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What is responsible for black tarnish on silver?
Silver on being exposed to air, forms a black film. Now my question is whether the film is silver sulphide (due to reaction with hydrogen sulphide in air) or silver oxide.
Both compounds are black.
In my view it would be silver oxide because oxygen is more abundant and, intuitively, silver seems react to oxygen faster than hydrogen sulphide- oxygen is more reactive than hydrogen sulphide.
But a little google search loosely hints it is silver sulphide. Any suggestions on this too?

I really like this question. I ran a cursory Google search too, and apparently the tarnishing is due to the trace quantities of hydrogen sulfide in the air (silver oxide constitutes a really tiny fraction of the black film). One way we could confirm this is to scrape off some of the film (you'll need a decent amount of it for this to work) and treat it with an easily available mineral acid (like HCl). If it is (largely) silver sulfide, then the reaction that ensues will liberate H2S gas which has a characteristic rotten-egg smell. [I don't have any tarnished silver-ware :( ]

It will help me if you list some sites you encountered which are in support of H2S @paracetamol. And isn't this counter intuitive, that it forms silver sulphide and lesser silver oxide? inspite of reactivity and abundance?

Wikipedia (under the "Properties section") https://en.wikipedia.org/wiki/Silver_sulfide and this: http://www.silvermagpies.com/2011/10/19/silver-tarnish/ and this http://scifun.chem.wisc.edu/HomeExpts/tarnish.html ...none very informative though :/

From above links, I went furthur to study about tarnishing, there is a link in H2S page in same section. It says tarnishing happens not only due to oxygen but at times due to sulphide. When and why, i can't figure out. https://en.wikipedia.org/wiki/Tarnish

Some excepts from https://theassayoffice.co.uk/tarnish_labexpert  "The tarnishing agent is most likely to be free sulphur or a volatile sulphur-liberating compound" and "Moisture may also be a vital ingredient. Paper/paperboard containing 0.5 or 2.5 per cent sodium sulphide tarnishes silver only in the presence of moisture. Similarly, various types of paper containing sulphur compounds which, on alkaline extraction and acidification, liberate hydrogen sulphide do not blacken silver unless they are moist."  It also says 0.006 per cent of free and dry sulphur causes tarnish stains. conclusions?

There is no such thing as "being reactive". Say, you have A and B; some compounds tend to react with A and not with B, and some would react with B and not with A. Which is more reactive? You can't tell; you can't even ask this question, for it's meaningless.
Black tarnish on silver is mostly silver sulphide, as confirmed experimentally. I'd even say that silver does not form a black film on being exposed to air, unless there is some source of hydrogen sulphide nearby (raw eggs, sewage pools, or something else).
It is just that $\ce{Ag}$ tends to prefer $\ce{S}$ over $\ce{O}$. Many other heavier metals are like that, and many lighter metals have different preferences. This is explained (kind of) by the hard/soft acids/bases concept.
So it goes.

I don't think it is necessary to have a hydrogen sulfide source. Silver gets tarnished pretty much everywhere.


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Getting rid of the stench
I recently cleaned my dogs kennel. My girlfriend has been putting her puppy in with my older dog.  The kennel was thoroughly dirty.  Much urine and fecal matter on the concrete. I used a solution of MR. Clean antibacterial and white vinegar to clean and try and deodorize the concrete.  Now there is a tremendously strong rotten egg smell in my garage. Any ideas how to get rid of this stench?

Bleach. The rotten eggs smell is hydrogen sulfide which will react to gjve non-smelly product with bleach.

A house air ozonizer might help as well.

So long as you don't mind some permanently discolored concrete, @Waylander , is right on target. Don't bother playing around with deoderizers and such and just destroy the scent-causing stuff at the source with copious amounts of bleach. Of course you'll want to dilute it a bit and be careful of splattering onto skin and eyes.

Not a good idea to put vinegar on concrete. The vinegar etches the concrete and makes the surface more porous.

It's probably a sulfide that's causing the stench.  If you have no luck with bleach, try hydrogen peroxide (3% to 6% is fine), which will oxidise the sulfur, and the products should be significantly less volatile.  It's the same thing that you'd do for a skunk spray.

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What compound is formed in this case?
I read from a source that the aqueous solution of ferrous sulfate ($\ce{FeSO4}$) gives an unstable brown compound on treating with $\ce{NaNO2}$ and dilute $\ce{HCl}$. Well, it is quite understandable that it is the $\ce{HNO2}$ formed in situ that further causes the reaction. But what is the reaction and what is that unstable compound formed?

Possible duplicate of What is the Structure of FeSO₄ • NO?

@Mithoron, it is not, cause I didn't know what it was.

Iron(II) reduces nitrite to nitrogen monoxide which forms a brown pentaaquanitrosyliron(II) complex with the iron(II) ions.
$$\ce{NO2^{-} + Fe^{2+} + 2H+ -> NO + Fe^{3+} + H2O}$$
$$\ce{[Fe(H2O)6]^{2+} + NO -> [Fe(H2O)5NO]^{2+} + H2O}$$
Iron(II) can also reduce nitrate to nitrogen monoxide. This is used in the brown ring test to determine the presence of nitrate in aqueous solutions.

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Carbonic acid in water
If we add a weak acid such as carbonic acid in water it weakly dissociates to form bicarbonate and hydrogen ions . Now water already has some hydrogen ions, having a ph=7.
So now indirectly we added hydrogen ions and base (bicarbonate ) ions to water which already has a hydrogen ion concentration.  The resulting solution is acidic . This means that in the addition of equimolar hydrogen ions and bicarbonate ions( base) hydrogen ions dominate.
The only reason which I can think of is this: lets say there were 1000 hydrogen ions in water, now addition of 100 hydrogen ions and 100 bicarbonate ions will cause a net increase in hydrogen ion concentration because bicarbonate wont be able to counteract the increase in hydrogen ions because bicarbonate, when acting as a base, will be in equilibrium with carbonic acid. I.e. 100 bicarbonate ions won be able to neutralize 100 hydrogen ions present in the solution, and because of their equilibrium they will ionize only say 50 and overall, and there will be an increase of 50 hydrogen ions.
Is my explanation correct?

To start off, here is the equilibrium we are talking about so we can refer to it later:
$$\ce{H2CO3 <--> H+ + HCO3-}$$
Since the pH of the solution is $\mathrm{-log[H+]}$, the pH of pure water will decrease upon the addition of carbonic acid.
Where I think you are getting confused is regarding the action of the bicarbonate that is produced. As this is an equilibrium reaction, bicarbonate will indeed act as a base, which is the reaction from right to left above. But lets say a molecule of bicarbonate reacts with hydrogen ion to reform carbonic acid. That just means another molecule of carbonic acid will dissociate to take it's place. This is all just a way of saying that the reaction shown above is an equilibrium, such that both the left to right and right to left reactions are occurring. But in the end there is never any net hydroxide produced or hydrogen ions consumed by this reaction, as there is always plenty carbonic acid available to counteract the basic action of the bicarbonate.
I hope I understood your question correctly. Feel free to use the comments below to ask for any clarifications.

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How does one properly do this alkali melt? (Pfleger's indigo synthesis)
TL;DR: Needing advice on the best procedure (e.g., preparation of reagents, and the order of addition) for an alkali melt/fusion used in Pfleger's indigo synthesis.
I need some advice on how to carry out the alkali fusion/melt step in Pfleger's industrial synthesis of indigo. SciFinder comes up with nothing but vaguely-worded patents and two horrible documents.
I would love to find a recently-written procedure on how it is made industrially (e.g., indigo for jeans, or however it would be made when ordering it from Sigma-Aldrich), but I can't find anything like that. All original documents (e.g., Pfleger's patent) lack important details.
This scheme (#3) outlines the synthesis that is used industrially; a melt of NaOH, KOH, and NaNH2 forms indoxyl out of phenylglycine, which then spontaneously (in the presence of oxygen) oxidatively couples using a radical mechanism to form indigo. The part that I'm having trouble with is finding a good procedure on how to do the melt.
I've seen arrow-pushing mechanisms that show NaNH2 being used for either deprotonation or for imine formation (depending on the literature source), while another source says that it's merely a dehydrating agent, Le Châteliering the reaction forward.
Since I don't want to waste my NaNH2 by having it react with water, I figured that I should dehydrate my NaOH and KOH first. So far, I have ground NaOH and KOH together and brought the temperature up to about 200 °C, at which point they become a liquid. Putting this on a vacuum, I see a fair amount of water come off, and when I remove the vacuum, the mix solidifies (but is still soft, probably because I haven't removed all of the water).
I'm assuming that I would then add the sodium/potassium salt of phenylglycine, then the NaNH2. In the Nigerian papers, they simply grind 5:5:1 NaOH, KOH, and NaNH2 (seemingly in open air), then add nitrobenzene and reflux. I'm leery about grinding NaNH2 in open air (safety guidelines suggest grinding it under a hydrocarbon solvent, though perhaps I'm being overly cautious?), and I'd expect that the NaNH2 would react with the water in the NaOH and KOH.
Assuming that nobody has a good literature procedure for this, I'm hoping to get some guidance on things like preparation of reagents, and the order of addition. I'm also interested in any other advice that anyone may have.

On academia, I found a reading of Industrial Dyes: Chemistry, Properties, Applications by Wiley (2003) that on first view sheds some light on the processes and chemistry.  Looks like an interesting secondary reference, rather than a Organic Synthesis.  I took some screen photos from there

and

and

however, the concluding bibliography may turn out sour

as many of these references are likely only in German.
Perhaps one may trace either in these, or more recent work (than 1890) citing them provide a more detailed procedure.
Addition: On archive, I found a "medium old" book about dye chemistry by Fierz, initially published in 1949, that comes pretty close to your intended alkaline melt; except omitting sodium amide:

Perhaps other old books dedicated to this topic equally tell more about these secrets, than modern textbooks?

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How to calculate the quantum density of states?
I've been thinking recently about Fermi's Golden Rule. Sparing the details of how to get Fermi's golden rule (you'll find like five derivations if you google it), the final result is,
$$\frac{\partial P}{\partial t}=\frac{2\pi}{\hbar}|\langle f|H'|i\rangle|^2\rho(E_f)$$which says that the probability per unit time, to first order in the perturbation from $H_1$, is a constant. Hence, this quantity tells you the transition rate. Notice, however, that it depends on the density of final states. Notice, if I'm not mistaken, that this is not for a specific state, but this is more accurately the density of all states above the original states (let's say we're in the ground state and ignore stimulated emission). I understand how to calculate the density of states for a classical, and hence continuous system. It's just an integral in phase-space.
Things become more complicated in a molecule, however, because the states are quantized. So, how do I go about doing this calculation of the density of discrete states? A really great answer would sketch out how one might do this calculation for a quantum harmonic oscillator.

I've seen a couple things online which do this kind of calculation for the motion of a particle in an infinite potential well on a solid. You basically just use the relationship between energy and the wavenumber, which is easy because you know the solutions to the Schrodinger equation for all states ahead of time.

If you really do want to calculate the density of states then you can try direct counting to get the total number of states N and then calculate the density of states $\rho =[N(E+dE)-N(E)]/dE$. This, of course, will only work if you have a few vibrational normal modes as all the combinations of quanta have to be included, e.g. in water with 3 modes, quanta (in the order of increasing energy) are: $(0,0,0), (0,0,1),(0,0,2),(0,1,0), (1,0,0),(0,0,3),(0,1,1)$, etc.  But for a molecule with just a few more atoms, it is quite probable you will get $10^6$ vibrational states /wavenumber at an excess energy of, say,  $5000 \pu{~cm^{-1}}$.
In the field of unimolecular reactions (RRKM models etc.) there has been much effort in calculating total and density of states. There is the Beyer-Swinehart algorithm; Beyer, T., & Swinehart, D. F. 1973, Commun. Assoc. Comput. Machinery, 16, 379 and Stein S & Rabinovitch B. J. Chem. Phys. v58, 2438, 1973. Whitten and Rabinovitch have also modified the classical sum of states to give good agreement with the direct counting (quantum) method. (J. Chem. Phys. 41, 1883, 1964, and 48, 1427 1968). Their modification $^{[1]}$ gives the total sum of states as:
$$ N(E)=\frac{(E+\alpha E_{zpe})^s}{s!\prod_{i=1}^s h\nu_i}$$
where s is the number of (harmonic) oscillators, $E_{zpe}$ is the zero point energy for the i th oscillator,
$$\alpha = 1-\beta w(E')$$
$$\beta = \frac{s-1}{s}\frac{<\nu^2>}{<\nu>^2}$$
and w depends on the range of $E'=E/E_{zpe}$. When $0.1 \lt E' \lt 1; w(E')=(5E'+2.73\sqrt(E')+3.51)^{-1}$, and when $1\lt E' \lt 8; w(E')=\exp(-2.4191(E')^{1/4})$.
The equation in N can be differentiated to find the density of states.
If, however, you want to calculate a rate constant using the Fermi Golden rule it is better not to start with the density of states separated out, but to calculate the Franck-Condon weighted density of states.
The rate constant has the Golden Rule form
$$ k= \frac{2\pi}{\hbar}|<av''|H|bv'>|^2\delta(E_{av'}-E_{bv''})$$
and then this is modified depending on what actual situation is to be analysed, e.g. thermally averaged over the Boltzmann distribution if the molecules are in solution.
In this case the best place to start is probably with the papers
(a)  Heller, Freed & Gelbart, J. Chem. Phys. 56, 2309 (1972),
(b)  Bixon & Jortner, J. Chem. Phys. 48, 715-726, 1968 and
(c) S. H. Lin; J. Chem. Phys. 58, 5760 (1973).
Each deal with non-adiabatic interactions causing radiationless transitions in molecules, i.e. from one vibronic level into a continuum of states, such as singlet to triplet intersystem crossing. There is a summary of different methods in Eyring, Lin & Lin 'Basic Chemical Kinetics' ch 7.
$^{[1]}$ equations given in Steinfeld, Francisco & Hase, 'Chemical Kinetics & Dynamics'. 2nd ed. p333.

Thank you for that very useful and clear answer! I'll get reading!


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Does anyone have information about Piquet and Calitet (chemists)?
I was given a set of names of chemists to research and one of them was actually a duo that I understand to have been French chemists (at least their last names suggest it). I have queried and queried not simply Google, but also academic databases (Wiley, ProQuest, Elsevier, etc.), yet I've found nothing.
I believe I am spelling the names correctly, but for clarity, phonetic expression of their surnames is:

"PEA-ket and CAL-ee-TET"

Hath anyone information on these names within chemistry (or similar natural science discipline)? It would allow me to end a nearly two month search – one way or another.
Most appreciative for any corrections if I have erred in spelling, nation of origin, etc.

Maybe it is not "Piquet" but "Picquet" (sounds for an English speaker very similar) as here http://www.icmub.fr/185-membres?r=185&action=view&id=17 -- organic chemistry?  or even Picqué / Picque like https://fr.linkedin.com/in/eric-picque-20012342?  Similar on Calité. In which context the names were given to you, may  some information be added (time / location / field of chemistry/history of science)?  Needn't be French, nor living / working in a French speaking country (Canada, Belgium, Switzerland, non-European countries, etc.), either (moving,  maiden / family name).

Can you exclude spelling errors? If not, there is

Amé Pictet, known from the Pictet-Spengler cyclization.
Claude Piguet, an inorganic chemist at the University of Geneva.

Both were/are is Swiss, not French.
However, the second name does not ring a bell for me.


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Addition of oxygen molecule to given aromatic compound
I did figure out that the hydrogen atom is reactive as it has three electron donating groups attached to it, I know it reacts. But the answer given is that 6 different compounds can be formed (minor and major; including stereoisomers). But I need some help figuring them out.

see https://en.wikipedia.org/wiki/Cumene_process for mechanism. I did not look at it too closely, but my intuition is that one of the three aryl groups can migrate, and each product has two enantiomers, so $3\times 2 = 6$ products in total.

Ahhh, I see what you mean. Yep, that's most probably the solution. I think i understand now. Thank you so much, i appreciate your help. But just to add something to the question, what would be the major product? does it have anything to do with the groups attached to the benzene rings and their +R effect? @orthocresol

The three hydrogen bond at -OCH3 would be reactive and
-CH3 at attached directly to ring would react so 6 products possible.

Well, what is the product?


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Are amines amphoteric?
An amine can react with an acid just like any other base to form a salt:
$$\ce{RNH2 + HCl-> RNH3+ Cl-}$$
According to my textbook (unreliable), amines also react with bases just like any other acid:
$$\ce{RNH2 + NaOH -> RNH- Na+}$$
Do amines (if not all amines, at least the ones substituted with a strong withdrawing group) react with bases as shown in the above reaction?
In  other words: Are amines amphoteric?

One easy way to verify the second reaction is to check if SA + SB -> WA + WB condition is satisfied. Here in the second reaction one of the products is $\ce{H2O}$ which is more acidic than $\ce{RNH2}$.  So the forward reaction isn't feasible.

This makes for a good read: https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/chapt16.htm

Their acidity is quite low and only very strong bases, like $\ce{CH3-}$ or $\ce{N^{3-}}$ can deprotonate amines.

Certainly $\ce{NaOH}$ isnt strong enough to deprotonate it, so that doesnr happen.

@NilayGhosh I believe that's incorrect. Zwitterions hve two functional groups, one acidic and one basic, and have both positive and negative charges at the same time within the ion. An amine cannot be a zwitterion.

Amines do not react with hydroxide to give amide ions - not significantly at least.
The typical $\mathrm pK_\mathrm a$ of an amine is $\sim 30$ (see Evans pKa table) and that of $\ce{H2O}$ is $\sim 15$. The equilibrium constant for the second equation you wrote is therefore on the order of $10^{-15}$.
In order to generate any amide ions you need a much stronger base than simply hydroxide. A typical choice is an alkyllithium (LDA is commonly made by reacting diisopropylamine with BuLi).
Back to the question whether amines are amphoteric: If your definition of an amphoteric compound is simply "can react either as an acid or a base", then yes, it technically is, since it reacts both with acids ($\ce{HCl}$) or bases ($\ce{RLi}$).
However, it may be useful to draw the line at reacting with aqueous acid and aqueous base. Otherwise, nearly everything with a hydrogen can be considered to be amphoteric (for example, $\ce{HF}$ can technically act as a base to form $\ce{H2F+}$). Under this definition amines are not amphoteric.


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Could you provide me a basic method or formula to solve this question in Physical Chemistry?
The series limit for the Paschen series of hydrogen spectrum occurs at 8205.8 $\dot{A}$
Calculate:

Ionization energy of hydrogen atom.
Wave length of the photon that would remove the electron in the ground state of the hydrogen atom.

The solution for this is:
1. 13.6 eV
2. 916  $\dot{A}$

Both questions are related. Use the following expresion:
${\displaystyle E_{2}-E_{1}={\frac {m_{e}e^{4}}{2(4\pi \epsilon _{0})^{2}\hbar ^{2}}}({\frac {1}{n^{2}_1}-\frac {1}{n^{2}_2})}}$.
For the energy of an 1s hydrogen electron then,calculate the difference between states 2 and 1 (where 2 will represent the ion and 1 the ground state).
The symbols are: $m_{e}$, the mass of the electron, $e$ is electron charge, $h$ is Planck constant and $\epsilon$ is the electric constant of vacuum.
Now, if n$_2$=$\infty$ represents a free electron (or the ion H$^+$) as it feels no atraction to the nucleus (although distance is not in the expression), and if n$_1$=$1$ represents the ground state, then replacing in the formula:
${\displaystyle E_{2}-E_{1}=E(ionization)={\frac {m_{e}e^{4}}{2(4\pi \epsilon _{0})^{2}\hbar ^{2}}}{\frac {1}{1^{2}}}}$,
Where i replace n=1 for ground state and n=$\infty$ for the removed electron.
Notice that this energy is positive, as expected.
Then, E(ionization)=h$\nu$=h$\frac{c}{\lambda}$, from which you get the wavelenght.
If you want $\lambda$ in meters, check the units of E(ionization) be in MKS system.
Hope it helps.


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What is the reason behind a good smell?
I know that volatile substance creates smell but how a good smell produce? Is there any common characteristics of the compounds having good smell?

I'm voting to close this question as off-topic because a "good smell" here is merely human perception and is not a concrete physical property of a substance. This is is better suited on the Bio.SE.

Hey there! That's a really interesting question you've put and I (personally), would love to see an answer to this. The issue, however, is that "good smell" is very crudely defined in Chemistry and that a decent answer to your question will 1) Delve into the evolutionary aspects of the development of human olfactory receptors which leads us to classify odors as "good" or "bad"; and 2) It will be uncomfortably long...and super long answers aren't what we're trying to generate on this site. (Also, I get get this feeling that this post will get get closed at the Bio.SE too...sorry)

See this: https://chemistry.stackexchange.com/questions/33823/what-causes-the-old-book-smell

There is a very wide variety of chemicals that have a good smell - and the flavor and fragrance industry has discovered and profited from thousands of them. To have a good smell, a compound first of all has to be low molecular weight and able to easily evaporate into air so it can be carried to your nose--i.e. a "volatile organic compound". Esters are a good class to look at for starters, and are responsible for many of the fruity odors. For example, amyl acetate is a simple compound, easily made in the lab, that smells like bananas. Another important class is terpenes: hydrocarbons made from 5-carbon units that include pine and cedar odors. Some odor components like those of chocolate are more complex, have many components, and are hard to reproduce exactly.
One way to think of good (and bad) smells is how over time animals and humans have evolved to appreciate the smells of some things (like fruit or cooking onions) as good and other things like rotting flesh as bad, since the former is good for you and the latter definitely bad. If you got it backwards, you would not survive for long. But flowers, for example, smell good but do not materially benefit us. Maybe we have some residual bee genes in us.


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Equilibrium and Free Energy

$$\ce{CO2(g) + Ca^2+(aq) + H2O(aq) <=> CaCO3(s) + 2H^+(aq)}$$
Find the equilibrium constant using a free energy table.

Finding the free energy and plugging it into the equation $\mathrm{G = -RT \ln (\mathit{K})}$ is easy, but I am confused about what $K$ it is. Is it $K_c$ or $K_p$, or does it even matter?
This is important because I am supposed to do equilibrium calculations with it later. I guess I have to use $K_c$, because only the concentrations of the aqueous solutions are given, plus $\ce{CO2}$ is in ppm. I think $\mathrm{G = -RT \ln(\mathit{K})}$ gives me $K_p$, so do I have to convert that into $K_c$ afterwards if I want to do equilibrium calculations?

I think it should be $K_c $ as the K in  the above equation you mentioned represents equilibrium constant for the reaction. Original equation has reaction quotient (Q) and at equillibrium it is just K.

K is properly defined as:
$$K = \dfrac{(a_C)^{c}(a_D)^{d}}{(a_A)^{a}(a_B)^{b}}$$
Activities involve relation of concentrations or parcial presures so K is dimensionless.
Let's find activities.
Ideal gases
$$a_{gas} = \dfrac{P}{P^o} $$  $P^º$ is 1 atm.
Solute in very dilute solutions
$$a_{solute} = \dfrac{C}{C^º}$$ $C^o$ is 1 M
Activities for solvents and pure solids are considered unity. $\ce{H2O}$ and $\ce{CaCO3}$ have unitary activity.
Answer
$$K = \dfrac{\ce{(a_{CaCO{_3}})(a_{H{^+}})^{2}}}{{\ce{(a_{H{_2}O})(a_{CO{_2}{_g}})(a_{Ca{^{2+}}{_{aq}}})}}}$$
$a_{CaCO{_3}}, a_{H{_2}O}$ are unity. It rests to use ideal gas equation and get $C_{CO2}$. $a_{Ca{^{2+}}{_{aq}}}$ and $a_{H{^+}}$ are usual concentrations.
EDIT
The way $K_p$ is converted to $K_c$ is:
$$K_p = K_c(RT)^{\Delta{n}_{gas}}$$
R: gas constant (0.082 L atm mol$^{-1}$ K$^{-1}$). T: temperature in Kelvin (K) at which the equilibrium reaction takes place.
To find $\Delta n_{ gas}$ , simply subtract the sum of the coefficients of the reactants from the sum of the coefficients of the products.
You can deduce that using $P=\frac{nRT}{V}$, $\frac{n}{V}$ is concentration.

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Monoepoxidation of 1,2-dimethylcyclohexa-1,4-diene using m-CPBA

What is the product formed when 1,2-dimethylcyclohexa-1,4-diene undergoes epoxidation with one equivalent of m-CPBA (meta-chloroperbenzoic acid)?

In the epoxidation mechanism, the π-electrons of the double bond attack m-CPBA:

So, methyl groups attached to the double bond should increase its nucleophilicity. On the other hand, since it is a nucleophilic attack, the methyl groups should also cause considerable steric hindrance. So what should the answer be?

The interaction of the alkene with m-CPBA is primarily nucleophilic/electrophilic in nature, i.e. the π MO of the alkene attacks the O–O σ* orbital. Therefore, the more electron-rich alkene (i.e. more highly substituted alkene) preferentially reacts.
With methyl groups there probably isn't too much steric hindrance. m-CPBA approaches above/below the alkene and doesn't directly attack over the methyl groups.
Indeed, the monoepoxidation of 1,2-dimethylcyclohexa-1,4-diene is reported to proceed selectively at the more electron-rich tetrasubstituted double bond (J. Org. Chem. 2006, 71 (4), 1653–1657):

How tolerant is this against aromatisation? You have a 1,4-cyclohexadiene, you look at it in the wrong way and it gives you a benzene ring.

@Martin-マーチン there's one record of this substrate being oxidised, with DDQ instead of mCPBA. I suppose it  may form 1,2-dimethylbenzene if the epoxide is opened  - form the diol then dehydrate


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What is the link between hydrogen production and potassium oxide?
Is there an industrial process that produces hydrogen with potassium oxide as a catalyst?
Or is there a process for the production of potassium oxide that has hydrogen as a by-product?

The Wikipedia page of potassium oxide states:

Potassium hydroxide cannot be further dehydrated to the oxide but it
can react with molten potassium to produce it, releasing hydrogen as a
byproduct.
$$\ce{2KOH + 2K <=> 2K2O + H2 ^}$$

Would you have a mixed alkali metal oxide if the following takes place: $$\ce{2Na + 2KOH ->[\Delta] 2NaOK + H2}$$

@Kav Maybe


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Producing methane, reactants and products
How do I balance this reaction when producing methane using the reactants:
$$\ce{CH3COONa(s) + NaOH(s) + Ca(OH)2 -> ?}$$
The products are partly
$\ce{CH4(g) + Na2CO3(s)}$
What else is missing in the products? Furthermore, how do I balance this?
Thanks in advance.

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

The answer is:
$$\ce{2CH3COONa(s) + 2NaOH(s) + Ca(OH)2(s) <=> 2CH4(g) + Na2CO3(s) + CaCO3(s) + H2O}$$
You have to get same number of equal atoms in each side of the reaction (this technique is called mass balance). If there were charges, they must be balanced also.
Example
There are $2\times 3 + 2\times 1 + 2\,$ hydrogen atoms in left side, and there are $2\times 4 + 2$ $\ce{H}$ in the right side. So both sides give $10$ hydrogen atoms.
I hope it helps.


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CO₂ molecule π bonds
Why does $\ce{CO2}$ make two π bonds? Why can't it make four σ bonds? Also, why do the π bonds not involve for repulsion?

"not involve for repulsion" What is that even supposed to mean?

that means,   according to the vsepr theory sigma bonds and lone electrons are the main reason for detect the shape of the molecule. Then that should be only sigma bonds and lone electrons are only the reason for repeal. Why pi bonds are not involving for that?

Recall the definition of σ-bonds. They are caused by the head-on overlapping of the orbitals.

$s$ orbitals always, and only form σ-bonds while the $p$ orbitals in the direction of bond axis (which is $p_x$), also form σ-bonds.
The other two $p$ bonds cannot form σ-bonds as they are perpendicular to the bond axis, and cannot overlap head-on. In such cases, π-bonds are formed, by the sideways overlapping of the orbitals.

The $p$ bonds not perpendicular to the bond axis involve in π-bond formation ($p_y$ and $p_z$).

@OsalSealaka Please dont use comments to say thanks. If this answer satisfies you, click the tick button. That lets everyone know you got an answer.

Actually σ bonds are formed when two hybridised orbital of two adjacent atoms overlap each other.
So considering an atom, there can be one orbital along one axis that would be overlapping with another while the other orbital may find a parallel orbital of adjacent atoms and for π bond so same is the case of $\ce{CO2}$.


text:
Why would rinsing the flask with acid in this experiment affect its accuracy?
The following question is #21 from this test form (PDF link):

The apparatus in Figure 4 was set up to measure the time taken for $\pu{20.0 cm3}$ of sodium thiosulfate solution to react with $\pu{5.0 cm3}$ of hydrochloric acid in a $\pu{100 cm3}$ conical flask at $\pu{20 °C}$. The timer was started when the sodium thiosulfate solution was added to the acid in the flask. The timer was stopped when it was no longer possible to see the cross on the paper.

What is likely to decrease the accuracy of the experiment? [1 mark]
A) Rinsing the flask with acid before each new experiment.
B) Stirring the solution throughout each experiment.
C) Using the same piece of paper for each experiment.
D) Using different measuring cylinders to measure the volumes of acid and sodium thiosulfate

The answer given as correct is (A). Why is rinsing the conical flask with acid likely to decrease the accuracy of the experiment? I would imagine that stirring the solution, option (B), introduces error since the rate of stirring is difficult to control.

Probably because the use of hydrochloric acid (which is a reactant involved in this reaction!) could result in some residual aqueous HCl in the flask, which may add to the HCl added later.
Anything you use to clean the flask should be inert/not involved in that reaction. If you really want to use HCl, then you should later wash the flask with a lot of water, but that's not said in the answer.
Also note that the volumes involved here are not so large (only 5 ml of HCl are used), so a "small" residue might have a large effect on the result.
As for the stirring, the colour of the reaction is caused by the sulfur precipitate. But I think that, if you stir in each experiment, the times might probably decrease, but you would still observe a trend with temperatures.
That answer would be wrong if it involved stirring in some experiments and not in others.


text:
Most nucleophilic oxygen base
What is the most nucleophilic known monoanionic oxygen base? I am thinking hydroperoxide ion or perhaps an alkylperoxy anion. But perhaps there is something I am missing.


text:
A Question on the Role of Triplet Oxygen in the Formation of Artemisinin
It is observed that whereas singlet oxygen is required for an ene reaction with an alkene to give an allyic hydroperoxide, triplet oxygen will react similarly with an enol to give an alpha-hydroperoxy aldehyde. (Example: see Chen HJ et al., Tetrahedron 69 (2013), 1112-1114.)
The drawing below shows two steps in the final sequence of reactions leading to the antimalaria drug artemisinin, as described in the cited article. The enol is formed by opening of the hemiacetal, and before it can tautomerize to the aldehyde, it reacts with atmospheric oxygen (in the dark, so presumably triplet oxygen) to give the alpha-hydroperoxy aldehyde shown. That in turn spontaneously folds up to give artemisinin (not shown). A concerted reaction of the enol with triplet oxygen, if it is a variant of the ene reaction, would be spin-forbidden.
Can anyone explain why this otherwise "forbidden" reaction works?

I offer the following thoughts on the paper you cite regarding the formation of artemisinin 9 from carboxylic acid 1. You are correct that the allylic hydroperoxide 2 is formed by an ene reaction with singlet oxygen (1O2). The stereochemistry of the hydroperoxide group is the result of the ene reaction occurring on the convex face of the unsaturated cis-decalin 1. The crude hydroperoxide 2 was exposed to an atmosphere of triplet oxygen (3O2) in a vigorously stirred mixture of petroleum ether and catalytic trifluoroacetic acid. The initial steps are 2 --> 6 via radicals 3, 4 and 5. Triplet oxygen forms hydroperoxide radical 3 which undergoes cyclization to 1,2-dioxetanyl radical 4. It is likely that radical 4 reacts with triplet oxygen prior to dioxetane fragmentation owing to the stereochemistry of the hydroperoxy radical in 5. The chain reaction is completed by the reaction of radical 5 with hydroperoxide 2. In the transformation 2 --> 3 the radical HOO was formed and can also intercede as a source of a hydrogen atom regenerating oxygen. Finally, acid-catalyzed cyclization of 6 leads to artemisinin (qinghaosu) 9. There is no issue of "forbidden" reactions.


text:
IUPAC Nomenclature of Cumene
Why is cumene referred as isopropyl benzene? And not sec propyl benzene, since the carbon of that chain is the middle one?

isopropyl is an accepted name, and is equivalent with sec. propyl., and "preferred" see e.g. https://chemistry.stackexchange.com/questions/63890/difference-between-isopropyl-and-isobutyl

@mykhal According to current IUPAC recommendations (2013), the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.

@Loong thanks for correction. I wanted to say that isopropyl is preferred over sec-propyl

Let me abstract your question to “Why is the $\style{font-weight:bold}{\ce{-CH(CH3)2}}$ group sometimes named ‘isopropyl’ but almost never ‘sec-propyl’?”
iso alkyls
‘Iso’, obviously as in ‘isomer’, in IUPAC organic nomenclature (last one is from 2013) evolved to non-preferred, retained names for several smaller alkyls, that are linear but specific single branching at the carbon next to the terminal carbon, by additional methyl.
These are (intentionally skipping our dour discussed $\ce{C3}$ branched alkyl):

isobutyl    $\ce{-CH2-CH(CH3)2}$
isopentyl   $\ce{-CH2-CH2-CH(CH3)2}$
(isohexyl    $\ce{-CH2-CH2-CH2-CH(CH3)2}$  , only in old IUPAC 1979 nomenclature, see here)

(However, in general naming of bulk chemical products, larger alkyls with different branching are named isoalkyl, e.g. ‘isooctyl’ $\ce{-CH2-CH(CH2CH3)-CH2-CH2-CH2-CH3}$.)
sec- alkyls
Obviously ‘sec-’ is for 'secondary’, meaning there's a branching into 2 carbons at the first carbon atom. The list of retained names has only one item:

sec-butyl $\ce{-CH(CH3)-CH2-CH3}$

but our candidate would fit in here as well.
Why only sec-butyl? A major reason could be that the next “sec-pentyl” would represent two different structures:

$\ce{-CH(CH3)-CH2-CH2-CH3}$
or $\ce{-CH(CH2CH3)2}$ ?

(Now one might ask, why there is not some naming system for alkyls with methyl branch at the first atom, similar to iso alkyls with the branching at the next-to-last carbon. I don't know.)
Why not sec-propyl? The $\ce{-CH(CH3)2}$ is a special case that can fall into both isoalkyl and sec-alkyl class. The better defined one, also resulting with simpler spoken name and simpler typography (no dash and italic) – isoalkyl, was once chosen. That's my humble hypothesis.
(And please note that the preferred IUPAC name for this group is propan-2-yl. Alternate name is 1-methylethyl. Isopropyl is still retained.)


text:
What types of ultraviolet light will ignite hydrogen?
I saw a reaction between hydrogen and chlorine ignite in the presence of UV light. I was wondering if the same thing would happen with something like near-ultraviolet light (around 400 nm). What kinds of UV light (if any) would start a reaction of stoichiometric amounts of hydrogen and oxygen gas in something like a balloon?

I guess that for a liquid mixture of hydrogen and oxygen read light would suffice (formation of singlet oxygen), whereas in the gaseous phase light with wavelengths < 240nm (formation of oxygen atoms) would be required.

"Read light" = "red light" maybe?

Hydrogen/chlorine needs only visible light to ignite. You can use a normal security light to do this because chlorine can be dissociated by visible light, $\lt 490$ nm which is of a bluish/greenish colour. The reaction is a radical chain reaction initiated by chlorine atoms.
A security light is suitable, they are cheap and bright and can produce a lot of radicals and thus an explosion. If you put stoichiometric amounts into a large plastic lemonade bottle with a rubber bung in the top it makes a good demo. But make sure to use a transparent screen between you and any audience as the bottle can sometimes shatter. The cork usually hits the ceiling and the bang is loud :)
If you want to do the same with hydrogen/oxygen, which is also a radical chain reaction, you will definitely need UV light as the dissociation energy of both is now far larger, oxygen $\lt 244$ nm, hydrogen $\lt 276$ nm. It probably won't work so well in a balloon as the rubber will absorb a lot of UV. (Also, UV light of this wavelength is dangerous to the eyes and skin).


text:
Why is carbon monoxide a "weak" lewis base?
A lewis base is substance that donates an unshared pair of electron.
In CO, a lone pair is present on both carbon and oxygen each. Moreover, lone pair of electron on carbon is localised so CO will act as lewis base.
What causes the CO to become weak lewis base?
Also tell, what "Lewis acidity of CO" would mean? Actually , I read that lewis acidity of CO is very small? But I am not able to understand it.

It isn't weak, just soft base and one of strongest pi acceptors.

$\ce{CO}$ is usually considered a Lewis base, as the lone pair on carbon readily donates it's electrons, for example forming transition metal coordination complexes. Although oxygen is more electronegative than carbon, the carbon is more rich in electron density in the $\ce{CO}$ molecule as it basically has a full formal negative charge. Additionally, products formed by the carbon bonding to Lewis acids are much more favorable that products that would be formed by bonding to the oxygen atom.
It is also possible for CO to act as a Lewis acid however. When acting as a ligand with certain low-valent transition metals, the metal can redistribute electron density back to the $\ce{CO}$ ligand in a process called back-donating. In this case the $\ce{CO}$ molecule accepts electrons via it's antibonding $\mathrm{\pi}$ orbitals, making it act as a Lewis acid.

By question was why it is "weak "base?

I edited my question.

@KaSikh , sorry I misunderstood that the reason for CO being a weak rather than strong Lewis base was the crux of your question. I'm glad you edited that into your question, though maybe it was just confusion on my part ;) I'll try to come up with a good explanation, but the cutoff between strong and weak for Lewis acid/bases is not nearly as well defined as that for Brønsted–Lowry. You need to have some point of reference; the degree of reaction with BF3 is one example I've seen. But there are general trends, mainly corresponding to polarizability, electronegativity, etc. I'll edit my ans

I appreciate it. It would be nice if you also explain lewis acidity of CO as mentioned at last in question.

I hope to get to it soon. Notice that the second paragraph of my answer discusses the Lewis acidity of CO.

Yes. But why lewis acidity of CO is small if CO is weak lewis base?


text:
Mechanism of radical cyclisation to form hydroazulene core

Recent studies on radical-induced cyclisation reactions have led to a simple, one-step method for the preparation of hydroazulenes from appropriately substituted cyclopentanone precursors. Treatment of 1, for example, with triphenyltin hydride and AIBN in refluxing toluene gave 2 in 79% yield. Give a mechanism for this transformation. (TES = triethylsilyl, SiEt3)

(taken from: McKillop, A. Advanced Problems in Organic Reaction Mechanisms; Tetrahedron Organic Chemistry Series, Vol. 16; Pergamon Press: Oxford, U.K., 1997; p 1.)

I know that triphenyltin hydride, Ph3SnH, can act as a reducing agent but I have no idea of its reduction mechanism. And I know that AIBN is a radical initiator:

The given conditions strongly suggest a radical pathway. However, looking at the product I also feel like there is a carbocation intermediate, as there is a ring expansion, but I don't know whether or how that happens.

This reaction is taken from Tetrahedron Lett. 1995, 36 (17), 3015. (DOI:
10.1016/0040-4039(95)00502-4)
The heat (provided by refluxing) is required to initiate the reaction by the decomposition pathway you have drawn. The cyanopropyl radical thus formed abstracts a hydrogen atom from the tin hydride, a consequence of the weak Sn–H bond:

The triphenylstannyl radical is the one that actually does interesting chemistry. Generally, there are a few options for radicals at this stage. One common pathway is to abstract a halide or chalcogenide from the starting material, but there is no such atom here (C–F and C–O bonds are stronger and rarely broken, so the OTES group is out of bounds). The other common pathway is to add to a π-bond. In this case the best place to add is the terminal carbon of the alkyne, which is least sterically hindered.

After this, the remainder is a series of addition-elimination steps. Note that in each case the ring size formed is 3, 5, or 6: these are all kinetically fast cyclisations.

In particular, note that:

Intermediate 4 is stabilised by electron donation from adjacent oxygen (similar to carbocations, radicals - which are also electron-deficient species - can be stabilised by adjacent lone pairs).
You are correct that a ring expansion is needed – in fact, two are needed – but this does not require a Wagner–Meerwein shift in a carbocation. Radicals are perfectly capable of doing this job, usually by a sequence of cyclisation (5 to 6) and retro-cyclisation (6 to 7). See also J. Org. Chem. 2012, 77 (19), 8634–8647 for another example of this motif, which is very common in radical chemistry.
There isn't tin in the final product, so we will eventually have to kick it back out again: this can be achieved via a β-scission mechanism, so we have to work towards an intermediate which has a radical β to SnPh3, i.e. 7.

The original paper supports this mechanism (although they skipped some steps, as is usual in journals):


text:
Difference in the acidity of ethanoic and methanoic acid
Peter Sykes attributes the difference in their acidity to differential solvation.

From what I understand, there is not much difference in how well the ions are solvated, since the Delta H is comparable, however, the change in entropy is much greater for ethanoic acid but I don't understand why?

On contrary it clearly says that difference is caused by solvation but between between formate and acetate

What an interesting way of looking at acidity! How I would look at it would be the more standard way which is to compare the stability of their conjugate bases...
The ethanoate ion has an methyl group, which is electron-donating, donating electron density towards the oxygens, concentrating the negative charge. This makes it a more unstable conjugate base.
In stark contrast, the methanoate ion, does not have the electron donating group and thus, does not have such a high concentration of negative charge. This makes it the more stable conjugate base.
The explanation proposed can also explain the differential solvation of the two carboxylic acids...
With a higher charge density, ethanoic acid would form a more "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is greater.
With a lower charge density, methanoic acid would form a less "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is not as great.
Consequently, the decrease in entropy for the dissociation of ethanoic acid is much greater than that for the dissociation of methanoic acid.
Therefore, the dissociation of ethanoic acid would be less favoured than the dissociation of methanoic acid, based on both considerations of entropy and conjugate base stability, which are actually linked.

But by the same argument, we could say that since the negative charge density is greater on ethanoate ion, it should be solvated to a greater extent, but as seen in the Delta H, they're both solvated to roughly the same extent. Why is the change is entropy so different for both?

Hmmm... Why is there necessarily a relationship between enthalpy change and solvation? In this case, when the water molecules form a more ordered hydration shell, only entropy is concerned. The decrease in solvent-solvent interactions and the increase in solute-solvent interactions is more or less the same. Thus, the enthalpy change would also be more or less equivalent for both. This is just my speculation.

I would think there would be a relation in enthalpy since the greater negative charge density would attract and form more hydrogen bonds with the solvent.


text:
Clarification about the difference of internal energy formula
I am trying to understand the concept of internal energy and it's difference. My textbook said that the difference in internal energy of a system is
$$
\Delta U=Q+W
$$
where $Q$ is the heat absorbed or released by system and $W$ is the work done by system to the surrounding or by the surrounding to the system. After that, it mentioned that the signs of $W$ and $Q$ are determined like this:

The work done by surrounding to system: $(+)$. The work done by system to surrounding: $(-)$. Same on heat.

After that general definitions, it started to explain the internal energy differences in various systems like isochoric system, isobaric system. In the section of isobaric system it says that:

If a system were free that can expand against external pressure, when heat is given to that system, it will give some energy by doing some work. So the internal energy difference is:$$
\Delta U=Q_p-W
$$ where $Q_p$ is the heat given to a isobaric system.

It is completly OK till the new formula, but isn't the new formula bizarre? Because if we give heat to a system and it will do work. As a result the difference would be like:
$$
\Delta U_{example}=Q_{p_{example}}-(-W_{example})
$$
So, does the work done by system increase its internal energy?

Your book has driven you crazy with its stupid sign convention.  If the gas is expanding against constant pressure, then the equation is still $$\Delta U=Q+W$$ but, in this case, the work done on the system by the surroundings is negative (i.e., W is a negative number, because the system is doing work on the surroundings).  So not all the heat added to the system is being used to increase its internal energy.  The increase in internal energy is less than if the gas was not pushing back its surroundings.

I can't upvote it due to low reputation but your answer is explanatory.


text:
What "physical method" does Feynman refer to in The Feynman Lectures on Physics (Vol 1)
In the introductory chapter, Atoms in Motion, of the Feynman Lectures on Physics (Volume One), Feynman alludes to a certain "physical method", which can be used to determine the structure of α-irone (and presumably) many other complex organic compounds.

Excerpt from page 1-8 (Chapter one, page eight):

How does the chemist find what the arrangement is? He mixes bottles full of stuff together, and if it turns red, it tells him that it consists of one hydrogen and two carbons tied on here; if it turns blue, on the other hand, that is not the way it is at all.
This is one of the most fantastic pieces of detective work that has ever been done—organic chemistry.
To discover the arrangement of the atoms in these enormously complicated arrays the chemist looks at what happens when he mixes two different substances together. The physicist could never quite believe that the chemist knew what he was talking about when he described the arrangement of the atoms. For about twenty years it has been possible, in some cases, to look at such molecules (not quite as complicated as this one, but some which contain parts of it) by a physical method, and it has been possible to locate every atom, not by looking at colors, but by measuring where they are. And lo and behold!, the chemists are almost always correct.

Knowing that this lecture was originally delivered in the early 1960s, what is this "physical method" Feynman's referring to and how does it work?

X-ray crystallography I guess.

He actually refers to "x-ray analysis" later in the chapter.

Crystallography had existed long before the 1940s, though. The timing seems suspiciously appropriate for NMR spectroscopy, even if the simplified description does seem to fit X-ray diffraction more.

The first proton NMR with resolved multiplets came in 1955, so he cannot be talking about NMR.

What method does he refer to in the first paragraph ? "He mixes bottles full of stuff together, and if it turns red, it tells him that it consists of one hydrogen and two carbons tied on here; if it turns blue, on the other hand, that is not the way it is at all."

In the mid-20th century, instrumental methods available to organic chemists were nuclear magnetic resonance (NMR), infrared spectroscopy (IR), mass spectrometry (MS) and X-ray diffraction (crystallography).
While it is probably impossible to figure out exactly what Feynman was referring to, it is possible to contemplate what was available (and most valuable as a day-to-day structural method) to an organic chemist in the 1940s-1960s. Also, there is some additional information related to the size of the molecule:

For about twenty years it has been possible, in some cases, to look at such molecules (not quite as complicated as this one, but some which contain parts of it) by a physical method, and it has been possible to locate every atom, not by looking at colors, but by measuring where they are.

The molecule he references is alpha-irone, which contains 14 carbon atoms. In the second part of the sentence, Feynman talks of locating atoms by measuring where they are. This has to be a diffraction technique (today, it could be AFM as well). On the other hand, X-ray crystallography was sufficiently developed to tackle molecules like vitamin B12 (cobalamin, contains 63 carbons), which Dorothy Hodgkin determined and received the Noble prize for in 1964. It is possible that the two parts of the sentence refer to two different techniques.
NMR
The first commercial NMR came out in 1952, a 30 MHz instrument from Varian. This might be a little bit late for the timeline here. It would be difficult to figure out the structure of alpha irone based on proton NMR alone.
IR
A low-cost IR-spectrophotometer came out in 1957. Because these instruments are cheaper than NMR, an organic chemist would be more likely to have one in the lab, monitoring reactions, while the NMR might be in a central facility.
MS
Mass spectrometry is an older technique which started to be applied to natural products in the 1960's. However, alpha-irone probably does not exceed size limitations, and seems sufficiently volatile to be a good mass spec molecule.
X-ray diffraction
Hauptman and Karle developed direct methods in the 1950s, allowing to solve crystal structures to up to about 1000 atoms. It is likely that at least the second part of Feynman's remarks refer to X-ray crystallography as this is the only method among those above that directly locates atomic positions (rather than infering which atom binds to which other atom).
Best guess
It is possible that the first part of the quote refers to NMR or IR (both had severe size-limitations in that era), and the second part of the quote refers to X-ray crystallography.


text:
Hyphenation of chemical names
When a word has to be split broken into multiple lines, it is almost always hyphenated. Chemical names can be considered words, or at least they consist of words. So the hyphenation should apply to them as well. However, hyphens often are integral parts of systematic chemical names. Therefore confusion might arise, whether such a hyphen emerged from hyphenation or if it is "mandatory". Are there any official rules for the (possible) hyphenation of the chemical names? I could not find any, yet. I am sure some journals could have it mentioned in their style manuals, but I am primarily interested whether this topic is mentioned by IUPAC.
In fact, I am undecided whether to represent inserted optional breaks in long chemical names (that have to fit into some fixed-width infobox) HTML(5) representation as  or <wbr>, which however means hyphenation, and non-hyphenation, respectively.

Not to mention the hyphens I add to words to fool my Autocorrect into not "correcting" them.

IUPAC project 2014-003-2-800 (started 2014-02-01; ended 2021-01-11) has been created to address this issue and resulted in a publication [1].
Reference

Dijkstra, A. J.; Hellwich, K.-H.; Hartshorn, R. M.; Reedijk, J.; Szabo, E. End-of-Line Hyphenation of Chemical Names (IUPAC Recommendations 2020). Pure Appl. Chem. 2021, 93 (1), 47–68. DOI: 10.1515/pac-2019-1005. (Open Access)

The IUPAC project has been completed in January 2021, so I took a liberty to update your answer by adding the reference to the released paper. Please feel free to add further details or correct my edit as you see fit.

I'd like to mention an interesting solution to a line-break hyphen replacement, that appears in Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (‘Red Book 2005’), page 215:

Other occurences follow in tables at pages 263–265, 268 (legend); 280 (legend); 282, 287–292, 304, 307, 309, 330. However, no containing section is dedicated to typography. We should not consider it as a recommendation of style.
The symbol  ⁐  exists as a Unicode character U+2050 CLOSE UP (editing mark). (However in the Red Book 2005 PDF, the technical realization is by rotated parentheses ().)
UPDATE:
I have spotted a similar mark, equal sign, in the INN Recommendation List, first apperance in the List 32. One example that features both hyphens and equal sign (tacrolimus definition):

This notation was since then used there for many years, but gradually vanished. Current practice in this publication seems to be to line-break the name only at the points where the hyphen is present, at the cost of early break.
This equal sign usage, or its meaning, seems to be somewhat related with the double hyphen. Which, however, is sometimes used in the opposite sense:

In Merriam-Webster dictionaries if a word is divided at the end of the line, and the division point happens to be a hyphen, it is replaced with a double hyphen to graphically indicate that the divided word is normally hyphenated, for example cross⸗
  country.

UPDATE:
Another occurence of similar notation, that I was able to find, even includes a definition. Handbuch zur Anwendung der Nomenklatur organisch-chemischer Verbindungen, 1979 (German):

Schreibweise chemischer Namen:
  (…) Wegen der besonderen Bedeutung, welche der Bindestrich im chemischen Namen besitzt, wird als Worttrennzeichen am Zeilenende ein „ = “ verwendet, wenn im nicht abgeteilten Namen an dieser Stelle kein Bindestrich steht. Ein Bindestrich am Zeilenende ist Teil des Namens. (…)

Sample example:

This notation seems to be used consistently throughout the book.


text:
Why don't electrons simply stick to the nucleus on account of being *oppositely* charged, instead of moving around it?
Why don't electrons stick to the nucleus on account of being oppositely charged charged (the nucleus being positively charged)? But contrary to my intuition, electrons seem to follow a whole bunch of wacky paths (orbitals) around the nucleus (as demarcated by orbital wave-functions). Why is this?
Do note:
I'm not asking "What keeps the electron in orbit". All I want to know, is why electrons don't simply find themselves attached to the nucleus, and why do they even bother moving around the nucleus (in orbitals) at all? My textbooks, unfortunately, make no reference to this issue...they wholeheartedly adopt the "current" QM model of the (hydrogen) atom.

The cheeky answer to this is "because something keeps the electron in orbit around the nucleus."

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

In a way, electrons do stick to the nucleus to form what we call an atom. There are no paths around.

eh, orbitals, orbits, all that stuff is just a mathematical approach to press quantum mechanics in a canvas our primitive brains can understand

Richard Feynman alludes to this exact problem in the first volume of the Feynman Lectures on Physics. I'll present his argument here (based on the Uncertainty Principle), albeit, in my own words ;)

The German physicist [since he dealt with atomic chemistry/physics...by all means, go ahead and call him a "chemist" ;) ], Werner Karl Heisenberg came up with what we now know as "The Heisenberg Uncertainty Principle"... a revolutionary idea that points to an "inherent fuzziness", that exists within quantum systems (which is a fancy word that refers to any region of space that is sufficiently "small" enough to introduce wave-particle duality), and becomes apparent when we try to "measure" the various parameters that constitute them.
Basically, what (one version of) the Uncertainty Principle states is that:

The product of the uncertainty in measurement of velocity and the uncertainty in measurement of the position of a particle can never be less than a certain constant; i.e- ħ/2

Rephrasing this

If we were to know a particle's position to a very high precision (i.e- small uncertainty in measurement of position) then the corresponding uncertainty in the measurement of the particle's velocity (or momentum, if you know its mass) increases greatly. The same holds true vice versa.

In other words,

You cannot know both a particle's position and momentum in a quantum system to a great precision simultaneously.

Here's a good analogy for this (Part 3: Atomic Structure - Analogy with a ballon)
From this point on, I guess the question can best be dealt with by "proof by contradiction".

First part of the question: Why don't electrons stick to the nucleus?

We know for a fact, that Heisenberg's Uncertainty Principle holds true (yeah, I didn't prove it...but for now, just take my word for it).
So, let's assume that electrons do stick to the nuclei of their respective atoms. In that event, we would know the electrons' positions and momenta (with respect to the nucleus) with almost absolute certainty...something that Heisenberg's Uncertainty Principle does not allow.
Seeing that this situation is impossible, it is obvious that our little "assumption" must be wrong (Knowing the the Uncertainty principle is indeed, correct).
So yeah, electrons don't just go and simply "stick" to atoms.

Second part of the question: Why do electrons follow a whole bunch of wacky paths?

The Dane, Niels Bohr postulated that electrons move in certain, fixed orbits around a nucleus...a fairly convincing (but not very accurate) model that is still often brought up in high-school and under-grad physics and chemistry courses. The Bohr Model was developed before the Uncertainty Principle.
But electrons do not move in fixed orbits ...the Uncertainty Principle triumphs again!
So if electrons don't stick to the nucleus, and if they don't move in fixed orbits around the nucleus, then where exactly are they?
We can never exactly point out an electron's position in an atom. However, we can determine the probability of finding electrons in a given region of space around the nucleus. These regions of space with a very high electron probability density are called orbitals.

This answer has been greatly over-simplified. Thought it does deal with the gist of the topic, I'm personally looking forward to seeing more detailed answers myself.

I wouldn't say it's oversimplified, but incorrect. Uncertainty Principle has no importance here.

@Mithoron Eh? But I recall hearing Feynman using the same argument: http://www.feynmanlectures.caltech.edu/I_02.html (Read page 2-3, Quantum Mechanics, 2nd paragraph). I think I'll trust Feynman on this one. Thanks for the down-vote by the way ;)

Feynman can also be wrong...

Electron degeneracy pressure not Uncertainty Principle.

The electron degeneracy argument is more to do with pushing atoms together and then what happens to energy levels via the Pauli Principle than with an isolated atom and its electrons and nucleus. Feynman's argument seem sound to me.

@Mithoron 1) electron doesn't fall into the nucleus even in a hydrogen atom, where degeneracy pressure doesn't exist, 2) degeneracy pressure comes from the uncertainty principle, as the very link you give also briefly mentions

I agree that the uncertainty explanation is correct. This principle determines what happens because as the electron gets closer to the nucleus and the energy gets lower the electron is confined into a smaller space, i.e. is closer to the proton, this means that the kinetic energy increases and so does the uncertainty in the momentum. So the energy is a compromise between the kinetic and potential energy and this applies to all levels. The existence of atoms would then seem to be a consequence of the uncertainty principle.

The non-local nature of the electron is due in general to its small mass and QM, not to fermionic character.

The planetary view of a a small electron orbiting a nucleus does indeed not give a satisfactory explanation; Bohr simply postulated stable orbits in his original model.
In QM, you could argue that the bound electron is the wave function [1], and therefore much larger than the nucleus, despite being much lighter. The closest thing to sticking together in this view is for the nucleus to settle in the center of that electron wave function.
For a more deterministic picture you might adopt Mills' view [2] which suggests the bound electron is a spinning charge distribution (orbitsphere) which completely encompasses the nucleus — think something like a soap bubble around a grain of salt. Angular momentum is an inalienable property of this electron, with or without a nucleus at its core; i.e. it is not so much that the electron ‘bothers’ to move around the nucleus, but rather it spins around its own axis all the time. (The size of a free electron in this model is the De Broglie wavelength, making it indistinguishable from a point-charge at high energies.)

[1] See for example https://gravityandlevity.wordpress.com/2015/04/11/how-big-is-an-electron/ option #1
[2] Randell L. Mills, "Classical Quantum Mechanics" *(Physics Essays, Volume 16: Pages 433-498, 2003)
http://www.brettholverstott.com/annoucements/2017/8/5/summary-of-randell-millss-unified-theory

This is only a comment on the answer by Pallas, which I upvoted:

Fig. 3-4. The wave function and probability distribution as functions of r for the n = 1 level of the H atom. The functions and the radius r are in atomic units
https://www.chemistry.mcmaster.ca/esam/Chapter_3/section_2.html
I visualize the electron as a ball of cotton candy (spun sugar), which is not a point, and which, as Pallas says, simply surrounds the nucleus. So the electron does, in fact, stick to the nucleus, as good as it can! An s-electron has a high probability of touching the nucleus. The only way to get closer would be to become a neutron - oops, but neutrons have a half-life of only about 15 minutes before decaying back to a proton and electron.
So the cotton ball with the nucleus in the middle is sticking, not travelling around. And maybe the uncertainty principle plays some part in why the electron isn't a point charge.
But I wonder why the electron is a trinity of charge: that is, why does it take the charge of 3 quarks to equal the charge of one electron? Surely, the lowest charge (a quark) must be the elementary unit. Mulliken would have agreed (I think).

To add to the plethora of recent answers: it's not that the electron cannot "crash" into the nucleus, it certainly can, the likelihood depending on the isotope in question. The process of fusion of proton and electron to form a neutron is called electron capture. None other than Luis Alvarez studied such processes early in his career. You might find people more knowledgeable on the subject over at physics SE.

A simplified explanation would be to think about a balance between minimising potential and kinetic energy. While an electron sticking to the nucleus would decrease potential energy due to electrostatic attraction, the spiked and extremely localised wave function would be high in kinetic energy (correlated with the curvature of the wavefunction).
I do understand that this is a simplified view but I believe it is a neat explanation.


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IUPAC names accidentally differing in hyphen only
Here is an example of two compounds

$$\ce{CH3CH2COOCH3}$$
  vs
  $$\ce{(CH3)2CHCOO-}$$

i.e.

vs

whose possible IUPAC names

methyl propanoate
vs
  methylpropanoate

accidentally differ only in presence/absence of a space. (Note that complete, unambiguous name of the second compound is 2-methylpropanoate)
Is there an (accidental) example of two IUPAC names for different compounds, that differ only in presence/absence of a hyphen ("-") instead?
(Question might sound hypothetical, strange and inappropriate, but is related to to this one)
UPDATE: Mentioned ‘methylpropanoate’ preferred IUPAC name is 2-methylpropanoate. There's a better example pair:

phenyl acetate ($\ce{CH3COOPh}$)
phenylacetate ($\ce{PhCH2COO^{-}}$)

Note that in some language localized official IUPAC nomenclatures, the names really differ by the hyphen instead of the space (e.g. in Czech: fenyl-acetát vs fenylacetát), but I was wondering about the English nomenclature.

The compound with the formula $\ce{CH3COOPh}$ is known as phenyl acetate, but the IUPAC name is: phenyl ethanoate.
The compound $\ce{PhCH2COO-}$ has the IUPAC name: 2-phenylethanoate.
The similarity is interesting, but the names are distinct enough, unless I'm missing something.

Note that a IUPAC name is not a synonym for a systematic name . preferred IUPAC names are acetic acid/acetate, while ethanoic acid/ethanoate are systematic IUPAC names.

What you say is almost 10 years old info, look at IUPAC 2013 Organic chemistry nomenclature (Blue book).


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Can you produce activated carbon from toast?
On the Skeptics stack exchange we have a question about burned toast being used as a treatment for poison, the idea being that burned toast acts as activated carbon.
We are skeptical due to the fact that gas activation requires heating to high temperature, but don't have the chemistry to rule out toast as a source of activated carbon.
My instinct tells me that:

Toast is not good source of carbon
Activation is not easily done in your average kitchen.

Can someone answer those 2 questions?

Is toast a good source of carbon
Can activated carbon be produced in your average kitchen without training or any special supplies?

Is toast a good source of carbon is quite a broad question without considering the type of toast we're referring to. Even if do stick to a particular toast, what makes a good carbon source sounds a little too vague.
As a side note, recent researches have led to the production of a low cost multi-functional carbon foam from bread. [Source]
As a rough estimate, let's consider we're talking about white bread. 100 grams of white bread consists of 49 grams (i.e. approximately 16%) of carbohydrates. [Source] For the sake of simplicity, let's assume all of the carbohydates to be monosaccharides, specifically glucose. (This definitely isn't the case). The molecular formula of glucose is $\ce{C6H12O6}$. If we calculate the percentage-by-mass composition of carbon in glucose, we get the value as 40%. 40% of 49 grams is 19.6 gram, or in other words, bread consists of (very roughly) 19.6% carbon.
I'm unsure of whether or not that makes it a good source. Reading this article tells me that a rough estimate for the carbon content in wood (it is about some conifers but nevertheless good for rough estimates) is about 45–50%.

The production of activated carbon is undertaken generally in two ways: [Source]

Physical activation: It requires an inert atmosphere of nitrogen or argon along with very high temperatures (600–1200 °C).
Chemical activation: Strong acids, bases or salts and a lower temperature is required (450–900 °C).

Theoretically the chemical method seems appealing but in a practical sense it is not advisable to carbonize materials in a home-kitchen as the fumes evolved contain carbon-monoxide which is extremely toxic.

So on a scale of 1 to 10, would you say that the average burnt toast would have properties of activated charcoal? vis adsorbtion powers?


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Alkoxide as a leaving group
In the reduction of esters to alcohols, the RO group is pushed off. However, the negative charge on the O atom is destabilized by the electron donating R group. Why is the RO group able to function as a leaving group given that the resultant alkoxide is unstable?
Link to reaction mechanism.

This is simplified mechanism, in reality this would be probably complex  with Al not bare ethanolate.

The intermediate before forming the alkoxide is a tetrahedral intermediate which is also unstable. You're going from a high energy species to another high energy species.

Sorry I don't understand what you mean by "going from a high energy species to another high energy species".

You're saying that the alkoxide is unstable. How that is different from the intermediate before. That is also an alkoxide except that it has an oxygen bonded to the central carbon. On collapse of the tetrahedral intermediate, you're going to form a new alkoxide and a carbonyl group, plus frequently two species. This is entropically and energetically favorable. You're focused on the alkoxide being "unstable", but the thing that comes before is also "unstable" and more so.


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Variation of current with time in gel electrophoresis
During gel electrophoresis, if the voltage of the power supply is kept constant, how would the current in the circuit change over time?
I was wondering whether the electrical resistance of the gel would change over time. There will be migration of ions present in the buffer and it will resist the change in pH. But how does the electrical resistance of the gel change? Can someone explain?

I think this might help you a bit:
https://www.researchgate.net/post/SDS_PAGE_should_be_run_at_constant_current_or_constant_voltage


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Deliquescence database
I am looking for a database or printed document providing the RH0 (relative humidity above which deliquescence will occur) for inorganic salts. I've looked everywhere on the internet can find nothing.


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Ethylene hybridization
In the carbon atom, there are electrons in 2s, 2Px, 2Py, and 2Pz. Out of these orbitals, only 2Px, 2Py, and 2s are hybridized. The 2Pz orbital is left. When making a bond there is a pi bond between two carbon atoms which is formed by the 2 Pz orbitals of the each carbon atom.  In $\ce{CH4}$, 2Pz orbital also hybridized.
The orbital 2Px and 2s always make sigma bonds. Each carbon atom connects with two hydrogen atoms in $\ce{C2H4}$. Therefore I like to imagine that 2py and 2pz orbitals can make sigma bonds with 2hydrogen atoms. Then only left orbitals are 2px and 2s. They can make double bonds between two carbon atoms. 2px and 2s always form as sigma bonds and then the double bond between carbon atoms, one is sigma bond and another also a sigma bond.
I know that according to the VSEPR theory there should be one sigma bond between two atoms. But atoms are hybridized for their comfortability. But according to my above imagination making 4 sigma bonds is more comfortable than making 3 sigma bonds and one pi bond for the $\ce{C2H4}$ molecule. Therefore, why is it not trying to make such a bond?

Such bonding produces an equivalent description of ethene.  See this earlier answer.

Suppose the carbon was $sp^3$ hybridised. The carbon would have 4 hybridized $sp^3$ orbitals which would point towards the corners of a tetrahedron; they are $109.5^\circ$ degrees apart from each other.
You have to form two bonds between the same two carbon atoms. Suppose you use one $sp^3$ hybridised orbital to form one $\sigma$-bond. How are you going to form another bond? The other bonds of the carbon are $109.5^\circ$ apart. This is not feasible.
The other option for the carbon is to assume $sp^2$ hybridization and have three hybridized orbitals stay $120^\circ$ apart. The one remaining $p$-orbital can be in any direction ($x$-axis or $y$-axis) except along the bonding axis ($z$-axis). In this way, you can form a double bond using one $sp^2$ hybridized orbital and one $p$-orbital.

@OsalSealaka Don't say "thanks", that's not what the conments are for. Accept and upvote it, that's how we say "thanks".


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What will be the possible major product(s) on heating this primary alcohol i.e. 9-fluorenylmethanol with (conc) H2SO4?

When the above compound is heated with (conc.) $\ce{H2SO4}$ the $\ce{OH}$ will probably be protonated and it will leave, producing a carbocation. After that possibly there will be a hydride shift. Now, after this step what could be the possibilities. I was thinking of ring expansion (which could potentially form phenanthrene) but again I doubt that a negative charge on a $sp^2$ carbon of the benzene ring would attack the positive charge after C-C bond cleavage. Or another possibility is formation of a double bond to form a structure like:

Which is the correct (major) product formed?

Ring expansion is possible and we get  an aromatic compound I suppose which is quite stable....

@SurajS For ring expansion we need to form a negative charge on the benzene ring which has to potentially attack the formed carbocation due to leaving of the $\ce{OH2+}$ group. Is that feasible?

I think this process favors thermodynamics, Phenanthrene might be the major product...

Sorry I don't get why - ve charge on the benzene has to be formed... Isn't it th 3° carbon making way for the rearrangement so that 1° carbocation stabilizes to 2°?

Which  "$1°$ carbocation stabilizes to $2^o$" are you talking about? I can't understand. Can you draw it and show? @SurajS

@SurajS A negative charge is formed on the benzene ring when the C-C bond between $\beta$ carbon and the $\gamma$ carbon of the $\ce{OH}$ group cleaves. That negative charge can attack the $1^o$ $\alpha$ carbon so that it can expand.

What I feel is that forming phenanthrene would be possible because of its high stability due to resonance. Initially, $\ce{-OH2}$ leaves out forming a 1°carbocation. Now the $\ce{\beta C-\gamma C}$ cleaves so as to make way for the better 2°carbocation which is also thermodynamically favourable.

Related: https://www.researchgate.net/publication/231517152_Dehydration_of_9-Fluorenylcarbinol_a_New_Synthesis_of_Phenanthrene


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How do I cool my water bottle immediately for an emergency use?
What are the chemicals I can use to cool my water-bottle at once if I'm stranded in a desert and I have only chemicals to cool water?
(I'm keeping those chemicals outside my water bottle and not inside it.)

Ammonium nitrate will do.

Maybe you have a liquid propane cylinder (or any liquified gas for that matter) , spraying the propane that is just expanded into vapour form on water bottle will be much helpful in cooling without compromising the water.

Ammonium nitrate, calcium ammonium nitrate, and urea are each used in instant-cold packs. The dissolution of any of these compounds in water is an endothermic process and will thus cool your container of water.
In a practical sense, it doesn't seem likely that you would sacrifice some of your water so the rest of it could be cold in an emergency situation. Of course, the water you use for cooling does not have to be clean and potable.

any water less chemical mix ? can it make ice

+1 for the second paragraph.  I bring in water for construction crews working on the library in my work office building.  It just sits there not cooled but the guys appreciate it nonetheless.  Moral of the story:  having water is more important than having it cold.

@murmansk Regarding your waterless idea, it is unlikely that any dry compounds would produce significant cooling on mixing. It is the heat of dissolution of certain solids into water that provide the cooling for most chemical cooling packs and such. As far as using a chemical cooling mixture to get ice, that would depend on the temperature of the water to start with, how much water you want to freeze and how much of the external cooling mixture you have. It would be difficult though, as the latent heat released when freezing water is quite large. Sorry so long to get back, hope this helps.


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Application of the Woodward-Hoffman rules to a [14+2] cycloaddition (follow up)
This is a follow up question to another question I made earlier.
I put some of this information into my own notes just so I can get my head around this.
So just to clarify. Does the Woodward Hoffman rule given below only apply to ground state reactions, thus the rule cannot be applied to the above reaction scheme because one of the components is in an excited state?
A ground-state pericyclic change is symmetry-allowed when the total number of (4n+2)s and (4n)a components is odd.
EDIT: (1) Correct labeling of the suprafacial components, (2) Removal of orbital phases, (3) Stereochemistry situated at the correct stereocentres.

Your diagram reveals some misunderstandings. (1) As you've drawn it both components there are suprafacial, not antarafacial. (2) You shouldn't be drawing orbital phases in a Woodward-Hoffmann analysis. Orbital phases are important for a Fukui FMO analysis, but not for W-H. (3) Stereochemistry in the product is missing, and the stereochemistry is important, because it tells you whether the reaction is s+a or s+s. If it's s+s (photochemically allowed), then the protons in the product must be cis to one another, like this: https://i.stack.imgur.com/gkOSu.png

Thank you for your help (othocresol & NotBaran). I've been having a difficult time understanding pericyclic reactions. (1) For some reason, I was defining the supra and antara based on the orbital adjacent to each other rather than between components (2). I figured out how to draw a non-shaded orbital in Chemdraw. (3) That was silly of me not to show the stereochemistry at the stereocentres.

The supra/antara is a common mistake when everybody starts out. I can confirm that it has confused me many times ;)

For reactions in the excited state, a modified statement of the Woodward Hoffman rule applies:

A pericyclic change in the first excited state is symmetry allowed when the total number of (4q+2)s and (4r)a components is even.

The only difference is that we're now looking for the total to be even rather than odd.


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Nanoparticles synthesis
I am following an article for synthesis of the nanoparticles.
There is a gold nanoparticles synthesis rout which is called UPD. In this rout, the underpotential deposition of Ag onto the surface of the Au particles is used. There are two metals (Au and Ag) in the obtained nanoparticles, but they aren't called bimetallic nanoparticles. Why is it called "gold nanoparticles synthesis" and not "gold-silver bimetallic nanoparticles synthesis"?

Its because the gold nanoparticles are the star of the show. The researchers were just looking at how surface modification using the silver would alter the morphologies of the gold nanoparticles. Probably the researchers wondered to themselves one day, "How could we possibly control the shape of these gold nanoparticles?" and decided to try with silver.
The bimetallic part would suggest that the researchers were aiming for gold-silver nanoparticles (agglomerated together, not one on the surface of another like in the research)


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Can the electrophile attack at the position of the NH2 group in nitration of aniline?
In the nitration of aniline, we observe the formation of a meta-substituted product (47%) almost as much as a para (52%).
However, can't the electrophile attack at the position where $\ce{-NH2}$ is already attached, so that after the attack, it will be stabilised by the mesomeric (+) effect of $\ce{-NH2}$? Maybe just after some time, the $\ce{-NH2}$ group will become $\ce{-NH3+}$ (acidic medium) and $\ce{NH3}$ would depart as a leaving group?

Can you try drawing arrows for it, because it doesn't really work that way. Ipso attack of an electrophile leads to a carbocation on C-2 which cannot be stabilised by the lone pair on N (and note that since the nitration is carried out in strong acid, that lone pair will be permanently protonated anyway). Next you cannot have "NH3" leaving and regenerating an uncharged benzene ring

@orthocresol I did try to. I meant to say if 'while' the electron rebound back to their position in the intermediate, -NH2 group protonates, and leaves?

Maybe just after some time, the $\ce{-NH2}$ group will become $\ce{−NH3+}$ (acidic medium) and $\ce{NH3}$ would depart as a leaving group?

No! You are confusing the leaving group oxonium ion with the ammonium ion. Simply put it, $\ce{R-NH2}$ does not leave as $\ce{R+}$ and $\ce{NH3}$, as you might expect from the hydroxyl group leaving mechanism.
But that does not mean there is no other way to make an $\ce{-NH2}$ group (which is a very bad leaving group) leave. You can instead make it leave as dinitrogen gas!
The way this works is by the reaction of $\ce{R-NH2}$ with $\ce{HNO2}$. If $\ce{R}$ is an:

aromatic group: a diazonium ion, $\ce{R-N2+}$, will be formed (stable at ice cold temperature). Further reactions such as Balz-Schiemann, Gattermann, Sandmeyer, etc. may be employed to replace the $\ce{-N2+}$ with another group, and make it leave as dinitrogen gas.
alkyl group: $\ce{R-N2+}$ will be formed but it will, being highly unstable (given that it does not enjoy the resonance stabilization of an aromatic ring), instantly decompose to release dinitrogen gas and form a classical carbocation $\ce{R+}$. This carbocation may rearrange, if necessary, and then receive an attack by a nucleophile.

This is how you make the bad leaving group, $\ce{-NH2}$, leave any aromatic ring or an alkyl group.

See the answer to this question  https://chemistry.stackexchange.com/questions/31644/is-the-ammonium-substituent-nh3-really-meta-directing-in-electrophilic-subst

@Waylander Thanks for the link, though I don't know how that answers this question, or offers a correction to the answer I've given. Can you please explain? Thanks again.

Simply because it addresses the question of the role of the anilium ion in nitration, and specifically makes no mention of any loss of NH3

@way I too have said that - without forming a diazonium ion - there's no way you could get the - NH2 group to leave the ring. So, I still don't see the problem...

With nitric acid as a nitrating agent, the difficulty is that the acid protonates the amino group and destroys its nucleophility.
However, other compounds may provide a nitryl ion moiety without a protic hydrogen, thus leaving the nitrogen open to nitration. A casual Google search does not turn up an example with aniline, but ammonia can be converted to nitramide at nitrogen by $\ce{N2O5}$:
$\ce{N2O5 + 2NH3 -> \underset{nitramide}{H2NNO2} + NH4NO3}$
The $\ce{NO2^+}$ portion of the $\ce{N2O5}$ nitrates a nitrogen atom on the ammonia molecule, and the displaced proton is scooped up by a second ammonia molecule to form a salt (ammonium nitrate) with the $\ce{NO3^-}$ part of the original nitrogen pentoxide.


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Why must NaOH be in excess if we wish to measure the effect acid strength has on the enthalpy of neutralisation?
A few different acids are given i.e $\ce{HCl}$ , ethanoic acid, $\ce{H2SO4}$, ethanedioic acid. We want to conduct an experiment in which we measure the temperature rise for neutralisation as a measure of the enthalpy of neutralisation for each acid (no calculations are involved other than simply noting the temperature change) while using $\pu{30 cm^3}$ of $\pu{2 mol/dm^3}$ $\ce{NaOH}$. The question states that $\ce{NaOH}$ should always be in excess, but why is this so? Why can the acid not be in excess provided that we keep the total volume constant?

Reaction has to be complete.

It will be complete even if the acid is in excess.

I think this is just to get the best possible results for your analysis. I mean, it would be better to have excess NaOH so you get a good reference because you know exactly when no more neutralization is occuring when all the acid gets used up. Better than having excess acid, though the idea is the same, but the excess is an uncertainty. Let's just say its more of a matter of being a technique that makes sense.


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Structural isomerism in iso-butyl and tertiary butyl alcohol
What type of structural isomerism exists between iso butyl alcohol and tertiary butyl alcohol?
I think that there exists only positional isomerism between them but the source from where I am studying says that they are both positional and chain isomers.

Positional isomers differ by change in locant of substituted carbon. Structural isomers differ in structure of chain.

I think the best way to describe this isomerism is constitutional isomerism because it takes into consideration of both position of functional group and structure at the same time.

Constitutional isomers may have different carbon skeletons (as in
  isobutane and butane), different locations of a functional group along
  the chain (as in isobutyl alcohol and tertbutyl alcohol), or different
  functional groups (as in ethanol and dimethyl ether)

Positional isomers have the same carbon skeleton and same functional groups but differ in location of such functional groups (carbon skeleton remains unchanged). On the other hand, structural isomers the atoms are in completely different order , they arise because of the possibility of branching
I think the reason why these isomers are considered both structural and positional is that they both have the longest chain of max 3 carbons (hence same carbon skeleton), at the same time different location of -OH group  (a different chain - structural)

Hope this helps
References
Organic chemistry Mc Murry
Chemguide: Structural isomerism


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When 3,4-dimethyl-2,4-hexadiene reacts with HBr what is the major product?
I found out from drawing it out that there are 2 products: 4-bromo-3,4-dimethyl-2-hexene and 2-bromo-3,4-dimethyl-3-hexene. I want to know which forms most of the time.
I used the fact in allyl carbons, the double bonds will travel to the carbon-carbon single bonds to form a double bond. I also used the fact that tertiary carbocations are more stable due to sigma bonds that can participate in hyperconjugation - that is delocalization of electrons by the overlap of sigma bond orbital with an empty p-orbital on an adjacent carbon.
I would like to see some of these reactions in a lab setting, but for now I'm learning organic chemistry reactions like as if it were math.

The second product you propose has four substituents, including one double bond, at C-3, but anyway HBr does not brominate alkenes or dienes just like that. Are you learning OC by yourself?

Yes I am. I'm stationed at a ship and my only resource is a book and me thinking.

Ah. Navy_Colors. Wow.  (Of course alkenes react with HBr, what was I thinking? The second product should be 2-Bromo-..., right?) Can you describe the first step of the reaction? Markownikow would be the reasonable thing to think.

Same question on slightly different substrate https://chemistry.stackexchange.com/q/59964/16683

@orthocresol Similar indeed, only C-3 here is too crowded to really allow the 1,2-addition, isn't it?

Conjugated dienes can undergo two types of additions with HX, they can undergo 1,2-additions and 1,4-additions.  A 1,2-addition occurs when HX adds to only one of the double bonds, not interacting with the other.  a 1,4-addition occurs when HX adds X to one of the double bonds, and because the system is conjugated, the hydrogen will be added three carbons away from X.  The 1,2-adduct and 1,4-adduct are either thermodynamically or kinetically stable.  The kinetic product will form quicker, but is less stable, it will dominate at low temperatures and short reaction times.  The thermodynamic product is more stable but takes more time to form.  It will be the major product at high temperature and long reaction times.  In this case, the kinetic product would be 2-bromo-3,4-dimethyl-3-hexene, the thermodynamic product would be 4-bromo-3,4-dimethyl-2-hexene, because the thermodynamic product has the bromine on the most substituted carbon.  Depending on the reaction conditions, either could form.  At low temps. and short reaction times, 2-bromo-3,4-dimethyl-3-hexene will form, at high temps. and long reaction times, 4-bromo-3,4-dimethyl-2-hexene will form.


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What causes polarization in noble gases?
According to my knowledge, polarization occurs when in a compound there is a difference of polarity its constituents.
But in the case of noble gases, there is a single component. What exactly is polarization of noble gases and how does it occur?

You've (or your textbook has) generalised the problem to the point where an answer makes no physical sense any more. Van-der-Waals forces have different components depending on the observed system (i.e. London forces, dipole-dipole interactions, etc.). Read and understand, don't try to memorise rules.

i too know van der waal forces  differ from point of observed system. but the issue is in case of noble gases i do not how dipole dipole interactions occur?

The answer is London dispersion: Induced dipoles. They are one central part of the van-der-Waals concept, which you seem to have missed.

You probably mean polarisability rather than polarisation. This is a quantum effect and due to the fluctuating electron density  any atom has. A transient dipole is formed as the electron density is, instantaneously, not symmetrical. Over time this transient dipole averages to zero but while it exists this produces an electric field that can induce a dipole in a nearby atom. It works out that the time average of the interaction between the two induced dipoles is not zero which generates an attractive (dispersion) force between them.  London's eqn. describes the interaction.

ok so if we consider this method then any atom can be polarized i guess?  but m sure the time period for which the noble gases remain a dipole would b vry small then. so is that enlarged by bringing a continous source of electric field?

Yes, all atoms can show this effect. The polarisation scales roughly as electron radius cubed, so approx as volume. Thus it is far larger in heavier atoms, iodine vs helium for example. The time period is very small but that does not matter as the important point is that on average the interaction is not zero.

Although this applies to all noble gases, consider helium. There are two electrons moving around the nucleus. The movement of those electrons is symmetrical and during a defined amount of time, the atom is not polarised. However, the momentary positions of those electrons can polarise the atom. This can happen when both electrons are relatively close, on the same one side of the atom. In this case a momentary dipole is formed, since there is separation of two charges, nucleus being positive and the pair of electrons being the negative pole.

but according to the atomic model electrons are present as electron clouds. their location cant be predicted according to heisenberg's uncertainty principle.

To Esha: You can think of the electrons as being present as clouds, but the cloud's density changes over space and time;  at some times, the two electron's clouds may both have high densities in the same region of space, which is analogous to saying the two single electron particles are close together. Whether you think about them as waves or particles, fluctuations in particle location/cloud density occur


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Selective bond breaking
Can a specific bond be broken by an engineered energy wave corresponding to the bond length?
For example, can we break the $\ce{C-N}$ bond and not the $\ce{C-H}$ in a hydrocarbon chain?

Arguably, the entire realm of chemistry is nothing but selective bond breaking. So yes, this can be done, though the light with specially selected frequency is a relatively rare tool to use.

@IvanNeretin ...and selective bond-making.

@SatwikPasani Yeah, right, that too.

In general, to break one bond vs another on demand in the presence of many others  is a very difficult thing to achieve. Not possible at all by heating molecules thus there has been much effort to do so using lasers.
But this becomes very difficult also. This is  because on exciting a molecule into an excited electronic state, such that it has enough energy to break a bond, and into the type of vibration involving the bond you wish to break,  the energy then leaks out to other different vibrational normal modes and can then end up anywhere. So it does not just stay where the energy was initially placed.
This 'leaking' occurs due to vibrational anharmonicity which provides a 'coupling' with which one normal mode can swap energy with another. The process can be very fast and even if femtosecond laser pulses are used energy still moves all over the place and competes with accumulating energy into the 'victim' bond effectively defeating the process. The area is, however, one of very active study so progress in this area is to be anticipated.

Thank you Porphyrin, do you happen to know which institutes lead the efforts or are active in these studies?

Look for chemistry  groups that use femtosecond laser spectroscopy and study 'coherent control'. There are also theoretical groups that study this. I'm sure that you will find several quite easily.


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If I mix an aqueous solution of hydrogen chloride with an organic solvent, will the organic solvent contain hydrogen ions?
For example, let's say I mix $36\ \text{cm}^3$ 2-propanol, $9\ \text{cm}^3$ methyl ethyl ketone and $15\ \text{cm}^3$ hydrochloric acid with a concentration of $1,0\ \text{mol}/\text{dm}^3$.
The hydrochloric acid is solved in water, and since the other compounds are organic and less dense than water they should form two layers. Would the upper layer be "acidic"?
This is an elution I've used in a thin-layer-chromatography lab and I'm trying to figure out what role the hydrochloric acid plays in affecting the compounds' ability to bind to the elution (mobile phase).

Are you sure the mixture you're describing has two layers? 2-propanol is soluble both in water and in methyl ethyl ketone. And if you used it as an eluent, didn't you check whether it was a single phase? Actually, can one use biphasic eluents at all for liquid chromatography?

$\ce {HCl}$ is so strong a proton donor that bases weaker than water could be protonated.  Alcohols and etherss, if they form a separate phase from water, are likely to be protonated.  Going the other direction, some organic compounds with little solubility in water might become more soluble in a hydrochloric acid solution if they are sufficiently basic.  The basicity of some cyclopropenones can be measured that way.  See (behind a paywall) http://pubs.acs.org/doi/abs/10.1021/cr60288a003?journalCode=chreay.