Why do sealed plastic bottles of pop collapse as the contents get old? A number of bottles of pop (two varieties) in unopened plastic bottles have sat for a few years in a room-temperature environment. All of the bottles have now collapsed a certain amount (all to about the same degree).

They all started out under pressure from the carbonation, but now, not only is that gone, but the internal pressure has dropped below ambient. They all still have an "air" space above the liquid contents; not sure if it is or is not more than when originally purchased. If one of these bottles is opened, it draws in air and regains its original shape.

I have the following questions:

What happened to the carbonation? Did it leak out, or did it undergo some sort of chemical reaction, or perhaps both?
How could the pressure inside the bottle drop below ambient? What chemical or physical process is causing it, and what is driving the process? Why would it not simply reach equilibrium with ambient?
What would be the ultimate result if these bottles were left even longer?

It is most likely the CO2, H2O and air in the pop bottle headspace diffused (leaked) out of the pop bottle.
You are correct: the pressure has reached ambient (not dropped below ambient). Because the pressure in a freshly bottle pop bottle was originally higher than the pressure in the room (normally atmospheric pressure), and the bottle's plastic walls were smooth and strong, any decrease in pressure inside the pop bottle would cause the pop bottle walls to collapse.
If the bottles were left longer, the CO2 and water would continue to evaporate from the liquid phase until the headspace contained approximately the same pressure as the air in the room. If somehow all the water evaporated, you'd be left with sugar and a bunch of other solids (that were dissolved). I doubt full evaporation of water would occur.

Also, if the bottle is made from PETE (polyethylene terephthalate), prolonged contact with water (including pop) will result in many by-products. Two of them are:

a. UV exposure, air and water (together) promote peroxide formation on polymer end-groups (OH, COOH). Another way of saying this is that the bottle walls degrade.

b. The PETE plastic could leach some trace antimony (if that was used in the polymerization process) and/or phthalates, suspected to be endocrine disruptors.

IF water permeates through the bottle walls, it can allow for microoganisms to diffuse in/out of the bottle. You might get some mould in there!

Those are just some things that could happen.


Mixing water and oil in microgravity outside of a container What would happen if I  where to mix water and oil in a vacuum in microgravity  where the temperatures  were low enough so that the oil and water remain liquids also, the mixture is not inside of a container.
 I would expect the water and oil two separate into two blobs, but wouldn't the London dispersion forces hold the two blobs together?

An experiment similar to this was carried out using a free-fall drop shaft to provide the microgravity environment in Japan in 2000.

From the abstract of the publication$^1$:

  An experimental study was performed to obtain the detailed information needed to provide a deep understanding of the combustion process and the secondary atomization of an oil-in-water emulsion droplet. The oil-in-water emulsion, which consisted of n-hexadecane as a base fuel, distilled water, and a trace of surfactant was tested. Photographic observation and temperature measurements were made of the suspended emulsion droplet during the heating-up and combustion processes under microgravity. The primary attention was toward the phase separation in the droplet, and the time histories of droplet temperature and the amount of water in the droplet, during the period of time prior to disruptive microexplosion. The results showed that the separation of the base fuel and water as well as their agglomeration and coalescence occurred with the lapse of time. The increase in the droplet temperature resulted in phase separation, and the formation of a single water droplet enveloped by a shell of the base fuel, prior to the microexplosion.

Interestingly, the phase separation was attributed to the increase in droplet temperature. It's my guess that the heating increased the rate of separation, but that given time the final result of "the formation of a single water droplet enveloped by a shell of the base fuel" would have taken place, and I think this would be the result of the situation described in the question; that van der Waals forces between the oil and water would prevent them from separating into two separate balls of liquid. The stronger hydrogen bonds of water were likely the reason for water forming on the inside of the droplet, and I would expect the same to happen in the case described in the question.

Summary, TL;DR:
Based on the results of the drop-shaft microgravity experiment described above, I would expect similar results for an experiment as described in the question. There should be an inner ball of water held together by hydrogen bonding, surrounded by an oil shell held together, and to the surface of the water, by the weaker van der Waals forces.

1) "Water-coalescence in an oil-in-water emulsion droplet burning under microgravity", Proceedings of the Combustion Institute, Volume 28, Issue 1, 2000, Pages 985-990.


Can Cyanide ion substitute methoxide? Question

Attempt
Since CN is better nucleophile than both OMe and Cl, I believe that it should replace both. But none of the options match my answer. Where am I wrong? Please help.

You need to reckon with how strong a base the nucleophile is compared with what you are trying to displace.  Only if the nucleophile is the stronger base can you expect the reaction to go based on thermodynamics.

Thus, both of the following reactions can go:

$\ce{RCl + OCH3^- -> ROCH3  +Cl^-}$

$\ce{RCl + CN^- -> RCN  +Cl^-}$

because both nucleophiles are stronger bases towards carbon than the chloride ion that would be displaced.  You can then say that the second reaction will go faster with a better (kinetic) nucleophile.  But if you have

$\ce{ROCH3 + CN^- -> RCN  +OCH3^-}$

where you are trying to displace the strongly basic methoxide ion rather than adding it as a nucleophile like the first reaction above, then cyanide ion is too weakly basic to push through.  Cyanide ion can be a good nucleophile only when displacing a weaker base or adding to an appropriate unsaturated compound (such as a carbonyl compound) without displacement.

Try it now.


Excited states and electron addition/removal energies within Hartree-Fock, CI, etc Coming from a DFT background, I'm used to the concept that the DFT eigenvalues do not correspond to excitation energies (i.e., the band gap, ionization potentials, etc.). To correct for these it's common to go to many-body approaches like the GW approximation to the self-energy, which is said to explicitly deal with quasiparticle addition/removal energies.

I'm curious how excited states are calculated within Hartree Fock or CI. Can one do a ground state calculation in HF or CI and then get excitation energies from the difference in these ground-state eigenvalues? Or is this similarly not possible and one needs to explicitly consider excited states in order to get the excitation spectrum (band gap, band eigenvalues for removal/addition of electron,etc.).

There are actually quite a few methods post-HF that you can do to calculate excitation energies. The simplest one you will want to look at is Configuration Interaction Singles (CIS). To do this, you express your Hamiltonian in the basis of singly-excited HF determinants and diagonalize it. It doesn't involve correlation and is roughly the same cost as a HF calculation. The matrix elements are given in Szabo and Ostlund,  Modern Quantum Chemistry pg 236.

You are already familiar that for DFT, the most common way to get excitation energies is through a time-dependent DFT calculation (TDDFT). You can do a time-dependent Hartree-Fock (TDHF) as well, though you may see this in the literature as the Random Phase Approximation (RPA). They are essentially the same thing. TDHF and TDDFT are pretty much identical save for the fact that TDDFT is based off Kohn-Sham density, and TDHF is based off of Hartree-Fock density. The equations are almost the same, and it isn't that much more expensive to compute.

If you want to include electron correlation in wavefunction-based methods to a greater degree, look for EOM-CCSD like methods. These are based off a HF, followed by a coupled-cluster ground state calculation, followed by an extension to get excited state energies.

My graduate research uses and develops a lot of these methods. In fact, I wrote about these methods not too long ago --- you can read more about it here: http://bit.ly/1anA9dZ


multi state or multistate or multi-state I am not sure if this question belongs to English language StackExchange.

In the context of multiconfigurational methods, what is the correct way to write multi-state character:

multi-state character
multistate character
multi state character

Personally I would go with option 1.

I prefer option (1), and cite Donald Knuth's reasoning as given here.

Knuth asserts that hyphenated versions of "newly coined" words morph to unhyphenated versions (e-mail becomes email, for example) as they gain acceptance and are more frequently used, which indicates to me that option (2) is not off the table, certainly because the word is not newly coined.

However, I argue that wide acceptance and usage of the wording in the context given here is not demonstrated and fall back to option (1). Finally, as pointed out by Martin in the comment thread, be consistent and use (or don't use) hyphenation regardless of which variation you choose.


Periodic table groups - which grouping is "right"? In searching online, I've noticed there are a lot of different ways to group the elements of the periodic table.

Take mercury in the two tables linked below, for example:

http://i.telegraph.co.uk/multimedia/archive/02101/periodic_2101916b.jpg
http://www.sciencegeek.net/tables/PToE_basic.pdf

On the first table, it's explicitly outside of the transition metals. On the second table, it's included in the "transition metals" group.

Mercury is an example of this inconsistent classification, but in general, why is there no canonical grouping for all elements, across the periodic table? I see some tables mentioning "metalloids," but others not doing so. Or, sometimes elements are classified in entirely different groups altogether (like zinc, cadmium and mercury in the above-linked tables). Or, the lanthanides and actinides being included sometimes in the transition metals group, and sometimes not.

So which classification is right? Or rather, which one is the "most right"? Really: which one should I focus my efforts on remembering?

There is no "most right classification scheme" for the elements of the periodic table, worth memorizing above all others.  This is because there is no single, 'universal' set of criteria that effectively, unarguably, and neatly divides the entire periodic table into groups. Depending on the properties of interest, different classifications/groupings are more or less appropriate.


Does boiling water alter the chemical properties of dishwashing liquid? An ex-housemate of mine claimed to be a former chemist, and once told me not to use boiling water (or recently boiled water) when leaving dishes to soak, as the temperature would chemically alter the dishwashing liquid, rendering it useless or at least significantly less effective. However I've never found anything in internet searches to support this assertion. Can you lovely peeps help give me a definitive answer?

It depends rather strongly on your brand.

Soap, by itself, is unlikely to be affected by the boiling water. They're relatively stable, especially for an organic molecule. However, many of us use detergents that include enzymes that may denature in the boiling water, which is what your friend may have been referring to: A denatured enzyme is not very effective at cleaning at all.

So, look on your bottle! If it mentions enzymes, it probably won't go over well with boiling water. If it's just soap, and, say, bleach, boil away!


How is the pka related to the ph of the solution for an amine? So was reading a magazine for competitive exams and this question is bugging me a lot.

How is pka values of a base associated with the pH of the solution. The solution for it is that the Nitrogens would remain this way and only the H from OH would be dissociated. That's obvious because the medium is basic and the acidic H would be dissociated, but what about the nitrogen atoms? How is the pka of a base represented? Is that the equilibrium constant for it's conjugate base?

The pKa's quoted for the amine groups are for the protonated forms of the amines. The protonated forms would add a +1 charge to the molecule in solution. Since amines are fundamentally basic, the protonated amine is a conjugate acid.


What is the differences between partial pressure and vapour pressure? Was looking at Henry's law and Raoult's law
constants and there seemto be lots of
equations involved.
Henry's law involves partial pressure and the latter involves the vapor pressure.
Wondering what the difference is?

In a mixture of gases, each gas has a partial pressure which is the hypothetical pressure of that gas if it alone occupied the volume of the mixture at the same temperature.

What does this mean? For example, if we have a mixture of gases $A$, $B$ and $C$ in an isolated room, then, according to Dalton's law, the pressure exerted by the gases will be the sum of their partial pressures :
$$P = p_A + p_B + p_C$$
where $p_A$, $p_B$ and $p_C$ are the partial pressures of each gas. Also, if we have a moles of $A$, b moles of $B$ and c moles of $C$, we can express the partial pressure of each gas as below:
$$p_A = \frac{a}{a+b+c} P$$
$$p_B = \frac{b}{a+b+c} P$$
$$p_B = \frac{c}{a+b+c} P$$

  Vapor pressure or equilibrium vapor pressure is the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

As the definition states, if we took an equilibrium as $\ce{A(l) <=> A(g)}$, where $A(l)$ and $A(g)$ are the liquid and the vapor state of the compound $A$ . The vapor pressure of $A$ is the pressure exerted by the gaseous state ($A(g)$) over the liquid state ($A(l)$).
Now, if we have a mixture of $A$, $B$ and $C$, we will have the equilibrium between the liquid and gaseous state : $\ce{(A, B, C)(l) <=> (A, B, C)(g)}$ and the vapor pressure $P_v$ (I noted $P_v$ so there won't be any confusion between it and the total pressure $P$ I mentioned before).
According to Raoult's law, the vapor pressure $P_v$ is the sum of the products between the vapor pressures of each compound taken separately and the molar fraction of the compound in the mixture (i.e. in the solution) :
$$P_v = x_A p_{v^A} + x_B p_{v^B} + x_C p_{v^C}$$

This may look just like Dalton's law, but there are different pressures involved.
The partial pressure of a gas is the pressure exerted by a gas in the volume occupied by a mixture of gases, while the vapor pressure of a gas is the pressure exerted by a gas over its condensed phase.
Although :

  The vapor pressure that a single component in a mixture contributes to the total pressure in the system is called partial pressure.

The vapor pressure still refers to the pressure of a gas over its liquid (or solid) phase in a phase transition (condensed phase$\ce{<=>}$non-condensed phase), while the notion of partial pressure can be extended to a mixture of gases only.

All the definitions are linked to Wikipedia, and they can really cause confusion throughout chemistry learners..


Why are temperature and pressure represented as the number 2 in Gibbs' Phase Rule? In our class we are learning about Gibbs' Phase Rule; $F=C-P+2$, in which $F$ stands for degree of freedom, $C$ for number of components, and $P$ for number of phases in the system. Our professor told us that the number $2$ stood for Pressure and Temperature, but I don't understand why they are treated as constants when Composition isn't.

I must be missing something obvious, but right now I can't see what it might be...

The underlying ah-ha moment is that the Gibbs phase rule works best in (constructing) a phase diagram. And those are usually 2-dimensional - with pressure along one axis and temperature the other. So the "2" simply reflects that 2 dimensionality. At 2 degrees of freedom, there are no constraints and the system can change either pressure or temperature and be stable without a phase transition. If you have 1 degree of freedom, then you either have a constraint on $P = k$ or $T = k$ or a "mixture" $f(P,T) = k$ (If P is changed, T must be a specific value) . At 0 degrees of freedom you are at a triple point $(P =k, T=k)$ or similar (there are a handful of named "points") - any movement in either direction means the universe is going to change the phases present or the components present.

So they are not really constants, but if the degrees of freedom are 0, then they are... "fixed" - or better yet "constrained".

The archetypical example is Ice-Water-Steam. There will be at least 3 lines in this diagram from 2 phases in equilibrium and the 1 component - and 1 point.

Out of the 3 components, lets take 2 and see what we get.

$$DoF = 1-2+2 = 1$$.

So we see that for 2 phases to be in equilibrium we have 1 constraint eating away at our DoF and we know then that there will be a fixed relationship between T and P here - we can change P and still have the equilibrium but then we have to change T as well with a specific amount. This creates a line in our T-P diagram. Since there are 3 ways to arrange this we get 3 lines (I+W) + (S+W) + (I+S)

So lets see what happens if we force all three phases to be at equilibrium:

$$DoF = 1-3+2 = 0$$.

So that is no wiggle room. There will be a point where this happens, a specific combination of P and T - where this equilibrium exists.

Now lets say I invent a new phase, and I call it Ice-9. Lets see if we can find a place where this coexists with all the others.

$$DoF = 1-4+2 = -1$$.

Nope. This is not stable. No equilibrium can exist which permits my Ice-9, regular ice, water and steam at the same time. So now we know that Kurt Vonnegut is just making stuff up, what a hack. The result can be seen here... 3 lines and a point. Don't mind the dotted stuff, that is because water is more complicated than what it is at first glance


Compensating for hardness in aqueous NaOH solution? I'm trying to strip the seasoning (polymerized oil) off a carbon steel skillet.

Following online instructions, I've prepared a lye solution, using tap water, 2.5% concentration by weight, which ought to produce results in a matter of days; two weeks in, it's still far from being done.

It occurred to me we have hard water here, 13 dKH/15 dGH, so I'm guessing the solution just isn't as basic as it was expected to be.

Heating it up isn't an option, so I'm planning to add more crystals of sodium hydroxide to speed the reaction up.

Assuming that my diagnosis is correct, and the instructions were based on tap water hardness that is typical for the US, how much more hydroxide should I be adding to compensate for the higher hardness?

The short answer to your question is that the dGH of your water is negligible compared to the amount of lye you are using, so no adjustment needs to be made on the basis of hardness. 15 dGH corresponds to 150 mg/L or 0.015% as CaO, which is trivial compared to your 2.5% lye solution.

Other possible reasons for the slow cleaning process include the way in which the oil has polymerized (as a result of what has been cooked and under what conditions) and the quantity of polymerized oil present. Regardless, the means by which you can speed up the cleaning using your method are to increase temperature and increase lye concentration. If heating really is not an option for you, then you could try a 10% solution (as noted by Waylander in the comments). Disposal of this solution is not a problem, just dilute the solution first or during disposal by running water and pouring slowly. Many household cleaning products use lye so this really isn't an issue.


A question about rotational symmetry of organic molecules and in particular, the cyclic Isomers of C3H4X2 I was thinking about an organic chemistry problem a few days ago but reached a dead end. There were 2 parts to this question.
The first was thinking about the isomers of 'C3H4X2' [where X could be anything like a halogen or a deuterium etc].
I came up with the following 7 isomers:

(I've really tried to make it as clear as possible! Hopefully it has worked!)
...but a website I found stated that there were only 4:
A -
B -
C -
D -
;with their 'D' representing a deuterium.
So this is the first aspect of my quandary - what am I doing wrong when I draw these isomers!
The second aspect to this question was about considering the symmetry of these organic molecules. In particular, considering firstly 'plane symmetry' [which I understand] and 'rotational symmetry' [which I thought I understand but clearly I don't!]
The same website says that molecule A (from above) plane symmetry and rotational symmetry, which is fine.
But it also says that molecule C has plane symmetry but not rotational. Why is this? Surely if I rotate this molecule 60 degrees I get the same shape?
And then finally, they say that molecules B and D have rotational symmetry but not plane symmetry which I think I understand but would appreciate any further comments if anyone had them.
Ultimately, I would be very grateful if someone could point me towards something which could help me understand rotational symmetry in organic molecules and in particular why these examples given above have their particular symmetries! Thank you very much, everyone!
Best wishes,

If the website diagram didn't clarify the situation immediately, the issue is not to answer questions about 1 to 7, but to understand why the question is wrong.
My organic chemistry course had us playing around with molecular models - 3D representations of reality, one type with a metal framework representing the multivalent nucleus and plastic tubes representing bond lengths, the other type with wooden balls and springs. They are a far cry from reality - but molecular models give an added dimension of understanding.
Lacking such 3D models, and lacking an answer sheet, one possibility would be to go to 3D images on the web: they are actually in 2 dimensions, but with perspective in the drawing, so it could be sort of 2.5 D, like this:

Only after knowing what is represented by the dotted lines and tilted bond can you understand what you are drawing. The proposed differences between structures 1, 2 and 3 indicate an unfamiliarity with what the bond representations actually mean. One look at the picture above should clarify everything. Perhaps one more comment should be added: the bonds between a carbon and its two substituents are never coplanar with the 3-carbon ring. Interestingly, I did not find any other similar images (of dichlorocylopropane). Everything was of the straight line or dashed line/tilted bond type. This just assumes that the lines/dashes have an obvious meaning. Sometimes a good picture is worth a thousand words. (This answer is worth 1254 words, if you include 1000 words for the picture.)


Non-zero dipole moment of hydroquinone Why does hydroquinone possess a non-zero dipole moment? The $\ce{OH}$ groups present at para positions on the benzene ring should cancel the effect of each other... if there is a plane change then exactly when do atoms in a molecule change planes?

According to J. Am. Chem. Soc. 1945, 67 (2), 322–324:

Planar configurations for resorcinol and hydroquinone are more acceptable on comparison of experiment and calculation than are the structures assuming either free or partially restricted rotation of the hydroxyl groups.

It states that the dipole is present because there is restricted rotation with the hydroxyl groups. The preferred conformation for hydroquinone is a flat molecule with the hydrogens in the plane.


Is it possible to form an alloy of gold and zinc? I know you can dissolve gold in mercury, but i do not have access to any. Would it be possible to dissolve it in molten zinc instead mercury.

Yes. There are alloys and intermetallic compounds up to $\ce{44\%Au:56\%Zn}$.  This phase diagram is from here.

That said, crystal structure changes with composition, so appearance on solidification might be hard to predict.

Are you trying to emulate Hevesy?


What is the refractive index of poly(methacrylamide)? I've been looking for the index of refraction for poly(methacrylamide). I searched several vendors and in https://polymerdatabase.com/ with no success. Is there any database where I can find this information?

The refraction index of the monomer would be useful too.

As reported in  Gooch JW (Jan W. Encyclopedic Dictionary of Polymers. Springer; 2011. doi:10.1007/978-1-4419-6247-8. Poly(N-methyl methacrylamide) has a refractive index of 1.5398


Coefficient in front of molecules in a chemical equation In a chemical equation representing a chemical reaction what do the numbers in front of the molecules of compound and elements indicate? Do they signify number of molecules or moles or volume?

Example : $$\ce{ 2H_2 +O_2  -> 2H_2O}$$

Here does the "2" before $\ce{H2}$ and $\ce{H2O}$ indicate number of molecules or moles or volume?
In a particular chapter in my textbook it was written that they indicate number of molecules(atoms) but in a later chapter the numbers were being treated as if they are number of moles or volume.

There's the trick: it's all of them.

The coefficients are a ratio. For every two unit of hydrogen molecule, you need one unit of oxygen molecule, and the result is two unit of water molecule.

That unit is often the particle itself, whether it's an atom, a metal ion, a more substantial molecule or even a protein or enzyme. It can also be the volume (for ideal gases under the same conditions), the moles used in reaction, or molarity in solution.

Simply remember that the coefficient applies to the item following it as a whole. As long as the ratio between coefficients is the same, you can use any unit that maintains it.


Free rotation of sigma bonds; diastereomer formation; thermodynamic and kinetic products For the reaction of 1,3-butadiene with DBr (D = heavy hydrogen), I found that several possible products could be formed.

Kinetic products

A pair of enantiomers is formed. One of the pi bonds attacks the heavy proton and the leaving group, bromide anion, comes back around to take care of the allylic carbocation. Because the bromide can attack from both above and below the plane of the carbocation, both enantiomers are formed.

Thermodynamic products

2) Now, because there is resonance, the carbocation possesses partial primary allylic carbocation character. As a result, bromide anion may also attack at a different position. This results in an internal alkene - the more stable alkene (the Zaitsev alkene).

Diastereomers are formed as well because single bonds can rotate, and the partial, terminal double bond can be "flipped" upward in the resonance depiction (I know that resonance depictions do not exist independently; hence "partial"). This allows the Z alkene to be formed in addition to the E alkene, as illustrated by my mechanism.

Questions

1) Are the products all valid and correctly categorized?

2) Is the E and Z isomer each produced, and in the way I think both are produced?

You have only considered reaction from the s-trans conformation of 1,3-butadiene, where the 2 double bonds are in a cis relationship about the central single bond.  The s-trans conformation is in equilibrium with the s-cis conformation.  The s-cis conformation can react also react to the terminal olefin, but also  (by the mechanism you described) it can react to produce the z 2-butene as one of its kinetic products.  Hence, all 3 butenes (1-butene and e and x 2-butene) can be formed as the kinetic products.


Density of d-block elements Something that confuses me slightly is the trends in density when comparing periods 4, 5, and 6 in the d-block. Looking at periods 5 and 6, the density peaks at group 8, with ruthenium and osmium, respectively. The density slightly drops at group 9 and falls further through group 10-12 and the p-block.
But in period 4, just from assumption you would expect iron, the first group 8 member to be the densest of period 4, but it's not at all. Sure, iron is the densest yet, but the trend increases up to copper and only starts dropping at group 12 and beyond. Why is there this discrepancy?
Furthermore, why does density seem to peak at group 8 anyway? What is it about the d-shells that create this general pattern?

Electron are added when you go to the right of a $d$ line, they have the same principal quantic number and so they are not making the atom bigger, but at the same time there is more protons and neutron resulting in more attractive force : the density increase because the volume is slightly the same but the masses of nucleus increase.

This phenomena start to be inverted at some point of the line because the "shell" (repulsive force due to other electron) of previous electron start to be important enough, the new added electron is not able to penetrate this cloud of electron and make the volume of the atom bigger, resulting in a tendency of decreasing density.

This phenomena of "shell" is explain by Slater and even have rule.

https://en.wikipedia.org/wiki/Slater%27s_rules?wprov=sfla1

EDIT : The explanation for the peak of density of the last group is explained by the lanthanide contraction. The $f$ orbitals have really low shield power and will cause a contraction of all the new added electron : it will be easier for them to penetrate the cloud of electron. This result in higher density

https://en.wikipedia.org/wiki/Lanthanide_contraction?wprov=sfla1


Why is fluorine deactivating while OH is not? Both have 2p-2p overlap and oxygen and fluorine have similar electronegativities but OH is strongly activating while fluorine is deactivating

F is the most electronegative element of the periodic table.Due to its -I effect, it shifts the electron density of any alkyl group towards itself, resulting in less density of electron cloud over the alkyl or phenyl group.This deactivates the ring for electrophilic substitution reactions.However , oxygen also being an electronegative element , has capacity to donate its lone pair electrons forming a back bond in case their is a deficiency in alkyl group.In case of cyclic rings that is in aromatic structures , the lone of oxygen is in  resonance with the ring during attack of the reagent , that is during formation of sigma complex.However this is not shown by fluorine due to its high electronegativity.Do not get confused with other halogens.Cl ,Br and I are ring activating groups as they also can form the same back bond as oxygen.


Are all lift-off resist negative photoresists? Are all lift-off resist negative photoresists? I'm a bit confused as I can't find anything in my textbook/lecture slides about the type of resist used for lift-off.

There are definitely positive photo resists, but there tends to be little incentive to use them. Usually you use a negative mask with a negative resist to create a positive part. The pattern to be made tends to be a relatively small area of the photo resist on the mask and thus it takes less time for the lithography machine to produce a negative mask which can be a bottle neck in an R&D setting. Also Positive resists tend to be perminate as they have chemically inter-reacted to stay in place not be dissolved later (like during lift-off), thus another reason positive resists are not used frequently.


How to calculate the percentage dissociation of phosgene using the pressure equilibrium constant? In phosgene gas reaction at $\pu{400 ^\circ C}$, the initial pressure $p_{\ce{CO}}= \pu{342 mmHg}$ and $p_{\ce{Cl2}}= \pu{352 mmHg}$ and the the total pressure at equlibirum is $\pu{440 mmHg}$.
  $$\ce{CO + Cl_2 = COCl_2}$$
  Calculate the percentage dissociation of phosgene at $\pu{400^\circ C}$ at $\pu{1 atm}$.

Let $x$ be the decrease in pressure of both $\ce{CO}$ and $\ce{Cl_2}$
\begin{array}{lcccc}
 & \ce{CO &+& Cl_2 &<=> &COCl_2}\\\hline
\text{Initial pressure of substance} & 342 && 352 && 0\\
\text{Final pressure of substance}   & 342-x && 352-x && x\\\hline
\end{array}

$$342 - x + 352 - x + x = 440 \implies x = 254$$

Using the formula for $K_p$ I found it to be $\approx 13$.
For the dissociation at $\pu{760 mmHg}$,
$$\ce{COCl2 <=> CO + Cl2} \implies K_p= \frac{1}{13}$$

Let $y$  be the decrease in pressure of phosgene which is proportional to the amount of substance of it lost or it's degree of dissociation.
$$\frac{1}{13} = \frac{y^2}{760-y} \implies y \approx 7.6$$

The value of $y$, that I have obtained, must be proportional to percentage dissociation, so percentage dissociation should be $7.6 \%$. But the answer given is $20.6 \%$.

I am not looking for the entire solution but would like to know where have I misapplied the concepts. I am doubtful about my second step, but can't figure out how it's wrong.

Here goes a possible answer, I tried to describe it step by step so you can see your (and my) possible errors. The first part of your calculation was conceptually good, but you make the calculation wrong.

As one of the comments say, $$K_p=\frac{(342-254)*(352-254)}{254}=33.95$$

So you have $K_p$. Then

\begin{array}{lcccc}
 & \ce{CO &+& Cl_2 &<=> &COCl_2}\\
\text{Initial pressure of substance} &0 && 0 && A\\
\text{Final pressure of substance}   & x && x && A-x\\~\\
\end{array}

With the condition that $x + x + A-x=760$, we have $A+x=760$.

Using our first calculation:

$$K_p=\frac{(x)*(x)}{A-x}=33.95$$
(Kp is the same at same $T$)

$$K_p=\frac{x^2}{760-2x}=33.95$$
here we get x=130.229, then:
dissociation of $A= 1-\frac{(A-x)}{A}$
we get dissociation of A=1-0793=0.207

or 20.7% (0.01% flew away)


Byproducts of LiAlH4 reduction of amides I know that $\ce{LiAlH4}$ is able to reduce amides into amines, e.g. benzamide into benzylamine. However, I am unsure what other by-products are formed as a result.

$$\ce{LiAlH4 + C6H5CONH2 -> C6H5CH2NH2 + H2O + ??}$$

Organic chemistry textbooks I flipped through usually don't focus on by-products, only noting that LAH ($\ce{LiAlH4}$) in dry solvent (THF, diethylether) is necessary for reduction, but when one converts amides to amines, an additional protolysis step is required:  e.g. from Loudon's Organic Chemistry, chapter 21B. "Reduction of Amides to Amines" [1, p. 1080]:

In the workup conditions, $\ce{H3O+}$ is followed by $\ce{^–OH}$. An aqueous acidic solution is often used to carry out the protonolysis step that follows the $\ce{LiAlH4}$ reduction (as shown in the following mechanism). The excess of acid that is typically used converts the amine, which is a base, into its conjugate-acid ammonium ion. Hydroxide is then required to neutralize this ammonium salt and thus give the neutral amine.

Although water itself rather than acid can be used in the protonolysis step, for practical reasons the acidic workup is more convenient. Thus, the extra neutralization step is required.

The latter $\ce{H2O}$-quenching step is the reason why $\ce{AlH3}$ is not remained among reactive species and, according to Brown, the brutto-formula of the metal-containing side-product is $\ce{LiAlO2}$ which is in agreement with stoichiometrical equation for reactants (amide : LAH = 1 : 0.5)[2]:
$$\ce{2RCONR'R'' + LiAlH4 -> 2RCH2NR'R'' + LiAlO2}$$
It is also worth noticing that if there is moisture, LAH readily reacts with water producing $\ce{LiOH}$ and $\ce{Al(OH)3}$. Also, there is a chance of side-reaction with the solvent, such as formation of lithium amide complexes with THF.
Recent review of catalytic and non-catalytic methods of amide reduction [3] also covers few more possibilities:

When less than stoichiometric quantities of $\ce{LiAlH4}$ are used for the reduction of primary amides, the corresponding nitrile may be formed:

Moreover, high selectivity toward the nitrile is also observed with amides containing electron-rich quinoline systems. If traces of water are present,
partial reduction of amides to aldehydes over $\ce{LiAlH4}$ can also comprise an undesirable competing reaction. Hydrolysis to the alcohol product has also been reported for secondary amide substrates.

GC analysis also shows presence of $\ce{LiOAlH3}$, $\ce{Li+AlH_xR_{4-x}-}$ in the system [4], though I would doubt that for benzamide $x$ can be less that 3.
References

Loudon, G. M.; Parise, J. Organic chemistry, 6th Ed.; Roberts and Company Publishers: Greenwood Village, Colorado, 2016. ISBN 978-1-936221-34-9.
Nystrom, R. F.; Brown, W. G. Journal of the American Chemical Society 1948, 70 (11), 3738–3740. DOI 10.1021/ja01191a057.
Smith, A. M.; Whyman, R. Chemical Reviews 2014, 114 (10), 5477–5510. DOI 10.1021/cr400609m.
Thiedemann, B.; Schmitz, C. M. L.; Staubitz, A. The Journal of Organic Chemistry 2014, 79 (21), 10284–10295. DOI 10.1021/jo501907v.


Why does the Gibbs free energy only correspond to non-expansion work? It has been defined as the energy available for work other than expansion work. Why can't it be used for expansion work

The association of the Gibbs free energy with “additional”, or “non-expansion” work, is simply a mathematical result of its definition: $G = H - TS = U + pV - TS$.

From the First Law, for a closed system, we have $\mathrm{d}U = \delta q + \delta w$.

For a reversible change, the Second Law tells us that $\mathrm{d}S = \delta q_\text{rev}/T$; so $\delta q_\text{rev} = T\,\mathrm{d}S$.

We can split $\mathrm{d}w$ into two parts, expansion work and additional work: expansion work is given by the formula $\mathrm{d}w_{\text{exp}} = -p\,\mathrm{d}V$, and for now we can just label additional work as $\mathrm{d}w_{\text{add}}$. In this context, additional simply means anything that isn’t expansion work. Putting all this together, we have

$$\mathrm{d}U = T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}}$$

and from the definition of the Gibbs free energy in the first paragraph,

$$\begin{align}
\mathrm{d}G &= \mathrm{d}(U + pV - TS)\\
&= \mathrm{d}U + (p\,\mathrm{d}V + V\,\mathrm{d}p) - (T\,\mathrm{d}S + S\,\mathrm{d}T)\\
&= T\,\mathrm{d}S - p\,\mathrm{d}V + \delta w_{\text{add}} + p\,\mathrm{d}V + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T\\
&= V\,\mathrm{d}p - S\,\mathrm{d}T + \delta w_{\text{add}}
\end{align}$$

Now, if a closed system is kept at constant pressure and temperature, $\mathrm{d}p = 0$ and $\mathrm{d}T = 0$. Therefore $\mathrm{d}G = \delta w_{\text{add}}$.

The simplest application of this interpretation of $G$ is in electrochemistry, where one could move $|\mathrm{d}n|$ moles of electrons through a certain potential difference $E$; the “additional work” done in this case is given by $-FE\,|\mathrm{d}n|$ (you could consult a physics text for details). This gives us $\mathrm{d}G = -FE\,|\mathrm{d}n|$, and (after several more steps) the familiar equation $\Delta_\mathrm{r} G = -zFE$.


Is 2-bromobutane chiral? My textbook says 2-bromobutane is chiral, but it look like it will form superimposition mirror image

For a molecule to be chiral it must have non-superimposable mirror images.  Here is a drawing of the two mirror images for 2-bromobutane.

The chiral carbon atom is denoted by an asterisk.  In the case of 2-bromobutane there are 4 different substituents attached to it. The molecule is chiral, you can't pick up one of the mirror images and superimpose it on the other.

Note to think about:

4 different substituents may make an atom chiral, but the molecule as a whole may still be achiral

For example meso compounds

image source

some molecules without 4 different substituents attached to an atom may still be chiral .

For example, compounds with axial chirality such as chiral allenes (1,3-dimethylallene), chiral biphenyls, chiral helicenes, etc.

image source

image source

image source


Potassium carbonate vs. Potassium hydroxide in Wood Ash I'm trying to get more concrete information about the chemical process of creating soap from wood ash.

What I’m confused about: Many soap-making materials say that you create potassium hydroxide directly from filtering water through wood ash. As far as I can tell this is false? You are actually getting potassium carbonate (and some small amount of other residue.)

To create potassium hydroxide from this, historically you would add a solution of calcium hydroxide (calcium oxide/burnt limestone dissolved in water). Calcium carbonate would precipitate out of the solution, leaving behind potassium hydroxide. From my understanding, a lot of modern soap-makers are simply misunderstanding the chemistry, or not realizing that the lime has to be added before they actually have potassium hydroxide.

However, some sources say that if the wood is heated hotter and given a more complete burn, thus combusting more of the organic material, then more hydroxides will be dissolved instead of carbonates, thus negating the need for lime. Is this true?

I have also read that if you take the potassium carbonate, dehydrated, and heat it to an (uncertain) degree, it will burn off carbon dioxide and create potassium oxide, which added to water will become potassium hydroxide. Is this true? How hot do you need to burn it, if so?

However, other sources say heating potassium carbonate only purifies it, creating pearl ash, which is still potassium carbonate.

Basically, I just need more information about potassium carbonate vs. potassium hydroxide from wood ash sources.

You can never get $\ce{KOH}$ directly out of wood ash. One can recall that potash $\ce{=K2CO3}$ is indeed pot-ash! So you are right that historically calcium hydroxide was needed to obtained $\ce{KOH}$ out of potash. The rest of the stories are all false. $\ce{KOH}$ is unstable in air with respect to carbonate formation. It is hygroscopic as well. Wood ash remains bone dry forever. I don't think there is even a trace of potassium hydroxide in wood ash.

Until and unless you don't have a furnace, which can generate temperatures above 1300○ Celsius chances of getting $\ce{KOH}$ in useful amounts are almost negligible.

Secondly, a soap made with tons of calcium salts will be nothing but literally scum (=insoluble calcium soaps). Calcium should not be a part of soap.

Here is a nice abstract on the composition of wood ash as a function of temperature. [1]

  The elemental and molecular composition of mineral matter in five wood types and two barks was investigated as a function of temperature using thermal gravimetric analysis, differential thermal analysis, inductively coupled plasma emission spectroscopy, and X-ray diffraction. Low temperature ash was prepared at 500°C, and samples were heated in a tube furnace at temperature increments to 1400°C. The dissociation of carbonates and the volatilization of potassium, sulfur, and trace amounts of copper and boron were investigated as a function of temperature. Overall mass loss of the mineral ash ranged from 23–48% depending on wood type. The mass of K, S, B, Na, and Cu decreased, whereas Mg, P, Mn, Al, Fe, and Si did not change with temperature relative to Ca which was assumed to be constant. Sintering of the ash occurred, but fusion of the ash did not occur. In the 600°C ash CaCO3 and K2Ca(CO3)2 were identified, whereas in 1300°C ash CaO and MgO were the main compounds. The implications for ash deposition in furnaces is discussed.

References:

Misra, M. K.; Ragland, K. W.; Baker, A. J. Wood ash composition as a function of furnace temperature. Biomass Bioenergy 1993, 4 (2), 103–116 DOI: 10.1016/0961-9534(93)90032-Y.


How does ferrous iron coordinate nitrogen in a porphyrin ring? In first semester biochemistry, my professor said that iron forms a coordination complex with nitrogens in the heme group.

I'm failing to understand what this even means since the charge on nitrogen does not change as a result of this bond. My organic chemistry textbook and biochemistry textbook do not even have a basic explanation of what is meant by this, and I can't find anything on the web.

How does iron form a bond with two of the nitrogens in the porphyrin ring, the histidine, and the oxygen?

This is something I have asked myself for a long time until I finally realized by myself. I have superficial knowledge on this type of chemistry but I believe this will be enough, so i will explain it in a very simple fashion. Well you have to look at Nitrogen first, nitrogen doesn’t change it’s charge just like you said, which seems weird at first since it forms four bonds. Now think of nitrogen in the lewis formula, what does it look like? Nitrogen has a free pair of electrons available. Think of the d-orbitals in Fe, their character and shape. Think of metals in general, having their electrons in a type of diffuse cloud. Now to the bond, the Iron is capable of grabbing these free electron pairs and use them in a bond, pulling them into it’s direction. This way the charge doesn’t change. The nitrogen “donates” it’s electrons which do not participate in regular bonding like you would see it in ammonia, but in complex bonds. Hope this is enough to clarify :)
Greetings from Germany


Why is it not practical for insoluble salts to be prepared from other insoluble salts? My textbook says that it is not practically feasible to make insoluble salts (say PbSO4) from other insoluble salts (say PbO), without explicitly mentioning the reason for the same.

My guess is: if the product is also an insoluble salt, it may deposit on the insoluble reactant, and stop further reaction. However, isn't this problem solved by simply stirring? (Is the constant stirring required the reason why such a method would not be "practically feasible"? But isn't stirring rather common for chemical reactions?)

So, what is the actual reason?

The comment "My textbook says that it is not practically feasible to make insoluble salts (say PbSO4) from other insoluble salts" is not precisely accurate. Limited solubility in water does not necessarily imply a lack of solubility in an acidic medium.

For example, I have prepared Silver acetate (moderately water-soluble, see Table) from the action of dilute acetic acid and 3% H2O2 in the presence of an electrolyte (a small amount of KNO3). Jumpstart the reaction in a microwave. The created Silver acetate is somewhat soluble in acetic acid (see this), hence there is no issue in the creation reaction.

The addition of moderately soluble PbCl2 to Silver acetate/acetic acid likely creates insoluble AgCl and Lead acetate (where the Silver acetate is also moderately water-soluble with respect to water). An explanation of solubility properties, per reference above, relates to a possible creation of an acetic acid salt.


What is the exchange interaction? As the wikipedia article for the exchange interaction so aptly notes, exchange "has no classical analogue."

How wonderful.

Exchange shows up essentially while enforcing the condition that two electrons should not be distinguishable.

That makes sense as a condition to enforce, but I don't understand how this should influence the energetics of the system.

For instance, when two neon atoms interact, there is a very small amount of energy contribution which is attributed to exchange. I understand that exchange is not a force because there is no force carrier, but there is indeed an interaction taking place which affects the energy of the system. What then am I meant to believe is going on? Am I supposed to believe that two electrons spontaneously switch places so as to maintain their indistinguishability? This seems ridiculous. Especially because such a process ought to happen over a finite period of time and thus be observable, but this doesn't seem to be the case (at least as far as I've heard).

What is exchange then? I understand this may not even be an answerable question beyond the fact that it is an effect which drops out of the math and simply is.

Hopefully someone has some interesting insight for me here.

Also, I have seen this question which I believe is referencing the same effect because exchange is known to play a role in ferromagnetism, but nothing ever really caught hold there probably because the question referenced a specific application and so was very difficult to answer well.

In quantum chemistry, probably the easiest way to understand the "exchange interaction" is within the context of the Hartree-Fock model.
$
\newcommand{\op}{\hat}
\newcommand{\el}{_\mathrm{e}}
\newcommand{\elel}{_\mathrm{ee}}
\newcommand{\elnuc}{_{\mathrm{en}}}
\newcommand{\core}{^{\mathrm{core}}}
\newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
$

To reduce the complexity of the electronic Schrödinger equation
$$
    \op{H}\el \psi\el(\vec{q}\el) = E\el \psi\el(\vec{q}\el) \, ,
$$
where
$$
    \op{H}\el
    =
    \op{T}\el + \op{V}\elnuc + \op{V}\elel
    =
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{\alpha=1}^{\nu} \sum\limits_{i=1}^{n} \frac{Z_{\alpha}}{r_{\alpha i}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
$$
we would like to separate the electronic coordinates from each other by writing down the many-electron wave function as a product of one-electron ones. Unfortunately, such separation won't work due to the presence of $\op{V}\elel$ term in the electronic Hamiltonian. But if instead of $\op{V}\elel$ potential, which prevents the separation of electronic coordinates, a model potential of the form $\sum\nolimits_{i=1}^{n} v_{\mathrm{MF}}(\vec{r}_{i})$ had entered the electronic Schrödinger equation, it would be reduced to a set of $n$ one-electron Schrödinger equations with the many-electron wave function being just a simple product of their solutions, one-electron wave functions $\psi_{i}(\vec{r}_{i})$.

More precisely, since we have to take spin of electrons and the principle of antisymmetry of the electronic wave function into account, the electronic wave function would be an antisymmetric product of one-electron wave functions $\psi_{i}(\vec{q}_{i})$ termed the Slater determinant,
$$
    \Phi
    =
    \frac{1}{\sqrt{n!}}
    \begin{vmatrix}
        \psi_{1}(\vec{q}_{1})  &  \psi_{2}(\vec{q}_{1})  &  \cdots  &  \psi_{n}(\vec{q}_{1})  \\
        \psi_{1}(\vec{q}_{2})  &  \psi_{2}(\vec{q}_{2})  &  \cdots  &  \psi_{n}(\vec{q}_{2})  \\
        \vdots                 &  \vdots                 &  \ddots  &  \vdots                 \\
        \psi_{1}(\vec{q}_{n})  &  \psi_{2}(\vec{q}_{n})  &  \cdots  &  \psi_{n}(\vec{q}_{n})
    \end{vmatrix} \, ,
$$
where the one-electron wave functions $\psi_{i}$ of the joint spin-spatial coordinates of electrons $\vec{q}_{i} = \{ \vec{r}_{i}, m_{si} \}$ are referred to as spin orbitals. Physically the model potential of the above mentioned form represent the case when electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons and given by $v_{\mathrm{MF}}(\vec{r}_{i})$.

Minimization of the electronic energy functional
$$
E\el^{\mathrm{HF}} = \bracket{ \Phi }{ \op{H}\el }{ \Phi } = \bracket{ \Phi }{ \op{T}\el }{ \Phi } + \bracket{ \Phi }{ \op{V}\elnuc }{ \Phi } + \bracket{ \Phi }{ \op{V}\elel }{ \Phi }
$$
with respect to variations in spin orbitals subject to constraint that spin orbitals remain orthonormal eventually results in the set of the so-called canonical Hartree-Fock equations which define canonical spin orbitals $\psi_i$ together with the corresponding orbital energies $\varepsilon_i$
$$
    \op{F} \psi_{i}(\vec{q}_{1})
    =
    ε_{i} \psi_{i}(\vec{q}_{1}) \, ,
    \quad
    i = 1, \dotsc, n \, ,
$$
where $\op{F} = \op{H}\core + \sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$ is the Fock operator with
$$
    \op{H}\core \psi_{i}(\vec{q}_1)
    =
    \Big(
        - \frac{1}{2} \nabla_{1}^2
        - \sum\limits_{\alpha=1}^{\nu} \frac{Z_{\alpha}}{r_{\alpha 1}}
    \Big)
    \psi_{i}(\vec{q}_1) \, , \\
    \op{J}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{j}(\vec{q}_2) } \psi_{i}(\vec{q}_1) \, , \\
    \op{K}_{j} \psi_{i}(\vec{q}_1) = \bracket{ \psi_{j}(\vec{q}_2) }{ r_{12}^{-1} }{ \psi_{i}(\vec{q}_2) } \psi_{j}(\vec{q}_1) \, .
$$

Here, the first part of the Fock operator, $\op{H}\core$, is known as the one-electron core Hamiltonian and it is simply the Hamiltonian operator for a system containing the same number of nuclei, but only one electron. The second part of the Fock operator, namely, $\sum\nolimits_{j=1}^{n} \big(\op{J}_{j} - \op{K}_{j} \big)$, plays the role of the mean-field potential $v_{\mathrm{MF}}$ that approximates the true potential $\op{V}\elel$ of interactions between the electrons. The first operator $\op{J}_{j}$ rewritten in the following form
$$
    \op{J}_{j}(\vec{q}_{1})
    =
    \int \psi_{j}^*(\vec{q}_{2}) r_{12}^{-1} \psi_{j}(\vec{q}_{2}) \,\mathrm{d} {\vec{q}_{2}}
    =
    \int \frac{ |\psi_{j}(\vec{q}_2)|^{2} }{ r_{12} } \mathrm{d} \vec{q}_2 \, ,
$$
can be clearly interpreted as the the Coulomb potential for electron-one at a particular point $\vec{r}_{1}$ in an electric field created by electron-two distributed over the space with the probability density $|ψ_{j}(\vec{q}_2)|^{2}$, and for this reason it is called the Coulomb operator.
The second operator $\op{K}_{j}$ has no simple physical interpretation, but it can be shown to arise entirely due to anti-symmetry requirement, i.e. if instead of a Slater determinant one uses a simple product of spin orbitals, termed the Hartree product, there will be no $\op{K}_{j}$ terms in the resulting equations (the so-called Hartree equations). It is for that reason that $\op{K}_{j}$ is called the exchange operator.

To quickly understand why exchange terms appear when using a Slater determinant instead of a simple product of spin orbitals look no further than at the Slater rules. For $\op{V}\elel$ which is a two-electron operator we have

$$
    \bracket{ \Phi }{ \op{V}\elel }{ \Phi }
    =
    \sum\limits_{i=1}^{n} \sum\limits_{j>i}^{n}
    \Big(
        \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{i}(1) \psi_{j}(2) }
        -
        \bracket{ \psi_{i}(1) \psi_{j}(2) }{ r_{12}^{-1} }{ \psi_{j}(1) \psi_{i}(2) }
    \Big) \, ,
$$
where the second (exchange) part would be absent if $\Phi$ were a simple product of spin orbitals, rather than a Slater determinant.1)

To the question on why do we interpret exchange terms in the Hartree-Fock as a manifestation of some sort of interaction. On the one hand, it must be clear from exchange terms $\op{K}_{j}$ being part of potential energy terms $v_{\mathrm{MF}}$, that they are indeed related to some type of interaction between electrons. However, it must be said that, unlike for the four fundamental interactions, there exists no force based on exchange interaction. Strictly speaking, I don't even think that the term "interaction" has a definite meaning in physics, unless it simply stands for the "fundamental interaction" in which case it is merely a synonym for "fundamental force".

Exchange interaction is not one of the four fundamental interactions, so the meaning of the word "interaction" here is a bit different than that for fundamental interactions. Oxford Dictionary defines interaction as follows,

  1.1 Physics A particular way in which matter, fields, and atomic and subatomic particles affect one another, e.g. through gravitation or electromagnetism.

And exchange interaction indeed is a way by which electrons (of the same spin) "affect one another". This can be readily seen as follows. Consider the simplest many-electron system, a two-electron one, and let us examine the following two important events:

$r_1$ - the event of finding electron-one at a point $\vec{r}_{1}$;
$r_2$ - the event of finding electron-one at a point $\vec{r}_{2}$.

Probability theory reminds us that in the general case the so-called joint probability of two events, say, $\vec{r}_{1}$ and $\vec{r}_{2}$, i.e. the probability of finding electron-one at point $\vec{r}_{1}$ and at the same time electron-two at point $\vec{r}_{2}$, is given by
$$
    \Pr(r_1 \cap r_2)
    =
    \Pr(r_1\,|\,r_2) \Pr(r_2)
    =
    \Pr(r_1) \Pr(r_2\,|\,r_1) \, ,
$$
where

$\Pr(r_1)$ is the probability of finding electron-one at point $\vec{r}_{1}$ irrespective of the position of electron-two;
$\Pr(r_2)$ is the probability of finding electron-two at point $\vec{r}_{2}$ irrespective of the position of electron-one;
$\Pr(r_1\,|\,r_2)$ is the probability of finding electron-one at point $\vec{r}_{1}$, given that electron-two is at $\vec{r}_{2}$;
$\Pr(r_2\,|\,r_1)$ is the probability of finding electron-two at point $\vec{r}_{2}$, given that electron-one is at $\vec{r}_{1}$.

The first two of the above probabilities are referred to as unconditional probabilities, while the last two are referred to as conditional probabilities, and in general $\Pr(A\,|\,B) \neq \Pr(A)$, unless events $A$ and $B$ are independent of each other.

If the above mentioned events $\vec{r}_{1}$ and $\vec{r}_{2}$ were independent, then the conditional probabilities would be equal to their unconditional counterparts
$$
    \Pr(r_1\,|\,r_2) = \Pr(r_1) \, , \quad
    \Pr(r_2\,|\,r_1) = \Pr(r_2) \, ,
$$
and the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ would simply be equal to the product of unconditional probabilities,
$$
    \Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2) \, .
$$
In reality, however, the events $\vec{r}_{1}$ and $\vec{r}_{2}$ are not independent of each other, because electrons "affect one another",
$$
    \Pr(r_1\,|\,r_2) \neq \Pr(r_1) \, , \quad
    \Pr(r_2\,|\,r_1) \neq \Pr(r_2) \, ,
$$
and consequently the joint probability of $\vec{r}_{1}$ and $\vec{r}_{2}$ is not equal to the product of unconditional probabilities,
$$
    \Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2) \, .
$$

Inequalities above hold true for two reasons, i.e. there are two ways electron "affect one another".

First, electrons repel each other by Coulomb forces, and, as a consequence, at small distances between the electrons $\Pr(r_1\,|\,r_2) < \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) < \Pr(r_2)$, while at large distances $\Pr(r_1\,|\,r_2) > \Pr(r_1)$ and $\Pr(r_2\,|\,r_1) > \Pr(r_2)$.
In the extreme case when $\vec{r}_{1} = \vec{r}_{2}$, the Coulomb repulsion between the electrons becomes infinite, and thus, $\Pr(\vec{r}_{1}\,|\,\vec{r}_{1}) = 0$ and consequently $\Pr(\vec{r}_{1} \cap \vec{r}_{1}) = 0$, i.e. the probability of finding two electrons at the same point in space is zero.

Secondly, as a consequence of the Pauli exclusion principle, electrons in the same spin state can not be found at the same location in space, so that for electrons in the same spin state there is an additional contribution into the inequalities between conditional and unconditional probabilities above.
This effect is relatively localized as compared to one due to Coulomb repulsion, but bearing in mind the relationship between probabilities and wave functions and taking into account the continuity of the latter, it is still noticeable when electrons are close to each other and not just at the very same location in space.

Now, for a Hartree product wave function
$$
  \psi_{\mathrm{HP}}(\vec{q}_{1}, \vec{q}_{2}) = \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) \, ,
$$
one can show that $\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ if two electrons are in different spin states as well as when they are in the same spin state2). But for a Slater determinant
$$
    \Phi(\vec{q}_{1}, \vec{q}_{2}) = \frac{1}{\sqrt{2}}
    \Big(
    \psi_{1}(\vec{q}_{1}) \psi_{2}(\vec{q}_{2}) -
    \psi_{1}(\vec{q}_{2}) \psi_{2}(\vec{q}_{1})
    \Big) \, ,
$$
$\Pr(r_1 \cap r_2) = \Pr(r_1) \Pr(r_2)$ will hold only for electrons of unlike spin, while when both electrons are in the same spin state $\Pr(r_1 \cap r_2) \neq \Pr(r_1) \Pr(r_2)$2). And this inequality is a clear indication of exchange interaction.

1) I leave it to OP as an exercise. Derivation for a simple product of spin orbitals is pretty trivial, while for the case of a Slater determinant OP can consult, for instance, Szabo & Ostlund, Modern Quantum Chemistry, Section 2.3.3.

2) I leave it as another exercise.


Why do we have an alkyl shift rather than epoxidation in a pinacol reaction? Why do we have an alkyl shift rather than epoxidation in a pinacol reaction?
I don't understand why the hydroxyl group doesn't attack the positive charge.

Excellent question.

When the first Pinacol rearrangements were being studied, there was a suggestion that the product of the reaction was indeed the epoxide (migrations/shifts weren't well known reactions at this point in time, and methods of characterisaton were limited). The reaction that was being attempted is shown below (source: Wikipedia), however at this point there was even dispute over the structure of acetone, further adding to the complications.

  One of the first reports of a Pinacol rearrangement

Today, we of course know that Pinacol and semi-Pinacol rearrangements occur when vicinal diols are treated with protic acid (or Lewis acids) to give the carbonyl product as a result of a [1,2]- alkyl shift.

  Source: Strategic Applications Of Named Reactions. Wiley.

Looking at the mechanism (below), we first have a rate-determining loss of water via protonation-elimination. This generates the all important cationic intermediate.

The simple answer to your question is that rearrangement occurs faster than epoxidation (which is never observed to any significant extent), this is intuitive if we consider that:

Under the acidic conditions, the alcohol required to form the epoxide is fully protonated, and hence not a very good nucleophile (in agreement with R-OH generally being a poor nucleophile except in basic conditions where we get the anionic alkoxide RO(-)
The migration generates a more stabilised carbocation by the influence of the oxygen LP donating into the carbocation to form the resulting ketone.

    Source: Strategic Applications Of Named Reactions. Wiley.

As an aside, if you'd kept reading through Clayden by another page or so, you would have discovered that epoxides also undergo Pinacol-type rearrangements, so realistically even if the epoxide did form, it would simply re-open under the acidic/Lewis-acidic conditions before undergoing a rearrangement reaction.

  Source: Organic Chemistry, 1 ed. Oxford University Press (Warren & Clayden)


alpha-hydrogen atom and keto-enol tautomerism The following is an extract from my book-

  For an aldehyde or a ketone to exhibit keto-enol tautomerism, it is essential that it must have atleast one $\alpha$-hydrogen atom. Thus, acetophenone, butan-2-one and propionaldehyde all contain $\alpha$- hydrogen atoms and hence show keto-enol tautomerism. In contrast, benzaldehyde, benzophenone etc. do not contain $\alpha$-hydrogen atoms.

What is meant by $\alpha$-hydrogen atom? How does it affect keto-enol tautomerism?

In general, the "alpha" hydrogen is a hydrogen attached to a carbon that is "alpha" (adjacent to) the substituent.

image source

So in this case, an "alpha" hydrogen is a hydrogen attached to the carbon alpha to the carbonyl.

image source

The following drawing represents a typical keto-enol equilibrium.  If we removed the last alpha hydrogen and replaced it with a substituent (alkyl group, phenyl group, bromine, etc.) then we could not form a double bond with that alpha carbon, it would have 4 non-removable substituents.  If we can't form a double bond that means we can't form an enol.

image source


MOs of Benzene using group theory I am trying to analyze the $\sigma$ orbitals of benzene molecule using group theory, doing the same thing that is done to the $\pi$ orbitals. The symmetry group of benzene is the $D_{6h}$; considering the $sp^{2}$ orbitals of the benzene:

and applying the symmetry operations to these orbitals, I can find the caracters of the representation with which they are transformed, and I can decompose this representation in the irreducible representation of the group. Doing this, I found

$$\Gamma = 2 A_{1g} \oplus 2E_{2g} \oplus B_{1u} \oplus B_{2u} \oplus 2E_{1u} $$

Now, I can construct the projection operators and with them I get the symmetrized orbitals for each representation. The molecular orbitals will be the linear combinations of these orbitals. My question is: In which function should I apply the projection operators? In the $sp^{2}$ function?

Your decomposition is for the hybrid orbitals that point from carbon to carbon, omitting the ones that point out towards the hydrogens, so you really want to project starting from one of the hybrid orbitals. First off we correct your decomposition:
$$\Gamma=A_{1g}\oplus A_{2g}\oplus B_{1u}\oplus B_{2u}\oplus2 E_{1u}\oplus 2E_{2g}$$
We only need half the operations of $D_{6h}$ because all the hybrid orbitals transform the same under $\sigma_h$, so we write out the matrices for the group operations of the subgroup $D_6$ and a set of matrix representations for each irreducible representation and the result of each group operation acting on $\phi_1$, first half:
$$\begin{array}{c|cccccc}R&E&C_6&C_6^2&C_6^3&C_6^4&C_6^5\\\hline
A_{1g}&1&1&1&1&1&1\\
A_{2g}&1&1&1&1&1&1\\
B_{1u}&1&-1&1&-1&1&-1\\
B_{2u}&1&-1&1&-1&1&-1\\
E_{1u}&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}
&\begin{bmatrix}-1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}\\
E_{2g}&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}
&\begin{bmatrix}1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}\\
R\phi_1&\phi_1&\phi_3&\phi_5&\phi_7&\phi_9&\phi_{11}\end{array}$$
Ummm... yeah, I assume the $x$-axis points right in your illustration and the $y$-axis points up. Also I number the hybrid orbitals $1$ to $12$ counterclockwise starting with the one just above the $x$-axis in the first quadrant. I will label the dihedral $C_2$ rotation axes $a$ to $f$ with $a$ parallel to the $x$-axis counterclockwise in $30°$ increments. So now we can produce the second half of our table:
$$\begin{array}{c|cccccc}R&C_{2a}&C_{2b}&C_{2c}&C_{2d}&C_{2e}&C_{2f}\\\hline
A_{1g}&1&1&1&1&1&1\\
A_{2g}&-1&-1&-1&-1&-1&-1\\
B_{1u}&1&-1&1&-1&1&-1\\
B_{2u}&-1&1&-1&1&-1&1\\
E_{1u}&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&-\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}
&\begin{bmatrix}-1&0\\0&1\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&-\frac12\end{bmatrix}\\
E_{2g}&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}
&\begin{bmatrix}1&0\\0&-1\end{bmatrix}&\begin{bmatrix}-\frac12&\frac{\sqrt3}2\\\frac{\sqrt3}2&\frac12\end{bmatrix}&\begin{bmatrix}-\frac12&-\frac{\sqrt3}2\\-\frac{\sqrt3}2&\frac12\end{bmatrix}\\
R\phi_1&\phi_{12}&\phi_2&\phi_4&\phi_6&\phi_8&\phi_{10}\end{array}$$
Then it remains to apply projection operators:
$$\Gamma_j^{(\mu,i)}=\sum_{R\in G}D_{ij}^{(\mu)}\left(R^{-1}\right)R\phi_1$$
(Note the $R^{-1}$ in the above formula)
$$\Gamma_{A_{1g}}=\phi_1+\phi_3+\phi_5+\phi_7+\phi_9+\phi_{11}+\phi_{12}+\phi_2+\phi_4+\phi_6+\phi_8+\phi_{10}$$
$$\Gamma_{A_{2g}}=\phi_1+\phi_3+\phi_5+\phi_7+\phi_9+\phi_{11}-\phi_{12}-\phi_2-\phi_4-\phi_6-\phi_8-\phi_{10}$$
$$\Gamma_{B_{1u}}=\phi_1-\phi_3+\phi_5-\phi_7+\phi_9-\phi_{11}+\phi_{12}-\phi_2+\phi_4-\phi_6+\phi_8-\phi_{10}$$
$$\Gamma_{B_{2u}}=\phi_1-\phi_3+\phi_5-\phi_7+\phi_9-\phi_{11}-\phi_{12}+\phi_2-\phi_4+\phi_6-\phi_8+\phi_{10}$$
$$\Gamma_{E_{1u},x}^{(1)}=\phi_1+\frac12\phi_3-\frac12\phi_5-\phi_7-\frac12\phi_9+\frac12\phi_{11}+\phi_{12}+\frac12\phi_2-\frac12\phi_4-\phi_6-\frac12\phi_8+\frac12\phi_{10}$$
$$\Gamma_{E_{1u},y}^{(1)}=\frac{\sqrt3}2\phi_3+\frac{\sqrt3}2\phi_5-\frac{\sqrt3}2\phi_9-\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2+\frac{\sqrt3}2\phi_4-\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{1u},x}^{(2)}=-\frac{\sqrt3}2\phi_3-\frac{\sqrt3}2\phi_5+\frac{\sqrt3}2\phi_9+\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2+\frac{\sqrt3}2\phi_4-\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{1u},y}^{(2)}=\phi_1+\frac12\phi_3-\frac12\phi_5-\phi_7-\frac12\phi_9+\frac12\phi_{11}-\phi_{12}-\frac12\phi_2+\frac12\phi_4+\phi_6+\frac12\phi_8-\frac12\phi_{10}$$
$$\Gamma_{E_{2g},x^2-y^2}^{(1)}=\phi_1-\frac12\phi_3-\frac12\phi_5+\phi_7-\frac12\phi_9-\frac12\phi_{11}+\phi_{12}-\frac12\phi_2-\frac12\phi_4+\phi_6-\frac12\phi_8-\frac12\phi_{10}$$
$$\Gamma_{E_{1u},2xy}^{(1)}=\frac{\sqrt3}2\phi_3-\frac{\sqrt3}2\phi_5+\frac{\sqrt3}2\phi_9-\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2-\frac{\sqrt3}2\phi_4+\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{1u},x^2-y^2}^{(2)}=-\frac{\sqrt3}2\phi_3+\frac{\sqrt3}2\phi_5-\frac{\sqrt3}2\phi_9+\frac{\sqrt3}2\phi_{11}+\frac{\sqrt3}2\phi_2-\frac{\sqrt3}2\phi_4+\frac{\sqrt3}2\phi_8-\frac{\sqrt3}2\phi_{10}$$
$$\Gamma_{E_{2g},2xy}^{(2)}=\phi_1-\frac12\phi_3-\frac12\phi_5+\phi_7-\frac12\phi_9-\frac12\phi_{11}-\phi_{12}+\frac12\phi_2+\frac12\phi_4-\phi_6+\frac12\phi_8+\frac12\phi_{10}$$


How can I keep homemade laundry detergent from disassociating? We've been using this recipe to make laundry detergent. Works fine, but after a few days it disassociates into a clear bottom layer and a pink top layer. It's mildly annoying to have to re-stir it every time. Can we do something about it?

You should use about 0.25 - 0.35% xanthan gum (thickening agent) per the total weight of the mixture. Using xanthan gum is kind of tricky. You want to thoroughly mix the gum with the baking soda and washing soda (maybe in a bag), and slowly add the sodas to the water while mixing. If you don't do this, the xanthan gum will form lumps that are difficult to manage.

Hydroxyethyl cellulose might also work as a thickener, but it could increase the sudsing effect. Also, I would try adding a dash of dish soap; only use about 1% of the total weight of the mixture though, or else the sudsing effect will cause the washing machine to foam over.


Why does meltwater taste different? I've noticed that water obtained from melting ice cubes (by keeping them in the open) tastes different from water cooled to the same temperature.

Furthermore, the taste goes away if you keep it in the open for a sufficient duration of time.

I can think of these possible explanations:

Dissolved gases being released
Something to do with the crystal structure of ice
Something to do with dissolved minerals

The Internet is giving me all sorts of varying answers.

Exactly what chemical differences exist between meltwater and cold water?

(not yet a fully grown answer: experiment still pending)

I heard that $\ce{CO2}$/$\ce{H2CO3}$/$\ce{HCO3-}$ is responsible for "fresh" taste of water.

@NickT: Chloramine has a very characteristic smell, so Manishearth could most probably tell us whether this was the smell/taste.

So, here are first questions for @Manishearth:

does water that has been boiled and then cooled to the same temperature taste similar to the taste you're asking for?
What about de-ionized water?

Of course, there's also that average-of-all-that-has-ever-been-in-the-freezer taste...


Tendency of formation of complexes in Group-1 elements My book states the following:

Down the group(Group-1), tendency to form complexes decreases due to decrease in charge density$\bigg(\frac{\text{charge}}{\text{radius}}\bigg).$ This results in decrease in ion-dipole interaction between metal atom and the approaching ligand.

My logic-
There is a sudden availability of inner d-orbital while moving from 3rd period to 4th period. This can help accommodate the incoming lone pairs. So the tendency to form complexes must increase.
So combining the two facts, the trend should be irregular- decrease down till sodium, suddenly increase in potassium, and then slowly decrease till Caesium.
Can you provide any insight into the theory I have given above?

The fallacy is the assumption that "inner $d$ orbitals" become "available".  Generally they are not, in the Group 1 and Group 2 metals.  There are rare occasions where some evidence of $d$-orbital bonding is found for heavier G2 elements, but the impact is small and not widespread.
In Groups 1 and 2, where there are inner $d$ orbitals they are so well shielded that an electron placed there is not tightly bound and thus in general tends not to stay put.  Somewhat ironically, $d$ orbitals become available to accept electrons from ligands just at the point where in neutral atoms they are being filled (i.e., the transition elements).  This is because we add protons to the nucleus as we add electrons to the subshell, and the electrons we are adding to the $d$ orbitals do not fully shield the remaining $d$-orbital states from the increasing protic charge.  It's as if the ligand electrons need to see their doppelgangers in the swimming pool before they figure out the water must be fine.
In truth, without readily available $d$ orbitals and with low charge density (except maybe lithium), the alkali metals are poor at forming complex ions.  That includes complexing with hydroxide ions, thus explaining the unique basic strength of the hydroxides of these metals.


Experimental difficulty with the iodine clock reaction I did the iodine clock reaction using sodium metabisulfite and potassium iodate. As I understand it, increasing concentration is supposed to speed up your reaction rate. I did 10 trials. The first 5 I varied the concentration of metabisulfite and kept the iodate constant. Trials 6-10 I varied the potassium iodate and kept the meta constant. I don't understand why as I increased concentration, the reaction rate became slower? For example in trial 6 I had 2 ml of sodium metabisulfite and 4 ml of potassium iodate, that only took 7 seconds. In trial 10 I had 2 ml of meta and 10 ml of potassium iodate and that took 17 seconds? Does anyone know what I did wrong? Increasing concentrations should have increased collisions and made the rate faster, but it did the opposite! Please help me figure out why things went wrong!

The theory is correct. Increasing concentrations increase the number of effective collisions per unit time and the rate increases. So accordingly even in the experiment it should happen like that.

So in visualizing your situation, it must have been an experimental error and  not the theory. (Say there must have been less amount of $\ce{H2O2}$ in the mixture).

In this reaction what happens first is the conversion of $\ce{I-}$ to $\ce{I2}$ (have to add $\ce{H2O2}$). This $\ce{I2}$ will then react with metabisulphite or (even sodium thiosulphate) to produce $\ce{I-}$. Since you have added a constant value of meta in trials 6-10 there is a definite amount of $\ce{I2}$ removed in each trial by meta then the remaining $\ce{I2}$ will react with starch to give the color change.


Typography of atomic orbital subscripts p_x It the notation $\mathrm{p_x}$, $\mathrm{p_y}$ and $\mathrm{d_{x^2{-}y^2}}$, are the subscripts variables (and therefore should be in italics) or are they labels (labels for directions, I suppose, and therefore should be upright)?

The subscripts that specify orbital axes are written in italic. This is mentioned in ACS style guide.$^{[1]}$ So for instance

$$\mathrm{d}_{x^2 - y^2}.$$

Be careful though. Other subscripts for orbitals are generally upright, e.g.,

$$\mathrm{t_{2g}}.$$

$[1]$ Anne M. Coghill, Lorrin R. Garson. ($2006$). The ACS Style Guide. Effective Communication of Scientific Information. American Chemical Society. DOI: 10.1021/bk-2006-STYG, ISBN: 9780841239999 (print), 9780841228306 (online). (p 256)


Unknown salt formed from electrolysis of urine I am not sure what I was expecting to happen with this but I created a circuit to electrolyze various household liquids. My electrodes are copper — or at least claim to be, I think they might just be zinc wires with some copper-colored coating.

Now I'm trying it on urine, which seems to eat through the wire rather rapidly. It produces some sort of gas with tiny bubbles. When these bubbles come into contact with the other electrode, they turn gray and fall to the bottom of the container. This gray material is almost green and that makes me think it is a copper salt rather than zinc. What are the bubbles and what is this salt?

Also, as a side question, how can I isolate this salt without potentially exploding whatever is in the bubbles?

If it's important: I'm using ~6 V, but I don't know the current.

The main components of urine are water and then urea.  Water makes up 90% to 95% of urine with urea making up about 50% of the remainder.  It's likely that your reaction is mainly the hydrolysis of water.  In that case, you are forming hydrogen gas at cathode and oxygen gas at the anode.  Also, some $\ce{NaCl}$ will be electrolyzed and form $\ce{Cl2}$.

As you stated, you are likely oxidizing copper, probably making $\ce{CuCl2}$.  Most zinc salts that you could be forming in your reaction would be white.  If urea is being electrolyzed, then you would also be forming nitrogen and carbon dioxide at the anode.  It's possible that at least some of the migrating bubbles contain $\ce{Cl2}$ gas which then could form solid $\ce{CuCl2}$.

You could do a quick test of some aspects of my answer by observing the hydrolysis of water using your system.  Just conduct your procedure exactly as you do for urine, but instead of urine use water.  The main salt present in urine is $\ce{NaCl}$, whose concentration varies quite a bit but is on the order of 3 g/L.  So just conduct your experiment on a 3 g/L $\ce{NaCl}$ in water solution and compare your results to the urine experiment.


What causes the stairstep structure in bismuth crystals? I've been interested in bismuth crystallization since I found out they existed, but have a question on their formation.

According to Wikipedia, the stair-step structure of bismuth crystals is "the result of a higher growth rate around the outside edges than on the inside edges."

What would cause this higher growth rate around the edges? Cooling?

Hopper crystals, that is, the "staircase" form, develop when a crystal forms so rapidly that material is used up before the interior finishes.

Table salt, $\ce{NaCl}$, also forms nice hopper crystals. If you want to try this, use "kosher" salt or CP $\ce{NaCl}$, which doesn't have anti-caking and iodide additives that would spoil the crystal structure.


Can I make UV-Vis calibrant in the lab? I have a Thermo Nanodrop 1000 spectrometer that I have reason to believe is out of calibration. After consulting the manual, it appears as though there is a way to check the calibration of the spectrometer using a premade CF-1 calibration solution. This solution itself appears to be aqueous potassium dichromate, but I cannot find the concentration and exact specifications of the solution online. I was curious if anyone has attempted to make the calibrant from reagent they had in lab, and if so, what exactly does one use to make the solution?

The last time I calibrated an absorbance spectrometer, it was one I made from parts (for an undergraduate Instrumental Analysis teaching laboratory) and I used a 0.0509 M holmium chloride solution I prepared. The visible region spectrum I obtained is here:

Holmium solutions are well known as UV-Vis calibration standards and you can purchase from, e.g., www.starnacells.com. Below is a screenshot from the Starna Cells web site, indicating that their ready-made sealed holmium reference standard cuvette cells have NIST traceability. Evidently, they are good down to 241.0 nm. I would have purchased one, if I had had the money at the time, but the solution I prepared worked well and was not particularly expensive. Erbium and praseodymium solutions also work well.

Sharp peaks in an aqueous solution at room temperature are always nice. Maybe this will help you!


Difference between sodium ion and a transition metal ion dissolving in water? In class, I learned that when $\ce{NaCl}$ is dissolved in water, the partially negative oxygen part of the water molecules surround the $\ce{Na+}$ ion and energy is given off (hydration enthalpy)

Likewise, when a transition metal ion like $\ce{Fe^2+}$ is dissolved in water, water molecules surround the ion. However, in this case, we say that a complex ion is formed.

Can someone please tell if there is really a difference in the two cases and if yes, could someone specify them.

There is a slight difference between solvated ions and aqua complexes: The nature of the bond.

When solvating a sodium ion, the electron pairs of the oxygen in water never really interact with the sodium in a way that creates a bond of some sort: It's all just electrostatic interactions.

However, when a transition metal ion gets surrounded with these water molecules, the lone pair electrons of the oxygen form a bond with the ion. This is because they interact with the $d$ orbitals of the transition metal (which usually also contains some electrons), thus changing the energetic state of the ion itself. First and second group elements don't have that energy-stabilizing effect because they don't have electrons in the orbitals that get split up by the formation of the complex.

It is not unimaginable that there then exist further electrostatic interactions between the transition metal aqua complex and the surrounding water molecules, thus leading to the solvation of said complex.


Is the application of Le Chatelier's principle wrong in this case? In the reaction:
$$\ce{N2(g) + 3H2(g) <=> 2NH3(g)}$$
Suppose the system has 3 moles of $\ce{N2}$, 1 mole of $\ce{H2}$ and 1 mole of $\ce{NH3}$ at equilibrium. $Q_\chi=K_\chi=8.33$ and $Q_p=K_p$.

Now I add 5 moles of $\ce{N2}$ in the container by keeping (i) pressure and (ii) volume constant (temperature constant in both).

According to Le Chatelier's principle:

If pressure is constant: equilibrium shifts to backward direction.
If volume is constant: equilibrium shifts to forward direction.

But if I look at the equilibrium in terms of mole fraction, $Q_\chi$ will be the same in both the cases and hence equilibrium should shift in the same direction i.e. in backward direction.

Why do they predict different equilibrium states?

[$K_\chi$ is equilibrium constant in terms of mole fraction, $K_p$ is equilibrium constant in terms of partial pressures].

Try doing the calculation using the fact that $K_x$ actually is not constant at a given temperature.  It depends on total pressure as well as temperature.  Assuming you maintain ideal gas behavior $K_p$, and only $K_p$, is constant when temperature is fixed.  In terms of that equilibrium constant you should find the reaction following Le Chatelier's Principle both of your cases.


How to analytically quantify a chemical's potential to intercalate DNA? This shows that polyphenols (like quercetin) have the potential to intercalate DNA, but at potency 100x less than that of ethidium bromide. How does one analytically measure/quantify "100x weaker"?

I'm also interested in how this could apply for the compound 9-fluorenol (hydrafinil) - and of the relative potency of 9-fluorenol and quercetin in intercalating DNA.

Our lab uses a thiazole orange exclusion assay to measure how well our polyintercalating peptides* bind DNA. The assay works by mixing thiazole orange with DNA so it becomes fluorescent. Then we measure fluorescence while titrating in peptide and the loss of flourescence indicates binding. This is because thiazole orange is only fluorescent when intercalated into DNA, and as the competing compound displaces it, the free thiazole no longer fluoresces. Note that this assay is not foolproof, polycationic compounds can condense DNA without intercalating, but this conformation change still forces the thiazole out and reduces the fluorescence, so loss of fluorescence implies binding, not necessarily intercalation. This method works for RNA too, but you'll need more thiazole to see signal because there are usually fewer double helices to intercalate into.

Another method is to measure viscosity, because intercalators make the DNA longer and the solution becomes more viscous, but you either need a very sensitive method to measure the change or a very concentrated DNA solution, and the viscosity can probably also be changed by condensing the DNA.

Another possible method MIGHT be fluorescent anisotropy, if your intercalator is fluorescent. It would spin rapidly outside the DNA, but slow down once it binds the double helix. However, conformational changes in the DNA can change its rotation speed and confound the results.

That paper doesn't really mention the assay, it's just to show off the peptides so you understand what I'm talking about a little more. Those peptides also bind by electrostatic interaction and condense DNA, so it's not a pure measurement of intercalation.


Why does silver oxide form a coordination complex when treated with ammonia? Part of the preparation of Tollens' reagent occurs according to the formula:

$$\ce{Ag2O (s) + 4 NH3 + 2 NaNO3 + H2O → 2 [Ag(NH3)2]NO3 + 2 NaOH}$$

I wonder how exactly the oxygen gets "unstuck" from the silver under the influence of ammonia. I've read about coordination complexes, so I understand what $\ce{[Ag(NH3)2]NO3}$ is, but I'm puzzled at what exactly happens to pull that oxygen apart from $\ce{Ag2O}$

I've done very little exploration of the issue, so this may be too basic a question - in which case just please give a hint on what to read up.

I found this (now deleted) comment:

  $\ce{NH3}$ dissolves silver oxide because the nitrogen has a pair of electrons that it can share with the silver forming a diammine complex. It has nothing to do with pH adjustment. – A.K. Jan 20 '16 at 17:17

Chemguide states this:

  This is made from silver(I) nitrate solution. You add a drop of sodium hydroxide solution to give a precipitate of silver(I) oxide, and then add just enough dilute ammonia solution to redissolve the precipitate.

It's quite elementary (perhaps). Silver (and gold) are known to show two main coordination numbers in its compounds, namely, $2$ and $4$. To start, we can consider $\ce{[Ag(OH)2]-}$ to be our initial complex as in aqueous solutions it can be thought of to be existing in an equilibrium with $\ce{Ag2O}$. See footnote [1].

A quick glance at the order for the strength of ligands tells us that $\ce{NH3}$ is a stronger ligand than $\ce{OH-}$. What I now intend to do is to compare the thermodynamic stability of $\ce{[Ag(OH)2]-}$ with that of the diammine complex $\ce{[Ag(NH3)2]+}$.

The strength of ligands is directly proportional to the thermal stability of a complex so yes, the diammine complex is more stable.

A similar case is seen with copper(I). Cyanide complexes of copper(I) are known to be extremely stable and hence it prefers forming a complex with cyanide over other ligands.

[1] Nevertheless, saying this may be highly controversial because Ellingham diagrams tell us that $\ce{Ag(OH)}$ being unstable gets converted into $\ce{Ag2O}$ which is a much more stable compound for silver to be in. However in aqueous solutions, an equilibrium of such a kind can always exist.


Long-term storage of Ascorbic Acid mixed with Sodium Bicarbonate Every day I mix two parts of ascorbic acid powder with 1 part of sodium bicarbonate powder and place the resulting powder into an airtight container. I have noticed that if the resulting powder is mixed with water within 30-60 minutes, it fizzles indicating the release of carbon dioxide as a byproduct of the acid-base reaction. However, if I leave it untouched (within a dry and airtight container) for 6 to 8 hours, it hardens. If I crush it (so it turns into powder again) and mix it with water, it no longer fizzles.

My questions are:

Why does it harden and no longer fizzles if mixed with water after a few hours? Does the acid-base reaction take place even without adding any liquids to the mixed powder?
Most importantly, does the reaction that causes the hardening and absence of fizzling cause any kind of degradation (e.g., oxidation) to the ascorbic acid? In other words, can I mix ascorbic acid with sodium bicarbonate (both in powder form), store the resulting powder in an airtight container and consume over a few months? Would that diminish the efficacy of the Vitamin C in any way?

Thank you in advance.

Update:

I have bought sodium bicarbonate from a different manufacturer and now after mixing it with ascorbic acid, nothing happens until water is added. It can sit for 24+ hours, without hardening and it will still fizz when mixed with water. I suspect the sodium bicarbonate from the previous manufacturer was slightly moist  and this one is completely dry.

The reason that your solid mixture is clumping and hardening inside your airtight container is that the reaction between the two compounds is producing water as follows:

$$\ce{C6H8O6 + NaHCO3 <=> NaC6H7O6 + CO2 + H2O}$$

where $\ce{C6H8O6}$ is ascorbic acid.

This reaction proceeds rapidly in the presence of excess water, which allows the reacting species to dissolve and thus have ready access to each other. This results in the observed fizzing from the release of carbon dioxide.

In the solid phase, the initial reaction is limited to the surfaces of the particles coming into contact with each other. As the reaction proceeds, the once purely solid material becomes wet via the acid-base reaction producing water, and the remaining materials can somewhat dissolve, facilitating migration of the reacting species so that on a timescale of hours all of your material has reacted.

Regarding the stability of storing the solid mixture vs. storing the solids separately, you are probably better off storing them separately until use. Both ascorbic acid and sodium ascorbate are susceptible to being oxidized by long-term exposure to air. Although water itself shouldn't act as an oxidant here, it could facilitate reaction with oxygen in the air in a similar manner in which it facilitates the acid-base reaction, as discussed above.


Temperature measurement Consider some liquid in a beaker. I am measuring its temperature using a thermometer. When i put my thermometer's bulb in the liquid, the particles of the bulb gain the same kinetic energy as of the liquid particles. Now, my question is that
Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid?
I have read somewhere that there is no effect of increase of rotational energy or vibration of molecules (as a whole) on temperature readings. Why so? Do the vibrations of the molecules don't change kinetic energy of the thermometer's bulb's particles?

Do the particles of thermometer's bulb attain the same translational, rotational and vibrational energy (vibration of atoms within a particle) as of the particles of the liquid?

No, but they will have the same temperature. Solids don't have any translational energy, the atoms time-averaged positions are constant. If you want to link the temperature to the energy of the particle, it is easiest to start out with a monoatomic ideal gas because here, there is a linear relationship.

Relationship between change in temperature and energy

Although under some ideal circumstances, particles have an energy of $\frac{1}{2} k T$ per degree of freedom, how much energy is needed to raise the temperature of a sample depends on the sample (and the temperature). The heat capacity is a measure of change in energy per change in temperature. So the heat capacity tells you the total energy going into all degrees of freedom when the temperature increases. The splitting up into translational, rotational and vibrational energy depends on the nature of the sample (for example, the physical state) and the temperature.


Specific Heat Ratio for a perfect gas mixture I am reading Rocket Propulsion Elements by George P. Sutton & Oscar Biblarz, 9th Edition. In the fifth chapter, I was introduce to the specific heat ratio k for the perfect gas mixture, Eq. 5-7. Without explanation the Equation was stated as follows:
(5-7) $k_{\text{mix}} = \frac{(C_{p})_{\text{mix}}}{(C_{p})_{\text{mix}}-R'}$
From Thermodynamics I know,
(a) $k_{j} = \frac{(C_{p})_{\text{j}}}{(C_{v})_{\text{j}}}$ ; Species specific heat ratio
(b) $R_{j} = \frac{R'}{\mathfrak{M}_{j}}; $ Species Gas constant equal to Univ. Gas constant over Molar Mass of species $\text{&}$
(c) $R_{j} = (C_{p})_{\text{j}} - (C_{v})_{\text{j}}$ ;Mayer's Formula for species
So, it seems to me that Eq. (5-7) should instead include the mixture gas constant $R_{\text{mix}}$ in the place of the Universal gas constant, $R'$.
I tried to derive Eq. (5-7) from Eqs. a, b, and c without success. Can someone clarify the derivation of Eq. (5-7) to me?

In general for a perfect or ideal gas,
$$C_p=C_V + R'$$
(using your notation) where the heat capacities are molar quantities.
It follows that for a perfect gas mixture $(C_p)_\text{mix}=(C_V)_\text{mix} + R'$.


Converting cyclic compounds to their corresponding aliphatic compounds Is there any reaction available to convert cyclic compounds to their corresponding aliphatic compounds?
eg. Converting cyclohexane to n hexane

The reaction you want to know about is "Cleavage reaction" in particular oxidative cleavage.

In this dihydroxylation, if the solution is warm or acidic or too concentrated, oxidative cleavage of the glycol may occur.

Then they can be reduced to aldehydes by reducing agents.Reduction of carboxylic acids to aldehydes is difficult because aldehydes are more reactive than carboxylic acids toward most reducing agents. Almost any reagent that reduces acids to aldehydes also reduces aldehydes to primary alcohols.

Lithium tri-tert-butoxyaluminum hydride, LiAlH(O-t-Bu)3 is a weaker reducing agent than lithium aluminum hydride. It reduces acid chlorides to aldehydes because acid chlorides are strongly activated toward nucleophilic addition of a hydride ion.

Under these conditions, the aldehyde reduces more slowly and can be isolated. Therefore, reduction of an acid to an aldehyde is a two-step process: Convert the acid to the acid chloride, then reduce using lithium tri-tert-butoxyaluminum hydride.

And now we can reduce it by the Wolff Kishner reduction.

Ethylene glycol, diethylene glycol, or another high-boiling solvent is used to facilitate the high temperature (140–200 °C) needed in the second step. I am assuming you can convert the cyclohexane to cyclohexene by any radical halogenation and then dehydrohalogenation.


How do we determine if some metal will undergo passivation under suitable environment? The passivity of certain metal fall in order:
$\ce{Ti->Al->Cr->Be->Mo->Mg->Ni->Co->Fe->Mn->Zn->Cd->Sn->Pb->Cu}$

Can we predict the order of passivation theoretically?

There is not much we can do. It is certainly not a property that is explained by a rigorously established single measurable parameter like oxidation potential. Passivation is a delicate interplay of multiple factors, some of which are not even related to the metal itself (remember, a metal may be passive in some environments, but not in the others).

However some trends can be observed. Surely, having a hard, high-melting, and chemically inert oxide does help a lot.