How do sodium-potassium pumps differentiate between sodium and potassium? I don't know if this should be Biology, but it seems pretty Chem to me.

Does the protein of the sodium-potassium pump tell the difference between sodium and potassium cations based on size, electron affinity, mass...?

$\ce{Na^+-K^+}$ ATPase (more commonly called the$\ce{Na^+-K^+}$ pump) is a transmembrane protein that consists of two types of subunits: The 1000-residue  $\alpha$ subunit contains the enzyme’s ATP and ion binding sites, and the 300-residue $\beta$ subunit facilitates the correct insertion of the  subunit into the plasma membrane.

The  subunit of this $160 \ Å$ long protein consists of a transmembrane domain (M) composed of 10 helices of varied lengths and, and three cytoplasmic domains, namely: the nucleotide- binding domain (N), which binds ATP; the actuator domain (A), so named because it participates in the transmission of major conformational changes; and the phosphorylation domain (P), which contains the protein’s phosphorylatable Asp residue.

The $\ce{Na^+-K^+}$–ATPase is an antiport that generates a charge separation across the membrane, because for every three positive charges exit the cell for every two that enter.

The key to how it functions lies in the precise phosphorylation of a specific Asp residue of the transport protein. ATP phosphorylates the transporter only in the presence of $\ce{Na^+}$, whereas the resulting aspartyl phosphate residue is subject to hydrolysis only in the presence of $\ce{K^+}$.

Thus, the "pump" has two conformational states (called E1 and E2) with different structures, different catalytic activities, and different ligand specificities.

Outline of the entire process

E1.ATP, which acquired its ATP inside the cell, binds three sodium ions to yield the ternary complex $\ce{E1.ATP.3Na^+}$.
The ternary complex reacts to form the “high-energy” aspartyl phosphate intermediate $\ce{E1-P.3Na^+}$.
This “high-energy” intermediate relaxes to one lower in energy, i.e $\ce{E2-P.3Na^+}$, and releases its bound sodium ions outside the cell.
$\ce{E2~P}$ binds two potassium ions from outside the cell to form an $\ce{E2-P.2K^+}$ complex. (structure pictured earlier)
The phosphate group is hydrolyzed, yielding $\ce{E2.2K^+}$.
$\ce{E2.2K^+}$ changes conformation to E1, binds ATP, and releases its two potassium ions inside the cell, thereby completing the transport cycle.

This enzyme appears to have only one set of cation binding sites, but they changes both their orientation and their specificity during the course of the transport cycle. The difference in specificity the two conformations have arises principally because the two ions differ in size.

The illustration given below, shows the ion binding site in this ATPase and one can easily note the difference in positioning of residues between the two conformations.

Particularly, I would like to call your attention to figure(s) a, b and c. The purple spheres represent sodium ions, and the green ones correspond to potassium ions. Fig a and b represent the same view of the ion binding site, and one can clearly see the slight positional differences of the residues which account for the difference in specificity.

Image Source(s):
Voet and Voet, Biochemistry
and http://www.nature.com/nature/journal/v502/n7470/full/nature12578.html


Entropy, enthalpy, and reversibility In my introductory chemistry course, we learned that for reversible processes, entropy is defined as $ΔS = q/T$. However, we did not spend that long on this before moving on to other things like Gibbs free energy. My questions are:
1. What exactly is a reversible process?
2. Why must we use things like standard molar entropies to calculate $\Delta S$ instead of just using $\Delta H$ according to this equation?

Your text will best explain reversibility.  The short answer is that a reversible process is one whose direction can be reversed by a microscopic change in external conditions.

For example a gas in a cylinder expanding against a piston will do so reversibly as long as the external pressure and the internal pressure are essentially identical.  Of course for expansion the internal pressure must be infinitesimally greater than the external pressure.

Next, entropy is really only defined for systems at equilibrium.

Next, the equation for the entropy above is not correct except at constant temperature.  In reality it is $dS = dq/T$, which generally must be integrated in order to determine the entropy by this method.

For the second question, there is some confusion here. $\Delta H$ is equal to $q$ only at constant pressure.

In fact there are lots of ways to compute the entropy change in a process, but this space is too small to go into them.  They should be in your textbook.


How pure gold is obtained back from dissolved state in Aqua regia? I know Aqua regia is used for purifying gold by dissolving gold in it, but how that dissolved gold is obtained back in the pure state?

I searched everywhere, but I didn't get the answer. So, I am asking here.

By electrolysis. In general electrolysis is a very common way of purifying metals if you have them dissolved, since you can selectively control the reaction by adjusting the potential current. Al and Cu are also obtained through electrolysis of their salt solutions.
Electrolysis is a redox reaction that occurs in the non spontaneous direction driven by a external potential. In electrolysis we have two electrodes (a cathode and an anode) submerged in a solution with the salt of the metal we want to purify. As it always happens in electrochemistry, the reduction reaction will occur in the cathode, while the oxidation in the anode.
Since in a gold solution obtained through Aqua Regia we have the gold forming a tetrachloride complex (AuCl4), the oxidation reaction we will have will be the production of chlorine gas from the chloride. So the global redox reaction that will be occurring will be:
$$\ce{2[AuCl4]- + 6e- -> 2Au + 8Cl-}$$
$$\ce{6Cl- -> 3Cl2 + 6e-}$$
And if you look at the standard potentials of the two reactions in this Wikipedia webpage you can see that you just need a 0.43 V external potential for the reaction to occur. The negative electrode should be made out of gold and it will grow during the process by the deposit.
Industrially this process is done by having hundreds of electrodes connected in series and submerged into the gold salt solution. In this image you can see a factory for Cupper purification.

You can see from the electrodes lifted by the crate that they have copper lamina that grow during the process. Electrolysis is normally used when you want very high purity in your product, like gold and copper destined for electric applications in which any oxygen impurities greatly decrease conductivity.
Another great thing about electrolysis is that it can be used to easily apply a thin layer of the metal over a piece, if the piece is conductive itself. This is how gold coated audio jacks are produced, by using the audio jack adapter as the cathode.


In pig iron, from where does carbon in it comes from? I've read that:

  "During extraction of iron another impurity appear which is carbon from coke in furnace."

However, I don't understand how there will be enough unreacted carbon that would blend with iron. As reaction of carbon and oxygen is very exothermic so there must be enough energy for all carbon atoms to react with oxygen.

A blast furnace is made of successive layers of charcoal and iron oxide mineral. In the beginning, the lower layer of charcoal is lit and the fire is maintained by air being pressed through the burning matter. The combustion produces first $CO_2$. But as the temperature is high and the charcoal is in excess, $CO_2$ soon reacts with $C$  producing $2$ $CO$  according to the equations :$$\ce{C + O_2 -> CO_2}$$ $$\ce{CO_2 + C -> 2 CO}$$ Then the CO gas touches the bits or pieces of mineral, which produces the following equation :$$\ce{3 CO + Fe_2O_3 -> 2 Fe + 3 CO_2}$$ This produces some metallic iron which is liquid if the temperature is high enough. This liquid flows between the bits of burning charcoal, and dissolves some charcoal during this flow. The obtained pig iron contains about 5% carbon. which must be later on burned in special furnaces, to get steel (containing about 1% dissolved carbon)


Why are all the phenyl protons in benzyl alcohol equivalent in the ¹H-NMR spectrum? Below is a $\ce{^1H}$ Nuclear Magnetic Resonance (NMR) spectrum for benzyl alcohol. The hydrogen in the alcohol group gives rise to a unique signal. The two hydrogens in $\ce{CH2}$ give rise to a unique signal with around twice the signal area. What I don't understand is why all hydrogens in the phenyl group are equivalent. My guess would have been there would be 3 signals from the phenyl ring. One pair of $\ce{H}$ that are equally distant from the methanol, another pair that is also equally distant, and last, a single hydrogen that is the furthest away from the methanol. Clearly this is not the case, and the 5 hydrogen give rise to a single signal only.

Why? What is a good way to think about equivalent and non-equivalent atoms in NMR spectrometry? I would rather avoid rules like the one that says "if you can substitute the hydrogen with another atom and it gives the same compound, the hydrogens are equivalent". I would rather like an intuitive way to think about it, if it's possible!

You're right, those five hydrogens are not magnetically equivalent. You would expect to see three signals for the aromatic protons.

If you look closer you'll see that the aromatic signal is not a perfect singlet, but is rather more complicated with an additional small peak and a broader base, it is a multiplet. You have three distinct signals, but they are so close in chemical shift that they overlap and result in this multiplet.


Are bi-cycles chiral molecules? I was studying how chirality influences the diastereoselectivity of reactions when I saw 1,4-dihydro-1,4-epoxynaphthalene.

I determine chirality by looking for C with 4 different groups. I see 2 C that accomplish this requirement (sp3 C bond to the O) but both are in a rigid structure, so I can't tell if they're (or not) chiral at all. If they are, can I use the same rules to determine R or S?

There are bi-cycles that are chiral, the one shown in the question is not. This can be shown, as pointed out in the comments, by superimposing the molecule on its mirror image, which is possible. Another fast way of ruling out chirality is finding the mirror plane within the molecule (in this case. perpendicular to the both cycles).


Effect of increasing product concentration on time of completion of reaction This was the question asked in our chemistry exam:

In the experiment of determination of equilibrium constant of esterification of ethanol, if , instead of using
clean dry bottle, if a bottle already containing slight amounts of ethanoate and water is used to carry out the
reaction under identical (remaining) conditions, what would be the consequence?

According to our lecturer it's answer is "The reaction will take lesser time to reach equilibrium". But according to Le Chatelier's principle, increase in concentration of products shifts the reaction backwards. So it should take more time to reach equilibrium.
Please help!!

In equilibrium reactions the equilibrium is approached exponentially in time. The lifetime of the exponential is the reciprocal of sum of the forwards and back rate constants multiplied by the equilibrium concentration of the species, thus for reaction $\ce{A + B \leftrightharpoons C + D }$ the lifetime of the reaction is $\tau = 1/[k_1 (A_e+B_e) + k_{-1} (C_e+D_e) ] $ where $k_1$ is the forward rate constant and $k_{-1}$ the backwards one, and subscript $e$ means equilibrium amount of a species. Thus if you add more material you change the equilibrium concentrations  and so the lifetime gets shorter as its a reciprocal in the concentrations

The lifetime is the time taken to change to $1/e$ of the initial amount, approx to 38%, whereas the half-life is the time taken to halve the initial amount. The lifetime is the reciprocal of the total rate constant for reaction which is why it is used instead of the half-life.

(The exception is the simple $\ce{A\leftrightharpoons B}$ reaction when the lifetime is independent of the equilibrium amounts ). In equilibrium reactions the lifetime is also called the relaxation time.


Refluxing with tilted condenser Can you reflux without any problems/dangers with a tilted condenser (e.g. because you don't have enough space in your fumehood)? I would say there isn't a problem, because both the surface area of the refluxed liquid and the length of the condenser (and thus the length of the trajectory of the water flowing through the condenser) remain unchanged. Is this thought correct?

This is my current setup

Also you should use countercurrent exchange right?

As long as everything is clamped and balanced - so that there isn't excessive weight on the joint connecting the flask to the condenser - the system should be OK.  After all, many rotary evaporators have the flask plus condenser tilted at an angle.


The Compressibility of a substance, dependent on My Chemistry book states that (talking about compressibility equation):

  "A sample with Volume V is subjected to a pressure increase, delta P,
  and the resulting change in volume, change in V, is measured.  The
  ratio is divided by V; thus, the tabulated value depends only on the
  substance being measured and not on the geometry of the sample."

I'm not sure exactly how they are drawing this conclusion.  How does dividing by (-) volume cause the compressibility factor to not be dependent on the "geometry" of the sample?

Thanks.

If you apply an external pressure on an object and analyze the deformational response of the object (say using Hooke's law in 3D), you find that the state of stress and strain within the object are completely uniform and independent of direction within the object (if the object has direction-independent properties to begin with), irrespective of the shape of the object.  This is sort of the same kind of thing that happens in thermal expansion, where the object strains uniformly in all directions, and the strains are independent of the object shape.


What is the meaning of 'mass action'? I understand the statement of 'Law of mass action', but I don't understand clearly what 'mass action' means.

I investigated it and found the following descriptions.

The law was originally suggested as a law about 'active mass'. [from Wikipedia]
The term 'active mass' is a historical term that professional chemists do not use anymore. 'Active mass' meant 'amount per unit volume'. [from Chemistry Stack Exchange #58141]

From these explanations, I understood that 'active mass' corresponds to 'concentration' in modern words, which is usually expressed in the unit 'mole per liter' today.

However, a question still remains for me. What does the term 'mass' intend to originally? I am not a native English speaker, so I cannot understand the nuance well.

The word 'mass' has some different meanings.

a physical quantity expressed by the unit kilogram,
a large amount of a substance,
a lot of people, etc.

Of all these meanings, which do native English speakers understand the word 'mass' as?

Background of this question

In Japan, 'Law of mass action' has been translated as '質量作用の法則' (shitsuryou-sayou-no-housoku). '質量' corresponds to 'mass'. Nowadays, however, this is accused as a mistranslation because the law states the nature of concentration and it does not relate to mass. On the other hand, if native English speakers also understand the word 'mass' as 'a physical quantity expressed by the unit kilogram', the translation is not a 'mistranslation' but just a historical word originally. So I would like to consult those who are familiar with both chemistry and English.

"Mass" in this context refers to the large number of particles and consequent behavior in a stochastic manner, as opposed to a small set of particles, where Maxwell's demon might have been able to influence outcomes.Of course, Maxwell didn't know know of Heisenberg's uncertainty principle.


Melting point of crystalline solids Why crystalline solids have sharp melting points but melting point of amorphous solids vary within a range?

Is it because the interaction energy is equal between atoms of crystalline solids so they require equal energy to break the bond or is it because the atoms of crystalline solid are similar and are equally distant so energy gets transferred to every molecule equally so all the bonds break simultaneously.

Remember that temperature is a macroscopic quantity, telling the heat content of system. So if the crystalline solid is at its fusion temperature, and every atom in the crystals have equal interaction energy with each other (ignoring all defects of lattice), the whole solid should fuse simultaneously, thus having sharp melting point.
On the other hand, amorphous solids don't have uniform lattice and interaction energies of atoms/molecules may differ. So even at the same temperature, the whole solid might not melt, so we say amorphous compounds melt at range of temperatures.


Forging steel inside a zero oxygen environment Intro: I have watched one too many blacksmithing videos and now I have a question: is it possible to forge steel (either just reshape or "forge-weld") inside a no-oxygen environment?

Scenario: considering the amount of material that turns into iron oxide during the forging process, would it be possible (or plausible) to fill a room with some non reactive noble gas (maybe helium or neon) and proceed to forge inside this room? Or would it be possible to forge in a complete vacuum?

If any of these two options are possible, what are the hindrances and requirements?

Extra info: I am not from the chemistry area and I haven't  studied chemistry in years. I took a look at some old textbooks of mine and also did some research on on Google but did not find anything specific to this question. This is purely out of curiosity, and since I don't know how reddit works (therefore can't do a more relaxed inquiry there) , I thought it was not a bad idea to ask here.

It is possible to forge iron in a low atmosphere environment, but it isn't worth it because there are few if any benefits to doing so.

This isn't true for some other metals, though, or even for some specialist steels. In many other metals and alloys the presence of oxygen or water vapour has a very bad effect on the quality of welds. But the common way to exclude air from these systems is to use an apparatus that flows some inert gas around the spot being welded rather than to fill the room with an inert gas. Some examples are given here.

This is mainly worth doing for many specialist aluminium, copper and titanium alloys where the metals are oxidised more easily than typical irons or steels and effect of impurities in the welds is much larger. It isn't as simple as just having an inert gas blanket though, as this brochure for commonly used gas mixtures shows.


The rusting of steel may be promoted by brushing with HCl followed by brushing with hydrogen peroxide solution. What is the chemistry involved? A welder showed me a way of promoting rust on steel by brushing on muriatic acid and then a hydrogen peroxide / water solution. What's going on, chemically speaking?

I understand the point of your question, here is a discussion of the associated science.
The process remains a classic electrochemical process where here H2O2 is a replacement for oxygen and the require $\ce{H+}$ is provided by the $\ce{HCl}$ which is also a source of chloride ions noted to promote corrosion as well as forming salts which are excellent electrolytes.
Some references, starting with a good depiction of the underlying electrochemistry, to quote:

Iron (or steel) rusting is an example of corrosion -- an electrochemical process involving an anode (a piece of metal that readily gives up electrons), an electrolyte (a liquid that helps electrons move) and a cathode (a piece of metal that readily accepts electrons). When a piece of metal corrodes, the electrolyte helps provide oxygen to the anode. As oxygen combines with the metal, electrons are liberated. When they flow through the electrolyte to the cathode, the metal of the anode disappears, swept away by the electrical flow or converted into metal cations in a form such as rust.

And some comments from ScienceDirect on the referenced chloride properties:

The chloride ions present in the water promote local corrosion such as crevice corrosion and pitting, and the higher the chloride content, the more corrosive the water.

I hope this helps.


How to clean a copper pan from black burned fat chemically? I have several copper pans which I use daily. I don’t have the time to always clean them. Now they have a thick crust of black burned fat. When I wash it with dish soda, nothing changes. And I don’t want to use abrasive mechanics (steel sponge), because I am afraid this will scratch the copper surface.

Is there a chemical trick to dissolve the black crust? What is that black crust chemically and why is it so strong?

The crust is a giant mess of hydrocarbons that have been burned, caramelized, and congealed into the surface of the pan.

I would recommend that you try the steel sponge to get some of the crust off, you may find that as you start breaking into the top layers of crust that it will fall apart. Another strategy would be to put the pan in the oven to the highest heat that is safe for the cookware and this might cause the pan to expand enough to create cracks in the crust or release it from the micro-cracks that were already in the pan surface. As a last resort you will have to chisel it off slowly.


Why are Friedel Crafts reaction not possible for molecules less activated than benzene? I know that Friedel Crafts alkylation reaction are not possible for aniline and phenol since they form complexes with Lewis Acids. But our teacher told us that Friedel Crafts reactions aren't possible for any benzene derivative which is less activated than benzene. Why does that happen?

Technically, the teacher's claim might be slightly modified.  Deactivated benzenoid aromatics do fail to undergo Friedel-Crafts reactions, but according to a summary description from the University of Calgary:

Deactivated benzenes are not reactive to Friedel-Crafts conditions, the benzene needs to be as or more reactive than a mono-halobenzene.

So, a monohalogenated ring which is only weakly deactivated, can still react.  However, stronger deactivation, for example with an acyl group, would kill the F-C reaction.
This property can actually be quite useful when we need to do the F-C reaction selectively.   Using acyl as the deactivating group, we can acylate an aromatic ring and be sure the reaction occurs only once.  The ring becomes too deactivated to react further.  In principle this can then lead to a monoalkylated product with good yield by subsequently reducing the acyl group, whereas direct alkykation which activates the ring could give polyalkylated products.


Energy at constant Entropy and Volume The system of constant entropy and constant volume will attain the equilibrium in a state of minimum energy. Why?

I know that:

$\Delta G= \Delta H-T\Delta S$
And at constant volume and entropy: $\Delta G =\Delta U$

But from above equations how can we justify the give statement?

A function such that the equilibrium state of a system is determined by its maximum (or minimum) is called a characteristic function. There is a whole bunch of those. All they are ultimately derived from the Second law after some fiddling around with differentials.

The said law basically postulates that entropy is one such function. But it is only good for closed systems, that is, those with constant $U$ and $V$. Other kinds of systems use other kinds of characteristic functions. To derive them, we typically go along the lines of Legendre transformations. Say, you have $f(x,y,z,\dots)$, which is a characteristic function in the variables $x,y,z,\dots$, and you want to switch from $x$ to $f'_x$ to use it as your independent variable. Then $f-xf'_x$ is a characteristic function in your new variables.

So let's begin with switching from $U$ to another variable. The new characteristic function is
$$S-U\cdot\left({\partial S\over\partial U}\right)_V=S-{U\over\left({\partial U\over\partial S}\right)_V} = S-{U\over T}$$
(Cryptic as it might seem, the identity $\left({\partial U\over\partial S}\right)_V=T$ is true, and moreover, is the definition of temperature.)

To bring this characteristic function to a more familiar appearance, we switch the sign (so we'll be looking after a minimum, and not maximum, as we did with entropy), then multiply it by $T$, since it is only going to be used in systems with $T=const$ anyway, and get the Helmholtz energy $F=U-TS$. That's what determines the state of equilibrium for fixed $(T,V)$.

Now let's make the final step. Luckily for us, $-S$ is the thermal derivative of $F$, hence the characteristic function in variables $(S,V)$ is
$$F-T\cdot\left({\partial F\over\partial T}\right)_V=F-T\cdot(-S)=F+TS=U$$
just as we were seeking to prove.

So it goes.

One more variable change from $V$ to $P$ will give us Gibbs energy, but that's another story.


Confusion with the Periodic Table The periodic table has 7 periods and they have 2,8,8,18,18... elements respectively from 1 to 7. But from what I understand, the periods each state the number of electron shells that the elements in that period has. So if that is the case, shouldn't period 3 have more elements, since it can hold up to 18 electrons, and therefore it can have up to 18 more protons from the largest atomic number element in period 2? (Since period 3 has a M electron shell, it can also have a p orbital, and in total therefore can have 18 elements.)

The same could be said for elements in group 4, group 5, group 6 and so on since they can have f,g, and h orbitals as well. Am I missing something here?

So if that is the case, shouldn't period 3 have more elements, since
  it can hold up to 18 electrons, and therefore it can have up to 18
  more protons from the largest atomic number element in period 2?

Indeed, elements of the same period have the same number of electron shells, but the "problem" is that in accordance with the Madelung/Janet/Klechkowski  rule, the $\mathrm{4s}$ orbital is occupied before the $\mathrm{3d}$ orbital. As a result, the $\mathrm{3d}$ orbital could be filled only when the $\mathrm{4s}$ orbital is already filled, i.e. only for elements of the 4th period.

Similarly, the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are occupied before the $\mathrm{4f}$ orbital. As a result, the $\mathrm{4f}$ orbital could be filled only when the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are already filled, i.e. only for elements of the 6th period.


BPS & BPA treatment with 80% Ethanol Background: BPA and BPS (Bisphenol A/S) are artificial estrogens and therefore toxic chemicals for us. They are used in the coating of thermal paper. Thermal paper is used to print in-store receipts. There is a growing demand for thermal paper that is free of BPA and BPS for the health and safety of the public.
Question: how do we test it? I heard from someone that both BPS and BPA dissolve in ethanol. If we apply 80% Ethanol solution (commonly found in hand sanitizer) on the coating side of the thermal paper then it reacts (and dissolves) with BPS and BPA; as a result, you notice black and greenish-black spots start to appear on the paper. If the paper is free of BPS/A then no spot or marks will appear on coat-sided paper.
Is this method of testing correct? Does it at least give some accurate idea? What could you suggest for testing?

I agree with M. Farooq, a test that might work is cutting small pieces of thermal paper and placing them in a tube with grain alcohol or a facsimile, and allowing them to soak for a few hours, then send the sample to a lab and ask them to test for bisphenol A/S. If you can, it's just better to ask the company.
Edit: you can just soak them in grain alcohol. It's hard to do a specific test for bisphenols, but I found one for phenols in general.
Procedure:

Prepare a dilute solution of ferric chloride in grain alcohol.
Using an eyedropper, add a few drops of the solution to a teaspoon of your solution to test.
If a color change is observed, phenols are present.

This is not ideal, but it's the best you can do.


Gaussian scan function help for constructing input file I have the cartesian coordinates for two molecules that are in a stacked position and I want to be able to calculate the potential energy at 0.2 angstrom steps from when the top molecule is 3.0 angstroms to 7.0 angstroms away in the z direction. My professor told me gaussian has a scan function to do this but I'm having trouble figuring some things out. I have the basis set/ correlation methods I need to use and the cartesian coordinates for all the atoms when they are arranged 3.0 angstroms apart. I tried following this site's advice;

 but I'm not sure I constructed the input file correctly. Here is what I have

.

From my understanding, this should just move up all of the atoms of the top molecule by 0.2 angstroms in the Z direction.

Questions:

Do I just throw this stuff anywhere into the input file? What about the original cartesian coordinate array? Surely it has to stay since the "step" array calls back to atoms 101-112. Do I have to change any other commands? The only parameters set are the resource limits and "#P BP86/6-311++G**"

Update:

I think I may have been unclear. I have already optimized my two molecules. One is a nanotube and the other is an arene and my task was to basically generate this plot. I'm trying to find the distance apart where the two molecules are stabilized. Before, I just ran the "#P BP86/6-311++G**"" with my arene @ 3.0 angstrom apart and made new input files changing the arene z coordinates in 0.5 angstrom increments but I can't capture that potential energy well. I decided to try smaller increments to increase the resolution but I have quite a few more situations to test so I wanted to just have Gaussian change the increments and recalculate the potential energy.

  Thankfully a full relaxed scan is not necessary in most cases. Oftentimes a nice little rigid scan will give you a decent result. Have a look at the scan keyword. The only downside to this approach is, that you need a z-matrix to make it work. And putting your brain to that task is also a bit tedious.

That's what I initially came across but with ~100 atoms, I just let Chemcraft generate the z matrix from my cartesian coordinates. The thing is, I'm not sure quite how to edit the z-matrix (relevant part below) with the scan commands.

Specifically I want to keep all bond angles the same but move atoms 101-112 in 0.2 A steps in the z direction, and calculate the potential energy.

I guess my question is now if this is a plausible approach. I'll try to look up how to construct the proper z-matrix if that is indeed the best way to do things.

Thanks @Martin

This is (probably) not going to work the way you intend it to work. Constructing a meaningful modredundant calculation is quite a difficult task, as you sometimes cannot anticipate what the program actually does. Using this interface you will perform a relaxed scan, i.e. you will step one variable, fix the values of the scanned coordinates, and run a full optimisation at this point. This allows you to systematically, brute force a potential energy surface. In you case you specify 12 coordinates to scan with 20 points each, resulting in a whopping $N=\frac{20!}{12!}\approx5\times10^8$ optimisations.

Here are a few pointers when constructing a modredundant input:

don't scan more than two parameters at a time, that roughly equates to a 3D graph of a PES
don't scan xyz coordinates, scan bond lengths and angles instead; scanning xyz coordinates results in fixing the coordinate system in parts disallowing the algorithm to rotate the molecule/ use coordination transformations to speed up the calculation
keep constant what you need not optimised; fixing things will result in a faster calculation

Let's make this a little bit more visual. As an example I'll choose $\ce{(CO2)2}$ and I am using the ridiculous low level of theory HF/3-21G.

First up optimise the equilibrium geometry.

%chk=opt.chk
#P HF/3-21G opt

title required

0 1
C       -1.345383226      0.000000000     -1.107741226
O       -0.793543854     -0.126832008     -0.084830947
O       -1.897222598      0.126832008     -2.130651504
C        1.651522395      0.000000000     -1.243964208
O        2.203361766     -0.126832008     -0.221053930
O        1.099683023      0.126832008     -2.266874487

The result will be the following geometry:

    6
scf done:  -373.128185
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

Now we want to move the two molecules closer together. If we would use the approach you have come up with, this would be really tedious. First we need to transform the coordinates in a way that we can understand which molecule to move where. Luckily this is not necessary at all. We can simply scan over the "bond length" of the carbon carbon distance. I chose five steps with an increment of -0.2 angstrom.

%chk=modered1.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2

That will give us the following graph and animation:

You can see that the OCO bond angles changes. The CO bond lengths changes, too, but that is not as visible. This is expected behaviour in a relaxed scan. Only one coordinate is fixed, which is the distance between the carbons. If you are interested in keeping at least the bond angles constant, freeze them:

%chk=modered2.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2
A 2 1 3 F
A 5 4 6 F

With the result, we'll see again something unexpected happening. There is a sudden rotation at 2.8 angstrom. For the given set of parameters, this is the lowest energy. This is a side effect of a relaxed scan. not everything will run smoothly and sometimes the calculation turns rogue and fails. It really is a delicate matter.

Just for shits and giggles, let's do one that will make us cringe. We will scan also both of the oxygen oxygen distances. Theoretically this should give us 20 data points. Spoiler alert: This will fail at step 8. Try it yourself:

%chk=modered3.chk
#P HF/3-21G
opt(modredundant)

title

0 1
 C    -1.546398     0.008833    -1.171593
 O    -0.884661    -0.109039    -0.227710
 O    -2.221473     0.124768    -2.098858
 C     1.852537    -0.008833    -1.180113
 O     2.527612    -0.124768    -0.252847
 O     1.190800     0.109039    -2.123995

B 1 4 S 5 -0.2
B 2 5 S 5 -0.2
B 3 6 S 5 -0.2

You can see, that with these small molecules it is already hard to predict what happens when you run the calculation. There are traps and pitfalls lining your way. More often than not it is trial and error.

Thankfully a full relaxed scan is not necessary in most cases. Oftentimes a nice little rigid scan will give you a decent result. Have a look at the scan keyword. The only downside to this approach is, that you need a z-matrix to make it work. And putting your brain to that task is also a bit tedious.

I have tried to convert the specifications to a z-matrix putting constants where needed and introducing the scan variable. I'm not sure I succeeded in the way I wanted to, but here we go:

%chk=scan.chk
%nproc=2
%mem=8000MB
#P HF/3-21G scan

title required

0 1
 c
xx   1  1.40000
xx    1  scan    2   90.000
 o    1 oc4         3   90.000      2  88.981
 o    1 oc5         3   90.000      2  -90.00
 c    3  1.93591    1   89.994      4 178.983
xx    6  1.40000    3   90.000      1  90.000
 o    6 oc8         7   90.000      3 180.000
 o    6 oc9         7   90.000      3   0.000

scan        2.794000  5  -0.20
oc4         1.152818
oc5         1.158751
oc8         1.152817
oc9         1.158750

The result is two rods, moving towards each other:

I expect that since you have more than 100 atoms, accomplishing a meaningful scan calculation is a very, very tedious task. It might be much simpler creating the points you want to scan by hand with a molecule editor. But without more information about the actual system, sound advice is difficult.


Why don't particles of a colloid settle? Why don't particles of a colloid settle but those of a suspension do?
- Is it because they are really small and so remain in constant motion which overcomes the force of gravity and prevents them from settling?
- Is it because they are charged and so they repel each other and stay suspended in the solution as they try to go away from each other?

Particles in colloid absorb either positive or negative ion preferentially from solution depending on there surface. This develops a potential on colloid surface called Zeta potential. Due to this potential colloid particles repel each other and can't coagulate.

Particle size also play a major role. With decreasing size surface to volume and hence surface to weigh ratio increases. So, same volume  of colloid  particles  have higher surface area than suspension. So, colloid surface have enough charge to keep them apart whereas suspension don't have.

Conclusion Both the point you've mentioned is responsible for prevention of colloid coagulation.


Finding compounds with B, Si, Ca, and Mg in differing amounts Not sure if my question is better suited to a software related group, but here goes:

Is there a software program that lets you add differing amounts of elements, minerals, etc., then tells you what the items added might make?  Specifically, I am looking for information about what boron, silicon, calcium and magnesium together (no idea about amounts for each) would make.

This would be an exceedingly difficult problem.

My suggestion might be to use a database like the Crystallography Open Database and perform searches with the elements you indicate.

For example, the search interface ensures B, Ca, Mg, and Si are all in the compound. I get 141 matches.

It would be incredibly hard to predict compounds with arbitrary stoichiometry. There are several genetic algorithm search packages that let one attempt to find low-energy crystal structures given a formula, including:

XtalOpt
GASP
USPEX

This is, however, an incredibly active area of research. I'm not sure if any of these packages can handle arbitrary amounts of different elements, or how they would handle missing elements. (For example, the elements you describe could be found in various borosilicate minerals given that oxygen is also allowed.)


Is sodium borohydride really a better base than an alcohol? I have learnt that sodium borohydride can reduce aldehydes, ketones and acyl chlorides into alcohols. However, it fails to react with esters or acid anhydrides.

On the other hand, alcohols — supposedly weaker nucleophiles — can successfully react with both acyl chlorides and acid anhydrides to give esters. Moreover, my book (Solomons and Fryhle) mentions that acidic conditions are not required for the mentioned reactions.

Why are alcohols able to do what sodium borohydride cannot, even in the absence of acidic conditions?

I have come up with a possible explanation, but I am not sure if it is correct: When an alcohol reacts with an acid anhydrides to produce an ester, a carboxylic acid is formed as a side product. Maybe this provides the required acidic conditions for further reaction?

In general, can a reaction be catalyzed by the formation of a side product initially not added in the reaction mixture?

General considerations

It’s not the basicity that is the important factor here, it is the nucleophilicity. The hydride ion, whether present as tetrahydridoborate or as a lone hydride is a very lousy nucleophile while an alcohol is an okay-ish nucleophile. The general mechanism of the addition is given in scheme 1.

Scheme 1: General mechanism of a nucleophilic attack on a carbonyl.

Now in the case of a hydride — a very weak nucleophile — this reaction is generally slow. It is faster, the more reactive — electrophilic — the carbonyl species is. If $\ce{X}$ is a carbon or hydrogen atom, then all we have is a localised double bond without any significant reduction in reactivity.[1] Reactions with those should thus be fast. If $\ce{X}$ is an element that has a lone pair, such as oxygen or chlorine, then there is a notable mesomeric stabilisation as shown in scheme 2.

Scheme 2: Mesomeric stabilisation by lone pairs next to a $\ce{C=O}$ double bond.

This stabilisation is not strong if a chlorine atom is $\ce{X}$ since chlorine is much larger and its lone pairs are more disperse. The stabilisation is stronger in the case of acid anhydrides; however, since a single oxygen has to stabilise two carbonyls it is still not too strong. In esters, carboxylic acids (and their anions) and most so in amides this stabilisation is very strong ad nucleophilic attacks difficult. This explains the general electrophilicity trend:

$$\text{aldehydes} > \text{ketones} > \text{acid chlorides} > \text {acid anhydrides} > \text{esters} > \text{amides}$$

Now let’s look at the reactions in detail. Throughout this post I will assume neutral conditions only; no acid or base catalysis.

The hydride attack

As I already noted, hydride is a bad nucleophile. It makes sense that it can attack the first members of the series — aldehydes, ketones and acid chlorides — but fails to do so for more stabilised members. For completeness, the mechanism of a hydride attack onto an acid chloride is given in scheme 3.

Scheme 3: Mechanism of a hydride attack onto an acid chloride.

We see another key point here: Once the carbonyl has been attacked, the oxide anion has a tendency to ‘rebound’ a lone pair and displace one of the bonded atoms. Generally, the displaced anion must be reasonably favourable for the rebound to occur. Chloride ions are good leaving groups and thus a displacement of chloride is likely to happen. Hydride ions are absolutely absymal leaving groups, so the hydride attack is not reversible.

After the inital attack and displacement of the chloride, we are left with an aldehyde which of course will react quicker and lead to full reduction to the alcohol.

Acid anhydrides are not attacked this way because the initial step does hot happen. If you accidentally got a hydride to attack, the $\ce{^{-}OOC-R}$ leaving group is great and thus a reaction would occur.

The alcohol attack

Alcohol hydroxy groups are still not the best nucleophile but they are better nucleophiles than hydride is. Thus, they even manage to attack acid anhydrides and — to a limited extent — esters and acids. The general mechanism of attack is shown in scheme 4.

Scheme 4: Reaction mechanism of an alcohol attacking an acid chloride including a potential second attack.

The key difference here is that the alcohol is a neutral nucleophile, not an anion like hydride is. Thus, when the rebound occurs, it is susceptible to redisplacement. This is shown in the final equilibrium of the mechanism above. This is also the reason why — under neutral conditions! — an alcohol may attack an aldehyde or a ketone but the attack is not productive. The alcohol will simply be rebounded back out since the hydride is a terrible leaving group.

There is a productive reaction mechanism for alcohols with aldehydes or ketones — the formation of acetals and ketals — however, that generally requires acid catalysis.

Finally, there is also the possibility of base catalysis when alcohols react. It would deprotonate them to give the alcoholate which is a much better nucleophile and even able to attack esters. However, base catalysis is outside of the scope of this question.


D/L and +/− labels of glucose I had just started to read about carbohydrates when I came across D-(+)-glucose. My book says that the '+' sign refers to its optical rotation measured through the polarimeter, and that it could either be + or −, but throughout the book D-glucose is always designated with the + sign.

Is it possible for there to be D-(−)-glucose?

For a given configuration there is a corresponding and fixed rotatory effect. However, the actual rotatory effect - that seen by a polarimeter - can either be + or −, independently of the D- or L-configuration of the stereogenic centre that shows up in the compound name (in the case of sugars, it refers to the carbon furthest away from the =O group).

In brief, D-glucose is always D-(+)-glucose.


Spectroscopic constants - explanation I'm a beginner in spectroscopy and I have troubles understanding papers describing values of spectroscopic constants. This paper, for example, describes constants $T_e, R_e, \omega_e, \omega_ex_e, B_e, \alpha_e, D_0$ and $D_e$.

I can see, that $R_e$ is an internuclear distance, but I have no idea, what are the remaining constants.

Could you provide some brief explanation? Preferably with picture of some potential energy curve(s), if possible, as I find it much easier to understand such concepts from schemes.

You can find plots and further explanations online (as I easily did). This is a homework-type question, yet I do feel that the information request for diatomic molecular spectroscopic constants warrants at least this partial answer. See this writeup for the "Vibronic Absorption Spectrum of Molecular Iodine" for more details and illustrations.

From NIST:

T$_e$ minimum electronic energy (cm$^{-1}$)

R$_e$ internuclear distance (Å)

ω$_e$ vibrational constant – first term (cm$^{-1}$)

ω$_e$x$_e$  vibrational constant – second term (cm$^{-1}$)

B$_e$ rotational constant in equilibrium position (cm$^{-1}$)

α$_e$ rotational constant – first term (cm$^{-1}$)

D$_e$ centrifugal distortion constant (cm$^{-1}$)

β$_e$ rotational constant – first term, centrifugal force (cm$^{-1}$)

From Wikipedia:

D$_0$ dissociation energy (cm$^{-1}$)


What are the boiling points of DMSO-water concentrations? What are the boiling points of DMSO-water concentrations with respect to wt% DMSO? I have found extensive information on the freezing point change of wt% DMSO but nothing regarding boiling.

The binary mixture of water and DMSO do not form an azeotrope (a constant boiling mixture, where the vapour produced has fixed percentages of each component). Thus, water will boil from the solution first at a bit above 100 ºC, followed by the boiling of DMSO at around 189 ºC.

The reason for the drastic change in freezing point is a little more complicated. Freezing point depression is a common phenomenon; for example, it's the reason sea water freezes at a lower temperature than pure water. In these cases, the reason for the freezing point depression are typically entropic.

However, in the case of DMSO and water, the depression is of particular note—mixtures can remain liquid down to -140 ºC, compared to the normal freezing points of 0 ºC and 19 ºC for water and DMSO, respectively. Quantum mechanical studies have suggested that the formation of DMSO-water clusters is at least partially responsible for the massive freezing point depression. These clusters are strongly hydrogen-bonded, but interact far less strongly with other clusters, making it more difficult to form a solid lattice (and thus a solid).

Kirchner, B.; Reiher, M. J. Am. Chem. Soc., 2002, 124 (21), pp 6206–6215


How is the light from a fire emitted? When you see a fire burning, you have wood (fuel) interacting with oxygen which (through a chemical reaction) gives off heat - and light.

But the light appears to emit at some point well-away from the actual fuel - i.e. the flames appear to rise inches (or more) above the actual fuel source where (I assume) the actual combustion/reaction is taking place.

I assume this obviously has something to do with the fact that heat rises - and there is oxygen up there, but there is no fuel up there.

I might conclude that the light is artifact of the elevated temperature of the oxygen, alone up there - but this wouldn't explain the fact that different elements burn different colors - i.e. color is related to the fuel.

So it would appear that the only logical explanations would be either:

a. Fuel becomes "detached" from it's (solid) fuel-source, rises with the heat and emits light after rising.

b. Light is emanated near the fuel-oxygen reaction but appears higher due to some sort of bending it it's travel path due to high heating of nearby air - like how light squiggles over a hot BBQ grill.

There is also the fact that in a very hot "coal" fire the light does emit from the fuel - but in a "roaring" fire the flames appear higher. (Is this because a "roaring" file has higher heat - which correlates to the difference in flame color?) Why would these flames appear "above" the fuel?

Okay - I hope I've illustrated my line of questioning sufficiently. So how does it work?

Light is emitted from flames by two primary mechanisms: one is small particles glowing incandescently because they are hot (the same mechanism that drives an incandescent light bulb); the other is from electronic transitions from specific energy levels in excited atoms in the flame produces as a by product of the combustion process (this is partly why flames can coloured specifically by elements present in the burning material).

But why do flames and their emissions appear so far from the primary materials that are burning? Clearly solids themselves can burn (eg charcoal in a barbecue is glowing from surface combustion of the carbon with atmospheric oxygen and the emission you see is because this makes the carbon hot enough to glow with heat).

But when things are that hot, plenty of other reactions can occur and this is particularly noticeable if the thing that is burning is less pure than charcoal (which is mostly carbon). Raw coal, for example, contains a lot of volatile impurities (smokeless coal is deliberately treated to reduce the volatile impurities). When the primary reaction of burning carbon happens in raw coal, those volatiles are often turned to gases and driven off the coal to combust as gases far from the solid coal. Of course this is happening at the same time as solid carbon in the coal is burning so the two effects are mixed up to give both glowing coal and gases burning some distance from the glowing coal. Old home chemistry sets used to contain experiments that allowed these effects to be separated. Raw coal is placed in a vessel connected to a tube allowing any emerging gas to be directed far away from the coal. The coal is heated with an external source (eg a Bunsen burner). The emerging volatile gases can then be lit without setting the coal on fire giving a flame a long distance away from the heated coal.

The same can be done with wood (which contains even more volatile and flammable components than coal).

So, when wood or coal burn in an uncontrolled way both the burning of the solid and the burning of the volatiles contained with the solid occur at the same time but the volatiles (as gases) can move far from the solids before they burn. This is why the position of the fame can be far from the burning solid.

Also worth noting

In addition to the volatiles being driven from the burning solid, partial combustion can also create gases like carbon monoxide which can then travel some distance away from the solids while burning (though CO tends to have not very bright blue flame). In fact a controlled version of this reaction (which sometimes also used steam to create hydrogen) was once the primary way of creating town gas (which was widely distributed to support gas lighting and cookers in cities.

If the primary goal is to create flames designed to illuminate rather than heat then burning can be designed to create a lot of small hydrocarbon particles in the flame. This is intentional for things like candles. Here the heat may come from burning the was but the light comes because that burning creates small incandescent soot particles which are heated enough to glow turning a lot of the energy into light.


Calculating pH of rain water The concentration of $\ce{SO2}$ in the atmosphere over a city on a certain day is $\pu{10 ppm}$. Given that solubility of $\ce{SO2}$ in water is $\pu{1.3653 mol L-1}$ and $\mathrm{p}K_\mathrm{a}$ of $\ce{H2SO3}$ is $1.92,$ estimate $\mathrm{pH}$ of rain on that day.

$\pu{10 ppm}$ means there is $\pu{10^-2 g}$ of $\ce{SO2}$ in $\pu{1 L}$ of air. This accounts to $\pu{1.25E-4 mol}$ of $\ce{SO2}$ in $\pu{1 L}$ air, which is less than its solubility, which means that all this amount will be dissolved in rain water to make $\ce{H2SO3}.$

Now, using the expression $\frac{4c^2α^3}{(1-α)}$ for $K_\mathrm{a}$ expression of $\ce{H2SO3},$ and substituting $c$ as $\pu{1.25E-4},$ I get $α$ to be approx. $0.99.$

Then $[\ce{H+}]$ is $2cα,$ which comes out to be $\pu{2.475E-4}.$ So $\mathrm{pH}$ is $3.606.$ But the answer given is $4.865.$ Why I am wrong?

The value $\pu{1.3653 mol L^{−1} atm^{−1}}$ is the solubility constant (or Henry's law solubility constant), not the solubility. The solubility is defined as the maximum possible concentration (the saturation concentration) of a solute under given solution conditions (e.g. temperature and pressure), whereas the solubility constant $H^{cp}$ defines how solute partitions between the gas (concentration in gas here given as the partial pressure in atmospheres — although this is not stated in the OP it can be inferred from the expected answer) and solution phases (concentration in $\pu{mol/L}):$

$$H^{cp} = \frac{c}{p}= \pu{1.3653 mol L^{−1} atm^{−1}}$$

Assuming atmospheric pressure of 1 atm then $\pu{10 ppm}$ translates into a partial pressure of $\pu{1e-5 atm}$ (if the atmospheric pressure is $p_T$, then the partial pressure of $\ce{SO2}$ is $p = p_T \cdot 10^{-5}$) and a solution concentration $\pu{1.365e-5 M}$  $\;(c=H^{cp} p)$. Proceeding from the definition of $\mathrm{p}K_\mathrm{a}$ the degree of dissociation $(\alpha=0.99886)$ can be computed and from this the $\mathrm{pH} = 4.865,$ which agrees with the expected result.

You should check solubility limits when evaluating this sort of problem, but since the solubility is not given you can ignore this and assume that the limit has not been reached.


Intrinsic Reaction Coordinate calculation-- both sides of TS lead to same minimum I am currently trying to find an addition transition state and verify it using an IRC calculation. I have found a good candidate, but the IRC calculation gives the same minimum on each side of the transition state. In my mind, this does not seem any different from a vibrational mode.

I think the TS is submerged, and I am sure this is complicating things, but it does not make sense that an IRC calculation would give the same minimum in both directions.

Any thoughts on what this signifies or how to remedy it?

Thanks

Be sure you have really obtained TS, that means stationary point, i.e. all gradients are zero. Than, you should have one imaginary frequency corresponding to the motion along the reaction coordinate. If this is true and you still get both IRCs going to same minimum, your assumptions regarding the shape of Potential Energy Surface (PES) could be wrong.

It is easier to find the minima first, so be sure you have well optimized reactant, product, and if by any means reasonable, also intermediates. It is possible that there exist some high energy intermediates, which complicate the whole story (if you are not aware of them).


Drying the solution of two salts What will be the arrangement of crystals if a solution of two salts, say, Sodium Chloride and Potassium Chromate, is dried? Will one salt crystallize before the other and become the crystallization nuclei of the other? Would both types of crystal form in a chaotic mixture? Or would they interact to form some new lattice?

As @Ivan Neretin states, it depends. For example, potassium sulfate and aluminum sulfate form "alum", with a single crystal structure.

Tartaric acid (and its salts) occur in different forms, and even though they have the  same chemical makeup, these stereoisomers can be separated because they form differently shaped crystals. One of Louis Pasteur's first studies was separating the isomers from a mass of crystals.

A good source on growing crystals is Crystals and Crystal Growing by Alan Holden and Phylis Morrison


What is the difference between dissolution and absorption? Let us first look at the definitions from Gold Book:

Absorption:

  The process of one material (absorbate) being retained by another (absorbent); this may be the physical solution of a gas, liquid, or solid in a liquid, attachment of molecules of a gas, vapour, liquid, or dissolved substance to a solid surface by physical forces, etc

Dissolution:

  The mixing of two phases with the formation of one new homogeneous phase (i.e. the solution).

I've re-read them again and again, but have been unable to come up with the terminology to exactly differentiate between the two. In fact, I've reached some interesting counter-examples.

Here's one: we always says "sugar dissolves in water" and never "sugar absorbs in water" (Ngram). But, going by the first definition which reads: "physical solution of a gas, liquid, or solid in a liquid", can't we say that the physical solution of solid sugar in liquid water is an example of absorption as well? The water is retaining sugar particles in its bulk after all, so why couldn't we call water an an absorbent and sugar an adsorbate?

Another one: we say that "cotton absorbs water" but never "cotton dissolves water" (Ngram). But, going by the second definition: "mixing of two phases with the formation of one new homogeneous phase (i.e. the solution)" can't we say that the solid cotton mixed with liquid water to form a new homogenous phase? After all, the wet cotton needs to be dried to separate the two components, so they clearly are homogenous.

Interestingly enough, the Gold Book definitions seem to have been created to deliberately introduce confusion. Notice that the definition for "absorption" mentions both the words "solution" and "dissolution". So, why did the Gold Book need to refer to dissolution to explain absorption?

Dissolution and absorption definitely are different from each other, otherwise some explicit relation should've been mentioned in the Gold Book. For example, the definition for "chirality centre" reads "A chirality centre is thus a generalized extension of the concept of the asymmetric carbon atom to central atoms of any element" clearly highlighting their relation (the former is a superset). Such isn't a case in the definitions for dissolution and absorption. So, they are different.

So, to summarize, my questions are:

What exactly is the difference between absorption and dissolution? And, as a direct consequence, given a physical process: how would I identify if it is a dissolution or absorption?
In the two examples (sugar and cotton) I highlighted, where is my reasoning wrong in applying "dissolve" in place of "absorb" and vice-versa?
Is there any subset/superset relation between absorption and dissolution? If so, how's that apparent from the Gold Book's definition?
Why did the Gold Book need to refer to dissolution to explain absorption? Couldn't it have explained them both in separate terms without creating confusion?

Casual use of "absorption" is different from the normal, technical, use of the term
The key difference between "absorption" and "solution" is confused in your examples because you use the casual use of the term not the technical one. Even so, the two processes are very distinct.
Solutions are homogeneous. But when things are absorbed, either in casual use or technical use, there are distinct phases. Absorption is usually a technical term for a surface phenomenon where, for example, gas molecules stick to the surface of a catalyst. This is a widely studied process as it is a vital part of many major industrial processes. Water absorbed onto cotton wool is a more casual use of the term absorption but the result is still not a homogeneous mixture. Liquid water sticks to the surface of the fine fibres of cotton wool by surface tension. There are still two phases: liquid water and solid fibre.
Only in the most careless and casual use should you ever describe the process of dissolving sugar in water (or sugar taking up water) as one "absorbing" the other.


Nomenclature of MnO4^-: is it an oxyanion? Having a real pain with nomenclature. Look at this substance:

$$\ce{MnO4-}$$

So, do we have an anion of manganese and oxygen, or is that just an anion of oxygen? Anyway, my book says that $\ce{MnO4-}$ is "permanganate".

Since it ends with "-ate", it seems my book thinks $\ce{MnO4-}$ is an oxyanion (because those are the ones that end with things like "-ate" and "-ite").

However, it was my understanding that an oxyanion is when you have oxygen and a non-metal. But manganese is a metal! What is going on here then?

And not just that. Let's assume that $\ce{MnO4-}$ is indeed an oxyanion called permanganate. From what I've learned, if you remove one oxygen atom (getting  $\ce{MnO3-}$), it should become just manganate, right? But, according to my book, manganate is $\ce{MnO2-}$. It had to lose two oxygen atoms and not one to become manganate. Why?

As far as I know, an oxoanion just refers to an anion which has oxygen covalently bonded to the central atom; the central atom can be pretty much anything apart from the Group 1 and 2 metals, since the only oxygen compounds those form are ionic oxides/peroxides/superoxides. For the 3d transition metals, you have $\ce{VO3^-}$ (vanadate(V)), $\ce{CrO4^{2-}}$ (chromate(VI)), $\ce{MnO4-}$ (manganate(VII)), and $\ce{FeO4^{2-}}$ (ferrate(VI)), amongst others.

I would not recommend reading too much into the nomenclature, because a lot of it is just what chemists are comfortable with after using for a long time - it's like a language that you just get more familiar with over time. But there are some general rules. The 'per...ate' name, such as perchlorate $\ce{ClO4-}$, implies a higher oxidation state than the normal '...ate' compound, which is chlorate $\ce{ClO3-}$ in this case. For chlorine, only certain oxidation states are available, namely the -1, 0, +1, +3, +5 and +7 oxidation states. Apart from 0, which occurs in diatomic chlorine $\ce{Cl2}$, you will notice that these are in increments of 2 - which means that you can generate the whole list of chlorine anions by adding one more oxygen, but keeping the charge of the anion constant: $\ce{Cl-}$, $\ce{ClO-}$, $\ce{ClO2-}$, $\ce{ClO3-}$, $\ce{ClO4-}$. This tendency to form compounds 2 oxidation states apart is very pronounced for the p-block elements, i.e. Group 15, 16 and 17.

However, for d-block elements such as $\ce{Mn}$, this isn't the case because single d electrons are easily lost and gained. So after permanganate, $\ce{MnO4-}$, which features $\ce{Mn}$ in the +7 oxidation state, the oxoanion with the next highest oxidation state is the manganate ion $\ce{MnO4^2-}$ which has $\ce{Mn}$ in the +6 oxidation state. (As far as I know the ion $\ce{MnO2-}$ is not called "manganate".)


Grignard alkene on a carbonyl I was given the following mechanism to prove today post-lecture:

and my attempt is given below.

Now, when the negative charge on the alkene attacks the carbonyl, the electrons are pushed onto the oxygen - which is understandable. My question is: why do these electrons come back down, rather than just be protonated to give an alcohol?

Nucleophilic acetyl substitution is common among carboxylic acid, acid chlorides, anhydrides, esters, and amides.

Two very common examples are anhydride decomposition and esterification of carboxylic acids. For example, if you look at esterification, carboxylic acid and alcohol first make an addition, giving tetrahedral intermediates 1 and 2. Intermediate 2 then collapses and displaces the water molecule as a leaving group to give the corresponding ester (see the scheme).

Your substrate is an amide. Like all other carboxylic acid derivatives, this molecule also undergoes a nucleophilic acetyl substitution with Grignard reagent, which is a strong nucleophile. The polarized carbonyl group has partially positive carbon, which is a good electrophile.

The nucleophilic carbon on Grignard reagent attaches to carbonyl carbon and gives tetrahedral intermediate as a metal alkoxide complex. But, this alkoxide can collapse to reform strong C=O bond, displacing the amino portion of the amide as a leaving group, in the form of the amide ion, $\ce{R1R2N-}$. This produces the ketone as a final product, assuming you have used only one equivalent of Grignard reagent (Channel reactions of Grignard reagent with esters).

  Why does the reaction continue after the tetrahedral carbon forms?

After nucleophilic attack on a carbonyl, the resulting tetrahedral species either gets protonated to form the final addition product, or it is an intermediate on the way to the substitution product. This depends on the type of reaction. For example, a geminal diol will usually dehydrate to form a ketone. There is a nice overview of the different scenarios, including some nitrogen chemistry, in this document.


Why is there a doublet instead of a singlet in this Wittig reaction salt configuration NMR? I did an NMR sample to my salt p-carboxybenzyl triphenylphosphonium bromide. I got a doublet at 5.26, and according to every article it belongs to the $CH_2$ next to $P^+$, but how come I didn't get a singlet for that?

Phosphorous exists entirely as $\ce{^31P}$, and $\ce{^31P}$ has a magnetic moment of 1/2 (just like a proton). Therefore, a single phosphorous will split any protons coupled to it into a doublet - as you observed. If you looked at the corresponding $\ce{^31P}$ NMR spectrum you would see that the phosphorous signal would be split into a triplet by the two protons.


Attack of amine on amide I am confused with the last step of clozapine synthesis:

How can the amine group attack the amide group to form the amidine in such acidic environment? Is the intermolecular reaction between an amine and amide group feasible under not too extreme conditions?

There reaction environment would have to be acidic to protonate the amide, the amide is not natively electrophilic enough to be attacked by the amine.

A possible point of concern is 'won't the nuclophilic amine be protonated?'  And the answer is absolutely.  But the protonation is a reversible reaction, and at equilbrium with a catalytic amount of acid (almost always TSA is used in catalytic quantities), the proton will be exchanged with various, some of which will lead to the condensation resulting in product.


Why two ( or more) gases have proportional volumes at constant pressure and temp.? (Avogadro's Law) I am in real confusion over Avogadro's Law..

In some of the problems I did on gas laws, there are the number of moles of the gas, the temperature. and the pressure. And it expects me to find the volume of gas.

But then in other problems there everything else is given including the volume of the container ( instead of the gas itself) and then something else is expected to be found.

The confusion starts here. If any gas occupy the volume of volumes, always, then why will volume change according the combined gas law?

In one of my previous post, I learned that volumes of different gases won't be proportional (meaning all of them will occupy equal space) unless the condition of constant tempt and pressure is reached.

But what really happen when constant tempt and pressure are reached? Do individual gases just separate themselves from one another? Because in order for their volume ratio to be equal to anything other than 1, their volume must be different.

you only have to answer the question in the title, the others questions are intended to help you understand what I mean. Sorry for the possible confusion

a link to my previous post

So, there are many questions, and I'm not sure if I understand you 100 %:

We have Avogadros Law, which applies for an ideal gas at constant pressure and temperature:
$$\frac{V}{n} = k \quad\Rightarrow\quad \frac{V_1}{n_1}=\frac{V_2}{n_2}$$
With $k$ as a constant.

And we have the ideal gas law:
$$pV = nRT \quad\Rightarrow\quad \frac{p_1V_1}{n_1T_1} = \frac{p_2V_2}{n_2T_2}$$

And this is probably what your first question is about. There are problems when you have a state $1$ and a state $2$, which differ.

For example: You have a container where everything ($p_1,V_1,T_1,n_1$) is given. Now you double the temperature - and the task is to calculate the pressure. Since the volume and amount of substance is fixed ($V_1 = V_2$, $n_1 = n_2$), you can just rearrange everything and calculate the pressure $p_2$.

In the ideal gas-model, gases do not seperate. I may be misunderstanding you, but if you put 1 mol $\ce{N2}$ and 2 mol oxygen in a container, you will have different partial pressures, but no demixing.

A relevant question was posted on physics.SE, that may help you if you have to dive deeper into this topic.


Mechanism of protein precipitation with TCA In practice, TCA (trichloroacetic acid) is widely used for quantitative protein precipitation, but so far I have failed to discover the exact mechanism of TCA action. I highly doubt it's just the effect of pH that precipitates proteins.
Any better guesses?

The paper "Trichloroacetic acid-induced protein precipitation involves the reversible association of a stable partially structured intermediate" (Rajalingam et al. 2009 Protein Science 18(5), 980–993.) may be relevant.

In the introduction, they mention the conventional explanation is that "TCA forces protein to precipitate by sequestering the protein-bound water" - that is, it may force protein precipitation in a related fashion to chaotropic agents or antichaotropic agents like ammonium sulfate.

Their experiments indicate a more complex explanation. They hypothesize  that the negatively charged TCA disrupts electrostatic interactions in the protein, resulting in a partially folded state which is prone to aggregation. The fact that TCA precipitation is less effective for intrinsically disordered proteins lends support to this hypothesis (as they don't have an aggregation-prone partially unfolded state).

The acidic nature of the TCA (the pH effect) helps with the rate of precipitation, but does not appear to be necessary, as the sodium salt of TCA is able to precipitate proteins in neutral solutions, although slower than acidic TCA.


Why does oxygen not like to be a double anion? The electron affinity of a neutral oxygen atom is −142 kJ (it releases this energy). The electron affinity for the now double anion $\ce{O^2-}$ is 710 kJ (work must be done on the atom).

My question is, why the large jump in electron affinity? We are taught that the electrons do not interact much with each other in the way of shielding. I think this to be incorrect, because otherwise, the effective nuclear charge on all of the valence electrons should be about the same, including the new ones added to ionize the atom. This would make the electron affinity about the same for each electron. Plus, doesn’t the oxygen atom want to complete its shell? So why is it putting up a fight?

The addition of an electron to any negatively charged ion will need to overcome the repulsive force of the negative ion. This is generally less favorable than adding an electron to a neutral atom, and the electron affinity will be positive. Even though the additional electron will complete the octet, as in the case of an  ion where the addition of electron will complete the  subshell, the electron affinity is a positive value.


Feasibility of Nucleophilic Substitutions Can $\ce{NH3}$ substitute an alcohol? The OH group will depart as a weaker base - some O is more electronegative - so will be a better leaving group than amine.
And what about using ammonium ions - such as in an ammonium salt? They will donate an $\ce{H+}$ to protonate the alcohol and make it a good leaving group, then show substitution followed by deprotonation to form an amine?
How is the the feasibility decided in case of nucleophilic substitution reactions? Is there any rule of thumb to know what happens, and what doesn't?

Short answer - No. See below for a workaround.

Consider your acids and bases:

  Substituting an alcohol with ammonia directly

$$\ce{NH3 + CH3CH2OH <=> CH3CH2NH2+ + OH-}$$

If this were an acid-base reaction (and it is, at least in the Lewis sense), then which side of this reaction would be favored by equilibrium? Which side has the stronger acid-base pair? Hydroxide is a stronger base than ammonia, so this equilibrium would favor the reactants.

  Using an ammonnium salt

The first step of this mechanism would be a proton transfer to form an activated leaving group and the $\ce{NH3}$ nucleophile.

$$\ce{NH4+ +CH3CH2OH <=> NH3 + CH2CH3OH2+}$$

Again, which side of this equilbrium is favored? Ammonia is definitely the stronger base compared with the alcohol. Estimates of the pKa of $\ce{ROH2+}$ are in the vicinity of $-2$, while the pKa of $\ce{NH4+}$ is $9.25$. The equilbrium constant of this reaction is on the order of $10^{-11}$.

  Workaround

If you are needing to do the transformation

$$\ce{ROH -> RNH2}$$

there is a workaround in converting the alcohol into a molecule like the tosylate ester, which can then react with ammonia:

$$\ce{CH3CH2OH ->[\ce{TsCl,base}] CH3CH2OTs }$$
$$\ce{CH3CH2OTs ->[\ce{NH3}] CH3CH2NH2}$$


How does one calculate the pH of an aqueous solution of a weak acid and a weak base? In order to create a simple program which calculates titration curves (also to provide some closure in the comments here) I am trying to come up with all the equations so I can calculate the equilibrium proton concentration for any given concentrations of any two given weak acids/bases.

This is a collection of what I've come up with so far and what I assume I still need. If someone could help me back on track I'd kindly appreciate it.

We're looking at the following reactions:
$$\begin{align}
\ce{HA &<=>[$K$_{\text{a, 1}}][$K$_{\text{b, 1}}] H+ + A- && (1) }\\
\ce{H+ + B- &<=>[$K$_{\text{b, 2}}][$K$_{\text{a, 2}}] HB && (2)}
\end{align}$$
A is the weak acid and B the weak base with the according equilibrium constants (the indices 1 and 2 refer to the acid and base respectively).

For the total masses I've got the following:
$$\begin{align}
\ce{[A]_{tot} &= [HA] + [A- ] && (3) }\\
\ce{[B]_{tot} &= [HB] + [B- ] && (4) }
\end{align}$$

The equilibrium constants are linked via
$$ K_\text{a, 1} \times K_\text{b, 2} = \frac{\ce{[A- ][HB]}}{\ce{[HA][B- ]}} $$

Furthermore:
$$
\ce{[H+ ]_{eq}} = \frac{\ce{[A]_{tot}} K_\text{a, 1}}{\ce{[A- ]}} - K_\text{a, 1}
 = \frac{\ce{[HB]}}{K_\text{b, 2} (\ce{[B]_{tot} - [HB]})}
$$

For the equilibrium proton concentration I've derived
$$ \ce{[H+ ]_{eq}} = K_\text{a, 1} \ce{[HA]} + K_\text{a, 2} \ce{[HB]} + K_\text{w} \ce{[H2O]}$$
where I neglect the influence of the water (the additional term $[\dots] + K_\text{w} \ce{[H2O]}$) since it is so small.

Now, I'm struggling with the following:
$$ \ce{[H+ ]_{tot}} = ? $$

I believe that with the last equation I will be able to solve for $\ce{[H+ ]_{eq}}$. But maybe I'm wrong there as well.

Thanks for any pointers.

Equations and variables

To calculate the proton concentration in a solution as given in the question the following five equations are used for the following five unknown variables: $\ce{[H+ ]}$, $\ce{[HA]}$, $\ce{[A- ]}$, $\ce{[HB]}$ and $\ce{[B- ]}$.

Ion balance:
$$ \ce{[H+ ]} = \ce{[A- ] + [B- ]} + \frac{K_\text{w}}{\ce{[H+ ]}} $$

Mass balances:
$$\begin{align}
\ce{[A]_{tot}} &= \ce{[HA] + [A- ]} \\
\ce{[B]_{tot}} &= \ce{[HB] + [B- ]}
\end{align}$$

Law of mass action:
$$\begin{align}
K_\text{a, 1} &= \frac{\ce{[H+ ][A- ]}}{\ce{[HA]}} \\
K_\text{a, 2} &= \frac{\ce{[H+ ][B- ]}}{\ce{[HB]}}
\end{align}$$

Solution

Combining the above equations we get:
$$
\ce{[H+ ]} = \frac{K_\text{a, 1} \ce{[A]_{tot}} }{\ce{[H+ ]}+K_\text{a, 1}} + \frac{K_\text{a, 2} \ce{[B]_{tot}} }{\ce{[H+ ]}+K_\text{a, 2}} + \frac{K_\text{w}}{\ce{[H+ ]} }
$$

After some reforming we get a fourth-order polynomial for the proton concentration:
$$\begin{align}
0 = & \ce{[H+ ]^4} + \ce{[H+ ]^3}(K_\text{a, 1} + K_\text{a, 2}) \\
  &- \ce{[H+ ]^2}(\ce{[A]_{tot}} K_\text{a, 1} + \ce{[B]_{tot}} K_\text{a, 2} - K_\text{a, 1} K_\text{a, 2} + K_\text{w}) \\
 & -\ce{[H+ ]}\left[ (\ce{[A]_{tot}} + \ce{[B]_{tot}}) K_\text{a, 1} K_\text{a, 2} + K_\text{w}(K_\text{a, 1} K_\text{a, 2}) \right] \\
 & - K_\text{w} K_\text{a, 1} K_\text{a, 2}
\end{align}$$


Finding root-mean-square speed of neon using kinetic energy Find the root-mean-square speed of $\ce{Ne}$ atoms at the temperature at which their kinetic energy is $\pu{6.24 kJ mol-1}.$

I tried using the kinetic energy formula
$$\mathrm{KE} = \frac{mv^2}{2},$$
but I don't really understand how to achieve the necessary values.
I tried to do it by converting mass of one atom to mass of a mole since the given energy is per mole, but I'm still not getting the answer. I understand that I need to do $\displaystyle\sqrt{\frac{3RT}{M}},$ where $R= \pu{8.3145 J mol^-1 K^-1}$ and $M = 20,$ but how would I get the temperature?

As hinted in the comments by @AngusTheMan we use the equation
$$\mathrm{KE_{avg}} = \frac{3}{2}kT,$$
but ultimately we are looking for the $v_\mathrm{rms}$, so we'll also use
$$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}.$$
You've been given your $\mathrm{KE_{avg}},$ $R$ and $k$ are constants, and $M$ is the mass of 1 mole of $\ce{Ne}$. We can set all of these equal and the solution will come from
$$v_\mathrm{rms} = \sqrt{\frac{3 \cdot R\cdot\mathrm{KE_{avg}}\cdot\frac{2}{3k}}{M}}.$$


Hofmann rule and deviation from E2 mechanism I've learnt that

  The best reaction conditions for synthesis of alkene by dehydrohalogenation are those that promote an E2 mechanism.

(From "Organic Chemistry" by TW Graham Solomons)

But, in the same book it is given that when 2-bromo-2-methylbutane follows Hofmann elimination with tert-BuOK as follows:

Clearly, it isn't following E2 mechanism.

When given such reactions, should I go with E2 mechanism or Hofmann rule, and why?

The difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium.

Let's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination.

Why not E1?

First let's consider the reaction as:

$$\ce{2-bromo-2-methylbutane + EtO^- (in EtOH) -> }$$

Clearly, this reaction is conducted in a highly polar solvent where the carbocation - formed when $\ce{Br^-}$ leaves - gets stabilised. The base then abstracts the hydrogen of $\ce{C3}$ to form 2-methylbut-2-ene by an E1 mechanism.

If you use the same substrate but this time in a highly non-polar solvent, the possibility of E1 decreases as carbocation formation won't get extra stability. But as the base is not sterically hindered, it can still abstract hydrogen to form the Saytzeff product, but there is a case where it can also form the Hoffmann product.

Why not E2?

The E2 mechanism follows an "anti-periplanar orientation of hydrogen and leaving group". Suppose the reactant's stereochemistry is such that there is no anti-$\ce{H}$ on the more substituted adjacent carbon for forming the Saytzeff product, but there is an anti-$\ce{H}$ on a less substituted adjacent carbon, which will lead to the Hoffmann product. In that case, the major product can be the less substituted alkene due to stereospecificity of E2 reaction.

Why E1cb?

Now consider your above given case. Here, the base $\ce{t-BuO^-K+}$ is sterically hindered and is a very strong base. So, prior to any carbocation formation, this base will abstract hydrogen from the less hindered position, and form a carbanion. Also, a less substituted carbanion is more stable than a more substituted one (due to a smaller +I effect). Both these effects lead to formation of a less substituted alkene via an E1cb mechanism.

So, depending upon which product is needed as a major one and the proper stereochemistry of reactant, you should choose whether to go with an E1, E2, or E1cb mechanism by choosing the proper base (bulky or less hindered) and proper solvent (polar or non-polar).


Molecular weight determination from freezing point depression Which is the better solvent for molar mass determinations by freezing point depression, benzene or cyclohexane?
  Which is the better method for molar mass determination, freezing point depression or boiling point elevation?

What is the correct approach for solving these?

In my opinion, resonance structures might be the key concept, but I don't know how it would help or hurt.
I think boiling has a decomposing effect that we must avoid, meaning freezing the compound(s) is preferable.

But, both of these statements are inadequate.

Molar freezing point depression
Benzene      5.12 °C/m
Cyclohexane  20.0 °C/m

Molar boiling point elevation
Benzene      2.53 °C/m
Cyclohexane  2.79 °C/m

I would base my argument on the constants, instead of waving hands.


Mechanism of enolether formation with ethyl orthoformate in estr-4-ene-3,17-dione (Djerassi's norethisterone synthesis) The second step of Djerassi's synthesis of 19-nor-17α-ethynyltestosterone (norethindrone), involves the protection of the ketone in position 3 via enolether formation, as in the picture:

What is the mechanism of this reaction? I couldn't find it anywhere. I suppose it should be similar to an ether synthesis, but I'm not sure about the precise mechanism.
(On a side note, I assume that the other ketone in 17 remains intact because of controlled reaction conditions.)

Usually a catalytic amount of acid is used in these reactions and they are run in the presence of alcohol (ethanol if you want to prepare the ethyl enol ether). Pyridine is typically added at the end of the reaction to recapture the acid catalyst.

The first step involves protonation of the carbonyl oxygen.  Protonation of the C-3 oxygen produces a resonance stabilized carbocation where the charge is delocalized over the carbonyl and the adjacent double bond (protonation at the C-17 oxygen does not have an adjacent double bond,  hence protonation at the C-3 oxygen is favored).  The nucleophilic alcohol oxygen reacts at the carbonyl carbon (highest positive charge density) to produce a hemiketal.  Loss of water gives rise to the enol ether product.  The eliminated water is captured by the orthoformate and generates more alcohol.

$$\ce{HC(OEt)_3 + H2O -> HCO2Et + 2EtOH}$$

Some argue that the reaction actually proceeds through the ketal (rather than the hemiketal) since ketals can be isolated in these reactions.  Others argue that just because the ketal is formed, doesn't prove it is actually involved in the enol ether formation reaction.  IMO, lacking definitive evidence, I opt for the simpler hemiketal mechanism.


What chemical reactions take place when mixing baking soda with citric acid and epsom salt and putting it into a microwave oven? When I mixed in a bowl baking soda, citric acid and Epsom salt, then put it in the microwave between 30-60 seconds, the mixture will start to fizz and puff up, kind of like dough rising. What is happening here?

I've done test with all of the ingredients individually in the microwave and nothing happens. I've tried various combinations of any two of the three ingredients and nothing happens. Only when all three are mixed together does the mixture fizz and puff up. It seems like the catalyst is the Epsom salt. What is in the Epsom salt that is setting off the other two ingredients?

I know that baking soda plus citric acid together with water will also fizz, as $\ce{CO2}$ is released. I was wondering when I have the three ingredients in the microwave and it is puffing up, is $\ce{CO2}$ also being released?

Furthermore, the resulting mixture that had puffed up, if I put it in water it will still fizz and create $\ce{CO2}$, just as if had I never put the mixture in the microwave and just put it straight in the water. What I'm also wondering about if the microwave had hurt the ability of the mixture to create fizz in the water later, i.e., would there have been more fizz in the water otherwise?

After asking my question I found this YouTube video that said heating up Epsom salt releases $\ce{H2O}$ gas. Which would make sense why my mixture is puffing up. Still I'd love to hear anyone else's opinion on this.

The solid phase reaction of baking soda and citric acid to form monosodium citrate, water and $\ce{CO2}$ takes place very slowly, so you don't get the visible foaming and bubbling that you do when water is added, or if you were to use baking soda with an acidic juice or vinegar (~95% water).

The molecular formula for Epsom salt is $\ce{MgSO4\dot\ 7H2O}$, meaning it's actually more than 50% water by weight. At $\mathrm{150^oC}$ it starts giving up some of those waters of hydration. Although water is a gas at that temperature, enough of the water is able to wet the baking soda and citric acid to get that reaction going, which then releases more water:

$$\ce{NaHCO3 + C6H8O7 → NaC6H7O7 + CO2 ^ + H2O}$$

Once the above reaction is initiated by the release of enough water from the Epsom salt, the additional water the reaction releases makes it somewhat self-sustaining and the $\ce{CO2}$ provides the foam and bubble show. The fact that additional reaction was observed after adding more water just shows that there was unreacted baking soda and citric acid remaining in the solid "crust".


Synthesis of 2-methyl-4-nitrophenol from benzene? Benzene first reacts with the reagents $\ce{CH3Cl/AlCl3}$, to attach a methyl ($\ce{CH3}$) group to the benzene ring.

Then I suggested reacting with $\ce{OH-}$. Because the methyl group is an activator, it will cause the $\ce{OH-}$ electrophile to attack either in the para or ortho positions. In this case I suggest it will attack the ortho position. Now we have two activators, hydroxyl and methyl, but because hydroxyl is a stronger activator, the position where the next molecule attacks will be controlled by the hydroxyl group.

The next electrophile that attacks is the $\ce{NO2+}$, through the reaction $\ce{HNO3/H2SO4}$, the nitro group will attack the para position of the benzene and we get our final product 2-methyl-4-nitrophenol.

However, I am not 100% sure if the methyl group should be introduced first to form toluene or the hydroxyl group to form a phenol.

There is a lot of good thinking in your proposed synthetic strategy.  The one major shortcoming is, as @Mithoron points out, $\ce{OH^{-}}$ is not an electrophile.

We need to introduce a methyl, hydroxy and nitro group onto a benzene ring.  If we look into our synthetic "toolbox" here are some thoughts that might come to mind:

A methyl group can be readily introduced using the Friedel-Crafts alkylation reaction with$\ce{CH3Cl}$. However, be aware that there are some general issues with the Friedel-Crafts alkylation reaction.  For example, if the alkylating agent can rearrange to a more stable carbocation it will likely do so ($\ce{CH3Cl}$ can't rearrange).  Also, the mono-alkylated product is more reactive than the starting material, so polyalkylation can occur (but in your case, we can use an excess of benzene - it's cheap - and get a good yield of the desired product toluene).  Finally, the Friedel-Crafts alkylation does not work well with deactivated aromatic rings. Therefore, the comment suggesting we run the alkylation after introducing the nitro group is incorrect; we must use the Friedel-Crafts alkylation before we introduce the nitro group.
There aren't any easy ways to electrophilically introduce a hydroxy group into an aromatic ring.  When I need an aromatic hydroxyl substituent, I think of the Sandmeyer reaction.
The nitro group can be introduced via straightforward electrophilic nitration.

There are a lot of ways to pull this synthesis together, but here is a relatively direct proposal.

The first step is our Friedel-Crafts alkylation to prepare toluene.
Next, we'll introduce our ortho hydroxy substituent.  To do this let's first sulfonate using fuming sulfuric acid (oleum) and block the para position. Blocking the para position will simplify our next step - we won't have to separate ortho and para isomers and we will get a higher yield of only the desired ortho (to the methyl group) product. Now the sulfonic acid group (meta-director) and the methyl group reinforce one another and will direct our next electrophile to the position ortho to the methyl group.
Now we nitrate using nitric and sulfuric acid and get only the product where the nitro group is ortho to the methyl group.
Next, we heat the reaction. The sulfonation reaction is readily reversible, by heating our reaction we can readily remove the sulfonic acid substituent since it has served it's purpose and is no longer needed.
Now it's time for the Sandmeyer reaction. We first reduce the nitro group with tin and hydrochloric acid to an amino group.  Then we diazotize it and decompose the diazonium salt in the presence of acidic water to introduce our hydroxyl group.
Finally we nitrate one more time to introduce the nitro group.  The directing effect of the hydroxyl group is much stronger than that of the methyl group and will primarily direct the nitro group into the position para to the hydroxyl.


can magnesium chloride and iodine react together? If everything is in a mixture of alcohol and water, would it be possible for magnesium chloride and iodine to react and produce something?

Magnesium chloride and Iodine with form an equilibrium.  This is a bit of a simplification of the reactions but the dissociated chlorides and free iodine will reversibly react to form iodide and triiodide ions and chlorine.
$$\ce{Cl- + 1/2 I2 <=> 1/2 Cl2 + I-} $$
$$\ce{MgCl2 + I2 <=> MgI2 + Cl2} $$
$$\ce{I- + I2 <=> I3-} $$
$$\ce{MgI2 + 2I <=> Mg(I3)2} $$
The amount of products of this reaction though will not be very appreciable, but yes a very literal sense magnesium chloride and iodine will react.


Entropy and Equilibrium Constant: Chemical Thermodynamics For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant $K$ in terms of change in entropy is described by:
(A) With the increase in temperature, the value of $K$ for exothermic reaction decreases because the entropy change of the system is positive
(B) With the increase in temperature, the value of $K$ for endothermic reaction increases because of an unfavorable change in entropy of the surroundings decreases
(C) With the increase in temperature, the value of $K$ for endothermic reaction increases because the entropy change of the system is negative
(D) With the increase in temperature, the value of $K$ for exothermic reaction decreases because the favorable change in entropy of the surroundings decreases

Now, I am having trouble understanding option (B) (which in fact, is one of the correct options).
Since the process is endothermic and the temperature is rising, the value of $K$ increases, so heat must flow from surroundings to the system to favor the reaction which implies increase in an unfavorable change in entropy of surroundings rather than decrease as given in the option.
Kindly clarify my doubt. Also, any insights (other than the ones already given in the links attached below) into other options are very appreciated.
This question might look Duplicate of the questions already posted here:

relation between equilibrium constant and entropy change

Effect of temperature on equilibrium constant in terms of entropy change

But since this question is not focusing on the same problem, so I hope, it will not be marked as duplicate.

I agree it's not a duplicate because, while you are asking about the same homework problem discussed in the other question you linked, the issue you are having is different.  Hence, because the last answer didn't address your specific confusion, you are asking a different question about that same homework problem.

Here's where you've got things mixed up:  Whether a reaction is exothermic or endothermic depends on the inherent nature of the reactants and products, not the equilibrium constant.  I.e., $\Delta H^{o}_{rxn}$ is the enthalpy change for a complete conversion of reactants to products at standard state, irrespective of the equilibrium constant.  It's not, as I believe you were thinking, the enthalpy change to go from pure reactants to the equilibrium mixture. Hence if you could, say, change $K_{eq}$ purely by tweaking $\Delta S^{o}_{rxn}$, then $\Delta H^{o}_{rxn}$ wouldn't change just because you've changed $K_{eq}$.

So B is correct.   If you raise T (and here we're assuming $\Delta H^{o}_{rxn}$ is independent of T), and $\Delta H^{o}_{rxn} > 0$, the reason the reaction shifts to the right is this: In calculating $\Delta S^{o}_{surr}$, we have to divide $q_{rev, surr} = q_{surr}$ by T, and  T is now higher.  Hence, for the same heat flow from the surroundings (which is given by $-\Delta H^{o}_{rxn}$), the resulting unfavorable entropy change in the surroundings is less.

[I've used standard state here, which means it's carried out at a constant external pressure of 1 bar, but what I've written applies at any constant external pressure.]


Why can't alcohols turn blue litmus red? I came across the fact that alcohols cannot turn blue litmus red. The fact confuses me as they are acidic and acids do turn blue litmus solution red. What prevents alcohols from turning blue litmus solution red?

Litmus is a weak acid, with a $\mathrm{p}K_\mathrm{a}$ of about 6.5.

Alcohols are extremely weak acids, with $\mathrm{p}K_\mathrm{a}$'s (typically) around 17.

The litmus itself is a much stronger acid than the alcohol. Dissociation of the alcohol won't produce enough protons to shift the equilibrium between the two colored forms of the indicator, so you'll see no color change.


What is the meaning of delta squared? I know that $\Delta x = x_2 - x_1$. This is well known
Therefore, it follows that $\frac{\Delta a}{\Delta b}$ = $\frac{a_2 - a_1}{b_2 - b_1}$ (Unless I missed an elementary math class!)

Now, I found an equation in chemistry, while reading, that turns this principle on its head. It proposes that when considering the second in a set of values:
$\frac{\Delta^2 a}{\Delta b^2} = (a_3 - a_2) \times \frac{1}{\Delta b}$

My issue with this is that $a_3$ hasn't been measured yet, we have only $a_1$ and $a_2$ so how can this be?

Secondly, they propose this:
$\frac{\Delta \left(\frac{\Delta a}{\Delta b}\right)}{\Delta b} = \frac{\Delta^2 a}{\Delta b^2}$

Can anyone please help with the logic behind the first equation and algebra behind the second one?

I hate to break it you, but they're hiding the math from you that is really going on behind the scenes.

It is very common in situations where the students are not expected to know calculus, that wherever a derivative would be needed, to replace, for example, the $dx$ with a $\Delta x$. This is because for many simple physical systems, taking the derivative results in simply measuring the average change in a property with respect to some other property.

For instance, consider a system where you know the position of some particle and how this position changes with time. If you want to know the velocity of that particle at any given time, you would take the derivative with respect to time, $$v=\frac{dx}{dt}$$but, in many cases you might be only interested in the average velocity and could thus skip the derivative part and write that $$v=\frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1}$$

But, and now we're getting to your question, let's take it one step further and say we wanted to know the acceleration of that particle. This would be the derivative of velocity which is also written as the second derivative of position with respect to time. Thus, we would write $$a=\frac{d^2x}{dt^2}$$Notice how similar this looks to your question about $\Delta^2$ if you simply replace the $d$ with $\Delta$.

Again, if we decided we were only interested in the average acceleration, then we would write $$a=\frac{\Delta^2 x}{\Delta t}$$
The reason why they say that $$\frac{\Delta(\frac{\Delta a}{\Delta b})}{\Delta b}=\frac{\Delta^2 a}{\Delta b^2}$$is again because that is true of derivatives. This would be written as $$\frac{d}{dt}(\frac{dx}{dt})=\frac{d^2x}{dt^2}=a$$

So, to summarize, the symbol $\Delta$ is often used in situations where you care about a change from one state to another (as with a change in energy) or when you only care about the average of some quantity.

As to why they are able to write$$\frac{\Delta^2 a}{\Delta b^2}=\frac{a_3-a_2}{\Delta b}$$ I'm not certain about. I would have to know what this dealt with specifically. My guess is that the quantity $a_3$ is somehow related to the second derivative of $a$ with respect to $b$ and they did some algebra behind the scenes.

Hope that helps.