When should we not draw a Lewis dot structure? My friend once told me that technically , we should not draw Lewis dot structure of molecules containing no central atom(for example for benzene,
acetic acid).
Is there an explanation for this?

Short answer: no, there's nothing preventing you from drawing arbitrarily complex molecules according to the conventions of Lewis dot structures. They are not special, and the rules for drawing them are settled by convention and have evolved since they were first developed in Lewis's original paper. They're just one type of diagrammatic rendering of molecular structure which (probably due to their simplicity and explicitness) has caught on in pedagogy. They don't see much use beyond that. Read on for further details.

It's worthwhile to precisely define our terms here. The IUPAC Gold Book provides the following definition for a central atom:

  The atom in a coordination entity which binds other atoms or groups of atoms (ligands) to itself, thereby occupying a central position in the coordination entity.

By that definition, the term is reserved almost exclusively for coordination complexes with a distinct metal atom at the center and ligands bound to it (with some exceptions for purely molecular Lewis adducts, e.g., amine oxides), where the electrons of the bond(s) originate on the ligands, rather than each atom donating an equal number of electrons in the formation of the bond(s). For non-coordinate covalent bonds, if the bonds were to be split homolytically (i.e., such that an equal number of electrons are distributed to each atom), the result would be separate, neutral atoms. Were coordinate covalent bonds to be split this way, that would not be the result.

Obviously, that definition is highly limited, and it's not the prevailing definition in my experience. The other usage of the term is essentially to refer to any non-terminal atom (i.e., one that has two or more bonds to separate atoms). There's nothing inherent to Lewis structures that prevents them being used to represent molecules containing multiple such central atoms. They're simply one representation of bonding. In practice, they don't see much use, chiefly because they're substantially less economical than condensed structural formulas or the skeletal bond-line structures (introduced in their prototypical form by Kekule), and because pure Lewis dot structures, as they were originally conceived, are exceedingly poor at representing stereochemistry.


Making a saturated solution of Na₂CO₃ I need to make a saturated solution of $\ce{Na2CO3}$ and according to the available data I should add $22\ \text{g}$ in $100\ \text{ml}$ of water at $25\ ^\circ\text{C}$.

I wanted to make $300\ \text{ml}$ solution, which would require $66\ \text{g}$. But, even after adding $110\ \text{g}$ of the salt, no turbidity appears which might indicate saturation. So I believe that the solution is still not saturated.

However, when I kept the flask overnight, almost the entire salt precipitated out in large crystalline chunks. I am a bit confused as to what the problem might be. Has anyone faced this kind of trouble?

$\ce{Na_2CO_3}$ forms several hydrates with decomposition point close to room temperature (30-40 Celsium). Moreover, the solubility of decahydrate decreases with temperature decrease, and solubility of anhydrous salt almost does not depend on temperature and is much higher than solubility of decahydrate.  The relevant pictures are googleable by words 'sodium carbonate solubility curve', see 'images' google service, not general search.

So, I believe you have prepared a solution of the salt at the higher temperature at day, but the salt mostly precipitated in form of decahydrate overnight, because the solution cooled significantly at night. Bonus points for this version if you have significant changes in temperature at your work during the nychthemeron.

Second version is that you indeed prepared seemingly oversaturated solution because it is not saturated for anhydrous $\ce{Na_2CO_3}$, but overnight it crystallized, because it is saturated for decahydrate. I'm a bit sceptical about such possibility, though, because significantly oversaturated solution crystallize even on slightest imperfections, and the still not-dissolved salt is a great collection of such imperfections.

In addition, if the solution stayed overnight open, it indeed may consumed some amount of atmospheric $\ce{CO_2}$ and partially transform into sodium hydrocarbonate with much lesser solubility.


Which is most reliable excitation energy: LUMO-HOMO energy or TD-DFT excitation energies? After running a TD-DFT , we can calculate the HOMO-LUMO difference and then calculate the the excitation energy.

Another approach is just simply pick the energies calculated in the TD-DFT and use them.

For me the latter makes more sense as that's the reason we do TD-DFT , otherwise we could access the HOMO-LUMO difference by any other methods but I have seen papers using the first concept. I also have noticed that the energies in second concept are closer to the experimental values.

Also when I run TD-DFT, I can see Oscillator Strength. Should I pick the energy that has the largest Oscillator Strength for calculation of the maximum wavelength ?
Here is an example. Should I pick number 8 for maximum wavelength calculation?

When you can, use TDDFT, in part because of the oscillator strengths

I like to point out to my group that optical excitations consist of two things consistently (e.g., in a UV/Vis spectra). There are the excitation energies (the x-axis) and the intensity or extinction coefficient (the y-axis). To some degree, the peak widths give a bit more information (e.g., into vibrational structure).

The example you give above is a great example. Most of the excitations are forbidden and have no oscillator strength. I'd pick the same excitation, although I'm curious what the molecule is - usually there are much higher oscillator strengths.

You're right that there are papers and groups that use the HOMO and LUMO eigenvalues. While some DFT methods (e.g. optimally tuned range-separated hybrids) do provide HOMO-LUMO energies that correspond nicely with experiments, there's nothing physical about Kohn-Sham orbital eigenvalues. Moreover, in general, unfilled orbitals are not handled well by DFT methods.

It's been a while since I looked for reviews, but general TDDFT excitation energies are ~0.2-0.3 eV from experimental absorption energies after correcting for solvent effects, etc.

In my group, we also do what I call a "consensus" model. That is, when you run a TDDFT calculation, you get the HOMO and LUMO energies anyway. Since there are slightly different random and systematic errors from each approach, you can calibrate both against experiment and use a multivariate linear regression. The result decreases the average error significantly.


Diffusion of carbon from ferrite to cementite in the eutectoid transformation Why does the carbon of an alloy diffuse from ferrite (0.022% C) to cementite (6.70% C) during eutectoid transformations? I know that the diffusion of a generic species (e.g. of atom A) carries A from regions where the amount of A is higher to regions where the amount of A is lower.

I found this sentence on page 337 of Callister (9th ed.).

Diffusion is perhaps more complicated than initially taught. Yes, all things being equal you will get diffusion from regions of high concentration to regions of low concentration.

But, in the case you mention, all things are not equal - there is a phase change occurring, and that means that there are chemical potential gradients as well as concentration gradients. Now you have an additional driving force for motion of impurities as they prefer to be in the phase with a lower Gibbs free energy.


Column order, signs of MO coefficients, and density matrix construction in Hartree-Fock I'm attempting to make a Hartree-Fock solver, using Python, following Daniel Crawford's guide.

I have been successful with the first four steps. However, I am having some trouble with this part in step 5:

$$
F^0_c C^{'}_0 = C^{'}_0 \epsilon_0
$$

I have the correct transformed Fock matrix (I can verify this as the site as step-by-step matrices that you can check against). My problem is that the MO coefficient matrix I get does not quite match the one on the site. I obtain it by multiplying the above equation by $C^{' -1}_0$ on the right:

$$
F^0_c = C^{'}_0 \epsilon_0 C^{' -1}_0
$$

Now it's in diagonalized form, which means $C^{'}_0$, the transformed MO coefficients, is the eigenvectors of $F^0_c$ and $\epsilon_0$ is a diagonal matrix of the eigenvalues. After I obtain $C^{'}_0$, I obtain the original untransformed MO coefficients $C_0$ using the next equation on the site:

$$
C_0 = S^{-1/2} C^{'}_0
$$

where $S^{-1/2}$ is the transformation matrix used to get the transformed Fock matrix from earlier. My transformation matrix, by the way, matches exactly the one on the site.

This is where my problem comes in. The $C_0$ matrix I get is this:

The one the website has as a reference is this:

If you look carefully, you will notice that mine and the reference one have the same columns; however, some are swapped around, and some have sign flips. This causes problems with the next steps, such as finding the density matrix.

I thought maybe the order should be such that $\epsilon_0$ is in order. The $\epsilon_0$ matrix I get is indeed out of order, and the same permutations and sign flips that will make it in order also makes my $C_0$ match the reference one in terms of where each column is, but NOT sign. Sadly the reference site does not have a $\epsilon_0$ matrix or $C{'}_0$ for me to compare with, only the final $C_0$, so I'm not sure if those are coming out of order also.

My first question is, will these sign flips matter? Is ordering the $C_0$ matrix in terms of getting $\epsilon_0$ the right thing to do?

I am carrying out these matrix operations with basic NumPy functions such as numpy.linalg.eig and numpy.dot. Nothing fancy there. It's worth noting that finding the transformation matrix, $S^{-1/2}$, requires diagonalizing in a similar way, but I didn't have any column switching problems there. It came out exactly as expected.

My second question relates to building the density matrix at the end of step 5. The website has this equation:

$$
D^0_{\mu\nu} = \sum_{m}^{occ} (C_0)^m_\mu (C_0)^m_\nu
$$

It says sum over the occupied MOs. The molecule being used in the example calculation is water, so two H's, and one O, so there are 10 electrons, so 5 orbitals. If I sum using the correct $C_0$ matrix I get on the reference website, from m = 1 to 5, I get the correct $D^0_{\mu\nu}$ result. However my textbook (Szabo and Ostlund, Modern Quantum Chemistry) has this equation (changing some of the notation to match):

$$
D^0_{\mu\nu} = 2 \sum_{m}^{N/2} (C_0)^m_\mu (C_0)^m_\nu
$$

Where N is the number of electrons. Well, 10/5 is 5, so I'm summing from 1 to 5 again, but now there's a 2 in front, so the sum will be doubled. Now, using the one on the website gets me the correct answer, but what could this 2 in the book mean?

In the future, this should really be two separate questions, but I will try and tackle both.

MO coefficient column ordering and signs

The electronic energy is invariant to orbital swapping within subspaces. If you interchange one occupied orbital with another, the total energy of a molecular won't change. The same goes for virtual-virtual. Now, you know that the MO coefficients are by convention shaped as $C_{\mu i}$, where the rows are the AOs and the columns are the MOs. What this means is that for $\ce{H2O}$ in a STO-3G basis, where there are 7 AOs forming 7 MOs where 5 are doubly-occupied and 2 unoccupied, switching the order of the first 5 columns should not change the electronic energy. The same goes for the final 2 columns.

As you say, the $\epsilon_i$ follow the same ordering as the columns in $C_{\mu i}$; this is because they are the matching eigenvalues and eigenvectors from diagonalizing the Fock operator. The eigenvectors are not going to be affected by a sign change, which doesn't change how the space is spanned.

Regarding why the ordering is not what you expect, there is no guarantee made by NumPy or the underlying LAPACK routine DGEEV that the results will be sorted, just that the vector/value pairs will be in the same order. If you want to sort the results, you can use argsort:

idx = energies.argsort()
energies = energies[idx]
eigvecs = eigvecs[:, idx]

Make sure the ordering of the MO energies is maintained along with the ordering of the MO coefficient matrix columns.

Constructing the density matrix

This is a very irritating difference in convention among many sources. See later, for calculating the electronic part of the dipole moment:

  $$
\left<\vec{\mu}\right> = 2 \sum_{\mu\nu} D_{\mu\nu} \left< \phi_{\mu} | \hat{\mu} | \phi_{\nu} \right>,
$$
  ...

  The factor 2 appearing above arises because the definition of the density used in this project differs from that used in Szabo & Ostlund.

This can also have an effect on whether or not there is a 1/2 in front of the $2J - K$ portion of the Fock operator.


Reducing character of group 15 hydrides It was given in my text book that reducing character of group 15 hydrides increases down the group. I am a bit confused about this. For $\ce{NH3}$ this is fine as I assume that:

$$\ce{NH3 -> N + H^+ + 3e^-}$$

My doubt arises in the higher elements of the group. In $\ce{SbH3}$, $\ce{H}$ is present as $\ce{H^{-}}$ and $\ce{Sb}$ as $\ce{Sb^3+}$. So would $\ce{Sb^3+}$ be oxidised to $\ce{Sb^5+}$ or would $\ce{H^{-}}$ be oxidised to $\ce{H2}$?  I know that for $\ce{Sb}$, $\ce{Sb^3+}$ is more stable than $\ce{Sb^5+}$ . So would  $\ce{H^{-}}$ be oxidised to $\ce{H2}$ ?

Also electronegativity of $\ce{As}$ and $\ce{H}$ is given tobe equal in my textbook. So what would be oxidation product of $\ce{AsH3}$ ?

For $\ce{BiH3}$ it seems like:

$$\ce{Bi^3+ -> Bi^5+ + 2e^-}$$

is more feasible as $\ce{Bi^5+}$ is more stable. Can someone please tell me the oxidation products of group 15 hydrides? Also can someone tell $\ce{E°_(H^{-}/H2)}$ value as I couldn't find it on net?

When your book says that reducing character increases down the group, it PRIMARILY meant in gaseous state. What you have described above are all ionic reactions, i.e., in some solution.

NOW in gaseous state, as we go down the group, the bond energy of hydrides reduces (due to increasing atomic size as we go down the group). So they can be more easily broken to release nascent hydrogen (see my note at the bottom), which is the cause of reducing nature. So reducing ability increases down the group in gaseous state.

NOTE : Nascent hydrogen is referred to the hydrogen which has just evolved post some reaction, and has not yet formed its molecule. This is not the same as hydrogen atom so keep that in mind.

Reducing property in a solution

This is the catch. In aqueous solution, reduction takes place due to the group 15 ion, and not hydrogen ion. So it depends on how easily the group 15 ion can lose an electron (reduction is gain of electrons)
Down the group, electronegativity decreases and hence it becomes easier to remove electrons for reduction. So reduction ability decreases down the group in solutions too.

Now to answer your question, Sb(3+) will be oxidised to Sb(5+) as it is what is causing the reduction. You should be able to answer your following question now.

Btw, the E0 value for (H-/H2)is 0.00. Note that all oxidation reduction reactions of hydrogen gas have this value as 0.00


Is there an efficient and effective way to clean oil from floor? I've recently had several jerry-cans of oil (awaiting disposal) topple over in the attached garage, and several rooms following influx of water following heavy rain. After the water level receded the floor is left with a thin layer of burnt oil.

Various sites on the net recommend using sawdust, baking soda etc. I'm talking close to 800sq.ft of area split into several levels with a step between different levels.

The question I have to ask, therefore, must be obvious now. How does one clean large volumes of burnt oil spilt distributed by water over cement, and stone floors?

EDIT: Nature giveth, and Nature taketh away. Another cloudburst, this time over 100mm of rain in a couple of hours, and muddy waters found their way indoors again. Now the waters have receded, and almost all the oil is replaced with a thin layer of mud. Hopefully washing with water will get rid of the most of it.

Assuming you only want chemistry-related answers, I'd recommend any type of detergent.

If you lift that restriction, however, then I'd recommend pressure washers.


Bond Type of Lead: Metallic or Network Covalent? I know that lead is classified as a metal, from its appearance, nonexistent band gap, and its position on the periodic table... etc. However, all of the elements above it (carbon, silicon, germanium, and tin) usually form covalent network bonding with four surrounding atoms. Why is lead suddenly different, engaging in delocalized metallic bonding when it has enough electrons?
Or is it different? Lead has a low melting point for a metal (600 K), and a lower melting point than, say, copper.

As you proceed down Group 14 the metallic bonding structure tends to become more stable than the tetrahedrally bonded structure we see in diamond, Si and Ge.  Tin is actually a transitional case.  At typical room temperature it is metallic (white tin) but upon chilling it transforms to the tetrahedral allotrope (gray tin).  By the time we get to lead we have fully gone over to the metallic phase.

Because the tetrahedral bonding structure requires each atom to contribute four valence electrons, this tetrahedral --> metallic transition may be seen as a form of the "inert pair" effect that emerges in heavy atoms.


Stereoisomers of Hexachlorocyclohexane What is the basis of naming the various isomers of hexachlorocyclohexane $\alpha$, $\beta$, $\gamma$, etc.? Here is a list of these isomers. Is it just random naming, or is it according to some convention?

I didn’t find anything definite. However, I found a PDF from the State’s Institution for Protection of the Enviroment in Baden-Württemberg (Landesanstalt für Umweltschutz) which gives me two or three relevant clues. Along with my intuition and a chat with a labmate, I conclude:

There is nothing systematic or conventionalised in the names of the isomers. They were likely named after the order in which they were isolated or their structures were determined.

Clues I used:

The $\unicode[Times]{x3B6}$, $\unicode[Times]{x3B7}$ and $\unicode[Times]{x3B8}$ isomers were synthesised last.
$\unicode[Times]{x3B1}$, $\unicode[Times]{x3B2}$ $\unicode[Times]{x3B3}$ and $\unicode[Times]{x3B4}$ were characterised first by a guy called Teunis van der Linden, after whom the active insecticide $\unicode[Times]{x3B3}$-hexachlorocyclohexane was named lindane.
Thereby concluding that $\unicode[Times]{x3B5}$ was discovered after $\unicode[Times]{x3B1}$ to $\unicode[Times]{x3B4}$.


Stoichiometry : dilution questions I need 500 ml of 0.2 mol/L  H2SO4 for a tutration lab. What volume in mL  of the 18 mol/L stock solution should you use?

I think this is a dilution question asking me to use C1V1=C2V2. I am confused because it sounds like s dilution question, but its asking for volume so i could possibly be using the (v/v) formula. If someone could help me out that would be greatly apprecaiated, trying to study for a test.

I gotchu.

It's definitely a dilution question. Just plug it into the equation you have, C1V1=C2V2.

$C_1V_1=C_2V_2$

$(0.2\: M)(500\: mL)=(18\: M)(x\: mL)$

Solve for x:

$x=5.56\: mL$

Therefore, you want to use 5.56 mL of 18 M stock solution. To make your 500 mL of 0.2 M sulfuric acid, just take the 5.56 mL of the stock solution, then fill it up with water until you have 500 mL total solution.

The idea behind the equation, by the way, is to preserve the number of moles in the solution while changing the total volume of the solution. Here, we find out that (0.2 M)(0.5 L) = 0.1 mol of sulfuric acid is needed in the final solution. So, we find the amount of stock solution needed to provide that 0.1 mol of sulfuric acid, by dividing 0.1 mol by the concentration of the stock solution.


Hyperconjugation involving other bonds Hyperconjugation is often illustrated with C-C and C-H bonds donating electron density to the p orbital in the adjacent carbon atom. Since the concept revolves around sigma bonding orbitals donating electron density to p orbitals, would other bonds such as Si-Si or Si-H bonds be able to take part in hyperconjugation as well? If they are not able to do so, why not?

Your description of hyperconjugation is a very minimalistic one. More generally, hyperconjugation should be seen as the interaction of occupied bonding MOs with unoccupied ones — these unoccupied ones can be both antibonding MOs or empty (cationic) p atomic orbitals.

In principle, $\ce{Si-H}$ or $\ce{Si-Si}$ bonds would also be able to donate electron density in a hyperconjugative matter. In pracitce, due to silicon’s low electronegativity and greater size these orbitals simply do not overlap as well and their energy levels are such that less stabilisation may be obtainable.

Carbon-silicon bonds are of course great at providing hyperconjugation since silicon — much like hydrogen — has a lower electronegativity than carbon, meaning the occupied orbitals have a greater contribution on carbon.


Whats the driving force behind simple diffusion? Is it just the count of molecules? If I were to have two separate containers with solutions of different concentrations with a small opening, most of the molecules would flow down the concentration gradient. Does this occur just because more molecules can pass through the opening or is there a force pushing them out?
My second question is if the molecules could pass through the opening under normal conditions even against the gradient?

All particles are constantly moving randomly due to their kinetic energy, and are thus constantly colliding into each other and the walls of their container (if any).
Particles in gases move rapidly and randomly and are spaced far apart, particles in liquids slide slowly over each other, and particles in solid vibrate around a fixed position.
Hence, diffusion can be defined as the overall movement of particles, from an area of higher concentration to an area of lower concentration, down a concentration gradient, due to their random motion.
Illustration of particles in a gas:

Source: wikipedia.org/wiki/Kinetic_theory_of_gases#/media/File:Translational_motion.gif
This explanation relies on Kinetic theory and Brownian motion.


Calculation of the specific rate constant (k) Suppose we found that the reaction of Red#3 with bleach is 1st order in hypochlorite. The observed rate constant ($k_\mathrm{obs}$) for one of the trials was measured to be $\pu{0.00400 s^{-1}}$. If the concentration of hypochlorite was fixed at $\pu{0.134 M}$, calculate the specific rate constant ($k$) for that trial.

I am very confused in which equation I should use to solve this problem. Am I to use this equation?
$$\mathrm{Rate} = k[A]^m[B]^n$$

If that is the case, for trial #1:

$$\mathrm{Rate} = \pu{0.00400 s^{-1}} \times \pu{0.134 M} = \pu{5.36e-4 mol L^{-1} s^{-1}}$$

But then what should I do next? This does not sound right. Please point out my mistakes.

We assume 1st order kinetics in hypochlorite and 1st order kinetics in the color. Putting this in the equation we get the following:
$$\frac{\text{d}[\text{Red}]}{\text{d}t} = -k\, [\text{Red}]^1 \, [\text{bleach}]^1 $$

Because you keep the concentration of bleach the same throughout the trial, you can write
$$k_\text{obs} = k\times [\text{bleach}]$$

Now, we can continue onwards as follows:
$$ k_\text{obs} = k \times [\text{bleach}] \Leftrightarrow k = \frac{k_\text{obs}}{[\text{bleach}]} = 0.0299~\mathrm{s^{-1}\; M^{-1}}$$


How come diluting a cell decreases the voltage? I just want to know why, if we dilute a cell with water, it decreases the voltage. Assume it is a galvanic cell.

Your assertion the voltage in a cell is decreased by diluting the cell  is not necessarily true. According to the Nernst Equation:

$$E_{cell}=E^0-\frac{RT}{nF}\ln(Q)$$

Where $Q$ is the reaction quotient relating $\frac{[\mathrm{metal~ion~formed~by~oxidation}]}{[\mathrm{metal~ion~reduced}]}$.

The value of $Q$ can be increased by: increasing the concentration of the metal ion formed by oxidation, or by decreasing the concentration of the metal ion that is reduced. By increasing the value of $Q$, the magnitude of the term $-\frac{RT}{nF}\ln(Q)$ is increased. This means that a larger number is subtracted from the $E^0$ of the cell, and that the cell voltage decreases.

The value of $Q$ can be decreased by: decreasing the concentration of the metal ion formed by oxidation, or by increasing the concentration of the metal ion that is reduced. By decreasing the value of $Q$ to a fraction, the value of the term $-\frac{RT}{nF}\ln(Q)$ is made positive. This means that this term is added to the $E^0$ of the cell, and that the cell voltage increases.

This is consistent with the Le Châtelier's Theorem. Essentially, increasing the relative concentration of your reactants drives the reaction forward (higher $\Delta V$), and increasing the relative concentration of your products drives the reaction backwards (lower $\Delta V$).


Finding the structure based on spectra I had to identify the structure of a drug based off the spectra I had. I suck at this. I gave it my best go and got some ugly structure, only to check it using a database and find out I was wrong. I gave up and searched for the chemical structure I was given, only to find out that the drug was diazepam and I was on the right track (my aromatic substitution was correct, and I got the correct degree of hydrogen defeciency, but I didn't know how to tie everything together and I was stuck with one more quaternary carbon that I didn't know where to place). I feel like the spectra I was given are hard to read (specifically the HSQC, HMBC, and COSY); do you all feel the same? I really want to know how to arrive at the answer.
The images are the spectra I was given. I was also given mass and IR strectra, but I cannot post too many pictures with my reputation. If you all need better resolution images, let me know. The molecular formula is C16H13ClN2O, and the groups I found with the IR were NH, C=O, and benzene. I am kind of afraid to ask the instructor at this point because I already found the answer. It wouldn't hurt to ask him anyway, would it? I just don't want to lose points.
EDIT: I took the original images down because I did not want the instructor to find it, if he hadn't already.

The most practical approach is to identify the structure of an organic compound by combination of spectroscopic techniques.  This is why you were provided with UV-Vis, IR, NMR and MS data, because it equally reflects the eventual approach in the lab after product isolation and purification, too.  Hopefully you have sufficient background what theses techniques assess (if not, consider consulting e.g., the spectroscopy section here on ChemSE, or on Chemistry LibreTexts).
Then, start training with simpler molecular structures in first place.  You may use, for example, Field's training booklet from the shelves (either print or/and electronic resource) of your school's library.  The obstacle here: for the majority of the sets, there is no answer in the book (the key is available only to instructors).  You may however list structures likely to be in agreement with the data per set with public references such as the SDBS spectral database.  You equally may train yourself picking an organic compound from Wikipedia and list the features a theoretical UV-Vis IR or MS or NMR spectrum would contain and again compare, e.g. with SDBS.
Yes, there are programs to predict the position and shape of NMR spectra, and you equally may find diazapam with a search via Hill formula on nmrshiftdb2.org, too.  They may help you to internalize how structural features (e.g., CH vs CH2) show in the APT spectrum because you may select either of the type of nuclei to be displayed, or to highlight their attribution of a calculated position of a signal to an atom in a given structure, or their indication of a cross peak e.g. in the HSQC.  Aside from an exam where these programs might not available to you, don't rely too much on them.  E.g., the position of signals in the $\ce{^1H}$-NMR spectrum depends at least on the concentration of your analyte, the choice of solvent (e.g. chloroform or DMSO) may influence if exchangeable protons on alcohols and amines yield a broad signal, or split into a multiplet (if there are coupling nuclei).
The only way to get better in this is training.  If the structure suggested is not in agreement with all of the spectral observations, the assignment is erroneous and one has to start again. It may be demanding, but don't give up.

As a starter:

N-NMR:  there are two types of N with different chemical environment
H-NMR:  there are two sets of alkyl-H; by integration possibly one $\ce{CH3}$ and one $\ce{CH2}$ or integer multiples of both.
APT-NMR:  again indication for a methyl, and a methylene group; count of CH: ... which are in the region expected for aromatic CH, count of C: ... etc.


How do I approach drawing all of the possible constitutional isomers? Does anyone have any tips on how to approach these types of questions?
A specific example would be:

  Draw all constitutional isomers of $\ce{C2H4O}$

I can get the easier ones like vinyl alcohol, but it's hard to get ethylene oxide and it takes a long time for me to get acetaldehdye. And how would I know if I got all of the possible constitutional isomers?

I approach questions like this by considering all possibilities of arrangements of the non-hydrogen atoms. Obviously, there's C-C-O, C-O-C, and then the cyclic ethylene oxide.

Then I make sure I've matched bond orders to get the hydrogen count.

So I get vinyl alcohol, acetaldehyde, and ethylene oxide. H2C-O-CH2 would only work for a biradical, which is probably not expected.

The systematic fashion then is to arrange the "heavy" atoms, first in all linear forms, then all cyclic forms, then all two-ring forms, etc. Of course with large formulas, the number of isomers goes up exponentially.


What is the topology of a phase diagram? This question is cross-posted from phys

Looking at various two-variable phase diagrams I was struck by that on every one I have seen so far all the phases formed simple connected regions; see, for example the phase diagrams of $\ce{H2O}$ or of $\ce{Fe-Fe_3C}$ in 1. Every phase is a connected region, no two disjoint pieces correspond to the same phase. Is this really true in general and if so why?

Going further, let us consider a three-variable phase diagram, say $p, T, y$, where $y$ maybe electric or magnetic field intensity, or some other controllable extensive/intensive variable. Is it possible that now a 3d region occupied by some phase is a kind of tube that makes a half turn so that if we fix, say, $y$ then the resulting 2d section would have two disjoint $p,T$ regions of the same phase? Can such phase "tube" exist or by some thermodynamic-geometric rule is it excluded?

A possibly more difficult question if a phase can form an annular  region, ie., homeomorphic to an annulus while surrounding another phase homeomorphic to a disk?

In short, what is the topology of the phase diagram and its sections?

Particularly since Fe was mentioned, I'll focus there. For pure iron, it is well known that $\alpha$-Fe, the stable phase at room temperature, is bcc. This is stable up to 1184K where it transitions to fcc $\gamma$-Fe. In turn, this transforms back to bcc $\delta$-Fe at 1667K before melting at 1810K. Yet, both $\alpha$ and $\delta$ have a bcc crystal structure.

If one looks at the Fe-Zn phase diagram:

one sees that the region of bcc phase, enclosed in dark blue, fully encompasses the green sliver of fcc $\gamma$. Using the Zn concentration as a free parameter, one can go from $\alpha$ to $\delta$ without changing phase.

In contrast, in the Fe-Pd phase diagram:

one sees a different picture. Here, Fe and Pd show complete miscibility in the fcc phase across the entire diagram. This means you can't go from $\alpha$ to $\delta$ bcc phase without going through the fcc phase. Yet, one can see that, by trading off Zn for Pd, one could manage to get around the fcc phase by some path in composition space.


How to choose a proper chromatography way to separate substrates? I'm a freshman and I recently learned how to do TLC and column chromatography. I also know a little about other ways to separate things by chromatography, such as affinity chromatography, gel chromatography, ion exchange chromatography, and gas chromatography, but without many details.

I am facing problems in solving a question requiring us to judge how to separate these combinations of substances by chromatography (if there was no such confinement, this will be much easier):

  N-phenylethanamide, N-phenylpropionamide
  methoxymethane, ethoxyethane
  Aβ36，Aβ42 (Aβ is a kind of peptide which is related to Alzheimer's disease, while the number, 36 and 42, refer to the amount of amino acids polymerized in the peptide)

Here are the specific issues I've met：

I've no idea how to deal with the first combination since their difference seems to not be significant, no matter in polarity, acidity (basicity), or molecular size. Is there any other chromatographic way, or did I get something wrong?
I think I can deal with the second combination by gas chromatography, since they are both volatile. But I'm not sure if it is a good idea. Is there a better way or not?
I've found some description for the separation of different Aβ, suggesting that the following ways are usually used: use antibodies (e.g. ELISA) or electrophoretic separation (e.g. PAGE). But seems that neither ways are related to chromatography. It is also said that reverse phase liquid chromatography after CNBr cleavage works, but this way is complex and time consuming. I'm wondering (a) if I can use antibodies or something else to deal with it by affinity chromatogrphy and (b) how CNBr works? (Now I know CNBr is used to cut proteins/peptides at where methionines are, but still not knowing how reverse phase liquid chromatography works.) Also I want to ask if there is a better way as well.

As you already mentioned, there are numerous types of chromatography, and they vary in several different ways. I can't tell you what would be best ways to separate each of these pair of substances, but I can give you some insights. Using chromatography always requires optimization of experimental procedures, and the first step is usually searching on papers and other references for information on how the specific separation is usually done.

N-phenylethanamide/N-phenylpropionamide and methoxymethane/ethoxyethane

Despite being very similar, the presence of one or two extra $\ce{CH2}$ group in each of the second molecules is enough to give the molecules different properties, specially considering they aren't big. For simple detection purposes, I'm sure it would be possible to find the appropriate solvent mixture to separate them using TLC or column chromatography.

A much more advanced technique, gas chromatrography is very powerful and could also be used to separate them, as long as the appropriate setup was used (injection and oven temperatures w or w/o gradients, column type and length, etc).

Aβ36/Aβ42

These would probably decompose in elevated temperatures, so gas chromatography is not a good idea. Peptides are, nowadays, mostly separated using liquid chromatography, usually in HPLC or UPLC systems using reversed phase columns. This type of chromatography will separate the peptides mostly by their polarity and size, and once again, reference search and experimental trials are essential to optimize conditions. HPLC and UPLC are used to separate mixtures of thousands of peptides (with varying efficiency, of course), but that should be a sign that two wouldn't be a problem, even if very similar.

Depending on the objective of the separation (identification and/or recovery), using CNBr or any other digestion step could be useful or not, but I don't think this would be the case for only two peptides.

PAGE and the variations of electrophoresis were for a long time the goto separation method for peptides and proteins, but recent developments in chromatography (and mass spectrometry) have made them fall off preference.

I'm not familiar with affinity chromatography (or the whole antibody thing, to be honest), so I can't say much about that, but, what I do know is that you won't always find or be able to manufacture the chromatographic column with the desired specificity.


What conditions can cause a material to exist as a gas? So my question is, how can something exist a gas. Is it when the pressure exerted by the gas is greater than the atmospheric pressure, then it is a gas?

And also how do manometers work, how can you measure the pressure of a gas if the atmospheric pressure is greater?

  Taking a look at the example above, I understand that the pressure
  exerted by the gas can be greater than the atmospheric pressure,
  because only then you have a gas. But what I don't understand is why a
  substance exists as a gas when the atmospheric pressure is greater?
  Shouldn't the atmospheric pressure crush the substance to a solid or
  liquid?

I know this is a lot, but please help.

Generally, when a substance is at > 0 K (0 Kelvin or -273°C) the atoms/molecules making up that substance possess kinetic energy, which means that each atom/molecule are [almost] constantly moving.  If that atom/molecule is in a vacuum, it would move in [almost] a straight line.  However, in the ordinary matter that we encounter everyday, atoms/molecules exist in macroscopic phases which are composed of a humongous number of particles.  Which means that each particle experiences frequent collisions with neighbouring particles, making the trajectory of each particle to appear zig-zag like.
The higher the temperature in the vicinity of an atom/molecule, the higher the kinetic energy of the particle, and the faster that particle's motion becomes.
Also, depending on the properties of the atoms/molecules in the substance, each atom/molecule experiences electrostatic forces (either attractive or repulsive) exerted by neighboring atoms/molecules.
At the interface between a liquid and a gas (e.g., water in a tumbler sitting on a table in a room at sea level), the surface of the liquid is constantly being bombarded by gaseous particles that have much higher kinetic energies than the molecules of the water in the tumbler.  This bombardment imparts enough kinetic energy to some of the water molecules at the surface.  Some of the bombarded molecules gain enough kinetic energy to enable them to leave the water surface to become gaseous water molecules.  Some of the bombarding particles (oxygen molecules, nitrogen molecules, water vapour molecules, etc.) from the air lose enough kinetic energy so that they become stuck into the liquid surface.  At equilibrium (maximum humidity saturation of the air in the room), the number of water molecules leaving the liquid surface is equal to the number of molecules entering the liquid surface.

It is the net result of factors 2, 3 and 4 that determines the phase of a substance.  For example, substances like water, which are composed mainly of polar molecules (molecules possessing an asymmetrical distribution of charge around it, making it behave like a bipolar/tripolar/multipolar magnet), have relatively high melting and boiling points.  Below the melting point, the molecules exist in a [3D] crystalline lattice.  Each molecule are in constant motion but are at the same time held together by electrostatic forces (hydrogen bonds).  Since in a solid, the kinetic energy or velocity of each molecule is not high/fast enough, those particles cannot overcome/escape from the electrostatic attraction exerted by their immediate neighbours.  Therefore, the molecules stay stuck in the crystal lattice, exhibiting their kinetic energy as vibrational motion.

When the melting point is reached, the kinetic energy of the molecules starts to overcome the electrostatic attraction from neighbouring particles, allowing them to move away from their neighbours.  This is the liquid state, in which the kinetic energy of the particles are enough to prevent the particles from being stuck in one place, allowing them to move about, making the phase fluid; however, the kinetic energies are not enough to allow the particles to move farther away from each other.  In the liquid state, the distances between neighbouring particles are almost the same as that in a solid.

For a liquid to become a gas, two factors must be overcome: (a) the intermoleculer forces in the liquid, and (b) the pressure above the liquid surface.  At the boiling point, the kinetic energies of the particles become high enough to overcome (a) and (b) to enable the particles to move farther away from each other, and we get a gaseous phase.

Factor (a) is overcome, for example, by heating the liquid in the container.

You can appreciate the effect of factor (b) when you go mountain-climbing (5000 feet or higher).  Take a thermometer with you to the mountain top.  You will observe that water at high altitudes boil at temperatures lower than 212°F (100°C).

Another way to experience the effect of factor (b) is through the following experiment:  Get a 50-mL (1 to 2-oz) syringe, remove the needle, draw water to about 25% capacity of the syringe barrel, plug the inlet with your index finger, then with your other hand, pull the plunger until it is almost at its maximum position.  You should see the water in the syringe boil without the application of heat.

This seems to be a simple question but a full answer can be appreciated only with some knowledge of statistical thermodynamics.

BTW, the pressure of a gas is the macroscopic result of a humongous number of particles impinging upon a surface (the diaphragm of a barometer, the surface of a manometer liquid, the walls of a container, etc.).


What would be the concentration of products in this calcium producing system? I wanted to figure out a way to make calcium metal with potassium oxide. $\ce{K2O}$ undergoes disproportionation at $\pu{350-430 °C}:$

$$\ce{2 K2O -> K2O2 + 2 K}$$

Now, if I mixed in $\ce{CaO}$ (1:2 molar ratio with $\ce{K2O})$ it would react with the potassium formed in the first reaction, forming calcium metal and $\ce{K2O}$ that would feed back into the first reaction. Eventually I should be left with mostly $\ce{K2O2}$ and calcium metal, but since these two react together, this can't be as simple.

Reaction of these two would lead to the same mixture as I started with. Is there a way to figure out what the concentration of Ca will be when the reaction reaches equilibrium?

T l, Dr:  despite what you see from the aqueous electromotive series, potassium does not reduce calcium oxide in this setting.

You may be assuming that potassium, which lies above calcium in the electromotive series for aqueous solutions, therefore lies above calcium in all other settings regardless of chemical environment or temperature.  It doesn't.  The electromotive series we see can change in different environments and at different temperatures.

For instance, in this Ellingham-Richardson diagram you see calcium oxidation at a very low oxygen potential, corresponding to a great tendency towards oxidation and thus a very high position for calcium in the "electromotive series" for anhydrous oxide formation.  Where is potassium?  Much higher up in free energy (starting around -650 on the free energy scale), corresponding to a less stable oxide and thus a lower tendency for potassium oxidation.  In the absence of water solvent and at elevated temperatures (or even without elevated temperature but still in the absence of water), potassium does not displace calcium, or even manganese, from oxide formation.  The order is very different from the aqueous one.

Apart from lithium, all of the alkali metals are notoriously poor at oxide formation, compared with what we expect from aqueous solutions.  Look again at the diagram; sodium isn't so hot compared with calcium, magnesium or aluminum either.

Calcium is made industrially by either electrolysis or reduction of calcium oxide with aluminum.  In the latter reaction aluminum still forms a less stable oxide than calcium, but unlike potassium it can form ternary calcium-aluminum oxides that are still more stable than just calcium oxide.  Thus part of the calcium from the lime is converted to these ternary compounds while the rest, not fitting in the ternary compounds, comes off as vapor.


Are there methods for iron content determination in spinach that do not require thiocyanate? I am looking for a way to measure the iron content of spinach for a chemistry experiment. Unfortunately I am not sure that thiocyanate will be available and I'm therefore trying to find an alternate method. The most common one seem to be a titration with potassium permanganate. The issue with this method is that other much more abundant metals in spinach would also react with the permanganate.
I had the idea to ash the spinach, dissolve the ashes in water. Then use a displacement reaction since iron is the least reactive metal in the chemical makeup of spinach. I am not sure exactly how but I thought it might be possible to use a compound to precipitate the more reactive ions out of solutions.
Do any of you have some ideas regarding this experiment?

There are iron testing kits in water which are available at an affordable price. I have not used them but read more about them here Testing iron in well water. You can test your spinach extracts for a yes no type answer if iron is present in significant quantities or not. Just keep in mind that PopEye's spinach is really not that rich in iron. Too much iron is a liver toxin.
I am afraid to share the bad news for you that with the resources you have, you cannot reliably determine iron content in spinach. Micronutrients or minerals cannot be determined by classical methods because they are present in a complex matrix and in very small quantities (parts per million range). They need instrumental analysis (absorption by molecular complexes, atomic emission or atomic absorption). I saw a couple of papers from ResearchGate which talk about the titration of spinach extract with permanganate. Nothing can be more wrong than this procedure in analytical chemistry. They are basically titrating all the organic stuff with it. Those methods are horribly wrong. Someone on SE Chemistry posted a query on how to determine iron in spinach by titration Iron in spinach...that is all incorrect. Yes, you will generate some numbers with those burette readings but those numbers have no significance.
Regarding your ashing method, that will not work either, because you need at several kgs of spinach to get a small amount (milligram) of iron. What should be the ashing temperature? How would you isolate iron from the ashes? Selective precipitation to get rid of other minerals? Nope, that is a bigger headache. For that you would need more reagents.
If this is a school project, make sure you have the resources after searching the literature. If this is an home experiment- classical methods will not work.


Calculating ratio of bicarbonate ion to carbonic acid in blood For the following reaction, $\ce{K_a} = \pu{4.2e-7}$ :

$\ce{H2CO3(aq) + H2O(l) <=> H3O+(aq) + HCO3-(aq)}$

We're given the pH of blood, which is $7.4$. We need to calculate the ratio ("$X$") of bicarbonate ion to carbonic acid in blood.

I know I need to convert the $\ce{K_a}$ to $\ce{pK_a}$, and did so as follows:

$\ce{pK_a} = -\log(\ce{K_a}) = -\log(\pu{4.2e-7}) = 6.4$

But I'm not sure how to proceed from here. Any help would be greatly appreciated.

For these types of problems, the general approach would be to use the Henderson-Hasselbalch equation. For a general weak acid/conjugate base system, this has the form: $$\mathrm{pH}=\mathrm pK_\mathrm a+\lg\left(\frac{\ce{A-}}{\ce{HA}}\right)$$ Since you are given $\mathrm{pH}$ and have found $\mathrm pK_\mathrm a$, it won't require much more work to obtain the ratio.


Refractive index of barium sulfate What is the refractive index of barium respectively where do I find it? I only find values for alpha, beta and x-ray radiation like on https://pubchem.ncbi.nlm.nih.gov/compound/barium_sulfate#section=Decomposition

I need it for all wavelengths or at least at 455 nm. Thank you in advance!

The index of refraction of $\ce{BaSO4}$ at 589 nm can be found in "Index of Refraction of Inorganic Crystals", in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.

Separate values are given for the indexes of the orthorhombic crystals:

$\begin{align}
n_x &= 1.6362\\
n_y &= 1.6374\\
n_z &= 1.6480
\end{align}$


why do we denature proteins before mass spectroscopy? I'm learning about mass spectroscopy in particular the use of it in analyzing protein structure. they said we have to use trypsinolysis to cut the protein to segments, but why does the protein have to be denatured first before trypsinolysis?

Trypsin is an enzyme, and has to bind to a target protein in a specific way in order to be active. Without first denaturing your protein, most of the protein will be conformationally/sterically inaccessible to trypsin, and you will only get partial digestion (or maybe no digestion at all).


Explain the difference between reaction intermediate and reactive intermediate Wikipedia definitions of the terms:

Reaction intermediate:

https://en.m.wikipedia.org/wiki/Reaction_intermediate

Reactive intermediate:

https://en.m.wikipedia.org/wiki/Reactive_intermediate

What I don't really understand :

Are they defined just based on their stability?

What I mean is that, if an intermediate is stable is it reaction intermediate and if unstable is it reactive intermediate?

(Or) is there any deeper difference between them?...After reading the entire page I have an intuitive feeling that there is more to them than just stability.

A reaction intermediate is nothing more than a plain old intermediate, which you probably understand. There is no stipulation as to how stable the intermediate must be. The IUPAC Gold Book defines intermediate as:

  A molecular entity with a lifetime appreciably longer than a molecular vibration (corresponding to a local potential energy minimum of depth greater than $RT$) that is formed (directly or indirectly) from the reactants and reacts further to give (either directly or indirectly) the products of a chemical reaction; also the corresponding chemical species.

I am not aware of a corresponding "official" definition for a reactive intermediate, but in my experience it simply refers to, well, intermediates that display exceptionally high reactivity. Typically these correspond to high-energy species such as carbocations, radicals, carbanions, or carbenes, which exist only transiently, and are difficult or impossible to isolate. A simple example is the carbocation intermediate in an SN1 reaction.

Obviously all intermediates are reactive to one degree or another – which is why they are intermediates and not simply thermodynamic products – but the term reactive intermediate is usually reserved for those which are particularly reactive. So, reactive intermediates are a subset of reaction intermediates.

As for an example of an intermediate which is not particularly reactive, consider the Staudinger reduction of an azide. The initial steps leading to loss of nitrogen gas are extremely fast and (very vigorous) effervescence is observed immediately upon addition of PPh3:

but the iminophosphorane, i.e. RN=PPh3, reacts with water only very slowly to produce triphenylphosphine oxide and the primary amine. Typically you need to heat the reaction mixture for a number of hours, and if you didn't add water it would probably be quite happy on its own. I would therefore consider the iminophosphorane to be an intermediate (or reaction intermediate if you prefer), but I would not refer to it as a reactive intermediate.


How can I tell if a molecule is polar or not? Consider $\ce{PCl3}$, $\ce{CH2Cl2}$, $\ce{HCN}$, and $\ce{SiO2}$. Which of these molecules are polar substances?

I appear to be lacking on the concept of "polar substance". I was just studying about Lewis structures and this question popped up in the book, but I'm not quite sure how is this related.

I think that a chemical bond is polar when one atom is more electronegative than another. But my basic understanding tells me that a chemical bond is non-polar only when the two atoms are the same (like $\ce{F-F}$ or $\ce{O-O}$). But all the above options have different atoms, so the answer would be "none", which is clearly wrong because that's not one of the options.

How can I tell if a molecule is polar or not?

Every sufficiently asymmetric molecule will be polar, but some more than others

The polarity of molecules is related to the polarity of bonds within the molecule, but just having polar bonds is not enough to create a polar molecule.
 Consider, for example, CCl4 and CHCl3. Carbon tetrachloride has 4 fairly polar bonds but they form a regular tetrahedron and the polarity of the individual bonds cancel each other out to leave a molecule which has no net dipole moment and is non-polar. But the chloroform molecule (which swaps a hydrogen for a chlorine) is fairly polar (~1.1 Debye–the unit used to measure dipole moments–usually abbreviated to D) because the symmetry is lower and the three C-Cl bonds now all point away from the C-H bond and their dipoles don't cancel out.

Similarly carbon dioxide is a linear molecule so the dipoles of the (fairly polar) C=O bonds point in exactly the opposite direction and so cancel out. Water, however is bent so the dipoles of the polar O-H bonds  do not cancel out leaving a polar molecule with a significant dipole moment (~1.8 D).

Saturated hydrocarbons are usually thought to be non-polar for most meaningful purposes. But those that lack symmetry still have measurable (but tiny) dipole moments. Propane $\ce {CH3CH2CH3} $ and isobutane $\ce {CH3CH(CH3)CH3} $, for example, both have small but measurable dipole moments (0.081 D and 0.132 D) so are, strictly, polar. In practice, though, these molecules have such low polarity that the effects of their net dipole are dominated by other non-polarity related intermolecular forces and can be though of as non-polar.

But the critical message is that the most fundamental feature that determines polarity is symmetry or the lack of it.

Also note that we are talking about discrete molecules here. One of your initial examples, $\ce {SiO2}$, is a network solid not a discrete molecule and the ideas won't apply.


Can vinyl alcohol react as a dienophile in Diels-Alder reactions? For my organic chemistry exam I needed to synthesize cylcohexanone and in one of the steps a Diels-Alder reaction was required. I chose to use vinyl alcohol followed by an elimination reaction to get the desired product.

My professor said that this could not work because a dienophile needs to be electron deficient and that vinyl alcohol is electron rich. Is there any argument I can make as to why the reaction would actually work? It was a reaction of diene + vinyl alcohol to yield cyclohexanol.

The actual answer he said was to use two dienes, one as the diene and the other as the dienophile followed by a few subsequent reactions ending with ozonolysis to get cyclohexanone.

Firstly, your Diels-Alder products are missing the double bond in the ring (formed between what was formerly C-2 and C-3 of the diene). You could always simply hydrogenate it down to the alkane, but I feel like this would present some selectivity issues in your professor's suggested synthetic route, especially since a terminal double bond is more likely to be hydrogenated than an internal double bond (sterics), and an internal double bond more prone to ozonolysis than a terminal double bond (more electron-rich).

Anyway, ron pretty much said what there is to say about "vinyl alcohol". It doesn't exist in any appreciable quantity, since the keto form acetaldehyde is much more stable (see another of ron's answers here):

Apart from what ron suggested (vinyl acetate), another alternative is to use 2-chloroacrylonitrile as the dienophile. The cyano group is electron-withdrawing, so the the usual requirement for an electron-poor dienophile is met. Treatment with hydroxide will unmask the ketone functionality, see for example the second and third steps in Corey's prostaglandin synthesis. (For a possible mechanism, see Mechanism for basic hydrolysis of α-chloronitrile to ketone?.) You can then hydrogenate the C=C bond chemoselectively to get the desired product, cyclohexanone.

2-Chloroacrylonitrile is what is called a "masked ketene". Ketenes don't undergo [4+2] cycloadditions such as the Diels-Alder reaction because they have a much greater propensity to react in a [2+2] fashion (even cyclopentadiene, an excellent diene in the Diels-Alder, reacts with ketenes via a [2+2]). So, these reagents are essentially surrogate ketenes: with one additional step (hydrolysis in this case), you can access what is formally a ketene+diene Diels-Alder adduct. There's a nice review of masked ketene reagents here.


Average Oxidation State - Utility? What is the purpose of calculating an average oxidation state for a certain element in a compound?

  Hence the “average oxidation state” of carbon in acetic acid is 0.

And

  the oxidation state value for carbon atoms in acetic acid is, however, computed as zero. This latter value represents an average of the values for the two carbon atoms derived by the rigorous assignment rules.

What does this tell me? Why not just stick with individual oxidation states? Is there something the overall oxidation state is supposed to tell me about acetic acid?

The only reason I can think of is that perhaps this average oxidation state of 0 for the carbon atoms tells me that acetic acid has some ionic character. Hence the oxidation state of 0 for carbon (as opposed to +4 for carbon in its elemental form).

It helps you when you want to balance a red-ox reaction by the mean of the oxidation number change. For example, if you want to balance the reaction of thiosulfate with iodine to tetrathionate and iodide, you realize that the average oxidation number of sulfur changed from +2 to +2.5. The oxidation half-reaction looks like this:
$$\ce{4S^{+2}-> 4S^{2.5+} +2e-}$$


Prostaglandin Nomenclature Prostaglandins are lipid substances derived from fatty acids, which act as powerful biochemical regulators.

In Wade's Organic Chemistry, they indicate that there are several families of prostaglandins (A, B, C, …) and that, according to this classification, a prostaglandin could be formulated as PGA1 or PGA2.

As an introduction it is very well explained, but I would like to go deeper into its nomenclature. I have searched on the Internet, but I have found nothing.

For example, the next prostaglandin is PGB1, but I don't know why. Nor would I know how to formulate one from its structure.

Each prostaglandin is named using the prefix 'PG' followed by a letter A to K depending on the nature and position of the substituents on the ring. PGA to PGE and PGJ have a keto group in respective positions on the ring which are further distinguished by the presence or absence of double bonds or hydroxyl groups in various positions in the ring(indicated by numbers or Greek letters*). PGF has two hydroxyl groups and PGK has two keto substituents on the ring. PGG and PGH are bicyclic endoperoxides. An oxygen bridge between carbons 6 and 9 distinguishes prostacyclin (PGI). Also, all prostaglandins have a hydroxyl group in the S-configuration on carbon 15 and a trans-double bond at carbon 13 of the alkyl substituent (R2).

*Numerical subscripts (1 to 3) are used to denote the total number of double bonds in the alkyl substituents, and Greek subscripts (α or β) are used with prostaglandins of the PGF series to describe the stereochemistry of the hydroxyl group on carbon 9.

So for example, if we write PGF2α, the F will indicate two hydroxyl groups in the cyclopentanone ring (F series), the 2 will indicate two double bonds, and the α will indicate that the hydroxyl grouping at carbon 9 is in the α configuration.

References

https://www.lipidhome.co.uk/lipids/fa-eic/eicprost/index.htm (@user55119's link)
https://www.glowm.com/section_view/heading/The%20Prostaglandins:%20Basic%20Chemistry%20and%20Action/item/312#2133
Norman A. Nelson, Journal of Medicinal Chemistry, 1974, 17 (9), 911-918
DOI: 10.1021/jm00255a001


Determine Cu+2 and SO4-2 concentration using Potentiometry? The title says it all. I've been researching as well but all I could come close was solid state ion selective electrodes. Could someone please help me out?

There should be two methods(or one) to measure the concentrations.

What is the range of the components? This is very important because this will determine if titration or photometry is used (or a combination of both).

Also interfering components should be known. For eaxample. Barium is known to react with a lot of other components (solubility table). Also EDTA reacts with oter components (Also depending on the pH (EDTA complexation constants)

Titration:

Copper and sulfate can be determined in one analysis using and copper ISE. This is done by adding an NH3 buffer (pH of solution = 10) and adding an exxess of Barium nitrate. Barium will react with SO4 and the free barium is titrated. Titrate the solution with EDTA. This first inflection point will be copper, the second one the free barium.

Keep in mind that EDTA reacts with a lot of other components. Also if the concentration difference (molar) between copper and barium is large, high titrant usages are to be expected

Photometry:

Copper can be determined using an variaty of chemical substances (e.g. cuprizone, Bathocuproine). The choise depends on range, interfering components ect..

Sulfate is possible. However, titration is preferred.


A chamber to control concentration of a vapor: Is there a name for this equipment? Mechanical engineer here so please excuse me if my question seems dumb to you!

I want to have a chamber $(\approx 40 \times 30 \times \pu{30 cm})$ to control concentration $(\pu{ppm})$ of a flammable material (e.g. alcohols or ketones) at a specific temperature $(T\approx \pu{24 \pm 10 ^\circ C})$ and pressure $(P\approx \pu{1 \pm 0.5 bar})$

What should this chamber be called? I need the the name to be able to search for it and also mention in my publication.

Candidate names so far:

Incubator --> mostly for biological entities
Reactor
Environmental chamber --> too expensive
Atmospheric controlled chamber
Autoclave --> for sterilization(high temperature/pressure)
Flash drum
Vapor inhalation chambers --> for animal testing

P.S. I have posted the question also on Reddit here and here. There are some good points mentioned there.

There is a book The Design of Controlled-atmosphere Chambers for the Study of Oxygen Toxicity, so I suggest "contolled-atmosphere chamber".


Liquid crystal current flow In a liquid crystal cell used for LCD the conductive substrates, usually made of ITO, are in contact with the liquid crystal. The question is: when a voltage is applied to the electrodes to reorient the molecules is there any current flowing through the liquid? The answer should be positive because LCs are conductive materials. But I know that the molecules should reorient themselves under an applied electric field for example between two plates like a capacitor even without being in contact with the LC. So why the electrodes are placed in contact with the LC and why LCDs consume electricity?

LCs materials are dielectric. But they are not perfect insulators and have some non-zero conductivity, allowing a leakage current to flow, slowly discharging the capacitor.
This is caused by mobile ions in the liquid crystal and inside the polyimide layer [2].
Infact a liquid crystal cell can be modeled electrically as this circuit:

               Rl
         _____WWW_____
         |           |
         |     Cl    |
     O--------||---------O

References:
http://www.lintech.org/comp-per/03lcd.pdf p.10-11
http://www.emdmillipore.com/showBrochure?id=200908.037 p.9


Predict photochemical decomposition of organic compound? Is it possible to predict wether an organic compound will photochemically decompose?

Also, is it possible to predict the products which it will decompose into?

Just as an experienced synthetic chemist can make an educated guess about how compound A and compound B will react, an experienced photochemist can usually do the same.  There are many, many well-known photochemical reactions that are mechanistically understood and sometimes even used in synthetic procedures, some commercially.  A few such reactions are the

Norrish Type I reaction

Norrish Type II reaction

Paterno-Buchi reaction

di-pi-methane reaction

photochemical induced halogenation, olefin dimerization and so many more.  So yes, an experienced photochemist can make reasonable predictions about which chemicals will react and what the possible products might be.


Is there an example where interpreting reaction coordinate as time leads to an incorrect conclusion? I have heard that we should not interpret the reaction coordinate in an energy diagram as time, but are there any situations in which we would arrive at an incorrect conclusion if we did so?

When I say interpreting reaction coordinate as time, I really mean as $\frac{\text{time so far}}{\text{total time for reaction}}$ (I would have written this instead of "time", but I'm not sure of a word or phrase for this).

Transition states are very short lived species, generally lasting less than a picosecond. At the single molecule level, crossing a transition state is a very fast event: it goes from reactants to products in less than a picosecond.

Some reactions are very slow because the probability of crossing the transition state is very low, but never because they spend a long time at the transition state. Thus, interpreting reaction coordinate as time is incorrect.

In a slow reaction, a small number of molecules will react early, but the majority will spend a long time in the reactants state, waiting to cross the transition state. If one could take a static picture of an ensemble of molecules, some would be in the reactants state, some would be in the products state, and it would be almost impossible to see any molecule at the transition state.


Gaussian calculation - PCM model for IRC compared to gas phase? Finding a reaction pathway thanks to Gaussian calculations is tricky exercise requiring a lot of resources for each step of this time-consuming process. Thus, man needs to wonder if each calculation has to be performed or not in order to gain as much time and resource as possible.
After achieving a 3-step mechanistic path for my reaction in gas, meaning finding all optimized structures, TS, verifications by frequencies of the stationary points and IRC, I now need to use the PCM model in my calculations to provide a solvent correction to my mechanism. However, I was wondering which calculations were required for this purpose.
From the 2 options I have in front of me, which is more consistent to avoid wasting time and resources ?

Perform new optimizations for the minima and TS in my pathway using the PCM model. Then calculate frequencies to be sure of my stationary points.
Perform new optimizations of the minima and TS AND perform new IRC calculations from the fresh new TS (after solvent correction) with PCM model.

The first option means that I perform an optimization from the reverse and forward structures (already further optimized) obtained from my IRC calculation in gas.
The second option means that I calculate a new IRC to obtain new forward and reverse structures that I have to optimize again.

Option 2 is the most scientific and rigorous. You should re-run the IRC in solvent again because the reaction path can be different from what you got in the gas phase.
Now, will it be different in solvent? Unless you are dealing with some exotic system, different reaction path with implicit solvation is unlikely (provided you have the same TS and minima with PCM). However, the reaction barriers may significantly change, which can cause two subsequent steps to become one single step (the TS between them forming a shoulder in the IRC). I have seen this happen.
What matters though is that different reaction paths can happen, so you should run the IRCs in solvent if you want to be sure of your result (also because you want to publish it).


What causes triboluminescence and what process occurs to produce light? I have observed this effect by rubbing together two pieces of quartz in a dark room but am unable to explain this effect. Would it be something to do with the way quartz is bonded as a covalent network?

I'm only a first year chemical engineering student, and we haven't really covered this topic in our course yet, but I will try and answer your question to the best of my knowledge.

(I am sure people more qualified me would provide better responses.)

When we apply certain stresses to a material such as a crystal, we separate charges (which is why it usually is asymmetric... the presence of anistropy allows charge separation.) These charges then can cause electric discharge, which ionizes the surrounding air and releases a flash of light.

There exist symmetric crystals which display this phenomenon as well... and it is suggested that (probably) impurities exist within such crystals which break the symmetry.

Anyway, this is my understanding of the phenomenon, and I don't claim is perfect in any way, but hopefully is helpful to you.


Uniform circular motion of electrons? Can we use the uniform circular motion formulae to calculate the velocity  of an electron along with the formula of radius of nth Bohr orbit? And using that can we calculate the kinetic energy of an electron?

Well yes you can if you assume the Bohr model is correct. However in reality, it is a little bit more complicated than that. In reality everything have both particle and wave like properties, which builds the foundation of quantum mechanics. The lighter the object, the more wave like properties it exhibits and the heavier the object, the more particle like properties it exhibits. So a ball actually has wave light properties but it is too heavy for it to be notices. Meanwhile a light wave actually has particle properties, however it is too light to be noticed. Electrons on the other hand turn out to a mass between these two extremes, hence both its particle and wave like properties are able to be noticed.Hence it is important to stop thinking electrons as small point particles.

Another important concept is the Heisenberg Uncertainty Principle which is the following formula: $$\Delta x \Delta p \geq \hbar$$ This means that the uncertainty of an object's position multiplied by the uncertainity in an object's momentum at best can only be equal to the constant value: $\hbar$ which equals $1.054*10^{-31}$ This equation can be applied to electrons. Hence, the important thing that this equation tells us about electrons in an atom is that rather electrons moving in a predictable circular orbit, they actually travel in any path that they want and we actually can't precisely determine position of the electron in an atom. Hence the quantum mechanical model of the atom actually consists of orbitals, rather than orbits.

Orbitals is the area which an electron is in 90% of the time. There are 4 main type of oribitals, s, p, d and f shown below.

So, finding the velocity of an electron is little more complicated than using simple classical mechanic equations. However its velocity can be determined in simplified conditions using quantum mechanics.


Significant Figures in almost-scientific notation For a homework assignment on significant digits, I have to find the number of significant digits in the number $0.40 \times 10^3$. I know that in scientific notation, the number of significant digits is equal to the number of digits in the stem. However, I've also learned that for a number to be in scientific notation, it must have a stem between $1$ and $10$. What is the rule for this situation? Is the number of Significant digits $2$ or $3$?

Most of the answers to your questions can be found in this wikipedia article. You are right if you say:

  I've also learned that for a number to be in scientific notation, it
  must have a stem between 1 and 10.

This is the convention in scientific notation. The number format of $0.40 \times 10^3$ is called exponential notation (at least it was called that way in a book about the programming language Fortran I'm reading at the moment): Here the number has a stem between 0 and 1 followed by a power of 10. This is quite common in computer science.
As to the significant digits: This wikipedia article (especially the paragraph "Scientific notation") contains the information you need. According to its information I would say that the number $0.40 \times 10^3$ has $2$ significant digits.


What is "mineral spirits"? I'm working in a project, and for a specific thing, I need to work on a possible recovery of $\ce{HCl}$ in aqueous phase. I found a patent on the net in which they use an amine to make a complex with $\ce{HCl}$.

They say in the summary:

  where a complex between the
  tertiary amine and the methyl chloride is formed, a hydrocarbon
  solvent, such as mineral spirits, is added to the phase containing the
  complex and the complex is then thermally dissociated.

I have never heard about that until yet.

I did some research on Sigma-Aldrich and I found that but I have not too much information, if you have ever heard about that can you please tell me a bit more about? Is that a molecule or a mixture?

Mineral spirits goes by many names, including "white spirit" and "paint thinner", and is basically just a hydrocarbon solvent mixture. According to this Wikipedia article:

  White spirit is a mixture of aliphatic and alicyclic C7 to C12 hydrocarbons with a maximum content of 25% of C7 to C12 aromatic hydrocarbons. A typical composition for mineral spirits is > 65% C10 or higher hydrocarbons,[6] aliphatic solvent hexane, and a maximum benzene content of 0.1% by volume, a kauri-butanol value of 29, an initial boiling point of 145 °C (293 °F) to 174 °C (345 °F), and a density of 0.79 g/ml.

There are also three different types based on whether or not it's undergone processing to remove sulfur, solvent extraction or hydrogenation. Then each type has a separate grade: low flash, regular and high flash grades. So with three different types each having three different grades, there are nine different sorts of mixtures that "officially" fall under the term white spirits, or mineral spirits.


Given the concentration and density of a sample, calculate moles of bleach? Given the concentration of bleach ($6\%\,\mathrm{m/m}$), density of bleach ($1.07\,\mathrm{g\,mL^{-1}}$), and molar mass of bleach ($74.44\,\mathrm{g\,mol^{-1}}$) how many moles of bleach are in a $4.0\,\mathrm{mL}$ sample? I don't understand where the concentration factors in, but the problem says to use it.

The key issue to address this kind of problems is to determine the mass of the pure compound in the given volume of the solution. So, we'll use first the density to determine the mass of the solution. Then, the mass concentration to determine the mass of the pure compound. After that, we can readily compute the number of moles of the pure compound using the molar mass.

The mass of $1\,\mathrm{mL}$ of bleach solution is $1.07\,\mathrm{g}$.
So, the mass of $4~\mathrm{mL}$ of bleach solution is $4 \cdot 1.07\,\mathrm{g}$

Or, each $100\,\mathrm{g}$ of bleach solution has $6\,\mathrm{g}$ of pure bleach.
So, the mass of  $4 \times 1.07\mathrm{g}$ of bleach solution has $\frac{4\cdot 6 \cdot 1.07}{100}\,\mathrm{g}$ of pure bleach.

The number of moles in  $4\mathrm{mL}$ of bleach solution is: $$\frac{4\cdot 6 \cdot 1.07}{100\cdot 74.44}= 0.0034\,\mathrm{mol}$$


When finding the liquid density which gives the most accurate results, a 25 ml Pycnometer or a graduated 10 ml cylinder? Its 10 ml graduated cylinder with 25 ml of pycnometer, and we are measuring the density of the liquid(pure liquid) in each of the materials and we are suppose to compare the densities found , in terms of accuracy

Excuse me guys am doing a a research project on density, these are the two methods i used to find the accuracy, been searching but i have no progress. Feel free to give suggestions.

Cylinder is always a "secondary standard" in volumetric measurements. It is never used in quantitative transfers. A pycnometer is designed to hold known volumes and it should always be used for density measurements at known temperature. Good quality pycnometers come with a certificate of quality assurance.

If you are interested in a deeper investigation, analytical glasswares come in two grades, Class A and Class B defined by so-called tolerances by various organizations. If you look closely you will see these labels on quality glassware. Similarly, look for TD and TC label on cylinders and other volume measuring glasswares. TD means to deliver and TC means to contain. If you don't see any A or B label, chances are that this glassware just good for educational purposes (or economical version). European union and other countries have set limits on volumetric accuracy.


Peroxide bleaching for bast fibres? Is peroxide bleaching ($\ce{H2O2}$) suitable for treating bast fibres? I read that to be successful, it needs to be combined with chlorine bleaching. Because of the end usage, I don't want to use this kind of bleaching.

Old patents of the 1930s, particularly from I.G. Farben often describe the use of hypochlorite/chlorine in the bleaching of bast.

However, from newer literature, such as this application manual for hydrogen peroxide bleaching from Evonik Industries, I would conclude that bleaching of bast can be performed without chlorine, supposed that the peroxide bleaching is performed

in alkaline medium
at elevated temperatures $(85-90\,°\mathrm{C})$


Are triglycerides soluble in polar solvents? In my textbook it is said that they are soluble in non-polar solvents (which makes sense), but also soluble in polar solvents (halogenoalkanes and some ethers like ethoxyethane) It is said that the polar solvents can induce a dipole in the fatty acid chain. If this is the case, why does water not induce a dipole in the chain?

Polar is not a black-and-white distinction. There are weakly polar compounds (like ether), medium polar compounds (dichloromethane) and strongly polar compounds (dimethyl sulfoxide). Fats (i.e. triglycerides) can dissolve in weakly and maybe moderately polar solvents but not in strongly polar ones.

While a dipole can be induced in triglycerides to some extend, most of the long nonpolar side chain can only give weak induced dipoles. Thus, weakly polar solvents may still dissolve these by dipole–induced dipole interactions but the triglycerides are just too nonpolar for strongly polar solvents.

Water is one of the most polar solvents and additionally able to hydrogen bond. While triglycerides can accept hydrogen bonds around the glycerin moiety, they are just not polarisable enough for water to act. Hence, water prefers to form its own phase.


Hydroxide as leaving group in Aldol Condensation in base My professor said that in the base-catalyzed aldol condensation, because there is a lot of hydroxide already in solution, hydroxide on the beta position on a ketone can act as a leaving group. What's the logic/reasoning behind this?

Quoting ORGANIC  CHEMISTRY by Jonathan Clayden  ,Nick Greeves , Stuart Warren ,these are elimination reactions .You cannot normally eliminate water from an alcohol in basic solution as hydroxide is a bad leaving group.It is the carbonyl group that allows elimination here: these are E1cB reactions, with a second enolization allowing the loss of OH−.

the base-catalysed aldol reaction sometimes
gives the aldol and sometimes the elimination product. The choice is partly based
on conditions—the more vigorous the conditions (stronger base, higher temperature,
longer time) the more likely elimination is to occur—and partly on the structure of the
reagents.

The key to what is going on is the carbonyl group.Negative
charges are stabilized by conjugation with carbonyl groups,The proton that is removed in this
elimination reaction is adjacent to the carbonyl group, and is therefore also rather acidic (pKa about 20).The anion that results is stable enough to exist because it can be delocalized on to the carbonyl group.

This next step is ,

The leaving group is not lost from the starting
molecule, but from the conjugate base of the starting molecule, so this sort of elimination,
which starts with a deprotonation, is called E1cB (cB for conjugate base).


Comparison of basic strength of benzylamine and ammonia According to me, ammonia should be the stronger base for two reasons:

$\ce{-C_6H_5}$ group shows $-I$ effect, due to which electron density on $\ce{N}$ atom of benzylamine decreases
steric hindrance in benzylamine is more than that in ammonia

But their $\mathrm{pK_b}$ values are: benzylamine: $4.70$ and ammonia : $4.75$. High $\mathrm{pK_b}$ value of ammonia shows that it is actually less basic than benzylamine, contrary to my reasoning. Why is it so?

Everything you said is correct, except that it is more relevant to consider the inductive effect of methyl on nitrogen: electron density is donated to nitrogen.

The inductive effect of phenyl is on the methyl, decreasing its ability to donate electron density to nitrogen, but apparently not completely.

I suppose that if you increased the negative inductive effect of the phenyl group with a couple of nitro groups, the basicity of the compound would decrease.


Effect of temperature on Hydrogen Bonding? We had this question in a test and I am not sure of the answer

  Q. Temperature changes don't effect the extent of inter and intra molecular hydrogen bonding. State true or false.

I think that raising temperature would increase the kinetic energy of molecules and would thus lead to weakening of inter molecular hydrogen bonds. But am not sure what will happen to intra molecular hydrogen bonds...

For e.g. if ortho-nitrophenol is heated, will the intra molecular hydrogen bonds between hydrogen of the hydroxyl group and oxygen of the nitro group break?

Any ideas/suggestions?

Yes. Temperature disrupts bond of all kinds. Heat up a protein hot enough and you can even disrupt its primary structure - the linear sequence of amino acids, and amino acids are held together through covalent bonds. This is why prion contaminated organisms and instruments must be heated to extremely, extremely high temperatures. The question is whether or not the bond will break; the question is simply what temperature.

Take for example nitrous acid, which exhibits two unique $\ce{O-N-O}$ bond angles. One of these bond angles (the smaller one) results from intramolecular hydrogen bonding between the hydrogen and the oxygen on the other end. This occurs because there free-rotation is permitted around the single bonds in the nitrous acid molecule. The energy difference between the two structures is calculated to be $0.45~\mathrm{kcal~mol^{-1}}$ or $1880~\mathrm{J~mol^{-1}}$.


Why do bubbles form when a reactive metal is placed in a less reactive metal's salt solution? What is the gas that if formed when a reactive metal is placed in a salt solution of a less reactive metal? Theoretically, it should just be a simple displacement.
I know that this question has been asked on a different page, but there Hydrogen formation was the accepted answer, which I don't think is the case (see observations).
Observations:
When a strip of magnesium ribbon was placed in a concentrated copper sulfate solution, the magnesium ribbon piece started bubbling.
Magnesium didn't bubble as violently when placed in plain water of the same temperature.
The gas was bubbled through a soap solution and the bubble was passed near a candle flame. It just popped and nothing happened.
Pure hydrogen gas was obtained by placing a Mg ribbon in acid, and bubbles were obtained. These exploded with a bang when passed over a flame (learnt the hard way not to do this!).
Regular distilled water was used everywhere.
Thanks a lot in advance for any help.

The bubbles are still hydrogen.  Transition metal salts are generally much weaker acids than whatever you put your magnesium ribbon into for your hydrogen control.  So you get less gas for the candle flame to react with when you use the transition metal salt, which I am sure you saw.  That the decreased amount of gas still popped, which is basically a low-intensity bang, should be deemed a positive test for hydrogen.


Stability of alkenes tri>di>mono, but how to explain tetra When we tell students about the formation of alkenes (by elimination for example), we often tell them that reactions will favour the thermodynamically favourable most substituted alkene. Zaistev's rule empirically described this:

  "The alkene formed in greatest amount is the one that corresponds to removal of the hydrogen from the β-carbon having the fewest hydrogen substituents"

The stability of the alkene with increasing substitution mono>di>tri is easily explained by hyperconjugation/qualitative MO theory (Phillip wrote a concise answer about that here). As an example to clarify, if theres a choice between eliminating to form a trisubstituted double bond over a mono substituted, the tri will predominate in most situations.

What I often struggle to explain in any detailed way is why tetra-substituted alkenes don't entirely follow the trend.  They are generally quite difficult to form (metathesis aside, but that isn't an elimination in the classical sense), and if formed, are generally quite reactive (on a similar order of magnitude to di-substituted alkenes if some studies on hydrogenation and some calculations are correct).

My general answer is 'sterics', based on the fact that 4 substituents around the same alkene is quite crowded, but I'm the first to acknowledge that this 1)doesn't really address all of the deviations from expected properties and 2)isn't really an explanation.

I agree with the OP and have seen numerous examples where tetrasubstituted olefins are indeed quite difficult to make, both in the literature and in my own experience, leading to the formation of Anti-Zaitzev (Hoffman) elimination products.
Many question writers are reluctant to make a solution where the tetra-substituted option is the correct answer. It is not as simple as the trend in carbenium ion stability where hyperconjugation and conjugation dominate around a trigonal planar (less sterically encumbered) center.
A tetrasubstituted olefin is too sterically encumbered due to 1) two A(1,2) interactions and 2) two exo substituents (gem-dimethyl) interactions.

Such proof for this failure in the stability trend is included in the original question as the kinetic stability of such an olefin is unusually and unexpectedly high. As far as the the thermodynamic stability, torsional strain makes effective hyperconjugation less possible in the transition state which varies only marginally upon the strength of base and the lability of the bond to the leaving group.

This trend: mono is less stable than exo-di is less stable than cis-di is less stable than trans-di is less stable than tri is....that is about all you can correctly say about the trend. Consider why cis-disubstituted is less stable than a trans-disubstituted olefin. Now consider a tetra-substituted alkene.

Zaitsev's "rule" like Markownikoff's "rule" is a just fast and dirty guideline for undergraduates and which both have numerous exceptions. It does not follow that tetrasubstituted must be more thermodynamically stable than tri-substituted and product ratios indicate this repeatedly in the literature. Stereoselectivity of E over Z is a non-problem compared to the difficulty of and abysmal yields when creating a tetrasubstituted olefin.

Simply put an ellipsis ("...") after the trisubstituted alkene as that is all that can be stated to an undergraduate class without presenting a falsehood.


Effect of concentration on rate of reaction I came across this problem:

  The $\ce{2NO + Br2 -> 2NOBr}$ follows the mechanism below:

  $$
\begin{align}
\ce{NO + Br2 &-> NOBr2} & &\text{fast} \tag{1} \\
\ce{NO + NOBr2 &-> 2NOBr} & &\text{slow} \tag{2}
\end{align}
$$

  If the concentration of both $\ce{NO}$ and $\ce{Br2}$ are increased two times the rate of the reaction would become

  A) 4 times
  B) 6 times
  C) 16 times
  D) 8 times

Now I know that overall order of a reaction is equal to the molecularity of the slowest step, hence order is 2 and since the concentrations of both the reactants are being increased by the same factor, i.e. 2 , the rate should be $2^2$ times irrespective of what is the order with respect to any reactant.

But the answer given is 8. How is this possible?

The fast reaction should be written as reversible reaction if $\ce{NOBr2}$ is a high energy intermediate:
$$\ce{NO + Br2 <=> NOBr2}$$

Then, the concentration of $\ce{NOBr2}$ can be estimated as:

$$[\ce{NOBr2}] = K_{eq} [\ce{Br2}] [\ce{NO}]$$

If the second reaction is an elementary step, the rate law follows the stoichiometry:

$$\text{rate} = k [\ce{NOBr2}] [\ce{NO}]$$

Substituting the concentration of the intermediate $\ce{NOBr2}$ into that rate law gives the overall rate law:

$$\text{rate} = k K_{eq} [\ce{Br2}] [\ce{NO}] [\ce{NO}]$$

and the dependency of the rate on doubling all reactants.

  Now I know that overall order of a reaction is equal to the molecularity of the slowest step

This is only true if the slowest step is the first step. Otherwise, you have to consider all the reactants that make the intermediates necessary for the slow step.

What if the first reaction goes to completion?
In this case, the concentration of the intermediate would depend on the limiting reactant. Changing all reactant concentration by a factor of two, the concentration of intermediate would increase by a factor of two as well (not like the factor of 4 if the first step is a pre-equilibrium).

What else do we need to know to solve this problem?
The problem description does not tell us the physical state of the reactants, intermediates and products. If $\ce{Br2}$ is a pure liquid, we would not be able to change its concentration. I would guess the intention is to have all species in the gas phase, though.


Reaction of formic acid with bridged bicyclic compound Write the mechanism for:

I've been able to add part of the product, but am having trouble with the ring formation. Here is my attempt:

I leave the arrows filling and details of last step to you since it is homework and online Marvin JS did not let me do arrows the way I liked to.


What happens when I use cyclohexanol to prepare cyclohexene? I've done a student experiment using cyclohexanol to prepare cyclohexene.

We mix $10.5~\mathrm{g}$ cyclohexanol and $5~\mathrm{mL}$ $85\%$ $\ce{H3PO4}$ in a $50~\mathrm{mL}$ flask, then distill while heating.

I want to know during this process what kind of side reaction happens because I notice there is something light yellow in the flask. After reaction, the liquid remaining in the flask is red and upon it there is something dark green and muddy.

I have a guess that there is some phenol in the cyclohexanol (since cyclohexanol may be prepared from phenol), and that during reaction the phenol oxidized, producing 1,2-benzoquinone (which is red) and 1,4-benzoquinone (which is yellow). Is this idea reasonable? If it is, why does the red color appear later?

And about the dark green thing: can it be produced by some kind of carbonized organic matter? I have been told that if the $\ce{H3PO4}$ and cyclohexanol were not well mixed, there would be some cyclohexanol carbonization.

(Can I ask one more question?) We are required to compare the chemical properties of cyclohexene and cyclohexane. So we reacted them with $\ce{Br2}/\ce{CCl4}$ respectively, under light or covered by aluminium foil. Should it be different when cyclohexene reacts under light or covered by aluminium foil? I notice the result is slightly different; the reactions in the light become a little opaque.

The acid-catalyzed dehydration of alcohols involves a carbocation intermediate, which can act as a Brønsted acid; the loss of a proton yields the desired cyclohexene. Furthermore, the carbocation may undergo other side reactions. In particular, the carbocation may be subject to a reaction with other molecules to give substitution products; or the carbocation may undergo carbocation rearrangement.

The coloured residue of the distillation is probably the result of cationic polymerization of cyclohexene. This reaction in combination with rearrangements may yield various oligomers.


Which gas is a oxidizing agent as well as a reducing agent? I think it's Hydrogen, because,

$\ce{PbO + H2 -> Pb + H2O}$

Here $\ce{PbO}$ is reduced to $\ce{Pb}$ and $\ce{H2}$ acts as a reducing agent.

But, in $\ce{2K + H2 -> 2KH}$, hydrogen gains a electron and undergoes reduction
Thus, $\ce{K}$ is oxidized to $\ce{KH}$ and hydrogen works as oxidizing agent.

My teacher says it's wrong but doesn't give any explanation.

Can someone explain why my reasoning is wrong.

I'm guessing your teacher is looking for sulfur dioxide as the answer, but I don't see how or why you're supposed to be able to arrive to this answer logically. Either you'd need to read about it specifically, or maybe you're supposed to stare at a table of standard reduction potentials and notice that $\ce{SO2}$ appears as both a reagent and as a product with comparable voltages:

$$
\begin{array}{}
\ce{SO2(aq) + 4 H+(aq) + 4 e- &<=>& S(s) + 2 H2O(l)} &  E^\circ = \pu{+0.50 V}\\
\ce{SO2(aq) + 2 H2O(l) &<=>& SO4^2-(aq) + 4 H+(aq) + 2 e-} &  E^\circ = \pu{-0.17 V}
\end{array}
$$

(The bottom equation is the inverse of what you would find in a reduction potential table)

Extrapolating from these aqueous potentials, it appears $\ce{SO2}$ is a modest reductant and a weak oxidiser, but both reactions can be accessed in reasonable conditions.

I think you're technically correct (as Ivan mentions), but what makes hydrogen gas a "less correct" answer is that it only rarely acts an oxidiser (in the presence of very strong reducing agents, such as alkali metals), whereas it can act as a reductant much more commonly. Compare these equations with the ones above:

$$
\begin{array}{}
\ce{H2(g) + 2 e- &<=>& 2 H^-(aq, extrapolated)} &  E^\circ = \pu{-2.3 V}\\
\ce{H2(g) &<=>& 2 H+(aq) + 2 e-} &  E^\circ = \pu{+0.00 V}
\end{array}
$$

(Reference for the hydride potential)

Note the much larger gap in potential between the two reactions. In this sense, the reducing nature of $\ce{H2}$ vastly overwhelms its oxidising nature, so most people will think of it as a reducing gas.


Electrochemistry molar conductivity Molar conductivity increases with decrease in concentration. This is because the total volume, $V$, of solution containing one mole of electrolyte also increases. It has been found that decrease in $\kappa$ on dilution of a solution is more than compensated by increase in its volume. Physically, it means that at a given concentration, $\Lambda_m$ can be defined as the conductance of the electrolytic solution kept between the electrodes of a conductivity cell at unit distance but having area of cross section large enough to accommodate sufficient volume of solution that contains one mole of the electrolyte. When concentration
  approaches zero, the molar conductivity is known as limiting molar conductivity and is represented by the symbol $E^\circ_m$. The variation in $\Lambda_m$ with concentration is different (Fig. 3.6) for strong and weak electrolytes.

Can anyone explain what does this sentence mean : "it has been found that decrease in k on dilution of a solution is more than compensated by increase in its volume"

Molar conductivity : $\kappa \times V$ ($\kappa$ is specific conductance), $V$ is volume of solution containing 1 mole of electrolyte.

I understand that molar conductivity increase with decrease in concentration. But at the same time this sentence is contradicting that because it says the decrease in "$\kappa$"(specific conductance) is more than increase in volume Which should decrease molar conductivity. I am confused.

I think you missed the word "compensated".

  The decrease in specific conductivity is more than compensated by increase in its volume.

This simply answers your question. The reduction in molar conductivity due to the decrease in specific conductivity is  overridden by the increase in volume.

It is saying that the latter overrides the former, and not just overrides, but overrides by a lot.


Is BF3 an electrophile? Boron has an empty $2p$ orbital.
But is it an electroplhile?
I know it is a Lewis acid.

Lewis acids are by definition electrophiles. Electrophiles love electrons, or negative charge. Boron has an empty 2p orbital and there exists a strong partial positive charge on the boron due to the extremely electronegative fluorine atoms covalently bound to boron. This strong partial positive character, coupled with a vacant orbital, makes BF3 a potent Lewis acid and thus an electrophile.

That being said, note that acidity and basicity are both thermodynamic properties, while electrophilicity and nucleophilicity are both kinetic properties. That is, while you can label a Lewis acid an electrophile, it would be incorrect to say, for example, that because a certain Lewis acid is a "strong" Lewis acid, then it must be a strong electrophile.

Formation Constants; Lewis Acidity