text:
How is 3-bromocyclohexene converted to cyclohexane?
My attempt is illustrated below:

or

Which of these methods is most viable?

You need acid (and probably heat) present for the top reaction to give elimination. It it also possible with the correct choice of catalyst to reduce the alkyl halide by hydrogenation.

For the last step in both reactions, ensure enough pressure of hydrogen, literally.  Otherwise, with cyclo-hexa-1,3-diene already in hand, you might drop into the formation of benzene.

you can directly get your desired product with $\ce{H2 + Ni + \Delta}$


text:
Probability of finding electron in each orbital
This may include some quantum physics.
Are there anyways to describe the wave function of Hydrogen's orbital (or that of a Hydrogen-like atom) in terms of the wave function of each orbital?
You can write wavefunction as $|\psi\rangle=\sum{c_i|\Phi_i\rangle}$ where $|\Phi_i\rangle$ are each possible states.
Suppose that the electron is in a pure state and orbitals are independent states, how can we find coefficient or probability to find the electron in each orbital?
You may consider certain parameters such as temperature in the calculation.
Thank you in advance.

Orbitals are the pure states. There is nothing to calculate.

You can choose the $\Phi_i$’s to be the orbitals, then the probabilities are trivially $|c_i|^2$.

Yes, each orbital is the pure state, as well as the electron's superposition between states. But I want to separate the electron's wave function into a superposition of each orbital i.e. solve for $c_i$.

@HanchaiNonprasart The inner product of the hydrogen wavefunction with an orbital yields the coefficient, $c_i = \langle \Phi_i|\Psi\rangle$.

Thanks for your answers and sorry for the confusing question so let make it clear. Can we assume that the electron of hydrogen is always in $1s$ orbital? If not, can we predict the probability of it without knowing its wavefunction beforehand?
Or more specifically can we predict coefficient $c_i$ and the wavefunction from the physical property of hydrogen sample such as it's average temperature?

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Isomerisation Reactions of Alkyl Halides
My organic chemistry textbook says:

Haloalkanes on heating at 573K or in presence of anhydrous at low temperature undergo molecular rearrangement to form an isomeric haloalkane.

I noticed a similarity in these two examples. In both cases, a primary halide was getting converted into a secondary hailde.
Now, can somebody show the actual mechanism of this isomerisation reaction? Also, does the phenomenon I mentioned have something to do with the mechanism?

Note that this occurs at high temperature or in the presence of a Lewis acid. The intermediate is the AlCl3X anion - alkyl cation ion pair (X = halogen). You get a hydride shift to give the most stable alkyl cation

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why were quasicrystals initially so controversial?
I am a mathematician. My (limited) understanding is that quasicrystals are structured as parts of aperiodic tilings of $\mathbb{R}^3$. Such tilings were already known when Shechtman first studied his alloy. So, I was wondering why there were so many accomplished chemists who resisted the idea of quasicrystals.
I know that the diffraction pattern that Shechtman obtained had a symmetry of order 5, which contradicts the crystallographic restriction theorem, but the crystallographic restriction theorem assumes that the crystal can be modelled as a discrete lattice, i.e. that it has translational symmetries, i.e. that it is periodic. Thus, it does not say anything against the existence of quasicrystals as I understand them.
Were chemists perhaps convinced that, for some chemical (rather than mathematical) reason, any crystal-like material must have a periodic structure?
Edit: I am afraid that I should elaborate further. I am aware of line groups, frieze groups, rod groups, wallpaper groups, layer groups, crystallographic groups, and Bravais lattices, and the history of both mathematical crystallography and tilings. What I am asking is this: Robert Ammann obtained a 3-dimensional aperiodic tiling in 1975. Dan Shechter published his paper in ... 1982? Is that correct? So, my question is, how is it possible for very intelligent people, including Linus Pauling, to still hold onto their assumption that crystals had to be periodic? Was there an actual chemical reason for them to think so?
I would like to thank everyone who is trying to answer this.

Anything new is controversial. The idea of non-translational long range order was new. True, it was known in math (thanks to Penrose), but that didn't matter much to chemistry. Those math people always come up with some weird stuff, you know.

@EdV And I think that OP asks if there was actual chemical reason and not if a not so great chemist was just stubborn.

There is perhaps also something to be said about the doubts that a 5-fold symmetric tiling which had really good diffraction patterns would emerge 'naturally' - the long range interactions seemed at odds with experience in fairly simple systems, much less all the pain and agony needed to get good crystals of, e.g., proteins, with similar long-range interactions needed.

@Joe Custer This is interesting. So, were some chemists negatively predisposed because Shechter's published diffraction patterns came out very clean? Was this just chance or did it have to do with the material?

I see there the parallel to the Special/General Relativity theory or Quantum Mechanics/electrodynamics. A "crazy" circle of  "crazy"  evidence ->, "crazy"  hypothesis/theory -> "crazy"  predictions -> "crazy"  evidence . Minds of many scientists refuse to accept such ideas at the time, but "reasonable" ideas have no clue.

@GeorgeKontogeorgiou - well, of course those materials were metal alloys, so chemists probably were negatively predisposed to begin with. I was a grad student at the time in materials science, and there was no negative predisposition, just surprise. It really didn't take long for folks to get comfortable with the the results since they were not that difficult to replicate. And then folks jumped in to figure it out. As I recall Neil Ashcroft did a lot of work making it mainstream.

So, my question is, how is it possible for very intelligent people, including Linus Pauling, to still hold onto their assumption that crystals had to be periodic?

Linus Pauling advocated for a hypothesis called twinning in which grains of periodic lattice were joined together at an angle, such that you got non-standard symmetries on a large scale that did not exist at the atomic scale. His argument was that there were plausible alternative hypotheses that had not yet been ruled out by the evidence.
While Pauling probably pushed his certainty too far, there is a sense in which this is how science is supposed to work. Confidence in scientific conclusions is only justified when a determined attempt to refute it by competent and motivated opponents has been made and has failed. A hypothesis gains in confidence only as all the alternative hypotheses get rejected. Thus, for any hypothesis to be justifiably accepted, you need systematic sceptics to try to attack it and pull it apart. It's like evolution by natural selection - only the fittest hypotheses survive. So in this, Pauling was acting as a good scientist, trying to pick holes in the theory, and challenge it for as long as it could be challenged.
You should also consider the psychology of people like Pauling. There is a tendency in human social interaction to fit in with the crowd, and follow the intellectual fashion. That's fine for learning and applying what's already known, but is an impediment to discovery. Scientific revolutions are made by mavericks who can stand against the consensus, and who can stick to their own beliefs against all the slings and arrows that the community can throw at them. Pauling was likely as successful in science as he was because he had such a stubborn, contrarian, self-confident personality. Of course, while such a personality is very useful to persist and push your theory against dogmatic opposition when you happen to be right, it can be a drawback for the individual on the occasions that you are wrong. Although, as I said, science needs such people to continually challenge and test the consensus, it is a reminder that we are all fallible, even the cleverest of us. And personality is not the same thing as intelligence.
It's common for a field to take some time to get used to new ideas. Mathematicians took a long time to accept negative numbers and complex numbers. Physicists spent many years after the invention of General Relativity rejecting the possibility of black holes - even Einstein once wrote a paper proving them to be impossible! The German physicist Max Planck said that science advances one funeral at a time. Or more precisely:

A new scientific truth does not triumph by convincing its opponents and making them see the light, but rather because its opponents eventually die, and a new generation grows up that is familiar with it.

I think I commented earlier to prevent such answer as that, wasn't too effective, I guess.

@Mithoron, the answer to the question was that Pauling advocated 'twinning'. That's 'an actual chemical reason', as requested. Why would you want to prevent such an answer?

I think the paragraph about 'psychology'—while perhaps an interesting point for reflection—is not needed, particularly on a site where Q and A are supposed to be factual rather than than speculative. In a way, the question sets itself up for anecdotal / speculative answers, but psychoanalysis may be taking it a bit too far. In my opinion, this paragraph doesn't really add much to your answer; the rest about scientific process etc. is more valuable.

@orthocresol The psychology is supposed to be addressing a more general point about the philosophy of science. Not only did Pauling have 'an actual chemical reason' for his opposition, but he was right to do so, and probably made the great achievements in chemistry that he did precisely because he stuck to his guns. This is probably a more important reason, in scientific terms, than the merely technical one. The importance of sceptical opposition to science is not anecdotal/speculative. And social dynamics are important to understanding why this behaviour constantly happens in science.

Hmm, well, ok. I honestly don't think it's answering the question (the 'chemical reason' OP sought is in your first paragraph and ends there), but I can also see why you feel it's important to include it. But let's leave this discussion; it's your answer, after all.

Sure. I can understand why people would object. I just thought that the "how is it possible for very intelligent people" part implied a particular view on the intellectual merits of opposition that also needed answering. Enough said.

While many radical ideas in science take a long time to find acceptance (eg plate tectonics) this is not one of those examples. There was resistance to quasicrystals, but their acceptance came quickly after the best known mathematical examples. The speed is, if anything, surprising when the acceptance involved rejecting the most fundamental assumption behind the theory of crystallography.

The basic problem was that chemists assumed that crystals had to be periodic.
It is probably easier to understand than you think. The chemistry of crystallography and the mathematics of symmetry are tightly related. And chemists always started trying to understand crystal structures based on the fundamental assumption that crystals had to be periodic in their structure. If it isn't periodic it isn't a regular structure, is it?
The problem wasn't helped by the mathematics either. Even for the simpler problem of tesselating the plane many mathematicians assumed that tilings had to be periodic. So, for example, we know that only triangles, rectangles, and hexagons can tile the plane but pentagons cannot (complicated shapes can but the symmetry falls into a small number of categories. I'm simplifying a lot!) Strictly speaking there are 17 possible "wallpaper" tilings with different regular symmetry.
That infinite tilings of the plane without periodicity could exist only emerged in the 1960s and Penrose didn't find the very simple ones we are now familiar with until the 1970s. Before that point chemists, who work in 3D not 2D, didn't even have simple examples of tilings that could exist without being periodic in 2D so they assumed that the same principle applied in 3D crystals. Given the theorems of how regular 3D symmetry can be built in the 3D case they assumed that crystals with, for example, 5-fold symmetry were impossible.
But as the mathematical theory of aperiodic tilings developed, it became more obvious that the false assumption was perfect periodicity. The 2D Penrose tilings show that a plane can be tiled in a way that does not precisely repeat. But does have an approximate symmetry (5-fold in the standard Penrose patterns).
Less than 10 years after the Penrose tiles were discovered, chemists started to recognise the 3D equivalent in crystal diffraction experiments. The big discovery was by Shechtman in the mid 1980s (which eventually won him a Nobel). The realisation that aperiodic structures could form well-defined diffraction patterns led to a change in the very definition of a crystal.
What had held chemists back was the assumption that crystals had to be periodic. But the real-world discovery of this didn't fall too far behind the mathematical theory so we can't just blame chemists for not noticing earlier examples in their diffraction patterns: they didn't have any theory to make sense of them given their assumptions about what makes a crystal.

@OscarLanzi I'm not sure that example works to create non-periodic plane tilings. And the first math example of a non-periodic plane timing was 1961 (a side effect of a proof of a different tiling problem) and had ~20k different tiles in it.

I am afraid that I should elaborate further. I am aware of line groups, frieze groups, rod groups, wallpaper groups, layer groups, crystallographic groups, and Bravais lattices, and the history of both mathematical crystallography and tilings. What I am asking is this: Robert Ammann obtained a 3-dimensional aperiodic tiling in 1975. Dan Shechter published his paper in ... 1982? Is that correct? So, my question is, how is it possible for very intelligent people, including Linus Pauling, to still hold onto their assumption that crystals had to be periodic?

See the history in Wikipedia: https://en.wikipedia.org/wiki/Quasicrystal

I mean, if this was just their mathematical intuition, it strikes me odd that they held onto it for some years. I would expect Dan Shechter to reference Ammann's work or something, and then other chemists would be like "oh, interesting, so there is nothing really compelling a crystal to be periodic, so maybe Shechter is onto something here. This is why I inquired whether there was some chemical reason, why they thought that a crystal should be periodic.

@GeorgeKontogeorgiou Ammann is roughly contemporaneous with the Penrose discoveries. Chemists had real world examples and knew their implications less than a decade later. That isn't a lot of resistance to the idea by chemists. Despite periodicity being a fairly fundamental assumption.

matt_black Perhaps it wasn't a long resistance, but it was quite strong. Dan Shechtman offered a few examples of people who were initially very hostile to the idea. Let us take the extreme case of Linus Pauling. Why do you believe that he so vehemently resisted the idea of quasicrystal? Do you know whether he had any arguments except for "we have always assumed crystals to be periodic, so this must be so, even if it's not mathematically necessary"?

Never mind my previous comments.  There's a subtle terminology I missed.

To extend matt_black's answer by a a few illustrations, imagine you want to tile the floor of your bathroom and have tiles of 2, 3, 4, 5, and 6-fold symmetry and you want to stick to one type only.
As soon as you select the tiles of five-fold symmetry, you will leave space in-between the tiles revealing the background:

These patterns are a simplification, but the same symmetry principles applies to more complicated tiles, too, e.g. the by Angelo Gavezotti:

(credit to IUCr teaching pamphlet #21)
or some of the tessellations by M.C. Escher:

(credit)
Note how these tiles' orientation are symmetry related.  For the classical pattern, you were allowed to move in integer steps from one equivalent position to the other (e.g., reptile of same colour) only by translation.  And this distance of translation would apply to all yellow, red, and green reptiles to meet again a yellow, red, or green reptile.

Thank you for the answer, but my question indicates that I know both the crystallographic restriction theorem and about translational symmetries. My point is, I also know that not all tilings must admit translational symmetries and obey the crystallographic restriction theorem, i.e. there are aperiodic tilings. And people of that time also knew this. So why were some chemists, including some accomplished ones, so hostile to the idea of quasicrystals?

@GeorgeKontogeorgiou It wasn't for giving a proper answer, rather than an extension of the other by a few illustrations.  Don't underestimate the power of dogma within communities in lines of «you belong to our school / line of thought if you follow concept A, refrain from touching concept B, and strongly disagree to this newly concept X by our (perhaps personally disliked) competitors». It takes courage to admit if a perspective previously taken was incomplete (or wrong) - irrespective if in the past, or in present time, irrespective if in science, or society.


text:
Exchange energy of d6 configuration
In NCERT Chemistry book, it is given as:

Exchange energy is responsible for the stabilization of energy state. Exchange energy is approximately proportional to the total number of possible pairs of parallel spins in the degenerate orbitals. When several electrons occupy a set of degenerate orbitals, the lowest energy state corresponds to the maximum possible extent of single occupation of orbital and parallel spins (Hund's rule). The loss of exchange energy increases the stability. As the stability increases, the ionization becomes more difficult. There is no loss of exchange energy at $\ce{(d^6)}$ configuration. $\ce{Mn+}$ has $\ce{(3d^5) (4s^1)}$ configuration and configuration of $\ce{Cr+}$ is $\ce{(d^5)}$, therefore, ionization enthalpy of $\ce{Mn+}$ is lower than $\ce{Cr+}$. In the same way, $\ce{Fe^2+}$ has $\ce{(d^6)}$ configuration and $\ce{Mn^2+}$ has $\ce{(d^5)}$ configuration. Hence, ionization enthalpy of $\ce{Fe^2+}$ is lower than the $\ce{Mn^2+}$.

But, there are 5 electrons of same spin in $\ce{Fe^2+}$. So, the number of exchanges would be $\mathrm{^5C_2}$ = 10. Then why is there no loss of exchange energy in $\ce{Fe^2+}$ ?

Possibly the loss of degeneracy in presence of ligands is being discussed.
The 5 d-orbitals of Fe which were degenerate will turn into 3 lower energied orbitals and 2 higher energied orbitals in presence of a ligand.

"4 unpaired electrons of same spin" - They don't have to be unpaired. There are 5 electrons of the same spin.

But still why is it zero?

Because you're looking at an ionisation process, and thus the key question is the difference in exchange energy between $\ce{Fe^2+}$ and $\ce{Fe^3+}$.

I used the same logic for ionization energy of Fe(II) and Mn(II), but I am unable to understand why is the exchange energy of Fe(II) is zero.

Where does it say that it should be zero? It says that there is no loss of exchange energy (upon ionisation).

I think I misunderstood this statement, "There is no loss of exchange energy at (d6) configuration". Can you explain what does it mean?

As @orthocresol points out, the key is that you need to compare the exchange energy before vs after the ionization process. Anything that is unchanged by ionization cannot affect ionization energy. This is a general idea for a lot of aspects of chemistry: close the thermodynamical circles carefully, since so much information can be gained from them.
So, let us compare the Fe$^{2+}$ to Fe$^{3+}$ ionization. Both ions have 5 electrons with the same spin,[1] leading to the same exchange energy. Additionally, Fe$^{2+}$ has an electron with the opposite spin, which does not contribute to exchange energy. All in all, no difference in exchange energy, so no effect of the exchange energy in the ionization energy. To reiterate: it is not that the exchange energy is zero, it is the before vs after difference (so, the effect) which is zero.
Now let us see what happens with the Mn$^{2+}$ to Mn$^{3+}$ ionization. Here we are going from 5 electrons with the same spin to 4 electrons with the same spin,[1] so here we are losing exchange energy between the states before and after ionization, and therefore there this is something extra to pay in the ionization energy.
[1] For simplicity of the argument, we will be assuming "high-spin" configuration, as is implicit in the question.

probably the exchanges happening in the same orbital contribute to the exchange energy. if we say that an electron is moved from 3d to 4s orbital then it is said to be excited(Exchanged is not suitable to be used in this case).
so when you ask me to talk about the Fe2+ ion here is its electronic configuration
**1s2 2s2 2p6 3s2 3p6 3d6**
so in this case as for stability, the electron tends to excite to the 4s orbital and hence we call this the excitation of electron and not exchange.
I hope this would clear your doubt.
cheers!!
edit1: you may feel doubted why the electron excites from 3d to 4s so as it gets 3d5 configuration which makes it more stable and more exchanges contribute to more stability.


text:
Website or Software for Molecular Orbital Rotations
Does anyone know what software or website allows viewers to see rotation of orbitals in a molecule such as conrotatory and disrotatory directions? I'm not an expert in this field sorry. I tried VR Molecules and Chemsketch but they don't go deeper in the molecular orbitals. I am studying pericyclic reactions, that's why.

Do not spend too much time identifying a program to visualize this like a movie.  Use your two fists and observe the relative move (take the thumbs as an indicator).  The closest example identified for a somewhat step-wise illustration (but lacking the orbitals) is a German tutorial already translated into English here.


text:
Why do fats have an even number of carbon atoms in their fatty acids?
Can betaoxidation and unsaturated fats in any way be used in explaining why fatty acids in most cases have an even amount of carbon atoms?

Oxidation is the reverse of what you need. Fats are not built by oxidation. As for the unsaturated fats, I don't think they are related to this question.

@IvanNeretin the exact question is: "Explain in your own words why most fats have an even amount of carbon atoms in their fatty acids. You can assume that it involves betaoxidation and unsaturated fats."                     I cannot either make sense of it.

All right, beta oxidation might make sense if we assume that the fats are built by the same process in reverse, that is, by adding C2 fragments. (I don't remember if that's true, but sounds plausible.) Unsaturated fats still puzzle me.

It says "it involves betaoxidation", so why don't you add what you know about that to your question. I know an answer to the exact question, but don't know how to explain it in your own words.

@KarstenTheis how would you have answered the question?

https://chemistry.stackexchange.com/questions/64348/do-unsaturated-fatty-acids-with-an-odd-number-of-carbons-exist


text:
is boiling point a colligative property?
I know that elevation in boiling point is a colligative property .But is elevation in boiling point a colligative property ? Boiling point should be a colligative property acc to me bcz it depends only on number of solute particles but many people say its not a colligative property . 1m Urea and 1 m Sugar solution have same boiling point as solvent is same (water)

Could you please explain in detail what exactly you are asking? It is quite unclear as of now.

The colligative property is 'elevation in the boiling point.' as you have correctly pointed out. The key idea to me is that the colligative property is a comparision between a solution and the pure solvent.
However, the concept of 'boiling point' is not a colligative property. It is considered a 'physical property.' Other examples pf physical properties are 'freezing point,' 'density,' and 'viscosity.' The concept of 'boiling point' isn't a comparison in the same way a colligative property is.


text:
Strongest acid that can exist in water
Fundamentals of Biochemistry: Student Companion: Life at the Molecular Level[1] states that,

The strongest acid that can stably exist in aqueous solutions is $\ce{H3O+}$.

This is restated in the Wikipedia entry about hydronium. In principle this makes sense for an acid $\ce{HA}$ that is stronger than hydronium since the forward direction in the following equation
$$\ce{HA + H2O -> H3O+ + A^-}$$
is favored. However, by adjusting the initial concentration of $\ce{HA}$, for a given $K_a$, I can ensure that the final concentration of the acid is high enough that it exists in solution in reasonable quantities.
What am I missing?
Reference:
[1]: Voet, D.; Pratt, C. W.; Voet, J. G. Fundamentals of Biochemistry: Student Companion: Life at the Molecular Level, 3rd ed.; John Wiley & Sons: Chichester, England, 2009.

Are you suggesting that concentration of a weak acid is a measure of acid strength?

@EdV No, simply that even acids with large K_a's can exist in aqueous solutions given high initial concentrations

So what? Nitric acid is a strong acid and acetic acid is a weak acid. Put 1 mL of nitric acid in a liter of water and very few nitric acid molecules will be present: the acid is said to be leveled down to hydronium and nitrate. For the acetic acid, you can still have plenty of acetic acid molecules in solution: that is what it means to be a weak acid. So the book and wikipedia are correct.

And if you have something like 99% pure nitric acid, then of course there are nitric acid molecules present: there is not enough water to yield hydroniums. But this is obvious and not what your book or wikipedia meant.

Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers.

@EdV Okay, I think I know where my confusion is coming from. I was assuming that acid strength is based on the K_a of an acid. So I was arguing I can have stable acids with K_a's higher than Hydronium's, or 1, that can exist in reasonable quantities in solution. But it looks like the textbook is defining a strong acid as anything that gives up almost all its protons in solution. In which case a strong acid by definition cannot exist in solution, and therefore the strongest acid that can is the hydronium ion. Did I get that right?

You can have acid dissociation constants much greater than one, e.g., nitric acid. So if you add it to water, with water in considerable excess, then it gets leveled: you get hydronium. But concentrated strong acid, like 90% nitric acid, obviously cannot be all leveled: there is not enough water to accomplish that. The textbook is just tacitly assuming the acid solution is not concentrated.

However, by adjusting the initial concentration of $\ce{HA}$, for a given $_$, I can ensure that the final concentration of the acid is high enough that it exists in solution in reasonable quantities.

What am I missing?

I think you are missing that an assumption of acid-base equilibria in water is that the $K_a$ values you are using tend to assume an infinite dilution in water.  It's the background phase, and it is (at room temperature) 55 mol / L in concentration.  If you did the calculation you suggest -- i.e. to find the amount of e.g. nitric acid you'd need to add to an aqueous solution in order to ensure that the final concentration of undissociated $\ce{HNO3}$ molecules is large -- you'll probably find that you need to add a large amount of acid, on the order of mols per liter or more.
At that point you arguably aren't even in water any more.  Now maybe the real question is what's the pKa of water in a nitric acid solvent.  In any case the whole $K_a$ value you've looked up for nitric acid at infinite dilution in water isn't going to apply.

That last paragraph is indeed pretty important. What's missing from this discussion is an analysis of the activity coefficient of the hydronium ions. In dilute strong acid solutions, all hydronium ions behave similarly as the activity coefficient is close to one. However, at high concentrations, this coefficient can deviate from 1 by several orders of magnitude depending on a multitude of parameters. Essentially, the same hydronium ion can become remarkably more acidic, "meeting" the acidity of any undissociated strong acid.

This only goes so far though - sufficiently strong acids will overpower the acidity of even "strengthened" hydronium ions. These kinds of superstrong acids form isolable hydronium salts when mixed with water a 1:1 molar proportion.


text:
Understanding equivalence point
I read this passage:
“It is important to emphasize that, while a strong acid/strong base titration will have its equivalence point had a pH of seven, the equivalence point does not always occur at pH 7. when titrating polyprotic acids or bases, there are multiple equivalence points, as each acidic or basic conjugate species is titrated separately.”
I want to make sure I understand what is being said. Only strong acid-strong base titrations have an equivalence point at a pH of 7, if a weak acid and/or weak base is involved (either as a titrant or titrand), that is not the case. Also for polyprotic compounds (as titrands?), there are multiple equivalence points, therefore the titration of a strong polyprotic acid or base won’t ONLY have an equivalence point at pH 7 (although it is possible to have one of the multiple equivalence points be at 7 but that would be a property of the compound, ie one of the pKas would be equal to 7).

Phosphoric acid is a weak acid and none of the three equivalence points are at pH 7. Carbonic acid, i.e., carbon dioxide dissolved in water, is another weak acid. Its two equivalence point pH values are shown here. Neither is at pH 7.

Thank you for correcting me on H3PO4. However, I was giving it as an example of a strong polyprotic acid. When I said one of the pKas could be 7, I was talking about polyprotic acids in general, as in it is possible for a polyprotic acid to have a pKa value of 7 (among other pKa values)

To clarify I’m talking theoretically. If a polyprotic acid has one pKa value equal 7, that’s the only way for its titration to have an equivalence point at pH = 7. Is that correct?

You need to look at this and also follow the link in the comment provided there by @Mithoron. Sulfuric acid is a bit different than you might have hoped.

I feel like the inclusion of an example is what’s making my explanation bad, so I’ll just remove it. I’m not trying to get into specifics, I’m just trying to understand equivalence points in general. I’m asking this: are the only possible ways for the equivalence point to be at pH = 7 with the titration of a strong acid with a strong base and theoretically, if one of the pKas of a polyprotic acid or base is equal to 7?

To directly answer the question you state in your second comment, the answer is no. Titrating HCl (aq) versus NaOH (aq) yields saltwater at the equivalence point, so the pH is 7. For sulfuric acid versus NaOH, the first equivalence point yields  a solution of sodium hydrogen sulfate, which is acidic. The second equivalence point yields a solution of sodium sulfate, which is basic. No pH 7.

Let us continue this discussion in chat.


text:
Effect of inert gas blanket on oxygen solubilized in wine
If applied to a volume of wine, could a blanket of argon become dense enough to remove oxygen from the wine? This is not a question of sparging the wine. Rather, could a blanket of argon over the top of a wine remove dissolved oxygen from the volume? No external pressure is applied, temperature is 55 degrees F.

Partially, it can, as there is direct proportionality at equilibrium between oxygen concentration in wine and oxygen volume percentage in the gas above. But it will take time until the equilibrium is established, unless forced mechanically.

At ambient conditions, the density of argon is greater than the of air; thus argon may displace air (one reason it is used as a protecting gas in chemistry and welding).  But in a barrel, it will just sit on top of the surface of the liquid.  Solvents for chromatography (a purification method) are sometimes degassed by placing the 2L bottles in an ultrasound bath to remove gasses dissolved in these liquids.  (There equally are ultrasound heads to immerse into containers, too.) I do not know if / how much such a treatment with ultrasound would harm (especially the taste of) wine.

A layer of added argon at atmospheric pressure would be inert and would not significantly alter the solubility of oxygen in the wine. Any effect would not be due to direct interaction between the gases (so certainly not chemical).
If the headspace of a wine bottle is flushed with argon (displacing the air above the wine), the oxygen in the wine will re-equilibrate into the headspace volume, resulting in a decrease in the amount of oxygen solubilized in the wine. The effect can be significant, but surprisingly small. For instance, for a 750 mL wine bottle with a 5 mL headspace (ullage), the ratio of oxygen in the headspace and liquid is ~1:5:
$$\frac{n_\ce{O2(l)}}{n_\ce{O2(g)}}=\frac{RT}{K_\textrm{c}}\frac{V_\textrm{l}}{V_\textrm{g}} = 0.032 \frac{V_\textrm{l}}{V_\textrm{g}}$$
This means that flushing the headspace will reduce the total oxygen in the bottle by only ~17%.
Note also: the density of argon will depend on the pressure. Since the argon is not pressurized above atmospheric pressure, the maximum pressure of argon will be approximately atmospheric pressure when all of the air is flushed out of the headspace, and lower otherwise. The density of argon in the headspace will be proportional to the pressure and will be a maximum if all of the air is flushed away. The remaining volume of the headspace will be occupied by air (including of course oxygen).


text:
Why does α-ᴅ-glucose form at a lower temperature than β-ᴅ-glucose?
β-ᴅ-glucose is more stable than its α-anomer due to the fact that all the −OH groups occupy the equatorial position. If this is the case, then why does α-ᴅ-glucose predominate at 30 °C while β-ᴅ-glucose does the same at 98 °C?
Here is an excerpt from PubChem's page for D(+)-Glucose  :

Glucose is a reducing sugar, i.e. it reacts with oxidizing agents such as cupric hydroxide ... In water, ... below a temperature of ca 100 °C, the stable, crystalline form of D-glucose is the alpha-form, which crystallizes as a monohydrate below 50 °C, above  100 °C, the beta-anhydrous form, is most stable (Ref. 1).
Below 50 °C, alpha-d-glucose hydrate, above 50 °C anhydrous form is obtained and at higher temp beta-d-glucose is formed (Ref. 2).

References

Schenck FW; Glucose and Glucose-Containing Syrups. Ullmann's Encyclopedia of Industrial Chemistry. 7th ed. (1999-2016). New York, NY: John Wiley & Sons. Online Posting Date: 15 Dec 2006

O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 824

β-ᴅ-glucose is more stable than its α-anomer due to the fact that all the −OH groups occupy the equatorial position.

This is specifically in aqueous solution. In the solid, you also have to consider the intermolecular interactions between glucose molecules.

If this is the case, then why does α-ᴅ-glucose predominate at 30 °C while β-ᴅ-glucose does the same at 98 °C?

At both temperatures, there is a mixture of the two in aqueous solution. I could not find how the equilibrium constant changes with temperature.

Why does α-ᴅ-glucose form at a lower temperature than β-ᴅ-glucose?

When you crystallize glucose, the least soluble solid will form predominantly (under thermodynamic rather than kinetic control). Solubilities are temperature-dependent. This property is used in the preparation of the solid alpha form (as a monohydrate) and the beta form (anhydrous):

The crystalization and purification of β-D-glucose from a mixture containing α and β anomers of the sugar illustrates how different reaction conditions can change the outcome of a process, for example, through changes in the relative solubilities of different anomers at different temperatures. In most commercial preparations of crystalline D-glucose, the α anomer predominates because it is less soluble at room temperature. However, at higher temperatures α-D-glucose becomes considerably more soluble and it is the β-D-glucose that is selectively crystallized as it becomes dissociated from the polar solvent.
Source: Reyes-de-Corcuera, J. I.; Teruel, M. A.; Jenkins, D. M. Crystallization of β-ᴅ-Glucose and Analysis with a Simple Glucose Biosensor. J. Chem. Educ. 2009, 86 (8), 959. DOI: 10.1021/ed086p959.

Counter-intuitively, you purify these crystals by recrystallization at low temperature (where the alpha form is less soluble). This works because the sample contains much more beta form than alpha form, so the beta form will crystallize. The low temperature prevents fast mutarotation in the aqueous state. The procedure is described below:

One hundred grams of the sugar are poured into 100 cc. of water at $0^\circ$. The $\beta$ form of glucose is very soluble and with vigorous stirring all dissolves in a few seconds. This solution is quickly filtered and 500 cc. of absolute alcohol are mixed with the filtrate. Seeding and stirring causes an immediate and rapid crystallization of $\beta$-glucose. By this method $\beta$-glucose has been recrystallized several times in succession and the specific rotation determined after each crystallization, the rotation being observed within one minute of the time when the sugar was dissolved in water at $0^\circ$.
\begin{array}{}
&\text{First experiment} &\text{Second experiment}\\
\mathrm{Original\,\beta - glucose} &+24.5^\circ & + 27^\circ \\
\text{After first recrystallization} &+ 21^\circ & + 21^\circ \\
\text{After second recrystallization} &+ 19.8^\circ & + 19.8^\circ \\
\text{After third recrystallization} &+ 19,5^\circ & + 19.9^\circ \\
\text{After fourth recrystallization} &+ 19.7^\circ & + 19.2^\circ \\
\end{array}
It appears that after the second recrystallization the $\beta$-glucose is free from the $\alpha$-isomer. In each of these experiments the initial specific rotation of the $\beta$-glucose which resulted from the fourth recrystallization has been determined by observing the progress of the mutarotation in water at $0.7^\circ$, where the rate is slow, and extrapolating the vales to the time of dissolving the sugar. By this method the initial specific rotation of $\beta$-glucose has been found to be, in the first experiment $+19.2^\circ$ and in the second one $+18.8^\circ$, giving an average of $+19.0^\circ$. This value agrees completely with that which C.Tanret records in his lates measurement. Roux found it to be $+19.8^\circ$.
Source: Hudson, C. S.; Dale, J. K. Studies on the forms of ᴅ-glucose and their mutarotation. J. Am. Chem. Soc. 1917, 39 (2), 320–328. DOI: 10.1021/ja02247a017.


text:
How is the molar weight of C-12 equal to 12 gram ? Why is "g" not considered while calculating molar weight?
So, I was just pretty confused about how we simply multiply the "mass" of total nucleons in Carbon by the Avogadro Constant, to get the molar "weight" as 12 grams, shouldn't it be the molar "mass", as we have not considered the effect of g anywhere ?

Technically, it is mass.  But, since the calculations aren't changed by using weight instead of mass (g, the acceleration due to gravity, is just a constant multiplier), "molecular weight" and "molecular mass" are often used interchangeably, where it's understood that "weight" means "weight under earth gravity".   But, yes, using "mass" would be more technically precise.  Essentially, it's no different from laboratory scales expressing their weight readouts as a mass (typically grams), when what they are really measuring is a force, which should instead be expressed in newtons.

2 g hydrogen atoms (H):  $\frac{2 g}{1 g/mol}= 2, mol$.   2 g hydrogen gas ($\ce{H_2(g)}$):   $\frac{2 g}{2 g/mol}= 1 ,mol$.

You are confused by the usage of the words, mass and weight. Try searching for their definitions.

This is a linguistic issue rather than physics or chemistry. In older literature of chemistry texts you will find that atomic weights was more frequently used, so the terms atomic mass and weights were interchangeably used by various authors. Recall that an everyday luxury called electricity was not available even to scientists  100 or 300 years ago. "Masses" of objects were determined by sensitive mechanical balances like these until the 1960-70s. On one side you had the object whose "mass" you were interested in and on the other side, a known weight was placed. You were basically comparing the weights (mass x acceleration due to gravity "g"), however, since "g" is same on both arms, it is mathematically cancelled, and we get mass of the object.

The International Union of Pure and Applied Chemistry has a committee called "Commission on Isotopic Abundances and Atomic Weights."
See IUPAC's atomic weights
Another minor correction in your post is that 2+2 is not equal to 4 at sub-atomic levels. The mass of 1 mol of C-12 atoms (not the sum of nucleons) is assigned by humans as 12 grams exactly. Recent revisions have slightly changed this value, but keep in mind that the sum of the masses of 1 mole of total nucleons of C-12 atoms does not sum up to 12.

Oh ok sir, but I still have some unanswered question. If I go by this explanation, I can use the term mass and weight interchangeably because they will "cancel out". So, I calculated say, that 1 mole of carbon dioxide should weigh 44 grams. I go to my laboratory and measure the gas, and come up with 440 grams for a certain amount of the gas, and hence write that I have 10 moles of the gas, or 10 times avogadro number of molecules. But - the reality is that "by mass" I still have just 1 mole and only avogadro times molecules !

Please clarify the point which is still not clear by adding a new edit in your original post.

If I go by this explanation, I can use the term mass and weight interchangeably because they will "cancel out". So, I calculated say, that 1 mole of carbon dioxide should weigh 44 grams. I go to my laboratory and measure the gas, and come up with 440 grams for a certain amount of the gas, and hence write that I have 10 moles of the gas, or 10 times avogadro number of molecules. But - the reality is that "by mass" I still have just 1 mole and only avogadro times molecules ! @M.Farooq

Abigail, Yes, mass and weight are generally used interchangeably but mass is what you get on balances not weight = mg. There is no role of "g" in the numbers you read on a balance.


text:
When is Zaitsev product formed in pyrolytic elimination reaction?
I recently came across a problem of choosing which product to be made in case of pyrolytic elimination reactions. I was taught that usually Hoffman products are formed in such reactions. For e.g.:

In this reaction, I sort of got the chills in my spine to make Zaitsev product because of the aromaticity obtained in the product but of course I obeyed the Hoffman rule that was taught to me and made the open ring structure of the 5 member ring of nitrogen connected at bottom. The answer is of course what I have illustrated but I need to understand when do we decide these. Another example is:

Can someone please explain how do I go about this i.e. how do I know when it is Hoffman and when it is Zaitsev?

I think you should also mention the temperature at which this happens. Cope elimination gives hoffman around $100^o C$ also relevant https://en.wikipedia.org/wiki/Ei_mechanism#General_Features the second example is a zaitsev product. There is clearly an $\alpha-H$ benefit threshold for the preferred product which is tipped by temperature

I think cope reactions exept xanthate esters occur at high temp

OK, but the 2 examples in the middle were in a research paper right? what was the temperature for those? they are definitely cope.

Yeah but we usually don't decide these products on basis of alpha hydrogens I have seen reactions that give hoffman even at 450°C  so I think temp is kot an issue

@davacd Kindly do mention the source of images. Thanks

@davacd yes xanthate esters require higher temperature for hofmann but they involve different leaving group so the balancing of thermodynamic and kinetic factors occurs in a different temperature range. Cope just happens to be in the 100 degrees range

Actually I got the images from a site you might know it , it's called master organic chemistry. Here's a link to the site. https://www.masterorganicchemistry.com/2018/09/24/the-cope-elimination/

@davacd On a general note, Hoffmann reaction product might be obtained if the leaving group is bulky enough.

It's at the bottom of the page when he menstions that all reactions arnt hoffman

@Rishi exactly they are in general obtained like that but I was lucky to find out a same example(that our teacher gave us) on a website on the internet furthur I will check in the research papers for this

@davacd it is not a cut and dry case. Hofmann isn't always obtained. Hofmann is only obtained in temperature ranges which vary case-to-case. But in context of competitive exams you should choose hofmann regardless unless +R effect is present. As they more often than not just want you to recognize the reaction and may not mention temperature.

@napstablook Okay and by +R you meant Resonance effect right and if -M then even better also can you check my comments on your ans bro thankyou

Cope elimination is a pericyclic reaction that always prefers syn-elimination. Due to hydroxylamine being a poor leaving group, the reactions often tend to prefer hofmann products.

However the reason for preference towards hofmann products is the stability of carbanion formed due to less $\ce{\alpha-H}$ and consequently less destabilization by +H effect. (since leaving group is poor, the acidic-H departs before the leaving group departs inducing negative charge at the $\ce{\beta-C}$)

The carbanion's stability is enhanced by the resonance effect in your first and last examples, and we know that resonance effect is often more stabilizing than hyperconjugation hence the preference of zaitsev product. Resonance effect almost always gets preference even if the reaction often produces hoffmann product. (since the leaving group can always be more bulkier there are edge cases)

Regarding your 2nd and 3rd examples they are probably giving zaitsev out of the temperature range for hofmann product in cope elimination. Where resonance effect is not in play, the determination between hofmann and zaitsev becomes a balance between the thermodynamic stability of the final product and the kinetic stability of the carbanion transition state.

All of the above examples at very high temperature will provide zaitsev product as thermodynamically favored product is formed. However in a certain temperature range these reactions give the hofmann product where the kinetic stability is dominating. Cope elimination produces hofmann product near $ \ce{T \approx 100^o C}$ pyrolysis of xanthates produces hofmann near $\ce{200^o C}$ and esters give pyrolytic syn elimination near $\ce{500^o C}$.*

The conclusion is that unless temperature is given, it is not exactly clear that you need to produce hofmann which is also stated in wikipedia

There are many factors that affect the product composition of Ei reactions, but typically they follow Hofmann’s rule and lose a β-hydrogen from the least substituted position, giving the alkene that is less substituted (the opposite of Zaitsev's rule).1 Some factors affecting product composition include steric effects, conjugation, and stability of the forming alkene.

The example of 5 member T.S. in wikipedia also supports zaitsev products. Basically the selectivity is not very good, and for correct answer we can only refer to data given to us. This is why temperature is important to mention.
*Source: Peter Sykes 6th edition section 9.9 (Pyrolytic syn elimination)

Thankyou for that informative answer, so to sum it up at higher temp we will obviously see thermodynamic stability before finalizing product and also if In an exam say temp is not mentioned for some reason then I think we should take decesion on the basis of resonance>hyperconj and that it should be minium as instability caused due to it(while simultaneously respecting the peri planar config needed) and also check for -I effect for stabalization of carbanaion

Can you confirm that in the case temp is not given then which should I give more prefferce inductive effect or less hyperconjugatuion? For eg if some withdrawing groups like halogens are there and there is also a choise of least hyperconjugatuion so which should I choose

Also can you please check that I have correctly understood your point while summing my thing up

@davacd from a purely exam point of view if temperature is not mentioned you should choose hofmann product and zaitsev if there is resonance to be had. If hyperconjugation effect is same for  $\beta-H$ options then go for checking -I effect. You shouldn't worry about inductive effect dominating hyperconjugative effect, that decision can only be made with the relevant data. So rigidly follow the order M>H>I .


text:
Equilibrium and endergonic reaction
If a reaction A <-> B has a value of dG°>0, then I know that A -> B is endergonic while A <- B is exergonic.
Now if I start with 100% B I could see how equilibrium is reached, B reacts to A until it reaches the stable situation (for whatever temperature/pressure we're in).
If I start with 100% A however, I would say nothing would happen, but if that's the case then this situation is already stable, so how can the equilibrium still be the same as the previous situation (since concentrations don't alter K)?
And even then, I know that that's not how it works (with only the exergonic direction working until equilibrium), because the equilibrium is reached when both reaction rates are equal, and that means that the endergonic reaction does indeed work at all time. Even in the first case in fact there's A reacting to B all the time, even though it's endergonic.
So I guess my question ultimately is: where does the energy come from?
(Silly thought: it doesn't take the energy from the exergonic way to "power" the endergonic way right? That would make every reaction be a net 0, and would also block the situation with 100% A at the start)

“If I start with 100% A however, I would say nothing would happen” - This is not correct. A will react to form B until the same equilibrium is reached.

This is explained in any textbook on physical chemistry or thermodynamics; but what you need to look up is the relationship between $\Delta_r G$ and $\Delta_r G^\circ$. The former indicates the direction in which any particular system will go (if it is positive then the reaction goes backwards; negative then forwards; and zero indicates equilibrium); it is applicable to any system whether in chemical equilibrium or not. The latter only indicates the final position of equilibrium; it does not tell you anything about systems that are not in equilibrium, or how they will react.

So the value of dG° doesn't have any "intuitive" meaning right?
Looking at dG, if I'm at 100% A then dG will be negative and the reaction goes A->B, correct?
Still, I'm left with the question of how the backwards reaction happens if it's unfavourable, even if A goes to B with negative dG, even at 99% A there's still some B going to A, until the rates of reaction are the same. How can that happen?

$\Delta_r G$ doesn’t say anything about rates of reaction or whether they both occur; it only tells you about the net change. Both reactions will always occur, the equilibrium is not dictated by which one occurs, but rather which one is faster. Note that the rates are also a function of concentration so if there is very little B to begin with, then the reverse reaction will have a very tiny rate, and the forward reaction will predominate. This precisely corresponds to the case of $\Delta_r G < 0$ which predicts that the net reaction will be in the forward direction.

So, and sorry for the insistence, but what does that mean at a single-molecule level? Does G have any meaning when talking about one molecule reacting? And if yes, how does a single molecule react "backwards" if it needs energy to do so?

No apology needed. I’m not sure that thermodynamics as a whole is suitable for the single molecule level. You need a different framework for that, probably QM, or specifically quantum dynamics.

With a single  molecule its either A or B, so you would need to count the number of times / sec either its A or its B to get the equilibrium constant. Thermodynamics works only for ensembles of molecules but if you measure a single molecule many times over you can approximate this.

Right, but if by going from A to B you require more net energy than you release (by bond breaking/building), where does the energy come from then? I thought that an endergonic reaction meant, on a single-molecule level, that the new molecule would be higher energy than the previous one. So by going from A to B we're going on a higher energy level, but there's no external energy input, and yet it happens continuously throughout the reaction... that's what I don't get.

Your molecules are not isolated, but in a solvent or on a surface so thermal energy f can occasionally provide enough energy to cause the reaction. The energy is 'given back' when the reaction returns the other way.


text:
When does bimolecular nucleophilic substitution reaction take place in phenyl group containing compounds?
We were studying SN Mechanism, and our professor gave us a question to identify the SN, mechanisms which would happen in the given compound.
They were $\ce{CH3Cl}$, $\ce{PhCH2Cl}$, $\ce{Ph2CHCl}$, $\ce{Ph3CCl}$.
So while attempting the given problem, I chose SN1 for both $\ce{Ph2CHCl}$ & $\ce{Ph3CCl}$,as they would face a lot of hindrance. But our professor told us that they both can also perform SN2 mechanism. Why would they both perform SN2 mechanism despite there being a lot of hindrance present?
My Reasoning: Although $\ce{PhCH2+}$ would be resonance stabilized but the hindrance present is too much for SN2 to proceed.

Your own reasoning, based on search and analysis, is supposed to be present to avoid the question closure for lack of own effort.

Honestly, it’s completely ridiculous to suggest that trityl chloride could undergo SN2, and also very disappointing to see such things being taught as if they were true. I’d wager that diphenylmethyl chloride also almost certainly favours SN1. With secondary halides it tends to be more nuanced, but here the groups are pretty bulky and the SN1 carbocation intermediate is pretty stable.

That line of logic is pretty useless. These molecules may also fall apart randomly into three phenyl radicals, a chlorine atom, and a carbon atom. The chances of that happening are basically zero as well. One could insist that that’s possible, which may well be true, but you wouldn’t learn anything useful from that assertion. Insisting that trityl chloride can undergo SN2 is technically correct, but pointless.

Any reaction capable of undergoing in two or more mechanisms has major and minor pathways. In your question the major pathway is obviously SN1 due to high steric hindrance and the resonance-stabilized carbocation, and the minor pathway is SN2. Since, your professor said they can ALSO perform SN2 mechanism, I think he was trying to say that the reaction doesn't only react via the SN1 pathway, but also via the minor SN2 pathway. However, the major pathway is definitely SN1 due to high steric hindrance and resonance stabilized carbocation.


text:
What is the IUPAC name for this compound containing a phenyl group attached to it?
What is the IUPAC name for the compound $\ce{(CH3)3CCH2CH(Br)C6H5}$?
The answer I got myself was (1-Bromo-3,3-dimethylbutyl)benzene, but from looking online most sources list the name as 1-Bromo-3,3-dimethyl-1-phenylbutane, even though the benzene group is supposed to take precedence, right? Why does butane form the suffix and not benzene?

Please note that names of chemicals (the 'bromo' in your suggested names) should not be capitalised unless occurring at the beginning of a sentence.

The compound whose name is to be determined is the following.

As you can see, we have two options for the parent hydride.

The butane chain which has 4 substituents - two methyl groups, one bromine atom, and one phenyl group.
The benzene ring which has one substituent.

Now, from Section P-52.2.8 of the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, IUPAC Blue book, we have,

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Therefore, we need to choose option for the preferred IUPAC name, therefore we now need to find the one substituent attached to the benzene ring.

In order to find the right numbering of locants, we use Section P-14.

Numbering
When several structural features appear in cyclic and acylic compounds, low locants are assigned to them in the following decreasing order of seniority.
(c) principal characteristic groups and free valences (suffixes);
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Furthermore,

The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference (…)

Therefore in this case, we have two possible locant sets:

4-bromo-2,2-dimethylbut-4-yl [2,2,4]
1-bromo-3,3-dimethylbut-1-yl [1,1,3]

We can remove the '-1-' in option 2 using section P-29.2, which states:

P-29.2 GENERAL METHODOLOGY FOR NAMING SUBSTITUENT GROUPS
[...]
(1) The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This
method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups and for the mononuclear hydrides of silicon, germanium, tin, and lead. Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

By section P-14.4 (c), option 2 is preferred, therefore the PIN of the compound becomes (1-bromo-3,3-dimethylbutyl)benzene

OK. One last thing, and sorry that I'm drip-feeding like this, but I notice it as we go along. The choice between butan-1-yl and butan-4-yl (note 'butan' not 'but', cf. P-29.2 again) is made not by locant set but rather by P-14.4(c), which comes before (f) and (g) and states that free valences (i.e. -yl) have priority for lower numbers. Again, I should probably have collated all these; sorry.

Hmm, in fact, I wonder if P-14.4 even comes into this at all; because by virtue of P-29.2/P-57.1.1.1 we already know the preferred prefix is going to be something-butyl. I get the feeling that P-14.4 is for deciding between butan-2-yl and butan-3-yl, for example. Ah, it does get pretty complicated sometimes.....

I would digress here, since I would say it's because of P-14.4(c) (now noticing) that the selection takes place and not via the set of locants rule.

@Safdar Faisal Thanks for answering my question! By the way, where did you get all those diagrams you used in your answer from?

I used ChemdrawJS. Another alternative is Chemdoodle @Shud. Consider hitting the green check mark if the question is answered.


text:
Why there is not a consistent pattern between the carbon chain length and the melting point of fatty acids?
I looked at the melting temperature of different saturated fatty acids and I expected to see that with a longer chain of carbon the melting point would be higher as after all the var der Waals interactions would be stronger, yet I found there is not a consistent pattern between the number of carbons in the carbon chain and the melting point:
C3H6O2: Melting point: −20.5 °C
C4H8O2: Melting point: −5.1 °C
C5H10O2: Melting point: −34.5 °C
C6H12O2: Melting point: −3.4 °C
C7H14O2: Melting point: −7.5 °C
C8H16O2: Melting point: 16.7 °C
C9H18O2: Melting point: 12.5 °C
Does anyone have an explanation for this?

You could have added acetic acid $\ce{C2H4O2}$, melting point +$16.6$°C.  This strange behavior must be related to the crystallographic structure of these acids in the solid state.

And why you neglect hydrogen bonding? It doesn't simply add to other interactions, but interferes.

Based on the formulas, those are saturated fatty acids, not unsaturated

@Andrew You are correct


text:
Why is Freundlich adsorption isotherm not valid at high pressures?
I have taken the following from Wikipedia :

Limitation of Freundlich adsorption isotherm :
Experimentally it was determined that extent of gas adsorption varies directly with pressure, and then it directly varies with pressure raised to the power $\frac{1}{n}$ until saturation pressure $P_{s}$ is reached. Beyond that point, the rate of adsorption saturates even after applying higher pressure. Thus, the Freundlich adsorption isotherm fails at higher pressure.

I have a question. Here is the Freundlich adsorption equation :
$\frac{x}{m} = Kp^{\frac{1}{n}}$
Now at very high pressures, we may put $\frac{1}{n} = 0$.  Then we have,
$\frac{x}{m} = Kp^{0}$
$\frac{x}{m} = K(1)$
$\frac{x}{m} = K$
which means that $\frac{x}{m}$ is a constant which means that the graph between $\frac{x}{m}$ and $p$ will be a straight line. This exactly matches with the experimental graph. As you can see in the image below, at very high pressures the graph is indeed a straight line and Freundlich adsorption equation beautifully explains it. Perfect! So, what is the problem here? Why in Wikipedia and even in my chemistry textbook it is written that Freundlich adsorption isotherm is not valid at high pressures? I just proved above that Freundlich adsorption equation beautifully explains that experimental graph at high pressures. Then why is it said that Freundlich adsorption isotherm is not valid at high pressures?

$n$ is meant to be independent on pressure, otherwise the equation has no meaning at all.

In part because monolayer coverage is a baked-in assumption. At high pressure, you start to get island formation (at least) and this isotherm is not valid.

No, @Nilay Ghosh, the answers in that question do not satisfy me.

But @Todd Minehardt, I just proved that the graph indicated by Freundlich adsorption equation at high pressures is a straight line which exactly matches with the experimental graph. So Freundlich adsorption equation is indeed valid at high pressures, right? I don't see what is the problem here.


text:
How many conformations are possible for ethane?
I am talking on the basis of observations made on the Newman projection of ethane.
According, to the various sources, that I found on many websites, the number of conformational isomers are 2. They are the gauche form and the staggered form, but according to the book I have there are three conformations. They are accounting for the inverted gauche form separately.
I am quite doubtful about it. By inverted gauche, I am talking about the gauche structure formed, by rotating the Newman projection structure by 180° degrees, along the sigma bond.

Ethane does not have a gauche form. The form which is neither skew nor staggered is called eclipsed.

How many conformations are possible for ethane?

It's the same like asking, "How many angles are possible between 0 and 2π?"
The answer is infinite!
Every slightly change in the rotational angle yields a new conformer. It's only makes sense to ask how many special conformers are possible for ethane. Special in the sense of them being recognised by a specific name.
Two types of conformers are named in case of ethane; staggered and eclipsed. Gauche form is not defined for ethane. You need unsymmetric substitutes on the carbon on either side of the bond being studied, like in propane.
The staggered conformer:

The eclipsed conformer:

Now in a full 2π rotation you will encounter both the staggered and the eclipsed form 3 times each, alternating every π/3 radians. But as all the substituents are hydrogen in ethane's case, all staggered and eclipsed form are equivalent respectively.
An excerpt from Conformations of Ethane[1]:

This process can be continued all around the 360°circle, with three possible eclipsed conformations and three staggered conformations, in addition to an infinite number of variations in between.

So to answer your question:
There are infinite conformers of ethane, out of which two are named and each of the named conformer is encountered three times during a full 2π rotation.

Reference:
(1) Conformations of Ethane
https://chem.libretexts.org/@go/page/67072 (accessed Jun 26, 2021).

yea I thought so too, but the answer provided is 3 ..that was the reason for my confusion...anyways thanks.. but if anyone has the reason for having three different conformations for ethane.. please do suggest

@sujayvivek Are you sure that ethane was the molecule in question.


text:
Formation of Aniline from Chlorobenzene
My textbook says:

When chlorobenzene is treated with aqueous ammonia at 473K under a pressure of 60 atmospheres in presence of cuprous oxide or cuprous chloride, aniline is formed.

Notice the reaction given. It shows Cuprous chloride ($\ce{Cu2Cl2}$) as a by-product. The question is, if cuprous chloride is used for treating chlorobenzene initially, the whole equilibrium should shift backwards, that is, the formation of aniline shouldn't be favourable, because $\ce{Cu2Cl2}$ is formed in the process.
Is my textbook wrong here in giving the statement? Or am I missing something?

Your textbook's text is correct but the reaction image is incomplete. Reaction of chlorobenzene with ammonia in presence of both the cuprous chloride and cuprous oxide forms aniline but via two different reactions. Your textbook only mentions one of the reactions in the image.
In the reaction involving the oxide, cuprous oxide is a reactant and is consumed during the reaction whereas in the case of the second reaction, cuprous chloride acts as a catalyst and isn't used up during the reaction. The mention of the second reaction can be found in Industrial Organic Chemistry, Klaus Weissermel and Hans-Jurgen Arpe[1], where the following reaction is mentioned,
$$\ce{Ph-Cl + 2NH3 + CuCl -> Ph-NH2 + NH4Cl}$$
The same reaction can also be found on The chemical thesaurus reaction chemistry database[2].

References:
(1) Rose, J. Industrial Organic Chemistry Klaus Weissermel and Hans-Jurgen Arpe 3rd Edn. VCH, Weinheim, 1997 Xvii + 464 Pages. £70 ISBN 3-527-28838-4. Appl. Organomet. Chem. 1999, 13 (11), 857–858.
(2) Leach, M. R. The chemical thesaurus reaction chemistry database
http://www.chemthes.com/rxn_dp.php?id=3489 (accessed Jun 25, 2021).


text:
Why is the oxidation state of Te in H2Te equal to -2?
Hydrogen has electronegativity of 2.2 whereas Telenium has electronegativity equal to 2.1. So since hydrogen is more electronegative than Telenium, as per what I have read, oxidation state of Te should be +2 in H2Te.
But actually it is -2, so where is my reasoning wrong?

Electronegativity doesn't work in this manner for elements lower than second period.

Oxidation states are merely a human convention.

@Nisarg could you cite the source of your answer or explain why it doesn't work for elements of period 3 and higher? thanks...

Electronegativity is not a constant, but is compound specific. It depends e.g. on the oxidation state.
There are multiple methods of electronegativity calculation, where some elements may switch the order.
The most used Pauling electronegativity is explicitly based on energies of particular bonds, but for small differences it is ambiguous about the difference sign.

The difference in electronegativity between atoms A and B is given by:
$$|\chi_{\rm A} - \chi_{\rm B}| = ({\rm eV})^{-1/2} \sqrt{E_{\rm d}({\rm AB}) - \frac{E_{\rm d}({\rm AA}) + E_{\rm d}({\rm BB})} 2}$$
where the Bond dissociation energy $E_\mathrm{d}$ of the A–B, A–A and B–B bonds are expressed in $\mathrm{eV}$, the factor $(eV)^{- \frac 12}$ being included to ensure a dimensionless result. Hence, the difference in Pauling electronegativity between hydrogen and bromine is 0.73 (dissociation energies: H–Br, 3.79 eV; H–H, 4.52 eV; Br–Br 2.00 eV)

$\ce{H2Te}$ is relatively strong acid with $\mathrm{p}K_\mathrm{a} = 2.6$, what gives a hint that in context of water $\ce{H2Te}$ dissociation, $\ce{Te}$ has higher electronegativity than $\ce{H}$. If the opposite had been the case, $\ce{HTe+}$ and $\ce{H- }$ would have been created by the dissociation, with following up $\ce{H- + H2O -> H2 + OH-}$.
Aside of that, oxidation state (OS) is a formal quantity. As a formal one, it is advantageous to keep the same OS for the analogical compounds along the group. For some , typically mostly non metal ones, hydrogen is kept at +1, for some others, like electropositive metals, as -1.
Another moot case is $\ce{PH3}$ and down the group. OS of these elements is conventionally kept as -3 here.
Be aware the OS is the tool to simplify things like compound categorization and redox equation enumeration. Keeping the strict  rule of universal electronegativity for OS would complicate things by making outlying exceptions.


text:
Acetal/ketal formation and deprotection
During acetal/ketal formation and deprotection, the same thing is used essentially which is acid and some sort of alcohol like ethylene glycol. And once deprotection occurs by the use of acid, ethylene glycol would be in solution so it would be essentially the starting material for the protection of a ketone or aldehyde. Is there some sort of external factors that favour acetal/ketal formation and others that favour deprotection or am I missing something here?

very similar to https://chemistry.stackexchange.com/questions/10722/what-makes-c-o-more-stable-that-coh%e2%82%82

I don't think it is the same because my question is about protection and deprotection, not essential about the bond stability of the two products of ketone/aldehyde and acetal/ketal

acidic medium promotes "deprotection" and basic medium vice versa. This is because H+ in acidic medium makes one of the oxygen linkage a good leaving group enabling the other oxygen to remove it since the carbocation is stable.

But aren't they both done in acidic medium?

no only "deprotection" is done in acidic medium. acetals are pretty unstable in acidic medium.

Wait then why do most mechanisms involve H3O+ in acetal formation.

The protection is done in anhydrous acidic media often with removal of water (D-S reflux). The deprotection is done in aqueous so that the molecules of alcohol/glycol that are removed are vastly outnumbered by the water present and there is very little chance of the acetal reforming.


text:
Are sudden adiabatic expansions reversible or irreversible? (Tyre burst problems)
Encountered Question:

The tube of the tyre of a car is filled with air at $\pu{27^\circ C}$ and  $\pu{2 atm}$ pressure. If the tube suddenly bursts, the final temperature of the air will be ($γ_\textrm{air}=1.5$) ____.

Approach here:
Given:
$T_1 = \pu{27 K}+ \pu{273K}=\pu{300K}$ ,$P_ 1 = \pu{2atm}$
in the process
$T_1^γP_1^{γ−1} = T_2^γP_2^{γ−1}$
gives
$T_2 = (2^{-1/3}) \pu{300 K}= \pu{240 K}$

Q: The pressure inside a tyre is 4atm at 27C. If the tyre bursts suddenly, the new temperature will be $(\gamma = \frac75)$_____.
Ans: $300(4)^{-2/7}$
Source

The above approach is correct if we assume reversible process but as bursting is very fast we should assume it is an irreversible process, hence $\Delta U = W_\textrm{on system}$ gives
$\frac{f}{2}nR(T_2-T_1) = -P_{ext}(V_2-V_1)$
Putting values gives $T=\pu{250 K}$.
So, why do we assume tyre bursting is a reversible process OR are above sources incorrect?

Yes you are right, bursting of tire should be considered irreversible.

Yes, the sources aren't correct. The first source is actually quite faulty and unreliable and in the second source it seems by seeing the webpage that the source is a website of physics (targeting IITJEE), also, in physics that is asked in IITJEE, every thermodynamic process is taken to be reversible one , that is why they assumed the process to be reversible adiabatic . But as @NisargBhavsar pointed out, the process should be irreversible adiabatic.

Thanks I will add answer soon and feel free to answer if you have better possible answer.

The adjective sudden does not fit well the noun equilibrium, which is a required state along paths of reversible processes. By other words, reversible system processes are those, where a closed cycle of system states can by achieved by the closed cycle of the environment state, so the closed path $\Delta S_tot=0$. ( Not to be confused with other meaning of reversibility as 2 way process.). In the first sense it is the opposite to what @ecneics said: Truly reversible process does not exist, as intermediate process states are not in true equilibrium, unless the process takes infinite time.

The second approach actually assumes that the expansion is irreversible, against the constant pressure of the surrounding atmosphere, and gets the right answer.

Dear Jay, please write the question in text here because it may happen in the future that the site linked get deleted.

Pl see this post @Jay

Conclusion from comments:

Sources are incorrect.
Second approach is correct assuming expansion is irreversible against the constant pressure of the surrounding atmosphere and hence gets the right answer.


text:
does entropy of half cell reaction change with stoichiometry
the standard potential of the following cell is 0.23V at 288K and 0.21V at 308K.
Pt|H2|HCl(aq)||AgCl|Ag
calculate $\Delta{S}$ .
I know that $\Delta{S}=nF(\frac{\Delta{E}}{\Delta{T}})$ here n is the number of moles of electrons exchanged in the reaction, but doesn't the number of moles of electrons exchanged depend on the stoichiometry of the reaction for example
$$ H_2+2AgCl\rightarrow2 Ag+2HCl$$
here n is 2
but if I write the reaction as
$\frac{1}{2}H_ 2+AgCl\rightarrow$ Ag + HCl
here n is 1 so wouldnt the entropy change with stoichiometry

Yes, it would change. $\Delta S$ should be calculated for the reaction as written.


text:
Units conversion of Henry coefficient in adsorption on solids
In a database of solid microporous materials I have the Henry coefficient in units of $\pu{mol kg^{-1} Pa^{-1}}.$ How to convert to $\pu{Pa^{-1}}?$
I tried to multiply by the molar mass of the sorbate (in my case $\ce{H2O})$ expressed in $\pu{kg}$ (so $\pu{18.02E-3 kg mol^{-1}}),$ but it turns out the same result of having a gas as solute and the water as solvent according to Sander [1].
Reference

Sander, R. Compilation of Henry’s Law Constants (Version 4.0) for Water as Solvent. Atmos. Chem. Phys. 2015, 15 (8), 4399–4981. DOI: 10.5194/acp-15-4399-2015. (PDF)

You wrote "How to convert in Pa$^{−1}$?" To what units?

Sorry, I mean "how to convert $\pu{mol kg^{-1} Pa^{-1}}$ to $\pu{Pa^{-1}}$?"

You cannot. How will you eliminate mol and kg?

What is your initial unit of Henry's constant and what is your desired unit?

I think you are confusing adsorbent and adsorbate (which you call "sorbate"). If water is adsorbing into the solid it is the adsorbate. The Henry's law constant allows you to convert between vapor pressure of adsorbate (water) and its concentration (as moles of water per kg of microporous material) in the adorbent (microporous material).


text:
Does ammonia exhibit hydrogen bonding in its gaseous state?
I was reading the properties of ammonia in my book (NCERT Chemistry Part I, Textbook for Class XII)
where it was written :

In the solid and liquid states, it (ammonia) is associated through hydrogen bonds as in the case
of water and that accounts for its higher melting and boiling points
than expected on the basis of its molecular mass.

My doubt is whether or not ammonia exhibit hydrogen bonding in the gaseous state.
Although I found a paper from 1987 which says that ammonia exhibits hydrogen bonding as an acceptor but not as a donor in its gaseous state, but it is 2021 so is there any improvement on the above statement?

Here is a 2020 paper: https://pubs.acs.org/doi/10.1021/acsomega.0c04274

just a fact that supports h-bonding in gaseous state:- HF has H-bonding that cyclises 3 HF molecules in a benzene like manner.

The paper[1] referenced by Nilay Ghosh says that,

it is well known that hydrogen bonds commonly exist in ammonia clusters and play an important role

and gives a picture of some of the clusters that were assumed to exist. So there seems to be evidence of significant covalent hydrogen bonding in possible ammonia clusters, at least from the calculations.
Calculated ammonia clusters:

Whereas the reference[2] given by the OP was less positive:

Surprisingly, no evidence has been found to support the view that $\ce{NH3}$ acts as a proton donor through hydrogen bonding,

but does describe ammonia as a powerful hydrogen-bond acceptor.
There is paper[3] which describes a hydrogen-bonded dimer that demonstrates how well ammonia can hydrogen-bond as an electron donor.
Spectrum of hydrogen-bonded dimer of ammonia:

But the question remains: "If ammonia is as basic, relative to water, as acetic acid is acidic, why the very different behavior?" According to a journal article[4], acetic acid is known to be significantly dimeric in the vapor phase: 50% of the vapor is dimer at $\pu{1 atm}$ and $\pu{400 K}$.
Compressibility factor vs Temperature for saturated acetic acid vapours:

One simple way to look at the difference is that according to chem.ucla.edu[5], hydrogen has an electronegativity of 2.1, nitrogen 3.0, and oxygen 3.5. The difference between the electronegativities of $\ce{H}$ and $\ce{N}$ generates a small protonic acidity. According to Chem LibreTexts[6] $\mathrm{p}K_\mathrm{a}$ of $\ce{NH3}$ is 34. Very small!
The difference in electronegativities between $\ce{H}$ and $\ce{O}$ is larger, and gives, for hydroxy compounds like $\ce{H2O}$ and $\ce{CH3OH}$, $\mathrm{p}K_\mathrm{a}$ values which are in something on the order of 14-15. So, while both $\ce{N}$ and $\ce{O}$ have lone pairs that are electron-rich, the polarity of the protons on nitrogen is not great enough to develop strong - polar - hydrogen bonding.
I suspect that covalent bonding was suggested in [1] as a means to bond the structures with molecular orbitals. Whether these structures or the calculations play a significant role in the behavior of ammonia is unclear.
Getting back to the quote by the OP, I'm not impressed with the claim that ammonia shows much in the way of association by means of its melting and boiling points:
$$
\begin{array}{|c|c|c|}
\hline
\text{Compound} & \text{Melting Point} & \text{Boiling Point}\\
\hline
\ce{CH4} & \pu{-182 ^\circ C} & \pu{-161 ^\circ C}\\
\hline
\ce{NH3} & \pu{-78 ^\circ C} & \pu{-33 ^\circ C}\\
\hline
\ce{H2O} & \pu{0 ^\circ C} & \pu{100 ^\circ C}\\
\hline
\end{array}
$$
Ammonia lies pretty much in between methane, with no hydrogen-bonding capability and water, with considerable hydrogen-bonding capability. If ammonia were to be considered to be capable of hydrogen-bonding (i.e., dimerization) in the vapor, then water should be even more so. The water dimer has been investigated[7] and values of bonding are suggested to be between $\pu{5-6 kcal/mol}$, but the proposed structure does not suggest any special stability:

Perhaps we could be more interested in water association!

References:
[1]: Wang, B.; Hou, P.; Cai, Y.; Guo, Z.; Han, D.; Gao, Y.; Zhao, L. Understanding the Hydrogen-Bonded Clusters of Ammonia (NH3) n (n = 3-6): Insights from the Electronic Structure Theory. ACS Omega 2020, 5 (49), 31724–31729.
[2]: Nelson, D. D., Jr; Fraser, G. T.; Klemperer, W. Does Ammonia Hydrogen Bond? Science 1987, 238 (4834), 1670–1674.
[3]: Wang, H.; Agmon, N. Complete Assignment of the Infrared Spectrum of the Gas-Phase Protonated Ammonia Dimer. J. Phys. Chem. A 2016, 120 (19), 3117–3135.
[4]: Tsivintzelis, I.; Kontogeorgis, G. M.; Panayiotou, C. Dimerization of Carboxylic Acids: An Equation of State Approach. J. Phys. Chem. B 2017, 121 (9), 2153–2163.
[5]: Illustrated glossary of organic chemistry - electronegativity
http://www.chem.ucla.edu/~harding/IGOC/E/electronegativity.html
(accessed Jun 27, 2021).
[6]: Libretexts. 21.4: Acidity and Basicity of Amines
https://chem.libretexts.org/@go/page/32545
(accessed Jun 27, 2021).
[7]: Wikipedia contributors. Water dimer
https://en.wikipedia.org/w/index.php?title=Water_dimer&oldid=996228917
(accessed Jun 27, 2021).

Kill two birds with one stone maybe? Acetic acid can form two hydrogen bonds in the dimer where the more restrictive geometry of ammonia or water allows oy one.

@Oscar Lanzi: Very good point! I wonder why we don't see very visible from ideal behavior. Or maybe azeotropes indicate unusual behavior: formic and acetic acids have a lot of azeotropes. I have a huge table from 1947.


text:
What is the reaction mechanism between acetophenone and ethyl phenylpropiolate to form a 2-pyrone?
I am looking for the mechanism of the following reaction (I believe it is a type of Michael reaction), which is in the textbook "Advanced Organic Chemistry Part B: Reaction and Synthesis" by F. A. Carey & R. J. Sundberg, 4th Ed, Ch.1, Practice question 12. This practice question does not appear at the 5th Ed.

I am having a hard time trying to understand how the will enolate attacks the alkynyl ester. If it attacks the triple bond, where should I push the pi-electron to? forming an allene intermedia? and how to process from the allene intermedia?

Your attempted mechanism is correct; you’re just missing two more steps. In the product there is a lactone. The OEt group in the starting material has been lost. Now how could that happen…..

At issue is how does base change the configuration of the double bond in your product.

@潘珈煒richie Random combinations and permutations in the reaction mechanism show that attack of O-Et(I suspect this won’t happen) towards the other C=O, further elimination, and removal of ethyl group by base then gives us the product. OR the resonance form -C(-O$\mathrm{^-}$)=O(+)-Et attacks at the other C=O and the elimination, and further removal of ethyl group by base gives the product.

@Rishi, I don’t think the ester will attack the ketone, because it is just not nucleophilic enough. Moreover, I draw the following step and I think the removal of the ethyl group is quite questionable.
I am thinking the ketone will attack the ester. So, consider the intermediate like a malonic ester, a base removes the proton in the middle, forms an enolate, and it uses the oxygen to attack the ester to form the lactone. But enolate usually uses the long pair at the carbon (or the pi-bond) to attack, not the oxygen. So, I double some of the previous steps of my proposed mechanism is wrong.

@潘珈煒richie it does usually but here I suspect that due to conjugated-diene stability it is possible for the oxygen to attack, the yield maybe not very impressive though (which may be why it got removed in the next edition)

The general idea in the answer above is correct, but some of the details need to be looked at more closely.
The first step is as drawn in the original post. Deprotonation of the ketone, followed by nucleophilic addition to the alkyne gives an allene-type intermediate. Note that the anionic intermediate only has one geometry, it doesn't have cis and trans forms as the other answer suggests. Reprotonation gives us an enone with unspecified double bond geometry (likely a mix).

The final cyclisation involves forming the enolate of the ketone and then a nucleophilic acyl substitution on the ester. But to get there, we first need to make sure of the double bond geometry: the nucleophile (enolate) and the electrophile (ester) have to be cis to one another.
The way to do this is through tautomerisation of the enone. This converts the original double bond to a single bond, which can rotate freely. (There is some barrier to rotation because it is part of  a conjugated diene; but the barrier is not that high, certainly nowhere near as high as a double bond, and can be overcome under usual reaction conditions.)

In general, if there were no other considerations at play, we would likely get an equilibrium mixture of isomers. However, only the correct geometry can go on to react here: so the 'correct geometry' will be continually depleted by forward reaction, and the equilibrium will shift; and eventually, all of it will be converted to the 'correct geometry'.
Once we've gotten there, it's just a matter of forming the enolate and kicking out the ethoxide.

Why does the enolate attack via oxygen rather than carbon? A simple explanation is that attack via oxygen forms a six-membered ring, whereas attack via carbon forms a four-membered ring, which is heavily disfavoured. Stereoelectronics also probably play a role: for the attack via carbon to proceed, the C=C π orbitals need to overlap with the C=O π* orbitals. This is almost impossible given the small ring size that would be formed, plus the extra double bond in the ring which makes the geometry of the entire molecule more rigid. On the other hand, the oxygen has lone pairs which point in all directions, so cyclisation is much easier.
NB: When drawing mechanisms, I would always suggest to use the resonance form of an enolate with the negative charge on oxygen. This is more representative of the actual electronic structure. For example, enolate geometry is an important consideration in organic chemistry; also, the stereoelectronic considerations above would be harder to pick out if we just thought it was a plain old lone pair on carbon. From an arrow-pushing point of view it's entirely equivalent, so there are no real excuses to draw a negative charge on carbon.

I was rechecking the process, wouldn't this reaction just never happen if it just ends up as a basic hydrolysis? https://www.google.com/url?sa=t&source=web&rct=j&url=https://www.sciencedirect.com/topics/engineering/catalyzed-hydrolysis%23:~:text%3D2.10.,-4.2%2520Mechanism%2520of%26text%3DBase%252Dcatalyzed%2520hydrolysis%2520of%2520ester,tendency%2520to%2520react%2520with%2520alcohol.&ved=2ahUKEwiYjZvWjLjxAhV8yzgGHXjUAf4QFjABegQIBBAF&usg=AOvVaw1wGNn2y2dSUgruLgRTiliZ since basic hydrolysis is irreversible? Can we say that the carbanion on the other reactant attacks fast enough?

The base is never specified. When you say 'hydrolysis' you're assuming that it's $\ce{OH-}$, but it could well just be sodium ethoxide or something. If ethoxide ion attacked the ester directly, it wouldn't lead to any (observable) reaction. This is a standard consideration when running Claisen condensations or similar reactions: if you have a $\ce{CO2R}$ ester you generally want to use $\ce{OR-}$ as the base. Or, alternatively, in the reaction above it could be a bulky base like t-butoxide which can't attack the ester.

Your attempted mechanism is correct. There is an active $\ce{\alpha-H}$ in the your final intermediate ($2$ $\ce{-COOEt}$s are present on the carbon (well one of them is attached indirectly, but it still shows it's $-M$ character so it is more or less the same)
The full version of the reaction should look like this (I couldn't find a primary source (like it is for a lot of practice problems)but this is what seems most probable to me ) :-

$\large \bf{ What~is~wrong~in~this~mechanism~and~rectification.} $

After being directed in the right direction by @orthocresol it comes clear that anionic transition state in step-3 is a miscalculation. I used this mechanism to overcome the problem of s-trans geometry being achieved after enolate formation in the third step.

However as mentioned by the accepted answer s-cis and s-trans do not have as much rotation barrier unless steric effects are excessive (here they are not as on $\ce{C=O}$ side only has lone pairs closely bound to oxygen)  here is an example where steric effects lock the molecule in s-trans conformation.

Also pointed by @orthocresol that in step-2 I have shown 2 intermediates having different geometry, this is not possible as the only structure in resonance is the allene intermediate you have drawn.

TL;DR : Step 3 is not required since the reaction can proceed without having to form a carbanion intermediate since the rotation barrier is not significant. Step 2 misrepresents resonance structures as different geometries and allene intermediate works best here (also the structures are irrelevant since after protonation it goes directly to step-4)
Addendum :- In response to the mention that enolates usually show nucleophilic addition with the carbon atom and not the oxygen atom (due to a mix of hydration and steric effects), I think here it is possible that nucleophilic addition is done by oxygen due to the stability of conjugated dienes which wouldn't have happened otherwise. Also the reaction by carbon's attack is not possible due to it forming a 4 member ring with a double bond.(However the yield of this reaction may not be a very impressive due to this reason but I am out of my depth regarding that, all I can see is that this should be the major product)


text:
How do I anodize bronze for thin film colors using electrolysis?
I am trying to change the color of the coins we have in Europe. I want to create patterns of certain colors.
They are made from "nordic gold" (89% copper, 5% aluminum, 5% zinc, and 1% tin).
I put them into a kitchen oven at about 250°C and noticed that they tend to change color. The colors were irregular and uncontrolled. See the picture.

I tried oxidizing it in a controlled way using electrolysis, like this guy on youtube. I used a two electrode setup with a soda solution as an electrolyte. The coin was my plus and had voltages up to 30V applied to it with no color change. It only started producing bubbles (probably hydrogen and oxygen). Low Voltages and low currents don't help as well. I scratched them with sandpaper and cleaned them with acetone before putting them into the electrolyte.
Do you know or have an idea on how to oxidize the coins in a controlled way?
EDIT: I couldn't believe that electrolysis had no effect, so I did it again; this time with salt-water as the electrolyte. at small voltages it just tarnished. At high voltages it becomes white, yellow, green and flakes are sticking to it when in solution.

Bronze will not anodize like aluminum ; aluminum forms a protective oxide layer , thicker with anodizing.  I worked in corrosion and never saw a reference to anodizing building a protective oxide on copper. It is probably removing surface corrosion to leave a clean copper or gold color.  I expect the aluminum is added for the strong hardening affect , not a color ,  The alloy is a first step toward a Manganese Bronze.

@blacksmith37 I haven't thought about that I might be corroding the aluminum out of the solid solution! That explains why it became so white.
I removed the coating and it had a strong copper color. So for the aluminum to stay inside or the copper just changing to a copper-oxide I suppose heating is my only option, as electrolysis seems to be too aggressive.
I found an article on copper thin film oxidation: http://www.microlabgallery.com/gallery/InterferenceFilmsonCopper3.pdf
It doesn't need to be the nordic gold, but it happens to conveniently have the right shape. I will try copper.

Copper only tarnished

Your reference is about very thin oxide film such as a few molecules of copper oxide. Much more commonly seen on stainless steels . For example look at stainless exhaust pipes on motorcycles ; gold at the cool end, to pink , to blur/purple at the hot end . A matter of the differencr of oxide layer a few molecules thick.  And carbon steel , thin is gold ( tempered at 300 F)  to thick is blue ( tempered at 700 F) ; (all in round numbers ).

Dealloying does occur under special circumstances but I doubt very much that you have extracted aluminum from the bronze.

I wish to create such thin oxide films. The reason I don't want to use steel is because it can rust. Titanium on the other hand is hard to machine. Based on this video it apparently also depends on the temperature: https://youtu.be/7g-azzYnMYo
So maybe I should use steel for simplicity and cover it with something later on. Does copper require temperatures higher than 100°C (212F water boiling)?

@blacksmith37 Why do you doubt that I extracted the aluminum? You mean growing such colors could be possible on bronze too? I found a paper that actually does a good job explaining how to make such films on copper: https://www.researchgate.net/publication/234863665_Transpassive_Dissolution_of_Copper_and_Rapid_Formation_of_Brilliant_Colored_Copper_Oxide_Films


text:
Electrophillic susbtitution of bromine on styrene and cinnamic acid in the presence of FeBr3, a Lewis acid
Bromine gas, in the presence of $\ce{CCl4}$, undergoes on addition reaction at the double bond in both styrene and cinnamic acid, illustrated below.
Styrene:

Cinnamic acid:

Bromination of benzene in the presence of $\ce{Fe}$ and bromine gas lead to electrophillic substitution rather than addition reaction just like given here.

If we brominate styrene or cinnamic acid in presence of iron, which reaction will happen: electrophilic substitution on the benzene ring or an addition reaction on the double bond of the side group?
I could not find references to the reactions given above - specifically, in presence of iron - anywhere for styrene and cinnamic acid.
According to me, the mechanism of bromination in the presence $\ce{FeBr3}$ should follow an electrophilic substitution pathway, because the ring is more electron rich so an electrophile would preferentially attack there. However, there still is a potential factor of steric hindrance because bromine is a large ion, thus I am not certain which way the reaction will proceed.

please mention reason of downvote so i can edit the question please

The text is a bit difficult to read. With some experience in the lab or / and look up in the literature, you see that tetra / $\ce{CCl4}$ (if you still allowed to use it) dissolves both the starting material, and bromine. If your question refers to illustrations, embed them here on chemistry.se (with credit to the source if not drawn by yourself); external links may vanish / move and thus render your question even less accessible and intelligible.

@Samardeepsingh I did not downvote your question BUT in general, you will attract downvotes and/or vote-to-close (VTC, and as I write this there are 3 of the 5 needed to do so) if you do not include your efforts/thoughts on an answer in a post. That would be in addition to any downvotes garnered for other reasons. I encourage you to amend your post with your efforts to avoid closure.

i edited my post  @ToddMinehardt thankyou for your nice feedback

In electrophilic aromatic substitution the activation energy is very high ( as it should be, since the aryl carbocation transition state is very unstable)
The role of $\ce{FeBr_3}$ or any lewis acid in general is not exactly promoting the electrophilic aromatic substitution as much as making a good electrophile. $\ce{AlCl_3}$ plays the same role much better* ($\ce{FeBr_3}$ is much weaker lewis acid than $\ce{AlCl_3}$). The evidence is in the mechanism:-

The lewis acid helps in creating an electrophile and later in extracting the hydrogen, it does the same thing for addition of $\ce{X_2}$ on alkenes.
Since the addition of $\ce{X_2}$ on alkenes reaction involves a less extreme transition state, it gets a very good lead against electrophilic aromatic substitution being kinetically driven
it would probably happen at room temperature. Of course both reactions will happen if you have more than $1$ equivalent of $\ce{Br_2}$
We see here that in doing addition of $\ce{X_2}$ on alkenes, we lower the stability of final product by blowing up the double bond involved in an extensive resonance stabilization, therefore at high temperatures, the molecules can attain the high activation energy required easily and hence prefer the aromatic substitution over addition of $\ce{X_2}$ on alkenes . Keeping the extended resonance intact.
*Source :- Morrison and Boyd section 16.13 (Halogenation of alkylbenzenes)

does this principle of kinetically favourable and thermodynamically favorable  product can be applied to both styrene and cinnamic acid ???

yes, this principle is applicable for both, I just used cinnamic acid because it has a longer conjugation.