text:
Forces of attraction of hydrated potassium ion in aqueous solution
I want to study the intermolecular forces (IMFs) in hydrated potassium ion, $\ce{K+ (aq)}$ in an aqueous solution of $\ce{KCl}$. According to my thoughts, among the 4 IMFs I know:

London dispersion forces (LDFs);
Dipole-Dipole interaction;
Hydrogen bonding;
Ion-dipole forces

LDFs are definitely included, but my question is whether dipole-dipole is included as well (because KCl is a polar substance)?

KCl is not a molecule at all.

KCl is a formula unit.

K+ is a spherical ion. The water molecule has a dipole. So which do you think are relevant?

Dipole-dipole interaction, specifically, will probably not happen, or they at least won't contribute significantly to IMFs. The thing with these different chemical bonds is that they are mostly based in identical features, and sometimes calculated differently, taking different things in account.
An ionic bond will be stronger than a polar bond, so that is why dipole-dipole interaction seems irrelevant, in comparison to ion-dipole forces.
All the kinds of interaction mentioned are derivatives of electrostatic interaction, also know as Coulomb interaction. LDFs are a direct derivative of Van der Waals bonds, but if you look more physically at all those bonds you can think of them as electrostatic interaction.

There are no intermolecular forces in KCl because it is an ionic compound. IMFs only exist in covalent compounds between molecules.

This is wrong because K+ is hydrated as stated in the question. There are indeed IMFs for ions in solution. Ion-dipole forces and LDF will dominate.

So an ion acts like an ideal gas in a solution???


text:
Michaelis-Menten rate law for enzyme which catalyzes two reactions: steady state?
Suppose an enzyme $\ce{E}$ can catalyze two reactions:
\begin{align}
\ce{S1 + E &<=> S1E ->  P1 + E} \tag{R1} \\
\ce{S2 + E &<=> S2E ->  P2 + E} \tag{R2}
\end{align}
I want to derive a rate law. Can I assume that
\begin{align}
\frac{d[\ce{S1E}]}{dt} &= 0 \tag{1}\\
\frac{d[\ce{S2E}]}{dt} &= 0 \tag{2}
\end{align}
like in the derivation of the Michaelis-Menten rate law?

Why do you think this assumption might be different in the two reaction case versus the one reaction case? (Note that the steady-state approximation relies on particular conditions, and isn't necessarily universally valid.)

@R.M.
I want the same conditions as given when the Michaelis Menten rate law is valid: high amount of substrate, only initial velocities etc.
The assumption in the one reaction case relies on "the concentration of the intermediate complex does not change on the time-scale of product formation " (citing wikipedia)
I do not know how to estimate if this still holds true for the two reactions case.

In the steady-state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. So in this case only if k2 >> k1 and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P.

\begin{align}
\frac{d[\ce{S1E}]}{dt} &= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}] - \ce{k2}[\ce{S1E}] = 0\\
\end{align}
Therfore:
\begin{align}
[\ce{S1E}] &= \frac{\ce{k1}}{k_{-1}+\ce{k2}}[\ce{S1}][\ce{E}] = K_a[\ce{S1}][\ce{E}]
\\
\end{align}
Similarly for the second reaction:
\begin{align}
[\ce{S2E}] &= \frac{\ce{k3}}{k_{-3}+\ce{k4}}[\ce{S2}][\ce{E}] = K_b[\ce{S2}][\ce{E}]
\\
\end{align}
The enzyme E is involved in both reactions and its total concentration(bound and unbound) is constant. This total concentration of enzyme $[E]_{0}$ is equivalent to the concentration of the free enzyme before adding the substrates. The concentration of the free enzyme at a certain time t is [E]:
\begin{align}
[E]_{0} &= [E] + [\ce{S1E}] + [\ce{S2E}]
\\
\end{align}
if you substitute $[\ce{S1E}]$ and $[\ce{S2E}]$ with the previous expressions then:
\begin{align}
[E] &= \frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}
Eventually:
\begin{align}
\frac{d[P_{1}]}{dt} = k_2[\ce{S1E}] = \ce{k2}(K_a[\ce{S1}][\ce{E}]) = K_p[\ce{S1}][\ce{E}] = K_p[\ce{S1}]\frac{[E]_{0}}{1 + k_{a}[\ce{S1}]+k_{b}[\ce{S2}]}
\\
\end{align}
and similarly for the rate of production of $P_2$

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text:
Is it possible to fractionally distill a compound made up of substances with very similar boiling points?
Let's say I have a solution that is made up of substances that have very similar boiling points, say less than 5 degrees apart. Is it even possible to fractionally distill those substances, say, using a vacuum still?
Which method is more precise, heating the solution and swapping out condensation vessels as the temperature increases, or having a column with the ability to fractionate in stages?

Yes, possible. Very good column's needed though.

What factors contribute to s good column

You have a SOLUTION not a compound

If the chemicals form an azeotrope, even with a wide separation of b.p. (e.g. ethanol + water), then it may not be feasible to separate by distillation. https://en.wikipedia.org/wiki/Azeotrope

There is the concept of a plate which is the difference between the liquid composition and the vapor composition. Each plate would have a slightly different temperature. So you'd need a column which in its mechanical construction would have a lot of plates.

@DrMoishePippik  I’m somewhat familiar with this concept, but does that mean that a portion of the distillate no matter beginning middle or end is partially both substances because the azeotrope forces both to have the same BP

Yes... for example, whatever proportion of ethanol and water a solution contains to start, as it boils down it tends to ~95.5% ethanol. Azeotropes  can be separated using a third chemical to entrain one of the ingredients, but not through distillation alone.

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text:
water in redox reactions
why water can't react as anode , with metals , in a redox reaction?
while studying electrochemistry, i passed through a table with certain metals , all having voltage (potential energy) less than water, which can react with water in redox reaction. in this case, the water react as oxidation agent.
my question, is why water cant react as reducing agent, with metals having potential energy bigger that it?

Water surely can and does react, given the right conditions. Could you please be more specific?

i will state from my lecture slideshow:
metals will react spontaneously with water if their E is less than -0.83V
my question is why ?
why a metal which has E>-0.83 V can't react with water?

Water can react as a reducing agent. However, you would have to look at the oxidation equation for water, which has an electrode potential of 1.23 V. If you wanted water to act as a reducing agent, you would have to react it with a metal whose E > 1.23 V. The equation that you are referring to (where E = -0.83 V) is the reduction of water. Therefore, you can only use it when you want water to act as an oxidising agent.
A note about spontaneity; it only refers to whether a reaction is likely to occur without any addition or removal of energy. In other words, the reaction would occur if the chemicals were brought in contact with each other. It does not mean that the reaction is impossible, because if you provided sufficient energy, the chemicals would react. This is usually supplied via a change in temperature. The reason metals whose E < -0.83 V react spontaneously with water is because the nature of the Gibbs free energy change (ΔG). For redox reactions, ΔG = -n x F x E cell where n is the moles of electrons transferred in the reaction, F is Faraday's constant and E cell is the cell potential. Reactions are spontaneous if ΔG is negative, meaning that E cell must be positive for the reaction to be spontaneous. When you add the half cell equations for metals whose E < -0.83 V and that of water, you get a positive value for E cell. Additionally, redox reactions are based on the idea that more reactive metals displace less reactive ones from compounds. Metals above water are reactive enough to displace hydrogen from water. Metals below water are not reactive enough to displace hydrogen and so cannot react spontaneously.

Water cannot act as a reducing agent to metals, simply because of the fact that metals do not react as oxidants, however metal-ions do. Below the values of the data of the standard electrode potentials (and the source site -- Wikipedia) is shown. Water is highlighted.
Data of standard electrode potentials
The table is sorted so that the strongest oxidants are placed highest in the table. As you can see, above water as a reductant, there are no metals reacting in half-reactions as oxidants. There are some metal-ions that would react as an oxidant. The reason behind why metals don't act as an oxidant, is probably because of their low electronegativity. The high electronegativity of some elements prevents them from becoming positively charged, and in a similar fashion, the low electronegativity of a metal prevents it from  becoming negatively charged.


text:
What is the range of photon energies can be effectively used to split a diatomic molecule like hydrogen?
I saw photo-ignition of hydrogen-chlorine mixture using UV light on YouTube.
Chlorine molecule bond dissociation energy is $\pu{2.51 eV}$ (bluish colour).
Bromine molecule bond dissociation energy is $\pu{2 eV}$ (orange colour).
Can I use any photon which has energies greater than $\pu{2 eV}$ to split bromine molecule,  like green or blue or UV light? Or should I use only photons which has energies around $\pu{2 eV}$?
My concern is about absorption spectrum of gases. And this question is not just for bromine gas, but for any diatomic molecule like hydrogen or Chlorine.

If the photon's $E=\hbar\omega=h\nu=\frac{hc}{\lambda} $ energy is greater than the energy of the bond which you want to break up it will break it up. It's always best to use higher energy photons but not too high because it may break up bonds which were unwanted to break up in your reaction . So the best for you is to use blue or higher energy UV light for your experiment.You need to deal with the rate of the reformation of bromine molecules, there will be an equlibrium between the production and decomposition of bromine molecules depending on the intensity, not the energy of the light. If the photons has higher energies than a given bond they will always break it up, this is why gamma radiation is so dangerous. The absorbtion spectrum doesn't have a lot connections to this phenomena. You can use the Maxwell-Boltzmann distribution
$$f_{v}(v_{i})=\sqrt{\frac{m}{2\pi k_{B}T}}exp\left[-\frac{mv_{i}^{2}}{2k_{B}T}\right]$$
to determine the Doppler-effect's impact on the absorption spectrum depending on the temperature and to construct an absorbtion function explicitly. The doppler effect will affect the percieved frequency and energy of the light by gas particles so if you want to be as accurate as possible, you need to calculate it.The Maxwell-Boltzmann distribution is only suitable in low temperatures (below several thousand Kelvins).


text:
Chemistry 101 - significant figures
So i had my first chemistry 101 exam and got 21/25 , but iam not sure about one question, it's about significant digits , can u guys check if my answer is true or false as shown in picture.
Thanks

We generally try to avoid just checking answers on this site. Usually, we want to see what your thoughts on the question were/what research you have done to try to arrive at an answer.

Truthfully this seems like a poor question. Either (A) or (B) could be argued.
(A) The ruler is marked in cm, so the measurement can only be made guaranteed to the cm. You can try to interpolate to get something better that a whole cm, but there is no guarantee that you'll be able to interpolate to 0.1 cm. It might be +/- 0.2 cm, or +/- 0.3 cm. So you get 1+ significant figures, but not quite 2.
(B) Although the ruler is marked in cm, the measurement can be interpolated to 0.1 cm. So there are 2 significant figures.
Now since (A) was marked incorrect, the teacher was obviously looking for (B). Such interpolation is not uncommon in a chemistry lab. For instance in reading a burette.

I'd like to augment this answer by saying that the ruler shown is clearly not to scale. Therefore, I have no idea how accurate the 1 centimeter ticks are even. In the absence of any other context, I don't think we can assign any kind of confidence to the base measurement let along try to interpolate a measurement to 0.1 cm. The context is really important here.

It looks like the answer should be B. A good rule of thumb to follow is that your number of significant digits goes one past what is directly marked on your ruler, graduated cylinder, etc. The significant digit should be the one which has some uncertainty. Choosing A implies that the error in your measurement is on the order of $\pm1\pu{cm}$ whereas B correctly shows that you can make a reasonable measurement on the order of $\pm0.1\pu{cm}$.


text:
Ionizing radiation and oxidation without oxidizing agents
My knowledge on this subject is a bit rusty, so excuse me if I'm asking the wrong question here. I was taking a look at my high school chemistry books from many years ago, and I noticed that reduction and oxidation reactions take part together; meaning that you can't have oxidation without an oxidizing agent.
My question is, that if we have a source of radiation, that emits alpha particles, which attract electrons, or beta particles which repel them, or a source of powerful electromagnetic radiation like gamma radiation, that can knock electrons off of atoms, and we bombard some gas with radiation, ionizing it, then would it count as oxidation without reduction? What exactly is going on redox-wise? We would have positive ions, but no oxidizing agent, or would we?

We would, but not for long. Soon enough, the ions will find their electrons and recombine with them, so the net result chemistry-wise will be "nothing but some heat". That is, unless we have alpha particles. With them, it is just plain redox reaction with well-defined oxidizing agent.

Ionisation is different thing then redox.

@IvanNeretin You mean, that He2+ particles can attract those electrons themselves and get reduced? I'm thinking, because of the mass of alpha particles, they would send electrons flying, and not attract them into an orbit. How would it be a redox then?

@Mithoron They sure are different. Oxidation is loss of electrons. In case of my question, it is ionization with positive charge, and loss of electrons, and thus, oxidation.

When they would just kick electrons out, that would not be a redox, precisely for the reason explained in my first comment. When they finally will attract them into an orbit, it would.

@IvanNeretin Oh, you meant the ionized gas itself! I thought you meant that alpha particles could get reduced. That does make sense now. But still my question remains. Do we have an oxidation without instant reduction, and then a reduction of positively charged gas, without oxidation? I mean, that they aren't simultaneous, and the gas, which is being reduced is the one that got oxidized a brief moment before that.

Yes I meant exactly that: alpha particles can and will get reduced. That would be a typical redox reaction, when one party gets oxidized and the other reduced.

@IvanNeretin But I still don't quite understand. Please correct me if I'm wrong, but I think alpha particles cannot absorb electrons into orbit, due to high velocity. Not until they slow down, and then they eventually become neutral He. What I thought was happening was that electrons would be scattered towards moving alpha particles, but due to their mass being roughly 7200 times (I'm just guessing) smaller than that of alpha particles, they would stick around and get absorbed by the ionized gas again. Isn't it what's happening in reality?

Well, that's mostly right: when the alpha particles move fast, they will just kick electrons out. Before these electrons reunite with their native atoms, they might briefly get absorbed by some other atoms. Calling that a redox would be somewhat of a stretch, as to me.

It's interesting that beta source Tc has been used as a corrosion inhibitor. "55 ppm of potassium pertechnetate... protects steel from corrosion, even if the temperature is raised to 250 °C (523 K)." https://en.wikipedia.org/wiki/Technetium Though this is a chemical effect, does beta emission have any effect?

@DrMoishePippik It is an interesting phenomenon indeed. Electrical current can prevent corrosion, and to be exact, "oxidation", buy supplying electrons to a solid metal. Beta particles are nothing but negatively charged electrons. If you bombard a solid metal with beta, you are in fact supplying it with electrons. In gas state, however, things can be different, and you can ionize some gas with beta. But, you know, you're dealing with radioactive materials, and the amount of electrons depend on the half life of the matter, and it's mass. Electricity would be a much more efficient way.


text:
Uses of ruthenium-106?
I read a news article recently about a cloud of ruthenium-106 being released over Europe. The article mentioned one of the uses for this specific isotope is in fuel cells for satellites. I tried to do some research but i couldn't find information on a fuel cell that used this specific isotope. What kinds of fuel cell or power generator use ruthenium-106?

https://en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator

@Mithoron for Ritegs alfa-emiters are usually used. Beta-emitters are usually used in https://en.wikipedia.org/wiki/Betavoltaic_device

Expanding @Mithoron's comment into an answer, $\ce{^{106}Ru}$ is used in radioisotope thermoelectric generators that power satellites. Plutonium-238, curium-244 and strontium-90 are the most often cited candidate isotopes for this purpose but other isotopes such as polonium-210, promethium-147, caesium-137, cerium-144, cobalt-60, curium-242, americium-241 and thulium isotopes have also been studied.
Apart from being an energy source, it also finds its application in medicine. $\ce{^{106}Ru}$ is used in cancer radiation therapy, especially to treat eye and skin tumors.


text:
In antacids like Eno that fizz up when mixed in water, why is the citric acid important?
I know that you can get rid of the acidity if you just take some baking soda added in water. I learned that these antacids that fizz up also have some citric acid and they react when added in water and produce $\ce{CO2}$ along with some other compound.
I like to know the use of citric acid in an antacid. Is it just for the feel of the fizz that subliminally reminds us of fixing the acidity in stomach?

To make bubbles...

It's the citric acid that reacts with the sodium bicarbonate. When the tablet hits the water, the citric acid and bicarbonate react, forming $\ce{CO2}$ and sodium citrate. This causes the tablet to break up, thereby much speeding up its dissolution.
The table will contain a stoichiometric excess of bicarbonate, with respect to the citric acid, so it's anti-acid properties remain present, even after bubbling/dissolution. Dissolving a tablet that doesn't disintegrate due to the acid-base reaction and its fizz would take much longer.
The fizzing effect probably also has some aesthetic/marketing value, separate from anti-acid properties of the tablet.

I expected the fizzing is more of a marketing thing here. So is there an actual use of the citric acid here in the medical point of view? Or is it just fine if you just ingest some baking soda to relieve yourself?

Are you saying that the only purpose of Citric acid here is to help dissolve soda better?

@Harsha: it helps dissolve and yes, there's an aesthetic/marketing aspect to it too.

and there is no medical significance to citric acid here? or to the product formed after the reaction?

@Harsha: not that I can think of. It's just a nice, weak and healthy acid, good for that kind of job. That it's a solid also helps, of course. And cheap.

It might also create a solution that is buffered against too large a swing in pH.

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text:
Waterproofing and recycling paper
I've recently begun studying about waterproofing paper. Through what I learned is that the reason why paper absorbs water so easily is mainly because of it's material composition. Paper is generally made from plant fibers and these fibers in turn are made of Cellulose. These Cellulose molecules make it possible for water to be absorbed by the paper easily due to adhesion.
Basically, the network of fibers in the paper are filled with empty spaces and when the paper comes in direct exposure to water, the water molecules eventually fill up those empty holes, attract themselves through the Cellulose molecules and penetrates through the whole material. Which soaks up the paper and degrades it.
One way to solve the problem of soaked paper is through waterproofing the paper. Coating it with wax or plastic and forming a thin layer of plastic or wax which prevents water molecules in penetrating the material when the paper is exposed to water.
This water proofing concept makes it look easy to water proof paper, but the problem comes in recycling or when the paper is disposed off. When recycling paper, every bit of its fibers is disassembled but if for an instance the coating of a water proofed paper is too tight, the paper then becomes unrecyclable and become of waste. Since disassembling tightly glued on fibers could eventually destroy the material.
One evident problem here is that the coating hinders the recyclability of the paper because of its reaction with it. In conclusion, the points to be made aware off is that in order to make an effective waterproof paper but at the same time recyclable as well. Is to coat a paper with a specific type of resin/coating that has an effect of tightly bonding the fibers but at the same time easy to disassemble.
Through this understanding, I've been researching on other coating materials that could eventually be effective in water proofing. I've been trying to research on organic resins such as from trees if they would be effective. Are there any effective chemicals or resins out there that would generally be good?

This doesn't answer your question directly, but there are other means of improving the wet strength of paper without fully waterproofing.

You shouldn't expect research quality material on stackexchange

@airhuff I've looked into it but I think having the paper soaked would alter its mechanical properties which would be different from waterproofing

@Vedant I just thought posting in to stackexchange could help me think of a material that would be ideal for the said research since I do not have wide knowledge about chemistry.

@CzarLuc not a problem, but usually when something hasn't been researched, people don't research it just for an answer :)


text:
Why there are just two Ramachandran angles in protein structure for two peptide bonds?
I really want to understand why there are just two ramachandran angles to understand the structure of the protein and why not three or many ? Can someone suggest some reference and explain this ?

https://chemistry.stackexchange.com/questions/40181/when-calculating-a-dihedral-angle-for-a-chemical-bond-how-is-the-direction-defi https://chemistry.stackexchange.com/questions/28167/how-to-get-a-proteins-dihedral-angles-ordered-by-variance https://chemistry.stackexchange.com/questions/485/what-is-c%ce%b2-c-beta-deviation

Is anything unclear for example, here: https://en.wikipedia.org/wiki/Ramachandran_plot This is rather trivial question.

Only two angles are needed because the -CONH- grouping  is assumed to be planar, which  generally holds true, so is a good assumption.


text:
Colour change acid-base titration
I titrated orange juice with NaOH. I used no indicator, but at around the point of inflection, the juice changed colour to a more mustard-ish colour. What is the reason for this colour change?

Many (arguably, most) natural dyes have some indicator properties.

E.G. tea (turns yellow with lemon) and particularly red cabbage. See https://www.thoughtco.com/edible-ph-indicators-color-chart-603655 for a convenient list of edible pH indicators. Also see anthocyanin.


text:
Why every atom of a radioactive element, though being indistinguishable in every aspect, doesn't decay simultaneously?
First of all, I know this (or, similar) question has been discussed previously here. But the discussion there couldn't help me completely getting rid of my confusion, hence this attempt.
In any radioactive sample of a particular element all the atoms involved in that accumulation of sample are obviously seem to be indistinguishable in every aspect. Then why all the atoms don't start decaying at the same time?
In practice we obviously know that they don't disintegrate simultaneously and hence my question arises that what is that x-factor that differentiates between the atom which decays first than the others?

For all we know, there is no x-factor. The atoms are identical, and then one of them just up and decays, while the other continues to wait for another hour, or maybe a year. It is just blind chance, as random as it gets.

The universe is probabilistic.

While probability is entirely sufficient to explain this, also consider that not all the atoms in your sample necessarily were formed at the same time - they may be the result of several supernovae etc.

@TAR86 Atoms don't have memory of their age, that's the point. Atoms are dumb.

@IvanNeretin - no, it is just the nucleus that is dumb (must be something to do with the strong force?) - electrons can solve Schrodingeer's equation in their sleep...

@Mithoron I have already clarified in the beginning of the question that it is similar but not duplicate!

Not enough. Also https://chemistry.stackexchange.com/questions/19564/why-does-radioactive-decay-occur

@Mithoron no offense, but I would request you to go through my question again, please.

That's only a more elaborate version of earlier one, your lack of satisfaction with earlier answers is supposed to be reason to not dupe it? Also it's not exactly about chemistry. Best answer is that's first order reaction which we don't have means to observe directly, unlike some chemical ones which were studied using femtosecond spectroscopy.


text:
Why is xenon hexafluoride not stored in glass?
Why is $\ce{XeF_6}$ not stored in glass containers? Isn't the reason glass is used for most acids is that it is non-reactive?

Aren't there lots of fluorides you can't store in glass?

$\ce{XeF_6}$ can't be stored in glass as it reacts with the silica that makes up the glass
$\ce{2XeF_6 + SiO_2 -> 2XeOF_4 + SiF_4}$
This can proceed further to eventfully make $\ce{XeO_3}$:
$\ce{2XeOF_4 + SiO_2 -> 2XeO_2F_2 + SiF_4}$
$\ce{2XeO_2F_2 + SiO_2 -> 2XeO_3 + SiF_4}$
Dry $\ce{XeO_3}$ is highly explosive ...
That is why both $\ce{XeOF_4}$ and $\ce{XeF_6}$ are stored in nickel containers at room temperature.
These are generally considered as acid-base reactions using the Lux-Flood definition, see https://en.wikipedia.org/wiki/Acid%E2%80%93base_reaction#Lux.E2.80.93Flood_definition
I don't have them to hand but any decent University level Inorganic Chemistry text book such as Housecroft and Sharpe or Greenwood and Earnshaw should cover this.


text:
What are some reactions which can be used to study "Effect of Catalyst on the Rate of Reaction"?
I wish to study the effect of catalyst on the rate of reaction as an investigatory project.
What are some reactions and ideas that I could use to study this?
One example to this is the decomposition of $\ce{KClO_3}$ with $\ce{MnO_2}$ catalyst, The time for which $\ce{O_2}$ is evolved can be studied while varying the amount of catalyst.
What are some similar nice reactions with which notable changes in the reaction with changing amount of catalyst can be observed ?
What can be other changes I could bring in the reaction to study different aspects of the reaction and this topic effectively?
Suggestions please.
I know this question is open ended but please feel free to suggest anything that comes to your mind that can be done in high school laboratory.

KClO3 is dangerous, and this question too broad.

@Mith Well, the OP does use the word "some" here, i.e- he's asking for a few, notable/striking examples. He doesn't appear to be asking "Give me a list of all such reactions". So not too broad IMHO O:)

@paracetamol It's open ended, everyone could add another answer, claiming his "some reactions" are better. It's opinion-based if reaction is "amazing". At least what OP considers "similar" to mentioned one should be clarified, and it should be something other then "can kill you if you won't do it properly".

@Mith "It's open ended" <--- True, that. But not the same as "too broad". And yeah, the "amazing" bit should probably go. As for "similar reactions", the OP probably meant "other reactions that proceed rapidly in the presence of a catalyst" (or something like that). Wanna make the edit? :D

@above I have made some edits, so you might want to suggest something now that I have added some details.


text:
Why can't a reverse fission reaction be automatically classified as a nuclear fusion reaction?
$$\ce{_2^4He + _13^27Al -> _15^30P + _0^1n}$$
Will this reaction qualify as a nuclear fusion reaction? The answer to this question in the test says that it won't, yet it seems to be the reverse of a fission reaction.
Why would it not be considered fusion?

While it is certainly a nuclear reaction, but nuclear physics folks would likely just call it an ($\alpha$,n) reaction.

Were you given a definition of nuclear fusion reaction in class? As far as I can tell, this would be an example of fusion.

Either the test answer is wrong, or the answer depends on some weird(?) definition of a fusion reaction. It sure seems like a fusion reaction to me.

Well, there are two elementary steps here - one is fusion, later neutron emission happens.

It may be that your teacher's definition of a fusion reaction is one that releases energy. This reaction requires significant energy input. Admittedly, that isn't a good definition of a fusion reaction.

The Wikipedia article on Nuclear Fusion starts off:

In nuclear physics, nuclear fusion is a reaction in which two or more atomic nuclei come close enough to form one or more different atomic nuclei and subatomic particles (neutrons or protons). The difference in mass between the reactants and products is manifested as the release of large amounts of energy.  ...

$\ce{^4He}$ and $\ce{^{27}Al}$ are certainly different than $\ce{^{30}P}$.
The masses of the products however are more than the masses of the reactants
$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\text{4-He} & 4.&\hspace{-1em}00260325415 \\
\text{27-Al} & 26.&\hspace{-1em}98153863\\ \hline
\text{total} & 30.&\hspace{-1em}98414188\\
\end{array}$$
$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\text{30-P} & 29.&\hspace{-1em}9783138 \\
\text{n}   & 1.&\hspace{-1em}00866491588 \\ \hline
\text{total} & 30.&\hspace{-1em}9869787 \\
\end{array}$$
So it would seem that the definition used would require the products to have less mass than the reactants. This seems like a weird definition to me...

Not really, fusion can make nuclei with less binding energy, but this energy has to be provided, much like in chemistry. See my comment^

@Mithoron - I agree with you. It seems like a fusion reaction to me. However in order for the reaction that the OP gave not to be a fusion reaction, per test answer, the mass deficit is the only reason that I can think of. It depends of course on how the particular professor defines a fusion reaction.

Okay, so between two such reactions, the one that mentions a release of energy is more appropriately a fusion reaction, correct?

@Shreya - I would consider the reaction given a type of fusion reaction. For various types see https://en.wikipedia.org/wiki/Nuclear_fusion#Methods_for_achieving_fusion I was trying to explain why the test answer was marked that it wasn't a fusion reaction. So you need to clarify the exact definition your instructor used for a "fusion reaction."


text:
Does electrostatic polarity in aluminum or alcohol based dye affect adhesion. If so, what would encourage the best adhesion?
I am trying to find the best method to encourage adhesion of a wood alcohol based dye to 6061 aluminum. The results I have gotten have generally been a somewhat blotchy appearance, and a uniform appearance is desired. Current Method:
Place about 2K parts in a rotating tumbler with a spherical vessel of 2 gallons that has a 3.25" throated opening.
Tumbler rotates a 40 rpms
inject 1 tsp dye. parts are usually coated within 15 seconds
once coated blow compressed air (50 psi) to dry parts.
Parts are usually dry within 30 seconds.
See results in picture
Would polarity affect adhesion? If so, how would one go about providing the correct polarity?
Would heating the parts affect adhesion?
Would hot air drying affect adhesion?

Certainly, but polarity of what? It might help to pre-treat he aluminium, or use a slightly different dye, or modify the dying/drying process. What options do you have? I'd just try it out, hard to predict the consequence of slower/faster drying, use of more or less solvent, etc.


text:
How do you write the ionization equation for calcium hydroxide?
This is a base that would ionize completely, and the dissociation equation would look like this: $$\ce{Ca(OH)2 <--> Ca^2+ + 2OH-}$$ but how would I write the Brønsted equation with water? $$\ce{Ca(OH)2 + H2O -> ?}$$
I know that water auto-ionizes to make $\ce{H3O+}$ and $\ce{OH-}$.

Rather Lewis, then Bronsted. Cation coordinates water molecules (probably 6 directly).

Neither the $\ce{H^+}$ ion nor the $\ce{OH^-}$ ion float around in water as the bare ions. Both are "understood" to be in some sort of coordination complex with water molecules. In fact there are probably several layers of water molecules around the ion which are not oriented randomly.

You don’t. You’re done after$$\ce{Ca(OH)2 (s) ->[H2O] Ca^2+ (aq) + 2 OH- (aq)}\tag{1}$$
You may want to write an equation corresponding to the hydroxide version of the Grotthuss mechanism, in which case you might add:
$$\ce{OH- + H2O -> OH2 + HO-}\tag{2}$$
or some permutation of that. But unless you isotopically mark one water molecule or you pinpoint one in solution, this equation does not correspond to a measurable process since reactants and products are identical.


text:
What is "physicist's water"?
There is a statement on page 60 of this dissertation:

the "physicist's water molecule" ($\ce{O-H = 1.1 \mathrm{\mathring{A}}, \angle HOH = 104^\circ}$)

I understand it to be a model similar to the "spherical cow in a vacuum,", in that it's meant to be an easy-to-remember approximation, but

What is the etymology of this phrase or definition?
Why this specific choice of parameters? It could just as easily be $\pu{1.0 \mathrm{\mathring{A}}}$, according to Wikipedia, and the angle could vary as well.

Googling the phrase "physicist's water molecule" yields the dissertation cited as the only hit. So it seem like a phrase the author invented.

That's a (odd) phrase, not a definition. Obviously you cannot define a water molecule, physicist's or other's, as it is given by nature.

@Karl I disagree; I think it is a definition in the same way that linear water or planar ammonia are definitions. It is supposed to invoke some well-defined picture in your mind, even if it isn't physically sensible most of the time.

(What would you do with a linear water molecule? Not even didactically useful.) I just checked the values for actual water, and they are 0.958 Angstroms and 104.45°. A spherical cow is a useful appoximation, as it simplifys modeling the cow. Using very imprecise values to stick into a computer simulation is however simply stupid imo, and using a wrong value like 1.1A is terribly stupid. I don't know what the author of that dissertation was thinking, but i wager it wasn't much. Tried and failed to sound clever, with that phrase.

(Or is he simply using those values as starting point to successfully refine better values? In that case I take back anything I said about stupid. Sorry, too lazy now to read the diss myself.)

"Not even didactically useful." That is incorrect. Walsh diagrams show why water is bent and not linear, and why ammonia is pyramidal rather than planar.

@pentavalentcarbon But you're no doing anything with it there, only assume it for a moment to show why it does not exist.

I suspect the author might refer to "physicist's water molecule" term in order to underline the fact the geometry and energy of this molecule has been introduced as is, and hasn't been affected by intra- or intermolecular interactions.
I believe there is an allusion to Ancient Greece's philosophical schools which treated natural phenomena such as air, fire, water etc. as the purest and simplest entities. The term "physicist" itself refers to pre-socratic philosophers (Ionians, mostly) which abandoned religious attempts to formalize the reality [1, p. 20]:

... Pre-Socratic philosophers [...] have to be called Physicists because they abandoned the civil religion, whose insistency deified the city’s eponym, the royal function or the legislator’s role, the valiant warrior or the fertile woman, because they disdained what the social projects into the religious, whereas they kept or invented a global religious, that divine that’s immanent to the universe.

For instance, based on what's been considered the most fundamental element, physicist’s water has originally been introduced by Thales of Miletus, just as physicist’s fire "belongs" to Heraclitus.
Reference

Serres, M.; Burks, R. Geometry: the third book of foundations; Bloomsbury Academic: London; New York, 2017,. ISBN 978-1-4742-8139-3. (Google Book)


text:
On the mechanism of the nitrogen-removal step of the Wharton olefin synthesis
In today’s reaction seminar, the Wharton transposition made an appearance. In our practice case it started from the epoxide as in scheme 1 below as on Wikipedia; Kürti/Czakó (page 482) begin from the enone which is oxidised by hydrogen peroxide to give the same epoxide.

Scheme 1: General mechanism of the Wharton olefin synthesis starting from an α,β-epoxy ketone to give the allylic alcohol.
The first steps are clear to me, I am interested in the mechanism of the final one. Our problem set’s solution key (taken from a Japanese book that I cannot reference for lack of understanding) draws a concerted [1,3] migration as depicted in scheme 2a. Kürti/Czakó depict a sequential release with an intermediate vinyl anion. Wikipedia makes no note on that step and does not discuss mechanistic proposals.

Scheme 2: Mechanisms given for the final step. (a) concerted pathway given by our problem set. (b) stepwise pathway given by Kürti/Czakó.
Kürti/Czakó also point out that:

Alternatively, the formation of a vinyl radical has been proposed.6
6: G. Stork, P. G. Willard, J. Am. Chem. Soc. 1977, 99, 7067–7068.

The concerted mechanism seems slightly unreasonable due to the strained four-membered ring that would be required in the transition state. Likewise, the stepwise mechanism requires the formation of a vinyl anion which — while possible — is generally not very stable. In its defence, the step would be rendered irreversible by the liberation of nitrogen gas and the subsequent anion immediately captured by any acidic proton around.
Which mechanism does the reaction most probably follow?

Reference:
L. Kürti and B. Czakó: Strategic Applications of Named Reactions in Organic Synthesis. Background and Detailed Mechanisms, Elsevier Academic Press, Burlington, MA, USA, 2005, page 482.

Based on Google, I suspect that Stork's paper has been the last insight into the mechanism, although Scifinder is undergoing planned maintenance right now so I can't verify that claim. Li/Corey Name Reactions puts forth the evidence, I'll just put a screenshot from Google Books: https://i.stack.imgur.com/0xSCc.png

The Wharton reaction is often run in the presence of acetic acid (pKa 4.76. A vinyl anion has pKa ~44. Survival not long enough to effect cyclization (Ohloff, ref. 1 in Stork (1921-2017) paper) as an anion. That is why Stork proposed a radical mechanism. See Taber, et al., Tet. Lett. 1992, 33, 903 for an update. For a discussion of these N2 liberating reactions, see ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/WolffKishner/…

@user55119 If you have insights that are the basis of an answer, please feel free to write up an answer, too =)

Water could act as a proton transfer agent, hence reducing the need of a four membered ring to a six membered ring in the transition state. But I'm guessing...

The Wharton rearrangement presents an opportunity to convert one α,β-unsaturated ketone to its regioisomeric allylic alcohol. To this end Ohloff, et al.,1,2 explored the transformation of α-ionone 1 into α-damascone 2. Epoxide 3 was exposed to hydrazine in methanol to afford bicyclic allylic alcohol 4 (30%) and the uncyclized allylic alcohols 5 and 6 (20%, ~1:1). The proposed mechanism (3 $\rightarrow$ 7 $\rightarrow$ 8 $\rightarrow$ 9) goes through the vinyldiazene 8, which loses nitrogen to form reactive species. Ohloff proposed a mechanism invoking the vinyl anion 9, which is either protonated by methanol to afford acyclic allylic alcohols 5 and 6 or undergoes cyclization with the endocyclic double bond followed protonation of the resultant secondary anion. The bothersome aspect of this mechanism is, can a vinyl anion in a sea of methanol survive long enough to avoid protonation and undergo cyclization? The same point can be raised about  the traditional Wolff-Kishner reaction run at elevated temperature in protic ethylene glycol.

Stork and Williard3 explored the scope of the cyclization under Ohloff conditions finding that both 5- and 6-membered rings can be formed. Cyclization of epoxyketones 10a and 10b formed 11a (85%) and 11b (60%), respectively, as mixture of stereoisomers. In the former example, little of the uncyclized product was formed. These authors suggested a radical-based mechanism because in a typical reaction, the ratio of cyclized-to-uncyclization product was not markedly affected by employing the more acidic β,β,β-trifluoroethanol or the less acidic tert-butyl alcohol as a solvent. The authors argue that if a vinyl anion were involved, the more acidic solvent would be expected to increase the amount of uncyclized product at the expense of cyclized material. An intense yellow color developed which faded with time upon mixing the epoxide and hydrazine independent of whether or not cyclization occurred. A strong UV absorption at $\pu{232 nm}$ and a weak one at $\pu{409 nm}$ were attributed to the vinyldiazene. A concerted vinyldiazene cyclization was considered but with less enthusiasm.

Taber and Stachel4 designed a set of experiments to resolve this dichotomy based on the known stereochemistry of radical and anionic cyclizations. Tri-n-butylstannane-initiated cyclization of bromide 12a gave an excess of cis-cyclopentane 13a over trans-14a in accord with the cyclization of bromide 12b to 13b and 14b (13b>14b).5 Alternatively, metal-halogen exchange of iodide 15a with t-BuLi gave more of trans-cyclopentane 14a over the cis-isomer 13a in an anionic cyclization in accord with the results of Bailey6 for iodide 15b. The Wolff-Kishner reduction of ketone 16a in triethyleneglycol (TEG) afforded an 11:1 ratio of trans-14a over cis-13a. This result led the authors to conclude that the Wolff-Kishner cyclization is anionic as opposed to radical in nature.

While it is not necessary that the Wharton rearrangement and the Wolff-Kishner reduction must proceed via the same mechanism, in my opinion there is ample opportunity for more definitive results. Should an occasion arise where only the radical or anionic mechanism would apply, then the definitive experiment has been conducted.

References:

G. Ohloff, G. Uhde, "Über eine ungewöhnliche Cyclisationsreaktion bei der Umsetzung von (+)‐Epoxy‐α‐dihydrojonon mit Hydrazinhydrat (English translation: About an unusual cyclization reaction in the reaction of (+)-epoxy‐α-dihydrojonone with hydrazine hydrate)," Helv. Chim. Acta, 1970, 53(3), 531-541 (DOI:  https://doi.org/10.1002/hlca.19700530309).
K. H. Schulte-Elte, V. Rautenstrauch and G. Ohloff, "Synthesen der stereoisomeren (±)‐γ‐Damascone (English translation: Syntheses of the stereoisomeric (±)‐γ-damascones)," Helv. Chim. Acta, 1971, 54(7), 1805-1812 (DOI:  https://doi.org/10.1002/hlca.19710540709).
Gilbert Stork, Paul G. Williard, "Five- and six-membered-ring formation from olefinic $\alpha, \beta$-epoxy ketones and hydrazine," J. Am. Chem. Soc., 1977, 99(21), 7067-7068 (DOI: https://doi.org/10.1021/ja00463a053).
Douglass F. Taber, Shawn J. Stachel, "On the mechanism of the Wolff-Kishner reduction," Tetrahedron Lett. 1992, 33(7), 903-906 (DOI: https://doi.org/10.1016/S0040-4039(00)91571-5).
A. L. J. Beckwith, I. Blair, G. Phillipou, "Preferential cis cyclization of 6-hepten-2-yl and related radicals. Example of orbital symmetry control," J. Am. Chem. Soc., 1974, 96(5), 1613-1614 (DOI: https://doi.org/10.1021/ja00812a066).
William F. Bailey, Timo T. Nurmi, Jeffrey J. Patricia, Wei Wang, "Preparation and regiospecific cyclization of alkenyllithiums," J. Am. Chem. Soc., 1987, 109(8), 2442-2448 (DOI: https://doi.org/10.1021/ja00242a032).


text:
If 360 rotation is possible, How Molecule exist with particular Configuration
I Have confusion, If we talk about Ethane Molecule, we can get different conformations. For ethane, the eclipsed and staggered are freely inter-convertible  and in between these two extreme conformations lie infinite number of conformations where the dihedral angle may be 1degree or 2 degree or 10 degree etc. Thus, a slight change in the overall 3-D orientation in space results in a new conformer.
Now Coming to question, If rotation around single Bond is So free, Why Glucose Exist as (+)-Glucose, Why not C4 carbon get rotated simply by 180 and Get converted to glactose (as its rotation is free)?
What I am missing here?

You are totally missing the point. Re read the answer, made some draw, buid a little model of few molecule

Alright, I get that. I was thinking on the basis of Fisher projection.  Thank you.

https://chemistry.stackexchange.com/questions/59124/what-is-the-perfect-definition-for-chirality

Geometric isomerism exists exactly because rotation can be blocked.

I think you are confused between arrangement of different groups around carbon ($C$) and free rotation about $\ce{C-C}$.
In conformers, the orientation of groups in space does not change, but only rotation about $\ce{C-C}$ occurs. Note that this will never change the spatial configuration of groups around any carbon!
So + glucose will not get converted to any other optical isomer on changing conformations about any $\ce {C-C}$.

isnt the rotation of C4 carbon by 180 degree will Give us Glactose ?

@Aman No, It will not change configuration of groups around any carbon. On each chiral carbon, there is only two configuration possible. Rotation does not change them from one two another!

That is my question. If it is possible in ethane or butane, Why it is not rotating in case of chiral carbon??

It is not possible in ethane or butane.

I mean, rotation is of course possible, but it doesn't change a chiral center into another enantiomer.

if you could rotate in 4 dimensions, then sure you could do it. can you rotate your left hand until it looks like your right hand?

@gilleain very good example, thats the way I was taught optical isomerism when I was noobie. The 4d rotation part is still a mystery.


text:
Why does Dead Sea water feel greasy?
The best explanation I found so far is this:

These salts (magnesium, potassium, sodium, etc.) are responsible for the "greasy" feel of the water.

But it doesn’t say which salts exactly are responsible, and why it feels so different from seawater.

Indeed. Concentrated mineral solutions are often even sold as "oils" (e.g., here, even though there's no actual oil present: "Though applied in a manner similar to beneficial natural oils, magnesium oil in its composition is actually a highly concentrated solution of magnesium chloride in water." I'm curious about this too!

Might be due to very low surface tension. I think we might be more "wetable" in Dead Sea than in sweet water. But I must admit I feel more greasy water while bathing in lake as compared to Mediterranean Sea.  Perhaps both are true. In the last case greasy should be due to the organic content. ...

@Alchimista The surface tension of salt solutions is regularly higher than pure water

Perhaps it  has nothing or little to do. I regularly feel bathroom pipe water more " greasy" than sea water so I thought is our "wetabiliy" involved.

I don't know what you mean by the term "greasy", but here's a possible explanation:
When Dead Sea water evaporates on skin, then salts will precipitate when their maximum solubility is reached.
The mixture of tiny salt crystals formed, and water might be responsible for a greasy sensation.


text:
What does this dashed and curved line mean in pyruvate molecular graph?

Hi all, I came across this while researching a gene sequencing assignment and I cannot find out what the curved/dashed line means. Can anyone tell me please? Is it something to do with the negative symbol? Thanks

I’m guessing there is a duplicate lying around somewhere. If there is not, please leave open, answer accordingly and turn this into the dupe target.

It actually hybrid form of the contributing resonance structures of the ion. The significance of the dashed line is that it shows delocalisation of negative charge over the two oxygen atoms.

Note that

Neither of the above two forms represent the actual compound
Actual compound is represented by the resonance hybrid :


text:
Concentration of mercury in body

Symptoms of mercury poisoning become apparent after a person has accumulated $\pu{20 mg}$ of mercury. If a person ingested $\pu{30 mg}$ of mercury, what concentration of mercury in parts per million, are in his body? Assume the person has a mass of $\pu{63.5 kg}$."

My thought process, using dimensional analysis, was to convert the amount of $\pu{mg}$ the person ingested into $\pu{g}$. Then I took it ($\pu{0.03 g}\ \ce{Hg}$) and divided by the molar mass of mercury to find the number of moles of mercury there are.
I then divided the amount of mercury by $\pu{63500 mL}$ because I assumed $\pu{1 g} = \pu{1 mL}$. I did not use the $\pu{20 mg}$ information that it got me at all. Thus what I got was:
$$\pu{0.03 g} \cdot \frac{\pu{1 mol}}{\pu{200.59 g}} \cdot \frac{1}{\pu{63500 mL}} = \pu{2.3e-9 mol L-1}$$
However, I completely messed up, because first, my answer was in $\pu{mol L-1}$, and I'm not sure if you're supposed to use the molar mass of mercury here. The answer was $\pu{0.47 ppm}$, but I don't know how to get to here. Could someone explain the thought process for this?

Welcome to chemistry.SE! If you have any questions about the policies of our community, please visit the help center. Best of luck.

There are 1,000,000 mg in a kg (this is, mg/kg = ppm), so all you need to do is divide the mass of $\ce{Hg}$ by the mass of the person as-is (the units work out to ppm):
$$\left({30\,{\rm mg}\over 63.5\,{\rm kg}}\right)\cdot\left({{\rm ppm}\over {\rm mg\cdot kg^{-1}}}\right) = 0.47\,{\rm ppm}\,\ce{Hg}$$

Where did mg x kg^-1 come from?

mg/kg is exactly mg x kg^-1, of "millgram divided by kilogram". Please upvote and accept the answer if you find it helpful.


text:
Alkene Electrophilic Addition Reaction Stereoselectivity When Two New Asymmetric Centers Form
I had a question about the stereoselectivity in electrophilic addition reactions (specific to alkenes) that form additional asymmetric centers in their products. I understand that when a reactant with one preexisting asymmetric center forms a product with an additional asymmetric center, the two possible product stereoisomers are a pair of diastereomers because the configuration for the preexisting asymmetric center will stay the same while the configuration of the  newly formed asymmetric center will vary. Since diastereomers differ in stability, their transition states differ as well, therefore the product with greater stability will come from a transition state with lower energy, which will require less activation energy to form. The product stereoisomer with greater stability therefore forms faster and becomes the predominant product.

However, the textbook also says that the electrophilic addition reactions that form products with two additional asymmetric centers from a reactant without any asymmetric center form four stereoisomers of which two pairs are enantiomers with each other.

It is also mentioned that this kind of reaction is not stereoselective. I don't understand how this is the case; the members that form enantiomers with each other might be energetically identical, but the stereoisomers that are diastereomers to each other must be energetically different, thus having different stability. If the logic from the first case applies the same here, wouldn't the reactions that form two new additional asymmetric centers also be stereoselective since they form diastereomers?

Stereoselective becomes a hand-wavy term as soon as you may end up with more than one type of stereoisomer. If you want to actually state what is going on, you would most likely use either of the terms enantioselective (selects for a specific enantiomer) or diastereoselective (selects for a specific diastereomer; this may or may not include both enantiomers of the diastereomer).
In the first case, there is only one selectivity that can be observed; you can only get diastereomers, no enantiomers. Therefore, there may be some reason in shortening the term to stereoselective. In the second case you should not resort to using stereoselective because it is not specific enough — and that is irrelevant of the question whether stereospecificity is observed or not.
The second reaction would likely indeed be diastereoselective to a certain extent while definitely not being enantioselective (because the product is symmetric). So calling it diastereoselective — and thus, by intrapolation, stereoselective — may be justified.
In practice, the selectivities of both reactions would be so low that no practicing organic chemist would call either ‘selective’.


text:
Is there a way to "electroplate" wood with copper?
Maybe wood can be coated with copper (II) sulphate first, or roasted in order to form a conductive charcoal (carbon) layer.

Galvanising usually means coating with zinc. But the short answer is: it would be very difficult to coat wood with copper metal.

@Gert Didn't know that, i am not a native speaker of english. I corrected for "electroplate".

Electroplate implies the substrate has some electrical conductivity. Wood's is very poor, hence the difficulty.

@Gert It's difficult indeed, hence the question.

I like the concept of roasting the wood, but I don’t see how you would get a uniformly conductive surface due to the organic structure of the wood. If you coated it with copper sulfate, you would end up losing your conductive layer as it the sulfate salt would likely dissolve in your electrolyte as soon as you put it in. That’s why algae resistant paints used on boats need a sacrificial anode to be effective.

The most common way to apply metals by electrodeposition to nonconductive materials is to apply a "strike" of underlying metal, usually nickel or copper, via a method like electroless plating. In electroless, the electrons for the reduction of the metal ions to the zero-valent state are supplied by a reducing agent in solution:
$$
\begin{align}
\ce{Red &-> Ox + 2 e^-} \\ \\
\ce{Ni^{2+} + 2 e^- &-> Ni^0}
\end{align}
$$
This "strike" coating is then used to provide the electrical connection necessary to apply the metal of interest—here, copper—by typical galvanic electrodeposition methods.

This solution came to  my mind, but would this "strike" actually hold on to the wood?

@Mithoron As long as the wood was dry and in good condition (not crumbling, rotting, etc.), I would expect the strike to hold quite well, actually. You'd want to use seasoned wood to minimize swelling/shrinking, and you'd have to have the wood submerged for the absolute minimum time possible, but I think it has a decent chance of working. The porosity and texture of the surface should be well suited to it. Of course, there could easily be complications I'm not anticipating.

Dip the wood in a solution of ascorbic acid (Vitamin C; orange juice will do) and soap and allow it to soak. Soaking time will vary by wood variant.  then dip the wood in a solution of copper sulfate or another copper salt and soap.  The ascorbic acid will reduce the copper, creating a conductive coating.  This may need to be repeated to achieve a continuous conductive coating. Finally, rinse and dry and you are all set.

Homemade electroless, nice! What does the soap do?

Has this been done before?

have you tried this? I would be amazed if orange juice has sufficient ascorbic acid to pull this off, and it has loads of stuff you don't need or want. In the UK you can buy a big tub of ascorbic acid powder from Holland and Barret health foods extremely cheaply. There used to be one called Linus Brand and Boots drugstores also used to do it. I know vitamins are far more difficult to come by in Spain, for example, where I have only seen them in licensed pharmacies.

I've done it on glass with pure ascorbic acid.

@A.K. I tried this, and it dit not work. I used solutions from tap water which were about saturated (some grains still visible in the vessel) at room temperature. I then dipped pices of beech wood, oak and balsa wood, dried them over night, and then dipped them into the other vessel, over several rounds. I left them in the bath for hours, until the weight hinted that the were fully soaked (it was small pieces). No effect thought. Just a light green colour, which also is what get if you mix the solutions directly.

@A.K.  I got the same result that HannesH did.  The soap I used was Fels-Naptha , a true soap, not a detergent.  The Fels-Naptha gave a pH of about 9 or 10, a good sign as the ascorbic acid reaction proceeds in a basic, not acid solution.  Further complicating things was that the copper sulfate/soap solution gave a flocculant. What "soap" did you use?  What was the pH of the ascorbic acid/soap solution?  Further details would be greatly appreciated.

Using Marble's Reagent to etch stainless steel, the wood stirrer got coated with copper. Marble's is $\ce{CuSO4}$, $\ce{HCl}$ and water, but I think I added extra $\ce{CuSO4}$.
This isn't technically electroplating, though.

So that is a mix of hydrochloric acid and copper sulphate? I wonder if the process here is that the wood undergoes hydrolysis from the acid, forming charcoal, which then is a reductand to the copper sulphate.

It might be possible to evaporate the copper in a vacuum so that it is deposited on the wood. No electric conductivity required.
Here is an example using gold.
YouTube: Gold & Casio Watch - Periodic Table of Videos
And a search found this random document about it
random document about vacuum evaporation

The question asks specifically about electro plating.

@pipe since wood is non-conductive, some way to coat wood with something conductive is a requirement for electroplating, and thus well within my question.

The word electroplate in the question is in quotes which I interpreted to mean "not necessarily literal electroplating".

Take finely powdered graphene and mix in a water based Polyurethane.
Put a copper nail (or drill a hole and epoxy a wire) into the wood in an inconspicuous area to be plated.
Paint a very light coating over the surface ensuring that you have surrounded the nail (or wire) and make sure the section you want plated has a uniform coating of graphene.
You should be able to plate on this now conductive piece of wood.
Alternatively, there are conductive silver-based paints, but these are relatively expensive.
You’re going to have to figure out a way to weight this down to keep it submerged in the electrolyte because it’s going to want to float.


text:
Why does pKw first decrease but eventually increase when temperature rises?

I understand the initial decrease in pKw.
$\ce{Heat + H2O (l) <=> H+ (aq)  + OH- (aq)}$
Temperature increases, reaction shifts to the right, so
$\ce{H+} \text{ and } \ce{OH}$   concentration increases.
Higher $\ce{H+} \text{ and } \ce{OH}$  concentration results in lower $\ce{pK_w}$, as $\ce{pK_w = - log ([H+][OH-])}$
But why then does the $\ce{pK_w}$ start to increase beyond $250$ degrees C?

I'm afraid your logic behind the decrease isn't exactly OK, or maybe it's just lacking details.

I think there is a lot you're missing. 25 mPa is about 24.7 atmospheres. At that pressure water boils at 224 C. There is a semi-empirical formula. See https://www.maplesoft.com/applications/view.aspx?SID=154073&view=html

The closer you come to the boiling point, the less energy the ions gain from hydratisation.


text:
Concentration of ammonium phosphate if dissolved in water

If $\pu{2.54 mg}$ of ammonium phosphate is dissolved in enough water to make $\pu{300 mL}$ of the solution, what is the molar concentration of the solution?

The answer is $\pu{5.68e-5 mol L-1}$. However, the answer I got using dimensional analysis was wrong:
$$\frac{\pu{0.254g} \ \ce{(NH4)3PO4}}{1} \times \frac{\pu{1 mol} \ \ce{(NH4)3PO4}}{\pu{144g} \ \ce{(NH4)3PO4}} \times \frac{1}{\pu{300L}} = \frac{\pu{0.00587mol} \ \ce{(NH4)3PO4}}{\mathrm L}$$
Could someone explain how they got this answer?

Please do read this section: How can I format math/chemistry expressions here?

Dimensional analysis? There's nothing to be analyzed. Most basic math is sufficient to solve this "problem".

Is there actually ammonium phosphate? It should lose ammonia as phosphate ion is a stronger base than $\ce{NH_3}$.

You missed the decimal point in $\pu{2.54 mg}$, mistaking it for $\pu{254 mg}$.

$2.54\pu{mg}=2.54\dot10^-3\pu{g}$. Also, $300$mL is different from $300$L. Also, the molar mass of ammonium phosphate is 149g.

First of all, ammonium phosphate formula is $\ce{(NH4)3PO4}$. Second, I would recommend to write down an expression first, and only plug in the numbers at the end. Third, mind the metric prefixes: $\pu{1 mg} \neq \pu{10^{-1} g}$. That's said, and taking into account correct molecular weight of ammonium phosphate $M(\ce{(NH4)3PO4}) = \pu{149.09 g mol-1}$:
$$C = \frac{n}{V} = \frac{m}{MV} = \frac{\pu{2.54e-3 g}}{\pu{149.09 g mol-1} \cdot \pu{0.3 L}} = \pu{5.68e-5 mol L-1}$$
where $C$ – molar concentration, $n$ – amount, $V$ – volume, $m$ – mass.


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Why is a Grignard reagent not protonated by chloramine?
When $\ce{RMgX}$ (Grignard reagent) reacts with chloramine ($\ce{NH2Cl}$) it forms $\ce{RNH2}$.
But why doesn't it form $\ce{RH}$ by taking a hydrogen atom from chloramine as the negative charge so developed on nitrogen will be more stable as nitrogen has high electronegativity and $\ce{Cl}$ being a electron withdrawing group further stabilise the anion.
So where am I wrong? Can anybody suggest a mechanism?

On nitrogen, chlorine is no longer electron-withdrawing. In $\ce{NCl3}$, chlorine actually bears the partial positive charge (demonstrated by reacting with chloride ions to form chlorine gas).

Still may be more electron wothdrawing/less electron releasing than hydrogen, though.

The $\ce{N-H}$ protons are relatively weakly acidic and this makes deprotonation kinetically less favored if an alternative nucleophilic attack mechanism is available.  With $\ce{NH3}$ no such alternative is there, due to lack of a decent leaving group, so deprotonation ultimately prevails; but $\ce{NH2Cl}$ allows displacement of a chloride ion and thus enables nucleophilic attack.


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Which electron is knocked off a Helium atom?
When a He atom is ionized, which electron is removed first, the one with the spin up or down? or is it just random?

An interesting question! Does the polarization of ionizing radiation interact with electron spin? Apparently, that has been investigated to some extent, and some spin information can be obtained from an entangled electron and photon.
So the answer to your question is not a clear-cut "yes" or "no", but depends on the parameters of the incident photon.

Very interesting, but this seems to be overthinking the question. Without additional constraints on how the ionization is achieved I believe that the answer is that it is just random.

@MaxW, agreed... without the (bolded) polarization, it's not meaningful, but with that constraint, a user might learn from it.

Hm, without an external field that gives orientation, the answer should be "can't tell", shouldn't it? Because if a single He atom is all the reference available, the photon doesn't even have a distinct polarisation.


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Order of stability of substituted benzyl carbanion

I think the order should be II > III > I. The +m effect should be more in case 1 than in 3, isn't it ? So, 1 should be more unstable than 3.
Someone please explain.

I'm not a chemist, but I doubt whether the mesomeric effect is "more in case I than in III" as you put it.
I'm under the impression that the mesomeric effects (alone) in both cases activate the benzene ring to the same extent (If you try drawing the resonance structures, you'd probably feel the same way).
However, I suppose in case I, the minor $-I$ effect of the $\ce{-OCH3}$ group makes I slightly more stable than III (because the negative charge is somewhat diminished).
So to me, $\mathrm{II > I > III}$ looks about right.

The answer is II > I > III because in II and III resonance effect will be same because resonance does not depend on the distance but since inductive effect depends on distance so -I of $\ce{OCH3}$ will draw more electron density than para position hence will become more stable. So correct order is II > I > III.

Isn’t resonance more important for stability than the inductive effect?


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Identification of a metal from group 2

A chemist adds $\pu{0.115 g}$ of a group 2 element to water and the resulting solution is made up to $\pu{400 cm3}$ by adding distilled water. The equation for the reaction is shown below. $\ce{M}$ represents group 2 element.
$$\ce{M + 2 H2O -> M(OH)2 + H2}$$
The concentration of hydroxide ions $\ce{OH- (aq)}$ in the solution is $\pu{6.56e-3 mol dm-3}$. Determine the molar mass of $\ce{M}$ to one decimal place. Use your answer to identify $\ce{M}$.

By definition molecular weight is mass $m$ divided by amount $n$, e.g. for the unknown metal $\ce{M}$:
$$M(\ce{M}) = \frac{m(\ce{M})}{n(\ce{M})} \tag{1} \label{eq:1}$$
From the chemical reaction of hydroxide formation one can deduce that $\pu{1 mol}$ of $\ce{M}$ results in $\pu{1 mol}$ of $\ce{M(OH)2}$. However, the amount of hydroxide-anions $\ce{OH-}$ in the solution (assuming complete dissociation) is twice as high as the amount of $\ce{M(OH)2}$. Hence
$$n(\ce{M}) = n(\ce{M(OH)2}) = \frac{n(\ce{OH-})}{2} = \frac{C(\ce{OH-}) \cdot V}{2} \tag{2} \label{eq:2}$$
where $C$ – concentration and $V$ – volume.
Plugging $n(\ce{M})$ from \eqref{eq:2} in \eqref{eq:1}, a final expression for the molecular weight can be established. Remember, before plugging in the numbers in make sure the units are unified.

 $$M(\ce{M}) = \frac{2 \cdot m(\ce{M})}{C(\ce{OH-}) \cdot V} = \frac{2 \cdot \pu{0.115 g}}{\pu{6.56e-3 mol dm-3} \cdot \pu{0.4 dm3}} = \pu{87.6 g mol-1} \tag{3} \label{eq:3}$$
 The unknown metal appears to be strontium $\ce{Sr}$.

Since as the molarity of the hydroxide is $\pu{6.56e-3 mol dm-3}$, by its stoichiometry the molarity of the metal must be half of that, i.e. $\pu{3.28e-3 mol dm-3}$.
We have $0.400\ \mathrm{dm^{-3}}$ of solution, so that contains $0.400\times \pu{3.28e-3 mol}$ of the element, which we also know weighs $0.115\ \mathrm{g}$ of that same element.
Dividing the mass of the element by the number of moles gives the molar mass to be: $87.6\ \mathrm{g/mol}$. That is strontium ($Sr$).


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Entropy, reversibility and equilibrium
Entropy is maximum at equilibrium, so if we have an equilbrium chemical reaction like:
$$\ce{HI <=> H+ + I-}$$
does it mean that the container of $\ce{HI}$ has maximum entropy or the universe?
Also, constant entropy implies reversibility but I don't understand how  the definition of reversibility applies to a system  of $\ce{HI }$ and $\ce{H+, I-}$ in a container. I mean, there's nothing to actually reverse, is there? By reversibility I mean that the system can be brought to its original state by an infinitesimal change(s) in it's condition.
Could anyone provide an explanation for this?
Note: I have edited my full question because previously it was very unclear.

Irreversibility is more a practical than a theoretical concept, in my opinion. 'In theory' all reactions are reversible. Take this highly schematised reaction:
$$\ce{A + B <=> C + D}$$
We define the equilibrium constant as:$$K=\frac{[C][D]}{[A][B]}$$
Using Nernst we can now establish a relation between $K$ and the left-to-right change in Gibbs Free Energy, $\Delta G^0$:
$$\Delta G^0=-RT\ln K$$
(Of course, as you know $\Delta G^0=\Delta H-T\Delta S$)
Evaluate this Nernst function to conclude that the more negative $\Delta G^0$ is, the higher the value of $K$:
$$K=\mathrm e^\frac{-\Delta G^0}{RT}$$
For very negative values of $\Delta G^0$, $K\to+\infty$ and by the equilibrium  constant equation, the concentration of the reagents at equilibrium is essentially $0$. Then we can write:
$$\ce{A + B -> C + D}$$
Such a reaction we would call irreversible. There is however no clear cut-off point and many reactions will have $K$ values that somewhat defy categorisation with respect to being 'reversible' or 'irreversible'.

There is no standard dividing line between reversible and irreversible reactions.  Even from IUPAC. It is simply the case that it is more convenient to treat some reactions as irreversible (for example when the reverse reaction doesn’t contribute meaningfully to the composition of the final reaction mixture).

does it mean that the container of HI has maximum entropy or the universe?

At equilibrium, the entropy of the universe is maximal. There is a lot of other stuff going on in the universe, so if you can isolate your reaction and the surrounding (no matter or energy leaving the surrounding), you can also say the sum of entropy in the system and the surrounding is maximal if everything is at equilibrium.
If you would like to talk about equilibrium based on thermodynamic functions of the system, the Gibbs energy of reaction ($\Delta_r G = \frac{dG}{d\xi}$) would be appropriate for a reaction at constant temperature and pressure. Under these conditions, the reaction reaches equilibrium when the Gibbs energy is at its minimum. In the absence of non-PV work, the minimum of the Gibbs energy of reaction corresponds to maximizing the entropy of the universe.
The entropy of the reaction $\Delta_r S$, on the other hand, typically is not at its minimum when the reaction reaches equilibrium (unless $\Delta_r H$ happens to be zero, as in some transport reactions).

I don't understand how the definition of reversibility applies to a system of HI and H+,I− in a container.

Reversibility means that a process is done at near-equilibrium conditions, i.e. with minimal increase in entropy. If you are considering a process that is at equilibrium, it is certainly happening under reversible conditions. The final state is equal to the initial state - "nothing" happened, so you can certainly reverse that "nothing" without entropy increase - in fact, you don't have to do anything at all.
It is different for a process like an expansion of a gas. Depending on how you run the process, going back to the initial state will result in a large increase of entropy (irreversible) or hardy any increase (approaching a reversible cycle).

You probably know this, but for the OP I think it is important to note explicitly that the Gibbs free energy of the system is minimized at equilibrium only under conditions of constant T and p (since, under those conditions, minimizing G_sys corresponds to maximizing S_universe).  That, of course, is why the Gibbs free energy is useful, since those are common reaction conditions.

How do I relate the reversibility that we study in chemical equilibrium to that we study in thermodynamics?

I think I now understand what you're trying to say. The root of your problem is that the word "reversible" in chemical equilibrium is different from "reversible" in thermodynamics. They're not same as you are thinking instead.
Note that a reversible, thermodynamic, physical process is better  known as a "slow" process. It is called "slow" because it takes infinite time to complete, because at each the system is in thermodynamic equilibrium. It is called a "reversible" because, if performed in the reverse direction in the same manner, you can retrieve 100% of the input energy that was input as work. This is different from "irreversible" processes, where some input energy is permanently lost.
A kinetic "reversible" process - the one you have shown - is better known as an "equilibrium" process instead, and is very different from a reversible, thermodynamic, physical process, as you may observe now.
Also, note that $\Delta S_\text{total}=0$ holds only for a thermodynamic, physical process (like a piston compressing a gas). Chemical reactions (like the one you've specified) instead always involve a change in entropy, even if they are isothermal.


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Why is it that energy of electrons in higher orbits is greater than the ones in lower orbits?
There is this picture I found online which indicates that the energy level is higher as we go up and more electrons are added to the atom.

What I'm having trouble understanding is why the energy level is higher and not lower. The electrons are further and further away from the nucleus, at infinity their energy would be zero. Is it correct to say that the electrons' energy is negative and as we go further it increases, going to zero at infinity?
That is the only way I can see all this make sense.

All energy values in the graph are negative, thus stabilisation. The zero energy of vacuum is somewhere over the top of it all.

What I'm having trouble understanding is why the energy level is
  higher and not lower. The electrons are further and further away from
  the nucleus, at infinity their energy would be zero.

You are correct. The innermost electrons are the most tightly bound to the nucleus and the outermost ('valence') electrons have the least energy, explaining why they can be removed (in chemical reactions e.g.)
But these energies are always negative and use the principle that infinitely far away from the nucleus the electron's energy would be $0$.
This way defined, the energy of the hydrogen electron in the ground state ($1s$) e.g. is $-13.6\ \mathrm{eV}$. The energy of the electron in excited states ($2s$, $2p$, $3s$ etc) is less negative, until for infinite distance (a free electron) it is zero ($0$).

Let's take a numerical example to demonstrate this. The ground state ($n=1$) of hydrogen's electron is $E_1=-13.6\ \mathrm{eV}$ and the first excited state ($n=2$) is $E_2=-3.4\ \mathrm{eV}$. What are the energies needed to remove the electron from these states?
Since as the energy at infinite distance is $0$, in the case $n=1$, then:
$$\Delta E= 0-(-13.6)=13.6\ \mathrm{eV}$$
And in the case $n=2$, then:
$$\Delta E= 0-(-3.4)=3.4\ \mathrm{eV}$$
So it takes less energy to remove an electron in a higher orbital than a lower one. That fits our observations.
Equally, the energy needed to excite the electron from $n=1$ to $n=2$ is:
$$\Delta E_{1\to 2}=-3.4-(-13.6)=10.2\ \mathrm{eV}$$

@Gert has given a great explanation and I like to give a small method for finding the answer for your question.
Consider an object which is 2Kg is on a level of 10m high from the considering level. It is drop from there and you know that it release its potential energy of mgh=200J. If it is on 20m high it release 400J. So we can see that
*when something drop from high energy level to a lower energy level it release its energy
*When it is drop from a more high energy level it release more energy than the other.
So, now we consider the Hydrogen spectrum of emission,
I will take some lines from that. Consider the line which is the reason for
3 →2
 4 →2,
5 →2
According to the energy release 5 →2>4 →2>3 →2. So we can understand that 5th level is in a higher level 4 is then and 3 is then. Now as the above example the 5th level has a great energy than the 4th level.(Such like 20m level has more energy than the 10m level).
The nucleus attraction also affect for the value of the energy of the energy levels. As it goes above the attraction is coming lower due to that the gap of energy levels is coming lower.
As a summery,
When go above energy of the levels be increase, but the gap of the energy levels be decreased.

The potential energy (given by mgh) should be ≈ 2kg10m10 ms^(-2) = 200 J, not 2kJ.

Thank you for showing it. I will edit it.  Sorry for have a such mistake.


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Beryllium iodide and solubility rules
The beryllium iodide $\ce{BeI2}$ as covalent compound is apolar, and thus, it should be water non-soluble. However, it is. Making a research on the web, I learned it is just in the middle of being ionic. Dos it happen with the $\ce{MgI2}$ of $\ce{MgF2}$ as well? I usually explain the following to my students:

Solubitity "general" rule: "Similar dissolves similar".
Solubitity is also affected by differences in electronegativity and bonding polarity.
Solubility is also affected by the capability to displace charge and solvation (that is, intermolecular forces also apply to some extent).
As a practical rule, we can consider ionic compounds those formed by G1 or G2 with G16 or G17, but this rule has exceptions, like the rules of electronic configurations.

Is all stuff correct? Solubility in water of $\ce{BeH2}$ and $\ce{XeF2}$ can be also considered with these rules?

General rules have the general tendency of falling apart in a generally rapid manner.

Be is very small, iodine very big, no wonder Be2+ prefers to be bound to water.

Just because something is covalent and has zero (permanent) dipole moment does not mean It's doomed to water insolubility.  There is zero permanent dipole in https://en.wikipedia.org/wiki/Hexamethylenetetramine, yet this is soluble in water up to almost a 1:1 mass ratio.

I know, similar dissolves the similar thumb rule is not general, as the principle of electronic configurations. I just came to remember the Usanovich theory, so to solvatate something, we have to be able to displace charge...Any way, the rules are to be improved...


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What is the mechanism of the reaction between hexane-2,5-dione and nitrous acid?
I don't understand the mechanism of the first reaction; I tried different ways but I can usually find how to make the ring without the double bonds, but it's obviously not the right thing; I think HNO2 reacts as the nitrosonium ion, bonding with one of the oxygens (since it has free electrons) but I can't never make it work.
EDIT:
Probably the acid is formed by nitric acid (like in one of the articles orthocresol linked), that I think is not written because it can be formed by the decomposition of nitrous acid; I still don't understand the mechanism of the formation of the ring.

Image: https://commons.wikimedia.org/wiki/File%3AIsocarboxazid_synthesis.png
by Nuklear (Own work) [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons

Where did you get this scheme from? Based on the papers I'm looking at, I believe it's wrong. The initial substituent should be COMe instead of CO2H, and the oxidation to CO2H is effected using nitric acid. See J. Heterocycl. Chem. 2009, 46, 909. Either that, or you use HNO3 instead of HNO2 to get to the carboxyl isoxazole in one step. See J. Med. Chem. 2004, 47, 3642.

What is "Bn" In that diagonal arrow?

I found it in a book I used : Vardanyan, R.S.; Hruby, V.J.; Synthesis of Essential Drugs; Elsevier (2006); I also found other sources online (like this image from Wikipedia) that uses that synthesis.

I think "Bn" is short for benzyl-, so it's benzylhidrazine

If you could give a link to the Wikipedia image, that would be good (mostly for some context, but also for giving some kind of credit to the original author)

You have misdrawn your final product, it should be the benzyl hydrazine not the phenylhydrazine

As you may know from diazotisation reactions, $\ce{HNO2}$ is a generator of the nitrosonium cation, $\ce{NO+}$. Electrophiles like these can react with the enol tautomer of hexanedione:

Following this, you tautomerise the nitroso compound to the oxime. The mechanism is exactly analogous to keto-enol tautomerism:

Note that I've elected to draw the oxime with a (E)-configuration. Both (E) and (Z) oximes can form, but the (Z)-configured oxime cannot cyclise in the following step to form an isoxazole:

Following this, to get to the compound in your question, you supposedly need to oxidise the acetyl substituent ($\ce{COCH3}$) to a carboxyl substituent ($\ce{COOH}$). Nitrous acid isn't an oxidant, which casts doubts on whether your source is veritable (perhaps it's just a typo).
In the literature, I find that this oxidation is accomplished using nitric acid, $\ce{HNO3}$, which is a perfectly fine oxidant. In one case (J. Heterocycl. Chem. 2009, 46, 909), the diketone is first treated with nitrous acid to get the 3-acetylisoxazole, and is then treated with nitric acid to form the 3-carboxylisoxazole. In a separate case (J. Med. Chem. 2004, 47, 3642), the diketone is simply refluxed with nitric acid to form the 3-carboxylisoxazole. I suspect that this works because nitrous acid is present in the nitric acid, so the mechanistic pathway is exactly the same (a similar effect has been observed with the nitration of phenols using dilute nitric acid).
Exactly how nitric acid oxidises an acetyl group to a carboxyl group, though, I don't know (and I'm not sure if it is known).

Edit: There is, in fact, a mechanism by which nitrous acid may oxidise the acetyl group to a carboxyl; see user55119's answer.

Can you use iodoform reaction here to convert acetyl to carboxylate? Just curious since there is a misstep in the question.

@MathewMahindaratne, I suppose you could, although my heterocyclic chemistry is nowhere near good enough to comment on whether the isoxazole will be stable under the basic conditions.

With my experience, they are stable. I'm worrying about its methyl substitution.:-)

Orthocresol did a great job describing how to get to the isoxazole methyl ketone. However, nitrous acid can be an oxidant. The methyl ketone in the presence of mineral acid, which you need to get the nitrosyl cation, can tautomerize the methyl ketone to the enol which reacts with the cation to get an α-nitroso ketone. The nitroso group tautomerizes to an oxime as orthocresol has shown. Acid-catalyzed dehydration to a nitrile leads to an α-ketonitrile. This species is similar in reactivity to an acyl halide. Aqueous hydrolysis gives the desired carboxylic acid with the loss of HCN. In nitrous acid (HONO), nitrogen is +3. In HCN, nitrogen is -3. This is a 6-electron reduction for nitrogen and a 6-electron oxidation of the substrate. The nitrosation is a 4-electron oxidation and the oxime dehydration is a 2-electron oxidation. Nitric acid is not required.

The truth depends on how the nitrous acid is made.  You don't have concentrated stock solutions of $\ce{HNO2}$.  You combine a nitrite salt and a strong acid in situ.   Use $\ce{HCl}$ for the strong acid and you have the mechanism here.  With $\ce{HNO3}$ you can use the alternate approach.


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How do scientists figure out the chemical structure of a given compound?
Please bear with me a bit while trying to explain my question as best as I can. I am a 4th year medical student so my command on chemistry is not too good, but I find it fascinating and intriguing, it is like magic.
My question is as follows:
How do scientists actually know the exact composition of a compound e.g. Caffeine ($\ce{C8H10N4O2}$), how do they exactly know that it has this exact amount of hydrogen or nitrogen atoms in a single molecule?
After knowing that, how do they know the precise geometry of the molecule (here I mean the hexagons and pentagons and how they connect).
And one last thing, how do they know the exact geometry and sequence of atoms in an enzyme?

The structure of Caffeine has been confirmed by synthesis i.e. the synthesised material is in all respects identical to the natural. As for the geometry of the molecule that comes from X-ray crystallography

The As to the Q you are addressed to  are all right but do not really satisfy your Q. Indeed,  chemists figured out the structure of many "pentagons and hexagons" by studying their reactivity and effectively synthesing them. Plus theory to rationalize and work out the details. AFM etc arrived later and can be seen as a celebration of the Genius.  You may try to read about history of benzene or chlorophyll, for instance.

A 'picture' only recently using atomic force and scanning tunnelling microscopes as given in answer to the question linked to above. But we determine a molecule's structure primarily  using a combination, as necessary, of x-ray crystallography, nuclear magnetic resonance  (nmr) and mass spectrometry . None of these produce a 'picture' directly but the data obtained is used to form a model of a molecule's structure.


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Phase diagrams and evaporation
Phase diagrams tell how a given substance changes it’s phase with pressure and temperature. While most sources discuss this in detail, none of these mention whether or not evaporation is taken into account when drawing the diagrams .
Any liquid can evaporate regardless of the temperature, given that the molecules have the needed kinetic energy. So, do phase diagrams take this into account?

Yes, pressure is taken into account. One way of measuring activities in molten mixtures is to measure the vapor pressure of the constituents over the melt. Since those activities determine the shape of the liquidus, clearly the vapor phase is considered/measured.

A phase diagram depends on compositions, pressure and temperature. Since such a diagram is a 2D drawing of a 3D system, typically either the pressure or temperature will be constant.


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What happens to the t-butyl cation in the TFA deprotection of a t-butyl ester?
I am a first-year chemistry graduate student and I am currently learning how to synthesize proteins. I have reached a stage in the process where I need to deprotect a carboxylic acid moiety protected by t-butyl using trifluoroacetic acid (TFA).
I went online trying to find the mechanism for this step, but in terms of step-wise mechanism with electron movement arrows, the only thing I was able to find was this:

(Taken from here, no references therein)
While this looks legit, I am wondering about the t-butyl cation and its fate.
My reaction is a bit simpler, and is presented below:

And my question is: is this truly where the deprotection ends? Does the t-butyl cation remains in solution as pictured?
(I know that a t-butyl cation is considered to be relatively stable on account of it being a tertiary carbocation, but still I was not sure that it can remain in solution like that.)

Probably gives isobutene, depending on what solvent you're using

The solvent is dichloromethane. What is its role?

In which case DCM plays no part as it will not react with tBu+, other solvents may react with it e.g. toluene

According to a post-doc colleague, step no. 4 IS the final state of the molecules in the solution (i.e. t-butyl can exist in the solution as a cation), but if water will be added to the solution, then the t-butyl cation will take an OH group and the TFA will become protonated. In case anyone was wondering...

It is worth remembering that when you cleave a tBu ester,  tBu+ can hang around and potentially react with e.g. electron-rich aromatics

Would tBu+ show up on NMR?

If you ran your reaction in CD2Cl2 and did an NMR of an aliquot I guess it might. If you concentrate your aliquot first before redissolving in your NMR solvent I think you would lose it as isobutene

I realize this question is old but the mechanism is nicely described here (steps 1-3): http://www.umich.edu/~chemh215/W06HTML/SSG2/ssg6/main_files/Page346.htm
Briefly, the released t-butyl carbocation is subsequently deprotonated by the anionic TFA, resulting in the formation of 2-methyl-propene. This regenerates the protonated TFA that is needed to continue the reaction, and thus the TFA is catalytic (i.e. not consumed in the reaction).

The reaction is, in principle, catalytic in TFA. If t-butyltrifluoroacetate is the by-product, then the catalyst is consumed. If only isobutylene is formed, catalytic is OK. Using excess TFA covers all contingencies. In the initial step, protonation is more likely on the C=O oxygen. Higher electron density.


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Synthesis Golf VII: (+)-Clopidogrel

(+)-Clopidogrel (sold as Plavix) is a common pharmaceutical used to help combat the risk of heart disease. The World Health Organisation lists Plavix on its list of 'essential medicines' due to its broad efficacy, low toxicity and low cost (less than 1USD/month in the developing world).
Many industrial routes rely on racemic synthesis and resolution, however syntheses of a single isomer have been reported.
The molecule makes an interesting target as it requires the synthesis of an unusual heterocyclic core, as well as the challenging stereocentre which is prone to racemisation.
Some ground rules:

the synthesis must include formation of the heterocyle (the whole thing, both rings), and a way for setting the stereocentre
you can start from anything you like (subject to the above condition), but it must be available in the Sigma Aldrich catalogue

Sorry this is late (again).

One answer :( - guess end of year makes people busy...

@NotEvans. when's the next challenge

This was a tricky synthesis. There were several issues. One is that the aryl glycine is possibly easy to epimerize, though the ortho substitution may help prevent that. One other major issue is that the thiophene is reducing and not compatible with any oxidative conditions and probably not compatible with electrophiles.
The amino acid moiety can be synthesized in several ways to set the stereogenic center. One is epoxide ring opening, but here I opted for an asymmetric Strecker reaction.
The six member ring was created via a [4+2] cycloaddition. This is probably the sketchiest portion of the synthesis.
The thiophene moiety was created using Paal-Knorr reaction with Lawesson reagent (B).
Detailed route:

imine formation
asymmetric Strecker with 5% PHTA (partially hydrolyzed titanium alkoxide) and 5% ligand (L). 1 eq tBuOH, tol. (Org. Lett., 2010, 12, 264.) Also, possible with Hoveyda type ligand: (JACS, 2000, 122, 2657.) This exact substrate is reported with good ee.
Hydrolysis with HCl. Similar route reported in (JACS, 1999, 121, 4284.) Though there, the authors Boc-protect at the end. We'll attempt to trap a different way...
formalin, toluene, pTsOH. reflux. Product imine is A.
LDA; TMSCl on methyl vinyl ketone to make electron rich diene.
Hopefully, a little heat will coax this [4+2] cycloaddition with A to take place.
Tsuji-Trost allylation is established for these types of TMS-protected enol ethers. Conditions are $\ce{Pd_{3}(dba)_{2}}$, DPPE, allyl methyl carbonate, THF. (Chem Lett., 1983, 1325.) We'll carry both diastereomers forward.
$\ce{O3}$; DMS.
Lawesson reagent (B).

nice one @Zhe I was thinking of asymmetric imine hydrogenation, but the Strecker Rxn kills two birds with one stone

@gannex I was thinking of that originally, too, but I don't have good access to lit searches, and I was worried about selectivity on the iminium ion...

I was thinking the same thing. I was researching it but then I went to sleep. Looks like that was a mistake :-P

Conceptually a nice idea. I am concerned about the formation of imine A under aqueous acid conditions (40% formalin).  How do you isolate it? Might you consider using gaseous  formaldehyde in the presence of the amine and diene. I entertained the thought of an isonitrile (from the formamide) as the dienophile. Couldn't find a precedent. If viable, you would have a ring with an extra double bond. After allylation and ozonolysis of the protonated amine (to protect it), the double can be removed by hydrogenation.

@user55119 Interesting. I also considered just using a solution of formaldehyde. Probably easier. My first attempt at the [4+2] used Danishefsky's diene, but I had problems manipulating double bonds after the cycloaddition. Using an isonitrile is interesting., but we'd need to be careful about epimerizing the stereocenter.

I'm new here. Are we allowed to use "Answer" to communicate ideas without providing a total synthesis?

@user55119 You probably shouldn't do that via an answer. I suggest adding a comment to the question for that.


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Is elastomeric better with more polymer, or with added calcium carbonate?
While comparing elastomeric roof coatings, I notice that one brand is 60% polymer, while the the other is 40% polymer along with 30% calcium carbonate.  Polymer is flexible.  Calcium carbonate is brittle.  With that consideration it would seem more polymer is better, since weather conditions challenge the membrane integrity for water resistance.  However, polymer degrades far more than calcium carbonate.  Perhaps inclusion of calcium carbonate provides some protection against degradation.  So is it better to have more polymer, or some of that replaced with calcium?  (Since polymer is more expensive than calcium carbonate, my instinct suggests there would be no reason to maximize polymer at its higher cost, if it did not improve performance.)

Stuff like calcium carbonate is generally used as a cost cutter, as it is far less expensive than even the cheapest elastomer (polymer). It allows the producer of the formulation to make a profit/make more profit, essentially.
Generally speaking, adding calcium carbonate to an elastomeric formulation will decrease the useful properties of said formulation, including its resistance to weathering. It's resistance to wear and tear (as represented by mechanical properties like tensile strength and abrasion resistance) will also suffer. Calcium carbonate containing formulations, all other things being equal, will also be less elastic and somewhat stiffer.
A polymer that has poor resistance to weathering (like SBR rubber) will generally also not benefit from calcium carbonate as a formulation ingredient. Instead, special anti-oxydant/anti-ozonants will improve this cheap polymer's fight against the elements.
But if a composition does contain significant amounts of calcium carbonate it is not automatically disqualified as suitable: that will largely depend on the end-user's expectations. In other words: cheap can work well too.


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Pka and buffers
I'm currently studying pKa in my chemistry class, and the mathematical definition $(-log(ka))$ doesn't really explain it tangibly to me. Can you please try to simplify what pKa means and how is it useful when studying buffers.

The notation $pSomething$ is somewhat old and probably dates back to the use of logarithm tables.
The $p$ notation simply means "Take the negative logarithm of what follows the p".
Example: let the acid constant $K_a$ of some (weak) acid be $1\times 10^{-5}$, then the $pK_a$ is:
$$pK_a=-\log  K_a=-\log \big(1\times 10^{-5}\big)=5$$
$5$ is of course a more manageable number than $1\times 10^{-5}$.
That is really all there is to it. We can take a $p$ value of anything, like a solubility product, or some equilibrium constant $K$. It is particularly helpful when the $Something$ is really small ($<<1$).
So taking the $p$ value is essentially a way of converting unwieldingly small numbers into something much easier to deal with. Buffers deal with weak acids and bases, so you'll find $p$ values frequently used in that field.
Of course you can convert a $pK$ value back to its $K$ value:
$$K=10^{-pK}$$

Well this is not surprising since chemistry/physics always do this kind of things, create a whole new something just to simplify calculations. Thank you though.

Yes. Although a bit old, $pK$ values still have their uses. I still use them a lot. Upvote? :-)

OP doesn't have enough rep. to upvote you, but can accept your answer.


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Change in enthalpy and entropy for the formation of nickel tetracarbonyl
I was working on an enthalpy/entropy problem and I had a question on predicting the signs of some reactions. For the reaction
$$\ce{Ni (s) + 4 CO (g) -> Ni(CO)4 (g)}$$
my textbooks says that when we predict the sign change in enthalpy and entropy, enthalpy is > 0 and entropy < 0. I'm not sure why this is since bonds are forming in this reaction and energy is being released (negative sign in enthalpy) and when a solid and gas react to form gas, it increases disorder (positive sign in entropy).
Am I understanding something wrong? Why is enthalpy > 0 and entropy < 0?

Wouldn't entropy be positive if the product is gaseous?

Not sure about the enthalpy part, but the entropy is easily explained once you remember that you are transforming four moles of gas into one.

What is the exact wording in your textbook?