text:
Not sure why, but I can't modify/balance this equation
I'm working on modeling a few reactions for my research. For example:
$$\ce{CO2 + H2S + CH4 <=> CH3SH + H2O}$$
was one I grabbed from the literature. However, this reaction is taking place in the aqueous phase around neutral pH, so most of the $\ce{CO2}$ will actually be $\ce{HCO3-}$. So in an attempt to model modified versions, I swap out the $\ce{CO2}$:
$$\ce{HCO3- + H2S + CH4 <=> CH3SH}$$
I've tried a redox-equation balancing approach but this hasn't really gotten me anywhere- sulfur doesn't change oxidation states.
I think my brain might just be fried, but can anyone lend any insight on this?

You can't just swap out the carbon dioxide. For starters you lost the water to balance all the oxygen atoms... The equation presented should be balanced as is. You're free to balance a different equation, but that's a different equation (try to make sure that one can actually be balanced).

If sulfur doesn't change oxidation states, then what does?

There's no way such reaction happens in such conditions! Where did you hear something like that?

Temp. could only have indirect influence -  it looks it's a complicated enzymatic reaction proceeding in vivo - saying it happens in aq. in mild pH is like saying you detonated a piece of TNT with a touch of a finger, when you actually used a button on a bomb.

@Mithoron- I see. Yeah, we are looking at the thermodynamics of the reaction- nothing kinetically/enzymatically, so it may not proceed quickly. I was just trying to modify the equation from literature that had used aqueous CO2 to actually use bicarb instead. Just in the modeling phase at this point- haven't carried this out in lab yet.

For starters, You have a problem because of a lack of charge balancing. You should also consider that at neutral pH $\ce{H2S}$ is present as $\ce{HS-}$ and $\ce{CH3SH}$ is present as $\ce{CH3S-}$.  Modifying the reaction to account for charge you get:
$$\ce{HCO3- + HS- + CH4 <=> CH3S-}$$
This will give you a start.  The other piece that will help is what the $\ce{HCO3-}$ molecule becomes. For the sake of being able to provide an answer I will assume formate ($\ce{HCO2-})$ is formed. Thus the reaction is now:
$$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2-}$$
We know that the molecule of $\ce{CH4}$ becomes $\ce{CH3S-}$ and the molecule of $\ce{HCO3-}$ becomes $\ce{HCO2-}$ thus to balance the lost oxygen of $\ce{HCO3-}$ and the lost two hydrogens of $\ce{CH4}$ we add water to the products.
$$\ce{HCO3- + HS- + CH4 <=> CH3S- + HCO2- + H2O}$$
This equation serendipitously requires no coefficients to balance.


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Why is there only one ionization isomer of [Cr(NH3)5(NO2)]Cl2?

How many isomers are possible for $\ce{[Cr(NH3)5(NO2)]Cl2}$? Consider all possible types of isomerism and draw each isomer.

My texbooks presents only three possible isomers: $\ce{[Cr(NH3)5(NO2)]Cl2}$ (the original compound), $\ce{[Cr(NH3)5(ONO)]Cl2}$ (linkage isomerism through $\ce{NO2}$), $\ce{[Cr(NH3)5Cl]Cl(NO2)}$ (ionization isomerism).
However, why is $\ce{[Cr(NH3)5Cl]Cl(NO2)}$ the only possible ionization isomer? Why can't $\ce{[Cr(NH3)4Cl(NO2)]Cl(NH3)}$, $\ce{[Cr(NH3)3Cl2(NO2)](NH3)2}$, or $\ce{[Cr(NH3)4Cl2](NH3)(NO2)}$ form as well.
The proposed ionization isomers were constructed by substituting one, or both, of the $\ce{Cl-}$ ions in the complex ion structure with either $\ce{NH3}$ or $\ce{NO2-}$; the answer seems to indicate that $\ce{NH3}$ cannnot under ionization isomerism with $\ce{Cl-}$.

Because $\ce{NH3}$ is not a counter ion and thus would not be bound to the complex if not bonded to the center ion.  In $\ce{[Cr(NH3)5(NO2)]Cl2}$  the $\ce{Cl-}$ ions do not form dative bonds with the $\ce{Cr^{3+}}$ ion but are attracted by the positive charge of the $\ce{[Cr(NH3)5(NO2)]^2+}$ complex.  If you were to form$\ce{[Cr(NH3)4C(NO2)]Cl(NH3)}$, the $\ce{NH3}$ would not be datively bonded to the $\ce{Cr^3+}$ center it would have no bond or attraction to the complex and thus would be a free ion, not part of the complex anymore. Thus the compound would now be $\ce{[Cr(NH3)4(NO2)Cl]Cl}$ as opposed to $\ce{[Cr(NH3)5(NO2)]Cl2}$.

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What is a quick and easy way to find an alternative axis of symmetry?

According to a question in my book (1R,3R)-1,3-dichloro-2,2-bis[(S)-chloro(2H1)methyl](1,3-2H2)propane has an S4 alternative axis symmetry. I tried trial-and-error method using an axis that passes through only the top group straight down, one that is a horizontal line across the screen and some others but couldn't find it. How exactly do I do this?And also this is a JEE practice question and I need to do it in at the most a minute or two.

The only way to get good at 3D visualization is to practice. Various books on group symmetry will have a flow chart, but I could never remember the thing. Typically we could use the "unmarked" character tables for the test in class.

@MaxW what is an unmarked character table? In my book actually just the definitions of plane of symmetry,centre, axis and alternating axis of symmetry is covered

"unmarked" means that you didn't write cheat notes all over it.

@MaxW is there a way of doing it without character tables? I'm not exactly sure what they are but I don't think they are in my syllabus

The S4c operation is a rotation by 90 degrees and reflection in the perpendicular plane. Have a look at  the answer https://chemistry.stackexchange.com/questions/58609/understanding-group-theory-easily-and-quickly/58613#58613 and also you can practice adding symmetry elements if you search for www.molecule-viewer.com and look at molecules under the  S4 point group for examples of the S4 operation.

Spatial perception is not an easy thing to teach you because there are some many cases. Building a model will likely start helping you to visualize the symmetry elements.

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Hybridization and bonding in ethenone H2C=C=O?
My book says ethenone is a planar molecule, so I was trying to understand its hybridization,I feel that like allenes ethenone too will not be planar. Supposing oxygen and the carbon near it to be in the xz plane and assuming that the pi bond between them involves py orbital of carbon, this implies the bond between that carbon and the adjacent carbon involves the pz orbital of that carbon. This implies that the end carbon's hydrogens will be in xy plane. What is wrong with my understanding?

You are right.the hibrydization is the same with allenes but in allenes there are 2 CH2 groups but in ethenone there is a CH2 and there's a O with two none bonding pairs. The hydrogens will be in the same plane as C=C=O but  the electron pairs on oxygen are not in the same plane

But all the atoms are in the same plane. Just to be clear.


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Discrete molecules
I came a across a passage in my textbook on Formula Mass topic which mentioned something about discrete molecules.
Some substances such as sodium chloride do not contain discrete molecules as their constituent units. In such compounds, positive (sodium) and negative (chloride) entities are arranged in a three dimensional structure.
What does the author mean by discrete molecules? What are the properties of discrete molecules? I came across this term for the first time. What does the author want to convey in the above context?

It's an ionic compounds. There are no molecules.

@Zhe Can compound be formed both ways? i.e. It can be either ionic or a molecule?

I'm not sure I understand what you are asking? If you are asking if there are covalent compounds and ionic compounds, the answer is yes.

@Zhe And both these are different concepts? Covalent compounds and ionic compounds?

@SakshamSharma Yes. There are a variety of different types of solids.

In a discrete molecule, you can know which atoms are bonded to each other.  like in methane; the hydrogens are bonded to a specific central carbon but do not freely ionize from it and interchange without an activation energy.  If I mix methane with acetic acid there will be no exchange of attoms.  In sodium chloride You do not know which sodium will be bonded to which chloride when melted/vaporized.  If I mix sodium chloride with silver nitrate in solution, atoms will exchange and I will get sodium nitrate and silver chloride.

What the author is pointing out is that not all solids are made up of distinct units (in his terminology "discrete molecules").
Consider the difference between the following solids: copper, ice, iodine, common salt, diamond.
In copper the units are copper atoms held together (in effect) in a sea of electrons but with a fixed arrangement of the atoms.
In ice, we have a rigid array of water molecules largely held in a 3D array in specific orientations by hydrogen bonds between the water molecules.
In iodine the structure consists of I2 molecules held together by weak dispersion forces.
In common salt the units of the crystals are individual ions of Cl- and  and Na+ ions in an infinite 3D array as described in the picture.
In diamond the structure consists of an infinite array of tetrahedral carbon atoms bonded to each other in an infinite network by covalent carbon-carbon bonds.
In ice and iodine the discrete units of the structure are molecules and when the solids fall apart the molecules retain their individual, discrete identity: ice melts to give water which also contains water molecules; iodine sublimes to give a vapour that still contains iodine molecules.
None of the other solids can be decomposed into molecules. If, for example, salt is dissolved in water, the result is a solution of the individual ions.
The point the author is making is that some solids contain units that are preserved when the solid is broken up and some do not. In many chemicals the units are discrete molecules that retain their identity when not in the solid state. In others the units that make up the crystal are either atoms or ions and there are no discrete components at any other level.

does the author mean to say that sodium chloride is not a molecule but a combination of metal and non metal? And this metal and non metal aren't molecules either? And hence it is an ionic compound?

@SakshamSharma Sodium chloride is not a molecule. it consists of sodium ions and chloride ions. It is made from combining a metal and a non-metal (an electron transfers from sodium to chlorine to create two ions). The resulting compound is an ionic compound. But be careful, some ions can be molecules as well, just not in this case.

can you give a simple example for the latter?

@SakshamSharma Sodium acetate consists of a sodium ion (as in salt) and an acetate ion (a molecule formed from the acetic acid molecule when it is deprotonated).


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Is there a physical explanation why benzene's pi molecular orbitals in a Frost circle are degenerate but pyridine's are not?
From here, the Frost Circles for both benzene and pyridine:

The $\pi$-MOs formed by 6 $\pi$-AOs of an aromatic ring are:

Listed with respect to the ordering in the above image, these MOs in benzene's $D_{6h}$ symmetry correspond to the group representations:
$\pi_6=B_{2g}$
$\pi_4 / \pi_5=E_{2u}$
$\pi_2 / \pi_3=E_{1g}$
$\pi_1=A_{2u}$
The $D_{6h}$ symmetry of benzene is reduced to $C_{2v}$ in pyridine. I believe the pyridine MOs transform in $C_{2v}$ symmetry as:
$\pi_6=B_2$
$\pi_5=B_2$
$\pi_4=A_2$
$\pi_3=A_2$
$\pi_2=B_2$
$\pi_1=B_2$
Onto my question:
Because the doubly degenerate $E$ representation does not exist in the $C_{2v}$ symmetry of pyridine, the corresponding $\pi$ orbitals are no longer degenerate. I could just stop there and accept that symmetry dictates degeneracy and there's nothing more for me to understand, but is there a way to dig a level deeper and connect the degeneracy to a physical explanation?
For example, I can imagine nodes breaking the extended conjugation of the $\pi$-atomic orbitals, which would explain why the $\pi$-MO with zero nodes is the lowest energy bonding orbital, and the $\pi$-MO with the most nodes the highest energy antibonding orbital. In this way, I can understand why the pyridine MOs climb in energy with increasing nodes. However, if it were solely about nodes and conjugation, then the benzene $\pi_2 / \pi_3$ and $\pi_4 / \pi_5$ MOs should not be degenerate; the one with more nodes would be higher lying in energy.
Or maybe the $\pi_3$ MO in pyridine is higher in energy because it has no $\pi$-atomic orbital situated on the nitrogen atom—the node is located on the nitrogen. Nitrogen being more electronegative than carbon, there should be a lower potential with some electron density localized in an atomic orbital on nitrogen. Therefore, the $\pi_2$ MO would lie lower in energy than its $\pi_3$ counterpart because nitrogen possesses an atomic orbital in the $\pi_2$ MO.
Is there a physical explanation why these orbitals in benzene are degenerate, and vice versa why the corresponding orbitals in pyridine are not?

My first thoughts are along the lines of your second-last paragraph. Here is a quick sketch of how to formalise it. From simple Hückel theory, you can obtain the coefficients of the AOs in the MOs:
$$|\psi_i\rangle = \sum_a c_{n,i} |n\rangle$$
where $c_{n,i}$ denotes the coefficient of AO $|n\rangle$ in the MO $|\psi_i\rangle$.
In Hückel theory the main difference between benzene and pyridine is that one atomic orbital is shifted in energy, so one of the $\alpha$ terms turns into, say, $\gamma$. Without loss of generality we will label the nitrogen with the index $n = 1$. In theory, the resonance integrals $\beta$ involving the nitrogen should change as well, but to a first approximation we assume they are the same.
In this case therefore the Hamiltonian matrix, when cast in the AO basis, changes from $\mathbf{H}_0$ to $\mathbf{H}$:
$$\mathbf{H}_0 = \begin{pmatrix}
\alpha & \beta & 0 & 0 & 0 & \beta \\
\beta & \alpha & \beta & 0 & 0 & 0 \\
0 & \beta & \alpha & \beta & 0 & 0 \\
0 & 0 & \beta & \alpha & \beta & 0 \\
0 & 0 & 0 & \beta & \alpha & \beta \\
\beta & 0 & 0 & 0 & \beta & \alpha
\end{pmatrix} \qquad \longrightarrow \qquad \mathbf{H} = \begin{pmatrix}
\gamma & \beta & 0 & 0 & 0 & \beta \\
\beta & \alpha & \beta & 0 & 0 & 0 \\
0 & \beta & \alpha & \beta & 0 & 0 \\
0 & 0 & \beta & \alpha & \beta & 0 \\
0 & 0 & 0 & \beta & \alpha & \beta \\
\beta & 0 & 0 & 0 & \beta & \alpha
\end{pmatrix}$$
Most of the Hamiltonian is unchanged, so we could consider this within the framework of perturbation theory, where we have a perturbed Hamiltonian $\hat{H} = \hat{H}_0 + \hat{V}$. The relevant perturbation is therefore
$$\mathbf{V} = \begin{pmatrix}
\gamma - \alpha & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0
\end{pmatrix}$$
or to condense this back into operator form, $\hat{V} = (\gamma - \alpha)|1\rangle\langle 1|$.
The proper mathematical analysis is slightly annoying because the (unperturbed) eigenstates that we're interested in looking at are degenerate, which then necessitates that you cast the matrix $\hat{V}$ in the basis of the degenerate states of interest. Here, the two states we're interested in are $|\psi_2\rangle$ and $|\psi_3\rangle$. The application of Hückel theory to benzene gives the formulae for these unperturbed states as
$$\begin{align}
|\psi_2\rangle &= \frac{\sqrt 3}{3}|1\rangle + \frac{\sqrt 3}{6}|2\rangle - \frac{\sqrt 3}{6}|3\rangle - \frac{\sqrt 3}{3}|4\rangle - \frac{\sqrt 3}{6}|5\rangle + \frac{\sqrt 3}{6}|6\rangle \\
|\psi_3\rangle &= \frac{1}{2}|2\rangle + \frac{1}{2}|3\rangle - \frac{1}{2}|5\rangle - \frac{1}{2}|6\rangle
\end{align}$$

Note that this choice of coefficients is not unique, as the two states are degenerate: any linear combination of these two MOs is also a permissible MO with the same energy. In any case, the relevant matrix elements of $\hat{V}$ in the new basis $\{|\psi_2\rangle,|\psi_3\rangle\}$ are given by
$$V_{i-1,j-1} = \langle \psi_i | \hat{V} | \psi_j \rangle$$
and if you assume that $\langle n | n' \rangle = \delta_{nn'}$ (i.e. the AOs have no overlap with each other, which is a key assumption of simple Hückel theory anyway), then it becomes very simple:
$$\mathbf{V} = \begin{pmatrix} \frac{\gamma-\alpha}{3} & 0 \\ 0 & 0\end{pmatrix}$$
Perturbation theory tells us that the eigenvectors of $\hat{V}$ are the "states which are stable towards the perturbation". What this means is that, since the degeneracy is lifted by the perturbation, you also remove the free choice of having any linear combinations be a permissible MO: the coefficients of AOs in the new MOs must be uniquely determined (ignoring the phase factor). Since $\hat{V}$ is diagonal in the basis $\{|\psi_2\rangle,|\psi_3\rangle\}$, it conveniently means that both $|\psi_2\rangle$ and $|\psi_3\rangle$ are the eigenvectors of $\hat{V}$.
More interesting here is the first-order corrections to the eigenvalues of $\hat{H}_0$, which are simply the eigenvalues of $\hat{V}$. In this case, since $\hat{V}$ is diagonal, the eigenvalues are just $(\gamma-\alpha)/3$ and $0$: that is to say, the MO $\psi_2$ is shifted in energy by $(\gamma-\alpha)/3$ and the MO $\psi_3$ is unshifted in energy.

On the bonus lesson of applying perturbation theory: Is there a name for what you did to derive $V_{i-1,j-1}=\langle \psi_i | \hat{V} | \psi_j$ ? I'd like to read more on how you knew to do that then maybe make a new question on constructing the full perturbation matrix for all states. I noticed the off-diagonal elements in the 2x2 matrix fell away because $\psi_3$ has no $n=1$ term. If the perturbative matrix included e.g. $\psi_1$, I'd expect a non-zero off-diagonal element for $\psi_2$ and $\psi_1$. Then the perturbative correction to $\psi_2$ EV wouldnt actually be $(\gamma-\alpha)/3$, no?

@Blaise well, the usual matrix representation of an operator in the basis ${|i\rangle}$ has $A_{ij} = \langle i | \hat{A} | j \rangle$ - that is standard QM. The only issue here is that our basis set begins its numbering at 2, so I had to minus one for consistency's sake.

Regarding adding in $|\psi_1\rangle$ into the basis set, after looking up my old notes, I'm fairly sure that the derivation of the results I cited requires that all of the basis functions have the same (unperturbed) energy. Hence, you're only supposed to deal with the wavefunctions within the degenerate set. You're correct that adding it in would lead to (unphysical) off-diagonal elements $V_{12}$. It's unphysical because if you then went on to diagonalise $V$ and find its eigenstates, you'd find that there's mixing between $|\psi_1\rangle$ and $|\psi_2\rangle$, which shouldn't happen...

... because mixing between states of different energy only happens (1) when the states are of the same symmetry which I don't think is the case here; (2) when you go beyond the first-order energy correction, to find the first-order correction to the wavefunctions themselves (ugly maths!). Sorry for the somewhat excessive rambling, but I like it!

In a benzene molecule draw the mirror line from one carbon atom, called C-1, through the opposite carbon.  For each degenerate energy level you can construct one orbital that's symmetric under reflection through this mirror, and a second orthogonal orbital that's anti-symmetric under reflection.  Note that the anti-symmetric orbital cannot have any electron density on the mirror itself; then, to keep the total electron density balanced when the energy level is occupied, the symmetric orbital must have maximum electron density along the mirror.
Now replace C-1, which lies on the mirror plane, with a nitrogen atom.  The molecule is still symmetric about this mirror plane, so molecular orbitals must still be symmetric or anti-symmetric under reflection about this mirror.  But whereas the anti-symmetric orbital has no electron density on the mirror and so is unaffected by the carbon/nitrogen trade, the symmetric orbital has maximum electron density right where we have replaced the carbon with nitrogen.  So the symmetric orbital is singled out as being most "vulnerable" when benzene is changed to pyridine even as the anti-symmetric orbital is not "vulnerable" at all.  This selectivity breaks the degeneracy.


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Heat of Combustion examples
What is the criteria to identify the order of heat of combustion among different organic molecules?
(like the main points to be noted while selecting the order)
The answer being Z>Y>X.
The method tried is that for the bonds to break easily , if enough of electrons are there it would break easily so order being X>Z>Y,
Please tell where I am wrong.

You have to consider the $3$D structure and sterics when dealing with cyclic compounds.  If you look at the geometry of cyclohexene below you can that the ring is not symmetrical, thus the placement of the vinical methyl groups in regard to the double bond matters.
In (X) when the vinical methyl groups are both on doubly bonded carbons they are forced into a coplanar arrangement which is unfavorable and maximally strains the ring's double bond and the methyl group bonds.  In (Y) when they are opposite the double bond the methyl groups are optimally spaced apart and steric strain on the ring and methyl bonds is at a minimum.  When they are not located on the double bond they are neither at a maximum or minimum, just in between so (Z) just falls in the middle.  There are of course other conformers of cyclohexene but the relation will still hold.
The question could have also thrown in a fourth isomer where one methyl group was on a double bonded carbon, but it would not be easy to rank it above or below (Z) without computational methods.


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What is the most stable structure amongst the keto-enol tautomers of pentane-2,4-dione?
I was assigned the following question and I cannot decide which is the correct answer.

Which is the most stable structure?
  (a) pentane-2,4-dione
  (b) 4-hydroxypent-4-en-2-one
  (c) (Z)-4-hydroxypent-3-en-2-one
  (d) (Z)-penta-1,3-diene-2,4-diol

I am confused between options (c) and (d).
Both seem to be stable due to resonance, but I think (d) should have more resonance as the oxygen's lone pair in the $\ce{-OH}$ bond will also participate in resonance making it more stable.
But the answer is given as (c).
Could you please explain why (c) is more stable than (d).

The correct answer is (a). Andy other structure is much less stable, even though you can measure the existence of (c) to a significant degree. Your book's solution is wrong.

My previous comment is somewhat wrong. The equilibrium is very solvent dependent. In water it is (a), in less polar solvents the answer is (c). There is a small article on Wikipedia about it, and some references within, which are quite interesting. I think this question deserves reopening and a more elaborate answer.

The compound in option (c) is highly stable because it forms a 6 membered ring by tautomerism and that ring is highly stable because all bonds in that ring have developed partial double character. Whereas in option (d) there is ring formation but only two bonds of that ring have partial double character. This makes (c) more stable than (d).

The C structure will have the highest stability.
Despite the keto form of tautomer having the highest stability due to strong C=O bond it would not occur so in the special cases of intra-molecular hydrogen bonding
The Intra molecular hydrogen bonding would stabilize structures b,c and d
However b tends to be less stable than c due to lesser alkene substitution which owes to the inability to form the maximum number of hyperconjugation structures such as c itself
The Structures c and d seem to be controversial but however the more electronegative oxygen would exist in the case of c due to resonance with the intervening double bond between the two functional group containing carbons and hence would have more electron density localized to oxygen, a highly electronegative element and hence would have higher stability.
However it is also correct to think of the structure d as something with more number of hydrogen bonding but it doesn't compensate the former reasoning.


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Bond lengths in naphthalene
As I was reading Clayden: Organic Chemistry 2nd Edition, I came across an image of naphthalene showing its varying $\ce{C-C}$ bond lengths, however, I was unable to rationalize the bond lengths by drawing naphthalene's resonance forms. I have searched for a good explanation for the bond lengths shown but I have not found anything remotely convincing.

The 3 resonance forms of naphthalene are shown below however, these resonance forms are inconsistent with the bond lengths displayed by Clayden. From looking at these resonance forms I would not expect the $\ce{C-C}$ bond fusing the benzene rings to be the shortest since it has less double bond character compared to some of the other $\ce{C-C}$ bonds in the molecule.

You are assuming that all resonance structures contribute equally.  The resonance structures on the right and left are equivalent but have one aromatic ring and one non-aromatic ring.  This is in contrast to the center resonance structure, which has two aromatic rings and is more stable, thus giving the bonds between carbons 1-2; 3-4; 5-6; 7-8; and 9-10 more double bond character than the other bonds and making them shorter.

Surely all the resonance structures are aromatic, since they all have a contiguous array of p orbitals containing 10 pi electrons, satisfying Hückel's "4n + 2" rule. Is it really correct to view two of the resonance structures as one aromatic ring connected to a non-aromatic ring?

Also, this wouldn't explain why the C-C  bond between carbons 9 and 10 is the shortest in length

Due to symmetry constraints, only the one structure shown makes actually sense. The way resonance is used here is wrong. One should start from localised bonds, as they are shown, and then use resonance to justify that the single bonds are shorter than expected, and the double bonds slightly longer.

@cяonus Naphthalene does not satisfy Hückel's rules; it is not a monocycle.


text:
Will a container made of ice hold an alcoholic beverage that is below freezing in an environment that is also below freezing?
This very specific question that came after saw I a travel episode where they served alcoholic beverages in ice glasses in an ice bar. I wondered if the drink contents would stay in the glass (of ice) without the glass (of ice) melting. I reason that the drink would be cold enough to not melt the ice thus allowing it to stay frozen.

Hmm, actually it's very normal that some of "glass" will melt just like in normal drink "on rocks", it's just that the process will be slow enough, you may not even notice it.

Depends on the temperature and alcohol content. At 0°C, the container will definitively melt.
Below, your container will melt from the inside until the liquor is diluted so much that its freezing point is raised to the current temperature. (Or the other way round, water ice precipitates from your booze, concentrating it until its freezing point has dropped to the temperature of the surrounding.)
You can easily calculate the equillibrium alcohol content for a given temperature, https://en.wikipedia.org/wiki/Freezing-point_depression#Calculation
(The result will be slightly off, because this calculation is derived for infinite dilution.)

So I assume there is a physical interaction that causes the solid ice to mix with the drink, but why do the water molecules “let go” of the other ones even in sub freezing temperatures?

Whenever there are foreign particles dissolved in it, the freezing point of a liquid drops, irrespective of the nature of those particles. The freezing point depression is (at least at low concentrations) just proportional to the number of particles.

@BenJohnson More to your question: A solid that swims in its melt is not static. There are always atoms or molecules freezing to the surface and others detaching. It's an equillibrium. An the ice in your experiment has two "melting points": The bulk melting point of solid ice (0°C), and the lower temperature where your alcoholic solution can dissolve the surface of the ice.


text:
Where does hydrogen evolution occur when aluminium amalgam is placed in water - aluminium or mercury?
Let's say a drop mercury is placed on clean aluminium sheet; mercury amalgam is formed. The solubility of aluminium in mercury and vise versa at room temperature is fairly low. If the entire sheet is now placed under water, hydrogen evolution is observed. Where is this hydrogen evolution from - the aluminium rich surface on the edges or the mercury rich surface on the drop?
(For reference, please look up videos of aluminium amalgam online, say on YouTube.)

And what do you think? This sounds very much an educational question, which means you should do your homework for it!

It is an interesting question, an aluminium or zinc amalgam in contact with water is a very complex and interesting system


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Isothermal expansion and compression for n steps
In my book, the graph for isothermal expansion for 2 steps and n steps is shown as  this

I can't understand the graph for both compression and expansion. Why is external pressure changing in going from  $V_1$ to $V_2$. I thought the external remained constant in irreversible reaction and why is external pressure different for both compresdion and expansion?

They are trying to show what happens if you expand or compress a gas in one or more steps at constant externally applied pressure (irreversible process), particularly when the number of steps becomes very large.  When you expand the gas in this framework, you suddenly drop the external pressure, and then hold it constant while you let the gas equilibrate at the new lower pressure.  When you compress the gas in this framework, you suddenly raise the external pressure, and then hold it constant while you let the gas equilibrate at the new higher pressure.  So, of course, the expansion steps are going to look different from the compression steps on the graph.  The idea is that, as you increase the number of steps (between two fixed end points), the cumulative amount of irreversibility becomes less, as you approach a reversible process.

Yes, you are right that external pressure remains constant in an irreversible process. But in this case expansion and compression is done in two step. In case of two-step expansion or compression, the process is carried out by passing the system through an intermediate state.
Let us first consider two-step expansion of $1\ \pu{mol}$ of an ideal gas from ($\mathrm{P_1,\ V_1,\ T_1}$) to ($\mathrm{P_2,\ V_2,\ T_2}$). Therefore to carry out the change in a two-step, at first step we put an external pressure  $\mathrm{P_i}$ over the piston which is less than $P_1$ but greater than zero. Now on the removal of the stops fitted upon the piston, the system experiences a sudden change and attains the state ($\mathrm{P_i,\ V_i,\ T_i}$). The external pressure remains constant in this step.
Now in the second step, the external pressure is made equal to $\mathrm{P_2}$ which is lower than $\mathrm{P_2}$ but greater than zero. Now here also on the removal of the stops, the system experiences sudden change and the gas molecules push the mass kept over the piston upwards. As a result, the system experiences a sudden change and the internal pressure becomes equal to $\mathrm{P_2}$ i.e the system attains the state ($\mathrm{P_2,\ V_2,\ T_2}$)  i.e. attains the final state. The external pressure remains constant at $\mathrm{P_2}$ throughout the second step.
Therefore in a two-step expansion, the external pressure in a particular step remains constant throughout that process.
But in the overall process, it doesn't remain constant. If the external pressure at 2nd  step remains the same as that of in 1st step, then the overall process comes to a stop.
Because expansion or compression takes place due to the urge of the gas molecules to attain the pressure equilibria between internal pressure and external pressure. Now in two-step expansion at 1st step the external pressure is Pi and internal pressure is $\mathrm{P_1}$. Due to this reason on the removal of the stops the gas molecules, due to the presence of the urge to attain the pressure equilibria push up the piston upwards to a certain height until the internal pressure becomes $\mathrm{P_i}$. After attaining $\mathrm{P_i}$ pressure, the system attains the pressure equilibrium i.e both internal and external pressure is $\mathrm{P_i}$.
Therefore in this stage, due to the pressure equilibria, the position of the piston remains same even on removal of the stops. So the process also stops. Now to proceed the process furthermore to achieve our desired final state, we have to disturb again the equilibrium between the external pressure and internal pressure so that the system regain its urge to attain the pressure equilibrium. Due to this reason in two-step expansion, the external pressure is changed after completion of one step.
In a similar manner, the variation of external pressure in two-step compression can also be explained.

Welcome to Chemistry.SE!  Please note the edits made especially the use of $\mathrm{}$ for equations and variables and $\pu{}$ for units.  We also use $ce{}$ for presenting chemical formulas and equations.


text:
Non central chirality in cheminformatics?
What are compact cheminformatics structural representations that allow to deal correctly with axial chirality, such as in binaphthyl compounds or metal complexes?

I like the question, but it might help to add where else you have looked for this info or any ineffective structural representations you have found, so that answers don't wind up repeating what you have already tried. @Anon

Nothing much, I just tried RDKit documentation and mailing list, found nothing.

I found out there is an XML based language to do this, called FlexMol. It's pretty old, and it is not implemented in any of the major cheminfo releases (as far as I know). Hope someone who's more into cheminfo than me can give some tip!


text:
Oscillator strength of a transition
I am trying to compute the oscillator strength in a molecule. Therefore I've got its Energy level values plus both wave functions $\Psi_i$ and $\Psi_j$ for the transition I want to calculate the oscillator strength for.
In general, the oscillator strength of a transition from $E_i \rightarrow E_j$ is defined as
$f_{ij} =\frac{4m_e\pi\nu}{3e^2\hbar}D^2$
where D stands for the transition moment integral which shall be
$D_{ij} = \int\Psi_j\hat{\mu}\Psi_i$.
Now I am wondering how the transition dipole operator $\hat{\mu}$ is defined, since I didn't find helpful.

So, the crux of your question is not about calculating $f_{ij}$ per se, but rather about how $\hat{\mu}$ is defined?

@orthocresol Well, I'd be extremely glad to get either an answer on how to calculate $f_{ij}$ or how the transition dipole operator is defined. Both would lead to an answer.

The $\mu=-e\cdot\bar r$ where the vector $\bar r$ is the sum of all charges on the molecule.  (The formula may also assume you  average over x, y, z directions as the radiation can be polarised along any axis so divide by 3.)

@porphyrin Thanks for your comment! This would mean that the transition moment integral becomes $D_{ij} = -e\int\Psi_j x\Psi_i dx$ if I only respect the x-direction, right?

yes where $x$ is the sum of charges in just the $x$ direction, i.e $ex$ is just the dipole component along $x$.


text:
Does HCl evaporate at room temperature?
So my question is...lets say that somebody pours hcl (30%) in the toilet and on bathroom floor.Does the acid evaporate after a while it is it possible that it's still present after like month?
Thank you so much for any answer.

Yes it does evaporate. There will be no trace of it after a month, even if there were no cleaning (but then again, in that case acid is not your biggest problem).

I would be less concerned with evaporation rate than with the possibility of reaction between the acid and the material on the floor.  If you cannot avoid acid getting on the floor, wipe it up immediately and rinse thoroughly down the drain.  Protective gloves, if you have them, would be good.


text:
Comparing acidity between groups

I tried to do this using the stability of conjugate base. I can see (2) will be most stable due to resonance however my confusion lies in (1) and (3).
While in (3), the negative charge is unstable but in (1) I can't tell if it will take part in resonance because in pyridine the lone pair doesn't take part in resonance so is the same thing happening here?

Pretty much. If 1 is deprotonated, the resulting lone pair is perpendicular to the pi system of the ring, therefore only indirectly resonance stabilised (i.e. the ionic configuration would contribute more). This position should be a bit more acidic, because nitrogen is a bit more electronegative than carbon.

The answer should be A.
Since 3 is vinylic hydrogen, therefore it is least acidic.
Both 1 and 2 are allylic and are at the alpha position with respect to the carbonyl group.
Nitrogen is quite electronegative, therefore hydrogen 1 is more acidic than hydrogen 2. Hence the result.

Answer - A
Explanation -
3 is least acidic as the negative charge developed after removal of hydrogen atom will not take part in resonance like other acidic hydrogens marked.
Both 1 and 2 will be resonance stabilized after removal of acidic H+ but 1 is more stable than 2 because the negative charge developed will be on Nitrogen atom which is more electronegative as compared to carbon. More electronegative atom would like to have electrons to complete its octet.
So, the order will be 1>2>3


text:
Numbering in IUPAC name of a trisubstituted cyclohexane

This is the organic compound. CHO - aldehyde group. Br - Bromine
How do name this organic compound (IUPAC Standard)? Specifically, how do you number the carbon atoms? I thought the answer was 1-Bromo-3-Methyl-cyclohexane-2-carbaldehyde.
But that is not the right answer. I think I might have gone wrong in numbering the carbon atoms. Could someone explain why and what is the correct answer?

You have already correctly identified the principal characteristic group of the compound as carbaldehyde. You have also identified the senior parent structure as cyclohexane, which is correct since the compound has only one possible parent structure. The remaining substituents are correctly expressed as prefixes. The last problem is the correct numbering of locants.
The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The lowest locant is assigned first to the principal characteristic group according to rule (c), i.e. 1-carbaldehyde.
Next, low locants are assigned to the other substituents as a set accoding to Rule (f). However, this could be ‘2-bromo-6-methylcyclohexane-1-carbaldehyde’ as well as ‘6-bromo-2-methylcyclohexane-1-carbaldehyde’ since both names correspond to the locant set ‘2,6’.
According to Rule (g), the lower locant is assigned to the substituent that is cited first as a prefix in the name, i.e. ‘bromo’. Therefore, the correct name is 2-bromo-6-methylcyclohexane-1-carbaldehyde.


text:
Order of a complex reaction
Is the order of a complex reaction determined by the slowest step always? Is it so that order of a reaction is determined by rate determining step and since in a complex mechanism, we write elementary steps hence the order should be the molecularity of the rds?
I am slightly confused in this since I saw that in E1CB the molecularity of the rds is 1 but the order of E1CB is 2.
Basically I know order is determined experimentally but for elementary reactions, don't we write order as the sum of stoichiometric coefficients (here I am assuming that it is not a pseudo order reaction and no products are there in the rate expression.)

You're confusing two concepts. The order of a reaction is determined by experiment. The rate of the reaction is limited by the slowest elementary step, which has its own molecularity (or order). But this order is not necessarily the same as the overall reaction order. And in fact, if the slow step is not the first step, they're usually different.

Related: https://chemistry.stackexchange.com/questions/15974/can-in-any-case-the-faster-step-of-the-reaction-be-rate-determining


text:
Can chiral atoms have same substituents but with different orientations?
So consider this molecule. We have to determine whether nitrogen atom is chiral or not.

MY IDEAS:
Now going by all I have read, for chiral atoms only criteria is substituents to be different. Here two substituents are same but as we can see orientations of the substituents is different.
So aren't the two substituents technically different because after all they aren't going to react the same way to plane polarized light?
I tried to break it into possibilities:

Now in first case I think we could rotate around one of the sigma bonds attached to nitrogen and form a conformation with a plane of symmetry.
So is in this case nitrogen not a chiral atom?

Now in the second case I do think we can't prove the two substituents to be same, no matter how much we rotate. So chiral?
CONCLUSION: I am a beginner in this topic and not really confident about my thought process
Can we really use conformers of molecules to support their symmetry because after all conformers are ever changing, not stuck in one orientation?
Also what's the logic behind checking the plane and centre of symmetry of molecules? How does that relate to rotation of PPL light?
->If two substituents of a sp3 hybridized atom are identical, we say not optically active.
 But both being identical will rotate PPL light in same direction, so a net rotation will be produced.
->So isn't it optically active?

So, where did you find this tetravalent nitrogen atom? ;-)

No, it is not an asymmetric centre. To test this, you can draw the two potential stereoisomers:

Note that isomers A and B are identical (by C2 rotation), so the nitrogen cannot be an asymmetric centre. If it were one, then these isomers would by definition be distinct.
The attached groups are identical, from the point of view of the nitrogen - both have (R) stereochemistry. If one were (R) and the other (S), the nitrogen would constitute a pseudoasymmetric centre - https://goldbook.iupac.org/html/P/P04921.html
Hope this helps.
Also, don't get confused about the difference between the nitrogen being a sterogenic centre, and the molecule as a whole being chiral. The molecule is chiral, since it is non-superimposable on its mirror image. Chirality is a molecular property, and although frequently people will refer to a particular atom in a molecule as "chiral", this is a misnomer. A better description of an atom with 4 different groups attached is "sterogenic". It is entirely possible to have a molecule containing stereogenic centres but the molecule is achiral (e.g. meso compounds).
The logic behind checking the plane and centre of symmetry is that planes and centres of symmetry are conserved on reflection in a mirror, so checking for these symmetry elements is a shortcut to checking whether a molecule satisfies the definition of chirality.

Please could you help with what's wrong with my ideas

Part of your issue is in distinguishing the difference between configuration and conformation. The configuration of a molecule is the description of which atoms are bonded to which, and in what 3D arrangement around each atom. You can only change configuration by breaking or making bonds. Conformations are just different arrangements of atoms achieved by rotating around the single bonds in the molecule. Stereochemistry concerns configuration only, not conformation.

In the current case, look at the two alcohol stereocentres and define them as (R) or (S) - you should find that both are (R). So, from the point of view of the nitrogen atom, it is attached to X, Y and two identical, (R)-configured groups. Since two of the groups attached to the tetrahedral N are identical, it is not a stereogenic centre.

But remember this is only to tell whether the nitrogen atom is stereogenic, it says nothing about whether the molecule as a whole is chiral. The gold standard test for that is always to draw its mirror image and check whether it is possible to superimpose the two.


text:
Does CF3COOH undergo decarboxylation ? Why or why not?

Does $\ce{CF3COOH}$ undergo decarboxylation ? Why or why not?

Attempt:
I feel it should decarboxylate because $\ce{CF3-}$ carbanion must be stable due to fluorine's -I (electron withdrawing) effect  , but I am not sure.

That's reported in the literature, but the inductive effect still doesn't mean it's very favorable from a kinetic standpoint.

Welcome to Chemistry.SE!  Homework questions are okay but must demonstrate an attempt to solve/understand the question. Please note that formulas can be better expressed with $\ce{}$ for chemical formulas/equations, $\mathrm{}$ for math term/equations, and $\pu$ for units.  Also, take a minute to look over the help center to better understand our guidelines and question policies.

@Abcd There really is no point adding the word attempt to the question body. If it is an attempt at solving the problem, that should be fairly obvious...


text:
Polycarbonate resistance to acids
If polycarbonate plastic is a polar material with ester bond linkages which usually hydrolyze in acids, why is it resistant to both weak and strong acids (as polar substances)?

Looking at what has been published about hydrolysis of polycarbonate resins to recover bisphenol A, typical conditions involve water, heat, optional base catalyst (or more rarely, acid, which probably further degrades BPA), and very often a cosolvent such as tetrahydrofuran that itself is stable to hydrolysis. So your intuitions about whether polycarbonate should be susceptible to acid hydrolysis are correct.
There has been a lot of attention paid to whether polycarbonate in household items (e g. baby bottles) releases bisphenol A into foods, causing health concerns- which it certainly can.
On the practical side, such as preparing dilute aqueous acids or bases in a polycarbonate container at ambient temperature (e.g.for lab use), degradation would be very slow due to the density, hydrophobicity,  and perhaps cross-linking of polymer chains into a tight matrix. This prevents the aqueous acid or base from physically contacting the bulk polymer, although slow surface degradation is still likely.


text:
How to interpret Gaussian output for excited states?
I have calculated the excited energies for benzene in Gaussian. The output therefore was:
 Excited states from <AA,BB:AA,BB> singles matrix:
 Excited State   1:      Singlet-A      5.5804 eV  222.18 nm  f=0.0000  S**2=0.000
 20 -> 22         0.41871
 20 -> 23         0.27308
 21 -> 22         0.27309
 21 -> 23        -0.41870
 Excited State   2:      Singlet-A      6.4315 eV  192.78 nm  f=0.0000  S**2=0.000
 20 -> 22        -0.27213
 20 -> 23         0.41722
 21 -> 22         0.41725
 21 -> 23         0.27213
 Excited State   3:      Singlet-A      7.9411 eV  156.13 nm  f=0.0000  S**2=0.000
 18 -> 23         0.49564
 19 -> 22         0.49673

Now I am wondering from where to where are these transitions. I know that the first one is from HOMO to LUMO. Thinking of QM I'd assume that the second transition is from HOMO-2 to HOMO+1 and the third one would be from HOMO-1 to HOMO+2. AM I thinking correct?

Saying that electronic excitations are the result of transitions from one orbital to another is kind of the 0th order approximation and quite primitive. Sadly, things are not that simple. Could you perhaps explain how  you got to your statements? What do you wish to conclude from them?

@Raditz_35 Firstly, I optimized the molecules ground state structure using DFT/B3-LYP/6-31G with only respecting singlets. After that I used TD-SCF for DFT with B3-LYP and 6-31G and singlets only again to obtain my excitation energies.
I want to calculate excitation energies for poly(p-phenylene) with n from 1 to 8 where n is the number of monomers. Then I want to compare the excitation energies obtained from Gaussian with the ones I calculated using particle in a box model. That's why I thought of transitions from one orbital to another.

The output is fairly obvious: The first excited state has excitations from MO 20 to 22, 20 to 23, 21 to 22, and a de-excitation from 23 to 21. So the first one is certainly not from HOMO to LUMO.// Please [edit] your question to include more details, especially the input files you have used to generate this. Also specify the version of Gaussian you are using.


text:
Why is bis(1,2,2,6,6-Pentamethyl-4-Piperidinyl) Sebacate used as a "bottom coat" for a superhydrophobic coating?
According to this MSDS sheet, a compound named bis(1,2,2,6,6-Pentamethyl-4-Piperidinyl) Sebacate is used in the "bottom coat" layer before a "top coat" of superhydrophobic NeverWet coating is applied.

The top coating composition can be inferred from this MSDS sheet, and is a dimethyl siloxane reaction product with silica (likely a fumed silica nanoparticle that's been reacted with dimethyl silanol).
Functionally, what bis(1,2,2,6,6-Pentamethyl-4-Piperidinyl) Sebacate must be doing is creating a reactive surface on the desired substrate for subsequent application of the siloxane nanoparticles.  It's not clear if the nanoparticles will be reacting with this bottom coat, or forming some sort of complex (and if so what the nature of this complex is).
According to Wikipedia, sebacic acid is a dicarboxylic acid with an 8 carbon linker, so my guess is that is the backbone of the base-coat molecule.  Apparently sebacic acid is a monomer widely used in the production of plastics, so there is likely a polymer or "pre-polymer" (made-up term) being deposited.
Any idea what the molecule might look like and what the mechanism and functionality might be for forming this base coat for the hydrophobic coating?
EDIT: looks like I found the molecule (CAS # 41556-26-7), and it looks like some sort of bifunctional crosslinker.  My question still stands as to how it functions and how it might be preparing a surface for addition of a superhydrophobic coating.

Note that the name bis(1,2,2,6,6-pentamethyl-4-piperidyl) sebacate clearly says how the molecule looks like. It should not be necessary to search it in a database in order to find that out.

Now it's obvious to me but I'm a bit rusty so lacked the confidence to be sure.

If I have to guess, the carbonyl is labile to attack by something , and that 1,2,2,6,6-Pentamethyl-4-hydroxy-Piperidine is a leaving group (labile protecting group).

Anyone have a greene book and/or a scifinder subscription?  Curious where this molecule has been used before and what properties the piperidine has as a carbonyl protecting group.

I am curious too. I found something about it here: https://en.wikipedia.org/wiki/Hindered_amine_light_stabilizers


text:
Molar entropy of vaporization

If a substance has a molar heat of vaporization of $\pu{3.05\times10^4J/mol}$ and a normal boiling temperature of $\pu{80.0^\circ C}$, what is the value of its molar entropy of vaporization $\Delta S_\mathrm{vap}$ in $\pu{J K^{-1} mol^{-1}}$?

I got
$$\frac{3.5\times10^4\ \mathrm{J/mol}}{(80+273)~\mathrm K}=86.4~\mathrm{J~K^{-1}~mol^{-1}}$$
however my texbook answer key says that the answer is $380\ \mathrm{J~K^{-1}~mol^{-1}}$. What am I doing wrong?

It seems like your textbook might have just done 3.05e4/80 = 381, which would be wrong.

@a-cyclohexane-molecule thank you so much for the clarification. I really appreciate it.


text:
How is laughing gas produced in the Nef reaction?
I can't think of a mechanism for this reaction to give laughing gas.
Attempt:
I used the base to take out the acidic hydrogen at alpha position with respect to nitro group, but that makes the intermediate formed even more unstable. What should be the proper mechanism?

https://www.organic-chemistry.org/namedreactions/nef-reaction.shtm https://en.wikipedia.org/wiki/Nitroxyl

The Nef reaction does not produce nitrous oxide directly. It produces nitroxyl as a reactive intermediate, which rapidly decomposes via a bimolecular reaction to nitrous oxide and water$^{[1]}$. You are correct that the base removes the alpha hydrogen; the nitro group is more than capable of accepting this negative charge. The crux of the Nef reaction is this:

The mechanism for the formation of nitrous oxide is as follows:

$[1]$ Wikipedia: Nitroxyl, Reactions


text:
Mechanism of photochemical hydroamination with LDA
I've recently come across a couple examples of intramolecular alkene hydroamination accomplished by treatment with LDA followed by visible light irradiation (tungsten light bulb):[1,2]

I'd assume that this proceeds via some kind of radical mechanism, with the initially formed amide ion losing one electron to form an aminyl radical which then adds into the alkene. Trost[2] also suggests that the addition is facilitated by an SET:

My question is – what species is taking the electron away, and how is that species generated? There doesn't seem to be any decent electron acceptor in this system, unlike, for example, Ru(II) in a photoredox mechanism. If one is aware of any mechanistic studies which have been carried out, it would be an added bonus.

References

Kou, K. G. M.; Pflueger, J. J.; Kiho, T.; Morrill, L. C.; Fisher, E. L.; Clagg, K.; Lebold, T. P.; Kisunzu, J. K.; Sarpong, R. A Benzyne Insertion Approach to Hetisine-Type Diterpenoid Alkaloids: Synthesis of Cossonidine (Davisine). J. Am. Chem. Soc. 2018, 140 (26), 8105–8109. DOI: 10.1021/jacs.8b05043.
Trost, B. M.; Tang, W. Enantioselective Synthesis of (−)-Codeine and (−)-Morphine. J. Am. Chem. Soc. 2002, 124 (49), 14542–14543. DOI: 10.1021/ja0283394.

Why not email the starred author of the first paper you reference and ask them?

I can't read the whole thing, but there is this review on hydroamination: https://pubs.acs.org/doi/10.1021/cr0306788

It just occurred to me. I wonder if the aryl ring in the substrate is actually serving as a chromophore...

Also, according to this reference, reference (2) has some reproducibility issues: http://iupac.org/publications/pac/pdf/2010/pdf/8209x1785.pdf

One more question, how many equivalents of LDA are used?

@Zhe excess iPr2NH and 6 eq nBuLi, I think

Probably last comment. I was looking around at random links, and this thesis suggests another mechanism where you form the radical cation.


text:
Rules of naming ionic compounds
While brushing up on my Chemistry basics, I came across a topic about naming ionic compounds. It is a question regarding Roman numerals.
It says "Roman numerals should not be included in the name when the metal can only form one cation."
The cation of $\ce{Zn^{2+}}$ has +2 charge, and $\ce{NO3-}$ anion has -1 charge.
which results in having $\ce{Zn(NO3)2}$
If Roman numerals are used for transitional metals and given that zinc is part of the transitional metals, why is this ionic compound named as zinc nitrate instead of zinc(II) nitrate?
Thanks.

Zinc, mercury, and cadmium are considered post transitions metals in some circles.  It's a blurred line for the group.

Zinc and cadmium are generally considered to be "almost invariably" in the oxidation state +2 (cfr. the introduction to this paper about one of the few exceptions), therefore the indication of the charge is unnecessary.
Zinc is usually listed as a "Type-I cation" (see Monoatomic ion), i.e. a cation that only appears in one oxidation state, or its case is emphasised as an exception among the transition metals:

Several exceptions apply to the Roman numeral assignment: Aluminum, Zinc, and Silver. Although they belong to the transition metal category, these metals do not have Roman numerals written after their names because these metals only exist in one ion.
-- Chemistry LibreTexts

(For the sake of completeness, the IUPAC Red Book section IR-5.4.2.2 "Use of charge and oxidation numbers" neither mentions "Type I/II cations" nor treats the zinc case separately)


text:
Copper Stoichometry Ratio Problem
I was given two equations as part of a problem into Copper Wiring production for planes:
$$\ce{4CuFeS2 + 9 1/2 O2 + 4SiO2 \rightarrow Cu2S + Cu2O + 2SO2 + 4FeSiO3}$$
$$\ce{Cu2S + 2Cu2O \rightarrow 6Cu + SO2}$$
This question later became the source of a Chemistry exam question for the English IB equivalent.
I was given the following information.

The previous equations are part of the manufacture of copper for wiring in planes.
  A passenger jet contains $4050\ \pu{ kg}$ of copper wiring.
  A rock sample contains $1.25$% $\ce{CuFeS2}$ by mass.
  Calculate the mass, in tonnes, of rock needed to produce enough copper wire for a passenger jet. ($\pu{1~tonne} = \pu{1000~kg}$)

Now, I tried to solve it as follows:
Adding both equations gave:
$$\ce{4CuFeS2 + 9 1/2 O2 + 4SiO2 + Cu2O \rightarrow 2SO2 + 4FeSiO3 + 6Cu + SO2}$$
Now, I can see that $4$ moles of $\ce{CuFeS2}$ from the ore would give us the required $\ce{6Cu}$ for the production.
However, the solution states that Cu and $\ce{CuFeS2}$ are in a $1:1$ ratio, which I couldn't explain why.

Using the equation I formed above, this idea didn't make sense.
Why is the ratio of Cu to $\ce{4CuFeS2}$ $1:1$?

No serious mining operation would just dump a third of their desired product.
 The question does say that the reactions are "part of the manufacture of copper." This does not preclude other reactions. With the given reactions alone we are not able to handle the surplus of $\ce{Cu2S}$.
$$\ce{\color{\red}{4CuFeS2} + 9 1/2 O2 + 4 SiO2 \rightarrow \color{\red}{Cu2S + Cu2O} + 2SO2 + 4FeSiO3}  \tag{1}$$
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O \rightarrow 6Cu + SO2 +\color{\red}{Cu2S}}\tag{2}$$
However we can infer from the first reaction that the sulfide can react with oxygen to form the needed oxide to balance reduction and oxidation.
$$\ce{ Cu2S + 1 1/2O2 ->  Cu2O + SO2}\tag{3}$$
The second equation can now become:
$$\ce{\color{\red}{2} Cu2S + 2 Cu2O +\color{\red}{1 1/2 O2} \rightarrow 8 Cu + 2 SO2 \tag{4}}$$
Thus in the first step $\ce{Cu2O \text{and} Cu2S}$ are produced according to quation 1.  In the second step they are reduced with aditional oxygen according to equation 4. To pove they balance first double equation 1.
$$\ce{8CuFeS2 + 19 O2 + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O} + 4 SO2 + 8 FeSiO3}  \tag{5.1}$$
Add $\ce{1 1/2 O2}$ to each side
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{2Cu2S + 2Cu2O + 1 1/2 O2 }+ 4 SO2  + 8 FeSiO3}  \tag{5.2}$$
Substitute the red products with equation 4 reactants.
$$\ce{8CuFeS2 + \color{\red}{20 1/2 O2} + 8 SiO2 \rightarrow \color{\red}{8 Cu + 2 SO2}+ 4 SO2  + 8 FeSiO3}  \tag{5.3}$$
This Gives a final balanced equation of
$$\ce{8CuFeS2 + 20 1/2 O2 + 8 SiO2 \rightarrow 8 Cu + 6 SO2  + 8 FeSiO3}  \tag{5.4}$$
As you can see on a molar eatio $\ce{CuFeS2\text{:}Cu}$ is 1:1 after accounting for the additional oxygen needed.

Suppose both reactions have gone to completion. Then, according to the first equation, $1$ moles of $\ce{CuFeS2}$ will produce $\frac{1}{4}$ moles of $\ce{Cu2S}$ and $\frac{1}{4}$ moles of $\ce{Cu2O}$. According to the second equation, $2$ moles of $\ce{Cu2O}$ and $1$ moles of $\ce{Cu2S}$ react together to produce $6$ moles of $\ce{Cu}$. Thus, $\frac{1}{4}$ moles of $\ce{Cu2O}$ and $\frac{1}{8}$ moles of $\ce{Cu2S}$ react together to produce $(6 \times \frac{1}{8}) =\frac{3}{4}$ moles of $\ce{Cu}$. Therefore, you need $\frac{4}{3}$ moles of $\ce{CuFeS2}$ to produce $1$ moles of $\ce{Cu}$. Thus, you need $\frac{4}{3} \times 63780$ moles of $\ce{CuFeS2}$ to produce $63780$ moles of $\ce{Cu}$ for the plane. Thus, if the rock consists solely of $\ce{Cu2S}$, mass of rock needed to produce this much $\ce{Cu}$ is:
$$\frac{4}{3} \times \pu{63780mol} \times \pu{183.5 \frac{g}{mol}} \times \frac{100}{1.25} \times \frac{\pu{1kg}}{\pu{1000 g}} \times \frac{\pu{1 tonne}}{\pu{1000 kg}} = \pu{1248 tonnes}$$

The problem I see here is the question has assumed that all of the $\ce{CuFeS2}$ will be converted into elemental copper at the end of the reaction. However whether or not this is possible also depends on the amount of $\ce{Cu2O}$ present.
If the rock consists solely of $\ce{CuFeS2}$, then the $x$ moles of $\ce{CuFeS2}$ will react to produce $x/4$ moles of $\ce{Cu2S}$ and $\ce{Cu2O}$. And then, $x/4$ moles of $\ce{Cu2O}$ will react with $x/8$ moles of $\ce{Cu2S}$ to produce $x/8$ moles of copper. Some of the copper compound $\ce{Cu2S}$ will be left unreacted because there is not enough $\ce{Cu2O}$ to react with it. As we can see, not all the copper compound present initially was converted to elemental copper. The answer has not considered the possibility that conversion of the copper compound may be incomplete.


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Why add NaBH4 slowly to 4-t-butylcyclohexanone in ether solution? [Lab Procedure]
$\ce{NaBH4}$ is to be added slowly to 4-t-butylcyclcohexanone in ether solution according to laboratory procedures, but I am curious as to why that is? 4-t-butylcyclohexanone undergoes reduction with $\ce{NaBH4}$, so is the reason behind adding NaBH4 slowly to ensure maximum product formation, or is there some safety concern behind this?

Are you sure this is $\ce{NaBH4}$ or it is really $\ce{LiAlH4}$?

I am sure it is NaBH4

"adding NaBH4 slowly to ensure maximum product formation" why would slow addition give higher turnover?

Sodium borohydride is insoluble in diethyl ether

It's generally good practice to add reagents slowly to prevent thermal runaway in exothermic reactions (where the heat given out by the reaction acts to accelerate it further, creating a feedback loop). Your reaction is in Et2O (B.P. 35C) so even a small exotherm would boil your solvent, hence the slow addition.

@PCK it is also good practice to render good answers as answers, like your comment above should be one.  Thanks!


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Rotational barrier of biphenyl in ground and excited electronic states
I'm studying how the energy of biphenyl varies with its dihedral angle. I know how to describe the dependence quantitatively, but still, I don't know how to explain why the rotational barrier of biphenyl in its excited state is distinctly higher than in its ground state. This is illustrated in the plot below, where you can see that the excited states have a much greater increase in $V(\tau)$ at $\tau = 90^\circ$.

Additionally, I've obtained the HOMO and LUMO of biphenyl at $\tau = 90^\circ$. Is it possible to explain this jump in energy using them?


text:
Reason for formation of bubbles at boiling point
As I understand it, boiling takes place when the saturated vapour pressure equals to the atmospheric pressure.
But, why does the vapour pressure need to be equal to the atmospheric pressure for water bubbles to form?
According to this video, the atmospheric pressure is cancelled by the vapour pressure at boiling, which allows for the bubbles to form (due to the zero total pressure).
But, I think the atmospheric pressure is the same. So, if we consider the atmospheric pressure to be 1atm at a time, the surface of the water feels 1atm pressure every time (I mean that atmospheric pressure is 1atm as same as the before, but the vapour pressure and the air pressure is acting as a partial pressure according to Dalton's partial pressure concept, it doesn't affect the total atmospheric pressure.)
So what is the real reason?

Related: https://chemistry.stackexchange.com/questions/38653/vapor-pressure-and-boiling-point-for-solutions-in-open-containers/38654#38654

@ Ivan Neretin : I read your answer at the linked page and I could understand the logic of forming the bubbles at the boiling point. But I am still arise the doubt that will the water surface feel 1atm or zero atm while the both pressures are same? Thanks

The bubbles form under the liquid surface where the liquid pressure is also 1 atm.  For a bubble to form under the liquid surface, the pressure inside the bubble must be high enough to push the liquid away.  The force balance on the bubble interface with the liquid below the surface tells us that the pressure inside the bubble, namley the vapor pressure, must match the atmospheric pressure in the liquid.


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Calculate mass concentration of Ca(OH)2 solution
Can anyone help me calculate the mass concentration of $\ce{Ca(OH)2}$ solution with molar mass $\pu{74.1 g/mol}$ at the $\pu{pH of 12.8}$.
My work so far:
From the pH I get the molar concentration of $\pu{0.063 mol/L}$.   Assuming we have $\pu{1L}$ of substance we have $\ce{n(OH-) = 2 \times n(CaOH2) = 0.126 mol/L}$
I put the results in $m = M \times n$ to get 9.35 g which is incorrect. Where did I go wrong?

you have $0.0315$ mol of $\ce{Ca(OH)2}$ when  it dissociates completely

$m = 0.0315 \times 74.1$ to get 2.337 g

As a previous comment said, calcium concentration is half of hydroxide concentration. It yields $0.0315$ then your procedure is correct!

Can someone please explain to me how the dissociation proceeds, so I understand how we arrive at the correct molar concentration

This is an ionic compound of$ Ca^{+2}$ ions and $OH^−$ ions. When an ionic compound dissolves, it separates into its constituent ions:
$$Ca(OH)2->Ca^{2+}{(aq)}+2OH^−_{(aq)}$$
Because ${Ca(OH)_2}$ is a strong base, this reaction proceeds 100% to products.

When we have the solution of $Ca(OH)_2$ with $pH$ equal$12.8$, meaning that $Ca^{+2}$ and $OH^-$ exist in solution. Perhaps $Ca(OH)^+$ doesn't exist in solution and solid state.

There is a general law about the solubility of salts containing doubly or triply charged ions, which is not often presented in textbooks. The law is the following : The measured solubility of such a salt or such a hydroxide is ALWAYS greater than the calculated values derived from solubility products, or from pH measurements. This is due to the fact that the solubility product and the pH data do not take into account the existence of intermediate ions.
In the case of $\ce{Ca(OH)2}$, there are of course ions like $\ce{Ca^2+}$ and $\ce{OH-}$ in its solution. But there are also ions like $\ce{Ca(OH)+}$, and non dissociated molecules $\ce{Ca(OH)2}$. Alas these ions and molecules are rarely mentioned in usual textbooks. But this explains why the solubility of $\ce{Ca(OH)2}$ is bigger than the values obtained from pH measurements. I know that the average reader is reluctant to believe me. I may be more persuasive if I use another example.
Let's speak about the solubility of $\ce{H3PO4}$. Any student in chemistry knows that the solubility is not related to the solubility product $\ce{[H+]^3[PO4^3-]}$, because of the existence of intermediated ions, like $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. These ions are accepted in the large chemistry family. Everybody accepts that the solubility of $\ce{H3PO4}$ is much bigger that the value derived from the solubility products, or from pH measurements.
That is the same for $\ce{Ca(OH)2}$ and all substances containing a doubly charged ion like $\ce{CaSO4}$, etc. And here the activity is not sufficient to explain this discrepancy. I know that some chemists think that the activity may be introduced to explain this discrepancy. But in the case of $\ce{CaSO4}$ for example, the concentration is so low that the activity factor may be considered as equal to 1.
For $\ce{CaSO4}$, the effect is more pronounced, because of the presence of two doubly charged ions. The solubility calculated from the solubility product is 0.0051 M. The measured solubility, obtained from direct titration is 0.015 M. Three times more !

How can we solve the above problem when we take into account the existence of intermediate ions? I appreciate your solution and calculation.

Hint: (1) Properly, the pH tells you that you have $0.063$ molar hydroxide ion.  (2)  Calcium hydroxide, $\ce{Ca(OH)2}$, could dissociate more than once.
I'm not giving it all away, so good luck.

The OP already figured out that there were two hydroxides. That isn't the OP's problem...

Not so, from the numerical calculations reported.  The molar concentration was not given correctly.


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Does the rate constant depend on pressure?
I have read in some books that when pressure increases, the number of effective collisions increases. Hence, $k$ depends on pressure. Is it true? If so, then why doesn't $k$ depends on concentration? The number of effective collisions should increase here also.

In general, equilibrium "constants" depend on temperature, pressure, and other factors. In many circumstances the "constants" are constant enough to provide reasonable values. For examples solutions are a lot less pressure dependent than gases. // If rate = k[A] then as [A] changes so does the rate. Thus there are more or less collisions.

The rate law already has an explicit dependence on concentration; $k$ is supposed to be the part of the rate law that doesn't depend on concentration.

Depend of pressure? No just as a-cyslohexane-molecule writes. Sometimes there is a confusion between rate and rate constant. Consider a first-order reaction such as isomerisation where only one species is involved.

According to hard-sphere collision theory of gas-phase reactions, the rate constant ($k$) for the elementary bimolecular reaction $\ce{B + C -> products}$ is
$$ k = N_\text{A} \pi (r_\text{B} + r_\text{C})^2 \left[ \frac{8RT}{\pi}\left( \frac{1}{M_\text{B}} + \frac{1}{M_\text{C}}\right)\right]^{1/2}e^{-E_\text{thr}/RT}$$
where $N_\text{A}$ is the Avogadro constant, $r_\text{B}$ and $r_\text{C}$ are the radii of the hard-spheres, $R$ is the ideal gas constant, $M_\text{B}$ and $M_\text{C}$ are the molar masses of B and C, and $E_\text{thr}$ is the threshold energy on a per-mole basis. Therefore, no pressure dependence.
For nonideal systems, the rate law for the same elementary reaction can be written as
$$ r = -\frac{d[\text{B}]}{dt} = -\frac{d[\text{C}]}{dt} = k_r [\text{B}][\text{C}]$$
The Brønsted–Bjerrum equation relates the apparent rate constant, $k_r$ with the infinite-dilution rate constant, $k_r^\infty$, and the activity coefficients of the envolved species:
$$  k_r = \left( \frac{\gamma_\text{B}\gamma_\text{C}}{\gamma^\ddagger}\right)k_r^\infty$$
where $\gamma_\text{B}$, $\gamma_\text{C}$, and $\gamma^\ddagger$ are the concentration-scale activity coefficients of B, C, and the transition state respectively. These quantities depend on temperature, pressure, and on the concentration of all species on the same phase. For this reason, the rate constant of nonideal elementary reactions can have a dependence on the pressure.

According to Arrhenius equation
$$k = A \exp\left\{-\frac{E_\mathrm{a}}{R T}\right\},$$
the rate constant $k$ depends on:

Activation energy $E_\mathrm{a}$: $k$ increases with the decrease of $E_\mathrm{a}$
Temperature $T$: $k$ increases with the increase of $T$
Pre-exponential factor $A$: This factor represents the total successfully oriented collisions, so a higher pressure will lead to higher concentration of particles, will lead to a higher the value of $A$, but the value of $A$ changes very little with pressure, we consider it as a constant value, so the effect of pressure on $k$ is very little.

Is it experimentally observed that A changes very little with pressure? Because I have also seen that in collision frequency expression (theoretical), it is directly proportional to pressure.

It is just analysis, The total successfully oriented collision that leads to a successful reaction when it has enough energy, and this number very large,  so $(A)$ increase with increasing number of collisions, but this effect very little on$ (K)$ in comparison of the pressure  effect on increasing successful collisions fraction, so we can neglect the effect of pressure on increasing number of collisions in this discussion .

@MartinThank you for your fruitful active effort, I like the formula  of the Arrhenius equation as it  written in my old textbook

It's so easy. You are to look deep at how actually a book represents the rate of a reaction in term of law of mass action.
For a general reaction
A  +   B ........>   P
 Rate >  [A] [B]

Rate = k   [A] [B]

K =  Rate /  [A] [B] .......1
Now look at equation 1. Here k is the ratio of rate of reaction to the product of concentration of reactants. It means when we increase the concentration of reactants the rate of the will increase to the same extent so that to keep the value of k a constant. Same is the case of homogeneous ideal gaseous system where we replace the concentration term by pressure.


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Can I expect 18O-hydroxide when adding slight amount of NaOH-16O to 18O-water?
In adding $\ce{^{16}O}$-sodium hydroxide to $\ce{^{18}O}$-water to get a $0.25\pu M$ solution, is the hydroxide in form of $\ce{^{16}O}$-hydroxide or $\ce{^{18}O}$-hydroxide?
is there an oxygen exchange between  $\ce{^{16}O}$-hydroxide and  $\ce{^{18}O}$-water?

Yes they will exchange, and pretty fast, too.

Yes, it will but it will be a hydrogen exchange, not an oxygen exchange.
$$\ce{2 H2^{18}O <=> \color{blue}{H3^{18}O+} + ^{18}OH-}$$
$$\ce{\color{blue}{H3^{18}O+} + ^{16}OH- <=> H2^{18}O + H2^{16}O }$$
The oxygen will mix with water and hydroxide to a near equivalent ratio of $\ce{^{18}O:^{16}O}$ in both $\ce{H2O}$ and $\ce{OH-}$ but not perfectly equivalent as isotopes do not have equivalent chemical potentials.

if only a bit of 16OH- was added into the 18O-water, can I assume that your chemical equations go to the right side infinitely? finally vast majority of the hydroxides are in form of 18OH-?? thank you.. could you please provide a link of a reference?

@LimingWang The reactions are reversible. Reference: autodissociation of water


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Will PAEK polymers hydrolyze in concentrated sulfuric acid?
One of my duties at the company where I work is to perform intrinsic viscosity (IV) experiments on PAEK (poly(aryl ether ketone)) polymers. These polymers don't dissolve well in most solvents, so the solvent for these experiments is concentrated sulfuric acid. I've found that if I leave the mixture to dissolve overnight and/or at too high of a temperature, the resulting IV is much lower (approximately half) than if I run the experiment as soon as all the material has dissolved at lower temperatures. The PAEK compounds I'm referencing are ones like PEK, PEEK, and PEKK (the results I've detailed above were seen primarily, maybe exclusively, with PEKK). The structures for these polymers is below.

PEK:

PEEK:

PEKK:

My thought on this was that with extra time and heat, the water in the sulfuric acid could attack the carbon alpha to the ether oxygen, hydrolyzing the polymer and creating shorter chain lengths. One of my colleagues disagrees, but to be honest, I haven't asked her to explain why.
Aromatic sulfonation can also occur when PAEK polymers are dissolved in concentrated sulfuric acid, as documented here, but I wouldn't expect that to have such a large impact on the viscosity (and I would also expect the effect of sulfonic acids to increase viscosity, not decrease it, though I'm less confident about that). My background is organic chemistry, not polymer chemistry/materials science, so I'm not super familiar with how different aspects affect things like viscosity. I do know shorter chains leads to a lower viscosity, though.
So here's my question: Is it likely that a significant amount of hydrolysis is occuring, or is it more likely some other factor causing this effect?

Well, to hydrolyse something, you need, well, water... and there's hardly any in conc. H2SO4

@Mithoron I realize it's low. We're using 95-98% sulfuric acid. Of course, most (approximately all) of that is bound up as hydronium, which is I realize is probably much less nucleophilic, but with enough time and energy, I thought the reaction might happen.

Not really, it dehydrates not adds water. You think about cleavage of chains, which might happen in other ways.

@Mithoron So, it could be forming cleaving to make organosulfates rather than  alcohols? That makes sense to me.

I resist concluding this but nothing else I can think of makes sense. The only way the polymer solution to become less viscous is through a cleavage of bonds reducing the MW of the polymer or dilution. Cleavage at the ether linkages would require reduction in an oxidizing environment or nucleophilic aromatic substitution which this polymer lacks the electron-withdrawing groups to effect, so those mechanisms should be eliminated (I concur with your colleague). Thus we can deduce that the cleavage must be occurring at the carbonyl groups.
To cleave the polymer at the carbonyl groups, the PAEK is likely undergoing an electrophilic aromatic substitution at the carbonyl groups. Though not commonly utilized, it is possible to substitute any electrophile on an aromatic ring in including acyl groups (especially with sulfuric acid which provides a high concentration of $\ce{H+}$ electrophiles) as shown in Figure 1 — essentially Friedel–Crafts acylation in reverse.

Figure 1. Electrophilic Aromatic Substitution Removing an Acyl Group.

How is cleavage at the ether bond a reduction? If H2SO4 protonates the ether oxygen and its conjugate base attacks the alpha-carbon, this SNAr reaction shouldn't result in an oxidation or reduction. Am I missing something in there? I like your reaction, though. Certainly a possibility.

@SendersReagent, nucleophilic aromatic substitution is possible, but requires a strong electron-withdrawing group(s), Sandmeier Reaction,  or breaking aromaticity see: WikiPedia Artice.

The ketone para to the ether is a decent EWG, but I admit I wasn't/am not sure it is strong enough to cause the SNAr reaction I've described. I have seen it done primarily with nitro groups in the past.

Actually, now that I think about it, PEEK is synthesized through an SNAr reaction (see this image). Of course, this is done under basic conditions, not acidic conditions. Protonation of the ketone will definitely affect its electron-withdrawing capabilities, but protonation of the ether oxygen will also make it a better leaving group for an SNAr reaction. Though the reactions are in very different conditions, I'm not convinced that an SNAr reaction is impossible for chain cleavage here.

That's a very good point. The only problem is that sulfate and bisulfate are poor nucleophiles.

That we definitely agree about. I'm trying to think of a way any SO3 that can form at elevated temperature could cause a reaction that could lead to the chain cleavage. As I said, sulfonylation can and does happen, but that doesn't explain the effect I'm seeing. The more I look at this,  the more I'm favoring your given reaction as the primary cause of reduced viscosity measurements. I totally get your starting comment about "I resist concluding this," but the effect is very consistent over several experiments.


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IUPAC Name for the following?

Need help with the IUPAC name for the following line structure.

Welcome to Chemistry.SE! Questions are the central focus of this site, but we ask that questions demonstrate an attempt to solve/understand the underlying concepts to maintain quality.  Please take a minute to look over the help center and tour page to better understand our guidelines and question policies.

In the image, each of the numbers with black ink represents the carbon atom of the main chain. How I got this chain? Simple, You just count until you have found the best possible scenario with the longest chain. In this case, it's the carbon chain with 6 atoms. Now in this chain, you can see that there are 3 branches. Each connecting to a carbon atom. Since it's a single bond, It's connected to a Methyl (CH3). If it had been a double bond then it would have been Ethyl (C2H5). Now, look at the position of the braches. Two branches are at 3, and one branch is at 4. so we get 3,3,4. The two 3's represent the two branches and the one 4 represents the single branch at position 4. Since, we have 3 Methyl we will use TriMethyl. Where Tri represents the number of methyl branching out of the main chain. The numbers give out the location and Tri tells us about how many there are. If There had been two methyls then we would have used DiMethyl. So far we have, 3,3,4-TriMethyl. We still haven't added the main chain. Notice that the main chain has 6 carbon atoms. 6 is represented with Hex. And notice how there are only single bonds so we will use ane. Hence, We have HexAne.
The answer is 3,3,4-TrimethylHexane
I tried my best to explain. Let me know If there is a way to improve this answer. I kind of suck at explaining. Feel free to ask me again if you still didn't understand this.


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Hydrogen venting: Why no flame arrestors?
I was reading the guidelines for safe venting of Hydrogen. One item says "no flame arrestors".
Just wondering the scientific reason why flame arrestors may be frowned upon when for venting of most other flammable vapors / gases flame arrestors seem standard.
https://www.eiga.eu/publications/eiga-documents/doc-21117-hydrogen-vent-systems-for-customer-applications/

Because they want the big fireball going outside as quickly as possible, and not back into the equipment.

@Jon: For arguments sake, say the gas was propane. Then one is ok having the big fireball inside? I mean which gases does an arrestor make sense & which ones not?

Hydrogen has a very broad range for combustion in air, with a LEL (lower explosive limit) of about 4% H2 in air and a UEL (upper explosive limit ) of about 96% H2 in air (or 4% air in H2 if you prefer) - well, actually a more recent reference from Matheson indicates 4% to 75% - I've learned something new today, thanks! It also has a very low ignition energy (energy input required to get it going). So, your exhaust system can fill up with lots of H2 and then get set off easily. This can quickly go from a small flame zone to deflagration in the exhaust system (sub-sonic) to detonation (supersonic reaction front), and the reaction will continue until you are outside of the LEL/UEL range.
Look over the Matheson reference, and you will see that there are very few other gases with such a broad LEL/UEL range. For example, methane (or natural gas) is only 5% to 15%, propane is 2.1% to 9.5%. So, if you fill up a space with methane or propane, only so much of it will burn, then it will quench as the mix goes out of the LEL-UEL limits.
The implication is that a hydrogen-air mixture can sustain detonation/deflagration across a much broader range, so it can (and by Murphy will) make a lot more flame and shock and awe. Bottom line, don't have anything in the way of all that excitement exiting the facility as quickly and completely as possible.
An article in Chemical Engineering discusses the issues briefly, although doesn't go into the engineering details (a bit odd for an engineering magazine?).

Thanks! So, which gases does an arrestor on the vent line make sense? And which ones not?

Flame Arresters are fitted to prevent the ingress of an external ignition entering the facility, not to stop a "big fireball going outside". Due to the wide ranging LEL-UEL limits explained above, it is even more essential that a flame arrester is installed.

You need a special flash back arrester for hydrogen service, the flame front for a hydrogen / air mixture moves far quicker than those of most fuel gas / air mixtures.
If you are building experimental equipment which uses hydrogen or something where hydrogen / air mixtures could exist inside it then I respectfully suggest that you obtain expert advice.
One enduring problem in light water reactors is that under accident conditions it is possible to form hydrogen as a result of the reaction of zirconium metal cladding with steam. The hydrogen thus generated was responsible for the explosive events at Fukushima.


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Balancing using oxidation number method

Balance the following equation using oxidation number method:
  $\ce{PbS + H2O2 -> PbSO4 + H2O}$

Attempt:
I have written down the oxidation states of all atoms and we can observe reduction of peroxide and oxidation of sulphur.
$\ce{Pb^{II}S^{II-} + H^{I}_2O^{-I}2 ->Pb^{II}S^{VI}O^{-II}_4 + H^{I}2O^{-II}  }$
The specific problem I am facing with this question is that oxygen is going from $-1$ to $-2$ in both the reaction products. How do I handle that?

related https://chemistry.stackexchange.com/questions/91469/h2o2-h2so4-2-ki-solving-the-equation

For sulfur we can start by identifying the number of electrons lost in the oxidation half reaction:
$$\ce{PbS^{2-} ->PbS^{6+}O4 +8e-}$$
Now we can balance the oxygens using water, which also leaves us with an excess of protons:
$$\ce{PbS + 4H2O -> PbSO4 + 8e- +8H+}\tag1\label{Oxidation}$$
Now consider reduction half reaction:
$$\ce{H2O^{1-}2 +2e- -> 2H2O^{2-} }$$
It will obviously require a proton source, so we balance with protons:
$$\ce{H2O2 + 2 H+ +2e- -> 2H2O }$$
Multiply by a factor of 4 to match the number of electrons needed by the oxidation reaction:
$$\ce{4H2O2 + 8 H+ +8e- -> 8H2O }\tag2\label{Reduction}$$
Combine the half reactions $(1)$ and $(2)$:
$$\ce{PbS + 4H2O2 -> PbSO4 + 4H2O}$$


text:
Change in internal energy for the heating and vaporisation of water

What is the value of change in internal energy at $\pu{1atm}$ in the process $$\ce{H2O(l,$323\ \mathrm K$)->H2O(g,$423\ \mathrm K$)}$$
  given $C_{\mathrm m,V}(\ce{H2O,l})=\pu{75.0 J K^{-1} mol^{-1}}$ and $C_{\mathrm m,p}(\ce{H2O,g})=\pu{33.314 J K^{-1} mol^{-1}}$
a. $\pu{42.91 kJ/mol}$
  b. $\pu{43086 kJ/mol}$
  c. $\pu{42.6 kJ/mol}$
  d. $\pu{49.6 kJ/mol}$

I tried to use the First Law here. For chemical reaction I found $\Delta n_\mathrm g$ to be $1$ and the heat of the reaction $\Delta H$ is given as $\pu{40.7 kJ mol-1}$. However, I am not able to use the specific heat data given in the question. What am I doing wrong here?
I am getting stuck in phase change reactions, particularly in using $C_p$ and $C_V$ here. I know the First Law will work, but how do we use $C_p$ and $C_V$ in such reactions?

@GENESECT I got you are started by replacing the image. That should help to format the other parts of the post. Also, option d was cut off in your image.

It seems like you might have to use the density of water and the ideal gas law to approximate the change in volume as water goes from the liquid to vapor state.

The easiest way to do this problem is to first determine the  change in enthalpy for this constant pressure process.  Do you think you can do that?  Then determine the change in PV between the initial and final states.

But why is heat of the reaction equal to change in enthslpy here... I don't understand the logic

Are you talking about the overall change, or the enthalpy change for vaporization?

You are aware that, for a constant pressure process, Q is equal to Delta H, right?

I got that. It's a constant pressure process so we can use ncp∆T.Will this be overall heat

No.  Heat is being added during vaporization at constant temperature and pressure, and this portion of the heat is equal to the enthalpy change for that step in the process.

Can you tell me heat in this whole reaction how to calculate it

To do this, you need to know the heat of vaporization of water at one temperature, in this case 373K = 100 C would be appropriate.  Do you know the heat of vaporization of water at 100C?  And, are you familiar with Hess' law?

Don't know about hess law but we are going to. Learn it

I know hess law now. I know the enthalpy of vaporisation but wouldn't the enthalpy be the heat added to water from 50 C to 150 C... Why are we just including the heat of vaporisation?

What do you get for the enthalpy change between the initial and final states, and how did you calculate it?  Please show the details.

I divided the reactions in three parts, prior to boiling, at boiling and after boiling. I cakculated the enthalpy seperately. I used the density of steam and calculated the enthalpy at vapourisation

The enthalpy of vaporization is tabulated, and does not require a calculation.

Sorry... I used it to calculate internal energy at boiling point

∆H=∆U+∆PV.i know ∆H here

But I am confused on what to take at ∆V

For $\Delta (PV)$, you use the pressure times the specific volume of the vapor in the final state and subtract the pressure times the specific volume of the liquid in the initial state.


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Limiting factor on Belousov-Zhabotinsky reaction
In the thin film Belousov-Zhabotinsky reactions (popular on Youtube/in chemistry demonstrations), concentric rings of different colors propagate across a petri dish for minutes to hours as the solution oscillates between the forward and reverse reaction.
I can't find an explanation online for why these reactions eventually terminate. Do the components degrade over time into a form that can't continue the reaction? Do they require heat input to continue? Or do they eventually find a non-oscillating equilibrium point because of small imbalances in the quantities of the two reagents?
Finally: is there a way to stage this reaction as a demonstration exhibit that can be restarted indefinitely, by stirring or another simple input, without constantly replenishing the reagents? What conditions would need to be met for this to happen?
(Assuming the potassium bromate/malonic acid version since the reagents are easy to find, but I'm open to other reactions that would be longer lasting)

Reaction goes forward, just in oscillatory way, with every oscillation it's closer to equilibrium.


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Properties of azeotropes

Consider two liquids with vapour pressures in pure form being ${P^o}_A$ and ${P^o}_B$. ${P^o}_B >{P^o}_A$ which implies liquid B is more volatile. If the liquids formed an ideal solution, then a straight line (drawn with red in the diagram) will be obtained for the vapour pressure of the solution. However, if they formed a non ideal solution with positive deviation, the red curve would have been obtained.
Now, we get a maxima on that curve. At that composition, the solution would be called a minimum boiling point azeotrope. I understand why minimum boiling point. It's because of the maximum vapour pressure but what I don't understand at all are these points:

The azeotrope boils at a constant temperature.
The composition of the azeotrope remains fixed while boiling.
The azeotropic mixture cannot be separated by fractional
distillation.

These points are really not explicit from the graph. I am not able to figure out the reasons for those points. Can someone please provide a mathematical/ intuitive view of azeotropes relating to these 3 properties?
Also, "in an azeotrope or constant temperature boiling mixture, the vapour has the same composition as the liquid." How can we confidently claim that the point of maxima is the point where the vapour has same composition as liquid?

https://chemistry.stackexchange.com/a/68641/59289

related https://chemistry.stackexchange.com/questions/5701/azeotropes-and-separation-by-distillation and a lot of other questions...

@Mithoron There are many related questions, but even I can't seem to find an answer specifically to these questions in a simple intuitive way.

@Mithoron Please see the picture now. Is there still any problem??

@Mithoron, not sure what you wanted here - the original photo was quite clear. Usually we don't like screenshots/pictures of lines and lines of text, but a diagram with a few words is generally okay.

@orthocresol It just looked bad IMO, new one looks better.

I will first enumerate the four points that need to be addressed.

The vapor and liquid phases have the same composition in an azeotrope.
The azeotrope boils at a constant temperature.
The composition of the azeotrope remains fixed while boiling.
The azeotropic mixture cannot be separated by fractional distillation.

I will assume the definition of an azeotrope to be a solution whose composition is such that it exists at a local extremum of a Pxy or Txy plot. I prefer to instead take 1. as a definition, in which case my original definition follows as a property of an azeotrope, but both approaches are equivalent.
"Proof" of 1., by  contradiction. Overlay the corresponding vapor pressure curve as a function of $y_A$ to make your plot a Pxy plot for species $A$. This new vapor pressure curve must touch the original curve at the azeotropic composition $x^*=y^*=z$, for otherwise a tie line would connect two vapor phases at some $P$, which is physically unreasonable.
Proof of 2. Refer to the Pxy plot from 1. The dew point and bubble point coincide at the azeotrope, so a phase transition happens only at one value of $P$ and hence only at one value of $T$.
Proof of 3. Refer to the Pxy plot from 1. The dew point and bubble point coincide at the azeotrope, and so must the compositions of each phase.
Proof of 4. Fractional distillation is based on the composition of the vapor and liquid phases being different. This is precluded by 3. in an azeotrope.

Thanks for the answer! What does this mean: "Overlay the corresponding vapor pressure curve as a function of  y_a to make your plot a Pxy plot for species A. " Also, what's y?

@Abcd, that quoted statement tells you to look at the Pxy plot for species $A$ (and provides a possible construction for doing so). $y_A$ is the mole fraction of species $A$ in the vapor phase, and $x_A$ the mole fraction of $A$ in the liquid phase.

Can you please elaborate on this bit: "This new vapor pressure curve must touch the original curve at the azeotropic composition x∗=y∗=z, for otherwise a tie line would connect two vapor phases at some P, which is physically unreasonable." ? (using a diagram if you don't mind)

@Abcd, you can verify my statement by constructing a Pxy diagram where the two vapor pressure curves don’t touch and drawing a tie line between them. This line intersects with two vapor phases.


text:
Why does Cu not react with dilute H2SO4?
$\ce{Cu + (dil) H2SO4 -> No Reaction}$
but
$\ce{Cu + (conc) H2SO4 -> CuSO4 + SO2 + H2O}$
What is the difference between the dilute and concentrated acid that one reacts but the other doesn't? They both are the same thing except that dilute one has more water than the concentrated one.

Dilute H2SO4 is an aqueous solution. Conc H2SO4 is not. Conc H2SO4 is a really reactive liquid.

It is not a bad question, often sulfuric acid is not regarded as oxidising while nitric acid is regarded as oxidizing


text:
Production of CTAB
I'm looking for the method/recipe/procedure for producing cetyltrimethylammonium bromide (CTAB). I didn't see it in Ullmann's, and I can't seem to find it online. Does anyone have a reference?

You should be able to find this on Reaxys/SciFinder, but looking at the structure it seems like the most obvious way would be to take the parent amine and quaternise it with MeBr (complicated by the fact that MeBr is basically a gas).

The following is what i consider to be the best/easiest preparation taken from here: Scheraga; Backus - Journal of the American Chemical Society, 1951, vol. 73, p. 5108
To the cetyl bromide was added a 30% excess of a 25%
solution of trimethylamine in methanol. The mixture was
refluxed on a steam-bath for one hour using a condenser
cooled with ice-water. The resultant solid was recrystallized
three times by dissolving in warm methanol and recovering
by an ether precipitation with cooling in an ice-bath. The
detergent was dried in a vacuum desiccator.
However, if this is for lab use (and not for literature review or relevant) i recommend to buy it since it costs only 15.8 £/100 g on Alfa-Aesar while the bromide itself is already at 14.7 £/100 g.

This article (Journal of Radiolabelled Compounds 33 (1993) 33 )  describes the synthesis of 14C labelled CTAB which should offer you a usable procedure