that KE will decrease unless you postulate the starting KE is zero.

Or put it like this: B + C --> A is not an elastic collision at all. It is again related to the previous Q you have mentioned.

@Alchimista There is no B + C -> A occurring. The issue is the decomposition A -> B + C. As you state, KE will decrease. I am asking how this can be reconciled with the conservation of momentum, given that decreasing KE will violate it.

I'm voting to close this question as off-topic because it's not about chemistry.

@Mithoron It's about the crossover between chemistry and physics. I could post it on the physics Stack Exchange, but what's to stop someone over there saying the same thing?
I'm asking what part of the behavior of molecules allows the laws of chemistry to agree with the laws of physics, because at the moment I can't see that they do. Sounds like a chemistry question to me.

Well, molecules aren't some balls, having only translational energy. Even your balls can rotate...

Your (a-5)^2 is the heat of reaction.

@Alchimista And it is impossible for (a-5)^2 to be less than zero, thus there are no solutions for the velocity a that allow kinetic energy to decrease while momentum stays constant. Therefore it appears impossible for both the reaction to be endothermic and the conservation of momentum to hold. You see the problem I'm having? I apologise if I come across as terse, but if you have an explanation, please post it as an answer rather than leaving cryptic comments that appear to misunderstand the question. I do appreciate your time in looking at this question.

Yes I see but I had no time to write an articulate answer as your problem could require. But perhaps it will suffice to say that you cannot put a disequaliy but rather a conservation of energy accounting KE terms AND energy of reaction. Let me know. Plus I would have other comments such as KE of your system does not relate to T as well as the moving in opposite direction is a certain situation. Let me know perhaps I could better answer if still necessary

@Alchimista In which way does KE not relate to the temperature? Thermal energy is a measure of the average KE of the system. If the temperature decreases, the average KE of the particles must decrease, so the total KE of the particles must decrease. That fact is independent of any individual situation or set of particle directions. The issue is that if KE does decrease, momentum must also decrease, which violates the conservation of momentum. I'm starting to think no-one knows the answer to this question, and the model is broken, or something.

Put delta H in the initial energy balance and redo calcs. For the other points T KE relationship is derived statistically and as such is valid for a system about Navog or so. If you have a standing still A the way to get the reaction is to input AH. If a is moving it crash to the wall and split taking energy from KE. No wall A goes with every velocity you want.  You can calculate every scenario but starting with the right energy balance.

Is air hotter when the wind is blowing?

check last paragraph of section 42-4 in Feynman lectures volume 1. http://www.feynmanlectures.caltech.edu/I_42.html

Try to do more than just provide a link to a source when you post an answer

While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference.  Link-only answers can become invalid if the linked page changes. - From Review

The important bit of the paragraph linked by Joseph Tang is as follows: "Incidentally, there is some trouble with such a reaction, A plus B, making AB, because we cannot conserve both energy and momentum when we try to put two objects together to make one that is more stable. Therefore, we need at least a third object C, so the actual reaction is much more complicated." Thus, as the accepted answer suggests, the proposed mechanism is too simple, and an outside influence is needed for the reaction to occur.

That's an excellent question, and your reasoning proves exactly what it seems to be proving: that such a scenario is not possible. A molecule can't just up and transform its own kinetic energy into some endothermic reaction within itself, since the conservation of momentum forbids that. Whether or not the molecule snaps in two (or maybe more) parts, the total kinetic energy must become lower, and the momentum must remain the same. There is no way to reconcile these contradicting demands.
How are endothermic decomposition reactions possible, then?
By looking at the bigger picture, that's how. Say, a molecule bumps into a wall or another molecule. Its momentum is lost; the constraint no longer applies. (The total momentum of the system is of course still conserved, but the system is now bigger than just one molecule). Now the molecule may bounce off like a tennis ball and fly away at the same speed (that's what we call an elastic collision). Or... it may react.
So it goes.

Actually you should correct the final paragraph as for momentum is always conserved.

And no, I want to make clear that I amnot he downvoter. Normally I would have just suggested to you a correction. Even more so after our opinion exchange.

OK, got it. Maybe I should have made it more clear.

Ivan your answer is clear.  Just look at the final paragraph

How do you measure temperature? The thermometer has to be in thermal equilibrium, so you need to send the thermometer with the object. But that means that temperature, for our purposes, is going to be invariant between different reference frames. No matter how you're moving, the reading on the thermometer is the same.
This is clearly not the case with translational kinetic energy of the bulk material. In the rest frame of the object moving with constant velocity, the kinetic energy is zero.
Therefore, you should not mix up the thermal kinetic energy of the system with the translational kinetic energy of the system. They can't be the same thing. In other words, when the exothermic system loses energy, you don't get to subtract that from the translational kinetic energy.
See also these discussions: here and here.
I would also like to point out that you've just described a rocket. And rocket science, while complicated, is still obeying conservation of energy and momentum.

Translational kinetic energy of the system is irrelevant. Look at one molecule (in an inertial reference frame, if you'd like); can it decompose?

@IvanNeretin I think we're saying the same thing. Translational kinetic energy is irrelevant when you consider the thermal energy, even if that's derived from random translational kinetic energy.

Temperature is not a vector quantity though the KE of an individual particle is
The apparent problem here arises because of a confusion between the bulk view of a reaction and the view from the perspective of individual molecules.
Individual molecules have kinetic energy and, in reactions, the kinetic energy may change if chemical energy is released or absorbed during a specific reaction (this is a conversion between potential energy in chemical bonds and kinetic energy). But this doesn't translate into the bulk perspective. Temperature is, in one sense, the average kinetic energy of the molecules in the system. Different molecules have different kinetic energies (Boltzmann distribution of them, in fact) but the net vector of those kinetic energies is zero as they are moving in random directions. Moreover, they are constantly exchanging their kinetic energy in random collisions with other molecules.
If the reaction vessel is not closed (so the products can escape) this point of view is incomplete. The net kinetic energy of the products will no longer be zero (the molecules can escape through the open part of the vessel taking their kinetic energy with them). The molecules can strike the closed end of the vessel and bounce off transferring momentum to the vessel and this is not compensated by by molecules striking the other end of the vessel as there isn't one. This is how rockets work, in effect translating the chemical energy released by a reaction into kinetic energy which is ultimately transferred into the kinetic energy of the rocket.

This answer is meant to put some order to my comments that, lets fragmented, appeared cryptic to the OP.
Moreover I try to answer without editing. The involved calculations just require second degree equations thus every reader can easily figure them.
The OP is forced to confusion because he assumed that the kinetic energy KE of one particle must be reflected in temperature T. This is a wrong assumption as for in a statistical ensemble of particles, a number of them have KE lower or higher than the average value, which is the one related to the macroscopic value T of the system.
See, for instance, https://physics.stackexchange.com/questions/65690/can-a-single-molecule-have-a-temperature
where all the answers deserve attention but the most voted one suffices to the current discussion.
The wrong assumption, once made, drives OP to think that the dissociation A --> B + C, being endothermic, results in a lower T of the products, or in other words the process takes energy at cost of the KE of the particles.
This is true for an ensemble in which collisions occur but obviously not in the system under consideration.
The only way for A dissociating is to input energy into the balance.
The system is easily treated supposing molecule A at rest, either isolated or in the ideal box as in the question.
Nothing happens to it unless the enthalpy of reaction DeltaH is inputted to the system in a way or another, let us assume is already in the box if we want to keep it closed and isolating.
In this case the conservation of momentum requires
mBvB = mCvC
and the conservation of energy
DeltaH = KEB + KEC.
There is not a disequality (forced by the vicious assumption endothermic process => T decreases => KE decreases) but just energy conservation.
In the case of A moving with an initial speed the treatment is the same. It suffices to treat the system in a new reference frame applying galilean relativity.
The thermochemistry and the physics of the process remains exactly the same.
In short, the term (a - 5)^2 sketched as example by the OP equals DeltaH.
A more realistic scenario is as follows. Let us open a small hole to the box and shoot A through it with a certain speed. As soon as A enter the box the aperture is closed and the box is isolating.
Now we can wait until A collides to the wall. Assuming KE of A is enough, the collision with the wall*** (not the following dissociation!) can be treated as an elastic one which solely triggers the dissociation reaction itself. In this case the DeltaH is taken at expense of kinetic energy, but indeed there is a collision transferring the required amount of energy. In a way the scenario is more similar to that of a statistical system of particles (equipartition and microscopic interpretation of T do not hold, tough and still).
The equations are now***
mAvA = mBvB + mCvC
and
KEA - DeltaH = KEB + KEC
Ideally B and C will recombine along their trajectory giving back A and so on.  However no work no heat can be extracted.
Although the real molecule(s) have other channels to store/transfer energy, the above arguments shall answer the question.
***for the treatment of the collision to the wall, and specifically for the reversal of velocities, the trick is as for the elastic bouncing of a small ball against a rigid and much more massive body.

Temperature is irrelevant. Ensembles are irrelevant. Look at one molecule. It has some kinetic energy and some momentum. Can it decompose?

@Ivan Neretin don get your comment.  It is basically my answer. I had to explain why OP was wrong in assumption.  Of course the molecule can't decompose by virtue of just kinetic energy , wasn't clear from my text?

@Ivan Neretin. I copy and paste from Answer above. " Nothing happens to it unless the enthalpy of reaction DeltaH is inputted to the system in a way or another,".

Then you answer contains a lot of unnecessary statements, and, while technically right, may still appear cryptic to a less knowledgeable reader.

@Ivan Neretin. My answer is just your answer worded as such to answer OP. Going through the wrong assumption he made and correcting his p and KE balances. You could have just not commented under it.

I wouldn't even call any of the assumptions wrong per se. Never mind. Let the OP decide which answer to accept.

Both of your answers are helpful, so thank you. However, I'm accepting Ivan's answer because it conveys the same understanding more concisely.

Ok. Just to remind you '" Or put it like this: B + C --> A is not an elastic collision at all. It is again related to the previous Q you have mentioned." Wasn't enough.


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Why doesn't 4 hydrogen and 1 carbon form a ionic bond? What is the limit of electrons that can be given away?
I understand that it's a covalent bond, but to my uneducated mind isn't it possible to have the carbon give 1 electron to each of the hydrogens and form an ionic bond instead? I read this response
Why can't carbon form an ionic bond?
which kind of makes sense, which leads me to my main question. Is there a limit to how many electrons(in this case) a carbon can drop/give before it's not possible? Does it depend on how badly the other(s) elements want to take/give electrons in said reaction?
As in, it has to give up 4 electrons, would it be fine it if had to give up 2? etc, hopefully my question makes sense.

I can understand your reasoning. But we need to consider many factors. When a compound is to be formed it should be stable enough. Or else the compound breaks down. Here the electronegativity of Carbon is 2.55 and Hydrogen is 2.2 (According to  Pauling's scale). So the electronegativity difference between these two atoms is 0.35. In the table given below...,

We can see that the nature of bonds between atoms of different electronegativity. As the difference is 0.35, the bond is Non-polar. And also for removing electrons from carbon are...,
Ionisation energy number & Enthalpy(kJ/mol)
1st------------------------->1086.5;
2nd------------------------->2352.6;
3rd------------------------->4620.5;
4th------------------------->6222.7.
So total energy required is 14282.3 KJ/mol. But electron affinity of hydrogen is 72.8 KJ/mol.So four hydrogens give 291.2 energy. So totally we need 13991.1 KJ/mol of energy to form methane.
While for removing successive hydrogen atoms from methane the bond-dissociation energies are 104 kcal/mol (435 kJ/mol) for D(CH3–H), 106 kcal/mol (444 kJ/mol) for D(CH2–H), 106 kcal/mol (444 kJ/mol) for D(CH–H) and finally 81 kcal/mol (339 kJ/mol) for D(C–H). The bond energy is, thus, 99 kcal/mol or 414 kJ/mol (the average of the bond-dissociation energies). Here the formation of methane is exothermic and is spontaneous while in the above situation it highly endothermic. All these factors give us why C-H bond is covalent but not ionic. Also you can do the same for either two bonds or three bonds anything. So we can say that $\ce{CH4}$  is covalent and cannot be ionic. And two bonds also cannot be ionic.
But in October 2016, astronomers reported that the very basic chemical ingredients of life—the carbon-hydrogen molecule (CH, or methylidyne radical), the carbon-hydrogen positive ion (CH+) and the carbon ion (C+)—are the result, in large part, of ultraviolet light from stars, rather than in other ways, such as the result of turbulent events related to supernovae and young stars, as thought earlier. So we can say these ions are in existence in our universe.
I hope it helps you.
I'm new to Stackoverflow so sorry if I've made any mistake in giving your desired answer
.

Your diagram isn't right. You simply can't have "100% ionic character" in chemical bond, because it wouldn't be chemical bond.

Sorry for it. I know that 100% ionic isn't possible. Forgive me for that error.

Here I tried to show that Carbon cannot form an ionic bond with hydrogen. So it will be good if you notice the mistake in the diagram and move on with the answer.

I think I understood your question.
In case of ionic bonding, an electron is "lost" by the one atom and gained by the other atom. So, it depends on how badly electrons are wanted or not wanted - quantified by electronegativity. More the difference in electronegativity, higher the ionic nature of the bond.
In case of covalent bonding, no atom wants electrons to be "lost". So, instead of absolute "give and take", they share. This is what happens in case of $\ce{CH4}$ as well.
Finally, the limit that you speak of does not appear as "how many electrons", but as "what difference in electronegativity". In practice, no bond is perfectly ionic and has some covalent character, but the ionic character is directly related to the difference in electronegativity.


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Clarifying understanding of shifting equilibrium in dynamic systems:
Firstly, let me confirm that my definition for right shift is correct: a right shift entails an increased ratio of product to reactant.
Say you have the reaction A -> B. The standard notion is that increasing concentration of the reactant shifts the equation to the right but that can obviously not be the case here. Explanation: Take two systems composed of A -> B respectively and add them together. You'll end up with 2A -> 2B with the ratio remaining the same.
However, if you have the equation A + B -> C, what will happen is that a greater percentage of B will react, and then the equation will shift to the right relative to B, but to the left relative to A, and to the left relative to the equation as a whole. However, if you increase both A and B by the same factor no shift will occur.
My overarching question is then whether I'm correct stating the above, and whether as I've described shifts left and right occur not relative to a side of the equation but rather relative to a specific part of it.

Your third paragraph is a bit unclear is " A + B -> C" in continuation with the earlier given paragraph, or are you considering a new traction altogether. Your arrows are not the ones for equilibrium.

Your understanding of Le-Chatlier's principle is wrong. Whenever there is an increase in the concentration of reactants (Either A or B in your case), equilibrium shifts towards products. And remember one thing Le-Chatlier's principle states that whenever a change has occurred in equilibrium, it tries to oppose the change. So here when you increase the concentration of reactant either A or B or both, the equilibrium shifts towards products.

Say you have the reaction A -> B. Standard notion is that increasing concentration of reactant shifts equation to the right but that can obviously not be the case here. Explanation: Take two systems composed of A -> B respectively and add them together. You'll end up with 2A -> 2B with the ratio remaining the same.
However, if you have the equation A + B -> C what will happen is that a greater percentage of B will be reacted and then equation will shift to the right relative to B but left relative to A. And left relative to the equation as a whole. Correct? However, if you increase both A and B by the same factor no shift will occur.

None of these statements are correct. In the reaction $\ce{A \to B}$, increasing the concentration of $\ce{A}$ does shift the equilibrium to the right. Mathematically,you look at this through the equilibrium constant,
$$K = \frac{[B]}{[A]}$$
Adding more $\ce{A}$ has to be followed by an increase in $\ce[B]$ to preserve the equilibrium constant constant. The logic of relating the species concentrations to their stoichiometric coefficients is flawed - and I am also kind of confused at what you mean.
As to the paragraph after that, there is no such thing as a reaction shifting relative to a certain reactant. The reaction either shifts towards products or towards reactants, there is no relativity.

The concentration of reactant A is not determined by its coefficient in an equation. The concentration of A is expressed in moles per liter and applies equally to situations where A-->B or 2A-->C. You measure (or set) the concentrations at the beginning of the experiment.
The equilibrium constant is different for each of those reactions, and indicates whether the effect of changing the concentration of A is first order (linear) or second order (concentration squared).


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Halogenation of Phenol
Why does only one $\ce{Br+}$ ion get generated in a non-polar solvent (such as $\ce{CS2}$), while more $\ce{Br+}$ ions are generated in aqueous or polar solvent (such as $\ce{H2O}$)?

The reaction of phenol with bromine is known as bromination of phenol. Solvent has great influence on the reaction. In different solvents, different products are obtained. The action of bromine on phenol can be explained as:
Reaction with bromine in water:—-

Phenol reacts with bromine water to give 2,4,6-tribromophenol.
In water, ionisation is facilitated. Phenol gets ionised to form phenoxide ion, which is even better ortho,para-directing. Bromine also gets ionised to a larger extent to form a large number of bromonium ions. And Bromine ions are highly stabilized in ionic solvents. So the formation of strong o,p-directing group and stabilization of $\ce{Br+}$ enhances the formation of tribromophenol.
Reaction with bromine in $\ce{CS2}$:—-

Phenol reacts with bromine in a  presence of Carbon disulphide to form a mixture of o-bromophenol and p-bromophenol. Among which p-bromophenol predominates. In $\ce{CS2}$ ionisation is not facilitated that much. As it is a non-polar solvent it doesn't the $\ce{Br+}$ on. Also $\ce{-OH}$ group is moderately o,p-directing. So as usual para-product is more formed.

sorry may I ask why is brominum ion produced form Br2 in aqueous solution? I know Br+ is more electrophilic, hence even more encouraging multiple halogenation. But isn't Br- bromide ion formed instead?


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Electromagnetic energy to Chemical energy
How can electromagnetic energy (photons) be efficiently converted to chemical energy? For instance, can water be efficiently split to hydrogen and oxygen using solar energy? As we are running out of non-renewable resources is there any way for us to use the solar energy so efficiently to generate power?

https://chemistry.stackexchange.com/questions/71172/what-might-serve-as-an-initial-starting-photocatalyst-for-this-large-water-split

You can certainly turn electromagnetic energy into chemical energy, but the key is 'efficiently.'
A photo-voltaic cell can be used to split water into Hydrogen and Oxygen. The Hydrogen can be burned or used in a fuel cell.
But the key is how efficient is 'efficient.'
Each step will lose energy, but if solar is the input, there is a huge supply. If a solar powered system were continually operating, and we had many of these systems, a huge amount of Hydrogen gas can be generated. Even considering the costs of delivering to where it is needed (which can be paid with the product itself) there is a huge and realistically limitless supply.
That being said, there is much 'waste' which hampers a calculated efficiency, but once established a system that relies on water for a raw material is relatively efficient because there is very little required for raw material acquisition.


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Using Beers Law and the volume/concentration/moles triangle, but unsure of how to proceed?
pH= 8
Absorbance = 0.997
Concentration mol/l = 6.01E-05
In this experiment I used Beer's law to convert absorbance to concentration with a known extinction coefficient (16.6*10^3) - so 0.997/16600 = concentration.
Next I want to find the amount of solute in the solution, which is that triangle formula, Amount in moles (mol) = concentration (mol/l) / volume (l), with a given volume of 8ml.
There are two stipulations: the given volume was 8ml not 8L, and I should be presenting the 'amount in moles' as micromoles. So 6.01E-05 moles is equal to 60.06 µmol, but it is still in µmol/L, whereas I want it µmol/ml. Therefore I multiply by 1000 again, giving 60060 µmol/ml.
Then to find Amount in moles using that triangle thing, I would multiply 60060 by 0.008 - right? This gives 480.48 moles, however I need µmol, so it's 480481 µmol?
This really seems wrong and I would appreciate some help!

Firstly,$ concentration (mol/L.)* Volume (L) = moles.$, is the correct statement.You have written that wrong.
Secondly, you have 60.06 $\mu$mol/L. For,converting that to $\mu$mol/mL. you should DIVIDE by 1000, because 60.06 $\mu$mol/L means in 1L i.e. in 1000 mL. 60.06  $\mu$moles are present.So, in 1 ml, $60.06/1000$ $\mu$moles are present.So, the concentration is now $0.06006 $ $\mu$mol/mL. Finally multiply by 8 to get $0.4808$  $\mu$moles are present in the solution.


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Crystal Field Theory - Operator Equivalent Methods
I am a bit confused by some of the points of operator equivalent methods in crystal field theory. I would appreciate any help I could get with any of my questions.
First, I am confused as to the difference in the operator equivalent method for the single electron case and the multiple electron case. I have read, for example, in Hutchings 1964 and Bleaney & Stevens 1953 that one considers $\mathcal{H}=\sum_{n,m}B_n^mO_n^m(L)$ where $L=\sum \ell_z$. However, this confuses me because for example, in $(3d)^7$, by Hund's rules, $L=3$. Therefore, in this operator equivalent method, one would expect a splitting into two triplets and singlet state. However, the d orbital has $\ell=2$, so one should have a five dimensional representation which splits into a doublet and triplet by group theory (as in Dresselhaus or many others). Indeed, I have seen conflicting diagrams from different sources. For example, in Section 6.2 here, it has $\ce{Co^{2+}}$ splitting from seven levels. However, in this paper, $\ce{Co^{2+}}$ splits from five levels, which makes sense to me because even if $L=3$ it is still the $d$ orbital. I think it may be that I'm not understanding what the levels mean in the operator equivalent method, but I'm not sure where to look.
Second, even if one determines the Hamiltonian using operator equivalent methods, I am not sure how to make good use of it. The operators tend to be written as some complicated function of $J$ and $J_z$ for rare earths (or $L$ and $L_z$ for the iron group), so I'm not sure how to do calculations familiar from more basic quantum mechanics, like finding expectation values of angular momentum or energies.


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Electron orbital filling order
In the p sub shell there are 3 orbitals,
m=-1, m=0, m=1
which are listed by axis x,y, and z.
Some show as -1=x,0=y,1=z
Another shows -1=x,0=z,1=y
Some pictures and text also show the filling order as ‘xyz’ and others state that the z axis is filled first m=0 and then the x axis as m=-1 making the filling order z,x,y
What is the correct filling order?
If I know this then I can figure out the d and f sub shells.
d~m=-2,-1,0,1,2
F-m=-3,-2,-1,0,1,2,3
Thanks, HJS

Your question makes no sense - orbitals don't have "direction" or "number" written on them.

@Mithoron Well, they sure don't have anything written on them per se, but they do have different orientations in space, don't they? For example, the three p orbitals have different orientation along the $x,y,z$ axes.

@GaurangTandon technically no. We can define a set of axes around a molecule and then write the wavefunction of the p-orbitals in such a way where it lines up with those axes, but this is more about how we choose to define our basis of states than something intrinsic about the orbitals. https://en.wikipedia.org/wiki/Atomic_orbital#Real_orbitals

@Tyberius You mean to say that if we remove the coordinate axes then all the orbitals are symmetric with each other. Alright, I get it! BUT how does this explain $d_{z^2}$ as compared to the other $d$ oribtals? Surely the former is not at all symmetric with the others even if  we remove the coordinate axes...

There is no order for filling atomic orbitals of the same azimuthal quantum number (l) such as px, py, and pz since they are degenerate ie. have the same energy (however they lose their degeneracy in a magnetic field, see Zeeman effect). Therefore, one electron in a p sub-cell can be in either 3 of them and the order of filling doenst matter. For when more electrons are added in degenerate orbitals look up Hund's rule and the Pauli principle.


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Question about the Law of Definite Proportions
If we know that:
$1g$ of carbon + $1.33g$ of oxygen = carbon monoxide
$1g$ of carbon + $2.66g$ of oxygen = carbon dioxide
My question is: why is that in the second case where we doubled the amount of oxygen we don't get double the amount of the first case (2 times carbon monoxide)
Or why don't we get the same amount of carbon monoxide and the remaining mass of the products is simply oxygen that didn't form any bonds.
Basically, what forces the reaction to occur in a certain way just because the proportions are different but the elements are the same.

In reality, there is not a sudden changeover of $\ce{CO}$ to $\ce{CO2}$ production when the oxygen supply is increased. This is the reason that all dwellings should have carbon monoxide detectors.
As an example, along with $\ce{CO2}$, gasoline (petrol) engines commonly produce 3-7% $\ce{CO}$ before additional air is injected.
The amounts you cite are how much oxygen is in pure samples of each gas, but producing either gas often requires filtering out traces of other gases that were also made in side reactions.
Producer gas is a fuel made by the incomplete oxidation of carbon (from coal or other source, sometimes with the addition of water) and contains both flammable $\ce{CO}$ and nonflammable $\ce{CO2}$ (and $\ce{N2}$ from the air supply). During fuel shortages, gasifiers were cobbled onto vehicles to enable them to run on this type of fuel.

Does that mean that if I have 100 grams of each element (Carbon and Oxygen) reacting and forming bonds with each other that I will have the formation of both carbon monoxide and carbon dioxide ? And if I have one of them I will have to filter them ? By the way I´m not caring about  the reactants that didn´t react.

Yes, the product is always a mixture of CO and CO2, though with a great excess of oxygen, CO2 predominates. And whether you need to filter the product depends on the applicati0n.

I dont want to push it, I´m already very thankful for your explanation but I got another question... Why does the CO2 predominates when there is a great excess of oxygen ? Is it because it´s a more stable molecule ? Or it´s another thing ?

More oxygen pushes equilibrium to the product that contains more oxygen. It's simply that there's more O2 available.

To supplement to DrMoishe's excellent answer, I'll tackle this part of your question:

why is that in the second case where we doubled the amount of oxygen we dont get double the amount of the first case (2 times Carbon Monoxide(2.Carbon Monoxide))

Suppose a reaction: $\ce{3A\+2B->C}$. To produce the double the amount of C, you'll need double the amount of each reactant, i.e., to produce $2$ moles of $\ce{C}$, you'll need $6$ moles of $\ce{A}$ and $4$ moles of $\ce{B}$. Simply only doubling amount of $\ce{B}$ won't work.
As an analogy, suppose you're baking a $1kg$ cake from $250g$ flour and $250g$ cream. Then, to produce $2kg$ of the cake, just using $1.5kg$ flour and $250g$ cream will not work! You'll need to use $250\times2=500g$ cream instead.
In your case also, you need to double the amount of both oxygen and carbon.

As DrMoishe Pippik has pointed out that the net reaction is different in reality.
However, I assumed that you were asking in theoretical chemistry. In this case, in low oxygen condition, Carbon reacts with oxygen to form carbon monoxide with the rate of 2:1 (mole of carbon:mole of oxygen). In free-Oxygen condition, firstly the surface area of outer carbon layer doesn’t have enough oxygen to react with, resulting in the presence of carbon monoxide. Then, there are still unused oxygen, both the remaining carbon and the new carbon monoxide react with oxygen to form carbon dioxide.
P.S: This is my first answer on StackExchange. Please point out any problem or misspellings, thank you!


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Acid in Imine Formation
Why is a strong acid like $\ce{H2SO4}$ used for imine formation here instead of a weaker acid like TsOH? I thought that if the pH was too low, the amine molecule would get protonated to an ammonium, which are not nucleophilic?

In short, this reaction has two steps: 1. Nucleophilic attack of nitrogen on the carbonyl carbon 2. elimination of water in acidic conditions

You need to protonate the oxygen in order to actívate the molecule towards nucleophilic attack by the nitrogen of Ethyl amine. pH plays a very important rule here. Too low pH gives an efficient protonation of the oxygen, but it also protonates the nitrogen of the Ethylamine(As ethylamine is weak base due to the lone pair of electrons on nitrogen in it), so the production of the imine results inefficient due to less amount of the desired reagent. On the other hand, at higher pH values, the oxygen does not get protonated enough to proceed the reaction as the electronegativity of oxygen dominates it from giving electrons to a low concentration of protons( As pH is high). So there is an optimum pH between pH 4.5 and 5, for this reaction to take place. So dil H2SO4 can give this pH level while weaker acids cannot. Also when you look at the mechanism for we need to dehydrate to get imine. Dil H2SO4 is good dehydrating agent.

For all the factors above diluted strong acids are better than weak acids.

Diluted H2SO4 isn't good dehydrating agent.

You can use any acid, since it will be immediately neutralized by the amine.
The p$K_a$ of $\ce{H2SO4}$ is $-9$ while that of TosOH is $-6.5$, but both will form $\ce{EtNH3^+}$ which has a p$K_a$
between 10 and 11.


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In the presence of an electron-rich species, why is Br2 the stronger electrophile than water?
In the electrophilic addition of aqueous bromine to ethene, the first slow step involves the electron rich (pi bond) of ethene attacking the bromine molecule to form a carbocation and Br-:

And, then water will react with the carbocation to form 2-bromoethanol. But, why does bromine react with ethene first and not water(which is of higher concentration)?
I think it's because Br2 is the stronger electrophile than water. But why is that? Oxygen is much more electronegative which makes H more electron deficient than Br and thus a stronger electrophile?

$\mathrm{p}K_{\mathrm{a}}$ value of water is $14$ while that of $\ce{HBr}$ is $-9$. Thus, $\ce{Br-}$ is very stable compared to $\ce{OH-}$.

sigh So you're asking why ethene won't get just protonated instead of binding Br+ ? Or what?

Short answer: you are correct that oxygen is more electronegative than bromine, but the oxygen's thirst for electrons is fairly well satisfied by 2 hydrogens, whereas the bromine -- aack! -- is attached to another electron-hungry bromine.
Bromine will snap up electrons from moderately active metals (even from water!), but water doesn't (or else does it much more slowly). Consider iron, zinc, magnesium - all react more rapidly with bromine than with water.


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Why are there no dihedral mirror planes in the D3h point group?
Consider the molecule $\ce{BH3}$, which belongs to the $D_\mathrm{3h}$ point group. Why are the three mirror planes in this point group labelled as $\sigma_\mathrm v$ instead of $\sigma_\mathrm d$? Don't they bisect the angles between a pair of rotational axis $C_{2}$ axes as shown in the diagram below?

Yes, the sigmas bisect the angles between the pairs of rotational axis C2 axes like in the picture. What is your question, really?

http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=603&option=4   In this website and others they list  $\sigma_{v}$ instead of $\sigma_{d}$. My question is why?

Got it. They call these planes $\sigma_v$ because they are vertical mirror planes. Why would you want to call them $\sigma_d$? Just being bisectors of certain directions of twofold axes is not enough justification for that.

The σ$_v$ planes in a molecule are a direct result of the C$_n$ main axis i.e. they are interchangeable if you perform the C$_n$ operation on them. On the other hand σ$_d$, which are called dihedral planes bisect the dihedral angles between members of the σ$_v$'s set. Therefore molecules with odd C$_n$ such as BH$_3$ wont have any σ$_d$ planes; only n (3 for BH$_3$) σ$_v$'s. In contrast a molecule such as benzene which has a C$_6$ main axis will have 3 σ$_v$'s (the ones including the carbon atoms) and 3 σ$_d$'s which are the ones passing between the carbon atoms. Of course it doesnt really matter which set is called dihedral and which is called vertical, it could be the other way round as well. Ref: Chemical applications of group theory by Cotton.


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In halogenation of ethene using bromine for example, why does the lone pair of bromine attack the C atom?
In the presence of the electron rich (pi bond) of ethene, the bromine molecule becomes a temporary induced dipole, one end being electron deficient and acting as the electrophile. (The pi bond in) ethene attacks the electrophile.
At the same time, one of the lone pair of electrons in bromine attacks one of the C atom. Why does this happen?

Normally, I'm familiar with the same process, but with the carbocation being formed without the lone pair of electrons of bromine attacking the other C atom:

Does this have something to do with resonance as shown by the diagram below?

The alkene's filled π orbital is higher in energy than the empty σ* orbital of the $\ce{X-X}$ bond. Also, the $\ce{X-}$ can attack the halonium ion but the result is just regeneration of halogen and the alkene: the first step of the reaction is reversible.

There are too many bond lines to explain the reaction; the question suggests that we have to explain the bond lines!
Let's start from scratch, with a different viewpoint. Ethylene isn't the attacker; it doesn't burst into flame in oxygen or do anything aggressive. It may be reactive, but in a "normal" range. Bromine, on the other hand, is a potent oxidizer and does go around interacting with relatively inert materials - it's a powerful bleach.
So, when bromine comes near the ethylene pi bond, it attacks the electron pair and grabs it, but the second bromine steals it away and departs as Br-. After all, bromine is more electronegative than carbon; the first bromine is just a link in the chain of events that allows the second bromine to escape with an electron. The first bromine winds up stuck to the C-C double bond; the two carbons and the bromine share the positive charge. The bromine certainly gets the lion's share of electron density, leaving the carbons with most of the positive charge. That's why another bromine can attack the carbons from the rear to give a dibromide. (If a Br- attacked the bromine attached to the double bond, it would just regenerate the reactants.) I bet that the second bromine to bond to the ethylene is not from the same bromine molecule as the first bromine. (But since all bromine molecules look the same, I can't prove it.) Doing the bromination in a solution with an overwhelming abundance of another nucleophile would probably give a mixed product.
A C-C double bond is about 1.34 Angstroms long; a bromine atom (in Br2) is about 2.29 Angstroms in diameter. Rather than envisioning the cation as a triangle of three grapefruit, I see it as a cantaloupe sitting on two large oranges. Just consider the empty p orbital of bromine and the pi lobe of the carbons; no need or benefit from making a three-membered ring.


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Resistance of Electrolytes
During electrochemical experiments, we consider the resistance of the electrolytes. What exactly is causing the electrolytes to offer resistance to flow of charges in them? Also, does the resistance also depends on the electrodes we use?

Related: https://chemistry.stackexchange.com/questions/75954/conductivity-of-water-and-chlorine/75959#75959

Ya it's good and answers my question.


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Why are ionic reactions very fast in aqueous medium?
(No, I am not talking about why Ionic Reactions are faster than Covalent reaction.)
This question has me stumped. I mean, at first it looks obvious; water is polar in nature, but what exactly does that help with?  Also, can ionic reactions even happen in non-aqueous mediums?

What do ionic and covalent reactions mean?

@orthocresol I think the asker means the difference between reactions involving only neutral species and reactions involving oppositely charged species.

Water has a permanent dipole character, which means it has a slight positive and negative charge permanently in it (due to electronegativity difference). So, in case of an ionic reaction, it helps in easier solvation of ions by pulling them with its charged centres (positive centre attracts negative ions and negative centre attracts positive ions)
Aqueous is a term specially for water as a solvent. Any polar solvent can support ionic reactions.


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Sulphate ion- confirmatory test
On adding $\ce{BaCl2}$ solution to $\ce{ SO4^{2-}}$ salt aqueous solution $\ce{ BaSO4}$ is formed as a white precipitate, which confirms the presence of the anion.
While doing the test in my school laboratory, we were given a list of procedures to follow; for $\ce{ SO4^{2-}}$ it read:
'Experiment: Aqueous solution of sample + dilute $\ce{ HCl}$ + $\ce{ BaCl2}$ solution
Observation: white ppt'
Why do we need to add $\ce{ HCl}$?

I should add that in the lab I was asked to add dil HCl and BaCl2 to a "sodium carbonate extract" instead of an aqueous solution of a salt.

The method consists
of slowly adding a dilute solution of barium chloride to a hot solution of the
sulphate slightly acidified with hydrochloric acid
It is  customary to carry out the precipitation in weakly acid solution in order
to prevent the possible formation of the barium salts of such anions as chromate,
carbonate, and phosphate, which are insoluble in neutral solutions; moreover,
the precipitate so obtained consists of large crystals, and is therefore more easily filtered
Reference: Vogel's Textbook of 'Quantitative Chemical
Analysis 11.72 Sulphate: Determination of sulphate as barium sulphate

Re: "Barium salts of such anions as chromate, carbonate, and phosphate, which are insoluble in neutral solution" You mean to say that the salt sample being tested may have additional ions such as chromate and carbonate, both also with a white ppt. A white ppt in a neutral solution will make it difficult to distinguish whether it was due to sulphate or carbonate ion. But, in acidic solution, it is clear that it must be only sulphate. Is my interpretation correct?

Ideally to eliminate false positives which would otherwise occur in the presence of these anions. They however are soluble in presence of  dilute mineral acids and(or bases) but  are insoluble in neutral solutions

Yes, that's what I meant, a false positive. Thanks for confirming!


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Why is the pH scale only from 0-14?
Why can't there be more hydrogen ion concentration in acids or lesser hydrogen ion concentration in bases? Why should there even be any hydrogen ion concentration in alkalis? (I know pH is calculated by the negative log of H+ ion concentration.)

From wikipedia,  "For concentrated solutions of acids, especially strong acids for which pH < 0, the $H_0$ value is a better measure of acidity than the pH." $H_0$ value is the Hammett acidity function. I don't know many more details to write an exact answer. I would suppose the $pH$ scale doesn't work very well for very strong acids. We might have to use the activity of  $\ce{H+}$ ions in solution instead of their concentration (as we do in ionic equilibrium calculations). Though I still don't know the complete answer.

I am voting to close this question as a duplicate of Is a negative pH level physically possible?. The question is based on a misconception; the definition of $\pu{pH}$ does not limit itself to the closed interval $[0,14]$. The proper definition is also based on activity, not amount concentration, molality etc. It is difficult to measure highly acidic and basic solutions with usual glass electrodes (due to acid and base errors, among other things) but this is a metrological issue.

It isn't.  We're just used to working in the lab where solutions are generally 1 M or less in concentration.  So we get no more than 1M (solvated) hydrogen ions where pH is 0, or at the other end no more than 1 M hydroxide ions corresponding to pH 14.
These 1 M limits are, however, just for our convenience and not any law of nature.  Right in our own labs we likely have concentrated acid stock solutions, well above 1M, where if we attempted to measure pH we would get less than zero.  Acid pickling solutions in the steel industry are concentrated enough to do the same thing, as manufacturers are more interested in getting hot mill scale cleaned off than in a convenient pH scale.  Bases like sodium hydroxide also have concentrated stock solutions with pH above 14.


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What material can be used as scraper in a chemical reactor?
I am having a 10L reactor built using SS316 for Plastic Pyrolysis.
TEMPERATURE : 550C
I would like a scraper to clean the char that forms on the reactor's inner layer.

Teflon scraper is available but it will not withstand 550C temperature.
So I am thinking about using flexible expanded graphite (Similar to graphite gland packing ropes) material. What do you think?

What other options do you suggest that I look into?

Not sure what you mean by scraper. What does it look like? Is it part of the reactor's stirrer whose role other than mixing is to shif the char that deposits at the sides of the reactor while in operation? Cant you use steel?

@vanadium My question is why are you not doing this in a continuous reactor to reduce char? The high temperature would lend itself well to this, especially if the plastic is melted. If it's milled/ chipped, there are continuous options for this too.

@AMM  Scraper is to remove the character which is formed in the vessel. It should make contact with the vessel to remove char effectively. In the picture, Teflon is used as scraper blade. Steel blade cannot be used as it will damage the reactors inner wall.

@Beerhunter I did consider having a plastic screw extruder and heating the extruder barrel to 550C. But am not sure on how effective it will be. Please do let me know if you can share more information about the continuous process.

I'm sure you can hire twin screw extruders for a trial. You might need to give it a heat boost (induction?) as the hottest I've seen is 300ish degrees C. The simplest method would be to melt/soften in a twin screw and pump into the subsequent high temp reactor for the pyrolysis step.

@Vanadium Ok, i see what you mean. So the scraper is the plastic dressed steel plates attached to the stirrer in the picture, correct? There are plastics around that can take such temperatures: https://www.professionalplastics.com/MELDIN7001 for example but looks too cumbersome and expensive to make such modification. I would try using bronze or any other soft metal instead of steel which shouldnt damage the reacton being a softer metal. Cant guarranty this would work though as i have never tried with the steel reactors i work with.


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What does the shaded area represent?

I'm not sure what the answer is, all of my friends and my chemistry teacher said it is A, I think it is obvious that the graph is talking about 2 catalysts so it could be D. But I have no idea what the E_a thing is, I've tried googling it to no avail. I got this question on an advanced Chemistry test. One thing for sure, it is not B because the word "additional" isn't there. And another thing, we can't say that one is the graph of the molecule without a catalyst and the other is its graph with a catalyst because according to the graph that would mean that sometimes the molecule without a catalyst is more active than without a catalyst which is obviously wrong.

Short answer: The two curves are the distribution of kinetic energy of molecules at two different temperatures. A catalyst won't change the distribution; only temperature change will change the energy distribution. The line marked Ea is the activation energy; a molecule has to have more than this to react. A catalyst could change the activation energy; since there is only one Ea, we can eliminate catalytic activity as an issue. All the molecules with less energy than this just bounce around, not reacting.
The curve with the higher peak represents the lower temperature. The one with the lower peak has a longer and higher tail out to higher energies, so the shaded area represents the greater number of high energy (reactive) molecules at the higher temperature. Answer: A.


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Use of the box method to determine lowest term symbol for lanthanides but not iron
I am trying to compare the predicted magnetic moment for Fe(CN)6 using the spin-only formula magnetic moment = $\sqrt{4S(S+1) + L(L+1)}$.
However, if I use the box method for Fe(CN)6 to predict the lowest term symbol of Fe3+, I get 6S. If L = 1, however, then the predicted value of the magnetic moment is similar to the experimentally derived value.
Can we not use this method because in the complex the octahedral field changes the energy levels of the d orbitals?
If this is the case, then why can we use the box formula to determine the magnetic moments of lanthanides using the formula magnetic moment = g$\sqrt{J(J+1)}$.


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What happens to the "leftover" Ions in the Mass Analyzer
I am very intrigued by what happens to the ions which do not make it to the detector in mass spectrometers, do they just pile up? I have never seen anyone clean our quadrupole, is this done? or do they just stay there indefinitely? I am talking of a triple quadrupole detector in a GC instrument, which deals with very small quantities, but after thousands of injections there must be a considerable mass of analyte on the analyzer. Even more so considering that most ions which go through the quadrupole do not make it to the detector.
 Thanks.

Well, MS is a dirty business. Never mind the source and quadrupole, who cleans the detector? ;-)

Most would end up being sucked out though the vacuum pump with the other residual gases. I assume that a lot of them collide with the sides of the vacuum chamber and are neutralized before sublimating in the vacuum.
The quads can become contaminated and require cleaning after several years of heavy use. Because there are such small quantities at that stage of the process, it's not uncommon for an MS to reach the end of it's service life before the quads need to be cleaned.


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Carbon 14 in organic chemistry
A small amount of natural carbon is $\ce{^{14}C}$.  This is radioactive and decays to $\ce{^{14}N}$.  This is exploited in radiocarbon dating.
This $\ce{^{14}C}$ is taken up by living organisms.  I know that natural biochemical processes can be sensitive to isotopes but I expect that the $\ce{^{14}C}$ can occur in any context where $\ce{^{12}C}$ could occur.  (Please correct me if I am wrong here.)
So, $\ce{^{14}C}$ might occur as part of the backbone of a fatty acid:
$$. . . \ce{-CH2-{^{14}C}H2-CH2 -} . . .$$
When the $\ce{^{14}C}$ decays, this will become:
$$. . . \ce{-CH2-NH-CH2 -} . . .$$
(I assume one of the hydrogens would be lost.)
Are any of these unusual compounds stable or do they promptly react in some way?
Are they studied?
Do they have any interesting properties?

I wonder if it can decay in-place i.e. in the middle of a compound the way you have depicted, without breaking the chain itself. The decay process is generally accompanied by a loss of heat (due to mass-defect) so the extra energy must manifest itself as kinetic energy of the daughter nuclei or heat, breaking the chemical bonds either way.

@GaurangTandon Will the result immediately break down: that's my question.  I am reasonably sure that the $\ce{^{14}C}$ does not know or care that it is inside a fatty acid and hence will be liable to decay.  I had in mind the $\ce{^{18}F}$-FDG used in PET scans though that is the reverse: an unusual molecule becomes a common one when the $\ce{^{18}F}$ decays to $\ce{^{18}O}$.

Related: What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays?

@Loong Thanks.  "The decay energy of 14C is unusually low" but, nonetheless, it seems likely that the bonds will be broken.  However, at least according to https://en.wikipedia.org/wiki/Fludeoxyglucose_(18F), the decay of $\ce{^{18}F}$ in $\ce{^{18}F}$-FDG does not break the molecule apart.

@badjohn I quickly scanned the reference given at wp, and I think it says nothing of the kind. https://www.ncbi.nlm.nih.gov/books/NBK23335/ Total own research, respectively own imagination of the wp author.

@Karl So, this part is probably incorrect: "After 18F-FDG decays radioactively, however, its 2-fluorine is converted to 18O−, and after picking up a proton H+ from a hydronium ion in its aqueous environment, the molecule becomes glucose-6-phosphate labeled with harmless nonradioactive "heavy oxygen" in the hydroxyl at the C-2 position."

@badjohn It's most probably imo incorrect, and definitively does not belong on wp without a proper reference. Same sentence can be found on de:wp. ;-)


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Why can we ignore the weak conjugate base/acid when calculating the pH in a titration problem?
Take the reaction 350 mL of 0.8M HCN and 0.6M NaOH for example. The problem asks us the calculate the pH when we add 500 mL of 0.6M NaOH into the 350 mL of 0.8M HCN (This is more than enough to neutralize the weak acid). However, the problem states that we can ignore OH contributions from water or any weak base. Can someone explain why we can do this?

This is a homework type of question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this. Good luck.

@airhuff I mean I did to the rest of the problem (ie solving the pH) but that little statement still stumped me.  If this still violates the policy then I'll gladly shut this down myself.

$\ce{NaOH}$ is a strong base. It is assumed that it dissociates completely in aqueous solutions, hence, it's contribution to the $\ce{OH-}$ concentration is the maximum as compared to other weak bases or water. Moreover, due to common ion effect, the presence of $\ce{OH-}$ ions in solution due to $\ce{NaOH}$ further hinders the dissociation of the already weak bases, thereby making their contributions negligible.


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How to change the Glass transition Temperature ( Tg ) of a Compund?
we wanna Increase our compound damping (loss) , we know that Damping properties of a polymer is dominated by glass transition...
So how can we change Tg In our PP/EPDM Compounds?

You cant change the Tg of a polymer without changing its composition  which means it wont be the same polymer any more. A specific compound or polymer will have a specific Tg. You could vary the Tg of say PE by controlling its degree of branching, but each PE you will get will be a different compound each with its own Tg although all of them will fall under the general category of PEs.


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How can a reaction exhibit first order kinetics when there is more than one reactant (the other being water)
This is for the hydrolysis of p-nitrophenyl phosphate by alkaline phosphatase, whereby the enzyme is in a solution of distilled water. We also add glycine buffer to the solution. It exhibits first-order kinetics, even though there is a second reactant (being water) - how is this possible?

It is because water is the solvent in the reaction and therefore its concentration remains practically constant throughout the course of the reaction. Hence, it is absorbed in the rate constant:
rate=k[p-nitrophenyl phosphate]$^1$[H$_2$O]$^x$=k'[p-nitrophenyl phosphate]$^1$
Hence, we say the reaction is peudo-first order. Also, in general, you could have a reaction with two reactants and it could still be first order overal since it could be 0 order in reactant A and first order in reactant B.

Why do you write $[\ce{H^2O}]^n$ ?

Because we dont really know the order in water due to it being the solvent. Stricktly speaking, you shouldnt have water in that rate equation at all. I just wanted to show the logic behind it and if i used [water] it would imply first order in water which is incorrect.  The correct form of the equation is the one on the right.

OK, makes sense. I'd recommend to replace $n$ by $x$. And really x is very close to zero, and then the equation is in good approximation correct. Right?

Actually, thinking about it again i would say that leaving n or x (where n or x is some number) there is ok. Say for example that we have a reaction in water, some kind of hydrolysis, in which we know the mechanism and the rate determinig step involves one molecule of substrate and one molecule of water. This makes the reaction second order overall. However because the concentration of water always remains constant being the solvent, the observed kinetics will be 1st order. Hence, pseudo 1st order.

$n$ implies an integer number, and that's surely not the case here, hence $x$.


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Conjugated diene t-buOK reaction
Why does the t-buOK abstract the following hydrogen. Doesn't it promote the Hoffman product so it should take the hydrogen from a secondary carbon instead of a tertiary one? Thank you

Because the resulting anion makes the ring aromatic.


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t-BuOK as a base
I thought $\ce{KOtBu}$ was used to form the less substituted alkene. Why is that not the case here?

That is because the benzylic hydrogen is more acidic.
When deprotonated, the anion is stabilized through resonance in the aromatic ring.
The base will take with the most acidic proton first.

Ahh, makes sense. Is this the same thing for allylic hydrogens, so if t-buOK had a choice between a hydrogen attached to a primary carbon or to a secondary but allylic carbon, it would choose the allylic carbon because it would form a more stable carbocation?

You seem to suggest that an acid base reaction is preferred over the
nucleophilic reaction. Is that a correct interpretation?

@SarahSmith That would be a carbanion

@GaurangTandon In this case it is. t-BuO is a very strong base but a poor nucleophile due to its large bulk which impedes its approach to the substrate.


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Synthesis of compound metal oxides
When certain oxides are combined in a pre-determined stoichiometry and synthesized with a ceramic-like synthesis process(power-in-tube), multiple phases may be obtained which not necessarily is same as the original stoichiometry desired. For example, in the superconducting compound $\ce{YBa2Cu3O_{7−x}}$, other ratios such as $\ce{YBa2Cu4O_y}$ (Y124) or $\ce{Y2Ba4Cu7O_y}$ (Y247) might exist in the same material.
So the question is:

What are some of the techniques of combining oxides to obtain desired major phase?

Online/book references or paper citations are also welcome apart from your valuable answers.


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How do they come up with a freezing point of ~-60F for CaCl2 & H20?
What am I missing?  I cannot reconcile the difference between what the internet says the freezing point of CaCl2 is vs the calculation, which means my calculation must be off, but I am not sure how.
Example Calculation:

A Solute of 1700g of CaCl2 would equate to ~15.32 Moles @ 110.98 Atomic Mass.
Mix in 1 Gallon of water, or 3.785 kg.
This represents a ~31% solution (1700/(1700+3785))
Molality would be 4.047 (15.32 / 3.875)
With a Van't Hoff of 3...
The Freezing Point Depression would be -22.52C (3*1.855*4.047)

I think all of that is right, but when I look at the tables online they show that a 31% CaCl2 solution would have a freezing point of ~-34C - ~-37C

http://meltsnow.com/wp-content/uploads/2011/04/image006.png
http://www.oxy.com/OurBusinesses/Chemicals/Products/Documents/CalciumChloride/173-01791.pdf

These sites shows a 30% concentration to have a freezing point of ~-46.1C:

https://www.engineeringtoolbox.com/calcium-chloride-water-d_1186.html
http://www.tetrachemicalseurope.com/getFile.asp?File_Content_ID=1624
https://www.dynalene.com/v/vspfiles/templates/210/datasheets/Dynalene_Calcium_Chloride_Series_Technical_Data_Sheet.pdf
http://www.phasediagram.dk/binary/CaCl2.PNG

They didn't come up with the number. They went out into the field and actually measured it. Chemistry is an experimental science, you know.
Your calculations are fine; it is just they were never supposed to work in the first place. When you calculate the freezing point depression like that, you use the linear approximation, and all approximations have their limits. You can't apply it to a concentrated solution; the error will be too great, just as you observed.
Look at them graphs by your links, they are pretty far from being linear; they are bent. (We may apply some hand-waving about molarity and molality not being the same thing, but it doesn't explain the effect away completely). Look at the left corner of the graph with a magnifying glass (related picture: Why do increasing concentrations of zinc chloride not linearly decrease the freezing point?). What would you call that section of the curve? Almost linear, shall we say? Well, that's where our approximation is almost precise.
So it goes.


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Grignard Reagent and Ester
Is this answer wrong? I feel like there should be a hydrogen instead of a carbon where I indicated in red. Thank you.

A possible answer could have been (a) though, especially since they have not specified that the $\ce{MeMgBr}$ here is in excess.

If e had a hydrogen there it would be the same structure as d.

Where would the hydrogen come from?

The answer is correct. After the first nucleophilic attack from $\ce{MeMgBr}$, the ethoxy group left and the ester is transformed into a ketone (Step 2 in the link), then another $\ce{MeMgBr}$ attacks and introduce an extra methyl group, resulting in a $3^\circ$ alcohol after hydrolysis. The process is well described in this page.


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Compound metal oxide stoichiometries
When certain oxides are combined in a pre-determined stoichiometry and synthesized with a ceramic-like synthesis process(power-in-tube), multiple phases may be obtained which not necessarily is same as the original stoichiometry desired. For example, in the superconducting compound $\ce{YBa2Cu3O_{7−x}}$, other ratios such as $\ce{YBa2Cu4O_y}$ (Y124) or $\ce{Y2Ba4Cu7O_y}$ (Y247) might exist in the same material.
From stoichiometry perspective,

Is there a limit on the number of oxides due to crystal lattice limitations and such?
What precautions can be taken during synthesis of these oxides when partial stoichiometries are involved?

EDIT: FCC for example, accommodates a max of 14 atoms per unit cell which YBCO does. But take for example Bi2223. https://www.researchgate.net/publication/225413281_Crystal_structure_and_superconductivity_of_Bi-2223
This has a crystal structure, unlike any traditional lattices.
The question I have is to what extent can one go to obtain desired stoichiometry? Bi2234, for instance, is hard to synthesize: https://en.wikipedia.org/wiki/Bismuth_strontium_calcium_copper_oxide. How do we know beforehand (educated guess) that a certain phase is not easy to synthesize?

So I've been thinking about how to answer this question appropriately for a while now. I'm not quite sure how deep into phase diagrams you want to go, however. In short, there is no limit on the number of possible oxide phases. Even 'simple' binary metal phase diagrams can have a dozen or more phases on them. As for precautions, well, that depends a lot on the particulars of the phase diagram and how you are trying to make the phase. How deep do you want to go?

I'm primarily interested in increasing the size of the unit cell while ensuring the synthesis produces a predominantly single phase so the unit cell satisfies the originally intended stoichiometry. For example, YBCO satisfies the ABC3 lattice. Increasing the number of oxides or their relative ratios under prime factors (3:7:11)  obviously gives fractional unit cell that satisfies some lattice structure but that's not the intent. We want the unit cell to be consistent with desired stoichiometry.

The question still boils down to how do you think you might guess different phases are not formed?

So are you looking to (1) add other elements to increase the unit cell size but keep it as 123, or (2) by adding other elements get to an ordered phase that retains the basic 123 structure but has multiple 123-type unit cells in its (larger) unit cell? By your edit it still isn't completely clear. However, I will try to gin up a fairly simple answer that may address some of your points.

@ViníciusGodim: yes, that's the fundamental problem. Associated with are the questions like, what oxide combinations, how to deal with partial stoichiometries. Also, Jon Custer, it need not be 123, I gave that as an example. To clarify, mentioned Bi2223 which isn't 123. It boils to the same question. What oxides to pick to grow the unit size (with higher stoichiometry) so a single phase forms?


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Feasibility of another product in peroxide effect mechanism of propene
In the Chain Propagation step of the reaction of propene  ($\ce{CH3CH=CH2}$) and hydrogen bromide ($\ce{HBr}$), why is $\ce{CH3-\dot{\ce C}H-CH2-Br}$ free radical formed when a more stable, resonance stabilized allyl free radical ($\ce{\dot{\ce C}H2-CH=CH2}$) is possible?

Source  : https://www.chemguide.co.uk/mechanisms/freerad/alkenehbr.html

Maybe because the electrons in the $\pi$ bond are higher in energy than the $\ce{C-H}$ $\sigma$ bond and radicals are so reactive that they just react kinetically instead of thermodynamically.

So, does it mean that allyl free radical is not formed at all or is it formed to a lesser extent?

I actually don't know. I was just trying to guess.

I will not write an answer because I don't know if my guess is correct.

There's no such thing as a kinetically stabilised reaction.

If you're confused, a good question to ask you would be: what is the mechanism of the initiation step?

Speculating, but the stability of the allylic radical may hinder further propagation. If the radical electron is delocalized over the whole molecule, it could be more difficult to abstract a hydrogen from HBr. @RinSam

The energy cost of bromine radical abstracting hydrogen from propene is 89 kcal/mole. The gain of forming $\ce{HBr}$ is 18-28 kcal/mol (for a $\ce{H-Br}$ bond from $\ce{H2}$  and $\ce{Br2}$; variation is due to solvent effects) + 23 kcal/mol (from the $\ce{Br2}$ which was fractured into 2 Br radicals by 46 kcal/mol bond energy); so the gain is 41-51 kcal/mol vs a cost of 89, or a net cost (net energy requirement) of 38-48 kcal/mol. This would be a slow reaction with an activation energy even higher than 41-51 kcal/mol. (All data from CRC Handbook)
The alternate reaction, of adding a Br radical (with 23 kcal/mol energy above ground state) to propene, loses the pi bond and its energy (84 kcal/mol) and gains a $\ce{C-Br}$ bond (70 kcal/mol). The energy cost of forming the bromoradical is 84 but the gain is 70 kcal/mol for $\ce{C-Br}$ plus the 23 kcal/mol from the Br radical, or 93 kcal/mol. Net gain is 9 kcal/mol. This reaction will go spontaneously while the first will not (rate determined by activation energies, of course).
It turns out that allylic stabilization is only part of the story, and is, in fact, rather small: the bond strength of $\ce{H - allyl}$ is 89 kcal/mol vs the bond strength of $\ce{H - n-propyl}$ which is 98 kcal/mol.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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Filtering fine particles before work-up (Tosylation)
I am trying on doing tosylation of PEG using Tosyl Chloride with pyridine. Some articles mentioned about filtering fine particles (pyridinium salts?) after diluting with DCM. But they did it with some filter-aid like celite which I don't have access to.
On my first try, I didn't filter it and during washing (HCl, NaHCO3, brine), there seemed to be a pretty intense emulsion. I am wondering if it was caused by those particles I didn't filter.
I also tried to quench it with water instead of dcm then extract with DCM, the emulsion thing still occured.
Therefore, I would like to ask if there is any other methods of filtering those fine particles beside filter-aid. Would syringe filter work?
Thanks a lot

Very likely the fine particles are causing your emulsion. Syringe filter worth trying.

You can try with a pad of silica gel. It won't be a column.  You go till no spot develop/is visible

Thank you! I will try both. @Alchimista regarding the silica pad, how do you prepare it? Is it by the same  fashion the celite pad is prepared?

Imagine to pack a column. You choose the largest column you have (in diameter! ) and you pack it with a minimal amount of silica. You can fix the silica even without slurry, just drop and press it and perhaps hold it with a cotton. Else you do it as with celite. Twice I used fltration through silica gel unplanned to recover products from accidental spill and sometimes to (with luck) removes trace impurity that prevented crystallisation from oily crudes. To say that after filtration you can washed/extract your eventually adsorbed product to minimize loss. But it depends on the scale and value.


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(RedOx) How should i continue?
i'm stuck on a redox, currently in the oxidation semireaction. The reaction is:
$\ce{B + NaOH = NaBO2 + H2 + Na2O}$
The reduction semi reaction is: $\ce{B = NaBO2}$
And i balanced it like this: $\ce{B + NaOH + H2O + 3e- = NaBO2 + 3H+}$
Problem is the oxidation semi reaction, i start with: $\ce{NaOH = H2}$
and balance it like: $\ce{2NaOH + 2H+ = H2 + Na2O + H2O + 2e-}$
Problem is that when i add everything up after balancing electrons and stuff (Multiply 3 with the oxidation semireaction and 2 with the reduction semireaction) Na won't be balanced. Plus, on a redox solver online, instead of balancing $\ce{NaOH = H2}$ he starts with $\ce{H2O}$, why is that?
What am i doing wrong?

I would start with H2O as well. But if you are positively certain that H2O is not there, then balance your second half-reaction without H2O on the right and without H+ on the left.

No H2O is mentioned in the text. Why should i start with H2O instead of NaOH?

Because we live in the world dominated by H2O. Well, if the products were specified this way, with Na2O and stuff, then probably they really mean the absence of H2O, so go with the second option.

Isn't the reaction supposed to be balanced in basic medium?

Problem is, they don't tell you if it's in acid or basic medium.. when i balance it with what i've done, the answer isn't right.. it has to do with that NaOH

I don't understand the emphasis on redox. True that the boron is oxidized, but the conditions are so unrealistic that I can't figure out what issue is being solved here.
The conditions are harsh: Na2O in the product means fused salt, so you are starting with molten NaOH, adding boron and getting H2 + NaBO2 and Na2O. Actually the molten mass would just appear to be Na3BO3 (sodium orthoborate).
The balanced equation would be 2B + 6 NaOH --> 3 H2 + 2 Na3BO3
(or --> 3 H2 + 2 NaBO2 + 2 Na2O using the products indicated in the problem).


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Comparing acidity of p-chloro benzoic acid and p-fluoro benzoic acid

I am really confused about comparing the acidity of p-chloro benzoic acid and p-fluoro benzoic acid.
The question here doesn't have a conclusion and the discussion here is confusing.

My Try:

-F and -Cl being +R groups create negative charge on o- and p- positions with respect to them.

-F having greater +R effect tends to produce greater negative charge on the positions ortho and para to it

-Cl having a greater +I effect tends to create greater amount of $\delta-$ charge than -F

The acidity of the molecule depends on the negative charge on the starred carbon atom (greater negative charge$\implies$less acidic):

My Confusions:

 Is inductive effect strong enough to act para to the specific group?
 Which among -F and -Cl will produce greater cumulative negative charge on the starred carbon atom, and how?

The answer given:
p-chloro benzoic acid is more acidic than p-fluoro benzoic acid, contrary to the order existing between p-chloro benzoic acid, p-bromo benzoic acid and p-iodo benzoic acid