text:
Half-life and shelf-life of second-order reaction
For integrated rate laws, I have attempted to find the shelf-life and half-life of second-order reaction:

However, the answers I have obtained (which are labelled with a red cross) are very different from the correct answers (bottom-right corner). I am wondering which parts were done incorrectly.

For one thing, how did you obtain negative values? As you're solving the problem, that should immediately cause you to think that something is amiss.

Also, when you're solving for $t_{10%}$, you need to substitute $[\ce{A}]=0.9[\ce{A}]_{0}$. The power of $1-n$ also applies to the 0.9.

Thank you so much Zhe! I did realise that I obtained a negative value (which is horribly wrong). I just could not figure out where went wrong. You were exactly right on my mistake - 0.9 applies to the exponent of 1-n as well.

Posting image of hand written text or formulas is highly discouraged, as it puts burden of reading, evaluating, referring and reusing on shoulders of responders. Furthermore, it cannot be indexed and searched for. Consider rewriting it.

Thank you for the critique @Poutnik - I was being quite lazy (and struggling to learn the formula function as well).

Your mathematics is correct until you calculated for specific cases. Let's go back to your $n$th order version:
$$\frac{[\ce{A}]^{1-n}}{1-n} = -k_nt + \frac{[\ce{A}]_\circ^{1-n}}{1-n} \ \text{where } n \ne 1 \tag1$$
At this point, since you are working on second order kinetic, it is easy if you substitute $n = 2$ in equation $(1)$:
$$\frac{[\ce{A}]^{-1}}{-1} = -k_2t + \frac{[\ce{A}]_\circ^{-1}}{-1} $$
Once simplify, it becomes:
$$\frac{1}{[\ce{A}]} = k_2t + \frac{1}{[\ce{A}]_\circ}   \tag2$$
For self life, substitute $t = t_{10\%}$ and ${[\ce{A}]} = 0.9{[\ce{A}]_\circ}$ in equation $(2)$:
$$\frac{1}{0.9[\ce{A}]_\circ} = k_2t_{10\%} + \frac{1}{[\ce{A}]_\circ}  $$
$$\therefore \; t_{10\%} = \frac{1}{k_2}\left(\frac{1}{0.9[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.9} - 1 \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{9} \right) = \bbox[yellow]{\frac{0.11}{k_2[\ce{A}]_\circ}}$$
Similarly, for half life, substitute $t = t_{50\%}$ and ${[\ce{A}]} = 0.5{[\ce{A}]_\circ}$ in equation $(2)$:
$$\frac{1}{0.5[\ce{A}]_\circ} = k_2t_{50\%} + \frac{1}{[\ce{A}]_\circ}  $$
$$\therefore \; t_{50\%} = \frac{1}{k_2}\left(\frac{1}{0.5[\ce{A}]_\circ} - \frac{1}{[\ce{A}]_\circ} \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{1}{0.5} - 1 \right) =  \frac{1}{k_2[\ce{A}]_\circ}\left(\frac{0.5}{0.5} \right) = \bbox[yellow]{\frac{1}{k_2[\ce{A}]_\circ}}$$

@Rodrigo de Azevedo: You can find the definition for standard state in any chemistry textbook, by the way.


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Effect of solutes on surface tension
What happens to the surface tension of a solvent, say, water when we mix a solute in it?
For the sake of concrete discussion, the examples that I encountered in a test were sodium chloride $(\ce{NaCl}),$ methanol $(\ce{MeOH}),$ and sodium alkyl sulphates $(\ce{ROSO3Na}).$
My guess is that since the surface tension can be defined as the force per unit length on a line drawn on the surface of the liquid, if the solute attracts $\ce{H2O}$ more than another $\ce{H2O}$ molecule, then the surface tension should increase.
Using this logic, I predict the following:

Addition of methanol should decrease the surface tension because $\ce{MeOH}$ forms lower number of hydrogen bonds than pure water.
Sodium chloride addition should increase the surface tension because ion dipole interactions are stronger than dipole dipole interactions or H-bonds.
The $\ce{ROSO3Na}$ kind of confused me. I assume they intended for me to perceive it as detergent and I know that decreases the surface tension. They actually asked to recognize the appropriate plot of variation in surface tension with concentration and the graph I guessed attained a constant value after a certain concentration. I thought it might be due to micelle formation.

If anyone can confirm or deny my intuition, or point me in the direction of any data that would be helpful.

Davies and Rideal1 published the following graph. This graph nicely describes the action of various solutes on a solution.

It's clearly visible that:

Addition of $\ce{MeOH}$ reduces the surface tension nonlinearly
Addition of $\ce{NaCl}$ increases the surface tension linearly, but by a small amount
Addition of a surfactant such as $\ce{RSO3Na}$ decreases the surface tension drastically, up until Critical Micelle Concentration is reached.

Smith and Gillham2 present a more detailed graph of the effect 1-butanol has on the surface tension of water.

References:

Hansen, Robert S. “Interfacial Phenomena (Davies, J. T.; Rideal, E. K.).” Journal of Chemical Education, vol. 39, no. 7, 1962, p. A552. doi:10.1021/ed039pA552.
Smith, James E., and Robert W. Gillham. “Effects of Solute Concentration-Dependent Surface Tension on Unsaturated Flow: Laboratory Sand Column Experiments.” Water Resources Research, vol. 35, no. 4, 1999, pp. 973–82. doi:10.1029/1998WR900106.

Thank you. This is exactly what I needed.

You're welcome. You can accept the answer if it answers your question completely.

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Pressure exerted by a particle of ideal gas
There is a step I fail to grasp in the proof that for a monatomic ideal gas, $U=\frac{3}{2}PV$, where $U$ denotes the internal energy of the gas.
The proof involves considering a particle with mass $m$ travelling with velocity $v$ inside a cube of side $x$. The particle is assumed to travel perpendicularly towards one of the sides before bouncing back off it. The force exerted by the particle at any instant is opposite to the force exerted by the side of the cube on the particle. The latter can be worked out using Newton's second law, prior to which we need to calculate the acceleration of the particle during the collision. Assuming the collision to be perfectly elastic, the change in velocity before and after the collision with the wall is $\Delta v=2v$. To work out average acceleration, according to me, one would need the duration $\Delta t$ of the collision, since assuming it to be instantaneous would imply infinite acceleration. However, the various proofs I've read assign to $\Delta t$ the time that is needed for the particle to bounce off the wall and return after colliding with the opposite wall, in other words $\Delta t=\frac{2x}{v}$. I really can't see how that $\Delta t$ is relevant to the force exerted by the particle on the wall during the collision.
[Edit :] If there are no considerations about duration of the collision, then theoretically, no two particles would ever collide simultaneously with the wall and therefore the wall experiences very frequent collisions with single particles.

The formula U=3/2PV cannot be true, as for the same P and V, U depends on on type of molecule and non translational active degrees of freedom that do not contribute to pressure.

@Poutnik  I edited it to say monatomic ideal gas, to address your comment.

Here's your confusion:
You need to consider two different things:

The momentum transfer per particle per collision.  There, since we assume an instantaneous collision, it doesn't make sense to try to figure out force from acceleration. [I suppose you could do this using limits, and maybe there are applications in which that does make sense, but adding that complication is completely unnecessary for introductory kinetic theory, given that there is a simple and direct way to calculate the effect of each collision.]  And the simple and direct way is to use the change in momentum, which is $2mv_x$.

Note: Your statement that "The particle is assumed to travel perpendicularly towards one of the sides before bouncing back off it" is incorrect.  Rather, in determining the momentum transfer per collision, we use the component of the velocity perpendicular to the wall, namely $v_x$.

The number of collisions per unit time.  Here is where the average time it takes for the particle to travel from collision to collision is relevant. This is where ${\Delta t}$ is being used -- not to calculate the momentum transfer/collision, but to determine the frequency of collisions per particle.  And you need the average momentum transfer/collision, and the frequency of collisions per particle, to determine the average force per unit area contributed per particle.

So, in summary: For momentum transfer per collision, we don't consider time, since we can get this directly from $2mv_x$.  For number of collisions per unit time, we need to account for the time between collisions.

What should I look up in order to understand how the frequency at which the wall induces a $2v$ change in momentum gives an idea of the pressure ? If there are no considerations about duration of the collision, then theoretically, no two particles would ever collide simultaneously with the wall and therefore the wall experiences very frequent collisions with single particles.

@JamesWell The average force on the wall over some finite timespan is equal to $\Delta p/\Delta t$, where $\Delta p$ is the total momentum transfer to the wall over the time $\Delta t$ from all collisions that may have occurred during that time. Increasing the frequency of collisions increases the total number of collisions during that timespan $\Delta t$, meaning that it increases $\Delta p$, so it increases the average force.

@JamesWell The question then becomes, "How much does the force fluctuate about the average value?" We can get an idea of this by assuming that the number of collisions on the wall in a particular time interval $\Delta t$ is roughly a Poisson distribution with mean $f\Delta t$, for average collision frequency $f$. Then the standard deviation is $\sqrt{f\Delta t}$, and the magnitude of the fluctuations compared to the mean is roughly $\sqrt{f\Delta t}/f\Delta t=1/\sqrt{f\Delta t}$.

@JamesWell So, if you have a very, very dilute ideal gas, or you measure over very, very short time intervals, you might notice some fluctuation in the force that you measure over time. But for most typical experimental situations, $f$ is very high (around $10^{20}$ collisions per second wouldn't be unreasonable) and $\Delta t$ is not that small; so it wouldn't be unreasonable to expect that the measured forces at different times will agree to a part per billion or better, certainly more precise than the precision of the usual devices you use to measure this quantity.

@JamesWell  You raise an interesting point.  It sounds like you're noting that, if each collision is instantaneous, then only one particle collides at a time, and the force on the wall must thus be constantly varying between zero and infinity.  And yes, that would be the case for truly instantantaneous collisions.  The resolution is that, in real gases, the collisions aren't instantaneous, and there are so many gas particles colliding per unit time that we typically will have many interacting with the wall at the same time.....

But if we had to actually account for the duration of collisions (i.e., make a more realistic model) that would make our calculations both much more complex, and less universal (gas-dependent). The nice thing about kinetic theory is that we can make these idealizing assumptions that allow the theory to be both mathematically and conceptually simple, while at the same time capturing the essence of why all gases exert pressure, and enabling us to calculate (for ideal gases) what that pressure is directly from a microscopic picture.

Also, This might help.  Take a look at this video:  https://demos.smu.ca/index.php/demos/thermo/91-boyle-s-law
The top wall is stationary.  Note how the collective colllisions of the beads against the top wall exerts a force.  As the top wall is lowered, L gets smaller, so $\Delta t$ gets smaller, so we have more collisions/unit time, and thus more force.  Here there are an incredibly tiny number of beads compared to how many gas particles we'd typically have, so the force fluctuates a lot.

In ideal gases no intermolecular forces, therefore no potential energy. Thus, internal energy is equal to total kinetic energy (KE) of the system. Consider $N$ monoatomic particles in a cubical box of side $\ell$ (assumption: ideal gasses consist of monoatomic point particles). The amount of ideal gas in the box is $\frac{N}{N_A} = n \ \pu{mol}$ where $N_A$ is Avogadro number. Suppose one particle travel in $y$-direction with velocity $v_y$. If mass of the particle is $m$, its momentum is $mv_y$:

If it collides on right $xz$-plane, it bounce back with same velocity but in but travelling opposite direction (assuming a 100% elastic collision). Therefore momentum change ($\Delta p_y$) is:
$$\Delta p_y = mv_y - (-mv_y) = 2mv_y \tag1$$
However, from Newton's second law of motion ($F = ma$ where $a$ is acceleration), if the particle's applied force on the plane is $F$:
$$F = ma = m \frac{\Delta v_y}{\Delta t} = \frac{m\Delta v_y}{\Delta t} = \frac{\Delta p_y}{\Delta t} = \frac{2mv_y}{\Delta t}\tag2$$
Here, $\Delta t$ is time difference between two collisions. That means the time required for particle to travel $2\ell$ distance with $|v_y|$ velocity that is $\Delta t = \frac{2\ell}{v_y}$. Applying that in equation $(2)$ gives:
$$F = \frac{2mv_y}{\Delta t} =  \frac{2mv_y}{\frac{2\ell}{v_y}} = \frac{mv_y^2}{\ell} \tag3$$
The pressure ($P_i$) applied by this single particle on one plane is $\frac{F}{A} = \frac{F}{\ell^2}$:
$$P_i = F = \frac{F}{\ell^2} = \frac{\frac{mv_y^2}{\ell}}{\ell^2} = \frac{mv_y^2}{\ell^3} \tag4$$
The total pressure on one wall:
$$P = \sum^N_1 P_i = \sum^N_1 \left(\frac{mv_{yi}^2}{\ell^3}\right) = \frac{m}{\ell^3}\sum^N_1 {v_{yi}^2} = \frac{Nmv_{y(m.s.v.)}^2}{\ell^3}= \frac{Nmv_{y(m.s.v.)}^2}{V} \tag5$$
where $v^2_{y(m.s.v.)} = \frac{1}{N}\sum^N_1(v_{yi}^2) $ and $V = \ell^3$ ($v^2_{y(m.s.v.)} =$ mean squire velocity in $y$-direction). Now, the equation $(5)$ cab be rewritten as:
$$PV = Nmv_{y(m.s.v.)}^2 \tag6$$
Yet, we still are considering only $y$-fraction of velocity. As a system, we have to consider effective velocity of all $x,y,z$ fractions. According to law of vectors, if the effective velocity is $v$, $v^2 = v_x^2 + v_y^2 + v_z^2$. In our box, the possibility of particle to travel all directions are same, thus:
$$v_{(m.s.v.)}^2 = v_{x(m.s.v.)}^2 + v_{y(m.s.v.)}^2 + v_{z(m.s.v.)}^2 = 3v_{y(m.s.v.)}^2 \ \Rightarrow \ v_{y(m.s.v.)}^2 = \frac{1}{3}v_{(m.s.v.)}^2$$
Therefore, from the equation $(6)$:
$$PV = Nmv_{y(m.s.v.)}^2 = \frac{1}{3}Nmv^2_{(m.s.v.)}\tag7$$
Note that $mv^2_{(m.s.v.)}$ is two times KE of the particle. Thus, $Nmv^2_{(m.s.v.)}$ is twice KE of the system. Since, KE of the system for ideal gas is its internal energy  $Nmv^2_{(m.s.v.)} = 2U$:
$$\therefore \ PV = \frac{1}{3}Nmv^2_{(m.s.v.)} = \frac{2}{3}U \ \bbox[yellow]{\Rightarrow \ U = \frac{3}{2}PV} \tag8$$

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Are platinum metal whiskers very unusual?
Over 20 years ago I pulled my first set of platinum spark plugs out of a car because the engine was missing badly. Two of the six plugs each had one whisker in the gap between electrodes. One crossed the gap and one was most of the way across. They started on the platinum center electrode and grew to the Inconel side (ground) electrode. They looked like straight, fine threads.
As I had worked in metallurgical R&D most of my life, I assumed they were whiskers, odd but no big deal, threw them away and put in new regular plugs. I was retired or I could have had a much better story with XRD and electron microscope data.
The reason I ask is that today I looked up metal whiskers in Wikipedia and found mostly info on whiskers formed on various solders and nothing on platinum.

Just curious, are those spark plugs made of pure platinum or is it a coating?

A "pure" platinum spark plug would obviously be too costly. It seems that a platinum disc is used on the plug tips instead of a plating.

Platinum isn´t really expensive. Iridium is! :-)) The price differences for spark plugs are amazing, anything from two to fourty euros apiece.

Whiskers growing on platinum are different to "platinum whiskers". The problem in you old spark plugs could be made of something entirely different.

Only the center electrode is Pt, Roughly 0.025" diameter , less than 2" long. Some have a few % alloy like iridium.  I have replaced 2 sets of 8 since then and they looked like new after 100,000 miles.


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Chain transfer agents and rate of free radical polymerization
In free radical addition polymerizations sometimes chain transfer agents are added. From what I understand chain transfer agents (e.g. XR' below) reduce the average molecular weight by transferring the free radical  from the growing polymer chain (P•) to another molecule (R'•).
P• + XR' → PX + R'•
My question is what effect would this have on the overall "rate" of polymerization itself, i.e say the consumption of monomer.
In other words the R'• species is then free to combine with more Monomer, right? And being a smaller molecule (in general) would it be expected to have a higher rate constant for the reaction:
R'• + M--> P•

Not really.
As long as you are in solution (=unreacted monomer or solvent), it makes no difference if the reactive chain end is diffusing slow or fast. The monomer molecules come by at the same rate.
The mobile chain end would help against a local depletion of monomer, but once your mixture gets so viscous that this could happen, neither a long or a short active chain can still diffuse well (and neither can the transfer agent).
The only general difference you will see with transfer agent is that the Trommsdorff-Norish or gel effect sets in a bit later, because the overall viscosity of short chains is lower. In that sense, the transfer agent even slows down the reaction a bit. Which in that case is very welcome.
Of course the transfer agent radical can be either more reactive with the given monomer or less. (See "copolymerisation parameter".) So the chain restart can either be fast (in which case it makes practically no difference), or it can eventually be much slower. That depends on the specific transfer agent / monomer pair.

Many thanks! So, if I interpret that correctly, then the effect on total rate of polymerization could be, in general, in either direction? Increase or decrease or not change? Would depend on the specific situation?

I believe I said that a speedup is rather improbable?! The reaction rate of a FRP isn´t even constant to begin with, but the successive stages of the interplay of initiator dissociation, recombination,  chain growth, chain transfer, gel effect, stationary state, etc. are generally well understood. Textbook stuff. ;-) Pick your favourite, read on, and don´t forget to do the exercises!


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Friedel-Crafts's alkylation of Toluene
In the Friedel-Craft's alkylation of Toluene with $\ce{CH3Cl}$ in the presence of $\ce{AlCl3}$, the major product was meta - substituted. My teacher gave it to me as an exception. Is there any particular reason behind this reaction ?

I could hazard a guess at an answer, but would rather not, as there are some things which really need to be clarified first. (1) Is this "major product" you're referring to obtained under thermodynamic control, or kinetic control? (2) Does this "major product" result from multiple substitutions, or only one? (3) Spellings: it's "Friedel-Crafts", "toluene", and chlorine is "Cl", not "cl".

Appreciate     Appreciate

Funny how, i don't see these questions when i look them up.

Really i dont think meta xylene is the major product here. atleast thats what my teacher taught me

@VAMSIKRISHNACHILAKAMARRI, I think you should also consider the points in my comment, minus the spelling one, of course.

@orthocresol Oh yeah the thermodynamically major and the kinetically major differ. However on general questions like this i think it normally means kinetically. As it is not specified I dont think thermodynamics is the case. However your comment could be the reason........


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Reduction of carbonyl compound by LAH (in anhydrous ether) followed by reaction by acidic water
What will be the final product upon reaction of $\ce{CH2=CH-CH2-C6H4-CH=CH-CHO}$ with $\ce{LiAlH4}$ (in dry ether) followed by acidic water?
LAH will reduce the carbonyl group to alcoholic group and the alkene group which is directly attached to carbonyl and aryl group to saturated bond. But it will not reduce the isolated alkene group.
Hence, the final product must be $\ce{CH2=CH-CH2-C6H4-CH2-CH2-CH2-OH}$

This is correct. LAH does not react with isolated double bonds.

@Waylander What would be the product in case of $\ce{C6H5-CH=CH-CH2-CH=CH-CHO}$? Is it $\ce{C6H5-CH=CH-CH2-CH=CH-CH2-OH}$?

No,  C6H5-CH=CH-CH2-CH2-CH2-CH2-OH. The double bond conjugated to the C=O will be reduced.

This answer may be applicable: https://chemistry.stackexchange.com/questions/87138/why-does-lialh4-reduce-the-double-bond-of-cinnamaldehyde-whereas-nabh4-does-not/87195#87195

@user55119 according to the link (and the link therein), the product will be $\ce{C6H5−CH=CH−CH2−CH=CH−CH2−OH}$ in case of $\ce{LiAlH4}$ and  $\ce{C6H5-CH=CH-CH2-CH2-CH2-CH2-OH}$ in case of $\ce{NaBH4}$.

The link only speaks to how cinnamaldehyde undergoes 1,4-reduction. Whether an alkyl group in place of a phenyl ring will give the same result is questionable.

@user55119 Yes, that is questionable but check this link (https://chemistry.stackexchange.com/questions/6977/why-does-nabh4-reduce-double-bonds-conjugated-to-carbonyl-groups-while-lialh4-d?noredirect=1&lq=1) and my comment.


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CAS id to SMILES
How can I convert a list of CAS ids to structures (SMILES). I have been trying (Chemspider API, PubChem and the chemical identifier Resolver) but there are still some ids I can't find the structure. For example 166177-30-6.

I've been consistently successful with CAS' at https://chem.nlm.nih.gov/chemidplus/

166177-30-6 resist to chem.nlm.nih.gov/chemidplus for example :)

PubChem has been a great resource for looking up structure (and other) information by CAS Registry Number. My search for 138683-67-7 yields a result: https://pubchem.ncbi.nlm.nih.gov/compound/2947 You can also search PubChem programmatically in python, for example, with just a few lines of code to retrieve the data from a list.

Yeah, PubChem seems to the most extensive. chem.nlm.nih.gov/chemidplus may also be relying on PubChem in the background. However, 166177-30-6 is not on PubChem.

I did a Google search of 166177-30-6 and found a single hit from a German specialty synthesis company. It is possible that this CAS RN is a custom medicinal chemistry intermediate. This could explain why it does not appear in PubChem and other resources. Other than going straight to CAS, options may be limited. A full-text patent search is a possibility. At least the CAS RN provides a distinct search term.

Here's the link: https://eoschemical.blogspot.com/2019/09/eos-med-chem-kundenspezifische-synthese_698.html?m=1


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Resonance of lone pair of nitrogen in 4‐nitroaniline and piperidin‐2‐one
While working out some questions on comparing the basic nature of various compounds, I came across the following two compounds:

In case of A I could see several resonating structures, and also the −M effect of the nitro group. So justifying based on the number of resonance structures, I could tell that the lone pairs are more delocalized in A than in B. Hence B is more basic.
However, on verifying my answer I found that A was more basic.
The only reason I could come up with is that in compound A, majority of the resonance includes a negative charge on carbon, but in compound B, the negative charge lands on the more electronegative oxygen, which better succeeds in stabilizing it.
But it doesn't seem right for me.
Can anyone please tell me the correct reason for this?

I dont know why 2 people downvoted this. Is there any problem with the question?

-1. It's unclear what you're trying to ask. Do you want to find out which compound is more stable? quality vs quantity is like apples vs oranges - you can't compare them.

But then on what aspects do we compare these compounds?

@aniruddhaDeb yes I have edited the question to make it more clear now

You might ask if the hydrochloride salts of these two compounds exist.

well that could be one way. But I want to know how the resonance of lone pair could be used to determine basicity here

One of these compounds has a basic amine nitrogen. The other is a non-basic amide.

Yes but by resonance how can it be explained?

Considering that you're referring to the difference in the number of resonating structures, you're right in that aspect. 4‐Nitroaniline does have more resonating structures than piperidin‐2‐one. But in organic chemistry, you should be knowing that for a phenomenon to happen, there have to be several contributing factors.
While this includes the number of resonating structures as one, you have to take into consideration the other factors as well. And this includes the other factor you have mentioned in your answer — the position of the negative charge.
Since carbon is an atom which has an electronegativity of 2.55 (According to the Pauling scale), a negative charge on it would make it highly unstable, lowering the resonance energy, thus reducing the resonance stabilization in it. Nitrogen and oxygen have electronegativity of 3.04 and 3.44, respectively, thus a negative charge on those atoms would increase the resonance energy, thus increasing the resonance stabilization.

@Firefox1921 Thank you for the answer, however isn't the negative charge going to oxygen in BOTH the cases? That is why i do not think it should be the main reason

Yes @VAMSIKRISHNACHILAKAMARRI, the negative charge is going to the oxygen in both cases. But it isn't going on carbon in both the cases, right?


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Gibb's energy change for system

Heat liberated by a given amount of an ideal gas is $\pu{1200 cal}$ at $\pu{300K}$. What is the Gibbs free energy change for the gas in this process?

My attempt: I know that for a reversible process the net Gibbs energy change of the system + surrounding will be zero. But I have no idea how to calculate the Gibbs energy change of the system alone. The only equation I know involving Gibbs energy is $ΔG=ΔH-TΔS$, but this equation will give me the net Gibbs energy change and not the change in free energy of the system alone.

Is that the exact verbatim statement of the question?

Well, if it happens at constant temperature, $\Delta H=0$ and $T\Delta S=Q=-1200$.


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Basic strength of pyridine vs. pyrrolidine
I want to compare the basic strength of pyridine and pyrrolidine . Now I can decide this in two ways.

Lone pair on nitrogen of pyridine is involved in resonance. We have to see whether or not nitrogen has lone pairs. In pyrrolidine nitrogen's lone do not involve in resonance therefore that would maintain the electron density on nitrogen making pyrrolidine more basic.
Now I also have other explaination. In pyridine and pyyrolidine if $H+$ is added to nitrogen to make a conjugate acid , nitrogen cation will be formed as it donates it's lone pairs to $H+$ . In pyridine the cation would be stabilised by resonance effect but in pyrrolidine there is no resonance effect. This would make pyridine more basic.

Can you please tell me which explaination is correct

A quick Google search turns up this: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch11/ch11-8.html

Your first argument is completely wrong. The second one is partially correct. Similar to benzene, pyridine is aromatic but the lone pair of $\ce{N}$  is not involved in the $(4n+2)$-$\pi$ electron system. Instead, the lone pair is in a $\mathrm{sp^2}$ hybrid orbital in the same plane away from conjugated $(4n+2)$-$\pi$ electron system. As a consequence, the $\ce{N}$ atom is weakly basic ($\mathrm{p}K_\mathrm{a} = 5.2$), because this lone pair of $\ce{N}$ can function as a source of electrons (in the Lewis base) and is in a $\sigma$-orbital. Therefore it is not a part of the aromatic $\pi$-system. Only the $\ce{N}$ atom is part of $\pi$-systemas it makes $\ce{C=N}$ bond with the system. Once protonated, it is still in the $\pi$-system and the pyridine nucleus maintains its cyclic, planar, conjugated, $6\pi$-electron system.
The lone pair of Pyrrole, on the other hand, is part of conjugated $(4n+2)$-$\pi$ electron system. Once protonated, pyrrole loses its aromaticity due to the loss of lone pair to proton ($\ce{H+}$). Therefore, this would result in protonation being unfavourable, and protonated pyrrole becomes unstable. As a result, pyrrole is a much weaker base than pyridine ($\mathrm{p}K_\mathrm{a} = -3.8$).
Pyrrolidine is technically tetrahydropyrrole. Unlike pyridine and pyrrole, the nitrogen atom of pyrrolidine is $\mathrm{sp^3}$ hybridized. Thus, it has more $\mathrm{p}$ character while $\ce{N}$ of pyridine and pyrrole have more $\mathrm{s}$ character comparatively. As a result, pyridine and pyrrole hold their lone pair electrons tighter than that of pyrrolidine. Consequently, Pyrrolidine has its lone pair readily available to incoming protons compared to that of pyridine and pyrrole. Therefore, pyrrolidine is the strongest base ($\mathrm{p}K_\mathrm{a} = 11.3$) among three.
Source of $\mathrm{p}K_\mathrm{a}$ values

I am sorry I mistakenly wrote pyrrole instead of pyrrolidine. Can you please edit your answer based on pyrrolidine.

@Akshat: I have changed my answer as requested.


text:
Creating a chemical solution with water and chlorine
I bought a swimming pool for my kids and also a measurement instrument for measuring the amount of chlorine in the pool in mg/l. The problem is now that you have to calibrate the instrument before using it. This means you have to know the current amount of chlorine in the pool.
Since the test strips are quite inaccurate and hard to read an exact value. My idea was to just create a perfect solution in a one liter water container.
Now I am not really sure how much of the given pool chlorine I have to put into the 1 liter water to get a 1.5 mg/l solution of chlorine. With that I would be able to calibrate the instrument perfectly.
What I am using is chlorine granules.
The instruction leaflet says it contains "CAS 51580-86-0 Troclosenatrium, Dihydrat (100g/100g)"
Registration: BAuA N-23822.
Would I just have to add 1,5 mg to 1 liter of water to get 1,5mg/l?
Is it maybe that easy?

Forget the test strips. Go to a public pool. Check the smell. Then go home and add your granules until you feel the same smell. That would be more accurate.

I'd go to a pool supply store and buy a colorimeter test kit. You get a tube full of pool water, put a drop of the test solution into it and match the color to a color scale. You'll need to check both pH and chlorine.

Is your measuring instrument made for measuring chloride or chlorine ? It is not the same. In your first sentence, you speak of chloride. Later on you speak of chlorine. Anyway, the best thing to do is to buy a bleach solution. The amount of chlorine is printed on the label. It is often $14$% chlorine $\ce{Cl_2}$. By diluting a sample (maybe $1$ mL) of such a solution in 1 liter water, you obtain a solution $0.14$ gram chlorine $\ce{Cl_2}$ per liter, and you can dilute it still more.

It is for chlorine. The tip with the bleach solution sounds good. I will try that!


text:
Comparing $\ce{PPh3}$ and dppe
The ligands $\ce{PPh3}$ and dppe (dppe = 1,2-Bis(diphenylphosphino)ethane) are both strong field ligands, but is one obviously a stronger field ligand than the other or are they roughly equal?
The reason why I ask this question as I am researching into $\ce{[NiCl2(PPh3)2]}$ and $\ce{[NiCl2(dppe)]}$ and I trying to find another other reasons for the geometry difference other than sterics.
My guess would be that dppe is a strong field ligand, but that is more of a guess than any real logic application.


text:
Will ipso substitution work in both cases of vanillic and isovanillic acids here or not?

What will be the product of the 2 reactions when bromine is taken in excess?

I think ipso substitution will work here and there will be two $\ce{Br}$ substituting in both products, but in the book it was given that there will be three $\ce{Br}$ in the second one. I cannot understand why.

Posting question as image or image link is highly discouraged for multiple reasons. Also, if a question is asked on Chemistry SE site, then, in contrary to sites like Quora, it is expected the author elaborates the topic in the question by at least basic own topic review and writes what he/she understands and what is the eventual stumble stone. The "quick questions" without explicitly expressed particular effort are not very welcome, and may be closed.

Related: https://chemistry.stackexchange.com/questions/91781/ipso-substitution-and-its-rules?r=SearchResults

@Poutnik: I have addressed that issue for now. But, OP should be aware of that for future perspectives.

Neither acid gives ipso bromination. Chem. Abst. has ring bromination of vanillic acid but no record of the same process with isovanillic acid, which I find hard to believe.


text:
Is the C-I bond polar?
Carbon and iodine have similar electronegativities. The reason quoted for the C-I bond having polarity is that iodine is more polarisable.
But what does this actually mean? Carbon distorts the electron cloud around ioidine, but how exactly does it distort the electron cloud? And how does this affect the polarity of C-I bond?
Does it, perhaps, distort the electron cloud of ioidine away. But why would it distort the electron cloud away and not towards the C-I bond where the electrons have the opportunity to experience the attraction of two nuclei as opposed to just one?
Edit: here is a quote from a textbook  (Organic chemistry, Fourth Edition by Paula Yurkanis Bruice):

"Carbon and iodine, however, have the same
  electronegativity. Why, then, does an alkyl halide undergo a
  substitution reaction? We know that larger atoms are more polarizable than smaller
  atoms. (Recall from that polarizability is a measure of how easily an
  atom’s electron cloud can be distorted.) The high polarizability of the large iodine
  atom causes it to react as if it were polar even though, on the basis of the electronegativity of the atoms, the bond is nonpolar."

Electronegativity is not a precise measure of anything.

If you ask this then edit out similar content from your earlier question.

I have a feeling you have an incorrect assumption - where did you get such quote and what was it exactly? What you quoted earlier doesn't say that, nor does truly mean what you implied.

@Mithoron Thanks for the feedback, I have added a quote to the question. I have also edited the other question like you have asked me to.

You know what? Maybe just read a better textbook... One thing is that it's using somewhat strange reasoning 2) You're getting to some weird conclusions further messing things up. Being polarizable  has like nothing to do with being good leaving group. Also iodine has a bit higher electronegativity then C - you're so literal minded that reading a table of values may be a better idea.


text:
Why basicity decreases for leaving groups, but nucleophicity increases for an overall compound, going down the periodic table?
Basically, I read that nucleophilicity increases going down the periodic table within a particular group, due to polarizability, but basicity decreases down a group (or family) in the periodic table, because those atoms can stabilize a negative charge, making them good leaving groups. Don't these two points seem to contradict each other, since increasing nucleophilicity should increase basicity?

https://chemistry.stackexchange.com/questions/20061/whats-the-difference-between-a-nucleophile-and-a-base


text:
Preparation of p-chlorobenzoic acid from p-chloronitrobenzene
I was reading my book where I spotted the preparation of p-chlorobenzoic acid from p-chloronitrobenzene.
The preparation method there was given as such:

$\ce{Sn/HCl}$
$\ce{NaNO2 + HCl}$ ($\pu{0-5 ^\circ C}$)
$\ce{CuCN}$
$\ce{H2O/H+}$

Now, the thing is that, I recall the first two steps from the process of forming diazonium salts but I have never seen the last two steps give the $\ce{-COOH}$ group.
Also I recall that $\ce{CuCN}$ gives us the nucleophile $\ce{CN-}$ where the negative charge is on the nitrogen atom and also that $\ce{H2O/H+}$ is used for hydration but I don't see how these two could help us form p-chloronitrobenzene from Benzenediazonium chloride.

The reaction of CuCN with the diazonium salt is a Sandmeyer reaction. The last step is a hydrolysis of the nitrile formed in the Sandmeyer reaction.

You recall incorrectly, the negative charge in cyanide is on the carbon

@user55119 But isn't the sandmeyer reaction only for halogens ,for eg NaCl and NaBr?

Read this  https://en.wikipedia.org/wiki/Sandmeyer_reaction

@Waylander But isn't it of the format of AgCN where the - charge is on the N atom?

No, in general cyanide is nucleophilic through carbon  https://en.wikipedia.org/wiki/Cyanide

@Waylander But we do have Ambident nucleophiles, for eg in NaCN and KCN charge in cyanide is on carbon whereas in AgCN , charge is on nitrogen.

Yes, we do. This answer explains why  https://chemistry.stackexchange.com/questions/58104/why-do-sn2-reactions-of-alkyl-halides-proceed-differently-with-kcn-and-agcn  but in general when you see cyanide ion - unless it is AgCN - assume that the nucleophilic reaction is through carbon.

CN- and NC- are different. One is called cyanide ion, other one is called isocyanide ion. Also, correct the spelling of nucleophile in your question.

@TheChemist Please don't delete \pu{…} macro in posts. See Undefined control sequence \pu in official SE Android app. The problem doesn't lie with the macro, it's the mobile app that you are likely using that hasn't been updated for a long time.

Oh ok didn't know that. Thanks for telling.

This is simply the summery of all comments involved:
$$\ce{Cl-C6H4-NO2 ->[Sn/HCl] Cl-C6H4-NH3+Cl-} \tag1$$
$$\ce{Cl-C6H4-NH3+Cl- ->[\text{a. } NaNO2/HCl \ (\pu{0-5 ^\circ C})][\text{b. } CuCN] Cl-C6H4-CN} \\ \tag2$$
$$\ce{Cl-C6H4-CN ->[H+/H2O][\Delta] Cl-C6H4-COOH} \\ \tag3$$
Note: The reaction $(1)$ is a typical reduction reaction, which reduce nitro group to amino group. The reaction $(2)$ is called Sandmeyer Reaction. Finally, The reaction $(3)$ is a hydrolysis condition to convert cyano group to carboxylic acid group. That hydrolysis can be done under basic conditions as well (then, final product is carboxalate).

Also note that even though it is not necessary, I put this answer up just to avoid people asking same kind of questions again and again. When they have a visual answer they might not do so.


text:
How does a cyclization occur with a carboxylic acid and (CF3CO)2O?
The product of the reaction below is unsaturated, bicyclic ketone, but I'm confused as to how (CF3CO)2O reacts with it to cause the cyclization. There is no base involved so I don't know how the carbon delta to the carboxylic acid attacks it. What is the mechanism?

The "base" or rather nucleophile is the alkene and the electrophile is the mixed anhydride. There are 4 degrees of unsaturation. Take it from there! You are capable of solving this based on our previous discussions.

@user55119 so the carboxylic acid attacks the anhydride to make its own mixed anhydride, and then the triflate part of the anhydride is eliminated by add-elim through the alkene? Then elimination to form a new alkene happens?

You have the mixed anhydride correct. Use the alkene as a nucleophile and lose trifluoroacetate wich you can use to deprotonate the tertiary cation via an E1 process.

BTW: Triflate is triflluoromethane sulfonate, F3CSO2.

Trifluoroacetate in mixed anhydride is a good leaving group. Alkene attack on carbonyl to make bicyclic system leave stable tertiary carbocation.


text:
Solubility of Organic Compound confusion
My teacher wrote the following on the board today:

A solute will only dissolve if the energy required to break the bonds
  in the solute and solvent is less than the energy released from the
  bonds formed between the solute and solvent.

I am aware that "like dissolves like" but not entirely sure what the sentence above means.
From the bond enthalpy chapter I learnt that, in an exothermic reaction the bonds formed are stronger due to lower in energy and thus more stable. Although I am not sure whether this is related to solubility, it seems to me that they are contradictory.
So can someone explain to my as to why bonds formed has less energy released than the broken bonds?

Where exactly do you see a contradiction?

suggestion: re read the quoted para again; you are confusing bond formation with bond breaking (in the solution)

@IvanNeretin Sorry for not being specific enough, it says bond broken is less energy, whereas in the context of bond enthalpy, bonds forms are less energy.

Got it. Well, then you oversimplify things down to complete nonsense. Indeed, it says that bonds broken mean less energy than something (what?), while bonds formed mean less energy than another something.

In addition to enthalpy, beware of entropic effects when molecules are very flexible.

There are two regions of bond energy to look at here:

The bonds IN a molecule: a carbon-hydrogen bond has about $\pu{100
    kcal/mol}$ of stabilization. An oxygen-hydrogen bond is a little
stronger, about $\pu{110 kcal/mol}$. Stronger means HIGHER in
stabilization energy; but we draw diagrams where the stronger bonds are
designated by lines that are LOWER on the diagram. We hope no confusion
results!
The bonds BETWEEN molecules of a pure material or a mixture. These
(much weaker) bonds between molecules of a substance can be
estimated by heats of fusion or vaporization. The heat/enthalpy of
fusion of $\ce{H2O}$ is $\pu{1.435 kcal/mol}$, and for $\ce{CH4}$, $\pu{0.223
kcal/mol}$. Hydrogen bonding (not the regular bond to hydrogen, but the polar
bonding between molecules) falls in this region of energy and is
frequently encountered when examining solutions.

The bonding between water molecules is strong enough to resist separation by methane molecules because the gain could be no more than the stabilization in methane ($\pu{0.223 kcal/mol}$), but the loss would be as much as $\pu{1.435 kcal/mol}$. The first sentence in the question refers to this region of bond energies, not to the much higher energy region (~ $\pu{100 kcal/mol}$) that would be extensively covered in a chapter on bond enthalpy.

You can think of the process of solubilization of a solute A as a reaction:
$$\ce{A(\text{solid})-> A(\text{soln})}$$
If the reaction results in a net formation of bonds - by which it is meant that the interactions in the product are stronger than in the reagent - then there must be an accompanying release of excess thermal energy to the solvent to balance the reduction in potential energy. If the reaction is performed at constant temperature and pressure (in contact with a thermostat) then that thermal energy will be transferred as heat to the surroundings, and we call the process exothermic.
Unfortunately however, the statement that your teacher wrote, while at times true, is not general. It is in fact common to dissolve a solute with accompanying net breaking of bonds, that is, with bonds among product being higher in energy (weaker) than in the reagent. As you might imagine, this requires an input of thermal energy, and if the temperature is kept constant the result is a transfer of heat into the system.
While the exothermic case can occur (up to a point) even when the system loses entropy, the later endothermic scenario requires some way of compensating the surroundings for the loss of thermal energy, since that loss results in a decrease in the entropy of the surroundings. This compensation takes the form of an increase in the entropy of the system.
A good example is a solution of $\ce{NaCl}$ (at normal TP). The process is endothermic ($\Delta_{\text{solv}} H>0$) but the entropy change compensates for the heat gained from the surroundings.


text:
Is pK_{In} = pH at equivalence point still true for weak basic organic indicators?
I am reading Ostwald Theory of Titration which says that indicators are organic weak acids or bases.
To prove the relation that $\mathrm{p}K_\mathrm{In} = \mathrm{pH}$, the textbook uses the following equilibrium as an example:
$\ce{HIn(aq) <=> H^{+}(aq) + In^{-}(aq)}$
which of course produces the the required result, but a thing I noted that it was for acid type indicator, so I decided to fill in the gap and do it for base type indicator myself, but it did not proceed as imagined:
Consider a weak organic base indicator $\ce{InOH}$, for it, the equilibrium will be:
$\ce{InOH (aq) <=> In^{+} (aq) + OH^{-} (aq)}$
and it's equilibrium constant then would be:
$\displaystyle{K_\mathrm{In} = \frac{[\ce{In^+}][\ce{OH^-}]}{[\ce{InOH}]}}$
then following the reasoning as done in previous case, for a sudden color change, $[\ce{In^+}] = [\ce{InOH}]$, so that,
$\displaystyle{K_\mathrm{In} = [\ce{OH^-}]}$
so that,
$\displaystyle{\mathrm{p}K_\mathrm{In} = \mathrm{pOH}}$
which inturn implies that,
$\displaystyle{\mathrm{pH} = 14 - \mathrm{p}K_\mathrm{In}}$
which doesn't seem to match with the previous result! Am I missing something here?
Note: I doubt that the condition $[\ce{In^+}] = [\ce{InOH}]$ might be the trouble, but I think it shouldn't be as in theory, the different color arises due to the respective indicator and their respective conjugate acid or bases.

It is true that acid-base indicators ($\mathrm{pH}$-indicators) are either weak acid or weak base. What I understood reading your question is that you have wrong impression about bases. To clear your view, not all bases contain $\ce{OH-}$ ions. Specially, most organic bases are weak and contain electronegative ion (e.g., $\ce{N}$ or $\ce{O}$) with at least one lone pair. This fact is well cleared by Poutnik's answer, thus, I'm not going to elaborate it more. Instead, I'd like to explain the action of a $\mathrm{pH}$-indicator (e.g., $\ce{HIn}$ in acidic form) to some depth.
Since Indicators ($\ce{HIn}$) are weak acids (or weak bases), in aqueous medium, it's in an equilibrium:
$$\ce{HIn + H2O <=> H3O+ + In-} \tag1$$
$$\therefore \ K_\mathrm{aIn} = \frac{[\ce{H3O+}][\ce{In-}]}{[\ce{HIn}]} \tag2$$
Taking log on both side followed by simplification gives:
$$-\log [\ce{H3O+}] = -\log K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right) $$
$$\text{Hence}, \ \mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{[\ce{HIn}]}\right) \tag3$$
This is a resemblance of Henderson-Hasselbalch equation. The equation $(3)$ shows that when $[\ce{HIn}] = [\ce{In-}]$, $\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$.
Let's consider the indicator, Litmus, which is we commonly use in labs. Litmus is a weak organic acid (see the structure in bottom box of the diagram; $\mathrm{p}K_\mathrm{aIn} = 6.5$). The $\color{green}{\text{green circle}}$ shows the $\ce{H}$ of $\ce{HIn}$ molecule. The un-ionized litmus ($\ce{HIn}$) is $\color{red}{\text{red}}$ (below $\mathrm{pH}$ 4.5), whereas the ion part $(\ce{In-})$ is $\color{blue}{\text{blue}}$ (above $\mathrm{pH}$ 8.2):

Usually color change happens within a range of $\mathrm{pH}$. It is a rule of thumb that this range falls between the $\mathrm{p}K_\mathrm{aIn}\pm 1$. That means, acid color persists if $[\ce{HIn}] = 10 \times [\ce{In-}]$ and base color dominates when $10 \times [\ce{HIn}] = [\ce{In-}]$. For example, apply these values in the equation $(3)$:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{[\ce{In-}]}{10 \times [\ce{In-}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log \frac{1}{10} = \mathrm{p}K_\mathrm{aIn} -1 $$
Similarly,
$$\mathrm{pH} = \mathrm{p}K_\mathrm{aIn} + \log \left(\frac{10 \times [\ce{HIn}]}{[\ce{HIn}]}\right) = \mathrm{p}K_\mathrm{aIn} + \log 10 = \mathrm{p}K_\mathrm{aIn} +1 $$
Hence, where $\mathrm{pH} = \mathrm{p}K_\mathrm{aIn}$ is the most intense point during color change. For litmus, color change happens between the range, $\mathrm{pH} = 5.5-7.5$ since $\mathrm{p}K_\mathrm{aIn} = 6.5$ for litmus.
Let's see litmus in basic solution where $\ce{In-}$ ions are predominant ($\color{blue}{\text{blue}}$ solution). Now you start adding acid to the solution and you can see what would happen to the solution (equation $(1)$) by using Le Chatelier's Principle. Since excess hydronium ions disturb the equilibrium, it'd shift to left hand side of equilibrium to reduce excess hydronium ions by reacting $\ce{In-}$ ions. Eventually, solution changes the color to $\color{red}{\text{red}}$. Opposite would happens if you added hydroxide ions to the $\color{red}{\text{red solution}}$ in equilibrium.
The scaning UV spectra in the upper box of the diagram shows how color change happens in solution with $\mathrm{pH}$. The indicator here is bromothymol blue, which is yellow in the acidic medium absorbing light at about $\pu{430 nm}$. When $\mathrm{pH}$ has changed systematically with time (by adding known aliquot of $\ce{OH-}$ solution with in fixed time), the solution changed the color accordingly, and bromothymol blue has changed to its basic color blue, absorbing light at about $\pu{620 nm}$. Note that the peak at $\pu{430 nm}$ decreases while the peak at $\pu{620 nm}$ increases while $\mathrm{pH}$ increases.

A basic indicator is usually a Broensted-Lawry base ( accepting protons ) rather than Arrhenius base ( releasing hydroxide ions ):
$$\ce{B + H2O <=> BH+ + OH-}$$
or $$\ce{BH+ + H2O <=> B + H3O+}$$
depending on if the indicator is used in its base form or conjugate acid form (the latter is usually more stable and more soluble).
Regardless of the applied indicator form, $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{ \frac{\ce{[B]}}{\ce{[BH+]}}}$$
where $K_\mathrm{a}$ is the acidity constant of the conjugate acid.

The indicators are never in the form "InOH" with a covalence between "In" and OH, producing $\ce{OH^-}$ ions in water. If a molecule contains one OH group attached to a Carbon atom, it would be an acid, an alcohol, en enol or a phenol. These sorts of molecules are never releasing $\ce{OH^-}$ ions in water. On the contrary, they are weak acids.


text:
Cyclopentanone is treated with piperidine
cyclopentanone is treated with piperidine in acid catalyzed conditions to form the organic product (A). (A) on reaction with benzyl bromide followed by hydrolysis gives another organic product (B) containing aromatic ring. If the sum of degrees of unsaturation of (A) and (B) is x and the difference of the number of carbon atoms in (A) and (B) is y, then find the value of x/y.
According to me compound (A) is an enamine, but I'm not sure about (B)!

So according to this, my answer is x/y = 10/7
But the actual answer is x/y = 4.5
I don't know where am I going wrong!

What reactions do enamines undergo? They are isoelectronic with enolates.

If you have a enolate and alkyl halide in a mixture, what product you expect to have?

A ketone with some extra carbons than the enolate!

Structure given for (B) is not quite correct. Your question says "(A) on reaction with benzyl bromide followed by hydrolysis gives another organic product (B)." Then what happens to the hydrolysis?

Exactly so, you regenerate the ketone

As per the reaction sequence what you said about A is absolutely correct and in B I think you didn't consider hydrolysis of the product. The product what you have written is correct before hydrolysis, and after hydrolysis the carbonyl reforms and the other component becomes piperidine again. So,the final product B is 2-(phenylmethyl)cyclopentan-1-one. Hope that helps!


text:
Why does H+ attack alkene first in this reaction?
The reaction is:
Why is the double bond attacked here even though it is in conjugation with the lone pairs of O thereby making it harder to extract. Also,don't the 2 lone pairs on OH have more electron density than the pi bond?

The cation is formed because it is in conjugation with the oxygen. BTW: How do you know the cation is formed? Is it because of some product that is formed down the line?

It's better to check the resonance hybrid whenever we have a confusion, so, in this case resonance takes place between lone pair of oxygen atom and the double bond and the electrons shift from oxygen towards the alkene. This creates a partial positive character on the oxygen atom and partial negative character on the alkene carbon which is farthest from the oxygen atom. Now we can easily decide that in between ether oxygen atom and alkene proton will tend to attack alkene carbon and that to on the carbon, which is farthest from the oxygen which results in the positive charge on the carbon next to ether oxygen (this cation is also stable because of octet completion due to resonance of oxygen (may also be called as backbonding)). Now deciding between hydroxy oxygen and alkene is a bit tougher task. Let us assume that proton attacked hydroxy oxygen and it formed oxonium ion, now it can be removed as water and a cation can be formed there. After formation of cation there is nothing to do and it will be waiting for the hydroxy component of water to get attacked by. But if the cation is formed at the alkene carbon it can go for rearrangement which increases the size of 4-membered ring to 5, which increases entropy of the system and also it will lead to formation of $\ce{C=O}$. That will tend the reaction to move forward (like pinacole pinacolone rearrangement). So,it's better to form cation at alkene site rather than forming oxonium ions.
Following is an arrow-pushing diagram of the above description:


text:
Why does Sodium Hydrogen Sulfate (Sodium Bisulfate) produce moisture?
In my chemistry lab, I have a vial of Sodium Hydrogen Sulfate (Sodium Bisulfate). I've noticed that when I take a spoonful using a spatula, the spatula gets covered in many tiny droplets, and the lid of the vial always has a layer of tiny droplets on the inside.
Does the Sodium Hydrogen Sulfate (Sodium Bisulfate) react in any way to produce this moisture? Is this to do with the salt being hydrous or anhydrous?

I guess moisture comes from the air.

Does the Sodium Hydrogen Sulfate (Sodium Bisulfate) react in any way to produce this moisture?

This process is called deliquescence (mouthful word). This process is not specific to your salt, even table salt, NaCl, would do the same in a humid environment e.g. on a rainy day. This has nothing to do with the state of hydrated or anhydrous salt. It is the nature of the compound.
The opposite of this phenomenon is efflorescence. The hydrated salt becomes dry when exposed to air.

I wouldn't categorize either sodium chloride or sodium bisulfate as deliquescent. If you want to see a good example of a deliquescent compound, put a teaspoon of calcium chloride (ice melt) in a petri dish and leave it out for a day. You'll find a liquid when you return. It dissolved in the water it absorbed from the air. When is the last time your salt shaker shook water? Table salt is nonetheless hygroscopic, thus why my grandmother put rice in the salt shaker as a desiccant. Just saying for the record.


text:
Why does a first order reaction only depend on the concentration of a single reactant?
I am new to kinetics so please explain it from basic, If there is a reaction  -
$$\ce{A + B + C -> D}$$
In first order reaction rate is given by-
$\mathrm R = k[\mathrm A]$ or $\mathrm R = k[\mathrm B]$ or $\mathrm R = k[\mathrm C]$
my question is why cant $\mathrm R = k[\mathrm A]^{1/3}[\mathrm B]^{1/3}[\mathrm C]^{1/3}$
 even though order of reaction is still $1$,----- is it by definition that 1st order reaction always depend on concentration of single reactant, or is there any experimental proof for this?.

because the definition of first order reaction is that it depends on concentration of single reactant

@chemstackisunhelpful  I am sorry but your  answer is unhelpful

pl explain it briefly

$\ce{A + B + C -> D}$ is not a first order reaction.

It can have "first order in A", though.

This reaction can have many intermediate steps. The rate of rate determining step is dependent on the concentration of one reactant only. So overall the reaction rate is dependent on the concentration of one reactant.

@Martin  but in my textbook all the formulas for 1st order, 2nd order and so on are derived by assuming the same reaction I have stated

@Manu but in rate determining step the reactants could have any reaction coefficient, that is  how much moles of reactants are reacting ,in the time of slowest reaction

@Martin and I think just by looking at hypothetical reaction it is impossible to tell the order of reaction

If you assert the hypothetical equation to be an elementary step, then the order is defined exactly with it. If the reaction has intermediate steps, then it is no longer an elementary reaction and you have to switch to empirically determined kinetics and everything can get a whole lot more complicated. Despite calculating every possible reaction pathway, you will no longer have any theoretical means of determining the order. And now I must admit I do no longer understand any part of your question. || If you were to add a citation for the book you're referring, that'd be awesome.

@Martin sorry but I am new to this thing, can you please state what part of question is not understood , because I have tried every possible way to explain my doubt in simple manner

Either the reaction equation is an elementary reaction, then it is not of first order; or it has an empirically determined rate law of first order, then it is not an elementary reaction. I have no clue what you are actually asking about. In the first case, the question cannot be answered. In the latter case, it is very dependent on what the reaction actually is, and the question cannot be answered in general terms. Please add a citation to the book you are using, it might help to understand your problem.

@Martin you mean to say if the reaction I have mentioned is elementary then rate would be k[A] or k[B] or k[C} but if it is complex reaction then it would become false?

No. If $\ce{A + B + C -> D}$, then the rate law would be $r = k[\ce{A}][\ce{B}][\ce{C}]$, if the rate law is determined to be $r = k[\ce{A}]$, then the reaction cannot be elementary.

I'm late but maybe I can be of some help. @martin is right, the distinction between an elementary reaction and a reaction with several elementary steps is very important. If the reaction is not elementary, the reaction orders with respect to each reagent can, at first, have any value. There are examples of first order elementary reactions, like the decomposition of ions in tandem mass spectrometry or different types of intramolecular reactions. Even if you have an elementary reaction of the type A+B-->C, it can be treated like pseudo first order in certain conditions.

@Blg90 Yeah I get it, the key to the question is if 1 of the reactant concentration in $A + B--> C$ is so greater that it almost becomes a constant,  it is excluded from rate law, even though this reaction is bimolecular, Thank you so much for help

As a purely mathematical definition, a first order reaction is one in which the rate is directly proportional to concentration of reactants. The thing you should look for is physical examples of this. This only happens if you have reactions where one compound is decreasing and the other one is abundance (pseudo - first order) , or , radioactive decay where the decay rate is proportional to number of atoms at a point.
The reaction you have shown with three reactants would most probably me not first order unless two of the three are abundant. So, the solution to the confusion is to think of the physically possible scenarios which the equation models instead of creating hypothetical situations and force the math to model that.

Yeah thanks for answer , after so long, I had already finished chemical kinetics !!

Did my answer help? ^^

Yeah you have written almost same, about what I thought after getting no help, But thank you for the answer it really helps a lot, to clear concepts


text:
I need help with Cahn-Ingold-Prelog (boat conformation)

I need to decide if the chiral carbon (with the attached hydrogen) is in R-configuration or S-configuration. To be able to do that, I need to know which of the atoms are pointing backward, forward and which are in planar. It looks like the molecule is in a boat conformation, but there is not clear drawing of which direction the atoms are pointing. Is there a specific rule when deciding which of the atoms on the chiral carbon is pointing backward, forward and which is in planar?
Thank you so much in advance! All answers are appreciated.

It's drawn like this in my workbook. If I'm not wrong; hydrogen is in the exo-position, but I don't know if that means it's pointing forward or backward. In chair conformations we usually have axial and equatorial positions. Does that apply to boat conformations too?

I tried drawing the molecule with help from the answers. Since hydrogen is in planar, I switch it with the atom pointing backwards. Based on the priorities of the atoms and their positions, this will give us R-configuration. However since we changed two positions the answer will be reversed and thus the chiral senter is S-configured. Does this sound correct?

In that case, see the left structure in my answer. The bond with the wedge is going back. It should be straightforward now.

As in hydrogen is pointing backwards? is there a spesific rule when it comes to boat conformations (e.g. equatorial vs axial positions in chair conformations)? Shouldn't there be two atoms in planar, one pointing back and one pointing forward? But in the drawing under there seems to be two pointing forward and two in planar.  I appreciate your answer, thank you.

Each carbon atom is tetrahedral. You treat it exactly the same way as if it weren't in a ring. The H and the C-C bond down are in plane. The C-N bond is wedge forward. The C-C bond of the ring is wedge back. Try building the structure if you're still confused.

I tried solving my question based on your answer in the official comment; does it look correct? :)

Welcome again! Unfortunately this is not really how Chemistry.se works. You can take the [tour] to find out more. The problem -- now -- is that the question and answer don't really go together. SE is not a forum, back-and-forth is discouraged as the system is not designed for it. We are rather looking for concise questions with concise answers.

You can't actually tell here because your drawing is abiguous. Where the bond going up from the nitrogen crosses the other bond, we need to know which one is in front. Usually, you show that by making a small break in the bond in the back and/or bolding the bonds that are in front.
Which one is it?

When I look at the OP's structure, I invariably see it as your left hand structure. I have to concentrate to see it as the right hand one. How about you? Would make an interesting survey question.

I had a situation at work recently with a silhouette that felt similar, but in this case, I can see both just as easily.


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Looking for advice regarding a larger scale soxhlet extraction
I have successfully completed a few soxhlet extractions at a much smaller scale using a soxhlet extractor with ~200ml capacity.  With that, I was able to extract from ~100g of pre-powdered, sonicated leaf with good results.
I'd like to extract a total of 1Kg, so I need to scale up. However, this is something I haven't done before. I'm currently looking into soxhlet extractors with a much higher capacity.  I found a Soxhlet extractor on eBay with a capacity of 1000ml with 34/45 lower and 71/60 upper joints.  I have also found a 16" 4-bulb Allihn condenser (71/60).
I plan first to defat the leaf with hexane, then extract with methanol since the alkaloids I'm after (Mitragynine) are highly polar.
Questions:

This condenser seems to have adequate cooling capacity, but I'd feel a lot better if someone could confirm or suggest a better option if there is one.
I'm also curious about what the optimal size flat bottom flask/amount of solvent would be.  After looking at similar apparatus, I'm guessing a 2000ml flat bottom flask with 1000ml of solvent.
Going from a 200ml to 1000ml capacity extractor, am I right to assume I can extract 5x as much powdered leaf?  In this case, ~500g?
This apparatus will be quite heavy, and the center of gravity not ideal.  How best to support it safely?

Thanks in advance!

Condenser capacity depends on the heat applied and the temperature of the coolant. You don't want to be above about one-third of the condenser. What is the volume before the Soxhlet siphons? Maybe aim for about three times that at least. Always use clamps with a stand too.

@Beerhunter thanks for your comment.  I'm pretty sure I found the perfect condenser for the job.  I calculated the volume of liquid that could be held in that space and it came out to roughly 1000ml, that's without adding the leaf to be extracted.  I also ordered a 3000ml RBF with two necks and a thermometer adapter for the second neck.  It would seem I estimated correctly.  I also got a heavy duty lab support stand, tripod style with 36" length, 1/2" diameter, and some heavy duty clamps that could accept the larger diameter of the upper apparatus.  I hope that's enough!


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Why is baking soda good at dissolving grease?
Can someone please explain why baking soda is good at dissolving grease?
Thanks,

It is not so much about dissolving as it is about surface tension.

@IvanNeretin How does surface tension come into play? Boiling baking soda in a dirty pan lifts black grease that could not be removed by detergents.

How is this off topic? How sodium bicarbonate could produce the same (or better) macroscopic results as a nonionic surfactant while not being a surfactant itself is definitely a chemistry question!

Baking soda is an alkali, so it dissolves grease very well.

Source

Alkalies are good grease removers because the alkalies form an emulsion, a mixture where oily or solid particles are held in suspension. The particles do not separate from the rest of the liquid; therefore, they are not redeposited on the surface being cleaned.

Source

Alkaline cleaners work well because they emulsify grease. Fatty acids are normally insoluble which is why they cannot be cleaned using water alone. The alkaline breaks down fat making them dispersible in water.

Source


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Why do some molecules show non-symmetric absorption and emission bands?
I'm looking at a molecule of [Ru(bpy)3]2+ in water and the absorption and emission bands are not symmetrical. I want to know why the mirror image rule doesn't apply here. Is this because of the MLCT, is this something to do with the solvent? Are there more likely to be processes involving non-radiative decay?

Mirror image spectra are only observed in solution and then only if the ground and excited state potential energies have almost exactly the same shape. Thus in rigid molecules such as anthracene a good mirror image is observed but not in more flexible molecules such as stilbene ($\phi-C=C-\phi$). The spectra of each vibrational level is wide in solution due to solvent interactions altering energy levels v. slightly, this does not happen in the collisionless gas phase where lines are sharp. In solution the width of lines from different vibrations can overlap making a broad largely structureless spectrum.

There are different effects, a stokes shift due to solvent causing lowering of excited state energy shortly after absorption, a displacement in position of minimum energy, due to different bond lengths in each state and anharmonicity in one potential vs the other. Even slightly different Franck Condon factors between states can mean that a phosphorescence spectrum is similar, but different to its fluorescence spectrum, see figure below, when correction is made for detector sensitivity vs wavelength.

The figure below (not to scale) shows how ideal mirror image spectra are formed. Notice that displacement of potential energy is needed and that vibrational frequency in ground and excited state can be slightly different.

So it's a combination of the structural factors (how "rigid"), solvent effects and electronic factors (Franck Codon)? Very well explained, thank you!

From the Jablonski diagram,

The mirror image is only true if you are talking about transitions from $S_0$ to $S_1$ (absorption) and $S_1$ to $S_0$. Also read about Kasha's rule.
Quinine is the most famous fluorescent molecule, its $complete$ absorption spectrum is not a mirror image of its emission spectrum.

In this case, is there a reason that the metal complex is excited to a higher state and not just to the S1 level?

The metal complex gets excited to all possible states, because why not.

@Abu, That is exactly what Kasha's rule says. Excitation can occur to various levels but emission always occurs from S1.


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Is the elimination reaction between 1 1 dibromo cyclopropane and phenyl lithium feasible in alkaline medium (and some heat)
I found this question in one of my question papers for the JEE ,
My sir gave me an explanation that due to resonance between lone pairs of the bromide and π bond formed.

This reasoning doesn't seem to be satisfying because of the cyclopropyl ring strain which is not lost even at the end of reaction.
Please give an alternate pathway of the above product is not correct.

The alkene formation is not correct under this conditions. This is an example of Doering–LaFlamme allene synthesis, a reaction of alkenes that converts to allenes by insertion of a carbon atom introduced in 1958 (Ref.1; see user  user55119's comment). In abstract, it states that:

In a two-step sequence, of which the first step involves addition of dibromocarbene to an olefin and the second involves reaction of the resulting substituted 1,1-dibromocyclopropane with magnesium or sodium, allenes are obtained. The overall structural change involves the insertion of a single carbon atom between the two of the original double bond and therefore represents a novel way of increasing carbon chain lengths by one atom.

In your example, the compound has undergone the insertion of additional carbon atom (between two $\mathrm{sp^2}$-carbon atoms of the original double bond). Therefore, it's now the rearrangement part assited by organo lithium compound, the mechanism of which is depicted in following diagram (Ref.2):

Accordingly, the first step of the gem-dihalocyclopropane (I) with alkyllithium can be depicted as a halogen-lithium inter-conversion with formation of a 1-lithio-1-halocyclopropane derivative II. The reaction is dependent on the nature of both halide and the lithium reagent, but its mechanism is not fully understood (Ref.2). For example, bromides react much more readily that the corresponding chlorides, and butyl lithium is generally much more reactive than methyllithium. It is expected that intermediate II would readily eliminate lithium halide as discussed in Ref.1. This may occur by two different mechanisms: (i) Pathway $\color{blue}{\bf{a}}$: Concerted elimination and ring openig to an allene; or (ii) Pathway $\color{maroon}{\bf{b}}$: An $\alpha$-eleminatin to the carbene intermediate III (Ref.2). The author comments that the pathway $\color{blue}{\bf{a}}$ may well be the way by which allenic product has been formed, although it appears that cyclopropyl anions have considerable stability. However, the formation of non-allenic isomers (byproducts) in a number of these reactions is difficult to explain (Ref.2).
References:

W. von E. Doering, P. M. LaFlamme, "A two-step of synthesis of allenes from olefins," Tetrahedron 1958, 2(1–2), 75–79 (https://doi.org/10.1016/0040-4020(58)88025-4).
Lars Skattebøl, "The Synthesis of Allenes from 1.1-Dihalocyclopropane Derivatives and Alkyllithium," Acta Chem. Scand. 1963, 17(6), 1683 – 1693 (DOI: 10.3891/acta.chem.scand.17-1683).

Sir , thanks for your wonderful support

But sir , how does phenyllithium affect the reaction (compared to any non aromatic organolithiums)??

There are two pathways that phenyllithium could follow - it could deprotonate, but there really not any reactive protons, or it could do lithium-halogen exchange with one of gem-bromines. This second pathway will lead to a carbene by loss of $\ce{Br-}$ and the carbene (a highly reactive intermediate) will rearrange to give allene $\ce{H2C=C=CH2}$.
This method is documented:

In Page 275: H. Maskill, "Chapter 1: Three- and Four-Membered Carbocyclic Rings," In Specialist Periodical Reports - Aliphatic, Alicyclic and Saturated Heterocyclic Chemistry, Volume 1, Part II: Three‐ and four‐membered rings (carbocyclic and saturated heterocyclic); W. Parker, Ed. (Senior Reporter); The Chemical Society: Burlington House, London, England, 1973, pp. 3-292 (ISBN: 0-85186-502-X).
In Pages 1492-1493: Armin de Meijere, Editor, Carbocyclic Three-Membered Ring Compounds, Houben-Weyl Methods of Organic Chemistry, Volume E 17 b - Cyclopropanes: Synthesis; Georg Thieme Verlag: Stuttgart, Germany, 1997,  p. 1492 (ISBN: 978-3-13-798204-3).

Can you show with an arrow pushing diagram how the carbene will rearrange to give allene?(just to verify my own ideas)

Yusuf Hasan: See this: https://en.wikipedia.org/wiki/Doering%E2%80%93LaFlamme_allene_synthesis

Sir , thanks for your wonderful support


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Is there a reason why we add (CH3CSNH2) but not H2S in qualitative group analysis?

Brief question: In $\ce{Cu}$-$\ce{As}$ group analysis we are expected to add thinoacetamide to make a precipitate with sulfide. Can we use $\ce{H2S}$ instead?

For safety reasons. Older generations of chemists who studied in 70s-90s will remember the stench of rotten eggs of the gas.

Please do not insert vertical pictures, which are hard to read. Put them in horizontal position.

Go and look up how toxic H2S is.

@M.Farooq Unfortunately hydrolysis of thioacetamide also stinks :(

@Mithoron, Still better than using Kipp's apparatus. Have you ever used one?

@Mithoron, So you don't belong to that club when qualitative inorganic analysis had all this stuff. Mercury is a no-no, so is Kipp's apparatus.

In my qualitative inorganic course they were called $\ce{H2S}$ group. It was also said that ‘use TAA but we do have a gas bottle in case you have a really pesky analysis.

@MathewMahindaratne And thanks for editing, sorry I didn't know how to..


text:
On mechanism behind Marcus theory
Regarding the mechanism behind Marcus theory, I was wondering if reaction occurs through tunneling when it does not have sufficient activation energy? Wouldn't expecting reactions to occur this way be like expecting random combustion to happen in the arctic?

Marcus theory doesn’t say that reactions only occur via quantum tunnelling. What gave you that impression?

Please read Marcus Theory for Electron Transfer, which gives a good description.

There is some more data here https://chemistry.stackexchange.com/questions/129172/electron-transfer-rates-in-marcus-theory

@orthocresol In electron transfer the electron does tunnel, but is aided by vibrational and solvent effects to equalise the energy levels first. This is why electron transfer is a solvent limited reaction, i.e. limited by the time it takes for the solvent to reorganise to equalise energy levels, see  R. Marcus, Biochimica et Biophysica Acta., 811, 1985, 265 - 322) for a figure. This crucial effect is not clear in a classical derivation of the rate constant. Experiment performed in viscous solvents shows that it is solvent limited.

@porphyrin, hmm, actually, maybe that's why OP (and I) seem a bit confused. Do correct me if I'm wrong - it has been a while since I had to think about this - but the reactant geometry has to distort to the TS geometry before the electron can tunnel, or "jump" from one PES to the other, and that reorganisation doesn't/shouldn't occur via tunnelling...? So the electron tunnels, but the atoms have to move effectively classically (except for occasionally H).

@orthocresol The reorganisation is just the extension/contraction of of bonds and most importantly solvent motion /reorientation to equalise the energy at which point the electron tunnels and relaxation occurs into the product. Reorganisation happens before the reaction, it is, as it were,  how the 'hill' is climbed to reach the transition state thats all. (This is the insight that Marcus had). Consequently it is the rate limiting process, even though it may take less than a picosecond in some reactions.  There is a nice diagram in the BBA paper.  I will shortly add this as an answer.

@porphyrin Fair enough! I was thinking of the "reaction" as being the whole process, including reorganisation & subsequent relaxation.

The reorganisation is just the extension/contraction of of bonds and most importantly solvent motion /reorientation to equalise the energy at which point the electron tunnels and relaxation occurs into the product. Reorganisation happens before the reaction, it is, as it were, how the 'hill' is climbed to reach the transition state thats all. (This is the insight that Marcus had). Consequently it is the rate limiting process, even though it may take less than a picosecond in some reactions.  After reaction, relaxation (reorganisation) also occurs so that the solvent reaches equilibrium with the new charges distribution.

Figure source; R. Marcus, Biochimica et Biophysica Acta., 811, 1985, 265 - 322)


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Did O₄ turn out to explain why liquid oxygen appeared to defy Curie's law? If not, what does?
Wikipedia's Liquid Oxygen says:

The tetraoxygen molecule ($\ce{O4}$) was first predicted in 1924 by Gilbert N. Lewis, who proposed it to explain why liquid oxygen defied Curie's law. Modern computer simulations indicate that, although there are no stable $\ce{O4}$ molecules in liquid oxygen, $\ce{O2}$ molecules do tend to associate in pairs with antiparallel spins, forming transient $\ce{O4}$ units.

The paragraph is a bit ambiguous and doesn't rule out that the "transient $\ce{O4}$ units" do not explain why "liquid oxygen defied Curie's law", so I'd like to ask:
Question: Did $\ce{O4}$ turn out to explain why liquid oxygen appeared to defy Curie's law, at least in some way? If not, or not completely, what does?

@andselisk I never used to for reasons mentioned in the answer there, but then one day a moderator reverted my title to MathJax which left me scratching my head. I'll try to keep track that Chemistry does it "the right way" ;-)

I found Lewis (1924): The Magnetism of Oxygen and the Molecule 0₄

I guess there is no right or wrong way in a sense that one cannot always find a decent workaround (Unicode or re-phrasing) for the expression written with MathJax, especially complex formulas often required on Physics.SE and Math.SE. So, the rules might deviate from site to site. Since you are quite active across the SE network, it might be hard to keep the track of these preferences, so no biggie, considering how literate your questions always are, MathJax in title is the least problem and can usually easily be fixed.

That wp article would better first explain how LOx "defies" Curie´s law, before making statements about hypothetical dimers.

It might be worth mentioning that Curie's Law is only an approximation - or, rather, it only works under certain conditions. One of the conditions is that the particles need to be magnetically isolated/independent from each other. As "in liquid oxygen, O2 molecules do tend to associate in pairs with antiparallel spins", this condition is not met.  The English Wikipedia Curie's Law article is a bit of a sad affair; it only mentions that the law only applied to "many paramagnetic material" (not all!). The German one is better. Technically O4 does not violate the law, as the law does not apply.

@Klaws thanks, I've changed it to "appeared to defy"

The observation that oxygen did not seem to strictly follow Curie's law (where $1/\chi$ is linear with temperature)  has been known since Kammerlingh Onnes pointed it out in the 1910's. By the late 1920's, it was known that diluting the liquid with nitrogen preserved Curie's law, suggesting that it took oxygen in close proximity to other oxygen to be the issue.
Fast forward some 80 years, and the computational tools available to us provide some key insight. As one such paper, Ab Initio Molecular Dynamics Investigation of the Structure and the Noncollinear Magnetism in Liquid Oxygen: Occurrence of O$_{4}$ Molecular Units takes a look at structure and properties of liquid oxygen. Quoting from their conclusion:

In  conclusion, we  performed ab  initio molecular  dynamics  to  investigate  the  noncollinear magnetism  of  a system with an evolving atomic structure. Application to liquid oxygen provides us with a picture in which the large majority of colliding O$_{2}$ molecules assume structural and magnetic configurations which closely resemble those in the O$_{4}$ molecule. Formation of truly long-living molecular O4 units  also  occurs but involves a considerably smaller fraction of O$_{2}$ molecules.

So, it would seem that O$_{4}$ units, whether long lived 'molecules' or temporary short ranged order is the answer to the deviation from Curie's law in liquid oxygen.
Several comments on these O$_{4}$ units - first, as noted in the paper above, 'long lived' means under a picosecond (which doesn't really seem long-lived to me, but...). Further, the authors above point to The Infrared Spectrum of Bound State Oxygen Dimers in the Gas Phase to explain what they mean by O$_{4}$, which they call a "somewhat rigid (O$_{2}$)$_{2}$ molecule".

When looking for papers did you find any experimental evidence for the structure of  O4 or similar dimers/trimers etc. ?


text:
How do I prepare the following compound from diphenylmethane?
I am having trouble with the preparation of this product from diphenylmethane as a starting material. One of my thoughts was to attach the butene into the structure, but I don't see how it is viable. So I would like some suggestions on how to proceed, and also with the name of the compound (butenediphenylmethane?).

Does it have to be that reactant? You'd probably have much easier time starting from the ketone and then doing a wittig reaction...

Diphenylmethanes deprotonate readily with alkyllithium

@Waylander how would you introduce the double bond after a reaction with propyllithium?

I would react the lithiated diphenylmethan with DMF, workup to give the aldehyde and do a Wittig

I'm a little concerned that that the product is very easy to convert to the 1,1-diphenylalkene.

Here's a mechanism that I thought of (It's more theoretical than practical):

Bromination in the first step will occur at the benzyl position due to a stable free radical. Sodium Propynide then adds more carbon atoms via the SN2 mechanism, which are then reduced with Lindlar's catalyst to give the cis-alkene.
The name of this compound should be (Z)-1,1-diphenylbut-2-ene, but as pointed out in the comments, the preferred IUPAC name (PIN) seems to be (Z)-but-2-ene-1,1-diyldibenzene (verified by ChemDraw)

Looks reasonable. Do you have a reference for the reaction with the sodium acetylide?

I couldn't find any papers, but this Is one source. Here's a similar reaction with Sodium Acetylide

It would be a SN1 reaction rather than an SN2 reaction because of the stability of the diphenylmethyl carbocation. Also, E2 won't occur as there are no $\alpha$-H atoms in the substrate

You can't generate carbocations in such a strongly basic medium, so SN1 is a no-no. Your reaction is quite probably OK, but a very brief literature search suggests that using the Grignard + Cu(I) is more likely to work than attempting a direct SN2. cc @Waylander

(one example at e.g. https://pubs.acs.org/doi/10.1021/jm4010434 where it's TMS-acetylene instead of propyne).

Maybe hydrostannylate the propyne then do a Stille with the bromodiphenylmethane

@orthocresol would a Corey-House synthesis with Lithium Dipropenyl Cuprate be a better alternative? (This would also remove the reduction step) I'm not that conversent with organometallics to follow Waylander's suggestion

I'm not sure if you can buy the stereochemistry (i.e. whether the vinyl anion is configurationally stable enough to be used directly in a synthesis). https://www.sigmaaldrich.com/catalog/product/aldrich/419540 doesn't seem to sell well-defined isomers. I think your original suggestion of using an alkyne anion is better. That concept is fine, my original comment was just an extra tidbit, because sometimes some conditions work better than others.

Thanks @orthocresol. I updated the answer to change the mechanism from SN1 to SN2 (no carbocations in a basic medium).

(Z)-1,1-diphenylbut-2-ene is not the preferred IUPAC name.


text:
Can geometrical isomers be enantiomers?
I have just been pondering upon the question above. I have found this example and I am sure they are enantiomers, but do they also count as geometrical isomers?

As far as I know, you have to call them enantiomers and not geometrical isomers since the right side is symmetrical. However, you can designate it as E or Z.. see this.

Please do not use geometric isomerism, it is obsolete and the usage strongly discouraged. Obsolete synonym for cis-trans isomerism.

So can i say them cis-trans Isomers or geometrical isomers clearly?

This issue has been discussed: https://chemistry.stackexchange.com/questions/58630/why-is-1-ethylidene-4-methylcyclohexane-chiral/96265#96265

The enantiomers 1a and 1b can have their double bonds defined by CIP rules 1a and 5. Rule 1a dictates that CH3>H while Rule 5 has R>S. Thus, 1a has an E-double bond  and 1b is of the Z-configuration as exemplified by the red bonds. This issue has been addressed previously on this site.

Addendum: While the enantiomers 1a and 1b (vide supra) have their double bonds labeled in upper case E/Z (ChemDraw), JSmol takes into account the enantiomorphic nature of the double bonds by using lower case e/z (vide infra). See the answer of @orthocresol.

That doesn't really address the question of whether these enantiomers are cis-trans isomers, too.

True enough but cis/trans for double bonds doesn't work here and this nomenclature has been supplanted by E/Z. Old-timers like myself may still refer to 1,2-substituted double bonds as cis/trans.

In general, I don't think there is anything wrong with cis/trans, especially when it is unambiguous. It is called cis-trans isomerism after all, and that is also pretty unlikely to change. But I believe the question wasn't so much about how to name the different compounds unambiguously, but rather whether enantiomers can show cis-trans isomerism or not. Hence my initial comment. I still like your breakdown though.

There is an advantage to having a descriptor for the double bond. By using the IUPAC name preceded by the three descriptors, the correct enantiomer may be drawn. Without employing e/z, one would have to describe the structure, say over the phone. The 2013 Blue Book is now 7 years old. By the next revision, e/z will likely be approved.

Technically, this double bond should not be labelled as (E)- or (Z)-.
It is what is known as an enantiomorphic double bond, for which the proper stereodescriptors are seqCis and seqTrans. This is described in P-92.1.1 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book).

Notice that for your compound, we have $(\mathrm a, \mathrm b) = (\ce{CH3}, \ce{H})$ and the two enantiomorphic ligands are the two sides of the cyclopentane ring.
The rules for this are briefly quoted in a previous answer:

P-92.1.3.5 Sequence Rule 5
An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seqTrans’.

As for whether it can be a "geometrical isomer", note first that the term geometrical isomerism has been deprecated in favour of cis–trans isomerism (IUPAC Compendium of Chemical Terminology, 2nd ed. ("Gold Book"), DOI: 10.1351/goldbook.G02620:

Obsolete synonym for cis-trans isomerism. (Usage strongly discouraged).

Also notice that the Blue Book (also P-92.1.1) describes cis/trans isomers as being in a separate category from the enantiomorphic double bond:

I think it is reasonable to conclude that the given compound cannot be considered as an example of geometrical isomerism, although the cynic in me says that this is mostly a matter of semantics.

After reading your view, I created the structures using JSMol and the double bonds are labeled in lower case e/z, presumably to reflect your point. However, ChemDraw uses the upper case E/Z designation.

@user55119 in general, many software packages do not fully implement the IUPAC recommendations (after all the book is ~1500 pages long (!!)). ChemDraw, as good as it is for drawing structures, is one of them. To me, not using E/Z makes sense, because E/Z is kind of in the domain of diastereomers (for example, we wouldn't use E/Z for allene enantiomers). That said, I think the whole thing is arguably more of a technicality than a real conceptual issue. Everybody including OP seems to be agreeing on the actual concepts.

You and others may be interested in these corrections to the Blue Book 2013.  I haven't checked to see if they have been implemented.
https://www.qmul.ac.uk/sbcs/iupac/bibliog/BBerrors.html
. Here is paper about revised CIP rules by Bob Hanson (JSmol). https://pubs.acs.org/action/showCitFormats?doi=10.1021/acs.jcim.8b00324

Edited version: I originally wrongfully assumed these two structures are not geometrical isomers of the double bond, because it is a trisubstituted double bond with at least two identical groups (although they have two different spatial arrangements as R- and S-designations). As a result, I conclude that the two compounds cannot be distinguished by (E) and (Z) prefixes. Therefore, I thought, it is safe to say they are just stereoisomers (according to the old rules when I was an undergraduate). However, rules have been changed significantly since then and according to Revised Cahn-Ingold-Prelog Rules - IUPAC 2013, (R)-configuration has higher priority than (S)-configuration in this situation. Therefore, actually (E/Z)-configurations exist in these molecules. Accordingly, the top structure has (E)-configuration while the bottom structure has (Z)-configuration (see the diagram below).
About stereoisomerism: Since two stereocenters are having R- and S-designations with identical groups (which are mirror images of each other if double bond is not there), we must check for the symmetric elements such as plane of symmetry. Because of one side of double bond is not symmetric (attached by $\ce{CH3}$ and $\ce{H}$ groups), the molecule is not symmetric as depicted in the diagram:

Keep in mind that, in any case, if  $\ce{CH3}$ group is replaced by $\ce{H}$ or vice versa, the molecule becomes meso-isomer, which is previously considered as not optically active. However, as Zhe pointed out correctly, revised Cahn-Ingold-Prelog Rules give rise to pseudo-chiral centers in meso-compounds. Nonetheless, the molecule in hand does not have a plane of symmetry, and hence have non-superimposable mirror images as shown in the image. As a results, the two structures given are enantiomers, one with (E)-configuration while the other with (Z)-configuration (see the diagram above).

An (R) stereocenter and an (S) stereocenter are not identical for the purposes for CIP. That gives rise to pseudo-chiral centers in meso compounds.

Now you get an upvote :)


text:
Safe reaction to maintain 30psi (2 bar) of gas, while that gas is released at random?
I'm going to use it to power a Lego Pneumatic Engine (LPE) (random example, not mine), that will be self-contained, and start and stop a lot using a throttle valve.  So the requirements are:

Use common materials to run the reaction

Might not even need to be a reaction.  Vapor pressure would do just as well.  See #5.

Be completely safe, both for indoor use with no special collection, and for the materials involved:

ABS Plastic for valves, cylinder walls, and plumbing hard lines
Rubber seals and hoses
Steel piston shaft

Fit in a small (0.5L) plastic water bottle

There's another bottle available, about 1/2 that size, that can be air-displaced into the main one through the standard rubber hoses and ABS hard lines (perhaps to start the process without losing a lot of gas), but I'd rather not use it if I don't have to.  It is available though, if needed.

Produce a large volume of gas (long runtime of the LPE)
Self-regulate to around 30psi (2 bar), so no worries about over-pressure when the LPE is not running

(this might be the tough one)

Not need special cleaning afterwards

My first thought was Baking Soda and Vinegar to generate CO2, with a pressure-regulated air pump to displace the vinegar into the baking soda.  But there are two problems with it:

The original reactants can get into the LPE as contaminants, which then need to be cleaned out.  (the reason for #6 above)
It doesn't stop at Lego's maximum pressure of about 30-35psi.  (the friction-fit hoses pop off around 37psi, and the reason for #5 above)

I have a design for a "smart" pump controller that accounts for most of the pressure coming from a finite-rate reaction and not the pump itself, but it seems a bit rube-goldberg-y to me, and still can't account very well for the expected run/stop pattern of the engine that uses it.

So, is there a different reaction that I can use instead, that can satisfy the requirements up top?

Some pictures may help:
The water bottle sits horizontally, under the black "hold-down bridge", with the bottom against the programmable brick.  (big yellow tower on the back)

"Steam" locomotive with Walschaerts valve gear.  And yes, it runs.  :-)  Both directions.
There's also an optional tender car with the other, smaller water bottle and the motorized air pump to displace something out of it, which is also controlled by the programmable brick.

Why not just a gas cylinder with a regulator?

@Andrew All the regulators that I've seen are huge by comparison.  The 0.5L water bottle would have to shrink to accommodate one.  Otherwise yes, that could be an option.

Without seeing the setup, it's hard to judge but what about grinding up citric acid and sodium bicarbonate and with a touch of water (literally some drops) make a pellet. Then can you have a small water feed that allows a drop in now and then? If you can make a hard pellet, it will slow the reaction rate giving off the CO2, but you may have to play with the hardness of the pellet.

The Linde MicroCAN with regulator is similar in size to an 0.5 L water bottle. Taller, but also thinner. http://hiq.linde-gas.com/en/images/MICROCAN%20Product%20data%20sheet_tcm899-90089.pdf

@Andrew Where would I buy one?  Their website looks like an industrial supply place where you're supposed to have your buying agent call their sales agent, form a corporate relationship, and then negotiate prices and quantities.  I'm just an individual with a hobby.

So, baking powder instead of baking soda?  And pump some water onto it to make it work?

Try calling and asking where you can buy. Explain that you want a microcan of noncorrosive gas (N2 or air) and regulator and hopefully they will put you in touch with a local distributor or sales rep. "HiQ" is the brand under which they sell the small cylinders.

@Andrew I found this today.  It's a bit more expensive than I originally had in mind, but I suppose that any high-pressure gas containment probably is.  Their system even has a way to refill it myself!  (12V pump or a hand pump)  So I don't have to keep buying pre-filled bottles.  2 bar is easily within the range of the final output, though not compared to the environment that it's designed to do it in.  But I wonder how easily it can be made to work anyway?  (I'm already using a sports inflator needle to adapt a bicycle pump to it.)


text:
Determining stoichiometry in ICP-OES data analysis
I have questions regarding data analysis for an inductively coupled plasma (ICP) experiment from a paper by Wang et al. [1].
The generic formula for the chemical compound is $\ce{Na_{2−x}Fe[Fe(CN)6]}$.
In the paper, it was mentioned that chemical formula and ICP data are
\begin{array}{lrr}
  \text{Formula} & \ce{Na} / \mathrm{ppm} & \ce{Fe} / \mathrm{ppm} \\\hline
  \ce{Na_{1.53}Fe[Fe(CN)6].{4.2}H2O} & 112300 & 357200 \\
  \ce{Na_{1.67}Fe[Fe(CN)6].{3.9}H2O} & 116100 & 339400 \\
  \ce{Na_{1.73}Fe[Fe(CN)6].{3.8}H2O} & 118400 & 332400 \\
  \ce{Na_{1.68}Fe[Fe(CN)6].{3.9}H2O} & 112000 & 332700 \\
\end{array}
My question is how they determined the $\ce{Na}$ number ($1.53$, $1.67$, and so on) on each cases based from the $\mathrm{ppm}$? I tried to make a $\mathrm{ppm}$ ratio of $1:2$ for $\ce{Na}$ and $\ce{Fe}$ however the number does not match with the paper.
Reference

Wang, W.; Gang, Y.; Hu, Z.; Yan, Z.; Li, W.; Li, Y.; Gu, Q.-F.; Wang, Z.; Chou, S.-L.; Liu, H.-K.; Dou, S.-X. Reversible Structural Evolution of Sodium-Rich Rhombohedral Prussian Blue for Sodium-Ion Batteries. Nature Communications 2020, 11 (1), 980. DOI: 10.1038/s41467-020-14444-4

Let R be the ratio of number of Na atoms to number of Fe atoms. Then, for the first compound, we have
$$\pu{R = }\frac{\pu{112300 ppm } /\ \pu{22.98977 g mol^{-1} } }{\pu{357200 ppm } / \ \pu{55.845 g mol^{-1} } } = \pu{0.7637  }$$
Since there are two Fe atoms in the compound, the Na number for the first compound is simply 1.527, i.e., twice the value of R calculated above.
Following the same procedure for the remaining three compounds, I get Na numbers of 1.662, 1.731 and 1.635. The last one is furthest from the value of 1.68 the authors published, but maybe 1.63 was read as 1.68.

Thank you very much for your concise explanation. It is really helpful.
I also fell a bit strange about the last number but I guess it might be a mistake.
Once again, thank you.

Thank you as well and glad to be of assistance!