Reaction Pathways: What to do when IRC geometries are relaxed (optimized) and the energy changes significantly The Story:

Reaction pathways can be difficult things to deal with using electronic structure methods.  Intrinsic reaction coordinates (IRCs) can be determined a number of ways (a discussion of these methods lie way outside the scope of this post).  Suffice to say, the reaction pathway is an 'energy' pathway of reaction, connecting reactants and products, usually through some intermediate and/or transition state (TS).

For the purposes of this question, imagine implementing some sort of IRC technique which begins with a previously resolved transition state (Figure 1).  The computation follows the reaction coordinate in both the forward and reverse directions from the TS energetically downhill until it reaches a (for the sake of simplicity) a minimum energy point with features which satisfy the method used.  In general, the structure and energy of this final IRC geometry should be very representative of the structure which is connected to the TS via the pathway determined.  If you were to optimize this final geometry, you should get good agreement between the optimized structure and the IRC configuration.

Now consider this scenario.  An optimization was performed on the IRC resolved structure B and the energy relaxation was significant and the structure significantly changed, giving rise to a new (and lower) energy.  One could no longer (within the formalism of IRC methods) say that the optimized structure B lies on the reaction pathway, and therefore is not a representative configuration at point B on the reaction pathway associated with the TS.

The Question:

Does anyone know of an example in the literature where this type of behavior occurs?  How does one 'deal' with a situation like this?

Other Information:

Let us assume the following:

That this scenario occurs with a few IRC procedures (not just one).
That the IRC computations were given a ridiculously large amount of
cycles and steps
That various stepsizes were sampled in parallel IRC jobs with no
significant difference in the results

(I will add more as I think of them)

Potentially Helpful Imagery

Imagine two somewhat complex molecules coming together to form a single covalent bond.  You run an IRC from the TS (which describes the bond formation) in the reverse direction and the intermolecular separation begins to increase.  This separation continues to increase until the IRC job terminates properly.  Now imagine running an optimization on these two (slightly complex) molecules which are separated by maybe 4.0 Å.  Obviously these molecules are going to interact because they are not infinitely separated.  You observe the (slightly complex) molecules moving toward each other during the optimization procedure (rotating, twisting, gyrating along the way) until you now have a minimum energy structure which is much lower in energy than the resulting IRC computation.  These two molecules interact via intermolecular interactions (no covalent bonds are formed between the molecules).  This optimized structure is analogous to a reactant complex (as opposed to separated reactants) and has an energy a few kcal mol$^{-1}$ lower than the corresponding IRC geometry.  How on earth does this fit in to the reaction pathway?

IRC is a gradient following algorithm that tries to follow the imaginary mode. Check the algorithm (several one implemented, so really depends on which you are talking about) and you will see that unlike in optimisation algorithms, there is a big emphasis on following the real gradient (and not just jump to the endpoint as fast as possible).

It works well around the TS where the gradient is large and has a well define direction. However getting closer to the optimised geometry, the requirements of following the exact gradient and having an effective optimization algorithm are contradicting. In analytical sense it may be not obvious, but when you have numeric gradients, small second derivatives (these are almost always very soft modes), the direction of the gradient may have big errors and you should keep a small stepsize to check it thoroughly.

In summary, the situation you wrote down is more or less common at some extent in almost all the IRC calculations due to the numerical nature of the problem. Harder imaginary frequencies may help.


Quantum Chemistry: Small imaginary frequencies I am running a freq calculation on a geometry optimized structure using the ORCA quantum chemistry program. The output of the freq. calculation contains small neg. frequencies. Normally, these are considered to be saddle points, and thus would lead to the conclusion that the geometry does not represent a ground state.

freq.             IR Int.
-17.6900          0.000
-16.8900          0.000
-10.6600          0.000
6.4000            0.000
11.7200           0.000
... more small frequencies with IR Int. 0
30.3700           0.000
36.7000           19.3544

However, I am wondering if this is really the case, as the output also assigns an IR intensity of 0.000 to these modes, and states that "The first frequency considered to be a vibration is 14", which is the first positive mode. I am a little bit stuck, as obviously a negative frequency shouldn't appear in the IR spectrum, so does ORCA just removes those frequencies from the IR, even though they are there? Or does the IR-Intensity of 0 imply that the negative frequencies are actually just a precision error of the calculation, and I can still assume the structure to be a ground state.

Normally, the criterion is "no negative frequencies". On the other hand, not much energy should be stored in these modes, so I wonder how this will affect the computed energies.

It's probably okay, but an indication you don't have a true minimum in the geometry.

It's important to calculate frequencies to know if you have a true minimum of the potential energy surface. Presence of imaginary frequencies are indeed indications of saddle points.

That's simple theory. Imaginary frequencies indicate problems in the curvature with respect to that degree of freedom - it's simply not a minimum.

If you want a true minimum, you'll need to improve convergence and numerical accuracy.

Now the practical side...

I'm not familiar enough with ORCA to know if the translational and rotational degrees of freedom are removed from the frequency list. Some programs list these, and some do not. Obviously these include 5 or 6 degrees of freedom which are inherently low energy and due to numerical instability or incomplete geometric convergence can sometimes be imaginary frequencies. Your list includes three very small imaginary frequencies, so they could be translations or rotations.

(I also mention this, because ORCA indicates the first vibration is at ~14 $\mathrm{cm}^{-1}$ and the code is likely removing the lowest frequencies as translations and vibrations.)

For even moderate-sized molecules, and particularly for large molecules, it can be very, very hard to converge geometries to true minima.

Consider your list - with multiple vibrational frequencies of 14-30 $\mathrm{cm}^{-1}$. Those are incredibly low energy breathing modes. Gradients and energy changes along those dimensions will be small and a great number of optimization steps will be required.

This is one reason that geometry optimization methods "give up" if the gradients get very small and displacements get very small. It may not be a true minimum, but in all likelihood, it's very, very close and would require a large computational effort to get the true minimum.

Note: Don't read this and think "oh, I don't need to bother." Your needs may vary. If you're doing accurate work, you absolutely need to push hard on geometry convergence.


What causes the "rotting fish smell"? Rotting fish seem to give off the same (very pungent) kind of smell, regardless of the kind (salmon, seabreen, tuna, etc).

What exactly is it that's responsible for this unique smell?

(Though I've answered my own question below, do feel free to post an answer of your own if you want.)

The "fishy" odor that you're familiar with is brought about by a whole bunch of compounds, and not any single one. Then again, if we were to narrow this down a bit, we could say that simple nitrogen compounds are the main culprits.

But suppose we want to blame only a single compound for the delightfully pungent odor of rotting fish, and we couldn't be bothered with the nuances of olfactory physiology or food chemistry to go through a list of compounds running into the hundreds (if not thousands). In that case, we can pin the blame on trimethylamine (TMA).

With regard to fish, the TMA is a result of the degradation of trimethylamine oxide (TMAO) present in the body fluids (blood, lymph, excreta) of the fish.

Just so you don't get the wrong idea, TMAO is present in the body fluids of all vertebrates ̶I̶ ̶c̶a̶n̶ ̶t̶h̶i̶n̶k̶ ̶o̶f̶, not just fish (albeit in much lower concentrations, which is why you don't mistake ground chicken for tuna).

Now why is TMAO more abundant in fish?

Ever tried to pickle a cucumber (or anything for that matter)? Notice that the cucumber shrivels up over time? It's losing water to the pickling fluid by a process known as exosmosis: the water is drawn out from a less concentrated environment (cucumber's tissue) to a more concentrated one (pickling fluid).

 Disclaimer: As user @lly pointed out in the comments, this particular bit has the potential to cause a bit of confusion. The inside of a cell (the cytoplasm I mean) is technically an aqueous colloidal solution (the pickling fluid is an aqueous solution as well). In the context of aqueous solutions, "concentration" is an indication of how much solute is present, not water. So in this context, if something is more concentrated, it has less water in it

Now the same thing is going to want to happen to a fish in salt water, which isn't so nice from the fish's perspective.

To counter the outflow of water, a fish has a significantly higher concentration of the familiar metabolite urea in its body fluids (as compared to terrestrial vertebrates). This establishes a situation where the fish's body fluids have roughly the same "concentration" as the salt water it swims in (the more correct term here would be "osmolarity", in place of "concentration")

That urea, if left to its own devices, will seek to destabilize  macromolecular structures and inhibit cellular functions; the TMAO present alongside the urea helps counteract those detrimental effects.

Of course, going by this logic, saltwater fish should have a much stronger "fish odor" than freshwater fish (which is certainly the case).

Lets not forget the classic kitchen tip: Rubbing fish with something acidic like vinegar, b̶a̶t̶t̶e̶r̶y̶ ̶a̶c̶i̶d̶ or lemon juice helps reduce the odor (the TMA degrades under acidic conditions).

Also, some of you might know that a fish-like smell is observed in rancid food. That, apparently, is the result of the oxidation of omega-3 fatty acids.


What's the chemical equation for the lithium iron phosphate battery The title says it all, I'm searching for the chemical equation to the lithium iron phosphate battery.

I know that the cathode is made of $\ce{LiFePO4}$ and that upon discharging, it is transformed to $\ce{FePO4}$. The Anode is made of graphite.

So I think that the reaction on the anode is:
$\ce{LiFePO4 -> FePO4 + Li+ + e-}$

Is this correct? If so, what's happening on the anode?

Close, the cathode reaction is:

$$ \ce{LiFePO4 -> FePO4 + Li^+ + e^-}$$

The anode is:

$$\ce{Li^+ + C6 + e^- -> LiC6}$$

The best description I can give is that the lithium ion intercalates into the graphite and "steals" some electron density from the six-membered conjugated ring.


Are the vibrational modes of CO2 active at room temperature? I was doing an experiment whereby I had to measure the specific heat ratios of certain gases. Carbon dioxide came out to be around 1.3, and checking with the accepted values this is close. My question is, the only way I can justify this answer is if $\ce{CO2}$ had 7 active degrees of freedom at room temperature. This implies that the molecule is vibrating.
I am not a chemist and I thought that molecules only store energy in this degree of freedom at high temperatures.

Yes, the vibrational modes are present at any temperature, including absolute zero where the lowest vibrational energy levels only are populated; the zero-point levels. The molecule is linear ( OCO ) so it has $3N-5=4$ vibrational modes for $N=3$ atoms. The symmetric stretch  corresponds to both CO bonds stretching in phase, and asymmetric stretch to one CO stretching while the other compresses (this has highest frequency), and there are two degenerate bends where the OCO angle changes (lowest frequency). If the z-axis is along the OCO bonds, bends are in the zx and in the zy planes. Wikipedia has lots of diagrams showing these vibrational normal modes.

I have calculated the heat capacity as shown in the figure. The translational value is 3R/2 at all but the lowest temperatures (a few K) and similarly for rotational levels, which have a value of R, again above about 50 Kelvin. (The rotational constant is 0.3902 wavenumbers). The vibrational heat capacity from the modes (1388, 667,667, and 2349 wavenumbers) gives the rising curve on the figure. At 300 K the value is about 3.4 R or 28 J/mol/K. It is clear that only a few vibrational quanta are populated at room temperature. The high temperature value of the heat capacity adds another 4R (each vibration counts R) and is not reached until about 4000 K. Clearly never obtained in practice.

Total heat capacity for CO2 vs temperature. The blue line is the total heat capacity, the lower horizontal line has a value R, the upper one R+3R/2. The top horizontal line has no particular importance and is the value at 300 K and is approximately 3.4R as mentioned in the comments.


Calculating the zero-point energy for the C2N2 molecule I shall calculate the zero-point energy for the $\ce{C2N2}$ molecule in the $^{1}\Sigma_{g}^{+}$ electron ground state with the following vibrational wavenumbers (symmetries in parentheses) which were found in IR and Raman spectra:

$v_1 = 2330~\mathrm{cm}^{-1}$ ($\sigma_g$)
$v_2 = 854~\mathrm{cm}^{-1}$ ($\sigma_g$)
$v_3 = 2158~\mathrm{cm}^{-1}$ ($\sigma_u$)
$v_4 = 507~\mathrm{cm}^{-1}$ ($\pi_g$)
$v_5 = 233~\mathrm{cm}^{-1}$ ($\pi_u$)

I have no clue about how to do it - maybe somebody can help me.

The adiabatic electronic energy only contains the energy of the electrons. You are missing the energy that the molecule has in form of the vibrational groundstate of the nuclei. This energy can be calculated trivially in the harmonic approximation. Within the harmonic approximation we can define normal modes and each mode corresponds to a 1D harmonic oscillator with its respective normal mode frequency. The eigenenergy of a harmonic oscillator is given by
$$
E_{n} = \hbar \omega(n+1/2), \ n\geq0
$$
Each normal mode of your molecule contributes thus such an energy when you include vibrational energy. The zero point energy refers to the system where each mode is in the groundstate i.e. $n_i=0 $, where $i$ is an index for the normalmode. The zero-point energy  correction is then simply
$$
E_{ZP} = \sum_i^{3N-5} \hbar\omega_i 1/2
$$
where $N$ is the number of atoms and $3N-5$ is the number of normal modes in a linear molecule. In a nonlinear molecule you would have $3N-6$ normal modes. $n$ is the quantum number of the oscillator state.
In your case, you have some degenerate eigenvalues due to symmetry. The $\pi_u$ and $\pi_g$ modes are actually sets of two degenerate modes, which means that their frequency would appear twice in the sum that i denoted. This also works out when we consider your example, we have the following number of modes
$$3\cdot 4 -5=7$$. Since we have to sets of doubly degenerate modes we see only 5 distinct frequencies although there are 7 normal modes.
Plugin all thatt into the formula yields,
$$
E_{ZP} = 1/2(\hbar \omega_1+\hbar \omega_2+\hbar\omega_3 + 2\hbar \omega_4 + 2\hbar \omega_5)
$$
with $\omega_i = 2\pi \nu_i.
$
It's sometimes a bit vague what people mean with zero point energy. Some people use the term for the correction that i defined above, but technically it should include also the electronic adiabatic energy. Which means that you have to add the electronic state energy to the $E_{ZP}$ value.
One might also want to include the contribution due to the rotational states of the molecule, but we can't do that without more information.


Why do we use LiCl as a source of chloride ions as nucleophile but not other alkali metal chlorides? $\ce{LiCl}$ has higher covalent character than other alkali-metal chlorides, so in polar solutions like ethanol, its ioniztion will be less, and so less chloride ions will be available for participating in substitution reaction.

So why don't we use sodium chloride or other alkali metal halides which are more ionic?

One reason is lithium chloride is more soluble in the organic solvents often used for such reactions, and getting the chloride source into the reaction medium is more important than having it be perfectly ionic.

Let's roll the tape.  Wikipedia identifies lithium chloride as soluble in a variety of polar solvents and gives numerical values for several of them.  The solubility reaches hundreds of grams per liter in some of them, including over 400 grams per liter in methanol.

Much less data are listed here for sodium chloride, but the solubility of sodium chloride in methanol is listed as 14.9 grams per liter under ambient conditions, about 30 times less on a mass basis than lithium chloride.  And the chloridecomponent shows an even bigger discrepancy because of the heavier mass per mole of chlorine in sodium chloride versus lithium chloride.

Clearly lithium chloride's solubility carries over better from water to polar organic solvents.  The reason for this is lithium chloride is more adaptable.  The small size and higher polarizing power of the lithium ion imparts greater covalent character in lithium chloride than in its heavier alkali metal congenres.  Alternatively, in a purely ionic model, it promotes the formation of ion pairs which are uncharged overall because of the equal but opposite charges of the constituent ions.  Either way, the lithium chloride adopts a conformation that resembles a polar covalent molecule, which fits better with the polar but weakly ion-solvating solvents organic chemists generally use than the dissociated ions favored by chlorides of heavier alkali metals.


Chemical equation for the reaction of Allura Red AC with NaClO I am trying to figure out the chemical formula for the reaction between Allura Red AC and $\ce{NaClO}$ (bleach). I have never met such reactions before in my study, and I was unable or maybe looked in the wrong places, to find the chemical equation for the reaction. The closest thing I have found is an image with the reaction but without the chemical equation. Could somebody please help me figure out the chemical equation for the reaction in the image?

Thanks for the help in advance.

The two molecules created by breaking the $\ce{N=N}$ bond have something wrong. They both have a final bond $\ce{-N-OH}$. This is impossible. It may be at least $\ce{-NH-OH}$. Maybe the $\ce{N2}$ group is removed, and a phenol or naphthol group is formed.


How many elements in food? A slightly deranged curiosity: if we had a 3D printer able to “print” anything starting from “cartridges” of pure chemical elements, how many of the 118+ possible cartridges (or of the 98ish for non-artificial elements) would we actually need to print most - or all - foodstuff?

(OK, one could have phrased the same question in a more mundane way, I agree.)

For the record, I am not a biochemist, a nutritional expert, or any sort of person who is required to know this for my profession.

That said, as far as I know, to get the vast majority of nutritional food substances, you'll need to be able to produce at least:

sugars and starches
lipids
amino acids
cellulose
various other non-polymeric organic molecules of various levels of complexity
salts
proteins

These together will give you a reasonable approximation of food which will be most of what you need to stay alive, without necessarily going into the complexity of actually producing a better analogue of dead tissue "under the hood".

Sugars and starches are entirely composed of carbon, hydrogen, and oxygen. The same goes for lipids, cellulose, and the majority of the organic components you'll need. For amino acids, and the other organic bits, you'll need to add nitrogen and sulfur.

You also need phosphorus for DNA, RNA, ADP, and ATP, as well as a whole bunch of other signalling and respiration-based molecules, and bone material.

You need sodium and potassium ions which are needed in living tissue for nerve signalling, and you also need calcium ions to make bones.

Many plant materials seem to contain small amounts of boron, though the biological role of it isn't concrete as far as I know.

As far as I know, the proteins are where the metallic awkwardness comes in. A lot of proteins have a reaction site containing a metal, metal-sulfur, metal-porphyrin, or another metal-organic molecule complex, which can come in various metallic flavours. For example, haemoglobin requires an iron-porphyrin complex, while chlorophyll requires magnesium-porphyrin, and a lot of respiration-based proteins seem to involve iron-sulfur complexes.

I'm fairly confident you could make reasonable analogues of most food with C, H, O, N, S, P, Na, K, Cl, and Ca, with relatively small additional quantities of B, Fe, and Mg. You might suffer after a while from micronutrient deficiencies, though.

(I'll probably find that I've forgotten something important somewhere.)


Irreversible process (Isothermal and Adiabatic) How exactly would we define an irreversible change for an isothermal process? Why can we say what the external pressure will remain constant? Please give an example.

Also, why can we not define both reversible and irreversible processes for adiabatic process simultaneously?
$$\Delta E_\mathrm{rev} = \Delta W_\mathrm{rev}$$
$$\Delta E_\mathrm{irrev} = \Delta W_\mathrm{irrev}$$
I got that if these both take me to the same point then $\Delta E$ will be same, but why must $W_\mathrm{rev}$ and $W_\mathrm{irrev}$ be different?

If you have an adiabatic reversible path that takes you from state 1 to state 2, any irreversible path from state 1 to state 2 will require removal of heat, so it won't be adiabatic.  This is a consequence of the 2nd law of thermodynamics, captured by the Clausius Inequality.


Non-stoichometric compounds I am looking for references (review articles, reports) about non-stoichometric compounds. Any idea of where should I look?

There are several resources available for this, for example, searching in Google Scholar yields over 37,000 articles (this is without any filters for age or specifics).

Organisations such as the American Chemical Society have a searchable publications link; related, CAS has a searchable reference list and ChemWorx which is a way to manage your research.  A similar set up is available with the Royal Society of Chemistry.

Individual universities also have searches/tutorials - for example, The University of California, Davis have a 'ChemWiki'.  Additionally, university libraries also are often accessible to the public and have search functions for journal articles, books.  Some universities, such as the University of Connecticut have Research Database Locator.

As mentioned in the comments, Wikipedia, particularly if a page is written well, have several links to journal articles and books etc.


Transition metals with closed shells From wiki page about valence electrons:

An atom with a closed shell of valence electrons (corresponding to an
electron configuration $s^2p^6$ for main group elements or $d^{10}s^2p^6$ for
transition metals) tends to be chemically inert.

My question is, which are the transition metals with $d^{10}s^2p^6$? Looking at the periodic table, I do not see anyone.
After this paragraph, wiki says that valence orbitals for d-block elements (transition metals) are (see table):

ns (n-1)d np

I though one of the characteristics of transition metals is that they do not have electrons in last p orbital.

Your premise Wikipedia's statement about transition metals seems to be flawed.
(1) It isn't possible to have a transition metal that has the electronic structure $\mathrm{(n-1)d^{10} ns^2np^6}$. Such an electronic structure would imply a noble gas.
(2) Gold $\mathrm{[Xe]4f^{14} 5d^{10} 6s^1}$ and platinum $\mathrm{[Xe]4f^{14} 5d^{9} 6s^1}$ are the most noble metals and neither even has a $\mathrm{(n-1)d^{10} ns^2}$ configuration.

Granted these configurations are for the free atom, not the bulk metal...


How to plate zinc evenly on a threaded bar I need to set a number of threaded bars into cement, to hold a new gatepost. When I cut the bar into shorter pieces, the metal near the cut became red hot, presumably hot enough to evaporate some or all of the zinc with which the entire rod was plated.

Now I want to plate zinc back onto the ends, and have had mixed success so far. I dissolved some zinc oxide with hydrochloric acid and hooked it up to a little PV panel I have, and after a while it seems there is a light-coloured deposit on the cathode. The pencil-lead (6B) anode is no longer as smooth as it first was, but it doesn't seem to be losing (much) of its bulk. I assume it to be zinc, because... well what else can it be? It plated thick enough that I had to (ab)use a nut to re-cut the thread, leaving a light-coloured shiny surface that gives me more confidence that it is, in fact, zinc.

Problem: the edges of the threads seem to be taking more zinc than the valleys. In some cases the thread's valley doesn't show any evidence of plating. The fractal-like deposit on the edges is quite pretty actually, pity I can't take a (good) closeup photo.

Is there anything I can do (perhaps using some other common(ish) kitchen chemicals?) that will result in a more uniform layer of zinc? Is it worth trying lower / higher current or pulsed current? Am I overthinking things - is it even necessary for the whole stud to have a layer of zinc?

I think the easiest way is to sputter coat the bars, if you can access one. One that can rotate your bars such that the rotation is vertical to the direction of sputtering would be the best. If not, switch sides manually.

Depends on the composition of your bars, you might want to coat something else before coating zinc. A wild guess: nickel.  I suggest so because I don't recall zinc being sticky. So the coating might come off easily.

And I'd try sputter slow but for a long time. I think low sputtering rate tends to produce more densely packed and uniform coating. I am not sure. But dense and uniform coating provides good foundation of the zinc coat.

For electric coating, I think low voltage and uniform coating is still nice. I don't think you want formation of gas at the bar electrode, as that disturbs the structure of the coating. Not sure what's influencing the nucleation sites on the bars. I would guess that the current is not uniform on the bars, say there more on the hill tops. Still consider coating something else first. I would try whatever is available. I would try to get the solutions as pure as possible though. Organic compounds can act as surfactants which tend to aggregate the particles.

Whichever your method is, search for some journal articles characterizing similar processes. That might help.


Why do liquid crystal displays (LCDs) visually distort under pressure? A venerable question, as old as LCDs themselves. I assume it has something to do with pressure-induced changes in the path length across the display, but that's just a guess. Of particular interest to me as I press my fingernail into my (TFT LCD) phone screen1 are the observations of:

A symmetrical bilobate distortion in the direction of movement of my fingernail.
Ripples of colour distortion that radiate outwards at about a centimeter per second.
A trail of slight colour distortion that with a persistence of a few seconds.

Any ideas as to the cause of this rich behaviour?

[1] Don't worry, it's a nokia n95 and as such will likely outlive the human race if not the solar system.

Liquid crystal displays work by using liquid crystals where the direction of alignment of the molecules in a plane rotates as you go from layer to layer. Usually this is done with twisted nematic crystals, however cholesteric LCs can also be used to achieve similar effects (I don't think smectic LCs are useful here as they don't align).

To create a liquid crystal cell, twisted nematic liquid crystal is filled between two crossed polarizers1, and a backlight is sent through. The important thing about liquid crystals is that the specific rotation is easy to control by varying the potential difference (more potential difference leads to more anisotropic disorder).

Now, if you make a grid of these and put color filters on them, we get an RGB LCD display.

What happens when you press on them? You change the concentration/length of the liquid crystal, as well as making it bulge out slightly2.. This changes the optical rotation it induces, causing the amount of light that comes out to change (as the amount of light is dependant on how aligned the rotated light is with the second polarizer)

I'm not too sure of the exact reasons behind the phenomena you list, but I can guess:

Firstly, the color changes are caused because each type of crystal cell is affected differently by the pressure. I think there's a difference in concentration between the different color cells to better suit human eyes. When I do it to LCD displays accessible to me, I only get a darkening, not a color change, so it seems to not be a universal feature of LCD displays. If the rotation of each color cell is affected differently by pressure, then pressure can easily cause color distortions as each color will not darken by the same amount.

I think the ripples are caused by high pressure where you change the optical rotation of the inner cells (the ones closer to your finger) so much that they go through a at least one full 360o rotation, leading to regions of both bright and dark light.

The persistence is because liquid crystals are viscous — they have to be, as they consist of long molecules. They seem to align quickly to potential differences (potential differences align the molecules without moving the liquid), but when we add pressure (where the liquid itself distorts) I think it takes some time to bounce back. The ripples move due to this same persistence.

1. If you're ever at a 3D movie, have a look at your phone screen with the 3D polarised glasses on. It looks pretty cool!

2. If this did not occur, then the rotation would stay the same as $\theta\propto lc$ and $lc$ does not change if there is no bulge.


Is it correct to include the concentration of water into the concentration equilibrium constant? For the reaction
  $$\ce{ C3Cl3N3O3 + 3 H2O -> C3H3N3O3 + 3 HClO3}$$
  write the expression for $K_\mathrm{c}$ for this equilibrium.

The mark scheme says it would be
$$K_c = \frac{[\ce{C3H3N3O3}][\ce{HClO3}]^3}{[\ce{C3Cl3N3O3}][\ce{H2O}]^3}$$

I thought $K_c$ shouldn't include water, is the mark scheme wrong? Or is this an exception that water is steam instead of liquid water?

How is one supposed to know that?

Here is another question I have just come across. (Different question, but still related.) Please take a look at the image below. Now, the mark scheme has seriously taken the concentration of $\ce{H2O}$ into account in (iii) and (iv).

1. Reaction of Trichloroisocyanuric acid with water:

Water is omitted from the equilibrium expression only if it is a solvent in that reaction because it is a pure liquid. We can't increase the concentration of a pure liquid or pure solid and hence they are omitted from the expression.

However water in gaseous state cannot be ignored. It would be included in the equilibrium expression.

The above reaction is actually mixing of Trichloroisocyanuric acid in water to work as a disinfectant. So in the above reaction H2O is basically pure liquid (not gas).

Though in the expression water concentration is mentioned but in the later steps it would be substituted as 1. Actually some authors tend to write the formula of equilibrium constant including all the reactants & products concentrations (including pure solids and liquids) first and then in the next step they substitute 1 in the place of concentration of pure solids and liquids and other given numerical concentration values in place of other reactants and products.

So that expression is just a formula and in the next step when you would be substituting the values, 1 will be substituted inplace of concentration of water.

2. Coming on to the reaction you have sent in the picture i.e. reaction of aldehyde and alcohol:

Here water is not going to be omitted.

Okay water is always not neglected from the equilibrium expression. When I said water concentration is assumed to be 1, I had actually assumed water to be a solvent. Remember if water is a solvent in your reaction, then you can neglect the water concentration term but if water is not a solvent, then water term needs to be included. In the above question, water is being created in the process .i.e it is a product. If water had been a reactant as well as product (i.e. a solvent), then you could have neglected it.

Water in the above reaction is not present/taken from initial situation. It was eventually formed during the reaction and hence it is a product ( not a solvent). Hence you need to include it in the Kc expression.

Do you know what actually happens, when water is a solvent, its concentration term appears on reactant as well as on the product side. Hence while writing Kc expression it is cancelled out from the numerator and from the denominator. But if water is a product (i.e. not a solvent), then reactant part wont contain any concentration of water term and hence it can't be cancelled out in Kc expression .i.e. it will appear in the numerator of the Kc expression.

Hope it helps.


How can a β,γ-unsaturated ketone be converted to 1-pyrroline? I am working towards a natural product. Presented are three intermediate compounds. After oxidation of the hydroxyl to a carbonyl, I want to introduce an imino-nitrogen atom into the structure. This way, an imino bond creates a five-membered ring, and the double bond disappears.

What reagent(s) should I choose for this type of reaction, and what does the reaction mechanism look like?

I suggest two possible approaches. Narasaka, et al. have effected similar cyclizations photochemically. I prefer to start with the allylic alcohol 1 to avoid complications with oxime formation. Formation of the bis acetate 2b is not a problem. However, given the greater acidity of oximes relative to alcohols does not preclude the formation of oxime acetate 2a. In the photolysis the naphthalene serves as a sensitizer while 1,4-cyclohexadiene functions as a hydrogen atom donor. The addition of hydrogen is likely from the convex face as shown. If not, after saponification of acetate 3b and Swern oxidation, base equilibration would likely lead to ketone 4.
Alternatively, a Barton-type radical cyclization (5 --> 3a) might be considered.
ADDENDUM: Direct photolysis of a structure similar to 5 without Bu3SnH has been achieved. See ref 2.

1) M. Kitamura, Y. Mori and K Narasaka, Tetrahedron Letters, 2005, 46, 2373.
2) F. Gagosz and S. Z. Zard, Synlett., 1999, 1978.


How do meth lab test kits detect the presence of a meth lab? Many houses in the USA are illegally used as meth labs to produce methamphetamines, which can cause many negative health effects in their inhabitants, even after the house is sold to the next unwitting owner.

The Denver Post estimates that 7% of single family and 11% of multi-family units that are foreclosed have traces of meth residue.

With these risks, many commercial test kits for meth lab residue have appeared. However, many of them are intentionally vague about their chemical mechanisms.

What chemical mechanisms do meth lab test kits use to detect the presence of a meth lab?

Simple self-made or commercially available tests use the Marquis reagent, which is a mixture of sulfuric acid and formaldehyde. For example, Marquis reagent can be prepared by adding 100 ml of concentrated sulfuric acid to 5 ml of a saturated aqueous solution of formaldehyde (40 % formaldehyde by volume or 37 % by mass).[reference]

The Marquis reagent reacts with many alkaloid drugs to form a variety of colours. Methamphetamine creates an orange dye (deep reddish orange to dark reddish brown).


Boiling point of Coca-Cola Inspired in part by the question boiling point of vodka - given that the exact ingredients of Coca-Cola are kept secret, based on what is known, what would a viable prediction of the boiling point of Coca-Cola be?

Would it be significantly different from related products (e.g. Diet Coke, Coke Zero, etc)?

Your best guess for the boiling point of any version of Coke would be 100 C, the boiling point of water.

Diet Coke is mostly water (the flavourings are a very small amount relative to the amount of water). The largest ingredient will be the sweetener but there will be only a fraction of a gram of that. It is unlikely you will notice any deviation from the properties of water.

Standard Coke has quite a lot of sugar in it. A standard can (~300ml) contains about 40g of sugar. To put it another way, the contents are more than 10% sugar by weight and the solution is about 1/3 mol/L of sucrose (other sugars will be slightly different). A standard calculation using the ebullioscopic constant for water suggests the elevation of the boiling point will be barely 0.2 C, so small you'd struggle to measure it without good instruments and a good experimental setup.


What chemicals are often confused for different chemicals? Recently I asked a similar question on Chinese SE: Why would Antimony (Sb) and Aluminum (Al) be confused?

It seems, though, that this is by no means a Chinese phenomenon.

The Wiki page for Tin foil mentions that:

  aluminium foil, [...] is still referred to as "tinfoil" in many regions.

What other chemicals are often confused for different chemicals?

I'm especially interested in things that have retained their incorrect names.

One especially serious example is confusing methanol with ethanol. Methanol is highly toxic and sometimes is deliberately added to or inadvertently present in bootleg alcoholic beverages, causing many deaths and blindness in survivors every year around the world. It would be a life saver for someone to develop a cheap dipstick test for methanol.


Memorize Transition Metal Electronic Ground State In the course of my undergraduate studies it has become clearer and clearer that it is quite helpful to know the ground electronic state of transition metals by heart, if only to speed up the process of analysing a reaction or to evaluate a compound in respect to its number of valence electrons.

Edit: Note that when I say ground electronic state, I don't mean the state in vacuum, but when the metal is bonding to ligands. As a consequence all the electrons are found in the $d$ orbitals.

My Method:
I have now put the $d$-Block of the periodic table as a wallpaper on my screen background so I get constantly exposed to the arrangement of the elements.

What is your way to memorize what metals have which $d^x$ ground state?

Wikipedia has a comprehensive list on chemistry mnemonics, with which I'm now memorizing the position of the transition metals.
For example, for the first two periods of the elements (excluding neon):

H He Li Be B C N O F
Ha. Healthy Little Beggar Boys Catching Newts Or Fish.

Or my favourite for the first period of transition metals:

Sc Ti V Cr Mn Fe Co Ni Cu Zn
Saint Thomas Very Carefully Made Five Completely New Copies of Zarathustha.


Proper IUPAC name of a substituted cyclohexenol Here's the molecule:

So far I've tried: 6-methylcyclohex-2-en-ol and 4-methylcyclohexen-3-ol but both are wrong. I'm rather confused as to where I should start: the alcohol or the double bond?

I think you made a counting mistake. It is 6-methylcyclohex-2-en-1-ol


Thermodynamic stability and Inertness The following is from my teacher's notes:

  The reaction (where the reaction mixture is thermodynamically unstable but kinetically inert) tends to go as indicated by a positive total entropy change, yet nothing happens because the rate of the reaction is too slow to be noticeable due to a high $E_\mathrm a$.

Does that mean that the reaction is taking place but we don't see it because of how slow it is, or that the reaction will not take place in the first place?

This is A2 chemistry.

The reaction can be thermodynamically unstable (favorable) not only due to positive entropy change but also negative enthalpy change, both result in negative free energy. The thermodynamic stability is determined by the difference between the potential energy of products and reactants. If the result is negative the reaction is said to be favorable and vice versa.

Yet this doesn't say anything about the rate of the reaction. The rate of the reaction or reaction kinetics is determined by taking into account an intermediate step in the way from reactants to products, the transition state. If the difference between the potential energy of the transition state and the reactants is too large, the reaction will be too slow. conversely if the difference is small the reaction takes place so fast.

Here arises a question: From where will reactants get the energy to reach and pass the transition state ?

The answer is that indeed a number of molecules has this energy but their population may be too low. This population is determined from the Boltzmann distribution. Heating may be used to supply some energy that increase this population of molecules.

Returning to your question: "Does that mean that the rxn is taking place but we don't see it because of how slow it is , or that the reaction will not take place in the first place?"

Answer: Although the term "kinetically inert" is only used for complexes, the same rules hold as in the other cases. The reaction does take place but we don't notice it. An example for this is turning from diamond to graphite or burning of paper (oxidation of a cellulose).

please, Don't hesitate to ask any question in the comments if something is unclear to you.

This is a useful link:
https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Equilibria/Chemical_Equilibria/Principles_of_Chemical_Equilibria/Kinetically_vs_Thermodynamically_Stable

Hope this helps.


Why does symmetry have to be maintained in molecular orbitals? Using the example of $\ce{XeF4}$:

What is the physical explanation enforcing the symmetry of the $\ce{1b_{1g}}$ orbital on the fluorine atoms? Why isn't the symmetry of a nonbonding orbital arbitrary?  If it's going to be nonbonding anyways, why can't we, for example, have a Fluorine p-orbital arrangement facing towards Xenon with three positive p-orbitals and one negative p-orbital?

To elaborate:

If I imagine a free Xenon atom in space, and the approach of four individual fluorine atoms, I would expect the bond formation to be randomized with respect to the orientation of the fluorine p-orbitals, and therefore for some arrangements to not be perfectly symmetrical, such as 3 positive p-orbitals, 1 negative, facing inward.  I understand bonds can't be made without symmetry between the Xenon and Fluorine orbitals; that makes physical sense because we can argue it by looking at orbital overlap that dictates bonds can only occur with appropriate symmetry.  But in a nonbonding case, such as $\ce{1b_{1g}}$ above, I don't understand why symmetry is also required.

Source for pictures:

https://scilearn.sydney.edu.au/fychemistry/calculators/make_mo.shtml?type=year1&theMolecule=xef4
http://www.chem.mun.ca/homes/cmkhome/SALCS&MOs.pdf

The orbital and geometrical symmetry are closely related. You know that $\ce{XeF_4}$ is square planar, therefore $\ce{D_{4h}}$ symmetric. That also means that the four fluorine atoms are indistinguishable.

So if you perform some manipulation of the molecule, e.g. rotation by $\ce{90^{\circ}}$, you must end up with the same picture (or just with opposite sign), therefore ruling out our suggestion of 3:1 different orbitals. This holds for all orbitals, not only bonding.

One could possibly argue, what is the physical meaning of the unoccupied orbitals, but this is out of scope of the question.


In HPLC, how does column diameter affect the retention time? I am using a smaller diameter column than my previous one keeping all other factors such as stationary phase, solvent, solvent gradient constant. (Obviously, flow rate will be less in smaller column). I would like to know will changing the diameter affect the retention time of my target?

I searched and got results like this and this But they do not mention about the effect of column diameter on retention time. Moreover, I am weak in understanding mathematical equations.

The key idea is that when you change column diameter, you keep the superficial linear velocity the same. For example if I have an analyte which elutes on a 10 cm x 4.6 mm i.d column at 7 minutes at 1 mL/min. Now we change the diameter to 3 mm, we have to use a flow rate of 0.425 mL/min to keep the retention time the same. Now we wish reduce the diameter further, say 2.1 mm i.d., now you should use 0.21 mL/min so that your analyte elutes at 7 minutes.

Search the keyword: superficial linear velocity, HPLC


Can electrostatic interaction affect nucleophilic attack by thiolate anion Suppose a protein side chain has a $\ce{-SH}$ group which is partially deprotonated to thiolate anion $\ce{-S -}$. This thiolate can perform a nucleophilic attack on a certain substrate.

Will the propensity of nucleophilic attack be reduced if the thiolate anion is in electrostatic interaction with a $\ce{NH3+}$ group of a basic amino acid? Or is the likelihood of nucleophilic attack independent of electrostatic effects?

P.S.- The $\ce{-S-}$ will probably be in eletrostatic interaction with H+ in the solution also. Maybe electrostatic interaction with $\ce{NH3+}$ will be more stable?

Core question: If a negative nucleophile is in electrostatic interaction, will it affect its nucleophilicity?

It would since the S atom is now negatively charged. However, S is a very poor hydrogen-bond acceptor as evidenced by its weak basicity (thiols are more acidic than alcohols). So this electrostatic effect will be greatly diminished if you compare with more basic nucleophiles like F- and O-. This is because sulfur larger and will be more polorizable making the localization of the negative charge more disperse around the atom and much weaker to form ion-ion interactions. It will still remain as a relatively powerful nucleophile, so you can expect side reactions.


Mathematical explanation behind dq = TdS Can somebody mathematically prove why $$\mathrm dq = T\,\mathrm dS$$ for a reversible process?

I realize I am supposed to provide some research but I can't find anything. The best I got is $\mathrm dE = T\,\mathrm dS$ which is basically $\mathrm dG = \mathrm dH - T\,\mathrm dS$ and it seems to be a circular argument (since for derivation of $G$ most proofs use the equation $\mathrm dq=T\,\mathrm dS$).

$\def\d{\mathrm{d}}\def\c{_\mathrm{C}}$Firstly, note that a more correct version of your equation is

$$\delta q_\mathrm{rev}=T\d S.\tag1$$

We use a Greek delta to signify that heat is not, in general, a state function. Its value during a change does not simply depend on the beginning and end points but also on the way which the change is traversed. Subscript '$\mathrm{rev}$' is for a reversible process.

Secondly, I should warn that the proofs I supply are in no way rigorous but should suffice for a standard thermodynamics course.

Thirdly, I have used a small $q$ for heat, reserving capital $Q$ for a partition function.

Classical thermodynamics

The classical definition of entropy in thermodynamics is due to German scientist Rudolf Clausius. He showed that, for a cyclic process, the following inequality holds:

$$\oint\frac{\delta q}{T} \leq 0.\tag{2}$$

  Lemma. For a Carnot cycle, the following line integral over a circular path equals zero: $$\oint\c \frac{\delta q_\mathrm{rev}}{T} = 0.\tag{L}$$

Proof

Recall and remember that in a Carnot cycle efficacy $\eta_\mathrm{C}$ of a system may be expressed in two equivalent ways, by a heat quotient and a temperature quotient.

$$\eta\c=1+\frac{q_2}{q_1}\tag{3a}$$
$$\eta\c=1-\frac{T_2}{T_1}\tag{3b}$$

Substrating one from the other yields

$$\frac{q_2}{q_1} + \frac{T_2}{T_1}=0\tag{4}$$

which, through algebraic manipulation, is the same as

$$\frac{q_1}{T_1} + \frac{q_2}{T_2}=0\tag{5}.$$ Hence for a Carnot cycle, reversible by definition, we optain

$$\oint\c \frac{\delta q_\mathrm{rev}}{T} = 0.\tag{L}$$

I will omit the subscript $_C$ for Carnot in the integral when I mention in the text or it is obvious by the same subscript on some other quantity.

Proof of Clausius integral inequality

Start with the first law of thermodynamics.

$$\d U = \delta q + \delta w \tag6$$

Therefore, for a Carnot engine,

$$w\c = -\oint \delta q_\mathrm{rev}.\tag7$$

Let us take some other engine, called engine $2$. Again from the first law of thermodynamics,

$$w_2 = -\oint \delta q_2.\tag8$$

Assume that, contrary to the theorem,

$$\oint \frac{\delta q_2}{T}>0.\tag9$$

Let us join these two cycles to form a composite engine. The work done by this system will be

$$w_\mathrm{total} = w\c + w_2 = -\oint(\delta q_\mathrm{rev} + \delta q_2) = -\oint\delta q_\mathrm{total}. \tag{10}$$

By our lemma $\mathrm{(L)}$, assumption $(9)$ and result $(10)$, we see that

$$\oint\frac{\delta q_\mathrm{rev} + \delta q_2}{T} \stackrel{(10)}{=} \oint \frac{\delta q_\mathrm{total}}{T} \stackrel{\mathrm{(L)},(9)}{>} 0 \tag{11}.$$

Now change the sign or direction of the Carnot engine, and its size so that no work is done in the composite cycle with engine $2$. (This should always be possible via Carnot's theorem which I will not prove here.) Then

$$w_\mathrm{total} = \oint\delta q_\mathrm{total} = 0. \tag{12}$$

If we break cyclic line integrals apart into $y-1$ smaller integrals that add up to the cyclic integral, we achieve

$$W_\mathrm{total} = \sum_{i<y} \int_{b_i}^{b_{i+1}} \delta q_i = \sum_{i<y} q_i(b_j) \stackrel{(12)}{=} 0,\tag{13a}$$
$$\oint \frac{\delta q_\mathrm{total}}{T} = \sum_{i<y} \int_{b_i}^{b_{i+1}} \frac{\delta q_i}{T_i} = \sum_{i<y} \frac{q_i(b_j)}{T_i} \stackrel{(11)}{>} 0.\tag{13b}$$

The terms $q_i(b_j)$ may be both positive and negative but in equation $(13a)$ cancel out to give zero. In equation $(13b)$ the terms are weighted by temperatures in such a way that positivity wins. Therefore, on the whole, negative $q_i(b_j)$ terms have been divided by larger temperatures whereas positive terms possess smaller temperature denominators. Physically, this implies that heat is transferred from a lower temperature reservoir to a higher temperature bath without doing any net work. This is in violation of the second law of thermodynamics (Clausius statement).

Therefore, our assumption $(9)$ that

$$\oint \frac{\delta q_2}{T}>0$$

is actually false. The result

$$\oint \frac{\delta q_2}{T}\leq 0$$

immediately follows.

  Corollary. For any reversible process, not just a Carnot cycle, $$\oint \frac{\delta q_\mathrm{rev}}{T} = 0.$$ In other words, temperature becomes an integrating factor in a reversible process.

Proof

By Clausius integral inequality $(1)$, we know immediately that

$$\oint \frac{\delta q_\mathrm{rev}}{T} \leq 0. \tag{14}$$

Assume again to the contrary that

$$\oint \frac{\delta q_\mathrm{rev}}{T} < 0. \tag{15}$$

Since this is a reversible process, we may change the direction of the engine. Only the sign of the integral will alter.

$$\oint \frac{\delta q_\mathrm{rev}}{T} > 0. \tag{16}$$

If one proceeds analagously to the proof of the Clausius integral inequality, this will lead to a contradiction. Thus, assumption $(15)$ is not valid, and by $(14)$ we are left with an equality.

Why bother with all of this?

We have now proven that for any reversible process,

$$\oint \frac{\delta q_\mathrm{rev}}{T} = 0. \tag{17}$$

This means that we may define a change in a new state function $S$ which we call entropy such that

$$\Delta S= \int_{S_1}^{S_2} \frac{\delta q_\mathrm{rev}}{T}\tag{18},$$

or in differential form,

$$\d S = \frac{\delta q_\mathrm{rev}}{T}. \tag{19}$$

This is the definition of a differential of entropy. There is nothing else to derive; what we showed above is that we may indeed define such a state function. To achieve your equation, multiply both sides of equation $(19)$ by temperature.

$$T\d S = \delta q_\mathrm{rev}. \tag{CT}$$

Statistical thermodynamics

Another way to go would be to define entropy in statistical thermodynamics. Then the formula $\mathrm{(CT)}$ may be derived.

Statistics in general

For a discrete random variable $X$, entropy is defined in statistics to be the function $\mathrm{H}$ such that

$$\mathrm{H}(X)= -\sum_k p_k\ln p_k \tag{GS}$$

where small $p$ denotes probability.

Boltzmann statistical thermodynamics

Say we have a large number $\mathcal{A}$ of isolated systems which we together call an ensemble. Suppose the system may be described as

\begin{align}\begin{split} \text{energies}\ \ &E_1\ \ &E_2\ \ldots &E_n\ \ldots\\ \text{number of such systems}\ \ \ \  &a_1\ \ &a_2\ \ldots\ &a_n\ \ldots\end{split}\end{align}

where

$$\mathcal{A} = \sum_i a_i.\tag{20}$$

The systems of different energies in this ensemble are distinguishable. The systems that have the same energy, i.e., are degenerate, are indistinguishable. If we supplied each system regardless of energy with an index of energy, the number of ways to order these indices would be

$$\mathcal{A}!.\tag{21a}$$

But as we said, the systems of equal energy are indistinguishable. The number of ways of ordering each of these is

$$a_j!.\tag{21b}$$

So to avoid overcounting we must divide:

$$K(a_1,a_2,\ldots,a_n,\ldots) = \frac{\mathcal{A!}}{\prod_i a_i!}.\tag{21c}$$

You may think of this $K$ as the total number of anagrams of words with $\mathcal{A}$ letters where each letter repeats itself $a_j$ times. Boltzmann used the big letter $W$ instead of $K$ and defined entropy to be

$$S = k_\mathrm{B}\ln W.\tag{BD}$$

The natural logarithm guarantees, among other things, that entropy is not a huge quantity, and that entropy for different (independent) systems is additive. In other words, if

$$W_\mathrm{total} = W_\mathrm{A}W_\mathrm{B}$$

then

$$S_\mathrm{total} = S_\mathrm{A} + S_\mathrm{B}.$$

Boltzmann entropy as a special case of statistical entropy

Begin with the statistical definition.

$$\mathrm{H}(X)= -\sum_k p_k\ln p_k \tag{GS}$$

Say each and every state in $W$ has equal probability. Then, since the sum of probabilities must equal $1$, we optain

$$p_j = \frac{1}{W}.\tag{22}$$

(You may prove that this constraint actually maximises entropy. This is what we expect from a measure of 'information' in a system where 'knowledge of the system' is smallest in the sense that each state has equal probability. See 'Intuitive explanation of entropy?' on math.stackexchange.com by An.Ditlev.)

Substituting

$$\mathrm{H}(X)= -\sum_k \frac{1}{W}\ln \frac{1}{W} = -W \frac{1}{W}\ln \frac{1}{W} = -\ln \frac{1}{W} = \ln W.\tag{23}$$

This shows that Boltzmann constant is actually unnecessary and a figment of historical definitions. We shall see this in one other way down below. We conclude that Boltzmann's definition of entropy is a special case of a more general entropy when probabilities are equal. In thermodynamics, this more general entropy is called Gibbs entropy.

$$S\equiv -k_\mathrm{B}\sum_k p_k\ln p_k\tag{GD}$$

This is our new definition of entropy. Here we have an absolute quantity; in classical thermodynamics (without the third law) we defined the differential of entropy.

Equivalence of Gibbs entropy and Clausius entropy

A partition function $Q$

Intuitively, we assume that the relative populations of system $a_m$ and $a_n$ are given by some function depending on energy.

$$\frac{a_m}{a_n} = f(E_m - E_n)\tag{24}$$

The difference in energy is a natural choice because any arbitrary zero for an energy state should cancel. Furthermore, the function $f$ ought to be transitive.

$$\frac{a_x}{a_n} = \frac{a_m}{a_n}\frac{a_x}{a_m} \iff f(E_x - E_n) = f(E_m - E_n)f(E_x - E_m)\tag{25}$$

The most obvious function with these properties is the exponential function. Thus we define

$$a_j = C\mathrm{e}^{-\beta E_j}, \ \ C\in\mathbb{R},\tag{26}$$

where $\beta$ is positive everywhere by convention. Its form is to be determined. We start by tackling the constant $C$. Sum both sides of equation $(26)$.

$$\mathcal{A}=\sum_i a_i = C\sum_i\mathrm{e}^{-\beta E_i} \tag{27}$$

Divide equation $(26)$ by equation $(27)$.

$$\frac{a_j}{\mathcal{A}} = \frac{\mathrm{e}^{-\beta E_j}}{\sum_i\mathrm{e}^{-\beta E_i}} \tag{28a}$$

In the limit, the left-hand side becomes probability of being in a state $a_j$.

$$p_j = \frac{\mathrm{e}^{-\beta E_j}}{\sum_i\mathrm{e}^{-\beta E_i}} \tag{28b}$$

The denominator is defined to be a partition function $Q$. Assumuming that $E=E(N,V)$ (number of particles and volume, respectively) we see immidiately

$$p_j(N,V,\beta) = \frac{\mathrm{e}^{-\beta E_j(N,V)}}{Q(N,V,\beta)}. \tag{29}$$

The first law of thermodynamics with partition functions

From the definition of expectation value for a discrete variable $X$,

$$\mathrm{E}X = \sum_ip_ix_i.$$

Let us identify internal energy with an expectation value of energy. Then

$$U = \sum_i p_i(N,V,\beta)E_i(N;V).\tag{30}$$

Take a differential of both sides.

$$\d U = \sum_i p_i(N,V,\beta)\d E_i(N;V) + \sum_i E_i(N;V)\d p_i(N,V,\beta)\tag{31}$$

This expression for probability and energy holds for a reversible process. If we associate the first term with work, the second term will be heat. Doing so will allow us to prove the equivalence but it will also lead to correct results for ideal gases if you do the maths. For brevity (ha ha), I will gloss over such discussions but in my opinion the rigorous explanation (if there is one) deserves its own question. (I am not aware of the rigorous proof.) Hence,

$$\delta q_\mathrm{rev} = \sum_i E_i(N;V)\d p_i(N,V,\beta).\tag{32}$$

Proof of equivalence

Recall the definition of Gibbs entropy.

$$S\equiv -k_\mathrm{B}\sum_k p_k\ln p_k\tag{GD}$$

Take the differential of both sides.

$$\d S = -k_\mathrm{B}\sum_k (1 +\ln p_k)\d p_k\tag{33}$$

The term $\sum_k \d p_k$ will equal zero by linearity and because probabilities sum to $1$.

$$\sum_k \d p_k = \d\sum_k p_k = 0\tag{34}$$

Thus,

$$\d S = -k_\mathrm{B}\sum_k \ln p_k\d p_k.\tag{35}$$

Substituting equation $(29)$ into the natural logarithm term in equation $(35)$ yields

$$\d S = k_\mathrm{B}\sum_k \left[\beta E_k(N,V) + \ln Q(N,V,\beta)\right]\d p_k.\tag{36}$$

Similarly to equation $(34)$ the final logarithm term cancels.

$$\sum_k \ln Q(N,V,\beta) \d p_k = \ln Q(N,V,\beta)\d \sum_k p_k = 0\tag{37}$$

We are left with

$$\d S = k_\mathrm{B}\beta\sum_k E_k(N,V) \d p_k(N,V,\beta)\tag{38}$$

which by equation $(32)$ translates to

$$\d S = k_\mathrm{B}\beta\delta q_\mathrm{rev}.\tag{39}$$

Note that $k_\mathrm{B}\beta$ is also an intgrating factor. From our corollary before and to complete the equivalence, we are forced to choose

$$\beta = \frac{1}{k_\mathrm{B}T},\tag{40}$$

and finally,

$$\d S = \frac{\delta q_\mathrm{rev}}{T} \implies T\d S = q_\mathrm{rev}.\tag{19,CT}$$

In conclusion, we have shown that Gibbs entropy is equivalent to Clausius entropy up to an additive constant. This is because a differential of a constant is zero. The choice of $\beta$ again demonstrates how superfluous Boltzmann's constant really is. We introduce it in the Gibbs entropy definiton, only to be forced to get rid of it. For more information on Boltzmann's constant, and how it relates to temperature and the zeroth law, see 'Is the Boltzmann constant really that important?' on physics.stackexchange.com by DanielSank. And so,

$$S_\mathrm{Gibbs} = S_\mathrm{Clausius} + r,\ \ r\in \mathbb{R}.\tag{41}$$

If the most classic form (zero entropy at absolute zero) of the third law of thermodynamics holds, and it always does not,

$$r = 0. \tag{42}$$

TL; DR

The reason why

$$\delta q_\mathrm{rev}=T\d S\tag1$$

holds depends on your starting point.

There are different forms of the first and second laws. Proving that a state function $S$ exists varies based on one's axioms.
Entropy may be defined in various ways. If a classical definition is used, there is only to prove that we may indeed define such a state function. If a statistical definition is applied, we must first prove the classical definition as a special case.
Note that the second law of thermodynamics only defines a difference or differential in entropy. Together with the third law, one may establish an absolute scale. However, the statistical definition is already absolute to begin with.
Boltzmann entropy is a special case of Gibbs entropy, not the other way around. Gibbs entropy agrees with Clausius entropy (in reversible thermodynamics) up to an additive constant.

$$S_\mathrm{Gibbs} = S_\mathrm{Clausius} + r,\ \ r\in \mathbb{R}\tag{41}$$

If the most classical form of the third law holds, $r = 0$.


Inability of alkyl fluorides to be reduced to an alkane in the following reaction Alkyl halides except fluorides on reduction with $\ce{Zn}$ and dil. $\ce{HCl}$ gives alkane. Example:

$$\ce{CH3Cl + H2 -> CH4 + HCl}$$

Why cannot alkyl fluorides be reduced?

Other answers may be more elegant, but it boils down to the carbon-fluorine bond being too strong, in fact stronger than carbon-carbon or carbon-hydrogen bonds.  The required bond breaking simply cannot be done under normal conditions.  Similarly Grignard reagents (formed with a much more powerful reducing agent) are much less common with fluorine than with heavier halogens, and the bad things chlorofluorocarbons do to our ozone layer come from breaking $\ce{C-Cl}$ rather rhan $\ce{C-F}$ bonds.


What does the slash in polymer nomenclature, e.g. Acrylates/ Beheneth-25 Methacrylate Copolymer, mean? I was given a task to find SDS-Sheets for polymers/microplastic particles in cosmetics and I am confused, since there are often two chemicals divided by "/", e.g. Acrylates/ Beheneth-25 Methacrylate Copolymer. So would it be correct to look for the SDS-sheets for the both of them separately? Is it just the mixture of the two? Maybe someone can provide me with some information of how I am supposed to read these kinds of names.

Acrylates/Beheneth-25 Methacrylate Copolymer is an INCI name for a copolymer

  … formed from the ester of 2-methylpropenoic acid and docosanol, ethoxylated, 25 mol EO (average molar ratio) and one or more monomers of acrylic acid, methacrylic acid or one of their simple esters

(Source: cosmetics.specialchem.com)

Also known by the trademark Aculyn™ 28.
Dow Chemical provides cumulative MSDS datasheet (pdf) for the entire Aculyn family, so I assume you can use it for your copolymer as well.


Is methanol really more acidic than water? The question Why is methanol more acidic than water? deals with the reasoning of why methanol is more acidic than water. However, as mentioned in the comments of that question, the acidity constant of water is $14.0$, as confirmed by two sources$^{[1][2]}$, with one of them offering a very convincing explanation. The Wikipedia page$^{[3]}$ for methanol quotes its acidity constant to be $15.5$.

So it would seem, water is more acidic than methanol. But an answer to the question mentioned above gives a good reasoning for why methanol is more acidic than water (it doesn't deal with the pKa of water, though). Also, the answer mentions that water would be much worse an acid in DMSO. Another question implies that methanol and other alcohols are more acidic than water.

On the other hand, it is also true that we regard alkoxide ions as being more strongly basic than the hydroxide ion (thus we use them primarily for elimination reactions).

This leaves us in a fix. So what is the ultimate truth?

    [1]: What is pKa of water? - Chemistry ChemLibreTexts
    [2]: Water (Data Page)
    [3]: Methanol

First, water's $\ce{pK_a}$ in water is $\ce{14}$ as explained here. This means that $\ce{H2O}$ is slightly dissociated in liquid form, such that $\ce{[H+] = [HO-] = 10^{-7}}$.

Second, I would say that caring too much about which compound is the best acid is a bit like arguing by definition. Acidity is a fuzzy concept used as a shorthand for deeper meaning. This means there is little point arguing about the shorthand when we have the actual properties it synthesizes.

In this case (sou rce):

$$
    \begin{array} {|c|c|c|l|}
    \hline
    \ce{solvent} & \ce{pK_a\ of\ water} & \ce{pK_a\ of\ methanol} & \ce{Interpretation} \\
    \hline
    \ce{water} & 14.0 & 15.5 & \ce{water\ is\ more\ dissociated\ than\ methanol} \\
    \hline
    \ce{DMSO} & 31.4 & 29.0 & \ce{methanol\ is\ more\ dissociated\ than\ water}\\
    \hline
    \end{array}
$$
So you can argue (as in the answer you linked) that methanol is more acidic than water in the abstract sense, because $\ce{DMSO}$ doesn't stabilize the anions by hydrogen bonding and so is closer to the "ideal" case (whatever that may be).

But what's the point ? You already have the $\ce{pK_a}$ values, any practical question you might have about the concentrations of theses solutes in $\ce{H2O}$ and $\ce{DMSO}$ is already answerable.


Why are bonding interactions so much stronger than interactions of other types? Why are bonding interactions so much stronger than all the other types of interactions like dipole-dipole, London dispersion, hydrogen bond etc. even though they are all of electromagnetic origin?

Bonding interactions effectively share electrons. In a gedankenexperiment, you can think of two isolated species with many unpaired electrons coming together, bonding and forming a species with less unpaired electrons. The exact nature of bonds is slightly different, involves quantum mechanics and orbitals, but the end of the story is that the electrons are in a less energetic state after bonding than before it. The energy gained by this process is rather significant, typically hundreds of kilojoules per mole.

These bonds that are created typically generate a molecule or some type of network solid (ionic, metallic or covalent). If you have generated a molecule, then you can rather easily separate a single molecule from all the surrounding molecules by physical methods such as high vacuum. It is not possible, however, to break the bonds by any simple, low-energy physical methods. (UV irradiation will do, but that is high-energy.)

The other interactions, London dispersion forces, hydrogen bonds, etc. are interactions between molecules. You already have two species that by themselves are in a rather stable, i.e. low energetic, state and you bring these two together. The energy that can still be gained is very minor compared to that obtained by bonding. This is also why this type of interaction is often called secondary interaction.

A second reason is the distance. As you note, most of these interactions can be explained electromagnetically. However, the distances between molecules are typically significantly larger than the distances between atoms of a molecule. Since electromagnetical interactions decline exponentially with rising distance, these forces are a lot weaker at higher distances.


Is it possible for the iodine atom in periodate ester to be octavalent? Recently, I was reading Clayden's Organic Chemistry (2nd ed.) and I chanced upon this diagram on p. 443 (as shown below) showing an octavalent iodine in the periodate ester. I found this rather puzzling as I thought that the maximum possible valency of iodine was seven. I am well aware that bonding schemes such as the 3-centre 4-electron bonding scheme allow central atoms to make more than the number we would expect the atom to make conventionally. However, this is the first time I am seeing octavalent iodine and thus, I would just like to confirm if it is in fact, possible for iodine to be octavalent.

Reference

Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

I would suggest that a reactive intermediate/transition state was formed, and not an actual compound. I'd say both $\ce{C-O}$ bonds are partially formed in the intermediate

The diagram in the book does not show where is the $\ce{Na+}$ during the reaction (see above picture), probably for the convenience (but therefore, somewhat misleading). Remember, a periodate ester is an ester of periodic acid.  The most common periodate esters are cyclic periodate esters, form as the intermediate in oxidation of vicinal diols such as sugars and glycols. For example: The oxidative cleavage with periodic acid has been successfully applied in the constitutional analysis of sugars (see: Chemgapedia article).

Experimental evidence for presence of cyclic periodate esters (rightfully with hepta-valence iodine) can be found in: Buist, G. J.; Bunton, C. A.; Hipperson, W. C. P. The mechanism of oxidation of α-glycols by periodic acid. Part X. The oxidation of pinacol, and a general discussion of the stability of periodate esters and their role in the mechanism of oxidation. J. Chem. Soc. B 1971, 2128–2142. DOI: 10.1039/J29710002128.

Also, two forms of periodic acids can be found in the corresponding Wikipedia article.


Why HCl is used in Lucas test? In Lucas test $\ce{HCl}$  is used for acid-base reaction between $\ce{Zn(OH)Cl}$ to give $\ce{ZnCl2}$ and $\ce{H2O}$ which is kind of good because $\ce{ZnCl2}$ is formed again and used in first step quickening the reaction. But why don't Lucas test use other acids like $\ce{H2SO4}$ (and add $\ce{ZnCl2}$ in bulk) or $\ce{HClO4}$ (stronger acid than $\ce{HCl}$ may faster the reaction). But why $\ce{HCl}$?

Is it had to do with acidic strength of $\ce{ZnCl2}$ and acid (I mean if stronger acid than $\ce{ZnCl2}$ may react with alcohol)?

The use of concentrated H2SO4 is not amenable. Per a source:

Concentrated sulphuric acid produces messy results. Not only is it an acid, but it is also a strong oxidising agent. It oxidises some of the alcohol to carbon dioxide and at the same time is reduced itself to sulphur dioxide. ... It also reacts with the alcohol to produce a mass of carbon.

On the other hand, per Wikipedia, to quote:

  The Lucas test in alcohols is a test to differentiate between primary, secondary, and tertiary alcohols. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides via an SN1 reaction:[3]...The differing reactivity reflects the differing ease of formation of the corresponding carbocations. Tertiary carbocations are far more stable than secondary carbocations, and primary carbocations are the least stable.

And further:

  An equimolar mixture of ZnCl2 and HCl is the reagent. The alcohol is protonated by this mixture, and H2O group attached to carbon is replaced by the nucleophile Cl−, which is present in excess. Tertiary alcohols react immediately with Lucas reagent as evidenced by turbidity owing to the low solubility of the organic chloride in the aqueous mixture. Secondary alcohols react within five or so minutes (depending on their solubility). Primary alcohols do not react appreciably with Lucas reagent at room temperature.[3] Hence, the time taken for turbidity to appear is a measure of the reactivity of the class of alcohol, and this time difference is used to differentiate among the three classes of alcohols:

Now, with respect to HClO4, there is at least one catalytic study of HClO4-SiO2 during the reaction of benzyl alcohol (1 mmol) and thiophenol (1.1mmol) at room temperature under solvent-free conditions (see).


What is the necessity of re-hybridisation of carbon in the transition state of SN2 reaction? In March's Advanced Organic Chemistry (The SN2 Mecahnism, page 374), it's stated:

  When the transition state is reached the central carbon atom has gone from its initial sp3 hybridisation to essentially sp2 with an approximately perpendicular p orbital.

I would like to know why and how this happens. Is there an underlying mechanism? How is this rehybridisation feasible?

What makes it difficult to understand is that SN2 reactions are concerted because bond formation and breaking occur simultaneously.

Check out this animation (for reference, this is material which accompanies the beautiful "Clayden - Organic Chemistry" book).

As you can see, there is indeed an intermediate step (the "transition state"), when the nucleophile and the leaving group are in a intermediate (and unstable, being highly energetic) state (nor attached or detached). Look closely at the geometry of the carbon and the hydrogens: that's the sp2 hybridization of the transition state carbon.

The mechanism (also shown on the page linked) involves an electron pair transfer from the nucleophile (here, bromide ion) to the carbon and eventually to the leaving group (chlorine).

The curved arrow formalism, of course, is just a representation of what goes on in the real scenario: as you can guess from the animation, the charged bromide transfers part of its electron density to the carbon (which carries a partial positive charge!). The chlorine atom withdraws the electron density as it is transferred from the bromide to the carbon, and eventually, when a carbon-bromine bond is formed (full orbital overlap between Br and C), the chlorine atom has gained a full electron pair, which allows it to leave as a chloride ion.

This is the how.

As for the why, it's a matter of energy. The reaction conditions are energetic enough to push the system towards the unstable transition state, and the equilibrium is shifted towards the products because the energetic state of the products is lower than that of the reactants (that is: the products are more stable than the reactants)


Why tautomers are considered to be the same chemical compound? Because of the rapid interconversion, tautomers are generally
  considered to be the same chemical compound. (Wikipedia)

But looking at the examples provided on that same Wikipedia article, it is obvious that different chemical reactions would take place in each case (for example enol vs. keto).

(source)

There are a number of types of rapid chemical equilibria that basically cannot be turned off. Tautomers are probably the best known examples, although fluxional behavior is observed in many inorganic and organometallic systems.

The origin for most tautomers is that protons are really quantum-mechanical. They can tunnel just like electrons can delocalize. Obviously different tautomers will have different stability, much like some electric resonance structures are clearly more stable.

The comments to your question above indicate that most chemists would just call it "acetone" but of course there's some prop-1-en-2-ol present too.

Yes, the reactivities might be different.

It's a partly a matter of the balance of the equilibrium (i.e., that for many tautomers, only one form predominates). There are exceptions, e.g.
phenols (from Wikipedia):

  In certain aromatic compounds such as phenol, the enol is important due to the aromatic character of the enol but not the keto form. Melting the naphthalene derivative naphthalene-1,4-diol 2 at 200 °C results in a 2:1 mixture with the keto form.

It's also a matter of the time-scale. For example, at room temperature if I take an NMR, do I see two sets of peaks? What about IR or other characterization techniques? If I can't tell the difference because there are one set of peaks, I think many chemists will consider there to be "one compound" even if multiple tautomers are present.

For an example of this (from the cheminformatics perspective), Marc Nicklaus of the NIH/National Cancer Institute gave a great talk at the 247th ACS national meeting "The NCI/CADD Group's InChI Usage and Analysis of Tautomerism for InChI V2".

The talk focuses on tautomers and what software is trying to do to derive simple rules (as opposed to lots of quantum calculations or experiments) to determine one standard tautomer. But they also measured several tautomers by NMR:

These would obviously be chemically quite different, but one form clearly predominates.


Convention in electrochemical cell notation While writing the cell notation for a cell, doesn't one have to write the two states in which a species exists in a sequence appropriate to the reaction in which they participate?
For example, in the Daniell cell reaction, shouldn't it always be written $\ce{Zn}|\ce{Zn^2+}$, and never, $\ce{Zn^2+}|\ce{Zn}$?
I didn't see anything in the definitions, and I saw a question asking for the cell reactions for this;
$$\ce{Pt,Cl2}|\ce{Cl^-(aq)}||\ce{Ag^+(aq)}|\ce{Ag}$$
Obviouly, chlorine doesn't react in that direction, but gets oxidised.

Good question. You can remember the cell notation with the mnemonic "reduction on the right."
So, on the right hand side, you should reduction process and on the left hand side you should show the oxidation process.
For example, for a Daniel cell, one can write
$$\ce{Zn|Zn^{2+}, SO_4^{2−}}||\ce{SO_4^{2−},Cu^{2+}}|\ce{Cu}$$
Now coming to your cell, $$\ce{Pt,Cl_2}|\ce{Cl^-(aq)}||\ce{Ag^+(aq)}|\ce{Ag}$$ is indeed written incorrectly because the writer seems to have the misconception that even in the cell notation, half cells must be written as a reduction.


Using helium to make home brewed beer I want to brew beer using helium instead of $\ce{CO2}$.

Most beers use about 2.5 grams of $\ce{CO2}$ per litre except stout like Guinness which uses nitrogen as it allows bubbles for much longer hence a frothy head on the stout.

So, perhaps a blend of helium and nitrogen would work? Helium is more soluble and if used in place of $\ce{CO2}$, as soon as the seal broke on a can it would escape rendering the cartoon voice idea useless.

How much helium can I squeeze into a litre of beer with or without compounding it to get the chipmunk voice?

To correct one point in the question, helium has the lowest water solubility of all gases. At one atmosphere and $\mathrm{25^oC}$, $\ce{CO2}$ is 100 times more soluble than $\ce{He}$ and $\ce{N2}$ is about twice as soluble as $\ce{He}$.  This is why deep-divers prefer helium to nitrogen.

At 45 psi, which is 3.1 atmospheres pressure, you could dissolve about 0.005 g $\ce{He}$ or roughly 30 mL when fully degassed.  A typical adult breath is around 500 mL, and of course the $\ce{He}$ will not degas instantly.

The bottom line is that pressurizing beer with helium will not make you talk like Donald Duck.  And the biggest difference you would see between $\ce{N2}$ and $\ce{He}$ would be in your bank statement.


Whats the Balanced chemical equation for heating sodium chloride and what type of a reaction is it? For my chemistry lab we had to heat sodium chloride and there was a colour change and a sizzling sound. Is it a decomposition or synthesis reaction, and what is the balanced chemical equation for it?

My guess would be that this was rather the dissociation of Sodium Carbonate or Sodium Hydrogen Carbonate.

The equations for the dissociation of Sodium Carbonate and Sodium Hydrogen Carbonate, respectively, are as follows:

  $$\ce{Na2CO3 -> Na2O + CO2}$$

$$\ce{and}$$

  $$\ce{2 NaHCO3 -> Na2O + H2O + 2 CO2}$$


Titration of H2SO4 w NaOH: Solving for the molarity of H2SO4? I am given $\ce{H2SO4}$ in a reaction vessel of about $50~\mathrm{mL}$. I need to solve for the molarity of $\ce{H2SO4}$. A $10~\mathrm{mL}$ sample of $\ce{H2SO4}$ is removed and then titrated with $33.26~\mathrm{mL}$ of standard $0.2643\ \mathrm{M}\ \ce{NaOH}$ solution to reach the endpoint. How do I solve for titration of the $50~\mathrm{mL}$ sample?

Obviously I can use the formula:
$$M_i \times V_i = M_f \times V_f$$

Which brings me to

$$M_i \times 10~\mathrm{mL} = 0.2643~\mathrm{M} \times 33.26~\mathrm{mL}$$

Thus:

$$M_i = (0.2643~\mathrm{M} \times 33.26~\mathrm{ml}) / (10~\mathrm{mL})$$

This will find the molarity of the $10~\mathrm{mL}$ sample of $\ce{H2SO4}$.

Now, how do I find the molarity of the $50~\mathrm{mL}$ sample of $\ce{H2SO4}$ from this? Would I just do five times the $10~\mathrm{mL}$ sample's molarity?

The molarity would be the same whether you have $5~\mathrm{mL}$ of $\ce{H2SO4}$ or a swimming pool full of it.

However, if you wanted to solve for moles of $\ce{H2SO4}$ in $50~\mathrm{mL}$, you would have to multiply the number of moles in the $10~\mathrm{mL}$ sample by $5$.

Note: Make sure you're working with molarity and not moles. I'm in analytical chem right now and often we're multiplying the number of moles in our sample by the total volume of the volumetric flask from which the sample was drawn, so we're doing calculations similar to this.


Reaction between chlorobenzene and sodium methoxide to produce anisole $$\ce{C6H5Cl + NaOCH3 -> C6H5OCH3 + NaCl}$$

So this reaction undergoes $\ce{S_N2}$ mechanism, but $\ce{C6H5Cl}$ is not likely to undergo such mechanism, so I doubt if anisole would be formed or not?

Chlorobenzene is inert towards most forms of nucleophilic substitution. In particular:

A $\mathrm{S_N1}$ mechanism is not likely here because the carbon is $\mathrm{sp^2}$ hybridised, the $\ce{C-Cl}$ bond is strong, and the dissociation of $\ce{Cl-}$ is therefore not very likely.
$\mathrm{S_N2}$ substitution doesn't occur because 1) the $\ce{C-Cl}$ bond is strong 2) the $\sigma^*$ orbital is in the middle of the ring and is inaccessible to a nucleophile.
The $\mathrm{S_NAr}$ mechanism could be followed if there is an electron-withdrawing group ortho or para to the chlorine atom, but in chlorobenzene, there is no such EWG.

I would like to suggest the possibility of a benzyne mechanism, i.e. an elimination-addition mechanism. The elimination step follows an $\mathrm{E1_{cb}}$ mechanism:

The benzyne intermediate is very electrophilic and nucleophilic attack by methoxide ion proceeds rapidly to give anisole.

Although this reaction is well-documented with both $\ce{NaOH}$ and $\ce{NaNH2}$, I can't find any real-life examples with $\ce{NaOMe}$. Most examples of chlorobenzene → anisole reactions involve some kind of catalyst. However, if it were to happen, it would probably occur via this parthway.


Why does formation of bond releases energy and where does that energy come from? I am having a hard time to imagine why does bond formation releases energy and where does that energy comes from?? It is easy to understand that breaking a bond requires energy because it can be viewed in many ways like breaking a wood into pieces requires energy, ripping your shirt into half requires energy, converting liquid water into gaseous water requires energy. These tangible concepts make it easy to understand why bond breaking requires energy. From these examples it answers the question why does bond breaking requires energy and where does that energy comes from (surroundings/thermal energy/mechanical force applied by the surroundings). But bond formation have so little comparisons and most of the examples associated with it are vague and hard to communicate with non science major. Thanks for the help.

It's a matter of perspective. You say you understand why bond breaking requires energy. In other words:
$$\ce{A2 + energy -> 2A}$$
But consider the reverse reaction:
$$\ce{2A -> A2 + energy}$$
This is also valid and can be understood in the following way: Bonds stabilize the electrons. In the unbound or atomic state, the electrons have a bit more energy. You can consider that atoms have some potential energy, like a ball on a hill. By forming a bond, the atoms "go down the hill" into a more stable state and thus lose this potential energy. This energy must go somewhere, and it is thus released in one form or another (heat, light, etc.) To break the bond is to push the atoms "up the hill" again, which requires energy.


pKa values for protonated nitrogens in ketoconazole Everywhere on the internet I see that the  two $\mathrm pK_\mathrm a$'s of ketoconazole are 6.4 and 2.3, but I don't think that is correct.

The $\mathrm pK_\mathrm a$ of imidazole is around 6.4 . But ketoconazole also has a piperazine ring and normally this $\mathrm pK_\mathrm a$ is around 9.8. There are electron acceptors but I still find 2.3 to be very low. Is this correct and which of the two nitrogens will be protonated?

The 6.4 pKa is close to the reported pKa of imidazole (6.95), so 6.4 is at the imidazole ring.

That leaves pKa 2.3 (or 3.64) for the other basic nitrogen. While that nitrogen is part of a piperazine, piperazine itself (pKa 9.8) isn't a good comparison. The nitrogen in question is also conjugated to an aromatic ring, so it would be better to look at aniline, with a pKa of 4.6. That's much closer to ketoconazole. Add in the inductive effect of the proximal amide, and a pKa in the 2-3 range makes sense.

Many common pKa's can be found on Prof. Dave Evans's pKa list.


Calculating Equilibrium Partial Pressures Given Kp and Mass For the reaction
  $$\ce{AsCl5(g) <=> AsCl3(g) + Cl2(g)}$$
  at $550$ K, the equilibrium constant ($K_p$) is $9.81$. Suppose that $3.150 \ g$  $\ce{AsCl5}$ is placed in an evacuated $600$ ml bulb, which is then heated to $550$K.

  What is the partial pressure of $\ce{AsCl5}$ at equilibrium?

So at the moment, I understand that
$$\ce{K_p=$$\frac{(P_{AsCl3})(P_{Cl_2})}{(P_{AsCl_5})}$$}$$
And that using a rearranged Ideal Gas Law I can get
$$\ce{P=$$\frac{n(0.0125mol AsCl5)*R*T(550K)}{V (0.6L)}$$}$$
Giving me $$\ce{0.94 atm}$$
but I'm not sure where to go from here. Can somebody help by pointing me in the right direction?

Say you start with $A$ moles of $\ce{AsCl5}$, and then you reach a temperature $T$, for which the equilibrium constant is $K_p$.

In order to achieve equilibrium, some amount of forward reaction ($\ce{AsCl5 -> AcCl3 + Cl2}$) happens to consume $x$ moles of $\ce{AsCl5}$. Based on the stoichiometry of the equation, you know that exactly $x$ new moles of the products are formed.

$$K_p=\frac{P_{\ce{AsCl3}}P_{\ce{Cl2}}}{P_{\ce{AsCl5}}}$$
Since for $V$, $R$, $T$ are constant, we can replace the above equation by:
$$K_p=\frac{RT}{V}\frac{n_{\ce{AsCl3}}n_{\ce{Cl2}}}{n_{\ce{AsCl5}}}=\frac{RT}{V}\frac{x^2}{A-x}$$
Solve this for $x$, and substitute in the ideal gas equation to get the answer.


Electronic configuration of period 3 element from its successive ionization energies Given below are the first five successive ionization energies $(\mathrm{IE})$ in $\pu{kJ mol-1}$ of an element in period 3.

  $$
\begin{array}{ccccc}
\hline
\mathrm{IE_1} & \mathrm{IE_2} & \mathrm{IE_3} & \mathrm{IE_4} & \mathrm{IE_5} \\
\hline
801 & 2427 & 3659 & 25022 & 32822 \\
\hline
\end{array}
$$

  Deduce the electronic configuration of the valence energy level of the element.
  Explain how you came to this answer.

I need clarification on this question. I have explained into using my own knowledge, therefore correct me if am wrong.

I believe that the element in period three would be Al, since there is a big jump from $\pu{3659 kJ/mol}$ to $\pu{25022 kJ/mol}$. For the first three electron $\mathrm{(3s)^2(3p)^1}$, they are all in the same energy level which is the third shell and that takes $\pu{801 kJ/mol} + \pu{2427 kJ/mol} + \pu{3659 kJ/mol}$ to remove them, after that we notice a big jump after the third ionization energy indicating the its entering a new level of energy which is the second energy level.

Please advice me if its correct or not.

Yes, your reasoning is logical and makes sense, but the question must have a typo because those IE values are the exact match for boron B, an element from period 2 right above aluminium Al (data from [1, p. 10-204], converted from $\pu{eV}$ to $\pu{kJ mol-1}$):

\begin{array}{lccccc}
\hline
\text{Element} & \mathrm{IE_1} & \mathrm{IE_2} & \mathrm{IE_3} & \mathrm{IE_4} & \mathrm{IE_5} \\
\hline
\ce{B} & 801 & 2427 & 3659 & 25022 & 32822 \\
\ce{Al} & 578 & 1816 & 2745 & 11580 & 14842 \\
\ce{Ga} & 580 & 1980 & 2965 & 6102 & 8299 \\
\ce{In} & 559 & 1821 & 2704 & 5210 & \\
\ce{Tl} & 589 & 1971 & 2878 & & \\
\hline
\end{array}

Another hint that it's not aluminium is the absolute value of the jump in energy between $\mathrm{IE_3}$ and $\mathrm{IE_4}$, which is too high for latter group 13 elements.
Probably the author meant group 13, not period 3, but that's just a speculation at this point.

References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data.; 2017; Vol. 97. ISBN 978-1-4987-5429-3.

Appendix

Source code for the plot, pgfplots:

\documentclass{standalone}
\usepackage{pgfplots}
\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
    \begin{axis}[
        title={Successive ionization energies of group 13 elements},
        xmin=1, xmax=5,
        ymin=0, ymax=35000,
        xlabel = $i$,
        ylabel = {$\mathrm{IE}_i$/\si{\kilo\joule\per\mole}},
        minor tick num = 1,
        mark = *,
        grid,
        legend cell align = {left},
        legend pos = north west,
    ]

    \addplot coordinates {
        (1,801)
        (2,2427)
        (3,3659)
        (4,25022)
        (5,32822)
    };
    \addlegendentry{Boron}

    \addplot coordinates {
        (1,578)
        (2,1816)
        (3,2745)
        (4,11580)
        (5,14842)
    };
    \addlegendentry{Aluminium}

    \addplot coordinates {
        (1,580)
        (2,1980)
        (3,2965)
        (4,6102)
        (5,8299)
    };
    \addlegendentry{Gallium}

    \addplot coordinates {
        (1,559)
        (2,1821)
        (3,2704)
        (4,5210)
    };
    \addlegendentry{Indium}

    \addplot coordinates {
        (1,589)
        (2,1971)
        (3,2878)
    };
    \addlegendentry{Thallium}

    \end{axis}
\end{tikzpicture}

\end{document}

Converted to PNG with the following command:

magick convert -density 120 G13_ionization_energies.pdf -quality 100 G13_ionization_energies.png


Preparation of sulfonyl chlorydes and mechanism of preparation of tosyl chloride After searching about the preparation of sulfonyl chlorides, I couldn't find much. I found that for preparation of tosyl chloride, chlorosulfonic acid can be used, as in:

Can chlorosulfonic acid be used to prepare sulfonyl chlorides in general? What are some other ways to prepare sulfonyl chlorides? If there is any.

The other part of the question, I'd like to know how the mechanism of this reaction would work. Having the equilibrium of the reaction:

$$\ce{3 ClSO2(OH) <=> SO2Cl+ + 2 SO3Cl- + H3O+}$$

How exactly is the $\ce{SO2Cl}$ formed? I didn't quite understand how the $\ce{Cl}$ became a cation. Apparently the benzene attacks that cation in order to form the desired product.

The way to understand this is to think of the autoprotolysis of water:

$$\ce{H2O + H2O <=> H3O+ + OH-} \tag{1}$$

In general autoprotolysis can be represented in the following form (you can recover eq. (1) by setting $\ce{A} = \ce{OH}$):

$$\ce{HA + HA <=> H2A+ + A-} \tag{2}$$

Now set $\ce{HA}$ as $\ce{ClSO3H}$ (or $\ce{(HO)SO2Cl}$, for clarity) and you have

$$\ce{(HO)SO2Cl + (HO)SO2Cl <=> (H2O+)SO2Cl + ^-OSO2Cl} \tag{3}$$

The cation produced in eq. (3) fragments to give water and the reactive chlorosulfonyl cation:

$$\ce{(H2O+)SO2Cl -> ^+SO2Cl + H2O} \tag{4}$$

The water in eq. (4) reacts with a third molecule of acid in an ordinary acid-base reaction:

$$\ce{H2O + (HO)SO2Cl -> H3O+ + ^-OSO2Cl} \tag{5}$$

The combined equation you have is simply the sum of eqs. (3), (4), and (5).


Can methanol be oxidised by PCC? Pyridinium chlorochromate (PCC) in $\ce{CH2Cl2}$ can be used to oxidise primary alcohols only to aldehydes. I can't however, find any reference of it oxidising methanol (which is strictly speaking not a primary alcohol). My question is, does PCC oxidise methanol to formaldehyde and if not, what can I use to achieve this reaction?

Look at the examples given here:

(Ignore the stuff written with pen; I have a habit of taking notes when reading)

PCC is taking two hydrogen atoms – one from the carbon atom attached to the -OH group and the other one from the -OH group itself, and then making a C=O double bond. Methanol has both of these. So, PCC should be able to oxidise it to formaldehyde.


Diastereomers - cyclic compounds Why in this problem there isn't a diastereomer with the $\ce{-CH3}$ in the back (dash) and the $\ce{-CH2CH3}$ in front (wedge)?

To answer this question I would first assign the R-S configuration to the original molecule and the molecule proposed by you and show you that both are identical. At the end I will also show you a simpler way to tell that both are identical.
Original molecule:

Proposed molecule:

As you can see, both the molecules have the same R-S configuration and thus are identical.
Now to see this easily and intuitively just assume you are looking at the original molecule from inside the screen. What do you see? You will observe that the original molecule will look just as same as the proposed molecule which means that both are identical. This method works because our frame of reference doesn't change the optical properties of a molecule until and unless we change the configuration of the molecule by doing some reflections.


Nuclear fusion reactions violate law of conservation of mass, but it has not yet been taken into account. Why? According to the law of conservation of mass, during any physical or chemical changes, the total mass of the products is equal to the total mass of the reactants. But in nuclear fusion reaction if considered an example, energy that sun emits in its core is due to collision of hydrogen nuclei and formation of helium nuclei. Here conservation of mass is not obeyed as certain part of mass is converted into energy. So, law of conservation of mass is violated here. Then law of conservation must be redefined as: during any physical or chemical change, the total mass of the products is equal to the total mass of reactants provided mass has not undergone conversion into energy.

Nuclear reactions appear to violate both the Laws of Conservation of Mass and Energy because mass is converted into energy or vice versa.

However, the concept of mass-energy equivalence that emerges as a consequence of Einstein's General Theory of Relativity makes the Laws of Conservation of Mass and Energy special limiting cases of the Law of Conservation of Mass-Energy. Since mass is now a form of energy, we can convert back and forth as long as there is no net loss between the two.

The equivalence of mass and energy comes from Einstein's famous $E=mc^2$, where $c$ is the speed of light (in a vacuum).


Standard reference for thermodynamic properties for octane where can I find thermodynamics properties for octane at saturated Temp.$T_s$?
I need this properties or where can i find all ?

$$liquid-density=\rho_l=?$$
$$vapor-density=\rho_v=?$$
$$latent-heat-of-evaporation=h_fg=?$$
$$liquid-kinematic-viscosity=\mu_l=? $$
$$liquid-conductivity=k_l=?$$

The NIST Chemistry Webbook maintained by the National Institute of Standards and Technology is a freely searchable database of chemical and physical data on a vast array of substances.

Everything you want for octane can be found there.


Why is beryllium transparent to x-rays? Beryllium has other fancy applications. It is transparent to x-rays, so it's used in the windows of x-ray tubes, which need to be strong enough to hold a perfect vacuum, yet thin enough to let the delicate x-rays out.

The quote is from the book The Elements by Theodore Gray. Why is beryllium transparent to x-rays?

The amount of X-rays absorbed by an element depends on the size of its atoms (its absorption cross-section, specifically, as affected by the size of core orbitals that contain electrons that can be excited by X-ray absorption, and the number of electrons in those orbitals) and how many atoms are packed into a given volume. Big atoms that are close together can absorb more X-rays than small atoms that are far apart.

Beryllium has very small atoms (it has a very small electronic core, with only two electrons in it), and very low density, so it will absorb relatively few X-rays compared to other materials.

According to this page, 37 % of X-rays with a wavelength of 0.56 nm will be absorbed by beryllium window that is 23.5 mm thick. If the window were made of aluminum, it would only have to be 1.4 mm thick to absorb the same amount of X-rays, and if it were made of lead it would only have to be 13 μm thick.


Finding major product when bromonium compound is acted upon by a nucleophile The original question is:-

  The major product obtained in the following conversion is:-

(Note that Me here means methyl group)

The initial step would be the formation of a bromonium intermediate. Which would then be attacked by a nucleophile $\ce{CH_3OH}$ as it is present in excess in solvent.

What I think is, $\ce{CH_3OH}$ would attack the carbon atom close to ketone, because that’s the carbon best able to stabilize positive charge due to resonance with keto group. This is what I think will happen

But the proposed answer was

Can anyone explain me why how and why this compound is formed?

I'd like to point out that the ketone would actually disfavor an attack at its alpha position:

Placing two positive charges on adjacent carbon atoms is much less favorable than not, especially when there is an oxygen perfectly poised to help stabilize this positive charge, as user55119 pointed out.