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Why and how does hydrogen peroxide decompose in the presence of light?
When exposed to UV light, hydrogen peroxide decomposes into $\ce{H2O}$ and $\ce{O}$. Why does this happen and more importantly how? Is the energy from light absorbed by the bonds which are specific to UV frequency and thus leading to decomposition?

please check the article under title:
"DECOMPOSITION OF HYDROGEN PERODIXE -KINETICS AND REVIEW OF CHOSEN CATALYSTS"
Acta Innovations •ISSN 2300-5599 •2018•no. 26: 45-52•45

Chemical reactions in which, a single substance splits up into two or more simpler substances are called decomposition reactions. These reactions are carried out by energy, supplied by different sources. The required energy can be supplied by heat (thermolysis), electricity (electrolysis), or light (photolysis).
Let’s talk about photolysis reactions (not photosynthesis):

Photolysis (also called photodissociation and photodecomposition) is a chemical reaction, in which a chemical (an inorganic or an organic) is broken down by photons and is the interaction of one or more photons with one target molecule. The photolysis reaction is not limited to the effects of visible light, but any photon with sufficient energy (higher than dissociation energy of the targeted bond) can cause the chemical transformation of the said (inorganic or organic) bond(s) of a chemical. Since the energy of a photon is inversely proportional to the wavelength, electromagnetic waves with the energy of visible light or higher, such as ultraviolet light, X-rays, and $\gamma$-rays, can also initiate photolysis reactions.

Like all other peroxides, hydrogen peroxide ($\ce{H2O2}$) also consists of a relatively weaker $\ce{O-O}$ bond, which is susceptible for light or heat. In the presence of light (the UV light from the sun catalyzes the reaction), $\ce{H2O2}$ spontaneously decomposes into water and oxygen.
The net equation for the reaction is:
$$\ce{ 2H2O2 -> 2H2O + O2}$$
The step-wise reaction mechanism is suggested as follows (Ref.1):
$$\ce{ H2O2 + h\nu -> 2 HO^.}$$
$$\ce{ HO^. + H2O2 -> HO-O^. + H2O}$$
$$\ce{ HO-O^. + H2O2 -> 2 HO^. + H2O + O2}$$
Using isotope studies ($\ce{^{18}O}$ labelled $\ce{H2O2}$), early work has been confirmed that the $\ce{O2}$ formed is cleanly derived from $\ce{H2O2}$ (Ref.2).
Notes: The rate increases rapidly in the presence of catalysts such as $\ce{MnO2}$ and $\ce{KI}$ (Ref.2). The rate of decomposition is slow at room temperature, but it increases with temperature. It is believed to be due to thermal decomposition of $\ce{H2O2}$, which seemingly accelerates the photolysis (Ref.3).

References:

J. P. Hunt, H. Taube, “The Photochemical Decomposition of Hydrogen Peroxide. Quantum Yields, Tracer and Fractionation Effects,” J. Am. Chem. Soc. 1952, 74(23), 5999–6002 (https://doi.org/10.1021/ja01143a052).
A. E. Cahill, H. Taube, “The Use of Heavy Oxygen in the Study of Reactions of Hydrogen Peroxide,” J. Am. Chem. Soc. 1952, 74(9), 2312–2318 (https://doi.org/10.1021/ja01129a042).
F. O. Rice, M. L. Kilpatrick, “The Photochemical Decomposition of Hydrogen Peroxide Solutions,” J. Phys. Chem. 1927, 31(10), 1507–1510 (https://doi.org/10.1021/j150280a004).

If the O-O is susceptible to decomposition wouldn’t the products be H-O and H-O (basically hydroxides)?

Yes, but those are radicals ($\ce{HO^.}$). That is the initiation reaction, which then contributes to propagation as described in next two equations.

Does infra red light react with hydrogen peroxide? You say visible light and above, so I was wondering if IR exposure is acceptable or at least makes for a slow reaction

Not quite a universally accurate reaction chain, in my opinion, as the $\ce{HO2^.}$ radical has a pKa of 4.88 and above this value, we have the superoxide radical anion ($\ce{O2^{.-}}$) as the active species whose chemistry is distinct from the $\ce{HO2^.}$ radical:
$$\ce{HO2^. = H^+ + O_2^{.-}  (pKa 4.88) }$$
Interestingly, as early as 1934 Haber and Weiss (Ref. 1) proposed that $\ce{HO2^.}$ is formed in the decomposition of hydrogen peroxide.
The subsequent reaction of the transient superoxide radical anion with hydrogen peroxide has been determined to also form the hydroxyl radical (Ref. 2). The higher pH version of the last reaction is, therefore, best represented as:
$$\ce{O2^.- + H2O2 ⟶ 2HO^∙ + OH- + O2}$$
And as  $\ce{H^+ + OH^- = H2O}$, the net product formation is not altered.
However, alkaline $\ce{H2O2}$ is well known to be less stable than acidic hydrogen peroxide (Ref.3) even in absence of light exposure, which accelerates its decomposition (which can involve radical pathways) liberating oxygen!

References:

Haber Fritz and Weiss Joseph, 1934, The catalytic decomposition of hydrogen peroxide by iron salts. Proc. R. Soc. Lond. A 147: 332–351
http://doi.org/10.1098/rspa.1934.0221
TOSHIHIKO OZAWA, AKIRA HANAKI, Reactions of Superoxide with Water and with Hydrogen Peroxide, Chemical and Pharmaceutical Bulletin, 1981, Volume 29, Issue 4, Pages 926-928. https://doi.org/10.1248/cpb.29.926
https://www.researchgate.net/figure/Effect-of-pH-on-the-decomposition-of-hydrogen-peroxide-H-2-O-2-0-800-mg-l_fig1_234110563, from Yazıcı, Ersin & Deveci, Haci. (2010). Factors Affecting Decomposition of Hydrogen Peroxide. https://doi.org/10.13140/RG.2.1.1530.0648.


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Boiling point of liquids
If we can have liquid - vapour equilibrium at any temperature what is the 'point' behind boiling point.Boiling point is just a special case of liquid - vapour equilibrium where partial pressure equals atmospheric pressure.Why is it told in my textbook that boiling point is where the liquid and vapour coexist. At any temperature we can have liquid and vapour in equilibrium.
During boiling process when vapourisation take place from the bulk of the liquid does that increase the vapour composition which alters the equilibrium constant without the change in temperature?..
If we look at the phase diagram boiling point lies on phase boundary between liquid and gas ...so does that mean boiling point is when liquid and gas coexist..

Well, yes, it means exactly that.

"boiling line" might be a more accurate term, but requires at least two variables, making it very hard to compare. Strictly speaking the same is true for melting, only the pressure dependence is less noticeable there.

Boiling liquid is not in equilibrium.

@Poutnik  It is, just not a nice spatially homogeneous or even static equillibrium. And its temperature needs to be above the boiling point, by a small but finite difference.

To be more precise, boiling is not thermodynamic equilibrium, but it is kind of dynamic equilibrium. There is need of continuous providing of thermal energy and eventually even liquid.

Additionally, the *boiling line" is primarily the evaporation line, as boiling is just a special mode of evaporation. At this line, liquid and gas can coexist with zero net mass transfer between phases.

At the boiling point, there is an equilibrium between pure vapor and pure liquid phase (at a certain pressure). A substance in the (pure) liquid phase can also be at equilibrium with a mixture of inert gas and that substance in the vapor phase (so the partial pressure of the substance would be lower than the total pressure). This happens at temperatures below the boiling point. These are two different scenarios (either pure gas phase or mixture in the gas phase).

Liquid and vapour can coexist at many combinations of temperature and pressure. But for most of them they are not in thermodynamic quilibrium.
For a given temperuture, there is only one (vapour partial ) pressure value with both phases in equilibrium.
For a given vapour partial  pressure, there is only one temperature value with both phases in equilibrium.
This temperature is also the liquid boiling point at the given external ( total ) pressure.
Standard boiling point is then temperature, when vapour partial pressure is equal external pressure 1 atm.
Note that boiling is not state of thermodynamic equilibrium, as there is net change of phase mass and thermal energy.
If partial pressure is higher than the equilibrium pressure for given temperature, the vapour condensates until the pressure reaches the liquid-vapour equilibrium curve of the phase diagram.
If partial pressure is lower than the equilibrium pressure for given temperature, the liquid evaporates until the pressure reaches the liquid-vapour equilibrium curve of the phase diagram.
If total pressure is lower than the vapour pressure for given temperature, the liquid boils until the total pressure reaches the saturated  vapour pressure.

But since we have vapour pressure for given liquid at different temperature does that constant vapour pressure defines equilibrium point at those temperature

No, the equilibrium vapour pressure is fully determined by temperature. If vapour pressure is lower than the equilibrium pressure, liquid evaporates. If higher, then vapour condensates. If total pressure is lower, liquid boils.

For all the use of equilibrium pressure I think you missed the most important part. By definition a liquid is at the boiling point when the vapor pressure equals the ambient pressure, whatever that happens to be.  The standard boiling point is when the vapor pressure equals 1 atmosphere.  The choice of 1 atmosphere is a completely arbitrary choice made by humans based on a collective experience. On the planet "Tatooine" for example the standard pressure would be different.

I have just omitted it as implicitly known but it is there.

Boiling point is when phase transition takes place so ,is it correct to say its the point when gas and liquid is in equilibrium rather than telling liquid and vapour is in equilibrium.

Equilibrium is when there is no net change in the phase amount and phases can coexist for unlimited time. That is not the case for boiling. If the liquid is kept in a closed volume and stops boiling when the pressure raises enough, then after this moment it is close to equilibrium.

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Is osmosis really a special case of diffusion across a semipermeable membrane?
A general idea we have about osmosis is that it is the movement of solvent particles through a semi permeable membrane from a region of high concentration to a region of low concentration. This seems correct in case of a gas. Gaseous particles spread through any container in which they are kept, which makes it seem to us as if they moved from high concentration to lower concentration. But in the case of a liquid how is this true? For example consider a bottle half filled with water, water  never moves from high concentration to lower concentration inside the bottle.
So is it wrong to explain osmosis as a special case of diffusion?
What actually makes solvent particles move from regions of higher concentration to regions of lower concentration? Is it solute attraction? Or a difference in pressure in pure solvent phase and solution phase?

Diffusion is not defined as the movement of particles from a region of high concentration to one of low concentration. However, diffusion is the explanation for why this happens. Diffusion is the random motion of particles which causes them to spread from their point of origin.

I've listed a series of points about diffusion and Osmosis.
The cause of osmosis is simply diffusion: the solvent can diffuse through the membrane and the solute cannot.
Thermodynamically, we say that the reason that the diffusion occurs is that the chemical potential of the solution is at a lower chemical potential than the pure solvent and so there is a tendency for the solvent to pass into its solution and to dilute it. The osmotic pressure is the excess pressure just needed to stop that flow. If more pressure were to be applied the flow of solvent would be reversed and the solution would become more concentrated.
Osmosis occurs because there is an increase in entropy overall when the solvent dilutes the solution. There is a decrease in entropy in the pure solvent side because its volume decreases but this is balanced by an increase in entropy on the solution side as the volume increases so this is not the reason osmosis occurs. The reason is the increase in entropy that is experienced by the solute now in a larger volume or equivalently more solvent molecules in which to dissolve. This means that there are more ways of arranging solute and solvent molecules compared to that before osmosis stared, and this represents an increase in entropy. (Left unchecked this will continue until the solution is so dilute it is indistinguishable from the solvent).
When we say that a solution has an osmotic pressure of, say, 5 atm, it does not mean that the solution exerts this pressure but that if it is in contact with its solvent through a semi-permeable membrane then an applied pressure of 5 atm is needed to maintain equilibrium.
It is possible for a solute in a solution to diffuse spontaneously from regions of higher to lower concentration so that uniform concentration is reached and the solution is at equilibrium.  Naturally, diffusion does not stop at this point but is always present both for solute and solvent.
In some cases, however, diffusion can occur form lower to higher concentrations. Such a situation is that when a solute is dissolved in two immiscible solvents in contact with one another, and in which the solute has different solubilities. Solute diffusion will occur until the chemical potential in each solvent is equal which can mean diffusion into the more concentrated solution. Thus at equilibrium there will be a large difference in solute concentration but no overall flow of solute due to diffusion.

A general idea we have about osmosis is that it is the movement of solvent particles through a semi permeable membrane from a region of high concentration to a region of low concentration.

You have to carefully specify which concentration you refer to. The net transport is from regions of high solvent concentration (low solute concentration) to low solvent concentration (high solute concentration).

This seems correct in case of a gas. Gaseous particles spread through any container in which they are kept, which makes it seem to us as if they moved from high concentration to lower concentration.

It does not just seem as if, that is what actually happens as net effect. The reason that more particles move out of a region of high concentration is because there is more of them.

But in the case of a liquid how is this true? For example consider a bottle half filled with water, water never moves from high concentration to lower concentration inside the bottle.

That would be an example of a phase change. Water molecules do move from the liquid to the gas phase and vice versa. However, because the water molecules interact in the liquid and don't interact in the gas phase, most water molecules are in the liquid phase. The reason that some are in the gas phase is that they are more dispersed in that phase and mix with other components of air, both of which is entropically favored.

So is it wrong to explain osmosis as a special case of diffusion?

No, it is perfectly alright to explain osmosis with diffusion. There is a higher concentration of solvent on the side of the membrane with lower solute concentrations, so more solvent molecules move away from that side than move to that side.

What actually makes solvent particles move from regions of higher concentration to regions of lower concentration? Is it solute attraction?

No, this also works for ideal solutions, where the solute:solvent interactions are very similar to the solvent:solvent interactions. It is a process driven by entropy, not enthalpy.

Or a difference in pressure in pure solvent phase and solution phase?

In a system where the liquids have room to expand, the pressure is the same. In a system where the liquid compartments are kept at constant volume, the pressure on the side of higher solute concentration increases (because the amount of solvent in that compartment increases through osmosis). So the pressure difference is caused by the osmosis, not the other way around.
There is a process called reverse osmosis used to purify water, where the pressure differential results in water going from the "dirty" compartment to the "clean" compartment.

So initially there is no pressure difference. I thought pure solvent have greater hydrostatic pressure than in solution phase because even though both solvent and solution have the same volume the concentration of solvent on solution side decreases hence the pressure and osmosis is to equalise this hydrostatic pressure.

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Gas Law - increasing temperature with constant volume held
I've set up a simple experiment to look at the ideal gas laws. My experiment is relatively simple in that I have a metal tube which is capped on one side. I am then pressurising the tube with air to 100 psi and locking it off.
My thought is that as the pressure increased, with volume held constant, the temperature of the tube should rise. However, after 30 minutes of observation, there is no rise in temperature. Am I missing something? I was expecting to see the temperature of the tube rise.
The tube is SS 316 with a ball valve capping it off at one end and using the pump to hold the pressure once it is reached.
Thank you.

Are you considering how well your system is/isn't? Consider enclosing the system and setting up a second temp probe to measure ambient temperature.

Thanks for the comment. The setup is not enclosed at the moment and in the workshop. I am using an infrared temperature sensor to measure the temperature of the pipe surface. I didn't think heat loss due to ambient outside air temperature would be too much of an factor, or I would at least see some temperature increase, however small.

Also, apologies, I didn't realise I posted this in the chemistry section.


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Two common fluids that don't mix inside a labyrinth
I'm not sure in which SE site put this question.
Imagine a typical child toy, where a bunch of spheres move by gravity through a labyrinth.
Now imagine a sealed small labyrinth with two fluids. I'd like to reproduce the concept of one fluid moving through another forming small bubbles, but I don't know which common fluids might be useful. Theose two fluids should not mix, the viscosity must be low (so the bubbles can move easily), both fluids must have different colors, they must have different density, the bubbles should not stick to the plastic of the labyrith, it should work at room temperature, and -if possible- I'd like to see many small bubbles instead of a single big bubble.
Which two common fluids might be useful? I tried air inside water (few bubbles and movement too fast), I also tested water inside kitchen oil, but it moves too slow and the colors aren't quite different.

If kitchen oil is too viscous, try kerosene, paint thinner, and so on. Don't bother about color. After all, any liquid can be colored anything.

You can also increase the viscosity of the water by addition of sugars

You can at least colour one phase after you identify a suitable couple.


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Why does H2O electrolysis produce O2 on the anode instead of H2O2
If the oxidation of $\ce{H2O}$ produces $\ce{OH-}$ ions and the anode separates electrons from them leaving neutral $\ce{OH}$ molecules, why don't they bind to each other covalently?

To an extent, they do.

Look at the redox potentials for oxidation of HOOH and H2O. The HOOH oxidizes much more easily, so any HOOH that is formed gets quickly converted to O2.

It does, as an intermediate product. The peroxide formed is on the surface of the electrode which changes it's chemical behavior.
Electrolysis is a four-electron process. In the first step, an electron is removed from a water molecule, producing a neutral OH adsorbed to the surface of the electrode. The second step removes another electron from the adsorbed OH, producing a neutral oxygen atom adsorbed to the electrode. Third, another electron is removed from a water molecule leaving an adsorbed OOH. Then the fourth electron comes from the OOH to produce O2.
My understanding is that the interactions with the electrode partially stabilize the unstable intermediate states. This is the reason some electrodes work better than others as catalysts.
This paper has more than you'll ever need to know: https://doi.org/10.1002/cctc.201000126


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Identifying an extra NMR peak in silyl protection reaction
I protected a phenol with the TIPSCl/triethylamine procedure. The product was isolated by column (hexanes/EtOAc) and shows an extra singlet at 1.05 ppm in $\ce{CDCl3}$, integrating to ~10 H. Whatever this is I cannot find in under UV or various TLC stains ($\ce{KMnO4}$, vanilin, PMA). The integration of the peak goes smaller after one more column. Any suggestions? Thanks!

Is there a corresponding bonus peak in the 13C?

Found two extra peaks in the aliphatic region, 18.1, 17.8, 14.6, 14.2,12.4. I know two of these are the isopropyl and a CH3 from an ethyl, did not assign them though.

I believe some of your triisopropylsilyl chloride may be hydrolyzed triisopropylsilylol, $\ce{((CH3)2CH)3SiOH}$. That may gives you $\delta$ 1.04 and 12.4 (for $\ce{C}$ attached to $\ce{Si}$) in $\mathrm{^1H}$- and $\mathrm{^{13}C}$-NMR spectra, respectively. Compare the spectral values in Ref.1 and 2 (see below; corresponding $\ce{^{13}C}$ values are in red):

References:

A. Goti, M. Cacciarini, F. Cardona, A. Brandi, “A convenient access to (3S)-3-(triisopropylsilyl)oxy-1-pyrroline N-oxide, a useful intermediate for polyfunctionalized enantiopure indolizidine and pyrrolizidine synthesis,” Tetrahedron Letters 1999, 40(14), 2853–2856 (https://doi.org/10.1016/S0040-4039(99)00310-X).
R. F. Cunico, L. Bedell, “The triisopropylsilyl group as a hydroxyl-protecting function,” J. Org. Chem. 1980, 45(23), 4797–4798 (https://doi.org/10.1021/jo01311a058).

This is what I'd originally thought (usually stuff like R3SiOH or R3SiOSiR3 are byproducts from silyl protection). What stopped me from putting it forward was the multiplicity, as the big peak for TIPS should be a doublet.

yes the singlet is what confuses me -- it looks like some other silylanes, maybe my TIPSCl have some TMS or TBDMS with it?

Take an NMR of TIPSCl.

@user55119: Good point. Even better, take another NMR with hydrolyzed TIP-Cl, if you are curious to know the by product.

@orthocresol: I agreed with you completely. Yet, both of sighted references reported methyl peak of triisopropyl silyl-O- as a singlet (may be steric effect cost the dihedral angle close to $90^\circ$?). JOC paper even counted 21H for possible 18H (possibly overlapping two signals for each isopropyl 7Hs). I edited my answer accordingly, putting the published spectral values for your convenience.

I appreciate your effort, and I'm not saying you're wrong, rather I don't know. I've never worked with TIPS groups, only TBS. Since I cannot judge whether the answer is correct or not, I refrain from voting.

FWIW, for TIPSOH, Chem. Lett. 2014, 43 (4), 429–431 says δH 1.07 (m, 21 H) and δC 17.7, 12.3. Most other references e.g. Angew. Chem. Int. Ed. 2011, 50 (33), 7533–7536 give similar values, but somewhat intriguingly,  Chem. Commun. 2016, 52 (70), 10625–10628 reports δH 1.43 (s, 21 H) and δC 30.3, 21.1. One wonders if they got the correct product...

@orthocresol: Good data. I think $\delta$H 1.07 (m, 21 H) and $\delta$C 17.7, 12.3 are more reliable as Tet Lett article sited, because the compound I posted should not have interfering peaks at that region. Thanks for the hard work. :-)


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Acetic Acid type of solvent
I want to ask a question about polar/non-polar (a)protic solvents.
The following information was given to me in the lecture hall today regarding solvents in organic chemistry.

I recall the structure of Acetic Acid as $\ce{CH3COOH}$ but surely given the $\ce{O-H}$ dipole, wouldn't that make Acetic acid a Polar Protic solvent i.e. the $\ce{H}$ attached to the $\ce{O-H}$ can be involved in hydrogen bonding?
Am I incorrect in this instance? Shouldn't $\ce{CH3COOH}$ be classified as a polar protic solvent?

Probably a typo.

It's not a typo. CH3COOH has particularly low dielectric constant. While it doesn't disqualify it from category of "polar solvents", it should hamper solubility of salts.

@Mithoron that's true. Typically, the importance of polar/non-polar, protic/aprotic solvents is taught in the context of SN1/SN2 selectivity and is presented primarily in terms of dipole.  Acetic acid has a dipole moment very similar to that of ethanol, and alcohols are commonly used to favour carbocation-forming reactions. That may be where the confusion is coming from.


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Are there more endothermic or exothermic reactions?
I'm wondering whether there are more endothermic or exothermic reactions.
Of course the amount of mass will stay the same, but I thought that not every exothermic reaction has an endothermic counter-reaction. Is this stream of thought correct and does that mean that there are more exothermic reactions than endothermic reactions?
In other words, does every reaction have a reverse reaction?

Well, if a reaction is exothermic, then its reverse has no choice but to be endothermic.

Answering "does every reaction have a reverse reaction?" (are there more endothermic or exothermic reactions):
You've actually hit on a topic people have been exploring for nearly a century:

Lewis. A New Principle of Equilibrium. PNAS, 1925, 11(3),179-183

The Law of Entire Equilibrium — Thus I am led to propose a law which in its general form is not deducible from thermodynamics, but proves to be compatible with the laws of thermodynamics in all cases where a comparison is possible. It may be called the law of entire equilibrium, and may be stated as follows. Corresponding to every individual process there is a reverse process, and in a state of equilibrium the average rate of every process is equal to the average rate of its reverse process. The rate at which one group or set of groups L goes over into another group or set of groups L' is the same as the rate at which the groups L' go over into the groups L. Moreover if there are various paths by which the first process occurs, there is an equal number of paths by which the second process occurs, and the rate is the same in both directions along every path. This will be true no matter how detailed are the specifications which define the several groups and the several paths.

The implications of this concept (and your question) have been well-elucidated by SE user Curt F. in two posts:

Is every chemical reaction in equilibrium?

Commenters suggest that "irreversible" reactions do not have an equilibrium. This is true, but tautological. In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree. To say otherwise would violate microscopic reversibility. A reaction that was 100% irreverible would have an equilibrium constant of infinity. But if =∞, then Δ∘=−ln would turn into Δ∘=−∞. So to get infinite energy we would just have to use 100% irreversible reactions! Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.

Spontaneous/Non-Spontaneous Reactions and Reversible Reactions

All reactions can be viewed as reversible from a mathematical standpoint, if the reacting system is "big enough". Reactions that are highly spontaneous in the forward direction mean that at equilibrium, the number of "reactant" molecules will be very very small (but not zero!).
Many times, "big enough" would mean astronomically big, such as bigger than the Earth. In these cases, reactions can be regarded as "irreversible", although there is no strict, universally agreed upon boundary between "reversible" and "irreversible".

This should implicitly answer your question, as a 'reversed' exothermic reaction is an endothermic reaction. If every reaction is fundamentally reversible (on some scale, whether or not it's reasonable, as Curt F. describes), then there are an equal number of exothermic and endothermic reactions.

Answering "which occurs more frequently: exothermic or endothermic reactions?" (are there more endothermic or exothermic reactions?):
This is an interesting way to look at things. Considering the ever-increasing entropy of the universe and the favorable relationship between entropy and heat, I would guess that there are more exothermic reactions. Exothermic reactions represent energetically favorable transformations and are more spontaneous than endothermic reactions, so it stands to reason that there are "more exothermic reactions."

Roduner &  Radhakrishnan. In command of non-equilibrium. Chem. Soc. Rev., 2016, 45,2768-2784

2.7 Heat death: the ultimate end of an evolving universe

From the second law of thermodynamics we know that the entropy of the universe will increase for all spontaneous processes. Since the total possible reversible heat flow is limited by everything having the same temperature, and when no chemical or nuclear reactions are available anymore which can produce further heat, entropy will reach a maximum but will not go to infinity. This extrapolated end of a universe that evolves further in time has been dubbed heat death. But how will this final state look like? This is a question that is complex to answer. Considering an extremely simplified model universe helps to get a rough idea.
We first consider a closed system consisting of 1 mol of glucose and 6 mol of molecular oxygen at 25°C and 1 bar pressure (see Fig. 18a). The equilibrium state at the same temperature and pressure consists of the system after complete combustion to 6 mol of CO2 and 6 mol of liquid H2O. The heat of reaction of 2805 kJ will be dissipated from the system...

5 Concluding remarks

Heat engines produce work that is typically used to drive processes which do not occur spontaneously. This is achieved by letting the machine operate between two temperatures, where energy is provided to the system at the high temperature and partly converted to work. This can be done only when an unavoidable remaining fraction of the energy input is degraded to lower temperature and dissipated as waste heat. It is shown here that this principle is not limited to heat engines but holds more generally for many other processes in which a subsystem of the universe is to be transformed to higher order, as in the separation by distillation or crystallisation, photosynthesis, solar cells, or self-assembly...
The simple model universe of Fig. 18, the combustion of glucose, predicted that the final state of the universe will be much warmer compared with today and consists of a higher fraction of atmospheric CO2 and H2O at the cost of O2. One might therefore ask whether the global warming that we see on our planet actually reflects that we are noticeably approaching the heat death of the universe as a consequence of combustion of fossil fuels with corresponding heat and entropy production. The answer is straightforward: the direction of the effect is correct, but it does not quantitatively reflect the approach of our planet to heat death. The solar heat incident on our atmosphere is about 5 orders of magnitude more than the total energy consumption by the human population. CO2 and water produced by combustion just increase the mismatch between incident and reflected solar energy by increasing the greenhouse effect of the atmosphere, thereby amplifying the direct effect by a considerable factor.

If we're to believe that heat death is a logical conclusion of the second law of thermodynamics, then there must be more exothermic reactions to result in an ultimately stagnant system where all available energy has been dispersed as heat and atomic components are at their lowest possible energy states — this outcome, by definition, represents an excess of exothermic reactions.

The problem with this heat death hypothesis is that there has to be another universe (the surroundings) for our universe to transfer heat to. Exothermic lacks meaning without a defined surroundings! But otherwise a fun answer :-)

Is that true? I thought the idea is that entropy drives our universe (the surroundings) to a point of equilibrium where nothing can happen within it. That doesn’t require an outside system, I don’t think — it would just mean that everything (all components) are reduced to their lowest energy state and all free energy is in the form of dissipated heat, so no changes can spontaneously occur in the system (our universe + its contents).

Ok, I agree with the energy dispersion, universal entropy maximization (heat death) theory, but still, exothermic means releasing heat to the surroundings. When you start talking about the entire universe as your system, there are no surroundings. That was my point.

But ok, maybe it was not a serious objection. Considering heat death, then, if molecules are small puddles of order (low entropy), one can imagine that entropy maximization would call for their destruction. But molecules minimize energy locally (in a vaccuum), and breaking bonds is endothermic, so what drives the destruction is an increase in the entropy of the system, not the surroundings, right? Anyway, I'll have to think some more about this cosmological stuff.

Your point about the remaining energy in bonds is interesting — I suppose it isn’t so much that all possible heat has been liberated, but that all available spontaneous (exothermic) reactions have been exhausted. Thought-provoking stuff!

Please try to not use edit statements. Q&A should read as one flow and changes should be merged. If someone is interested in edits, the history of these is available for every post. (Obviously this is difficult for a chameleon question like this.)

@Martin-マーチン I almost never would, but this question (at least to me) appeared to be three different questions over the course of a few days. I’ll remove the “edit” terms and clarify which reading of the question corresponds to each answer so future readers aren’t confused.

Indeed that's why I added the part in the parenthesis (see also meta: Exit strategies for "chameleon questions"). It was only meant as friendly note, since I wasn't able to deduce how familiar you are with SE. (Something I should have also hinted towards the OP.)


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Qualitative Comparisons of Dielectric Constants of Organic Solvents
This question is inspired by the discussion around this earlier question.
It is relatively simple to do a qualitative comparison of the polarities of different molecules (for example by analyzing the presence of polar bonds alongside the geometry of the molecule).
Now, ethanol and acetic acid have almost identical dipole moments ($\mu$=1.68 for acetic acid, $\mu$ = 1.69 for ethanol) but dramatically different dielectric constants ($\mathcal{E}$ = 6.15 for acetic acid, $\mathcal{E}$ = 24.5).
Is there a similar qualitative analysis that can be done for dielectric constant of a solvent?

Acetic acid easily forms a dimer, that largely eliminates it's dipole moment.

Somehow related: What is the order of the following compounds based on polarity?

Yes, but it is much more complicated. The molecule does not have to have a dipole, in which case the (static) dielectric constant (relative permittivity) depends on the molecule's polarisability and if it does have a dipole it depends on both factors. The equation is $\displaystyle \epsilon_r=\frac{1+2b}{1-b}$ where $b=\left(\alpha+\mu^2/(k_BT)\right)N/(3\epsilon_0)$ and $N=N_A\rho/M$ where $\alpha$ is polarisability, $\mu$ dipole, $\rho$ density,$M$ molar mass, $\epsilon_0$ permittivity of free space.  See Atkins, & Friedman, 'Molecular Quantum Mechanics'. The dielectric constant also depends on frequency of the applied electric field, but the value quoted is normally the static value measured by comparing the capacity of a capacitor with and without the solvent. The name relative permittivity arises from the potential a charge produces. In a vacuum at a distance $r$ this is $\displaystyle \phi=\frac{q}{4\pi\epsilon_0 r}$, in a dielectric medium of permittivity $\epsilon$ this becomes $\displaystyle \phi'=\frac{q}{4\pi\epsilon r}$ and $\epsilon_r=\epsilon/\epsilon_0$.
As an aside you can see why a polar molecule is generally insoluble in a non-polar solvent. If the permittivity is small (say hexane) the electric potential of a charge is higher at any given $r$ than for a solvent with a large $\epsilon$, such as acetonitrile. This means that polar molecules are less soluble in low dielectric solvents than they are in high dielectric ones.


text:
Whether a combination reaction goes to completion
There is a sentence in my textbook:

When a combination reaction results in a gas that is allowed to be
  released, a precipitate that drops to the bottom of the container, or
  a nonionizing product, the reaction will go to completion.

What is meant by "the reaction will go to completion"? Does it mean all of the reactions will be converted to products and no reverse reaction occurs?

It generally means that the reaction is irreversible (no reverse reaction occurs), or that the extent of the reverse reaction is so small as to be of no consequence.  This is the case for many precipitation reactions.  They are technically reversible in that some of the precipitate is in equilibrium with solvated ions, but the relative amounts are small.


text:
How do I write a procedure for creating a buffer?
I need to write an original procedure with exact numbers for creating a buffer:

Create a buffer using $\ce{CH3COOH}$ and $\ce{CH3COONa}$ that has a $\pu{pH}$ of exactly $3.75$.
A $\pu{50 mL}$ sample of your buffered solution has to be able to withstand the addition of $\pu{25.00 mL}$ of $\pu{0.10 M}$ $\ce{NaOH}$ solution.
"Withstand" here is defined as less "than $0.5$ change in $\mathrm{pH}$."
The buffered solution will break after the addition of no more than $\pu{35.0 mL}$ of the $\pu{0.10 M}$ $\ce{NaOH}$.
We are given $\pu{0.10 M}$ $\ce{NaOH}$ and $\pu{6 M}$ acetic acid.

Where I've got so far:
I took the $\mathrm{p}K_\mathrm{a}$ of acetic acid $(4.75)$ and the desired $\mathrm{pH}$ $(3.75)$ and put it into the Henderson-Hasselbalch equation:
$$3.75 = 4.75 + \log\frac{[\text{salt}]}{[\text{acid}]}$$
$$\log\frac{[\text{salt}]}{[\text{acid}]} = -1$$
So
$$\frac{[\text{salt}]}{[\text{acid}]} = 0.10,$$
which means we want the ratio $[\ce{CH3COONa}]:[\ce{CH3COOH}]$ to be $0.10$.
I think we can prepare a solution containing $\pu{0.10 M}$ $\ce{CH3COONa}$ and $\pu{1.00 M}$ $\ce{CH3COOH}.$ But with what we're given $(\pu{0.10 M}$ $\ce{NaOH}$ and $\pu{6 M}$ acetic acid), I don't know how to proceed and find exact numbers for a procedure.

It seems to me as almost exact copy of a recent question

https://chemistry.stackexchange.com/q/115839/35806

And as requirements here are even more strict than there, neither here the resolution exists. The withstanding and breaking requirements are even more contradictory than before. Other buffer system must be used, with maximum capacity near pH=3.75. Or perhaps even lower.

Initial note
The specific problem with your scenario is, that you need a buffer of buffer capacity progressively decreasing with increasing $\mathrm{pH}$, but capacity of simple buffers is for $\mathrm{pH} \lt \mathrm{p}K_\mathrm{a}$ increasing.
As consequence, its "breaking" capacity is too big, compared to "withstanding capacity".
General analysis
Let there is given the initial $\mathrm{pH_{ini}}$ and the acid dissociation constant
$\mathrm{p}K_\mathrm{a}$
Let there there is given "breaking volume" of $\ce{NaOH}$ $V_\mathrm{br}$, where the buffer losses it's buffering ability.
Let there is given "withstanding volume" $V_\mathrm{ws}$ when $\mathrm{pH}$ changes by value $x$.
Then:
$$\mathrm{pH_{ini}}=\mathrm{p}K_\mathrm{a} + \log {R}$$
where
$$R=\frac{ [\ce{A-}]_\mathrm{init}}
{[\ce{HA}]_\mathrm{init}}=10^{\mathrm{pH_{ini}}-\mathrm{p}K_\mathrm{a}}$$
$$\mathrm{pH_{ini}}+x=\mathrm{p}K_\mathrm{a} +\log  \frac {R+V_\mathrm{ws}/V_\mathrm{br}\cdot (1-R)}{1-V_\mathrm{ws}/V_\mathrm{br}}$$
If we consider
$$R_1=10^{\mathrm{pH_{ini}}+x-\mathrm{p}K_\mathrm{a}} $$
$$V_\mathrm{r}=V_\mathrm{ws}/V_\mathrm{br}$$
then
$$R_1=\frac {R+V_\mathrm{r}\cdot (1-R)}{1-V_\mathrm{r}}$$
$$V_\mathrm{r}=\frac {R_1-R}{R_1-R+1}$$
Goal
The desired $$V_\mathrm{r}\ge \frac{25}{35}=0.714$$
Acetic acid case
If we consider acetic acid with $\mathrm{p}K_\mathrm{a}=4.75$ then $$R=10^{3.75-4.75}=0.1$$
If the change should be $\Delta \mathrm{pH}=0.5$:
$$R_1=10^{3.75+0.5-4.75}=0.316$$
$$V_\mathrm{r} =\frac{0.316-0.1}{0.316-0.1+1}=0.177\lt 0.714$$
If the change should be $\Delta \mathrm{pH}=1.0$:
$$R_1=10^{3.75+1.0-4.75}=1$$
$$V_\mathrm{r} =\frac{1-0.1}{1-0.1+1}=0.475 \lt 0.714 $$
Formic acid case
If we consider as a replacement e.g. formic acid with $\mathrm{p}K_\mathrm{a}=3.75$ then $$R=10^{3.75-3.75}=1$$
If the change should be $\Delta \mathrm{pH}=0.5$:
$$R_1=10^{3.75+0.5-3.75}=3.16$$
$$V_\mathrm{r} =\frac{3.16-1}{3.16-1+1}=0.684\lt 0.714$$
If the change should be $\Delta \mathrm{pH}=1.0$:
$$R_1=10^{3.75+1.0-3.75}=10$$
$$V_\mathrm{r} =\frac{10-1}{10-1+1}=0.9 \gt 0.714$$
Summary

Acetic acid is unuseable for making a buffer to fit your requirements.
Formic acid can be used to make a buffer with "withstanding" parametr $\Delta~ \text{pH} \lt +1.0$, seen in the other question.
Formic acid very tighly missed the "withstanding" parametr $\Delta~ \text{pH} \lt +0.5$, so seems about usable.
Even better results could be achieved with an acid with $\mathrm{p}K_\mathrm{a} \lt 3.75$ .


text:
What is the standard way to know the price of a chemical substance?
I've encountered many vendors with broadly varying prices for the same chemical substance, including amazon and Alibaba. I wonder if there is a standard way to know the universal/average price of a chemical substance.

Universal price is not a thing at all. As for the average price, you can get it by averaging the prices of all vendors, as simple as that.

But be careful that you're looking at the same "grade" or "quality" of the chemical. There is a vast difference in price between muriatic acid and ultra-pure hydrochloric acid.

Thank you for the comments,@MaxW I mean the same grade.

@IvanNeretin I'm asking here because the prices are very different which prompted me to ask about a standard or trusted way to know the price of a chemical as any other good that has an average price, economically speaking.

Well, it is pretty much the same as with any other type of goods on the market. There is nothing chemically specific about it. Take it or leave it. Or look around and hopefully you'll find a better deal. Or maybe not.

The price is the convergence of what you're prepared to pay for it and the seller is prepared to sell for ...  Sometimes a published "price" may be a useful guide but ... your question is one of economics rather than chemistry.

Cost-benefit is a part of any science, and the working scientist who is usually budget -bound is the best advice giver in this regard. isn't it ?@user1945827

@IvanNeretin it is doubtful f you see a vendor offering a good with, say 100 USD, another offering for 20 USD and a third for 280 $ ?

Aldrich is a very well-known and reliable supplier for laboratory chemicals. They may not have the exact material you want, but if they have it, it is probably a good value. Start there and see if some other company can do better (and how quality differs). ChemicalStore.com is another supplier.
If you are seeking larger quantities (55 gallon drums or 50 pound bags), Spectrum is a possibility, or you may be interested in a larger (local) chemical distributor.
The best supplier is reliable and this may cost a bit. Establishing an account and being a continuing customer could gain you some help when you need it. Saving 5-10% on an item may not be worth it if it causes you to price-shop - you will be shopping and hopping all the time.

What is the meaning of price-shop?

Price-shopping is shopping for a better price for the same or similar product, without considering other factors like quality, availability (how soon), reliability, service, and perhaps manufacturing location.

I found this useful answer on quora:For laboratory amounts a good chemical supply catalog, such as Aldrich,
For commodities it is more difficult. A starting point is the ICIS Chemical Price Guide: Chemicals A-Z


text:
How does equalizing voltage for lead acid battery depend on the temperature?
I am using advanced lead-carbon batteries (these ones) in an off-grid solar system. The spec specifies equalizing voltage 14.1 V for 12 V block @ 25 °C (77 °F). I keep the batteries in an outdoor shed, and the temperature where I live tends to hang around 0 °C (32 °F) these days.
So, for lead acid chemistry, would the equalizing voltage go up or down with temperature going down? I feel like it would go up (more voltage needed to do the same job at lower temperatures). How does this reasoning stand? What are the underlying scientific considerations?

If the batteries are commercially available, I would ask the manufacturer. They surely have more information than what is in the specs.

@KarstenTheis I am interested in the underlying scientific considerations too, not just in the number to configure my solar system.

I cannot confirm the accuracy of this entry in the wikipedia but: "There are no known, independently verified ways to reverse sulfation. There are commercial products claiming to achieve desulfation through various techniques (such as pulse charging), but there are no peer-reviewed publications verifying their claims. Sulfation prevention remains the best course of action, by periodically fully charging the lead-acid batteries." https://en.wikipedia.org/wiki/Lead%E2%80%93acid_battery#Sulfation_and_desulfation

@BuckThorn sulphate crystals may not be completely or substantially removable, but if they were not at all (i.e. reaction not reversible) there would have been no lead acid batteries. I think the question can be generalised as "how does temperature affect the voltage that is required to dissolve lead sulphate at the same speed".

The mechanism of reversing sulfation appears to be to exceed the nominal charging voltage by a small amount, in order to counter internal resistive losses. This strikes me as a problem that is specific to your batteries and their history and in a sense too broad to give you a detailed answer. As a starting point however you can take Loongs or similar answer which gives you the baseline voltage you need to exceed. as computed from thermodynamic principles. In practiceyou may want to measure the current during charging and see how that changes.

@BuckThorn I am not trying to calculate equalizing voltage at 0 °C. I just want to know whether it is higher or lower than at 25 °C. I am not too convinced yet that it is lower just because the output voltage goes down with temperature. The answer "goes up" or "goes down" should be the same for any lead acid battery.

Generally, the electrode reactions are both based on lead in different oxidation states. At the negative electrode, $\ce{Pb}$ is oxidized to $\ce{Pb^2+}$ during discharge.
$$\ce{Pb <=> Pb^2+ + 2 e-}$$
At the positive electrode, $\ce{Pb^4+}$ is reduced to $\ce{Pb^2+}$.
$$\ce{Pb^4+ + 2 e+ <=> Pb^2+}$$
For a classical lead–acid battery, the overall cell reaction is approximately
$$\ce{Pb + PbO2 + 2H+ + 2 HSO4- <=> 2 PbSO4 + 2 H2O}$$
As long as $\ce{Pb}$, $\ce{PbO2}$, and $\ce{PbSO4}$ are available at the electrodes, the equilibrium cell voltage depends only on the acid concentration (that's why equilibtrium cell voltage can be estimated based on measured acid density).
The dependence of the equilibrium voltage on concentration is given by the Nernst equation:
$$U_\text{cell}=\left(1.931+0.0592\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
This equation applies to one cell; a battery, however, has six cells in a row. Thus the total voltage is
$$U_\text{battery}=6\left(1.931+\frac{RT}{ZF}\log\frac{a_{\ce{H+}}\cdot a_{\ce{HSO4-}}}{a_{\ce{H2O}}}\right)\ \mathrm V$$
where
$R$ is the gas constant,
$T$ is temperature,
$z$ is the number of electrons transferred in the cell reaction, and
$F$ is the Faraday constant.
We are only interested in the dependence of voltage $U$ on temperature $T$. All the other parameters may be taken as constants. Thus, our equation can be simplified to
$$U=\left(11.586+T\cdot k\right)\ \mathrm V$$
where $k$ is a constant.
The exact parameter values for a classical lead–acid battery might differ a bit from a modern battery with lead–carbon electrodes; however, we can calibrate our model using the data given by the manufacturer, i.e. $U=14.1\ \mathrm V$ at $T=25\ \mathrm{^\circ C}=298.15\ \mathrm K$
$$14.1\ \mathrm V=\left(11.586+298.15\ \mathrm K\cdot k\right)\ \mathrm V$$
and we find
$$k=0.008432\ \mathrm{K^{-1}}$$
Therefore, our complete equation is
$$U=\left(11.586+T\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V$$
For a new temperature of $T=0\ \mathrm{^\circ C}=273.15\ \mathrm K$, we get an estimate of
$$\begin{align}U&=\left(11.586+273.15\ \mathrm K\cdot 0.008432\ \mathrm{K^{-1}}\right)\ \mathrm V\\
&\approx13.9\ \mathrm V\end{align}$$

So far this tells how battery's voltage output depends on temperature. But the question is about equalizing voltage which is voltage that needs to be applied to the battery in order to remove sulfate crystals that build up on the plates over time. At 25 ∘C 14.1 V do the job. Are you sure that lower voltage is needed at lower temperatures to remove the same sulfate crystals?

Wouldn't the standard cell potential also be temperature dependent? What is the reaction enthalpy of the process? Also, the conductivity and the kinetics of various steps in the process would change, potentially affecting the required equalizing voltage.

@Greendrake Whatever the value of 14.1 V given by the manufacturer is referring to – I only made an estimate for the temperature correction of this value.

@Loong Sorry but this does not answer the question at all then. I did not ask how voltage output depends on temperature.

@Greendrake You asked how the equalizing voltage (14.1 V at 25 °C) depends on temperature. And I tried to answer that.


text:
Can Acids have pH of more than 7?
Consider a solution of $\text{HCl}$ with a concentration of $10^{-10}$ $\text{M}$. Now, if I find it's pH:
$$\begin{aligned}\text{pH} &= -\log([\text{H}^+])\\
\text{pH} &= -\log(10^{-10}) = 10\end{aligned}$$
At room temperature, $\text{pH} + \text{pOH} = 14 \implies \text{pOH}$ of the given acid is $4$. This means that the $[\text{OH}^-]$ ions is $10^{-4}$ $\text{M}$. But from where do these $\text{OH}^-$ ions come?

At this dilution rates you can no longer get away with using concentration instead of activity, and you have to calculate $\mathrm{pH}$ as $\mathrm{pH} = -\log a(\ce{H+}).$ Note that $\mathrm{pH}$ also heavily relies on solvent/medium.

I object. At dilutions this great, you can more than ever  rely on using concentration instead of activity. Now to the point. What is the pH of pure water?

@andselisk right answer, but to another question ;-)

In one word: Autoprotolyis

^ That said, I think this is a useful duplicate for search, concisely worded.

In acid base reactions in water there is always the reaction $\ce{H2O <=> H^+ + OH^-}$ occurring and this cannot be ignored. The equilibrium constant $K_w=10^{14}$ so the $\ce{[H^+]}=10^{-7}$ from this reaction far exceeds that from $10^{-10} $ molar acid. see answer https://chemistry.stackexchange.com/questions/100346/calculate-ph-of-a-mixture-of-a-strong-base-and-acid-knowing-only-the-ph-wt-v/100355#100355


text:
Reconciling these different solubility values
In a review article [1] on the solubility of CaCO$_3$(calcite), the solubility is reported to be on the order of 1 gram/litre.
Since CaCO$_3$ has a molar mass of about 100, this solubility is equivalent to $10^{-2}$ M.
The solubility product is the square of this, $K_\mathrm{sp}=10^{-4}$.
However, the solubility product of calcium carbonate (according to various sources, like Wiki) is actually $\sim 10^{-9}$ which is 5 orders of magnitude smaller.
What am I missing?
[1] Miller, John P. "A portion of the system calcium carbonate-carbon dioxide-water, with geological implications." American Journal of Science 250.3 (1952): 161-203.

1 g/l is an insanely high figure for CaCO3; I won't believe it even if I see it appear in glowing red letters on a wall, written by an invisible hand. But there's a catch: what I said only applies to pure  water. Add some acid, and you'll easily dissolve this much, and maybe a good deal more.

Perhaps the influence of carbon dioxide?

These experiments don't seem to contain any acids that would account for it. According to the literature (Plummer et al, American journal of science 278.2 (1978): 179-216) the range of pCO$_2$ might account for like 1 order of magnitude variation, not 5. I can also cite papers where pH has no more than 1 order of magnitude effect.

Reading the title of the reference gives the hint.

But what does the "CRC Handbook or Chemistry and Physics" say? Because it is the Bible for chemical and physical constants

@Poutnik Are you referring to the CO2? Because I address this in the above comment^.

@lemon Yes, I do. As presence of CO2 can raise content of calcium in water a lot, in form of carbonate hardness CaCO3 + H2O+CO2->Ca(HCO3)2. Having dissolved hundreds of miligrams of calcium in bicarbonate mineral water is common. One has to take into account equilibrium constants for CO2(g)<>CO2(aq)<>H2CO3<>HCO3-<>CO3^2-


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

The rearrangement is a series of shifts.

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

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

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

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

Advanced Organic Chemistry , FIFTH EDITION ,Part A: Structure and Mechanisms
by Francis A. Carey and Richard J. Sundberg.
Quoting from page no.446 it says
"The preferred alignment of orbitals for a 1,2-hydride or alkyl shift
involves  coplanarity of the p-orbital at the carbocation ion center and the
orbital   of the migrating group "

1,2 shift pathway is seen.

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

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

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

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

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

Ring-Expansion Reactions

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

Therefor ,the above sources have all talked about 1, 2 shifts only .
So my question is , is 1,3 alkyl shift as given in the post (Which ring expansion in cyclobutyl(cyclopropyl)methanol is favourable?)
  allowed ?

This may be a stupid question, but have you checked the literature? And if you can't, can we get a volunteer?

@ Zhe i donot have axcess to litrature as i have encountred a paywall.I needed to clarify regarding 1,3 shifts as shown.Then we may have to urge authors to rewrite books so that such information is accessable (imagine asking clayden to change).Please SHARE THAT INFORMATION.That should help every one

That urging authors to rewrite part might be taking it too far. Just because a paper says something for limited scope, does not mean it is true everywhere. Hell, it doesn't even mean it's true at all. Also, this would be absolute hell for beginning students.

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

then the mechnism given in https://chemistry.stackexchange.com/questions/116143/which-ring-expansion-in-cyclobutylcyclopropylmethanol-is-favourable/116152#116152 is wrong .

I see my answer is the same as Lautman's, Zhe's comment notwithstanding. Didn't scroll down far enough. The point is why release strain from the cyclobutane ring first?

In the answer above, as well as my answer to the linked question, the strain is released from the cyclopropane first.

@Michael Lautman: I belatedly saw the duplication after I posted it. If you are concerned about the duplication, I can delete mine or link to your comment.


text:
Quantitative Transfer of Volatile Standards
(Edited w/ updates) Okay, long story short I work in a lab which has been working on cannabinoid analysis (in the licensed Canadian market) and have had a lot of issues surrounding the standards/response factors varying over time.
They are only available in pure methanol or ACN, are very expensive, and only come in $\pu{1 mL}$ quantities. I currently micropipette $\pu{150 \mu L}$ of these into a vial and then dilute, though the vapour pressure of the solvent causes it to immediately push droplets back out of the pipette tip, meaning the volume changes slightly based on the speed at which you transfer, and I think this (among other things, like slow evaporation of retained stocks in storage) is causing variation. Relative response factors between cannabinoids vary up to 30% between different preparations on different days, which is unacceptable.
So I need to both conserve the liquid I use and find a way of transferring that can reduce the effect of volatility. I proposed weighing the ampule, transferring quantitatively to volumetric flask and weighing again, then back calculating the mass amounts using density, as the concentrations come per volume. I was confused about what my manager's issue with the idea was at first but clarified and it was about potential effect of the dissolved cannabinoids on the density of the solution via partial molar volume effects on the manufacturer's end. Aka how much would this affect density and are the concentrations they gave us reliable? They're FDA exempt, meaning they're unregulated and there is no one "gold standard" preparation or test method.
I wrote a pseudo-proof for this effect being negligible (TL;DR because 2.5mg solute of any kind out of 786mg solvent would produce a variation of <0.5% in density even when wildly exaggerating the partial molar volume effects) and our chief scientific officer reviewed and signed off on it so I'm going to be doing the gravimetric transfer.
Ampoule chips were not really a concern, they very reliably break cleanly and any chips are likely too small to be seen/matter much on that same ~782mg scale during my preparation.

What are you using the standard for?

If you're transferring the entire contents to a volumetric flask, you don't need to mess about with density.  You will have the mass from the ampoule weighings.

@Beerhunter: The ampule could be chipped and could loose some weights.

@Night Writer UPLC assay of cannabinoid content from a whole whack of different products. I'm responsible for developing methods to extract and test new dosage forms. Actually get something new in almost every week.

@Beerhunter Funnily enough, I kept thinking this also. I've since clarified the issue w/ manager and concern was about the expansion or contraction of the stock on the manufacturer's end due to partial molar volume effects. I ended up writing a proof that the effect would be laughably negligible even in a "worst case scenario" and TL;DR we're doing it my way now.

This might be a stupid question, but is concentration that important in UPLC? Are you assaying for concentration? Can you use a different reference if so?

though the vapour pressure of the solvent causes it to immediately push droplets back out of the pipette tip, || You can cool the pippette and the ampule in a freezer...

I am not an analytical chemist, so sorry if the suggestion could be inapprioriate, but do you use the standard without dilution, or you have to dilute it? If you need to dilute it, what about preparing a solution of the sample in a solvent with lower volatility, such as ethylene glycole? (I mean, in order to obtain your standard as a mixed solvent solution, in methanol/glycol)

How do you detect the cannabinoids?  By UV absorption or by MS or by something else?

In addition to cooling, when you pipette the 150 uL, first pipette up and down a few times in the source container.  This will fill the headspace of the pipette tip with solvent vapor and reduce evaporation (and thus droplet ejection) in the final step where you do transfer.  Also make sure you are using the right pipette size (e.g. a 200 uL pipette with smaller tips is better than a 1000 mL pipette with bigger tips and bigger aperature for this job.)


text:
Hydrogen peroxide and glycerol combination
Is the mixing of store-grade hydrogen peroxide 3% and glycerol in the ratio of 1:2 (V/V) at room temperature affecting the decomposition rate of hydrogen peroxide?

If you want to ruin both reagents, then you can go ahead and mix them. Hydrogen peroxide oxidizes glycerol to various reagents. A possible reaction pathway can be depicted as follows (Ref.1):

This reaction is catalyzed by transition metals and other substances (Ref.1-3).
References:

G. Wu, X. Wang, T. Jiang, Q. Lin, “Selective Oxidation of Glycerol with 3% $\ce{H2O2}$ Catalyzed by LDH-Hosted Cr(III) Complex,” Catalysts 2015, 5(4), 2039– 2051 (https://doi.org/10.3390/catal5042039).
D. Mandelli, W. A. Carvalho, L. S. Shul'pina, A. M. Kirillov,  M. V. Kirillova, A. J. L. Pombeiro,  G. B. Shul'pin, “Chapter 19: Oxidation of Glycerol with Hydrogen Peroxide Catalyzed by Metal Complexes,” In Advances in Organometallic Chemistry and Catalysis: The Silver/Gold Jubilee International Conference on Organometallic Chemistry Celebratory Book; A. J. L. Pombeiro, Ed.;  John Wiley & Sons, Inc.: New York, NY, 2014, pp. 247– 257 (https://doi.org/10.1002/9781118742952.ch19).
P. McMorn, G. Roberts, G. J. Hutchings, “Oxidation of glycerol with hydrogen peroxide using silicalite and aluminophosphate catalysts,” Catalysis Letters 1999, 63(3-4), 193– 197 (https://doi.org/10.1023/A:101907312).

Hmm, interesting, I would not think oxidation stops at formic and oxalic acid.

@Poutnik: I was surprized too. But this is s peer-reviewed papar so put my trust on the work. :-)


text:
What is the total number of isomers, considering both structural and stereoisomers of cyclic ethers with the molecular formula C4H8O?
According to me, it should be 11 and this is how I have worked out as shown in the attached diagram.
However, I am not sure whether all these 11 isomers can be considered as those belonging to cyclic ether.
I had erroneously uploaded incorrect diagram earlier. Now I have uploaded the correct diagram. My specific query is whether the 11th Isomer shown in this diagram, wherein Oxygen is not part of the ring, can be considered as Cyclic Ether.

I would not include the last structure, even if it was clever of you to think about it.
I think that the term "cyclic ether" restricts the scope to "cyclic compounds which have an ether moiety inside the ring". Your last compound would be considered as a linear ether with an alkyl and a cycloalkyl substituents.
Oh and your 6th structure does not have 3 isomers but just 2 because the (R,R) an the (S,S) isomers are actually equivalent by rotation of the molecule. This compound is said to be "meso".

Thanks for your views. Though most of us are of the same view as yours i.e. "cyclic compounds which have an ether moiety inside the ring", should only be termed as cyclic ether, however, I have not been able to locate any standard reference book which would explicitly define it so. I shall highly appreciate, if you can guide me to any reference book wherein this has been explicitly deliberated upon.

Unfortunately, I don't think there is a general rule for that. It merely comes from usage, i.e. the chemical function must be part of the cyclic structure, EXCEPT if it is a monovalent moiety (e.g. alcohol, halide, carboxylic acid...) which can never be inside a ring. For example in the Gold Book (https://goldbook.iupac.org/ ), lactams are defined as "cyclic amides"


text:
How to find the strongest base using stability arguments

Answers: The most acidic is D and the most basic is A.
So first off I compared A with B:

I'm kinda confused on this. I looked at the base NH one and saw that when it was protonated it lost its aromaticity. So as a result, the NH²⁺ one must be an unstable acid, thus it is the stronger acid. The stronger acid must have the stronger base on the same side, thus I concluded that A is stronger than B. Is this the correct way to think about it?
Then, I compared A with D:

I'm unsure about this as well.
When I compare the two bases on each side (the O⁻ and OH⁺ one), I see that the O⁻ one is more stable (on the left side), thus it is the weaker base, meaning the OH⁺ one (on the right side) is the stronger base. However, that is not right as A is the strongest.
However, when I compare the two acids on each side. I see that the OH one (right side) is more stable, thus it must be the weaker acid, so its conjugate base which is the O⁻ one must be stronger. This way of thinking got me the correct answer. However, I'm still unsure if this is the reason why as I was always told to compare the charged species.
Also E doesn't have a lone pair to donate, so I neglected that in the conversation of the strongest base. For C, it just didn't seem likely to me.

welcome to Chemistry SE.  Thanks for showing your work.  A few points:  You are correct about the first comparison.  A is a stronger base than B. Your  comparison of A with D is wrong (check the structure of D).  As for E, you are right, but cations are almost never basic.

You have to be careful with the "stability" argument. If the species looks "ok" and looks "instable" when protonated, it is not a good base. If the species looks "instable" but looks "ok" when protonated, it is a good base. If both conjugate base and conjugate acid look "wacky", all bets are off.

@MichaelLautman Ethylenediamine?

@KarstenTheis you got me!  I should have been more specific.  I wasn't thinking about diamines.  The ammonium group isn't basic.

@MichaelLautman Well, you said almost never. And I did not come up with a situation where a cationic functional group accepts a proton. On the flip side, there are many anions that can lose another proton (sulfuric acid, phosphoric acid).

Would one of the close voters please comment on their vote? I'm at a loss how this doesn't comply with HW policy.

A vs. B
I will label your first reaction scheme, using the letters A-E for the deprotonated (i.e. as shown in the problem) form and the symbol $\ce{AH+}$ etc for the protonated form (after acting as a base and accepting a proton):
$$\ce{A (stronger base) + BH+ (stronger acid) <=> AH+ (weaker acid) + B (weaker base)}$$
Your argument about aromaticity is great; imidazolium ($\ce{BH+}$) is the stronger acid compared to cyclohexanol ($\ce{AH+}$), so the alkoxide ($\ce{A}$) is the stronger base compared to imidazole.
A vs. D
The protonated tetrahydrofuran (D) is highly reactive ("unstable") in water because it would want to get rid of a proton. That does not make it a stronger base. In your reaction scheme, you add another proton to a species that already is a strong acid (not sure if it even exists, see Why doesn't H₄O²⁺ exist?).
$$\ce{A (stronger base) + DH+ (stronger acid) <=> AH+ (weaker acid) + D (weaker base)}$$
The cyclohexoxide ($\ce{A}$) is the stronger base than protonated tetrahydrofuran ($\ce{B}$) because we know phenol ($\ce{AH+}$) exists, but we don't think twice-protonated tetrahydrofuran ($\ce{BH+}$) exists.

the O- one is more stable (on the left side), thus it is the weaker base, meaning the OH+ one (on the right side) is the stronger base

You have to specify stable in terms of what reaction. Protonated tetrahydrofuran is very stable against protonation, so it is not a base.

However, when I compare the two acids on each side. I see that the OH one (right side) is more stable, thus it must be the weaker acid, so its conjugate base which is the O- one must be stronger.

Now you are comparing the two in terms of either one losing a proton, which is the right thing to consider when comparing two acids.
C and E
You are right about E, the nitrogen already makes four bonds, so it can't accept another proton. I'm not sure where a proton would go in C. In any case, it is not a good base.

I see, now if I compare A and D by saying which is the strongest acid first.

AH+ (weaker acid) + D (without the H on the O, so the base form of it) --><-- A (stronger base) + D (stronger acid). In this way, D is more unstable than AH+, so D is the stronger acid, thus weaker base. However, I was always told to compare charged species. A with an H is neutral. So does it not matter if you compare a charged species with a neutral species?

@Zach This sounds like a new question: Are you asking whether A or D is the stronger acid? Or are you mixing and matching - trying to protonate A and deprotonate D? "D stronger acid thus weaker base" does not work. It is "DH+ stronger acid thus D weaker base". If you want to compare A and D, they have to compete for a proton, i.e. appear on different sides of the chemical equation.

@Zach About "I was always told to compare charged species". You might run into trouble with that rule. What if both acid and conjugate base are charged? What if the neutral species is highly reactive and the charged species is what you usually find in solution (e.g. HCl in water)?

@Karsten Theis: You mentioned A as a phenolate ion, but is not. It is the conjugate base of cyclohexanol.


text:
Why aren't there 2 products formed?

The question is asking for the major organic product. This seems like an $S_N1$ reaction to me as there is a weak base in $\ce{CH3OH}$ and heat. If it is, shouldn't there be 2 products. One with the $\ce{OCH3}$ in the equatorial position (shown in the answer key here) and another with the $\ce{OCH3}$ in the axial position. I think there are 2 products because the C attached to Cl is an asymmetric center, thus there will be an $\ce{OCH3}$ that is wedge and an $\ce{OCH3}$ that is dash.

The acetal has 2 conformations, one with the CH3O equatorial and the other with it axial. Adolph von Baeyer had a similar problem with cyclohexane carboxylic acid in a chair conformation. He assumed that there were 2 compounds, not realizing that they are in rapid equilibrium and not 2 different compounds.

It is not very likely to have a weak base in the medium when there is sulphuric acid around

Wait, is it because the major organic product would have OCH3 in its equatorial form as it is unstable to have OCH3 in its axial position. Even though there is some of OCH3 in its axial position conformation, more of the OCH3 in its equatorial position will be favoured?

The axial conformation is preferred known as the anomeric effect. https://pubs.acs.org/doi/pdf/10.1021/ja00437a007

A lone pair on the ring oxygen forces Chlorine leave. The anomeric carbon in the resulting oxocarbenium ion is sp 2 hybridized, so that part of the molecule is planar. (An oxocarbenium ion has a positive charge that is shared by a carbon
and an oxygen.)
When the alcohol approaches from the top of the plane, equitorial isomer is forrmed. Similarly if alcohol apporoches from bottom of sp2 hybridised oxo carbenium ion , then ,axial isomer is formed .So the given chloro compound should give two isomers (as shown in figure below 2& 3)..
Which is the major product ?
Substituents on a cyclohexane ring prefer the equatorial position to relieve steric strain. cyclohexane ring in chair form with substituent in equitorial position ring-flipps to chair in which the substituent is axial.

When an endocyclic heteroatom ( oxygen, nitrogen, or sulfur) is  at a position adjacent to the methoxyoxy group,the hetroatom substituent now prefers the axial orientation (despite of the associated steric strain).The equilibrium preference of most such compounds for the alpha-isomer is referred to as the anomeric effect.

Such behavior is explained as

In α-methyl glucoside, the non bonding HOMO with a pair of electrons on the ring oxygen is antiperiplanar to the antibonding LUMO of C-O bond in methoxy group. This allows hyperconjugation between them and thus by stabilizing the α-form.
In axial position,such interactions are possible.Whereas, in equitorial the methoxy group is at equatorial position and cannot involve in hyperconjugation since it is not antiperiplanar to the lone pair on ring oxygen. Therefore  equitorial isomer is less stable than the axial isomer.

Other Explanations:

The  dipole interactions of the two ether functions are  in the equatorial conformer are nearly parallel;while, in the axial conformer these dipoles are oriented in opposite directions.

image source :https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/carbhyd2.htm#anomr

Electrostatic Repulsion: Decreased electrostatic interactions are favored

All the above combination of factors result in equitorial isomer being less stable than the axial isomer.
Referance
https://www.scripps.edu/baran/images/grpmtgpdf/Krawczuk_Nov_05.pdf
Organic Chemistry ,SEVENTH EDITION , Paula Yurkanis Bruice

For the record, the equAtor circles the earth as do equAtorial groups cyclohexane. ;)

"When the alcohol approaches from the top of the plane, equitorial isomer is forrmed." - Why?


text:
How can the potential energy at the infinite energy level in an atom be zero?
I have recently learnt about atomic structure and the Bohr model of the atom and have observed a discrepancy between it and my previous knowledge based on physics.
For simplicity, assume the atom is of hydrogen.
Total energy of electron $$E= -13.6 \left(\frac{z^2}{n^2}\right) \pu{eV/atom}$$ where $z$ is atomic number, $n$ is energy level.
Thus, at $n=\infty$, $E=0$.
Now, by the relation Total Energy = potential energy/2 (from $TE=-KE=PE/2$),
$$PE=0/2$$ so $$PE=0$$
Now, the infinite energy level ($n=\infty$) occurs at a finite distance from nucleus (as the difference in distance between the energy levels decreases at higher energy levels).
Thus, $PE=0$ at a finite distance from the nucleus.
But, I have previously studied that $PE$ of a system of a negative charge (electron) and positive charge(nucleus) is $=0$ only at an infinite distance.
Thus, the discrepancy.
Please let me know where I have gone wrong.

You incorrectly assume that when $n=\infty$ the electron and proton are at a finite separation.

Let's forget for a moment that the Bohr model is pretty rudimentary. Why do you think that the orbital radius for n-> infinite is finite? Can you explain this a little more so we can answer?

Please see the following image:  https://www.google.com/search?q=orbit+of+n%3Dinfinity&source=lnms&tbm=isch&sa=X&ved=0ahUKEwjKsc7e7sfiAhUe4XMBHQx5CNYQ_AUIECgB&biw=1280&bih=578#imgrc=LUbuupHc87igmM:

Let me know if it makes my question clear...

Yes your question is clear, and clearly based on a wrong premise. You know the formula for energy; you plug $n=\infty$ and correctly deduce that $E=0$. But you don't know the formula for radius and can't plug $n$ there, so you have to rely on obscure googled pictures instead. That's the root of all evil.

See page 10 in https://users.physics.ox.ac.uk/~smithb/website/coursenotes/qi/QILectureNotes2.pdf

The picture is not obscure, it is a great representation of the relationship between n and E in the H atom. However, it has nothing to do with radius. The distances shown between the bars are not spatial distances, they represent the energy differences. The vertical axis is energy.

@electronpusher i think you have the best point...i have got a solution...thanks


text:
How does molarity change with dilution?
If I have a 1M solution of NaOH and I want to obtain a 0.2M solution. Then according to the formula $$ {M_1V_1 = M_2V_2}$$ I would get the answer as 200 mL. In my textbook they have said that "Thus, 200 mL of 1M NaOH is taken and enough water is added to dilute it  to make it 1 liter. Note that the number of moles of solute was 0.2 in 200 mL and it has remained same , i.e., 0.2 even after dilution as we have just changed just the amount of solvent and have not done anything with respect to NaOH. But keep in mind the concentration.". So my question is, by adding water to the 0.2M solution to make it one liter wouldn't the molarity change?

You have reversed the steps. You have answered your question first and then you have asked the question.

The answer was in my textbook and I'm unable to understand it

What have you done to understand it ? It is based on very trivial math of a basic school.

I have so far understood that 200 mL of a 1M solution gives me a 0.2M. But if we add water to make it one litre then won't the molarity change?

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.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
They are not banned, but we require a significant a priori effort. Otherwise, such a question may get closed. Please [edit] in your full reasoning or thoughts on this.
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1M does not mean 1 mol. It means 1 mol/L. 200 ml of 1M solution is still 1M solution.

This question shows one reason why the quantity "concentration" should not be called "molarity".

I think you have confused amount of substance with molarity of the solution. For example:

$\ce{NaOH}$ solution with $\pu{1M}$ in molarity means $\pu{1L}$ of solution contains $\pu{1mol}$ of $\ce{NaOH}$.
You can calculate amount of $\ce{NaOH}$ in any volume of the solution by multiplying its molarity and required volume in $\pu{L}$:
$$\text{Amount of }\ce{NaOH}\text{ in } \pu{mol} = \text{molarity of the solution} \times \text{volume in }\pu{L} = MV$$

Now see what happens when you measured $\pu{200 mL}$ of $\pu{1M} \; \ce{NaOH}$ solution:
$$\text{The amount of }\ce{NaOH}\text{ in } \pu{mol} = \text{molarity of the solution} \times \text{volume in }\pu{L}$$
$$= \pu{1 mol\:L^{-1}} \times \pu{200 mL} \times \frac{\pu{1L}}{\pu{1000 mL}}=\pu{0.2 mol}$$
If you diluted this $\pu{200 mL}$ volume to $\pu{1L}$ what happens to its molarity?
That solution still contains $\pu{0.2 mol}$ of $\ce{NaOH}$, because you add nothing but water.
Now, $\pu{1 L}$ of new solution contain $\pu{0.2 mol}$ of $\ce{NaOH}$. Thus, by definition (see the first bullet above), its molarity is $\pu{0.2 M}$ (or $\pu{0.2 mol\:L^{-1}}$).


text:
Nucleophilic Substitution
when I add $\ce{RCl + H2O -> ROH + HCl}$
but it's also true in reverse $\ce{ROH + HCl -> RCl +H2O}$
however this is not in case of $\ce{FeSO4 +Zn -> Zn + FeSO4}$
$\ce{ZnSO4 + Fe -> no rxn}$
How can a reaction be true both ways ?

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.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
They are not banned, but we require a significant a priori effort. Otherwise, such a question may get closed. Please [edit] in your full reasoning or thoughts on this.
See Homework

my mistake i will correct it

What does FeSO4 + Zn have anything to do with nucleophilic substitution?

If you insist on plain text, what is discouraged in Chemistry SE site, at least write the equations properly. Asking the question the way it is rather presents disrespect to readers.

It should be better to write equations as
$$\ce{RCl +H2O <=> ROH +HCl}$$
$$\ce{Fe^2+ +Zn <=>> Zn^2+ + Fe }$$
with the latter having the equilibrium shifted to far right.
But is case of high concentration of $\ce{Fe^2+}$ and very low concentration of $\ce{Zn^2+}$, the net reaction would be ongoing leftwards.
If the left side of equation below is smaller than right one, the net reaction in inclined toward iron being replaced by zinc. And vice versa.
$$E^{\circ}_{\ce{Zn/Zn^2+}}+0.0295\log {a_{\ce{Zn^2+}}}=E^{\circ}_{\ce{Fe/Fe^2+}}+0.0295\log {a_{\ce{Fe^2+}}}$$
Note that forward and backward reaction are ongoing simultaneously, having the same rate at the equilibrium.

what should be done to carry rxn in foward reaction


text:
Rationale behind Barton's rules for catalytic hydrogenation
Organic Chemistry by Morrison and Boyd,Seventh Edition says -

When cyclohexanones are used as substrates, the stereoselectivity of the reduction is found to be dependent on the acidity or the basicity of the medium. The observations about this stereoselectivity are summed up in the form of a few empirical rules, called Barton's rules, which are as follows:
(i) For unhindered cyclohexanones, in the catalytic reduction with Adam's catalyst($\ce{PtO2}$) in acidic medium, the less stable axial alcohol predominates in the product
(ii)In a basic or a neutral medium , however, the same catalyst gives the thermodynamic product called the equatorial alcohol as the major product
(iii) But for hindered cyclohexanones, the less stable axial alcohol is formed in all $\ce{pH}$

As far as I know, catalytic hydrogenation is a syn addition which is performed by delivering the hydrogen from the same side when the substrate is adsorbed on the surface of the catalyst. From what I know about the theory of heterogenous catalysis, it is the nature of catalyst which governs all aspects of the product formed here.So I feel that the medium's nature shouldn't have much effect here.
My question is: What is the basis/rationale for these "Barton's rules"?
The latest downvote has called my attention to this question,I guess maybe I wasn't specific enough when I had asked this question. What I wanted to know is the relation between variation of pH and the subsequent change in the stereochemistry of the product.
My ideas on the subject On giving it some more thought, I guess the acidic medium may generate a protonated ketone, which would be more electrophilic than the regular ketone. So maybe,the reaction happens quickly and therefore,the kinetically controlled product i.e. the axial alcohol is obtained. In the basic medium, the basic cleavage has to happen in a somewhat more controlled fashion on the cyclohexanone, and hence the thermodynamic product is obtained. However,this is purely my conjecture.

Considering it's been a year, are there any other self-advancements on what the reason may be? Or was it just a tag edit?

@Safdar No, I haven't really had much success with this question since. It was mostly just a tag edited which I didn't find suitable for the question


text:
How to convert -OH to terminal -SH group in nucleolipids:
I am recently working on some synthesis of Nucleolipids and I want to transform $\ce{-OH}$ to terminal $\ce{S}$ group.
Can anyone help?

The first thing that comes to my mind is replacing a $\ce{-OH}$ group with a $\ce{-Cl}$ group using $\ce{PCl3}$ (though remember that this would change the configuration) and then replacing the newly made Halogen using $\ce{HS-}$ may be from Sodium hydrosulfide , I really don't know about the yield of reaction and ease of doing reaction but indeed this is a way .

Mesylate, react with thiourea, hydrolyse

I've also successfully converted benzylic OH groups to SH groups using thiourea without the mesylation step, if the system has electron-donating groups in ortho and para positions. I used thiourea (2 eq.) in acetone/water, followed by hydrolysis with 3 eqv. NaOH on a 2,4-dimethoxybenzyl aclohol to get the thiol in 70% yield. Alternatively, you can try Mitsunobu reaction using thiobenzoic acid, Ph3P and DIAD/DEAD/DCAD, followed by hydrolysis. Take care not to oxidise your compound during synthesis! If you end up forming disulfide, keep TCEP handy ;)

An alcohol can be converted to a corresponding thiol using Lawesson's reagent (LR; 2,4-bis(4-methoxyphenyl)-1,3,2,4-dithiadiphosphetane 2,4-disulfide). However, the reaction is messy and could give you a mixture of products in addition to intended thiol, including eliminated product, alkene (Ref.1). For example, treatment of acyclic 1,4-diols with LR gives the 1,3-dienes (Ref.1).
However, the substitution versus elimination outcome with LR is depends on the structural features of the substrate (Ref.2), and hence, worth considering. The computational mechanism is depicted below:

Alternatively, you can use highly efficient Mitsunabu type reaction with diisopropyl azodicarboxylate (Ref.3). The abstract summerize the reaction as:

Various alcohols were converted to their corresponding thiolacetates by treatment with triphenylphosphine and diisopropyl azodicarboxylate in the presence of thiolacetic acid. The overall conversion was both highly efficient (89–99% yields) and stereoselective (99.5% inversion).

References:

T. Nishio, “Direct conversion of alcohols into thiols,” J. Chem. Soc., Perkin Trans. 1 1993, (10), 1113–1117 (DOI: 10.1039/P19930001113).
M. A. Chiacchio,  L. Legnani,  P. Caramella,  T. Tejero,  P. Merino, “Pivotal Neighboring‐Group Participation in Substitution versus Elimination Reactions – Computational Evidence for Ion Pairs in the Thionation of Alcohols with Lawesson's Reagent,” Eur. J. Org. Chem. 2017, (14), 1952–1960 (https://doi.org/10.1002/ejoc.201700127).
R. P. Volante, “A new, highly efficient method for the conversion of alcohols to thiolesters and thiols,” Tetrahedron Letters 1981, 22(33), 3119–3122 (https://doi.org/10.1016/S0040-4039(01)81842-6).

Lawesson's stinks like hell and is a serious pain to work up and don't underestimate how much of nuisance getting rid of triphenylphosphine oxide is.

@Waylander: Tell me about it! :-)


text:
Is there a relationship between loss of electrons and EM wave reflection?
Is there a relationship between the loss of electrons in a cation and the amount of electromagnetic spectrum, a compound of this cation would subsequently reflect or transmit?
For example, any compound with $\ce{Fe^3+}$ ions is dark orange. This must mean compounds with $\ce{Fe^3+}$ reflect the longer wavelengths of the visible light spectrum more strongly, so the range of $\approx\pu{635–590 nm}$.
While a compound with a $\ce{Fe^2+}$ cations is light green. This, I think means that compounds that house $\ce{Fe^2+}$ ions reflect wavelengths of the range $\approx\pu{560–520 nm}$ more than others.
Now,  I am not aware at all of any link between atomic electron loss  influencing the transmission/reflection of wavelengths at all, so it could very well be something else entirely that may instead be influenced by the gain/loss of electron. I do not know.
So is there a relationship between the two? If so, what is it?

The is a direct link, but the topic is very complex, covered  by UV-VIS molecular spectroscopy and quantum chemistry.

Actually, I think you have it backwards. The color you see is due to the colors of light which are removed from white light when traveling past the iron. That is, if you go and find an absorption spectrum of iron cations in different oxidation states, you will find they primarily absorb the complementary colors of the color you actually see.
As far as the relationship between charge and the color, the ions with a smaller charge will likely have a larger HOMO-LUMO gap which means you will see colors of shorter wavelength, which is what you report. This property will depend very sensitively in the presence of ligands and other environmental changes, so it is not always possible to guess the color of a compound based only on charge.
This is typically how colors work for systems that absorb light. It is, however, possible for the we see to be due to reflection. Pure reflection, though, is independent of wavelength, so the phenomenon you describe is actually scattering, which does depend on wavelength. This type of phenomenon is responsible for the color of the sky.
Finally, you sometimes see the color which is emitted after absorption, as is the case in fluorescence and phosphorescence.
Hopefully that clears up your confusion

I thought that x coloured objects would reflect x wavelength of light. for example red objects would reflect red light, are you suggesting that the light is absorbed as opposed to reflected?

@Ubaid Hassan I am saying that it depends on the material. For instance, a wall that is painted red will have some material in it which primarily absorbs blue light. Thus, the light that is reflected back is going to be a mixture of all the visible wavelengths other than blue. Therefore, you will tend to see red light.

See, for instance, this website, especially the picture that shows the absorption spectrum of chlorophyll. Chlorophyll is what makes leaves appear green, but this is because chlorophyll absorbs blue and red light, not because it reflects green light. You are not wrong to say a particular color is reflected. That is a necessary but not sufficient condition for an object to appear a certain color.


text:
Can Iron(III) oxide oxidize elemental iron in a nitrogen environment under room temperature without any catalysis?
Will there be electron transfer to allow this process to happen:
$\ce{2 Fe(III) + Fe(0) -> 3 Fe(II)}$

In what situation? Also what speed?

What are the counter ions?

Whether this qualifies as "without catalysis" is a matter of perspective, but it certainly does happen under ambient conditions.  One mechanism of atmospheric rusting involves the $\ce{Fe(III)}$ bearing rust reacting with fresh iron to form $\ce{Fe3O4}$, which then oxidizes in air to form more rust.  See Section 1.6 here.
We can capture the $\ce{Fe3O4}$ stage by merely placing the rusty material in water with a little electrolyte, such as rinse water after an acid pickling process, where air cannot easily penetrate.  The rust, being converted to $\ce{Fe3O4}$, turns black.  This emphasizes the fact that the iron reacts with the $\ce{Fe(III)}$ in the rust without oxygen, once the rust has first formed.
I've actually seen the $\ce{Fe3O4}$ formation happen on steel products during rinsing after pickling, giving black stains or stains bearing a black component.  To prevent it, we had to design and maintain the rinsing (and drying) processes to prevent the rusting.

Thanks for answering this question! I guess I should rephrase my question: if solid Fe2O3 and elemental iron were mixed in powder form, and stored under anhydrous environment, will there be electron transfer from iron(0) Fe2O3, leading to the formation of Fe(II) species?

Reactions between solids are generally slow because diffusion only goes so fast. However, if you produced the two materials in nano size to increase surface area and contact, mixed thoroughly (in absence of air) and then perhaps under pressure (density of $\ce{FeO}$ is 5.745, density of $\ce{Fe2O3}$ is 5.25) with an increase in temperature ($\ce{FeO}$ is stable above 575 °C: https://en.wikipedia.org/wiki/Iron%28II%29_oxide) after some time (how long depends on temperature and pressure), you could get complete transformation of $\ce{Fe2O3 + Fe}$ to $\ce{3 FeO}$.
If magnetite ($\ce{Fe3O4)}$ forms as an intermediate product, its low density (5.175) suggests it would not survive.
But on the other hand, the heats of formation do not favor $\ce{FeO}$: for $\ce{FeO}$, 65 kcal/mol; for $\ce{Fe2O3}$, 200 kcal/mol; for $\ce{Fe3O4}$, 268 kcal/mol (CRC Handbook).
It's hard to say. The electrochemical data do not take oxygen or solid states into consideration. But if you squeeze the mixture hard enough, I'm sure it will proceed.

Theoretically, yes. Just look at the following redox reaction:
$$
\begin{align}
\ce{Fe &<=> Fe^2+ + 2e-} &\quad \mathrm{E^\circ} &= \pu{0.447 V} \tag{Oxidation}\\
\ce{Fe^3+ + e- &<=> Fe^2+} &\quad \mathrm{E^\circ} &= \pu{0.771 V} \tag{Reduction}\\
\ce{2Fe^3+ + Fe &<=> 3Fe^2+ } &\quad \mathrm{E_{cell}^\circ} &= \pu{1.218 V} \tag{Total redox}
\end{align}
$$
Accordingly, $\mathrm{E_{cell}^\circ}$ is a positive value, and therefore, the reaction is spontaneous.

Thanks! But I thought this is enthalpy calculation, am I wrong? I am not familiar with the notation E0 here. Do you have a reference article about this process? Or probably a book chapter I can read?

The values I used are from https://sites.chem.colostate.edu/diverdi/all_courses/CRC%20reference%20data/electrochemical%20series.pdf. Yet, some certain values you can find in any General Chemistry book.


text:
Ammonium chloride in aqueous solution
When Ammonium chloride is in an aqueous medium its cation part ammonium undergo hydrolysis. Since ammonium is conjugate acid of a weak base it readily release one $\ce{H+}$ and converts back to $\ce{NH3}$. But its given in my textbook that hydrolysis of Ammonium chloride gives ammonium hydroxide. Why Ammonium hydroxide is formed instead of staying as ammonia? Because according to my understanding ammonium hydroxide is formed when ammonium ion and hydroxide ion attract each other to have an ionic bond, and since ammonium ion converts back to ammonia how is this possible?

No, there is not ammonium hydroxide, except as a very minor and non-isolable component.  The hydrolysis product is overwhelmingly just plain ammonia.

When ammonium chloride is hydrolyzing, there is established equilibrium
$$\ce{NH4+  + H2O <<=> NH3 + H2O+}$$
with $$\mathrm{pH}=7-\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$
$$\mathrm{pOH}=14 - \mathrm{pH}=7+\frac 12 ( 4.75 + \log {c_{\ce{NH4Cl}}} )$$
Concentration of ammonia is several orders higher than concentration of hydroxide ions, with origin of water dissociation, so no ammonia hydroxide there.
When ammonia is dissolved, there is established equilibrium
$$\ce{NH3  + H2O <<=> NH4+ + OH-}$$
with $$\mathrm{pH}=14-\frac 12 ( 4.75 - \log {c_{\ce{NH3}}} )$$
Equimolar concentrations of $\ce{NH4+ + OH-}$ are sometimes formally called ammonium hydroxide, but using this term is deprecated and it cannot be isolated.


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Catalysts to accelerate foamy polyurethane reaction
This is the reaction that I am referring to:  https://www.youtube.com/watch?v=EoRUSo0BWC0
I am interested in finding catalysts to accelerate this fun chemistry demonstration. I have heard that some amide catalysts can accelerate CO2 production leading to faster foam rise, but I have not found any sources confirming this. Does anyone know of any catalysts to accelerate this fun chemical demonstration?

In those preparations, the foam arises from the reaction between water and isocyanate, which leads to the production of CO2 via formation of a carbamic acid, which upon decarboxylation generates the blowing agent, CO2.
The trick would be that of making water more reactive towards the isocyanate, without (as expected by a catalyst) partecipating actively in the reaction.
Several compounds of this kind have been reported and are sold, for instance: https://patents.google.com/patent/WO2009048807A2/en mentions N,N,N',N',N",N"-pentaethyldiethylenetriamine and morpholine derivatives, and commercially (I won't link any product, but googling "Blowing Catalysts" might help you) catalysts based on  N-methylmorpholine, Bis[2-(N,N-dimethylamino)ethyl] ether, DMEA and PMDTA are available.
This said, for practical applications the formulations have probably the best ratio of reactants, surfactants and catalyst for obtaining a "good" foam (with good physical properties, and possibly with negligible smell), but if your goal is that of making the reaction go faster, maybe an excess of (any?) tertiary amine, and possibly some drops of water in excess should work in making things faster (with the possible drawback of obtaining a low quality foam, maybe with "bubbles" larger than intended, or with bad "texture").
Additional note: if speed is what you are looking for, gently heating the mixture should work without adding anything else