text:
"Nitrogen (gas) is physisorbed over iron (metal) at room temperature": Is this for real?
My copy (PDF) of Levine's Physical Chemistry (6th ed.), page 570-571, section: "Adsorption of Gases on Solids", states:

Physical adsorption is nonspecific. For example, $\ce{N2}$ will be physically adsorbed on any solid provided the temperature is low enough.

and

... N2 is chemisorbed at room temperature on $\ce{Fe, W, Ca,}$ and $\ce{Ti}$ but not on $\ce{Ni, Ag, Cu,}$ or $\ce{Pb}$.

Alright, fair enough, thanks for the examples.
But
A bunch of (notoriously error-prone-citation-deficient) course books issued by my school, all unambiguously concur on the contrary:

At $\ce{83K}$, $\ce{N2}$ is physically adsorbed over iron metal. Raising the temperature to $\ce{773K}$ then results in the nitrogen being chemisorbed over the metal as atomic nitrogen.

Which in other words is: "At room temperature, nitrogen is physically adsorbed. The nitrogen is chemisorbed after the temperature's raised to 500 degrees Celsius".
But this contradicts the information in Levine's book.

Who's right here: Levine or my book? (References are welcome) O:)

EDIT-
LinearChristmas has correctly pointed out that the temperatures at which the physical adsorption --> chemisorption transition takes place depends on pressure as well as the nature of the iron surface.
In light of this, I'll modify my question:
Q- My coursebook and Levine's lists (implicitly in Levine's case) the transition from physical adsorption --> chemisorption at different temperatures. Between the two, which one lists the more improbable temperatures?

Unfortunately, I don't think this can be answered as is. This probably isn't entirely your fault; general textbooks tend to be secretive when it comes to experimental details. Firstly, what are the pressures and how were measurements done on each? It is also incorrect to assume an iron as being equivalent to some other iron. Surface properties, including porosity, are strongly dependent on how the material has been handled. I  further recommend adding references for both sources. Finally, is it also possible $773\ \mathrm{K}$ was simply a typo for $273\ \mathrm{K}$? (just in case)

@Linear A good point! Not sure about the iron surface/crystal, but I just assumed this is carried out at atmospheric pressure (unlikely, but I wouldn't know any better). As for the 773/273 typo, it could be a typo, but  three different booklets (all issued by our teacher) carry 773 in it. :'(

Okay. Sadly, we still have to know more about the iron to make an informed comparison. For example, 10.1039/TF9595502166 reports on some iron catalysts having chemisorbed N2 at measurable rates from $473\ \mathrm{K}$. Their own sample, ‘highly reduced singly promoted iron catalyst’, reported similar results for a temperature of $273\ \mathrm{K}$. There is another caveat. The second source in OP: chemisorbed over the metal as atomic nitrogen. Note that there are other ways of nitrogen chemisorbing, though mononitrogen could be most common (I'm unsure).

@Linear < Nods sadly > This seems more painful than I first thought. I suppose then, any potential answer would be aiming to dismiss either one of the two claims (in my post) as "improbable" rather than outright "incorrect". Thanks for the insight! O:)

Yes, I'm afraid so. I do see a scenario where this is closed as primarily opinion-based. To be sure, we may wait for other responses since this is a relatively new thread.

I think you are misinterpreting the course book issued by your school since it never mentions what happens at room temperature.
For me, none of the books is (seems) wrong, because they do not contradict each other. I will mention each affirmation including the textbook which said it (Levine, or Your book):

Levine says: N2 is physically adsorbed on any solid at low temperature. This agrees with your text saying N2 is physically adsorbed at $\pu{83 K}$ on Fe.
Levine says: N2 is chemisorbed at room temperature on Fe. Your book does not mention about room temperature. No contradiction here.
Your book says N2 is chemisorbed at $\pu{773 K}$. Levine does not mention this temperature, but if it already occurred at room temperature it will also occur at higher.

The only 'gray area' is at which temperature does N2 get chemisorbed (room temperature or not?). In the article of de Boer et al. mentioned by @Linear Christmas above, they measure chemisorption of N2 even at $\pu{23.9 ^\circ C}$.
They clearly state:

Hitherto it has been reported in literature$^{1,2}$ that an iron catalyst starts to chemisorb nitrogen at a measurable rate at about $\pu{200 ^\circ C}$. The chemisorption activity of our intensely reduced sample was considerably higher; already at $\pu{0 ^\circ C}$, nitrogen was chemisorbed at a measurable rate.

Reference:
J. J. F. Scholten, P. Zwietering, J. A. Konvalinka, J. H. de Boer, "Chemisorption of nitrogen on iron catalysts in connection with ammonia synthesis. Part 1.—The kinetics of the adsorption and desorption of nitrogen," Trans. Faraday Soc. 1959, 55, 2166-2179 (https://doi.org/10.1039/TF9595502166).

"Raising the temperature to 773K then results in the nitrogen being chemisorbed..." <--- That sounds like my book says that at 773K and above nitrogen is chemisorbed, meaning at lower temperatures (including room temp) it is physically adsorbed... so I'm not very confident about your 2nd and 3rd bullet point. First bullet is okay.

Being chemisorbed above 773 K does not exclude being also chemisorbed below.


text:
Why does copper form bubbles in vinegar in this situation?
I watched this video in which the host tries to show the principles behind the voltaic pile.
The host first immersed zinc in vinegar and bubbles are observed forming around zinc. I've learned that this is because hydrogen ions in vinegar get electrons from zinc atoms and become gaseous.
Then the host take a conductor and attach at one end the piece of zinc and at the other a piece of copper. Bubbles start to form around copper too! (fig1)

Why does this happen?
Why do some electrons travel through the conductor instead of being attracted immediately by the hydrogen ions in the vinegar?
What if the conductor instead was very long and the pieces of copper and zinc far apart? (fig2)

When you put a zinc rod in a acid solution containing $\ce{H^+}$ ions, the zinc piece loses $\ce{Zn^{2+}}$ ions to the solution leaving electrons behind. These electrons attract hydrogen atoms to the zinc rod, and an electron transfer occurs, resulting in bubbles of hydrogen forming on the surface of the zinc. However, the evolution of gaseous hydrogen from zinc surface is rather slow, and that is why pure zinc reacts slowly with dilute acids.
The situation changes when a copper,or any other metal less electropositive than hydrogen is attached to zinc, by a wire. The copper can now be considered as an extension of the zinc rod: a place form where hydrogen can take electrons quickly and become gas. As the rate of evolution of hydrogen from the copper piece is faster, hydrogen bubbles out from the copper piece. This does not mean, however, that the zinc stops reacting with acid; it does, but at a very slow rate.
A common mistake is to think that the copper rod participates in the electrochemical reaction taking place. The copper is merely a pathway to release electrons to H+ ions in the solution. The only reaction taking place is
$\ce{Zn + 2H^+ -> Zn^{2+} +H2\uparrow}$
Copper has no place in the reaction.
So, to answer your questions,

Why does this happen?

This happens because copper can do away with the electrons quicker than zinc, and when electrons get a pathway, they will take the pathway.

Why do some electrons travel through the conductor instead of being attracted immediately by the hydrogen ions in the vinegar?

Well, some electrons do attach to H+ ions from vinegar directly, and hydrogen bubbles should form on the zinc plates, although they are less in amount. And I believe my earlier explanation answers this question.

What if the conductor instead was very long and the pieces of copper and zinc far apart?

It does not have anything to do with distance. If you take a large solution, the reaction will certainly occur. On the other hand, if you separate the poles into two solutions the copper rod will not give off electrons as the circuit is not complete. Think about it this way, if the reaction was to occur, the solution with zinc will become progressively positive and the solution with copper will become negative. This can never happen.
Footnotes:
(1) As per your comment below, I now attempt to answer the question "What makes copper better than zinc in releasing electrons?"
The answer lies in a factor called overpotential, sometimes referred to as overvoltage. There is one type of overpotential which is important in this case— activation overpotential. This is caused due to the fact that when hydrogen atoms are produced on the surfce of a metal ($\ce{H+ +e-> [H]}$), they take some time before they can combine to form diatomic hydrogen and some more time before the hydrogen molecules can be released as gas. This results in the formation of a layer of hydrogen on the surface of metal effectively reducing the rate of reaction and halting it; this effect is known as polarization.
As you can see from the data chart in the Wikipedia page, Zn electrode has -0.77 V overpotential for evolution of hydrogen while there is no overpotential for copper electrode. This means that copper is not polarized easily.
(2) I mentioned in the beginning of my answer that pure zinc reacts with dilute acid slowly. Commercial zinc, on the other hand, reacts quickly with acids, owing of the same effect of polarization. In commercial zinc, impurities of other metals form cathode and the pure zinc forms anode in the acid medium, essentially producing a galvanic cell, where the anode and cathode are connected. Thus the reaction occurs quickly, without the hindrance of polarization.
(3) You should read this paper : http://pubs.acs.org/doi/abs/10.1021/ed078p516. It has some facts which you may find interesting.

Hi! This is a perfect answer, it just need to clear me up on some details and it will be accepted. I did not understand why electrons prefer to travel and reach copper, what makes copper better than zinc so that electrons prefer to move to copper even though they can stay in the zinc and still react there. Why the majority of electrons mek this choice?

Moreover, so if we take a very big pot of vinegar and a very long and thin wire (kms) the reaction still happen, electrons still prefer to travel for kms to reach copper than react in the zinc? Instead if we use 2 different glasses, shouldn't it happen (the reaction) for a while until everything is electrically balanced and then stop?

@GabrieleScarlatti , I will edit the answer. Please wait for some days. I have not had the time to write a detailed explanation.

@GabrieleScarlatti , How will the reaction happen for a while until everything is electrically balanced. As I mentioned in my answer, if the reaction were to happen for two different solutions, one solution will become positive while the other solution becomes negative. Take an analogy: If you have two objects of different temperatures in contact, what would happen? Should more heat transfer from cold body to hot body, making them progressively hotter and cooler? or should the hotter body give up heat so that both attain equal temperature?

@GabrieleScarlatti , Edited post to include the answers to your questions

Copper is cathodic when the galvanic circuit is formed.  In the acid solution hydrogen ions are reduced at the cathode.
The zinc anode is more complicated.  The anodic reaction is zinc oxidation to ions which dissolve.  At the same time there is hydrogen evolution, the usual chemical reaction.  The chemical reaction and the anodic reaction are competing for the available zinc.
To set up the galvanic cell and make hydrogen evolve from both ends you have to connect the electrodes through the electrolyte (or via a salt bath).  Figure 2 does not have a complete circuit, so it gives no reaction except zinc dissolving chemically.

Hi! Thank you for your help! Btw it's still not clear to me, why electrons  decide to travel through the conductor from zinc to copper instead of beaing attracted to the ions immediately, dissolving only the zinc(fig1),and why fig2 is not a complete circuit? What makes it different form circuit 1?

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text:
Quadrupole MS: mass as a function of voltage
Every time I see an article about Quadrupole MS (e.g. here), they give formulas like:
$$ a_x = \frac {8eU} {m r_0^2 \Omega^2} $$ and $$q_x = - \frac {4eV} {m r_0^2 \Omega^2}$$
I suppose the instrument changes voltage and keeps the rest constant, but I failed to find any article that would show direct formulas on how to calculate m/z from that. I could rearrange it like this:
$$ \frac {m} {e} = \frac {8U} {m r_0^2 \Omega^2 a_x} $$ But this doesn't really help because $a_x$ itself is unknown and is derived from mass.
PS: this question is about Quadrupole MS though it looks to me that Ion Trap MS has very similar concepts and math.

?!?  You don't seem to be asking about a Quadrupole Mass Analyzer but rather a Quadrupole Ion Trap.

My confusion is in part because you refer to Quadrupole MS which I infer to mean a Quadraople Mass Spectrometer.

My impression was that both Q Ion Trap and and Q Mass Analyzer (aka Filter?) use similar algorithms since all articles mention Mathieu equation. So I decided not to distinguish between them at the beginning. If I'm wrong, I'd greatly appreciate if you could explain the difference. At the moment though I'm mostly interested in the Q Mass Analyzer. As for MS - to me Spectrometer is a synonym of Analyzer (e.g. https://chem.libretexts.org/Core/Analytical_Chemistry/Instrumental_Analysis/Mass_Spectrometry/Mass_Spectrometers_(Instrumentation)/Mass_Analyzers_(Mass_Spectrometry)).

And here is an example of source where they refer to the aforementioned equations in Quadrupole Mass Analyzer, and then discuss similar math in Ion Trap Analyzer: http://www.chromacademy.com/lms/sco36/Fundamental_LC-MS_Mass_Analysers.pdf

These parameters arise from non-dimensionalization of the Mathieu equations for ion motion in exogenous, periodically varying electric fields (i.e. oscillating potentials).  You can't calculate $m/z$ directly from $a$ and $q$.  For a fixed $U$ and $V$, you can regard $a$ and $q$ as non-dimensional parameters that correspond to $m/z$.  (Conversely, at fixed $m/z$ you can regard $a$ and $q$ as non-dimensional versions of the electrical potentials $U$ and $V$.)  The reason that those parameters don't tell you everything is that you still have to solve the Mathieu equations.
An analogy might be to the Reynolds number for characterizing fluid flows.  It, like $a$ and $q$, is a non-dimensional number that depends on properties of the molecules under study and on the system geometry.  Knowing just the value of the Reynolds number is useful for understanding some of the general characteristics of a flow, but if you want to compute fluid motion, i.e. the velocity profile in a pipe, you still have to solve the Navier-Stokes equation.
For quadrupole mass analyzers, only certain regions of $a$-$q$ space result in stable ion trajectories (i.e. where ions don't collide with the surfaces of the four rods in the quadrupole).  This map shows which values for $a$ and $q$ result in stable solutions to the Mathieu equations, and thus is called the Mathieu stability diagram.  The same document you linked to shows an example of such a diagram:

Thus, to compute $m/z$, we need to impose the criterion that $a$ and $q$ are in a stable region of the stability diagram, more specifically in the region "A" shown in the diagram.  By rough eyeballing, the "middle" of this diagram corresponds to about $q = 0.75$ and $a = 0.07$ or so.  You can use either of those values to estimate $m/z$, but a more precise estimate, including line-widths etc., would require knowing both.

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text:
Calculate the atomic radius of iridium atom
I was delaying with this question as seen below.

Calculate the radius of iridium atom, given that Ir has an FFC crystal
  structure, a density of $\pu{22.4 g/cm^3}$ and atomic weight of $\pu{192.2 g/mol}$.

Let me show you how my teacher solved this question.
Givendata:
$$\text {Weight of crystal} = \pu{192.2 g/cm^3}$$
$$\text {Density} = \pu{22.4 g/cm^3}$$
$$\text {Volume of crystal} = V = \frac {192.2}{22.4}$$
$$V = \pu{8.58 cm^3/mol}$$
In FFC, there are 4 atoms in one unit cell.
$$\text {Volume of one cell} = 8.58 \times \frac {4}{6.023 \times 10^{23}}$$
$$ = 5.7 \times 10^{-23}$$
$$V = a^3$$
$$\therefore a = 2.83r$$
and
$$V = (2.83r)^3$$
$$5.7 \times 10^{-23} = 2.83r^3$$
$$r = 1.34 \times 10^{-8}$$
Where did that $a = 2.83r$ come from? That's what I didn't get.

Iridium has a face-center cubic (FCC) structure.
You assume that each iridium sphere has a radius $r$. But the edge of a unit cell has the dimension $a$. The volume of the unit cell, which is a cube, is $a^3$. The relationship between $a%$ and $r$ is that $a = 2.83r$.

FOLLOWING THE TEACHER"S(?) WORK...(or how you copied it?)
The following step is ok...
$$V = (2.83r)^3\tag{1}$$
But the second step here is wrong. The 2.83 should be cubed too.
$$5.7 \times 10^{-23} = 2.83r^3\tag{2}$$
(3) is totally wrong if we accept (2).
$$r = 1.34 \times 10^{-8}\tag{3}$$
From (2)
$$r = \sqrt[3]{\dfrac{5.7\times 10^{-23}}{2.83}} = 2.72 \times 10^{-8}\tag{4}$$

Starting over...
$$V = (2.83r)^3\tag{5}$$
$$5.7 \times 10^{-23} = (2.83r)^3\tag{6}$$
$$r = \dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = 1.36 \times 10^{-8}\tag{7}$$
After calculating the the volume of the unit cell, I'd solved for $a$ by taking the cube root, then divided by 2.83 to get $r$.
You have to be careful with significant figures too. It is not $5.7\times10^{-23}$ but rather $5.70\times10^{-23}$. Using a calculator I'd carried 5 significant figures in all the calculations then rounded to 3 at the end. In my day I used a slide rule so everything was to 3 significant figures.

So, is it constant?

Yes the 2.83 is a constant.

That was what I was looking for! Also,  shouldn't the answer be $1.36 \times 10^{-8}$? I mean this step  $5.7 \times 10^{-23} = 2.83r^3$

My teacher solved this two times. How did you get that answer?

You don't understand the difference between equations (2) and (6) ?!?

My teachers has solved it and both got $1.34 \times 10^{-8}$

Again, don't you see the difference between equations (2) and (6) ???

I see it. However, why did they get that answer?

The difference between 1.34 and 1.36 is some sort of rounding error. Is that what you're hung up on?

So, isn't answer 1.34?

The critical part is that $(2.83r)^3 = 2.83^3r^3 = 22.67r^3$ so going from eq (1) to eq (2) is WRONG.

What do you mean?

I showed my third teacher as well, He've found $1.34$. Is it really wrong?

As I said it is a rounding error. You are not supposed to trim significant figures until you get to the final result. So if  you calculate $$\dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = \dfrac{3.8\times 10^{-8}}{2.83} = 1.34276\times10^{-8}= 1.34\times10^{-8}$$ but if you calculate $$\dfrac{\sqrt[3]{5.7 \times 10^{-23}}}{2.83} = \dfrac{3.8485\times 10^{-8}}{2.83} = 1.3598\times10^{-8} = 1.36\times 10^{-8}$$ So yes, they are wrong.

PS - Your teacher also used the wrong number for the Avogadro constant. The accepted value is now $6.022140857(74)\times 10^{-23}\text{ mol}^{−1}$ and rounded to 3 decimals is $6.022\times 10^{-23}\text{ mol}^{−1}$

This is how I would do the calculation...
Given data:
$$\text {Weight of crystal} = \pu{192.2 g/mole}$$
$$\text {Density} = \pu{22.4 g/cm^3}$$
Ok, the density is good to only 3 signifant figures so the answer shouldn't have any more than that. But doing the whole problem, I'll carry 5 significant figures throughout all the intermediate calculations to try avoid rounding errors within  the multiple calculations. I'll round the final result to 3 significant figures.
$$\text {Volume of crystal} = V = \frac {192.2}{22.4}$$
$$V = \pu{8.5804 cm^3/mol}$$
In FFC, there are 4 atoms in one unit cell.
$$\text {Volume of one cell} = 8.5804 \times \frac {4}{6.0221 \times 10^{23}}$$
$$ = 5.6992 \times 10^{-23}\text{ cm}^3$$
A unit cell is a cube with each side being $a$
$$V = a^3$$
$$\therefore a = \sqrt[3]{5.6992 \times 10^{-23}} = 3.8483\times 10^{-8}\text{ cm}$$
But the FCC crystal, spheres of radius, $r$, packed inside a cube with side, $a$, the geometric relationship between $a$ and $r$ is: $$a = 2\sqrt{2}r = 2.8284r$$ so
$$r = \dfrac{3.8483\times 10^{-8}}{2.8284} = 1.3606\times 10^{-8}\text{ cm}$$
now rounding $r$ to 3 significant figures gives
$$r = 1.36\times 10^{-8}\text{ cm}$$


text:
Do mixtures have a boiling point?
Let's assume a mixture which can be heated without decomposing. This site explains that while pure substances have a sharply defined melting point and boiling point, mixtures tend to have a melting range. But what about the boiling point of mixtures? Is it always a point, always a range, or sometimes a point and something a range? If it is "sometimes", is one of the cases the rule and the other the exception, or are both pretty common?
To keep the question simple, we can stay with atmospheric pressure.

If there is a eutectic, then at that composition it is a single melting point. Quite often for some random mix of elements A and B it will be a range. The binary phase diagram will show what you need to know.

Assuming the components are miscible, a mixture has a distinct boiling point, which depends on the precise molar fraction of the solution. Look up vapour-liquid diagram.

See e.g. https://www.chemguide.co.uk/physical/phaseeqia/nonideal.html

Mixture is a more general term. Solution can be of liquids or a liquid(s) and solid. Each case is different tough the relevant one is mentioned in the comment by Gert.


text:
(S)-glycidyl benzyl ether NMR problem
Can someone explain the appearance of the resonances at $\pu{4.6 ppm}$ (spectrum taken from here)?

Is it due to long range coupling? But the coupling constants are too high for such long range coupling. I assumed, according to the shift, that this is coming from the two protons nearest to the benzene ring?

The pattern is neither a true quartet nor a doublet of doublets: it is two doublets. Many thanks to long for pointing out that it is also frequently described as an "AB quartet".
The peaks arise due to the benzylic protons, i.e. the $\ce{CH2}$ between the ether oxygen and the phenyl group. Due to the existing stereocentre on the epoxide, these two protons are rendered diastereotopic: due to their different position relative to the stereocentre, they experience a different net magnetic field, and hence have a different chemical shift. Therefore, in the absence of coupling, one expects to see two singlets.
If we add in spin-spin coupling, then we see that both protons are capable of coupling to each other. The magnitude of the geminal coupling $^2J$ is generally somewhere between 10 to 15 Hz. In this case the zoomed-in diagram shows that the coupling is roughly 12 Hz. So, we now expect to see two doublets.
Why, then, are the intensities of the two peaks in each doublet different? This is due to so-called "roofing": when the difference in the resonance frequencies of two protons is comparable to the coupling constant between them, the shape of the multiplets will be such that they have a "slant" towards each other. In this case, the difference in resonance frequencies is roughly 20 Hz (measure the distance between the centre of each doublet), which is on the same order of magnitude as the coupling constant.
This can, to the best of my knowledge, only be explained with quantum mechanics: see What is the origin of 'tenting' or 'roofing' in NMR spectra? (the question has more examples).


text:
Negative phases don;t exist for 1s orbital why?
!]3
what is the dot on the two s orbitals?
IF NEGATIVE PHASE DOESN'T EXIST WHAT IS THE S ORBITAL IN BLUE IN THE FIRST PIC?

Short answer: the dot is the nucleus (not to scale). The negative phase is a result of LCAO (linear combination of atomic orbitals) with a negative coefficient. The two different phases contained on one orbital for p and higher is something else entirely.

A $1s$ orbital can have either positive or negative phase.  Which it is, is our label: all the experimentally accessible features are the same with either choice.
Now consider two such orbitals.  Either one can have either phase, but there is one more factor:  the relative relationship between the signs of the phases.  They can combine with the same sign to produce a bonding orbital, which is the same bonding orbital whether we render the common sign positive or negative.  Or they can assume opposite signs and combine to make an antibonding orbital; this orbital is the same whether the first orbital is positive and the second negative, or vice versa.  All the experimentally accessible information is dependent on relative signs of the contributions to a molecular orbital, not how we render the absolute sign of any orbital.


text:
Saytzeff and beta elimination
I am not able to understand the difference between Saytzeff's rule and beta elimination reaction. Also I am writing two reactions below:
\begin{align}
\ce{CH3-CH2-CH(Br)-CH3 + \underset{\text{(aqueous)}}{KOH} & -> CH3-CH2-CH(OH)-CH3} \tag{1} \\
\ce{CH3-CH2-CH(Br)-CH3 + \underset{\text{(alcoholic)}}{KOH} &-> \underset{\text{(major)}}{CH3-CH=CH-CH3} + \underset{\text{(minor)}}{CH3-CH-CH=CH2}} \tag{2}
\end{align}
Are they correct? If they are, then what are they called?

The first one would be a bimolecular substitution reaction following the SN2 mechanism. The second one however , is an elimination reaction. There is a net decrease in the no:of substitutions on carbons 2 and 3 in the second reaction which means some of the substituents (the bromide and proton) here have been ‘eliminated’.

*in the major product of the second reaction, while for the minor it is the first and second carbons that are involved in elimination.

Could you elaborate on what exactly you find confusing about the saytzeff’s rule and beta elimination?

okay now i understand.

If the reagent is alcoholic KOH, then the reaction mechanism followed is E2 elimination (if steric/other conditions permit)
However, if the reagent is aqueous KOH, it is simply a substitution (SN) type reaction in which the Br atoms are replaced by -OH groups (not to mention, if conditions permit, i.e. SN is actually possible in the given molecule*)
*example: SN type reactions are usually forbidden in compounds like PhBr (bromobenzene) as the π cloud of the benzene ring repels the incoming nucleophile (SN² not possible) and formation of phenyl carbocation is not thermodynamically feasible due to its instability (SN¹ forbidden)
As far as your main question is concerned, i.e. difference between Saytzeff's rule and Elimination -
They are two completely different things, one is a rule to correctly predict major products of elimination reactions and the other is a reaction mechanism, which defines the pathway for the reaction to proceed through.
In fact, Saytzeff's rule states:
'In an elimination reaction, the most substituted product will be the most stable, and therefore the most favored.'
This is not always true, however. Major products are formed when the reaction intermediate is stabilized (product stability is usually secondary)
In case of eliminations, especially if the reaction is to proceed through an E1 pathway which involves formation of a carbocation, stability of the carbocation becomes a very important deciding factor in the product percentage. If the carbocation so formed (assuming reaction proceeds via predicted pathway. Practically, it is not fruitful to talk about such intermediates as they have a very small lifespan and reaction may proceed through other more favourable pathways) is highly destabilised even though it is more substituted, for example, carbon attached to three nitro groups (or any strong electron withdrawing groups for that matter) then Saytzeff's rule is violated and corresponding Hoffman product is found to be major. Of course, there are several other cases in which Hoffman product is found to be major and corresponding Saytzeff product is minor. I believe that discussion deserves a separate post altogether, so I won't elaborate on it here.
Hope it helped!

The answer might not be very different from that given by user55439 however, I guess a little more clarity could be given.
Beta-elimination of a proton to result in a pi-bond is a class among the many pathways that give an unsaturation. The saytzeff rule which states that ‘the unsaturation occurs between those carbons which are more substituted’ ,acts like a decision maker , when there are multiple beta-eliminations possible. In the above reaction that you have given, the major product has the unsaturation between the two secondary carbons while, the minor product has the unsaturation between one secondary and a primary carbon. Now, as saytzeff says , the unsaturation formation between the more substituted carbons is favoured.
In short, saytzeff determines the statistically more probable elimination pathway adopted by a reactant , when there are multiple pathways possible.

We know that beta elimination means elimination of beta hydrogen and hydrogen will give its electron to its carbon so that it can make double bond with another carbon . When there is two beta hydrogen in a chain then at the time of beta elimination the position of double bond will be different ( see the reaction you have given , two product came with different position of double bond ) then now we use satyzeff rule to identify which one produce more i.e main product .
If  starical hindrance is more then that product will remain as a main product as its starical hindrance is more other compound cant attack it( in your reaction you have given under the product major)and that compound will be produced more . In your reaction one product have double bond between 2 and 3 number carbon as double bond is in middle of it will have more starical hindrance than the one having double bond in 1 and 2 carbon since in 1 and 2 carbon double bond it have alkyl group in one side only and alky group is main factor here because it creates starical hindrance .

 Photo shows which one have more starical hindrance ( example)
N.B > My answer doesnot represent what is saytzeff rule , i just showed application and mecanism why product is more .

The actual reason is hyperconjugation.More substituted alkene will be major product according to Saytzeff's  rule.Steric hindrance is not the reason.

But why should the alkene with more steric hindrance be more stable and hence the major product? Only hyperconjugation explains it.

@SourabhYelluru and i did not mean what is saytzeff we all know saytzeff says  that more branched alkene will be the major product . I just showed application and mecanism why it is major product

@SourabhYelluru bro i have learned that alkene with more starical hindrance is a major product because percentage of product is more (butene-2 is 80%) because due to starical hindrance other compound cant react with it . In butene -1 percentage is 20% because starical hindrance is low so it react with other compound and and give us less product.

I might be the only one, but I wouldn't consider bro a nice way to address people.


text:
Coordination number 4 corresponds to tetrahedral shape yet NH3 has trigonal pyramidal shape why?
I understand the reason for the trigonal pyramidal shape (presence of lone pair) but if there's a relation between coordination number and geometry of molecules it's violated here isn't it?

It has a tetrahedral geometry but a trigonal pyramidal shape.There is a difference between geometry and shape.

If you count all you end up with a distorted tetrahedrun , Isn't?

NH3 has a coordination number of 3...

$\ce{NH3}$ has a trigonal pyramidal shape because the molecule isn't planar. You only look at the position of the atoms present so the lone pair on the nitrogen atom is irrelevant.
$\ce{NH4^+}$ has a tetrahedron shape with the N in the center.


text:
What is wrong with the Aufbau principle that results in so many exceptions?
From what I have observed in the periodic table is that many transition elements do not follow the Aufbau principle. And it's not just the elements like $\ce{Cu}$ and $\ce{Cr}$ that differ from predicted configurations because half-filled and completely filled d-orbitals are more stable, their are many odd exceptions like Nb ($\mathrm{[Kr]}4d^4\,5s^1$), Pd ($\mathrm{[Kr]}4d^{10}\,5s^0$). What reasons lead to such configurations?

More stability.

Don't answers to respective question explain it? https://chemistry.stackexchange.com/questions/2469/how-can-one-explain-niobium-s-weird-electronic-configuration?rq=1 https://chemistry.stackexchange.com/questions/151/why-do-elements-in-columns-6-and-11-assume-abnormal-electron-configurations

https://physics.stackexchange.com/questions/103718/aufbau-principle-in-modern-quantum-theory

The main conceptual problem is that you're assuming that simple principles lead to simple systems. They do not. The atom is a complex system based on very simple quantum mechanical and physical rules. In other words, the Aufbau principle is way too simple to capture the complexities of what you're trying to describe.

It's basically because you can't just add up electronic energies using a shell-by-shell, electron-by-electron approach.  As soon as you have two or more electrons in close quarters, they interact with each other in ways that cannot be described by simple electrostatic laws because of the quantum mechanical nature of the electrons.  The comments (specifically Mithoron's) refer to other questions whose answers explain this much better than I can.  Please look at those.

This is the first answer I am answering after joining with this forum, I read all the answers in the above and I try to add few more details rather them.

The aufbau principle states that in the ground state of an atom or
  ion, electrons fill atomic orbitals of the lowest available energy
  levels before occupying higher levels

As the present you can understand there are much of deviations now, that's why after that
*Pauli's exclusion principle
*Hund's rule of maximum multiplicity
such principles were generated. But as you say there were such deviations in some element's electron configuration(eg; chromium, copper, niobium, molybdenum, ruthenium, rhodium, silver, platinum) in chemistry, they have made reasons for them.As the @Fawad said they are for the more stability. Most of the d orbitals shows these deviations because of
*s orbitals has more geometrical stability than d orbitals
*d orbitals has shared 5 orbitals, but s has only 1 orbital, so it has much numeric stability
They are in two energy levels but due to such reasons the d orbitals has exceed the energy than the 4s. So when we are understanding the periodic table we should give more attention for the stability also.


text:
Rate of Decarboxylation of pyridinecarboxylic acids
What would be the rate of decarboxylation between the molecules described as pyridine molecules having $\ce{-COOH}$ group at ortho-, meta-, and para-positions?
In which molecule's decarboxylation would be fast or slow?

Put the $\ce{-COOH}$ group each at ortho-, meta-, and para-positions of pyridine to make 3 different molecules, then determine the rates.

What are your thoughts on the subject?

I know that a carbanion is formed in decarboxylation mechanism. So the rate depends upon the stability of the carbanion formed. But I can't decide how to judge the stability of carbanions so formed.

You are dealing with compounds that are both amines and acids. Think zwitterions NOT carbanions. Provide a diagram of your efforts if you have the capability/

The rate of decarboxylation of picolinic acid 1 is greater than isonicotinic acid 2. Zwitterion 4 has been proposed as the intermediate (Ref.1). Note that the electron pair is in the sigma framework. An electrostatic stabilization is proposed that requires that the charges be proximate. A competitive reaction in base (no zwitterion) demonstrated that 1 decarboxylated more than 100 times faster than 2 which was recovered after 20 days. Since nicotinic acid 3 is inexpensive and wasn't studied by the researchers, I suspect that it is stable also.

References:

G. E. Dunn, Gordon K. J. Lee, Harold Thimm, "Kinetics and Mechanism of Decarboxylation of some Pyridinecarboxylic Acids in Aqueous Solution," Can. J. Chem. 1972, 50(18), 3017-3027 (DOI: https://doi.org/10.1139/v72-480).

Why would 2 be better than 3 ? I suspect that carbanion would be better stabilized by inductive effects of nitrogen on 3 ?

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


text:
Finding final temperature for adiabatic expansion

An ideal gas, molar heat capacity $C_V=\frac{5}{2}R,$ is expanded adiabatically against a constant pressure of $1\ \mathrm{atm}$ until it doubles in volume. Given initial temperature of $25\ \mathrm{^\circ C}$ and initial pressure of $5\ \mathrm{atm}$, calculate final $T$ (etc.).*

The author uses the First Law to calculate $T_2$ as $274\ \mathrm K$.
My question is, why can't we use the relation $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$?
It appears we cannot, since $\gamma =C_p/C_V$ and so, since $V_2=2V_1$, $T_2=274\ \mathrm K$, $T_1=298\ \mathrm K$,
$$\log_2\frac{T_1}{T_2}+1 =\gamma = \frac{C_p}{5/2}\rightarrow C_p\approx 2.8, $$
so assuming $C_p - C_V = R$, giving $C_p = \frac{7}{2}R$, does not seem to work.
Can someone explain why the highlighted equation doesn't apply?
*This is adapted from an example in Castellan, Physical Chemistry.

The equations you are proposing to use are valid strictly for an adiabatic reversible expansion. In the present adiabatic expansion, the external force per unit area is dropped suddenly from 5 atm to 1 atm, and held at that value until the volume doubles.  This takes place rapidly and spontaneously, and is not reversible, so the equations for an adiabatic reversible expansion do not hold.  For an irreversible expansion, the ideal gas law (or other equation of state) cannot be used to describe the force per unit area acting on the piston face, because the equation of state only applies to a system in thermodynamic equilibrium (or, since a reversible expansion consists of a continuous sequence of thermodynamic equilibrium states, to a reversible expansion).  In an irreversible expansion, the force at the piston face (where work is occurring) is determined not only by the amount of volume change, but by the rate at which the volume is changing.  This is because of viscous stresses that are important in rapid expansions and compressions.  In addition, the mass (inertia) of the gas comes into play, causing the pressure and temperature to be spatially non-uniform within the gas during its deformation.  So there is no single uniform value of the pressure or the temperature that can be assigned to the gas.

I wonder if the definition of "adiabatic" used to include reversibility? Fermi's text, originally published in 1937, says on page 25: A transformation of a thermodynamical system is said to be adiabatic if it is reversible and if the system is thermally insulated..."

Interesting.  This surprises me.  Incidentally, Fermi's text is excellent in specifying that, in using Clausius' Inequality, the temperature to employ in dQ/T is required to be that at the interface between the system and its surroundings (typically a reservoir) where the heat transfer is occurring.

It's a great text and I assume he meant to restrict reversibility to the example at hand (which uses the boxed relations above, so it clearly required reversibility).


text:
How does the dx2-y2 orbital bond with four ligands in square planar metal-metal complexes
I understand that the complex [Re2Cl8]2-  has a bond order of 4 between the ruthenium ions. The bond comprises of sigma interactions between
Dz2 orbitals, two sets of pi interactions between dxz and d yz, and  delta interactions  between the dxy orbitals of the two ruthenium ions. My textbook says that the dx2-y2 orbitals remain for the metal ligand bonding.
How do you fit 16 electrons from 8 ligands into two orbitals,
I thought you could only fit two electrons in an orbital.
Are the dx2-y2 orbitals forming hybrid orbitals with the empty 4s orbitals ?

Eight of the 16 electrons are in the quadruple bond.

If you understand how a normal triple bond between 2 carbons has a bond order of 3, then it should be pretty clear why adding a $\delta$ bond in addition to this creates a bond order of 4.

Zhe, sure I understand that part, but its not quite what im stuck on

gsurfer04, yeah I agree that 8 electrons are in the quadruple bond, but we have to account for 24 electrons, 4 from each rhenium ion and 2 from each of the 8 ligands. If the dz2, dxz, dyz and dxy are used to form the quadruple bond, where are the remaining 16 electrons going, as all we have left over are two dx2-y2 orbitals which can hold a total of 4 electrons. So where do we put the other 8?


text:
Epoxide ring opening via decarboxylation

The answer for the above multiple choice problem is option (C), from which I can understand that at some point in the mechanism the molecule must've undergone decarboxylation.
Also, I think the first step should be protonation of the epoxide oxygen. (No other choice, isn't that the most nucleophilic site?)
I'm not able to understand how the ring opening takes place (guessing it is via decarboxylation, but how?)

You are correct about protonating the epoxide of the glycidic acid 1 (by some other molecule of glycidic acid). Ring opening gives stable, tertiary cation 3 and NOT cation 6, which is destabilized by the carbonyl group. Collapse of 3 gives CO2 and the enol 4 which tautomerizes to the aldehyde 5.

Cool! Thanks a lot. Another small question, could you tell me about other 'unusual' or 'unexpected' epoxide ring openings that come to your mind? I prefer calling this one unusual because it's very rare (first time for me) to see an epoxide ring being opened via decarboxylation.

A similar issue is the reaction of some epoxides like 2,2-dimethyl oxirane with a Grignard reagent (RMgBr). The Lewis acid Mg++ coordinates with the epoxide and causes rearrangement transferring  hydride to the tertiary site yielding aldehyde 5. It then reacts with the Grignard reagent to give a secondary alcohol. This rearrangement competes with direct additions of RMgBr to the epoxide.

@user55119 Could you please explain how exactly 3 "collapse"d? I mean, how can we predict that a carbocation is going to collapse? And that too only by decaboxylation?

@GaurangTandon: Think of this as an E1 elimination with the loss of CO2. I don't know how general the procedure is but look at the uses of the glycidic esters from the Darzens condensation: http://www.organic-chemistry.org/namedreactions/darzens-reaction.shtm

To be honest, the E1 elimination does still feel "weird" to me. I have only ever seen it in Hoffmann and Saytzeff elimination of halohydrocarbons. Considering I am a high schooler, I guess there's no better way to explain this to me, is there?

A close example is the addition of Br2 to alpha-methyl acrylic acid (CH2=C(CH3)CO2H).  Heat CH2Br-CBr(CH3)CO2H with pyridine (weak base), you get deprotonation of the CO2H, loss of CO2 and the terminal Br. The product is 2-bromopropene. This one is like an E2 elimination.  And I am well, well, well beyond my high school years. That translates to a lot of experience.


text:
Condensation Reaction
In the aldol condensation reaction of a ketone with an aldehyde, both having acidic alpha hydrogens present; what is the major product?
Is it formed from enol of the ketone, or of the aldehyde? Is it from the enol of the ketone, because aldehydes are more reactive towards nucleophilic addition? Confirming, just to be sure.
Furthermore, which aldol condensation products are usually neglected?

Depends on the identity of the aldehyde and the ketone, and how the reaction is ran. Practically speaking, we often first enolise one component irreversibly first before adding the other (usually the aldehyde)

The answer to these questions literally fills entire chapters of reference books...

The aldol condensation is usually conducted in base. We should be talking about enolates, not enols (acid medium). As Zhe has said, this a broad topic. Provide specific examples.


text:
Applying Huckel's rule to non-planar compounds

Which of the following compounds is aromatic?

From the 4n+2 rule, it must be either option 2 or 4. The answer given in my book is option 4. Why? I think It might be due to one being planar and the other not, but to me both seem to be planar. Is there a way to check if one is planar or not?

Number 2 is not aromatic because transannular $\ce{C-H}$ interactions cause the molecule to be non-planar although it satisfies $4n+2$. This problem was solved by the synthesis of number 4, which, while a little less than planar but rigid, displays aromatic properties (NMR, etc., Ref 1).
References:

E. Vogel, H. D. Roth, "The Cyclodecapentaene System," Angew. Chem. Int'l. Ed. 1964, 3(3), 228-229 (DOI:  https://doi.org/10.1002/anie.196402282).

Yeah, there was also analogous ether IIRC.

If you have a molecular model kit, or a computer simulation that plays a similar role, use it to build both molecules.  You discover that neither (2) nor (4) is entirely planar: (4) is not planar because the internal methylene group is out of plane (the only way to get realistic bond angles at the central carbon), and you find (2) is not planar because of the internal hydrogen atoms pushing each other out of the way and distorting the ring accordingly.  But (4) comes much closer to being planar.  Thus (4) is more likely to show at least some of the conjugation which, with an optimal $4n+2$ electron count, makes the ring aromatic.


text:
Etymology of saturation degrees (-ane, -ene, -yne) in aliphatic compounds
I like to know where the names for saturated, unsaturated double, and unsaturated triple bonded aliphatic compounds came from.
What is the reason/etymology behind the suffix -ane, -ene, and -yne?
I suppose it parallels something like the inflection-based umlauts in Germanic languages (which may be seen rudimentary in modern English for example in words like mouse/mice) but that is utter speculation. Yet I did not find anything in the web (I think some good extensive Chemical Etymological Encyclopaedia is really missing).

I found the following information through a website linked to Yale University.
The naming structure seems to have evolved from some of the early nonsystematic names given to hydrocarbons. It started with Dumas who in 1835 came up with the name Methylene for the $\ce{CH2}$ radical. The "-ene" suffix is a Greek feminine patronymic which was used to denote that Methylene was the "daughter of" (derived from) the "spirit of wood" or what we now call methanol (Dumas believed methanol had the formula $\ce{CH2\cdot H2O}$).
The naming structure caught on for ethylene since it was related to ethyl(even though the meth- in Methylene came from a Greek work for wine and not the numerical prefix used in methyl).
When the use of these feminine patronymics (-ene, -ine, -one) became common during the 1860s, Hoffman proposed making a consistent nomenclature that followed the standard progression of vowels: a e i o u. This led to the suffixes -ane, -ene, -ine(-yne), -one, and -une to refer to increasingly unsaturated hydrocarbons.
Since -one was already in use for ketone structures, only the first three suffixes caught on.

This answer probably gets pretty close to the truth (+1). A small comment. The word methylene (Fr. methylène) was probaby co-coined by Dumas and Péligot ($\to$DP) in 1834. As you note, -ene is from Greek; however, the interpretation is a bit debatable (imo). DP believed that methylene formed the essential radical of methanol (or wood spirit). It would make more sense for wood spirit to be the 'daugther' than the other way round. In Flood's reference: '-ene ... is a name-forming suffix; it has no real meaning in itself.'  $|$ ref: W. A. Flood, 1963, The Origins of Chemical Names

@LinearChristmas Petit Robert (1996) notes: "méthylène, n.m. – 1834; du gr. methu «boisson fermentée» et hulê «bois»", or (freely translated to English:) "noun (male gender), first referenced 1834; derived from Greek methu (fermented drink) and hulê (wood)".  Which would be in line with alcool de bois (French) and wood alcohol as sometimes seen description for methanol (contrasting to alcool vinique (French) and spirits of wine about ethanol).

August Wilhelm Hofmann, of rearrangement and elimination fame, suggested the hydrocarbon systematic suffixes in 1866.
http://chem125-oyc.webspace.yale.edu/125/history99/5Valence/Nomenclature/alkanenames.html
http://www.tandfonline.com/doi/full/10.1179/1745823414Y.0000000006

You should try to avoid answers that are mostly link-only. Usually, you want to give a little more detailed a summary of the links, especially in cases where they aren't accessible or may not be in the future.

The question was answered. One should be careful about "paraphrasing" other's work. The author of the site is extremely knowledgable.

But I would argue it doesn't necessarily answer the question. Without going into why he chose those suffixes, it doesn't really inform why Hoffman didn't choose any other possible notation for the suffixes. I would also say we can't really avoid paraphrasing in science. Almost every scientific article I have read has had some citation of prior work that requires summarizing that work and it's implications. You definitely still have to cite your reference in some way, but just pointing to a reference on its own usually isn't enough.

Fair enough. Use caution.

I find the linked article (2nd link) a very useful complementation to the answer of Tyberius.


text:
Number of Sigma Bonds in this Structure Containing a Ring
(This question was on my last gen. chemistry exam.)
I counted 12 sigma bonds, but the answer was 13. I'm taking this course online so there is no explanation given. Could anyone offer a quick explanation on how this structure could give 13 sigma bonds?

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

You might also want to give the site's homework policy a read. In this particular post, simply counting the number of bonds would suffice as "effort" IMHO (I don't see what more could be done). Do bear in mind though, that homework-type questions (posted without any effort taken towards solving it) tend to get closed. Welcome to Chemistry.SE, cheers! :-)

The simple answer is that all bonds (in this molecule) contain $\sigma$ orbitals but a few also contain $\pi$ orbitals!

There're are 13 sigma bonds there (count the red lines),

You've clearly tried to attempt this yourself, so good on you (and by virtue of that, it's equally clear that you counted the bonds too hastily).
I counted (quickly, because it looked simple enough) and got 12 too at first. Counted a little slower the second time (I was actually going to attempt to prove to you that the number of bonds is 12), turns out the ring is not entirely conjugated, so one carbon atom in it actually has two hydrogen atoms bonded to it. ;-)


text:
Cone voltage on mass spectrometers
Why is the cone voltage on mass spectrometers positive for positive ions?
I would have thought that the positive ions would be attracted to the
cone by a negative potential. Is the voltage, perhaps, indicative of something else (and not of the degree of attraction between ions)?

A mass spectrometer typically uses a beam of electron to produce positively charged molecular ions or molecular ion fragments. Since the ions are  positively charged a negative potential would be used to accelerate the ions.

Thats what I thought MaxW but all the references i have refer to a voltage

of about 10 - 100 volts and there is no mention of this being a negative voltage

Short answer: one should not consider the absolute value of a potential as long as the potential difference drives the ions in the appropriate direction.
Longer answer:
The potential difference is needed to accelerate ions in the appropriate direction of the mass spectrometer.
One should consider that in an atmospheric pressure or moderate vacuum ions need to be driven by the electric potential difference towards the MS high vacuum region by a constant decrease in potential (due to loss of kinetic energy through collisions with the gas) whereas in the high vacuum region, there is no need to keep this evolution true as long as the potential difference remains positive relative to the initial (entrance, near thermal kinetic energy) potential.
Just as an illustration, considering a simplified ESI atmospheric pressure introduction with only an ESI capillary, a cone to rough pumping region and a skimmer to the UHV region, one can have:
ESI Needle: + 4,000 V
Cone: + 100 V
Entrance skimmer: 0 V
Or
ESI Needle: + 4,200 V
Cone: -100 V
Entrance skimmer: -200V
The ions will see the same electric field in both cases, as it is the potential difference which creates the electric field and not the absolute value of the potential. There is just one issue: sometimes it is easier in instrument design to have one region of the instrument operated at ground potential (or close to ground potential), and it would be this part of the design which will decide on the absolute values of the potentials that will be applied.


text:
Which substance has the fastest rate of evaporation at room temperature?

Which substance has the fastest rate of evaporation at room temperature?
A) $\ce{CH3OH}$
B) $\ce{C7H16}$
C) $\ce{C6H14}$
D) $\ce{C4H10}$

Really?!? This is homework. You can't google the boiling points of these?

@MaxW I'm trying to understand it conceptually. thanks.

The boiling points would give you a good indication. The vapor pressure at room temperature would be exact. B,  C, and D have a isomers. Truthfully figuring out the isomers and looking up the data will help you learn something about chemistry that us just giving you the answer won't teach you.

The rate of evaporation depends not only on the equilibrium vapor pressure but also on the diffusion coefficient of the species in air (assuming that the temperature is somehow held constant).

Why don't you just look up the vapor pressures and diffusion coefficients (in air) of these substances at room temperature and do a mass transfer analysis to estimate the rate of evaporation of each?

OK, I started a table for you. I think all these chemicals would be in Wikipedia. Not sure if all the data is there, but I'd guess so. As I said the three alkanes have multiple isomers, so you need to copy lines in the table.
\begin{array}{cccc}
\hline
\text{Formula} & \text{Name} & \text{Boiling point} & \text{Vap. press.} \\
&&&(\pu{kPa @ 20 ^\circ C}) \\
\hline
\ce{CH3OH} & \text{Methanol} & 64.7 & 13.02 \\
\ce{C7H16} & & & \\
\ce{C6H14} & & & \\
\ce{C4H10} & & & \\
\hline
\end{array}

Let's start with easy one. How many isomers does $\ce{C4H10}$ have? What are the names?

A) CH3OH has hydrogen bonds therefore is much stronger than the dispersion forces in the other molecules, so wouldn't it have the slowest rate of evaporation?

$\ce{CH3OH}$ is so different than the other molecules that such comparison is pretty useless. For isomers I can take an educated guess at to which isomer will have greatest and lowest volatility. // I just know the answer because I've handle a lot of these chemicals. But If I just tell you then you won't learn anything. If you go to the trouble of figuring this out you will really learn something. // Again - How many isomers does C4H10 have? What are the names?


text:
Unsaturated ketone coupling pattern
Can someone explain to me the coupling pattern by proton $5$? Why is it triplet of quintet? Is it long range coupling?
I'm assuming that the proton $5$ couples to two other protons on carbon $4$. But what's next?

Sure, it couples with two other protons on C4, hence the triplet.

It's a triplet of septuplets. It's hard to see, but you can see the seventh peak sticking out.

You are close. The triplet part is the coupling of the enantiotopic hydrogens at C4. The width of the C5 blowup is ~20 Hz in width which means J = ~7 Hz. The smaller coupling is due to allylic coupling with the methyls (C7 and C8). Assuming J5,7 and J5,8 are the same, then the six hydrogens produce a septet. You just can't see the outer peaks because they are too small. If coupling to the two methyls is different then the septet will appear broadened because of lack of resolution. A larger blowup of C5 would reveal the outer peaks (one and seven).


text:
Why does [H3O+] of a buffer change very slightly when diluted?
I am confused as to why the [H3O+] of a buffer changes very slightly when diluted (at least in an intuitive sense).
Consider:
$\ce{HNO2 + H2O <=> H3O+ + NO2-}$

[$\ce{HNO2}$] = 0.45 M and [$\ce{NO2-}$] = 0.30 M yields [$\ce{H3O+}$] of $6.8 \times 10^3 M$. After solving: $4.5 \times 10^{-4} = \frac{x(0.30+x)}{0.45-x}$ where the ionization constant of nitrous acid is $4.5 \times 10^{-4}$

If these concentrations were to be divided by two (the solution is diluted by a factor of 2)

[$\ce{HNO2}$] = 0.225 M and [$\ce{NO2-}$] = 0.15 M yields [$\ce{H3O+}$] of $6.7 \times 10^3 M$. After solving: $4.5 \times 10^{-4} = \frac{x(0.15+x)}{0.225-x}$

Mathematically this makes sense, but I can not intuitively reason why a buffer would behave this way in the "real world." Why can buffers resist changes in pH so effectively when diluted?

In the real world it doesn't quite work out that way. The equilibrium depends not on the concentration but the activity of the chemical species. In dilute solutions the two are pretty much equal. But about 0.1 molar and above the activity is less than the concentration. In essence the ions are not just floating around the solution bare. Because water molecules are polar, each ion has multiple spheres of water molecules oriented towards the ion. So in essence, at higher concentrations, the water molecules get used up and the oriented spheres start to interact with each other.

Buffers also don't stay constant when greatly diluted because of the autodissociation of water. But that leaves a great range where they are effective. Also a buffer doesn't typically need to maintain an exact pH, but rather +/- 1 pH unit.

The solutions of the equations are wrong. They should be $x = 6.725 \cdot 10^{-4}$ for the first one, and $x = 6.700 \cdot 10^{-4}$ for the second one, both corresponding to a pH of about 3.17. The mathematical reason is pretty clear I suppose (essentially, $x = [ H_3 O^+ ]$ is much smaller than the acid and anion concentrations you're considering). But IMO that's also the intuitive reason: if on dilution the ratio between the concentration of the acid and anion species stays ~constant, from the Henderson-Hasselbach equation you know that $[ H_3 O^+ ]$ will also stay ~constant.

BTW, if you want to do a more accurate calculation that is valid for lower concentrations too (but not 100% exact, due to the activity as MaxW observed), you need to write the mass balance for each species, the charge balance (including $[OH^- ]$) and the equilibrium constants (including water), and eliminate all unknowns except $[ H_3 O^+ ]$. If you do things right, you will be left with 1 equation in $[ H_3 O^+ ]$ and known constants, which you can solve (numerically) to obtain the pH. This will show you that below a certain concentration, dilution does change the pH a lot.


text:
How to form 6-31G basis set; specific instructions
I'm new to basis sets. I read that for 3-21G, given the following basis set for carbon (from Basis Set Exchange):
C    S
    172.2560000              0.0617669
     25.9109000              0.3587940
      5.5333500              0.7007130
C    SP
      3.6649800             -0.3958970              0.2364600
      0.7705450              1.2158400              0.8606190
C    SP
      0.1958570              1.0000000              1.0000000

You form three 'atomic orbitals' (is that the right term?) as:
let f(alpha) = (2*alpha/pi)^(3/4) * exp(-alpha*r^2)

O1 = 0.061 f(172) + 0.358 f(25) + 0.7 f(5.5)
O2 = -.39 f(3.66) + 1.21 f(.77) + 1 f(.195)
O3 = 0.23 f(3.66) + 0.86 f(.77) + 1 f(.195)

I truncated the numbers to make it easier to type...

Assuming that is correct, that makes sense to me. But when I look at the 6-31G basis set data for carbon:
C    S
   3047.5249000              0.0018347
    457.3695100              0.0140373
    103.9486900              0.0688426
     29.2101550              0.2321844
      9.2866630              0.4679413
      3.1639270              0.3623120
C    SP
      7.8682724             -0.1193324              0.0689991
      1.8812885             -0.1608542              0.3164240
      0.5442493              1.1434564              0.7443083
C    SP
      0.1687144              1.0000000              1.0000000
C    SP
      0.0438000              1.0000000              1.0000000

From the naming nomenclature, I thought the only difference would be 6 primitive gaussians for the inner shell which I see, but . . . I was not expecting that third SP line. I don't know how to use it.
I want to write:
O1 = .0018 f(3047) ...
O2 = -.119 f(7.86) + ... + 1.0 f(.168)
O3 = 0.068 f(7.86) + ... + 1.0 f(.168)

Just like I did for the 3-21G case.
Maybe I could add:

O4 = -.119 f(7.86) + ... + 1.0 f(0.0438)
O5 = 0.068 f(7.86) + ... + 1.0 f(0.0438)

But I don't think this is right because I read online that there should be 9 orbitals formed (although they didn't explain how to get them).
Thanks for the help!
Update
Ok, so I looked up 6-311G instead of 6-31G. However, I still don't see how to form 9 orbitals from 6-31G:
C    S
   3047.5249000              0.0018347
    457.3695100              0.0140373
    103.9486900              0.0688426
     29.2101550              0.2321844
      9.2866630              0.4679413
      3.1639270              0.3623120
C    SP
      7.8682724             -0.1193324              0.0689991
      1.8812885             -0.1608542              0.3164240
      0.5442493              1.1434564              0.7443083
C    SP
      0.1687144              1.0000000              1.0000000

I really need to see all 9 orbitals written out explicitly.
Update 2
Ok, with more research I finally found this
So it looks like I'm missing a function. I need to define:
f1(alpha) = (2*alpha/pi)^(3/4) exp(-alpha r^2)
f2(alpha, x) = (128 alpha^5/pi^3) x exp(-alpha r^2)

Then the 9 orbitals are:
O1 = .001 f1(3047) + .014 f1(457) + ... + .362 f1(3.16)
O2 = -.119 f1(7.68) + -.16 f1(1.88) + 1.14 f1(.54)
O3 = 1.0 f1(.168)
O4 = .068 f2(7.68, x) + .316 f2(1.88, x) + .744 f2(.54, x)
O5 = 1.0 f2(.168, x)
O6 = .068 f2(7.68, y) + .316 f2(1.88, y) + .744 f2(.54, y)
O7 = 1.0 f2(.168, y)
O8 = .068 f2(7.68, z) + .316 f2(1.88, z) + .744 f2(.54, z)
O9 = 1.0 f2(.168, z)

Can anyone confirm this? Or, if not confirm, specify what's wrong?
Thanks!

Could you possibly be looking at 6-311G?

Tyberius is correct - in Pople notation, you're looking at a 6-311G basis set. That said, for a 6-31G, there will be 10 atomic orbitals (not 9, as you state).

@Tyberius : I just went back to Basis Set Exhange, and yes, it seems that I inadvertently downloaded the 6-311G  set instead of the 6-31G.

@ToddMinehardt : you're right, I had 6-311G instead of 6-31G. But still, I only see how to form 3 atomic orbitals: 1) the 'S' data 2) the first set of coefficients of the first 'SP' + the second 'SP', 3) the second set of coefficients of the first 'SP' + the second 'SP'. I don't know if that is right...
Could you please explain how you get 10 orbitals?

My mistake - it is 9, not 10.

I believe you have not correctly interpreted the first basis set, which should read:
O1 = 0.061 f(172) + 0.358 f(25) + 0.7 f(5.5)
O2 = -.39 f(3.66) + 1.21 f(.77)
O3 = 1 f(.195)

Note that these are only the s-orbitals. Using the coefficients in the right-most column, two triplets of p-orbitals are also present. By doing it this way for the second basis set, which is 6-31G, you would obtain 3 s-orbitals and two triplets of p-orbitals, making it nine AOs in total. Note that all orbitals in a tuple must have the same contraction scheme, otherwise, the calculation would not be rotationally invariant (to within integration accuracy).
Aside: Why does this basis specify s- and p-orbitals this way?
Depending on the integral scheme, one can reuse certain intermediates of some integrals for others and save some computational time. However, modern programs tend to work better with segmented basis sets (i.e. basis sets that do not allow for reuse, because the $\alpha$ exponents are all different). The results are also of higher quality. IMHO, the Pople basis sets are only of historical interest or for hobbyists with very limited resources.


text:
Why do we get slightly more inversion product than retention product in SN1 reaction?
Theoretically, in $\mathrm{S_N1}$ reaction we should get a racemic mixture as the product because the nucleophile can attack from either side of the formed carbocation.
However, many textbooks claim that there's slightly more inversion product than retention product.
Is there any specific reason for this to happen.

It is well known that SN1 reactions often give incomplete racemisation:

Although many first-order substitutions do give complete racemization, many others do not. Typically there is 5–20% inversion, although in a few cases, a small amount of retention of configuration has been found. These and other results have led to the conclusion that in many SN1 reactions at least some of the products are not formed from free carbocations but rather from ion pairs.

We can explain this by realising that the SN1 and SN2 mechanisms are not really discrete, as often taught, but rather two extremes of a 'mechanistic continuum'. This has implications for both the stereochemical outcome of many substitutions (i.e. proposed SN2 are not always as stereospecific as one might expect), and also the kinetics (i.e. measuring the rates of substitutions won't always give data consistent with purely SN2 or purely SN1 mechanisms).

At the SN1 extreme, there is no covalent interaction between the reactant and the incoming nucleophile during the transition state (TS) for the cleavage of the leaving group.
At the SN2 extreme, the TS has covalent interactions in which there is concerted formation of a new bond between the incoming nucleophile and the reactant , with cleavage of the bond to the leaving group.

This mechanistic continuum was first explained by Winstein, who studied both the kinetics and resulting stereochemistry of nucleophilic substitutions at saturated carbons.[1] This model is widely accepted, though rarely taught at the undergraduate level.
Winstein's theory proposes the presence of several additional intermediates along the reaction coordinate- these are known as 'ion pairs' and represent varying levels of association between the reagent and the leaving group. The energy levels between these ion pairs are small, and nucleophilic attack can occur at any stage (hence the continuum).

 Fig1: Winstein model for nucleophilic substitution. Taken from Carey's Advanced Organic Chemistry. Part A. /2
The Winstein model fits with what is commonly taught, namely that SN1 gives racemisation and SN2 gives inversion:

SN2  reactions occur when the nucleophile intercepts the contact ion pair and give stereospecific inversion of configuration since the leaving group is still associated with the carbocation, essentially shielding one face from attack
SN1 reactions occur when the nucleophile intercepts the fully dissociated free carbocation and gives complete racemisation as both faces of the planar carbocation are equally accessible

Which mechanism operates is largely determined by how stable the carbocation is. Primary carbocations are highly unstable and as such there is little (if no) dissociation. Tertiary cations are stable and as such complete dissociation can occur.
The interesting case is where nucleophilic attack occurs on a solvent separated ion pair - we see neither complete inversion nor complete racemisation, as some association still exists at this stage (partially, but not completely blocking one face of the cation).
Modern Physical Organic Chemistry[3] gives an overview of all of the possibilities, though the figure is a little hard to follow:

References

Winstein, S.; Clippinger, E.; Fainberg, A. H.; Heck, R.; Robinson, G. C. Salt Effects and Ion Pairs in Solvolysis and Related Reactions. III.1Common Ion Rate Depression and Exchange of Anions during Acetolysis2,3. J. Am. Chem. Soc. 1956, 78 (2), 328–335. DOI: 10.1021/ja01583a022.

Carey, F. A.; Sundberg, R. J. Advanced Organic Chemistry, Part A: Structure and Mechanisms, 5th ed.; Springer: New York, 2007.

Anslyn, E. V.; Dougherty, D. A. Modern Physical Organic Chemistry; University Science Books: Sausalito, CA, 2006.

As we know, $\ce{S_N1}$ reaction is stereochemically non specific. This means, that it does not force the formation of one stereoisomer, as $\ce{S_N2}$ does. In $\ce{S_N1}$ reaction, presence of carbenium ion (=carbocation) as intermediate, results in the feasibility of attack from both sides of the ion, resulting in the formation of both stereoisomers(R and S).
In an ideal condition, where a "free" carbocation is generated, the amounts of R and S isomers of the products will be exactly 50% each, provided that there is no steric hindrance in one side that makes the formation of one isomer advantageous. In simple words, if both sides of carbocation is similar, then 50% of each isomer should be produced.
Howewer, in reality, completely "free" carbenium ion is not formed. The leaving group, in most cases, remains weakly attracted to the carbenium ion due to electrostatic forces, so it occupies the front side of the carbenium ion, and as a result, backside attack becomes more advantageous and less sterically hindered than front side attack.

In practice, few $\ce{S_N1}$ reactions give the expected $\ce{50:50}$ mix of enantiomers, and inversion of configuration exceeds retention to an extent of up to $\ce{\approx20\%}$. This probably occurs because the 'real' mechanism involves a transition state in which the carbenium ion and the leaving group are still loosely bound by electrostatic interactions, making attack from the side opposite to the leaving group less sterically hindered than attack from same side. In other words, the 'real' $\ce{S_N1}$ mechanism involves a transition state that lies somewhere between the two extremes that we have described for the $\ce{S_N1}$ and $\ce{S_N2}$ pathways.
[-An Introduction to Organic, Inorganic and Physical Chemistry. 4th edition. Catherine E. Housecroft, Edwin C. Contsable. Pearson Education Limited. 2010]

Therefore, the amount of product showing inversion is more than the amount of product showing retention of configuration. The actual percentages depend on the reactants involved.


text:
Why is free radical chlorination exothermic whereas bromination is endothermic?
In the energy profile diagrams below, it can be clearly seen that the free radical chlorination of an alkane
$$\ce{RH + Cl2 -> RCl + HCl}$$
is exothermic. However, the corresponding bromination
$$\ce{RH + Br2 -> RBr + HBr}$$
is endothermic. Why is there a difference?

That's because both reactions products, hydrohalogenide and halogenoalkane are more stable with chlorine. The orbital overlap is not as good in the case of the larger bromine.

This earlier answer may be helpful.

It looks like your energy diagrams are for the first propagation steps not the full reaction [RH + X(dot) ---> R(dot) + HX]. You also need R(dot) + X2 ---> RX + X(dot) for the second step. Chlorination is more exothermic than bromination but both are exothermic.

First off, we can see that both reactions are exothermic If we do the thermodynamic analysis.
Given that your starting materials have the same heat of formation:
$$\Delta H^\circ_f(Products) = \Delta H^\circ_f (\ce{R-H}) + 0, \quad{\Delta H^\circ_f(\ce{Cl}) = \Delta H^\circ_f(\ce{Br})= 0}$$
$$\ce{R = Et}\\ \Delta H^\circ_f (\ce{EtH}) = \pu{-84 kJ mol^{-1}}$$
$$
\begin{array}{cccccc}
\hline
\text{Product 1} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \text{Product 2} & \Delta \text{H}^\circ_f [\pu{kJ mol^{-1}}] & \Sigma\ \Delta \text{H}^\circ_f\  [\pu{kj mol^{-1}}] &\text{Result}\\
\hline
\ce{HBr} & -\text{36.45} & \ce{Et-Br} & -\text{97.6} &-\text{134.05}<-\text{84} &\text{Exothermic} \\
\ce{HCl} & -\text{92.31} & \ce{Et-Cl} & -\text{137} & -\text{229.31}<-\text{84}  &\text{Exothermic}\\
\hline
\end{array}
$$

Back to your question:

Why is there a difference?

The graphs you are looking at only consider the reaction:
$$\ce{R-H + X^. ->  R^. + HX}$$
For which we get:
$$
\begin{array}{cccccc}
\hline
\text{X}
& \Delta \text{H}^\circ_f(\ce{X^.})
& \Delta \text{H}^\circ_f(\ce{RH})
& \Delta \text{H}^\circ_f(\ce{R^.})
& \Delta \text{H}^\circ_f(\ce{HX})
& \Delta \text{H}^\circ_{rxn}
& \text{Result}\\
\hline
\ce{Br} & -\text{96.94}^{**} & -\text{84} & -\text{119}^\dagger &-\text{36.45} &\text{25.49}&\text{Endothermic} \\
\ce{Cl} & -\text{121.29}^{**} & -\text{84} & -\text{119}^\dagger & -\text{92.31} &-\text{6.02}&\text{Exothermic}\\
\hline
\end{array}
$$
*all units in $[\pu{kj mol^{-1}}]$ all values from Wikipedia unless otherwise stated
** Table of Bond Dissociation Energies
$^\dagger$ Values from NIST WebBook
Here the enthalpy of radicalization of the alkane is different for chlorine or bromine, but the halogenation in total is exothermic


text:
What would be the correct mechanism for hydration of ethene?

Given above is the mechanism that I drew but my textbook puts it a bit differently. It says that the ethene is dissovled in the concentrated $\ce{H2SO4}$ to form ethyl hydrogen sulphate $\ce{CH3-CH2OSO3H}$ which when diluted with water reacts to form ethanol and the acid
$$\ce{CH3CH2OSO3H + H-OH -> CH3CH2OH + H2SO4}$$
So which mechanism is true?

I think that the first mechanism occurs when a solution of sulphuric acid in water is reacted with ethene. The second mechanism occurs if you first add conc. H2SO4 to ethene, and then add water. I am not fully sure, though. I will answer if I get a reference to back my claim.

At first I thought your mechanism was very comprehensive but upon closer inspection I think it has some flaws because in this reaction conc. sulphuric acid is used as a catalyst and it must be regenerated at the end and in the above mechanism only ethanol is generated.

This must be the mechanism shown in your book.

Hmm, yes but no but yes but no …

The sulphuric acid is indeed regenerated in the proposed mechanism, no?


text:
Rust removal ended with black oxide on the metal?
I have tried to remove common red iron rust from a small piece of metal that looks like steel.
I submerged it into hot water with a bit of citric acid, and let it sit overnight. In the morning I have no rust, but the thing is now covered in dark-black oxide.
Was that possibly steel that I was cleaning or something else? How do I remove that black oxide?
How should remove rust in situations like this in the future?!
left - piece from the same metal that has no rust, and was not cleaned, right - piece that was cleaned

A picture would really help IMHO.

In the absence of oxygen iron reacts with water forming $\ce{Fe(OH)_2}$, which can react with rust forming $\ce{Fe_3O_4}$ via the Schikorr reaction.  See https://en.m.wikipedia.org/wiki/Schikorr_reaction.
You should use a stronger acid, which will dissolve the rust and bring the iron ions into solution.  You will, of course, lose the original metal that rusted instead of recovering it, but you can't make such a recovery by "home remedies" anyway.  With such an aggressive approach rethink soaking overnight.

I used the fourth method from https://www.apartmenttherapy.com/how-to-clean-rust-off-old-loaf-141206 What did I do wrong?

Did you try the scrubbing step?  Also I see no indication of how heavy a rust stain can be dissolved versus ending up reacting as above.


text:
Reimer–Tiemann-like reaction of catechol with diiodomethane
I have studied the Reimer–Tiemann reaction of phenol with chloroform in presence of aqueous strong alkali. It takes place with intermediate dichlorocarbene.
I have also studied the Reimer–Tiemann "carboxylation" reaction, which requires carbon tetrachloride, and proceeds with intermediate carbocation.
I am however unfamiliar with what the following question requires from me:
$$\ce{Catechol + CH2I2 + NaOH -> ?}$$
Can anyone please state the step-wise mechanisms through which this reaction occurs? Thanks!

Methylene iodide appears to be at the wrong oxidation to give a Reimer-Tiemann product. The more likely product is benzo[d][1,3]dioxole aka methylenedioxy benzene. The catechol  dianion does  a double SN2 displacement on CH2I2. The second displacement is intramolecular.

@user55119 I've understood what you're saying in the last 2 sentences. And yes, this is exactly the product given in my textbook! But, the book does not mention any mechanism. Why is the mechanism SN2 here? I mean, in the above two reactions, there was Riemer Tienman reaction. But, in this case, it is not. You gave the reason in your first sentence but sorry I couldn't understand it.

In methylene iodide the carbon has a charge of zero. In chloroform carbon is +2. The carbon in CCl4 is +4. Thus, CCl4 gives a carboxyl group as you have noted. Chloroform leads to an aldehyde group. But CH2I2 undergoes displacement faster than a carbenoid species can arise from CH2I2. One might imagine the carbene "CHI" forming. If if it reacts with the negative oxygen, it is the same intermediate as the first SN2 displacement. If the carbene  reacts with a ring carbon, the best you can hope for is a benzylic iodide and then a benzylic alcohol in the presence of base.

@user55119 Thanks! You're mentioning that "Methylene iodide appears to be at the wrong oxidation to give a Reimer-Tiemann product." So, from what I can infer from all your comments, is it a fact that only certain oxidation numbers can give certain products? Is there a rule through which I can predict which oxidation numbers lead to which intermediates?

Go by the oxidation number of carbon as a general rule.

A mechanism similar to https://books.google.co.in/books?id=PLHRtYzD3b8C&pg=PA186&lpg=PA186&dq=dioxole+formation+using+CH2I2&source=bl&ots=5U68ke71iP&sig=7vNNWfcydwUG13uqYb0PFwrBSzo&hl=en&sa=X&ved=0ahUKEwid1tXWx4LYAhVGtI8KHRHCDOUQ6AEIKDAA#v=onepage&q=dioxole%20formation%20using%20CH2I2&f=false

@EashaanGodbole Thank you for helping out. Unfortunately many sources that may be accessible to you via google books are not accessible to others. It is therefore always better to include a human-readable reference, including the page number, so that one could look up this via alternative means.

From the comments I assume that the reaction is an exercise from a textbook. It would probably be helpful to give the complete reference, so that it could be viewed in context. It is also always a good idea to include your own thoughts about the problem in the post.

@Martin-マーチン Thanks! I am not sure you can view it from here, but this book. Page 166 under the heading "preparation of piperonal". It uses 3-4-dihydroxy-benzenecarbaldehyde instead of catechol, but the reaction is same, with CH2I2 and NaOH. (cont)

(cont) (Note: The question I posted above based on catechol came from my coaching textbook but (a) it is not available in market (b) the answer it gave is the same as given by user above)

@Martin-マーチン I was referring to the only page that is visible as Gaurang pointed out.

@EashaanGodbole Sorry for not replying; yes I can view your link. The reason why I did not reply because I thought you were leaving your comment as a "note" for anyone interested in posting a detailed answer. You  didn't detail anything in your comment about the mech you linked to, so I thought your tone was "I am leaving this link here in case anyone wants to have a look". if you've more detail to add to wrt that linked page, please do so!

@EashaanGodbole The problem is: It was and is not visible to me. So I cannot tell what you were referring to.

This is not really an answer to your question, but I do not think that it is possible to answer it. From one of the many comments we learn that the reference is actually the "preparation of piperonal". It uses 3-4-dihydroxy-benzenecarbaldehyde instead of catechol, but the reaction is same, with $\ce{CH2I2}$ and $\ce{NaOH}$.
I was able to find this reaction described in Chemistry of Natural Products
by Sujata V. Bhat, B.A. Nagasampagi, Meenakshi Sivakumar on pages 307f.

4.22.B. Synthesis of Piperine
Piperine was synthesized (Ladenburg, 1894) by the reaction of the piperic acid chloride with piperidine, which confirmed the structure of the molecule (Figure 4.57). The synthesis of piperic acid was achieved starting from piperonal 4.288, which was obtained from catechol using Reimer-Tiemann reaction followed by the condensation with diiodomethane in the presence of a base. Piperonal was condensed with acetaldehyde in the presence of sodium hydroxide and the product obtained was then heated with acetic anhydride and sodium acetate to yield piperic acid.

I find this a lot more in accord from what you would expect. I cannot imagine any way in which a Reimer-Tiemann (RT) like reaction would occur given the starting reagents. Instead the first step from catechol to 2,4-dihydroxy-benzenecarbaldehyde occurs via an RT reaction. This is very well discussed by H. Wynberg (Chem. Rev. 1960, 60 (2), pp 169–184, DOI: 10.1021/cr60204a003). The mechanism can also be found on Wikipedia.
The second step then involves diiodomethane and forms the acetal. I suppose after deprotonation of the phenolic hydrogen, it can attack at the diiodomethane in nucleophilic substitution fashion, further following another deprotonation and an intramolecular SN step to close the ring.

Unfortunately I could not find anything to further back up my suspicion, since the original synthesis only starts at 4.288. See here (in German): A. Ladenburg,  and M. Scholtz, Ber. Dtsch. Chem. Ges. 1894, 27, 2958–2960. DOI: 10.1002/cber.18940270356.
However, since there is proof that the reaction with 2,4-dihydroxybenzenecarbaldehyde forms the acetal, it would be highly suspicious if it were not doing a similar reaction with catechol.

Catechol does react with diiodomathane in presence of $\ce{NaOH}$ as the base. But it does not perform Reimer-Tiemann reaction, but instead it forms methylenedioxybenzene (2H-1,3-Benzodioxole is the preferred IUPAC name according to Wikipedia). As Martin - マーチン notes in his answer, this reaction is most commonly found in the synthesis of piperonal, but the same reaction can also be used for catechol.
The mechanism is similar and instead of diiodomethane, you can also use dichloromethane, dibromomethane, methylene sulphate and methylene chlorobromide. As this paper explains:

The ionizing base for catechols has always been an alkali hydroxide, alkoxide, or carbonate; the methylenating agents have variously been methylene iodide in alcohol or acetone, methylene bromide, 'methylene sulphate', methylene chlorobromide, and methylene chloride. The best yield (69%) of benzo-1,3-dioxole was obtained with methylene iodide, methanolic potassium hydroxide, catechol, and copper bronze, heated and shaken together at $\pu{100-110°C}$ for $\pu{18 hr}$ in a sealed tube. The yield from methylene chloride with these conditions was 53%.

From the same paper:

The formation of benzo-1,3-dioxole probably proceeds in two steps:

A bimolecular nucleophilic displacement of halide ion from the methylene halide by catechoxide dianion (when the ionizing base is strong enough to form the dianion); yielding the haloguaiacol anion (III).

An intramolecular displacement of a second halide ion from (III)

Source: W. Bonthrone and J. W. Cornforth, "The Methylenation of Catechols," J. Chem. Soc. (C) 1969, 1202-1204 (DOI: 10.1039/J39690001202)

Another paper also mentions the reaction between catechol and diiodomethane with methanolic $\ce{KOH}$.
The reaction mechanism is $\mathrm{S_N2}$ so the mechanism provided by Martin - マーチン in his answer should be correct. In case his answer is deleted or modified in future, I am providing a simple mechanism with catechol and diiodomethane only:
The final product is methylenedioxybenzene (1,3-Benzodioxole), i.e. methylenated catechol.


text:
Reaction between silver(I) hydroxide and an alkali
I've read about this in a booklet, but I can't find its stepwise reactions or its derivation:
$$\ce{AgOH + NaOH -> Ag2O}$$
And how does a base ($\ce{AgOH}$) react with another base ($\ce{NaOH}$)? Could someone please explain?

$\ce{AgOH}$ is rather unstable and forms $\ce{Ag2O}$ even upon hydrolysis, and it's also amphoteric. I don't know what makes you uncomfortable here, but I suspect that you assume that presence of $\ce{OH}$ in formula would always mean it's a base and a base only. Well, it's not true: amphoteric hydroxides aside, there are alcohols, and even sulfuric acid's formula can be written as $\ce{HSO3OH}$, which doesn't mean it's a base either.

I have no idea, that wasn't me. Perhaps it's considered as a homework question due to the lack of research from your side. On average the votes are decently fair here though, try to participate more actively, if you want your xp to grow up:)

What effort am I supposed to show in a straight forward reaction? And yes, I wasn't blaming you. Sorry.

No need to apologize, asking for a reason why you receive down-votes is fine. For the future, I would've refered to wikipedia page where formation of $\ce{Ag2O}$ is described, and explain that you understand this and this, but that particular thing is unclear to me. It would show your level of understanding and that you did some research beforehand, plus it would help to answer the question more precisely and not just give common comments like I did:)

When silver(I) hydroxide is reacted with sodium hydroxide, the excess of hydroxide ion result in formation of hydroxo complex predicted to be $\ce{Na[Ag(OH)2]}$. But the reaction is not so simple. This complex is actually a simplified form of the product that is formed. But it is actually a non-stochiometric form of silver oxide. It is unstable and decompose to form silver oxide hydrate and further to silver oxide.  It is explained here:

The situation with silver ions is somewhat complicated. When silver
  comes in contact with hydroxide, and no complexing agent is available,
  then something is formed, which sometimes is referred to as
  $\ce{AgOH}$. This is not correct. But $\ce{Ag2O}$ also is not correct.
  $\ce{Ag2O}$ is a black solid, when silver (I) is added to hydroxide,
  then a fairly light brown solid is formed. What happens is that an
  hydrated form of silver oxide is formed, so a better formula is
  $\ce{Ag2O.xH2O}$, but the real structure of the compound is very
  complicated. It really is another compound than $\ce{Ag2O}$, but it
  also certainly cannot be written correctly as $\ce{AgOH}$. I think, in
  reality, it will be a complex structure with indeterminate
  stoichiometry, containg $\ce{Ag+}$ ions, $\ce{OH-}$ ions, $\ce{O^2-}$
  ions and water molecules, and the only thing which can be said is that
  it has average formula $\ce{Ag2O.xH2O}$.

Similar question: What is solubility of silver oxide in NaOH and KOH


text:
How long does it take to replicate Friedrich Wohler's synthesis of urea?
I’m an author, writing a story where a doctor from modern day gets dropped into the late 1800s in California. They need urea to help a patient.
Friedrich Wohler's synthesis of urea happened before the time of my story and the necessary chemicals, Silver Cyanate and Ammonium Chloride, appear to be part of a normal pharmacist chemicals of the time.
So my questions are:

With the help of a local pharmacist, could they synthesize urea?
And if so, how long would it take?

Thanks for the help.

It's a fairly easy experiment to replicate, though not on an industrial scale. As a demonstration to chemistry students the whole thing can be done in a few hours. The original Wohler paper is readily available (though in German).

Related: https://chemistry.stackexchange.com/questions/81501/was-wöhler-s-urea-synthesis-carried-out-with-or-without-oxygen/81527#81527

OK, I am not answering the question but the author Kyros might be interested in how Professor Benjamin Silliman, Sr. (1779-1864) described Wohler's synthesis in his Elements of Chemistry (1831) (Vol II, pg.600) just 3 years after the landmark experiment.
Although Silliman conceded that the properties of the "artificial" urea had the properties of urea, he appears to have been skeptical. Wohler wrote to his mentor J. J. Berzelius in Sweden about his discovery on how he, Wohler, could make urea without the need of kidney, be they animal, human or dog. Berzelius was delighted because the reaction was an example of isomerism, a term Berzelius had coined.

Preparation. — By double decomposition, by muriate of ammonia and cyanate of silver; or by single decomposition, by cyanate of lead and liquid ammonia. In this second process, the oxide of lead is liberated and a compound is obtained, in colorless rectangular prisms. In their properties, they are identical with urea, and their composition is exactly the same; they give the same results in analysis as cyanate of ammonia, provided 1 equiv. of water is supposed to exist in that salt.
Still the artificial urea, although from the mode of its formation it would appear that it contains only cyanic acid and ammonia, yields neither, by chemical agents. Potassa does not evolve ammonia, nor do the stronger acids give carbonic acid and evolve cyanic acid, with the peculiar odor which appears when the cyanates are decomposed by a stronger acid. The artificial urea gives no precipitate with salts of lead or silver.*

I don't know why this has been downvoted; even though there is no direct answer, this information is interesting and highly relevant. Just in case I OCRed the picture with the text to make this searchable (maybe this was the cause for a down-vote); please feel free to rollback if you think what I did was unnecessary.

@andselisk: Are you addressiing me?

Regarding the rollback - yes, it's up to you whether to accept it or revert; but my main intent was to improve your answer, and I just generally expressed my opinion regarding downvoting of this answer, that's all.

Not clear about the meaning of rollback (I'm new) and what did you change.

Oh. You need to read this: What is a 'rollback'? What I changed is in my 1st comment. It's always better to post text as, well, text, and not a picture/photo/scan to make the post searchable and maintain uniform formatting.

It's fine what you did. I am in favor of teaching moments even though the info does not directly answer the question. Perhaps a no-no on SE given the down vote.

Let us continue this discussion in chat.


text:
IUPAC Naming. Which one is correct?

I have named this according to the numbering shown as
3-ethyl-6-(1-methylbutyl) oct-2,4-diene-7-ynenitrile.
But my book's answer key says
3-ethyl-6-ethynyl-7-methyl deca-2,4-dienenitrile
This means it has taken the triple bond as a side chain. But IUPAC rule says that the chain with maximum number of double and triple bond is taken even if it is shorter than a chain with lesser double or triple bonds. Which answer is correct and why, for the given example?

“But IUPAC rule says that the chain with maximum number of double and triple bond is taken even if it is shorter than a chain with lesser double or triple bonds.” Such a rule had been given in old recommendations that are now obsolete. According to current IUPAC recommendations, the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

Related: How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond?

As Loong said, the up-to-date IUPAC recommendation stipulates that the chain with the longest continuous chain of carbon atoms should be the parent chain. For this reason, I would agree with your textbook's answer.

Can you back that up with some more details and cite the rules necessary to understand it. Otherwise it is just a restatement of a comment and not necessarily a real answer.


text:
Reduction potential of a complex
I am given the reduction potential of $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$ as well as the dissociation constants $\ce{K_D}$ for a $\ce{Fe^{3+}}$ complex and a $\ce{Fe^{2+}}$ complex (effectively, the ratio of concentrations of the $\ce{Fe^{3+}}$ complex and $\ce{Fe^{2+}}$ complex, if I divide one constant by the other).
I need to find the reduction potential of the $\ce{Fe^{3+}}$ complex.
How would I go about this?
Is it possible that I could use the reduction potential of $\ce{Fe^{3+}}$ to $\ce{Fe^{2+}}$ as an estimate for that of the $\ce{Fe^{3+}}$ complex, and then use the
Nernst equation to plug in the concentrations of the two complexes?  Or is there no guarantee that these two potentials would be close?
Specifically, these are iron-siderophore complexes.

You should try to see if you can write a combination of the reactions you are given that added together will give the reaction of reducing the complex.

Thank you!!!  I don't know why I couldn't see the solution to this!!

no problem! If you are working with redox problems, you should always be on the lookout for ways to combine equations.


text:
Chain reaction vs single reaction
I am considering the following chain reaction:
$$\begin{align}
\ce{A_0 + B <=>[$N\cdot k_+$][$k_-$]A_1\\
A_1 + B <=>[$(N-1)\cdot k_+$][$2\cdot k_-$]A_2\\
A_2 + B <=>[$(N-2)\cdot k_+$][$3\cdot k_-$]A_3\\
...\\
A_{N-2} + B<=>[$2\cdot k_+$][$(N-1)\cdot k_-$]A_{N-1}\\
A_{N-1} + B<=>[$k_+$][$N\cdot k_-$]A_N} \\
\end{align}$$
Initial condition are $A_0 = 1, A_1=0, ... ,A_N=0$ and $B$ can vary from 0 to 1, $k_+$ is faster than $k_-$.
I have noticed (trough extensive numerical simulations) that result of this chain: $A_N$ is identical with the product of the following reaction in the power of N.
$$\begin{align}
\ce{A_0 +B&<=>[$k_+$][$k_-$]C} \\
\end{align}$$
Meaning that
$$\begin{align}
\ce{ $ [A_N]=[C]^N$} \\
\end{align}$$
I am wondering how to check this result analytically. I would like to know what to start with.
Thanks!
PS. I have looked at the ODEs and so far my calculations failed to prove the point, but I am not sure about my directions.

Just a product of the reaction between $A_0$ and $B$, where $A_0$ and $B$ are exactly the same as in the initial condition for the reaction chain above.

The problem is that you seem to be using capital letters to denote both chemical species and their concentrations.