What is the chemical formula of KIT-6? I have seen KIT-6 used in literature as a catalyst support, but cannot find the chemical formula for it.

According to original paper (1), KIT-6 is a highly ordered cubic $Ia3d$ silica ($\ce{SiO2}$) templated by Pluronic P123 ($\ce{[EO20PO70EO20]}$) block-copolymer surfactant. As a source of $\ce{Si}$, tetraethoxysilane (TEOS) or sodium silicate are primarily used.

(1) Kleitz, F.; Choi, S. H.; Ryoo, R.; Chem. Commun. 2003, 0 (17), 2136–2137. DOI: 10.1039/B306504A


Protoporphirinogen IX formation in Heme synthesis This picture shows one of the step of heme synthesis. In this step, two propanoate groups of Coproporhyrinogen III are decarboxylated to from protoporphyrinogen IX. The enzyme used is oxidase. Oxidase enzymes catalyze the removal of hydrogen in the presence of oxygen to from water or hydrogen peroxide.

So, will a water molecule form in this step. 2H+ will come from the two carboxylic groups and oxygen will combine with it to form water. Will water molecule form in this way?

So, will a water molecule form in this step. 2H+ will come from the two carboxylic groups and oxygen will combine with it to form water. Will water molecule form in this way?

No.

Firstly, note that there are 4 more hydrogen atoms in the reactant than in the product (one from each carboxylic acid group and one where each double bond forms).

However, at physiological pH, the two protons on the carboxyl groups will not really be there.  The equation in the question also ignores the 4 electrons transferred.

As for the other two protons, a 5’-deoxyadenosyl radical abstracts a hydrogen atom (proton + electron) from each propionate side chain.

See Coproporphyrinogen III Oxidase Enzyme and references cited therein for more information.


Lattice Energy (enthalpy) I know that when making a bond the reaction is exothermic (positive)
and breaking a bond the reaction is endothermic (negative).

Lattice enthalpy or lattice formation enthalpy is defined as the enthalpy of formation of 1 mole of an ionic compound from gaseous ions under standard conditions, which is endothermic (negative).

My question is ..
Why lattice formation is negative? It should be positive when making the ionic lattice because we are releasing energy.

Is bond making/breaking different than lattice making/breaking?

You have to consider the system. If energy is being added to the system the sign is positive, if energy leaves the system the sign is negative.


Why can’t I use solubility product for normally soluble compounds? Solubility product only applies to sparing soluble ionic compounds (chemguide)

I read this in a website. But I don’t see what this should be the case. Taking $\ce{NaCl}$ for example, there’s an equilibrium established as it dissolves. Why then does solubility product not apply to this scenario?

In addition to Mithoron's and Poutnik's comments, the solubility product may not be relevant to your solution chemistry anyway.  Solubility products enter into solution equilibria when the precipitates are present, as in this answer where relative solubility products predict whether mostly magnesium or mostly iron will be dissolved in water when their hydroxides are present.  With readily soluble materials, you likely do not get the precipitate so solubility products, even if you can compute then on an activity basis, have no bearing on the equilibrium.


How were x-ray diffraction patterns deciphered before computers? I know the general idea behind x-ray crystallography is to take a high quality crystal and place it in the path of an x-ray beam. Areas of high electron density will diffract the beam and lead to spots on a detector screen where photons have constructively interfered with each other (and nothing appears where destructive interference has occurred). The crystal is then rotated and another diffraction pattern image is taken and the cycle repeats until there are enough views to reconstruct the atomic structure of whatever you crystallized. My question is what is the general process for doing this without computer software? How was this done in the 20s? The only equation I know for x-ray diffraction is Bragg's Law but is this the only equation used to interpret the data? Surely there must be others? How do you translate the spots on a detector to electron density plots using Bragg's law?

I agree with @andselisk that this question is quite broad. I will focus on two specific questions asked

  The only equation I know for x-ray diffraction is Bragg's Law but is this the only equation used to interpret the data? [...] How do you translate the spots on a detector to electron density plots using Braggs law?

Apart from Bragg's law (which tells you where the diffraction spots are for a known orientation of a crystal with know unit cell), it was also known that real space (electron density) and reciprocal space (diffraction pattern) are related by 3D Fourier transform. For structures containing one or two atoms, just knowing the unit cell parameters and the symmetry is enough to get the entire structure. For anything slightly more complicated, the Fourier transform of amplitudes or later, of intensities (Patterson methods, http://reference.iucr.org/dictionary/Patterson_methods), had to be used. The first crystal structure analyses were of crystals with centrosymmetry, where the phase problem is easier to solve.

  My question is what is the general process for doing this without computer software?

Diffraction data was measured on film, with gray-scales to assess intensity of signals. To calculate a Fourier transform, pre-computed tables were used, such as the Beevers-Lipson strips. As Andselisk commented, Fourier transform was used late in the 20s, and initially for problems that were one- or two-dimensional. There is a nice account by P.P. Ewald available here: https://www.iucr.org/publ/50yearsofxraydiffraction/full-text/structure-analysis
This was written at a time when the myoglobin structure had just been solved, after "50 years of X-ray diffraction".


Why is the carbon dioxide we exhale transparent? Why is carbon dioxide that comes from dry ice opaque and smoky while the carbon dioxide that we exhale transparent?

Essentially, because the carbon dioxide sublimates from solid (dry ice) to gas at a very low temperature (roughly −78 °C at 1 atm), it causes water vapour in the air to condense, causing a visible fog. Thus what you are seeing is not carbon dioxide, but rather water.

When we exhale and it is reasonably warm, the carbon dioxide expelled is roughly body temperature (36–37.5 °C) and so we do not see any fog forming.

What happens when we exhale in cold weather is a bit different. Our breath contains a high percentage of water, higher than the percentage of water that can stay in gaseous form at a low temperature. Thus the ‘extra’ water condenses into droplets.

For reference, here is the phase diagram of $\ce{CO2}$:

And here is the phase diagram of $\ce{H2O}$:

And here is the saturation fraction of water at various temperatures at 1 atm (sea level):


Why does ice cream make soda fizz? I've noticed that adding a chunk of ice cream to soda makes the soda fizz slightly near the soda-ice cream interface. I thought it was a physical effect due to the temperature, but adding ice has no effect.

It still can be a physical effect due to solubility, or it may be a chemical effect.

I'd like to know which it is, and the details of the mechanism.

Soda contains $\ce{CO2}$. Over time the gaseous $\ce{CO2}$ comes to the surface. The ice cream increases the rate of this  $\ce{CO2}$ from the soda causing increased fizzing. This increased rate is a result of the solvation of ice cream particles in the liquid. The particles serve as nucleation sites for the growth of gas bubbles. The bubbles get large quiet fast thus you observe fizzing.

After a bit of experimentation I found the following:

Effect of adding milk — There is a bit of extra bubbling and a curdy precipitate is formed. (Couldn't dare to drink it.)
Effect of adding salt (yes ice cream does contain salt) — Lots of bubbling around the salt. After the salt settles to the bottom, bubbles collect around it become big and then come to the surface.

Only thing left to try is sugar and vanilla essence but I ran out of soda.

So we come to know that the fizzing effect may be due to milk or salt or a combo of both.


Dependence of steric hinderance on enol content The following is extract from my book-

  To illustrate this, let us consider the enol contents of acetyacetone and $\alpha$ - methylacetylacetone. Although both the enols are stabilized by H bonding the enol form of $\alpha$ - methylacetylacetone is destabilized to some extend by steric repulsion due to the presence of $\alpha$ - methyl group.

What kind of H bonding (sorry I know i am being stupid, the figure in my book showed a bond connecting H and oxygen and dotted line connecting another oxygen and the previous hydrogen. How can hydrogen form two bonds?) is being talked about and what is steric repulsion? How does the steric repulsion destabilize the enol form?

How can hydrogen form two bonds?

In the case of an enol, hydrogen does not form two full bonds.  What the drawing below is trying to show with the enol hydrogen is that one bond (the O-H with the solid line) is elongated, partially broken, while the other bond (the O..H with the dotted bond) is also elongated and only partially formed.  It's as if the partially broken OH bond and the partially made OH bond are each "half a bond", so all together the electron sharing is such that there is still just one full bond to hydrogen

  How does the steric repulsion destabilize the enol form?

Look at the bottom part of the drawing where we have inserted an alpha-methyl group into the 1,3-diketone.  Notice the steric interactions indicated by the arrows.  The situation looks similar to that seen with cis-2-butene.  The methyl groups are large enough (larger than a hydrogen) and close enough to one another that their electron clouds "bump" into one another causing what we often call "steric destabilization."  Remember that trans-2-butene is more stable than cis-2-butene because the trans isomer does not have the adverse steric interaction present in the cis isomer.  It is that same type of steric interaction that occurs here with the enol when an alpha-methyl group is present, and just like in cis-2-butene, it destabilizes the enol.  Therefore less of the enol will be present at equilibrium when an alpha methyl group is added to the enol.


Oxidation state of Halogens So the question is which is the most common oxidation state of halogens and I have to justify my answer. Is my following thought correct? Do I have to add something?

The most common oxidation state of halogens is $-1$. Halogens are the most electronegative elements of the periodic table. Their outer electron configuration is $n\mathrm{s^2}n\mathrm{p^5}$. If chlorine, for example, gains one more electron, the outer p orbitals are completely filled (resulting in a full octet). Since halogens are electronegative they can easily remove an electron from the nearby atom. So then the halogen becomes isoelectronic with a noble gas and it is more stable in this state.

There is not one single most common oxidation state across all halogens. While fluorine is the most electronegative element of which compounds are known — meaning that it can only exist in the $\mathrm{-I}$ and $\pm0$ oxidation states — the electronegativity continually reduces down the group to iodine. Iodide in the oxidation state $\mathrm{-I}$ is easily oxidised to $\ce{I2}$ or $\ce{IO3-}$ with the oxidation states $\pm0$ and $\mathrm{+V}$, respectively.

What is true, however, is that all halogens except fluorine show oxidation states from $\mathrm{-I}$ to $\mathrm{+VII}$ as can be predicted from their electronic structure $n\mathrm{s^2}\,n\mathrm{p^5}$.


Could diatomic helium exist? I know that $\ce{He2}$ (homonuclear diatomic helium) does not exist because its bond order is zero. Since the bond order of $\ce{He2+} = 1/2$, that means that the positive $\ce{He2+}$ ion exists, but how does the positive ion exist if the neutral molecule doesn't?

Although we know that the helium dimer has a tiny bond dissociation energy, we don't consider it a molecule, as the "bond" does not survive a molecular vibration, even rotation of the molecule is enough to cleave it.

From an answer by Lighthart on "Molecular orbital theory & predicting the stability of a molecule?" I took the liberty to borrow the MO schemes:

In the scheme you can clearly see how we arrive at the bond order of zero for $\ce{He2}$. Both, the bonding $\sigma$ and the anti-bonding combination $\sigma^{*}$ are doubly occupied. Dihelium is only held together by weak van der Waals forces.
In the case of the dihelium cation $\ce{He2+}$, one electron is removed from the anti-bonding orbital. There are now more electrons in bonding than anti-bonding orbitals. This effectively results in a net energy loss or bond energy gain, and the overall bond order increases to 0.5. In a more phenomenological context, three negative charges are now in a field of four positive charges, this gives rise to Coulomb attraction.

Reference: Wikipedia article on dihelium and references therein.


What causes reactions to react the way they do? We were taught that there are primarily four kind of reactions, synthesis, combustion, single displacement, and double displacement. When the reactants are given to you, the reaction can be predicted.

For example:

If the reactants are AB + CD, it will be a double displacement (two pairs of compounds)

If the reactants are AB + C, it will be a single displacement (one pair and one compound/element)

If the reactants are A + B, it will be synthesis.

Why do the "format" of the reactants determine the type of reaction? Why can't, for example, AB + CD synthesize to create ABCD? Or AB + C synthesize to create ABC? Why is it assumed that if the reactants are in a specific "format" a specific reaction will occur?

If the reactants are AB + CD, it will be a double displacement (2 pairs of compounds)

An example of this might be two salts that dissociate into their respective ions in a solvent, and then the cation and anions that were not previously paired find their way to each other, if they form a salt that is not easily dissociated in this solvent, we have a new ionic compound (salt) which precipitates out as a visible solid in the mixture container. This might not always be the case, when we write net ionic equations we leave out anything that was not involved in creating the precipitate.

If we have the net ionic equation: $\ce{A^{+}_{(aq)} + B^{-}_{(aq)} -> AB_{(s)}}$ occurring when the solutions mix, then $\ce{AB}_{(s)}$ is said to be the precipitate. Notice that we leave out the other ions C and D because they were observers in the chemical equation, perhaps their attraction to each other is not strong enough to overcome the solvent. We should not always be able to tell whether this will happen or not unless we refer to experiments in which this substance had been seen to precipitate out before, perhaps even measuring its solubility in some experiments, the extent to which it will dissociate can also be calculated from experimentation.

  Why can't, for example, AB + CD synthesize to create ABCD?

If you are trying to determine if a covalent bond is compatible, one method is to use the octet rule and see what combinations might occur. For example, you have reactants of silicon and fluorine. What will form? Looking at the periodic table to see the electrons they contribute, draw a lewis structure for fluorine and silicon. Silicon contributes 4 electrons, and 4 fluorine atoms contribute the rest to give all elements full octets. Silicon tetrafluoride seems to be the most stable using this model. This is one method, but often times reactions occur the opposite way too. However, perhaps for some reason it is more likely for the reactants to be present in a given reaction, even after it seems to have stopped on a macroscopic scale (even thought it really has not stopped activity), we need to come up with more things to explain why this behaviour might happen. So as scientists, we devise new models.

This is a less than obvious and bigger topic than you might realise. A general chemistry textbook might have a chapter called chemical kinetics or reaction mechanisms which describe this ten times better than I can. Some things that will help you in understanding some models of reactions and their mechanisms are things like rate laws and how to determine rate constants in the laboratory, or how to determine the activation energy proposed by Svante Arrhenius in 1880's which basically means that two chemicals need to have at least a certain amount of energy (the activation energy) to overcome their repulsive behaviors and other limiting factors (the energy required to actually break apart a molecule's covalent bonds). For a collision of two particles to produce a different particle, there are other things that determine if a new molecule is formed from a collision of two molecules. First, like I said they must overcome repulsions with kinetic energy but they often need to be aligned spatially for a certain molecule to form in some proposed mechanisms. A mechanism is a proposition that an in-between step occurs in the reaction, but it is very unlikely that two hydrogen molecules will smash into a carbon dioxide molecule and create methane and oxygen. We must propose intermediate steps to make it sound more probable. although this would represent the reverse reaction of combusting methane, and will likely never occur because the potential energy of the system would increase, and we know how much the universe hates that kind of behaviour, if not then read on.

Some concepts that will also help you understand "why reactions happen" are some laws like the 2nd law of thermodynamics which states that the entropy of the universe is always increasing. This explains why a reaction in which a reaction of type that you describe $\ce{ABC + D -> ABCD}$ is actually quite possible. The problem is that the reverse reaction also takes place with a higher probability, these molecules might actually break apart more than they combine. This two-way speciality is why this reaction will likely be written as such: $\ce{ABC + D <=> ABCD}$ indicating that it does possibly happen, but can go both ways, and most often goes backwards. The reason for this is that there is more entropy (aka: disorder) when there are more moles of molecules present. In your examples the reactions are unlikely because the reactants ($\ce{ABC}$ and $\ce{D}$) are twice the number of moles than the products ($\ce{ABCD}$). This is a limiting factor for the reaction to commonly take place. It is useful to note that this reaction might occur once in awhile, but it is more often that molecules fluttering about will break apart than create larger and more complex molecules. Indeed, it is more likely that this reaction will be reversed and proposed as a mechanism of a larger process, if at all, where D breaks off and combines with something else, lets say E and ABC still remains along with a very unbreakable compound called ED which has a triple bond that will likely not be broken at the given temperature.

Lets talk more about entropy: You have a snowball you just rolled up in your hand (it is composed of many flakes of ice). Think of each ice particle that makes up the snowball as an atom in a larger molecule (the snowball). Chances are when this snowball starts moving (like molecules tend to do) it will run into something, and when that happens it will become many different smaller flakes and chunks (atoms and smaller molecules). This is because the universe prefers disorder and chaos. If you threw your clothes freshly out of the dryer into your closet, will they miraculously fold themselves neatley and in order? No, because that would make them more orderly and less chaotic, instead they will flail about in the air and land in a disorganized manner, this then makes sense by the phenomenon of the second law of thermodynamics.

Look back on your equations now, why can't $\ce{AB + CD -> ABCD }$ occur? Hopefully this is more obvious to you now. There is more disorder in the reactants than the products. This reaction might still occur, but normally balanced chemical equations represent the net affect that tends to occur. Perhaps occasionally someone will win at the game BINGO, but that does not necessarily mean that the same person will keep on winning at BINGO, more likely it will be someone else in the room, and then someone else, etc. With that BINGO analogy, the person who won at bingo played again, and lost. The same way ABCD might have been created, but then almost immediately was destroyed again.

Chemical engineers might desire to create your molecule $\ce{ABCD}$ and to do this what they might do is add a large amount of $\ce{AB}$ molecules into the reaction chamber. Due to the overwhelming number of these molecules, the reaction is more likely to take place. Not everything in the container will eventually be $\ce{ABCD}$ molecules, but there will be more of them that there was before. These kinds of reactions are actually called equilibriums meaning that the reaction can occur both ways. We write the equilibrium reaction as $\ce{AB + CD <=> ABCD}$ to explicitly show this. The quotient of reactants to products will not be 1:1. Sometimes the rate constant is very large for a reaction implying that it will go to completion, in which case we simply write the reaction in a single direction as you often see with chemical equations.

That is far as I am willing to explain because I am merely repeating snippets of what any chemistry book will explain in a chapter on chemical kinetics, hopefully my repetitive explanation will help you get a better idea of what is actually going on here, but there is still a lot more that you could learn or I will bother to explain.

Food for thought: In the equilibrium reaction $\ce{AB + CD <=> ABCD}$, A chemical engineer decreases the pressure of the reaction chamber, will the reaction shift more in favor of the reactants or the products? These are some questions that chemistry books will ask you after chemical kinetics are discussed. But I do not expect you to be able to answer it yet.


How do you change atomic valence in Avogadro? I have a bridging hydroxyl group between two metals, so the oxygen is triply coordinated in this structure.

How do I change the valency to three in Avogadro so that these bonds to the metal are displayed?

Perhaps it might be necessary to also play around with the van der Waals radius.
Or is there a direct way to enforce a drawn bond between two atoms?

While there are some automatic valence features in Avogadro, you can always turn them off. Bonding is mostly a matter of appearance.

For example, let's import $\ce{IF7}$.

Notice that the central iodine doesn't show enough bonds. So we'll draw some in. Switch to the draw tool and turn "Adjust Hydrogens" off.

Now just draw connections between the atoms. Voila!


Why don’t we have a list of all compounds yet? With only three basic particles making up atoms, and their interaction fairly well understood (we can even predict the chemical properties of elements we have not yet discovered) atoms seem to be like Lego bricks: only fitting together in a limited number of ways. So how come, in the age of super computers, we have not yet invented a program that can predict every single possible chemical compound (lets start modestly: of around fifty atoms or less)? Would a “super version” of such a computer program- one that would create a “master list” of all possible compounds and their reactions - theoretically make the field of experimental chemistry obsolete? I realize the vast number of possibilities, but just as with a chess game, the number of actual possible boards is far less than the theoretical number of all possible chess pieces and positions. Many simple compounds could be ruled out as “impossible” and a dead end, thereby considerably shortening the list of possible compounds. Am I missing something obvious here?

Because the combinatorial complexity is far to big for practical computation
One of the simplest possible cases that illustrates why this is impossible is given by restricting the choice of atoms to just carbon and hydrogen, forbidding any double bonds and disallowing any rings. This becomes equivalent to enumerating the number of possible mathematical trees of a certain structure a calculation interising in graph theory as well as chemistry.
Graphs are not exactly the same as hydrocarbons as there are spatial constraints on the molecules that can exist in 3D space and also because the molecules are in 3D, some variants with the same graph may be chemically different due to chirality (see this answer for more details). But Some calculations based on just 50 carbons put the number of possible variants at >1021 (see this wikipedia page) which is a ridiculously big number.
And this is is with just two elements from the hundred or so that could be used and it is ruling out most of the interesting structural features that could add a great deal of extra complexity. All this complexity arises from the carbon skeleton alone and it is already beyond any computational capacity we can contemplate having (just using one byte per compound would use a great deal more storage than at least one estimate of the total world capacity to store information.
So, if you still think this sounds like an easy task, you haven't done those calculations or appreciated the gargantuan scale of the resulting numbers.


CO : CO₂ ratio when given amounts of carbon and oxygen react with each other By the reaction of carbon and oxygen a mixture of $\ce{CO}$ and $\ce{CO2}$ is obtained. What is the composition of the mixture by mass obtained when 20 grams of $\ce{O2}$ reacts with 12 grams of carbon?

I wrote the equation as

$$\ce{4C + 3O2 -> 2CO + 2CO2}$$

and identified $\ce{O2}$ as the limiting reagent then the masses of $\ce{CO}$ formed will be 11.66 grams and mass of $\ce{CO2}$ formed will be 18.33 grams, which gives the ratio as 7:11, but the answer is given as 21:11.

Can somebody please tell me where am I going wrong?

You should have written:

$$\ce{C + O2 -> CO2}$$

and

$$\ce{C + \frac{1}{2}O2 -> CO}$$

Then you should have determined what fraction of the carbon reacts by by the first reaction and what fraction of the carbon reacts by the second reaction.


What can we learn from the energies of molecular orbitals? Molecular orbital theory is successful in calculating the structure of molecules (minimizing the total energy with respect to atomic positions). The solutions of an MO calculation are useful in deriving molecular properties such as dipole moment or vibration frequencies. These do not rely on energies of electrons in distinct orbitals, i.e. even though the theory is based on combining atomic orbitals to molecular orbitals, the somewhat arbitrary separation into orbitals is not used in these calculations.

On the other hand, many conceptual explanations of chemical properties and reactivity are based on orbital energies, such as HOMO-LUMO considerations, rationalization of electronic spectra, nucleophilicity etc. This answer suggests that the orbital energies of occupied orbitals are somewhat arbitrary and those of empty orbitals are almost meaningless. In the answer to a different question, there is a claim that you can apply orbital energies in just one way, and there is no guarantee how good the answer will be:

  Basically the only interesting quantity that you can get from an orbital energy is the first ionization energy based on Koopmans' theorem, which also is just approximately true.

So, what can you do with the orbital energies (either the numeric values or perhaps the ranking by energy)?

Not a full answer, but some general thoughts:

In general, what we can learn from individual orbitals (and their energies) is very limited, since orbitals are just an abstract tool to represent the many particle electronic wave function. This is manifested by the need for methods beyond Hartree-Fock (DFT, post-HF, etc) and the fact that the orbitals in these methods become even more abstract: Kohn-Sham DFT considers the orbitals to be non-interacting and moves the electron-electron into an external potential. In post-HF the ansatz for wave functions is extended in different (yet similar) ways, but ultimately still based on orbitals. Thus, an exact theory does not consider individual orbitals, but always combines them. In the context of such a theory, it is thus very hard to relate a molecular property to orbital energies only.

However, there still seems to be some rough correspondence, as MO diagrams are widely used to explain things like stability and reactivities. But I think this is all just empirical. Also, the fact that orbitals are an approximation means it does work on some scale. For example, we can distinguish between core and valence electrons based on the orbital energies, as those are typically well separated. (the cutoff energy may however depend on the system). But arguing which valence electron is more strongly bound is more difficult, as their orbital energies are more similar. But this is not really helpful, since we can separate core and valence electrons without applying quantum mechanics.

But we can extend that idea a little bit: Looking at differences in orbital energies, we can estimate the distribution of the electrons and predict a high-spin or low-spin situation of unpaired electrons (e.g. in transition metal compounds). Or the stability of radicals. Or the stability of bonds, e.g. in $\ce{C=C}$ bonds, the $\sigma$ bond is more stable than the $\pi$ bonds. We can even explain the more stable bonding $\ce{N2}$ compared to $\ce{O2}$, although we are comparing two systems with different nuclear charges and number of electrons here. But these arguments mostly work without looking at the actual numbers for orbital energies. It is mostly just qualitative.

Ultimately, we still know something about the orbital energies that we can use. The values are just very fuzzy (high uncertainty, large error bars), therefore we have to be careful with any conclusion. This may actually be the reason for how a large part of chemistry actually works: Often there are concepts which kind of work, but are not applicable in every situation, e.g. the HSAB concept. There may be other competing theories which cover other situations, but a full exact theory (other than elaborate quantum mechanics) rarely exists. In chemistry everything is a bit fuzzy and often it comes down to just know or test the properties of a certain compound. Physics on the other hand can be much more systematic.


Why does a real gas with attractive forces between particles experience an increase in internal energy when expanding isothermally? I am trying to understand at the molecular level why a real gas with attractive forces between particles experiences an increase in internal energy when expanding isothermally. Here's my thinking:

For a gas, the path-independent differential form of change in internal energy is

\begin{equation}
dU = \frac{\partial U}{\partial V}{ \delta V} + \frac{\partial U}{\partial T}{ \delta T}
\end{equation}

For an isothermal expansion,
\begin{equation}
{ \delta T} = 0
\end{equation}

so \begin{equation}
dU = \frac{\partial U}{\partial V}{ \delta V}
\end{equation}

(For an ideal gas, internal energy is proportional to temperature, so an isothermal expansion doesn't change the internal energy.)

The way I'm looking at the real gas situation is that:

When the real gas expands, say by free expansion into an evacuated chamber, the Δ U = Δ P.E. + Δ K.E.
The increase in volume causes the average distance between particles to increase and therefore increases the potential energy.
The increase in potential energy comes as the expense of a decrease in kinetic energy, because Δ U= 0 = Δ P.E. + Δ K.E., in the same way that throwing a ball upwards into the air causes it to lose K.E.
However, in the case of the gas, this is a very short-lived decrease in K.E., because the decrease in K.E. of the particles lasts no longer until they reach a wall.
Because the walls of the evacuated flask are maintaining an isothermal environment, they transfer heat energy to the particles inside, either by contact conduction, or radiation, thereby restoring K.E. to initial value.
U final = final P.E. (higher than initial conditions) + initial (restored) K.E., and is therefore increased.

(In the case of a volume expansion of a real gas by doing work against a piston, I believe that there would be an even larger transient drop in K.E., immediately counteracted by the introduction of correspondingly more heat because of the isothermal constraint.  So the end-of-pathway conditions are still the same, increased P.E. and restored K.E.)

Is this reasoning correct? The version of Atkins I have asserts that for a real gas
\begin{equation}
\frac{\partial U}{\partial V} > 0
\end{equation}

but doesn't provide a molecular explanation.

The internal energy change with temperature can be found from one of the Maxwell equations and is

$$ \left (\frac{\partial U}{\partial V}   \right)_T = T\left (\frac{\partial p}{\partial T}   \right)_V -p$$

and  a real gas can be represented  with the empirical van-der-Waals equation $\displaystyle (p+a/V^2)(V-b)+RT$ where a represents the attractive potential and b the molecular volume.

Performing the calculation gives

$$ \left (\frac{\partial U}{\partial V}   \right)_T = \frac{a}{V^2}$$

which shows what you note in your question.

As the gas expands work has to be done to separate the molecules because they have an attractive  potential between them, thus the internal energy has to increase  because the temperature is held constant. If the gas was thermally isolated during expansion it would cool and this is observed in the Joule-Thomson effect.


Why is maltose a reducing sugar but not sucrose, even though they're both disaccharides? I came across this on Wikipedia:

  Many disaccharides, like cellobiose, lactose and maltose, also have a reducing form, as one of the two units may have an open-chain form with an aldehyde group. However, sucrose and trehalose, in which the anomeric carbons of the two units are linked together, are nonreducing disaccharides since neither of the rings is capable of opening.

https://en.wikipedia.org/wiki/Reducing_sugar#Aldoses_and_ketoses

I took the following Haworth projection of sucrose for comparision.

That's all fine and dandy, but the thing is, I can't seem to find any source that (explicitly) explains why this whole 'ring-opening' business even happens in disaccharides like maltose, but not in sucrose.

The Wikiedia article probably implies that this has to do with the fact that both the monomers of sucrose are bound through their anomeric carbons, which is not the case with maltose. A quick look at the structures of maltose and sucrose confirms this, however; pardon me if I'm being dense here (I can't seem to think straight today...don't ask me why, I'm clueless myself), but I don't see how that piece of information helps.

Can anyone explain this disrepancy?

The difference in stability between maltose and fructose boils down to the different structural elements their aldehyde/ketone groups have been turned into during the formation of the disaccharide.

In maltose, you combine two glucose units using the 1-hydroxy group of one and the 4-hydroxy group of the other. The 1-hydroxy group has been created in a process called acetal formation by the attack of what used to be the 5-hydroxy group and is now the ring oxygen. This is shown in equation $(1)$, the explicit carbon atom being carbon-1; the aldehyde.

$$\ce{R'-OH + RCHO <=>> R'-OH^+-CHRO^- <=>> R'-O-CHR-OH}\tag{1}$$

Notice that there are two oxygens bound to that carbon after the reaction; if we wanted, we could label it $\ce{RCH(OR)(OH)}$; this structure is called hemiacetal. If we use that hemiacetalic hydroxy group to bond to a second glucose unit,[1] we have replaced the hydrogen with an alkyl residue. Like the difference between alcohols ($\ce{R-OH}$) and ethers ($\ce{R-OR'}$), this group has a different reactitivy and is termed full acetal or O,O-acetal.

Hemiacetals are rather easy to break down as two mechanisms are available: you can either initially protonate the other oxygen or you can use a base to facilitate proton removal of the hydroxy group — ideally perhaps while offering the other oxygen a proton by another base. Full acetals, however, are not as easy to break down; the require protonation, implying a sufficiently strong acid.

Coming back to maltose, remember we have two glucose subunits. We used one to generate a full acetal but the other one (on the right-hand side of your scheme) is still hemiacetalic and can reconvert back to an aldehyde. This is why you preserve the reducing abilities; remember that equation $(1)$ is fully reversible and according to Le Châtelier removing aldehyde (e.g. by oxidation) will enhance the reverse reaction.

In sucrose, on the other hand, carbon 1 of glucose and carbon 2 of fructose are connected. These are the two carbon atoms that form hemiacetals in isolated glucose/fructose. Because of this, there is no hemiacetal group $\ce{RCH(OR)OH)}$ in sucrose — only full acetals — and no way it can easily revert back to an aldehyde or a ketone. Therefore, it has lost its reducing ability.

This is the key information of your quotation by the way. The anomeric carbon is the one that will be a hemiacetal. If two anomeric carbons are connected, they have formed a full acetal. Therefore, identifying the anomeric carbons and seeing if they still have a free hydroxy group (i.e., if they are hemiacetals) allows quick determination whether a disaccharide is reducing or not.

Note:

[1]: While formally correct; the actual reaction mechanism involves the loss of this hemiacetalic hydroxy group.


Are words on my T-shirt removable? I wonder if the  words on my T-shirt are removable? If yes, what kind of (chemical or physical) ways are there?

They look like paints on the cloth, as the letter "C" has become blurred in the way shown in the picture over the time when I was wearing it. But I admit that I don't know anything about how words are printed on clothes generally.

As you say, it appears to be a kind of paint? So perhaps if you get a razor or something sharp you could try scraping and it will probably come off eventually, that would be one physical way. On the other hand if you want to go crazy try putting some paint thinner on there and see what happens, but don't blame me if you ruin your shirt experimenting on it, I have never used paint thinner before but I am guessing it consists of a non-polar compound and dissolves similar compounds (paint) so that it might be washed off with water. This is all hypothesis of mine, I have no idea what will happen except for what a razor can do.

Edit: I believe that Mr. Bean may have already demonstrated this experiment for us.


Why mustn't the sample be dipped in the mobile phase in paper chromatography? As the title says it, I am curious about why the sample (ie, an extract of plant's chlorphyl transferred onto paper) must not be dipped into the mobile phase (ie, the solvent which goes up the paper)?

My first guesses are that it is because the sample would dissolve into the solvent and only a small portion of the sample would go up the paper, but I am really keen to know if there is more to it than just this.

Dipping sample to a mobile phase would cancel all the purpose of chromatography as a separation technique.

Imagine properly performed chromatography as the athletic race, where all runners start at the same point and time. The finish order is by their average speed.

Now imagine a crazy version of a New York marathone, where runners start in  long sequence of runner batches. But nobody would  think about registration of runner starts, so there would be no idea which runner was the fastest to be decorated as the winner.

Back to chromatography, instead of chromatographic spots ( PC, TLC) or peaks (HPLC, GC), there would be useless, smeared, overlapping long bands.


Can i breathe pure oxygen gas? Sorry for the irrelevant tag because I was unable to find An appropriate one .Suppose there is a vaccum chamber and i filled completely with O2 without any traces of any other gases . Can a normal person breathe in it comfortably. I suspect it as pure O2 may act as oxidiser but since we need it for breathing . Can our organs manage that.

Breathing pure oxygen is potentially toxic. At reduced pressures, pure oxygen can be inhaled for days, or weeks. It is used to treat individuals suffering from stroke, heart attack, etc. and can be comfortably used for up to 24 hours. The body can be overwhelmed by oxygen during long periods or higher pressures, free radicals in the oxygen cause damage through oxidative stress which can result in tissue damage and death.


Which is most stable conformation of optically inactive butane-2,3-diol? The most stable conformation of optically inactive butane-2,3-diol is:

(A)
(B)
(C)
(D)

I chose C as the correct answer but, in the answer key it is given to be B. C is the staggered form and the most stable one. It can also have hydrogen bonding between -OH and -H groups. B can also have hydrogen bonding but there would be repulsion between the two Me groups. So, which one is correct?

Conformation B is probably correct because of the hydrogen bonding, if we assume this substance is either pure or in a polar solvent. The Me-Me gauche interaction is on the order of +3.8 kJ/mol (see most textbooks, or it is one half of methyl's A value, which involves two gauche interactions), while the hydrogen bond can be on the order of −5 to −25 kJ/mol. According to the Wikipedia article on hydrogen boning, the typical $\ce{O-H-OH}$ hydrogen bond has an energy stabilization of −21 kJ/mol. Since these groups are not quite properly aligned, a safe assumption is that the hydrogen bond interaction would be about half of that ideal value.

However, in a nonpolar solvent, there might be a benefit to conformation C, which has a net-zero molecular dipole moment. The entropic repulsion between the nonpolar solvent and B would have to be more serious than the two Me-OH gauche interactions (probably about +1.8 kJ/mol each). However, without that information, conformation B should be correct.

The correct answer cannot be A or D since they are representing a chiral diastereromer.


How are the hydrogens attached to the nitrogen atom in the ammonium ion? Naively attaching hydrogens to a nitrogen atom's electron orbitals results in a picture like this:

However, there are two problems with this drawing that I can see.

First, I have heard that the ammonium ion has a tetrahedral geometry so the hydrogens would have to be moved around and the orbitals would have to be stretched around or distorted for this picture to make sense.

Second, while I have drawn each of the hydrogens attaching to different external orbital I don't actually know which orbitals the hydrogens would attach to or why.

How are the hydrogens actually attached to the nitrogen atom in the ammonium ion?

I suppose one possibility might be to have the molecule be a hybrid of various $\ce{NH3 + H^+}$ configurations. I guess this could be tested by analyzing the charge distribution of ammonium and to see if the positive charge is more distributed around the outside or the inside.

The nitrogen atom, on both ammonium ion ($\ce{NH4+}$) and ammonia ($\ce{NH3}$), is $\ce{sp^3}$ hybridized. That's because we have 4 sigma bonds on ammonium ion and 3 sigma bonds + 1 electron pair in ammonia. The most stable orbital configuration for this set up is a tetrahedral shape.

Nitrogen can have a $\ce{sp^2}$ configuration if it has 3 sigma bonds and a pi bond, like in $\ce{H2C=NH2}$. The $\ce{sp^2}$ orbitals are "used" in the sigma bonds and the remainder $2p_z$ orbital is "used" in the pi bond. This gives a triangular plane shape to the molecule.

Finally, nitrogen can also have a $\ce{sp}$ configuration if it has 1 sigma bond and 2 pi bonds, in $\ce{N2}$ for example. In this case, the $\ce{sp}$ orbital is "used" in the sigma bond and the remainder $2p_y$ and $2p_z$ orbitals are "used" in the pi bonds. This gives $\ce{N2}$ a linear geometry.

Molecular orbital theory and hybridization are a little hard to process (at least for me it was and it's still being), so take your time to read some more about it, but at least with those examples you should have an idea of what to expect from some molecules geometries.


Replacing Salt Bridge with Platinum Sheet What would happen, if in a galvanic cell, the salt bridge is replaced by an inert platinum sheet?

Would charges accumulate on the sheet? Would the cell still work, and produce less emf than earlier, or stop working at all?

P.S.
I know why a salt bridge is necessary for the functioning of a galvanic cell - my question is rather specific to what would happen if I were to use a platinum sheet instead, and why we can't use it.
(Hence, not a duplicate.)

In continuation of the discussion above, I have another opinion:

  What would happen, if in a galvanic cell, the salt bridge is replaced
  by an inert platinum sheet?

  The cell would stop working in most cases. Opposite charges would accumulate in the two > half cells, opposing the reduction and oxidation half reaction.

I am afraid the statement is not correct. The set up in which Cu and Zn are partitioned by a platinum sheet, is equivalent to two galvanic cells connected in series.

The two cells
(a) Cu-Pt cell
(b) Pt-Zn cell

which are connected via conductor. In this case, the conductor is also platinum.

I just was reading LeBlanc's historical textbook on electrochemistry (lovely book from German translation) for another purpose. He shows a nice picture of two cells connected in series.

Consider a similar situation in case of Pt. However we may not be able to calculate the potentials because Pt is not dipping in Pt ions in either beaker. Instead, one end is in copper ions and the other end in the zinc ions. It still may show a positive potential (Ecell) just like how a lemon battery does (=Zn and Copper rods immersed in a citric acid solution).

This has been discussed before in physics. https://www.physicsforums.com/threads/copper-wire-as-salt-bridge.146868/ The cell still works! Charges can never accumulate like static electricity in a galvanic cell in the presence of a conductor.


A gram of salt added to pure (distilled) water introduces around $2 \times 10^{22}$ ions? I am currently studying Practical Electronics For Inventors, Fourth Edition, by Scherz and Monk. Chapter 2.5.2 Resistivity and Conductivity, claims the following:

  Adding an ionic compound in the form of common salt ($\ce{NaCl}$) to water increases the ion concentration within solution -- $\ce{NaCl}$ ionizes into $\ce{Na^+}$ and $\ce{Cl^-}$. A gram of salt added introduces around $2 \times 10^{22}$ ions. These ions act as charge carriers, which in turn effectively lowers the solution's resistance to below an ohm per meter. If we use the solution as a conductor between a battery and a lamp -- via electrodes placed in solution -- there is ample current to light the lamp.

How did the authors calculate that a gram of salt added introduces around $2 \times 10^{22}$ ions?

I would greatly appreciate it if someone would please take the time to clarify this.


Why do we use dG < 0 to describe a spontaneous process? With something like a reaction or phase change, all sources use the criterion that $dG < 0$ for the reaction to be spontaneous and then substitute an appropriate expression for $dG$ to the specific application. For phase change my book stats $(\mu'-\mu'')dn' < 0$ where $'$ is phase A and $''$ is phase B, $\mu$ is the chemical potential and $dn' > 0$ means gaining of molecules in phase A.

However, reactions require activation energy and phase change require nucleation (I think this is kind of like an activation energy); both of these processes require an increase in $G$ before a larger decrease. Why do we not use $\Delta G$ (between initial and final equilibrium states) instead?

Edit:

I am confused by the use of dG (a rate differential) as opposed to $\Delta G$ (a net change between end states) because I'm not sure how one would integrate this for say, a phase change, since it must pass through a potential barrier where G must increase. Is the integral just $G_2-G_1$ between the end states and thermodynamically the process ignores the barrier (since thermodynamics deals with equilibrium only)? A typical thermodynamic process I'm imagining is quasistatic compression, where we can get work from the integral of PdV and P is defined as it goes through an infinite number of equilibrium states; I'm not sure if it's possible to draw a parallel here for free energy and phase change/reacting systems.

Edit 2:

From this graph it looks like G increases then decreases to stable equilibrium after phase change. If we find dG/dr from this graph and integrate dG(r), should we get the same $\Delta G$ as from thermodynamics (before and after phase equilibrium)? If so, this seems to me like finding work of a rapid piston-cylinder compression of an insulated system; PdV is undefined since we are not passing through a set of equilibrium states but we can still find work from $\Delta U$. Difference is that in this case, our initial state is technically not in equilibrium/stable since the system favours a phase change, so we are beginning from a state that technically does not exist on the phase diagram (like if we have a supercooled/superheated liquid).

Maybe its best to see how we generally derive the Gibbs energy... If you think about this then the answer should be clear ;)

The Gibbs potential is a function of $P$, $T$ and $N$. That is to say:
\begin{equation}
G=G(T,P,N)
\end{equation}
Where $T$ is the temperature, $P$ is the pressure and $N$ is the number of particles.
\begin{equation}
dG=\frac{\partial G}{\partial P}\bigg|_{T,N}dP+\frac{\partial P}{\partial T}\bigg|_{P,N}dT+\sum _{i=1}^{I}\frac{\partial G}{\partial N_i}\bigg|_{P,T,N_j\neq i}dN_i
\end{equation}
Use some Maxwell relations (Look these up):
\begin{equation}
dG=VdP-SdT+\sum ^{I}_{i=1}\mu _i dN_i
\end{equation}
By definition we state the following formula at constant temperature and pressure:
\begin{equation}
dG=\sum _{i=1}^{I}\mu _idN_i
\end{equation}
Therefore you are actually just expressing the Gibbs energy in a different way when your book quotes the chemical potential. In fact the very definition of the chemical potential is:

\begin{equation}
\mu _i=\bigg(\frac{\partial G}{\partial N_i}\bigg)_{P,T,N_{j\neq i}}
\end{equation}

I don't know your mathematical ability but if it helps...its pretty clear that this is just a simple Legendre transform of the internal energy to switch between $U$ and $G$.
(As an aside), with a little product ruling we get:
\begin{equation}
dG=\sum _{i=1}^{I}\big(\mu _idN_i+N_id\mu _i\big)
\end{equation}
This leaves us with the Gibbs-Duhem relation:
\begin{equation}
\sum ^{I}_{i=1}N_id\mu_i=-SdT+VdP
\end{equation}
Which is a handy result.

Also just to add on the end here about nucleation points... lets imagine a precipitation in solution....As the size of the nucleus grows (as more bits "clump" together) the Gibbs energy of the particle will rise ... this is due to two competing effects:

1) the interface interaction (increasing energy)

2) the volume free energy (decreasing energy)

The total energy is just the resultant of these processes ... at small size the surface interactions outcompete the volume energy liberation .... as the radius of the particle increases we see a switch.


What happens to an enzyme after it breaks down a starch molecule? I am a master distiller at a distillery and I am about to make some whiskey. First adding corn, cooking it and releasing it's starches. After that, will add in some malt, packed with enzymes to break down those starch molecules into sugars. My question is, what happens to the enzyme after it breaks down the starch? can 1 enzyme breakdown multiple starch molecules?

Enzymes are a form of catalyst. Catalysts do not get consumed in a chemical reaction. I believe in reality, enzymes do degrade overtime and you need to replenish them overtime to keep the reaction efficient.
Can one (1) enzyme break down multiple starches? Sure.
I did not find specific literature on fermentation, but I found one (1) carbonic anhydrase in our tissues and bloodstream can hydrate 1,000,000 carbon dioxide molecules per second. So your enzyme probably has a similar magnitude of catalytic rate.
Reference
Berg JM, Tymoczko JL, Stryer L. Biochemistry. 5th edition. New York: W H Freeman; 2002. Section 8.1, Enzymes Are Powerful and Highly Specific Catalysts. Available from: https://www.ncbi.nlm.nih.gov/books/NBK22380/


What is the best way to detect the presence of tear gas? I realise that tear gas can be several different things (Pepper spray / CS Gas / CR gas, PAVA, etc)

Is there a good way to detect some or any of these in the air?
Ideally using an electronic sensor that can be reused?

Owlstone make an electronic multi-agent detector - I don't know how much it costs but I would guess somewhere in the range \$10K - \$100K. There might be cheaper single-agent electronic sensors or chemical tests.


Synthesizing 2-bromoaniline from bromobenzene I am trying to synthesize 2-bromoaniline from bromobenzene. I came up with a reaction path, but I want your thoughts on whether this is feasible and can produce a decent yield.

Sulfonation of bromobenzene (to block the para position)
Nitration of the product from (1). This should produce 4-bromo-3-nitrobenzenesolfanic acid.
Reduction of the nitro group via Zn(Hg) and HCl (to turn the NO2 group into NH2).
Reverse sulfonation (to get rid of the sulfonic acid group).

And the final product from this should give 2-bromoaniline. Should this theoretically produce a decent yield even if bromobenzene is slightly deactivated?

It will be much quicker to run the direct nitration of bromobenzene and isolate the ortho nitro product by chromatography. It is the minor nitration product, but your overall yield will be better by cutting out two steps which will give only moderate yields.
Nitration and chromatography procedure here 1


Is Beer-Lambert law also valid for fluorescence and luminescence? When I measure optical density in a microplate reader the absorbance is proportional to the concentration and the height of the liquid. Is Beer-Lambert law also valid for fluorescence and luminescence?

Let's say I prepare a fluorescent solution and measure e.g. the signal of 100 μL and of 50 μL. Do I get the half signal for 50 μL?

No the Beer Lambert law  ($I_{trans}=I_0\exp(-\epsilon[C]L)$) does not apply as this shows that the transmitted light intensity decreases exponentially with concentration as it passes through a solution.

Fluorescence/luminescence is directly proportional to the concentration of molecules excited provided that the solution  is dilute so that dimers or excimers are not formed and that self-absorption of the fluorescence light by another part of the same solution does not occur.


Drawing Constitutional Isomers Are these all valid isomers for the formula $\ce{C_3H_6O}$? I drew the first five; my friend came up with the last three, which have double bonds. Are these all isomers or different? More importantly, are there any more?

Also in general, how can one come up with all the isomers for a molecule? I heard of graph theory, but I think that's a little bit too advanced for me. I guess the only way is to draw out as many as you can and name them to weed out redundant structures?

They are all constitutional isomers.  If you are counting enols (your friends #3, it is the enol of your #1), then include the enol of acetone as an additional constitutional isomer.


Bonding and anti-bonding orbitals in the light of time-dependent Schrödinger equation? In organic chemistry, people draw 2p orbitals like this:

and then they explain how the orbitals combine to non-bonding (π*) or bonding (π) molecular orbitals, like this:

depending on whether the orbitals overlap out-of-phase (blue + white) or in-phase (blue + blue, or white + white).

But in reality, the orbitals are time-dependent solutions of the time-dependent Schrödinger equation, so they are not really like the first image above, but more like this:

So, what does it mean, when the orbitals overlap in-phase or out-of-phase? As far as I understand, the phase difference of the orbitals of two random atoms could be anything from 0 to 2π, and the cases of perfect overlap (0 phase difference) or perfect anti-overlap (π, or half-cycle, phase difference) should be quite improbable.

As F'x has indicated, the sum of atomic orbitals $\phi_k$ (AO) that form the molecular orbital $\psi_j$ (MO)
$$ \psi_j = \sum_k c_{jk} \phi_k $$
is just a solution (an approximate solution) to a self-consistent variational procedure to solve the time-independent Schrödinger equation, obtained after choosing a particularly well suited basis to represent the solution. They approximately solve this equation:
$${\sf F}\psi_j = \epsilon_j \psi_j$$
where ${\sf F}$ is the Fock operator or if you wish, an effective one-electron energy operator and $\epsilon_j$ is an effective "one-electron" energy that gives a good approximation to the ionization potential from that orbital.

Then you have to construct your overall n-electron wave function $\Psi({\bf x}_1,\ldots,{\bf x}_n)$, which in the simplest case (the one that follows from the minimal molecular orbital theory) is a Slater determinant. You can notice that for this wave function the coefficients $c_{jk}$ are not unique. Following the general properties of determinants, by a similarity transformation of the basis you can get a new set of coefficients $c'_{jk}$ that will not solve the Fock equation, but after which you obtain the same overall electronic wave function.

In everything that I have said so far, there are no time-dependent phases: we only have one "real" (observable) energy, an eigenvalue of the time-independent Schrödinger equations for the electronic Hamiltonian
$$ {\sf H}_{elec} \Psi({\bf x}_1,\ldots,{\bf x}_n) = E \Psi({\bf x}_1,\ldots,{\bf x}_n)$$
for a fixed geometry. Now is where you can put the time-dependent Schrödinger equation into work. In principle you could create electronic superposition states combining $\Psi({\bf x}_1,\ldots,{\bf x}_n)$ of different energy. These are the ones that would evolve in a manner similar to what you have assumed, with time-dependent evolving phases.


Is the QST2 method suitable to find the transition state of an enatiomeric R-S interconversion? I am trying to calculate the energy barrier between the two possible enantiomeric components of a polymer with Gaussian. I wonder if I am able to locate a transition state between two of those structures with the keyword opt=QST2?

Maybe, but more likely no. Depending on the complexity of your molecule the guess from QST2 is not good enough. Additionally you might run into some unforeseen coordinate system troubles.

As a small demonstration I'll use $\ce{CHFClBr}$ and the DF-M06L/def2-SVP level of theory and Gaussian 09 Rev. D.01.

I'll optimise one conformation and then just exchange the $y$ with the $z$ column.

%chk=m06ldef2svp.chk
#p M06l/def2SVP/W06
gfinput gfoldprint iop(6/7=3)
symmetry(none)
opt

title: optimisation

0 1
C        0.000000000      0.000000000      0.000000000
H       -0.530433945      0.045629756     -0.953288460
F       -0.905627118      0.105968719      1.035862034
Cl       1.161565217      1.332282245      0.093221601
Br       0.942449903     -1.690284863      0.135370837

That gives the following file for the transition state search. It is important, that the ordering in the two input blocks is identical, otherwise the program will fail horribly.

%chk=m06ldef2svp.chk
#p M06l/def2SVP/W06
gfinput gfoldprint iop(6/7=3)
symmetry(none)
opt(QST2, maxcycle=50)

title: optimised conformation

0 1
 C     0.004249     0.005882     0.022885
 H    -0.510890     0.037500    -0.945427
 F    -0.880186     0.102687     1.016533
Cl     1.139127     1.361087     0.088671
Br     0.915653    -1.713561     0.128503

title: yz exchanged

0 1
 C     0.004249     0.022885     0.005882
 H    -0.510890    -0.945427     0.037500
 F    -0.880186     1.016533     0.102687
Cl     1.139127     0.088671     1.361087
Br     0.915653     0.128503    -1.713561

In this case the calculation will produce the following error:

 GradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGradGrad
 Berny optimization.
 Bend failed for angle     2 -     1 -     5
 Tors failed for dihedral     2 -     1 -     5 -     3
 Tors failed for dihedral     2 -     1 -     5 -     4
 FormBX had a problem.
 Error termination via Lnk1e in /path/to/g09/l103.exe at {a date}.

The problem is that in any case of switching stereo centres, you most likely have have an $\approx180^\circ$ angle at one point and then the automatic generation of redundant internal coordinates will fail.
You can (maybe) prevent this at a much higher computational cost by optimising in Cartesian coordinates. Alternatively you specify the redundant coordinates by yourself, but even then it is probably best to use QST3.
In some rare cases, such calculations can actually give you a starting structure for a real TS search.

In any case, a little chemical intuition is necessary and often a single step mechanism is much, much higher in energy than a step-wise path. Sometimes it is better to use your imagination and guess a transition state (which you can then optimise with very low level tools like force fields or pm6), and refine later on.

The example I chose is certainly an extreme one. In this case isomerisation will likely happen via $$\ce{^{$S$}CHFClBr <--> [CHFCl]+ + Br- <--> ^{$R$}CHFBrCl}$$ or similar, and a transition state won't be easy to find.


Synthesis of oxalyl chloride that doesn't involve phosphorus I have been synthesizing TCPO (bis(2,4,6-trichlorophenyl) oxalate) for chemiluminescent reactions, using 2,4,6-trichlorophenol and oxalyl chloride.

Since I am tired of paying for oxalyl chloride (and hazard class 3 shipping), I wish to find a synthetic route to produce my own stock.

The most common synthesis of oxalyl chloride uses oxalic acid and phosphorus pentachloride. This would reduce the cost, but is also problematic because phosphorus chlorides are considered to be drug precursor (21 CFR 1310.02), still too expensive, and hazardous to ship.

Can anyone recommend a synthetic route to oxalyl chloride that does not use phosphorus? Or an alternate (safer / easier / less expensive) route to TCPO?

Most amide coupling agents are suitable for esters. Feast your eyes on: Amide bond formation - beyond the myth of coupling reagents.

This lists old favourites such as CDI, carbodiimides, etc. A lot of it is overkill. The water soluble carbodiimides might be preferred for ease of urea removal later.

The likes of CDI or $\ce{Ac2O/DMAP}$ can react with your phenol when over-stoichiometric amounts are used.

Alternatives could be: boronic acid derivatives e.g.  phenylboronic acid catalysis; scandium or yttrium triflate catalysts (all requiring water removal, but a Dean-Stark could be used if a water immiscible solvent is used); dimethyl oxalate (already suggested)  but in conjunction with Takasago's ZnTAC24 catalyst. MeOH removal by distillation is required, which is quite trivial for such a volatile component.


Does ethanol "decompose" or "remove" phosgene from chloroform? A question I was asked was: "Why is ethanol added to chloroform bottles?". I know that chloroform can be slowly converted to poisonous phosgene gas, and that ethanol is added to prevent phosgene levels from rising. However, the options I was given were:

It decomposes phosgene by converting it into ethyl carbonate.
It removes phosgene  by converting it into ethyl carbonate

Is there a difference between the two wordings, and is one more correct than the other? The answer is supposedly "decomposes".

It's a weird way of phrasing it, but I think they are trying to lead you to this

Basically, chloroform is stabilized with ethanol or amylene, but if a sample is dried, it will no longer contain stabilizer. To actually remove phosgene, you can wash it with NaHCO3. This would essentially leave you with entirely harmless products.

This is my best guess as to what they are trying to get you to understand. Please let me know if this helps.


Can gamma radiation cause transmutation? In irradiation of food for sterilisation, is gamma radiation absorbed by the food? If so, can it theoretically cause production of radioisotopes? Or does it ionise atoms in the food?

Yes.

See Photo-Fission in Heavy Elements (1947) Phys. Rev. 71, pages 3-10 :

  fission should be possible for all heavy nuclei which lie well
  beyond the minimum of the packing fraction
  curve, provided sufficient excitation is provided
  to produce the necessary deformation of nuclear
  fluid which precedes division of the nucleus.
  Such excitation can be provided by particle capture or by absorption of a $ \gamma $-ray into the
  nucleus.

"Packing fraction curve" is an older term for the nuclear binding energy curve, so the reference is saying that gamma ray induced fission is possible for nuclei that are substantially larger than nickel.

On the low end of the binding curve, there is a different process called photodisintegration.

  For example, magnesium-25, upon absorbing a photon of sufficient energy, emits a proton and becomes sodium-24.

Even the most stable nuclei such as iron and nickel can be disintegrated by energetic gamma rays.

See Photoneutron Thresholds Phys. Rev. (1951) vol. 84, pages 387-394 for quantitative information concerning how much energy a photon must have to photodissociate a given nucleus.


Does space in between have any significance in IUPAC nomenclature? The correct IUPAC name of the following compound is:

(a) 2,4,6-trimethyl cyclohexanone
(b) 1,3,5- trimethyl benzophenone
(c) trimethylcyclobutanone
(d) 2,4,6-trimethylcyclohexanone

I was confused between option (a) and (d). I opted for option (a), but our teacher told that ans is an option (d). The only difference I see between these two options is that there is space in (a). Does space have any significance in IUPAC nomenclature?

$\boxed{\textbf{d}}$ is correct. It is the standard IUPAC name.

However, "2,4,6-trimethyl cyclohexanone" will often be understood as 2,4,6-trimethylcyclohexanone, similar to how, in theory, neopentane, 2,2-methylpropane and 1,1,1,1-methylmethane are all equivalent.


Orbitals and Pauli Exclusion Principle without Slater determinants I'm trying to understand how to apply the pauli exclusion principle to the Born-Oppenheimer approximation of molecular structure.

Let me preface this by saying I'm not a chemist nor physicist. I'm a mathematician by trade which has the side affect of understanding electronic structure as solutions to a PDE.

So given some set of fixed nuclei and a given number of electrons I can construct the Schroedinger equation:

$$ \left(-\frac12\sum_{i=1}^N\Delta_i^2 - \sum_{i=1}^N\sum_{A=1}^M\frac{Z_A}{|r_i - r_A|} + \sum_{i=1}^N\sum_{j>i}^N \frac{1}{|r_i - r_j|} \right) \Psi = \varepsilon \Psi $$

I've read in Szabo and Ostlund that the pauli exclusion principle manifests itself by saying the wavefunction $\Psi$ must be antisymmetric in spin+location coordinates.

The next thing most people do is to introduce Slater determinants from 1-electron orbitals. This is an approximation to the solution space of wave functions, evidenced by the fact that FCI procedures must assume a sum of slater determinants for more accurate solutions. I want to keep this theoretical, so for now I don't care much for computational considerations.

The problem becomes as such, when talking about slater determinants it makes sense to append a spin variable and to talk about spin-orbitals of single electrons. Without slater determinants, single electron orbitals don't make sense (we only have the N-electron Schroedinger equation). Appending a spin coordinate also doesn't make sense because anti-symmetry of $\Psi$ and the Schroedinger equation don't see it.

So please help me out here, assuming I can solve the electronic Schroedinger equation exactly, what conditions do I need to place on the wavefunction so that the Pauli Exclusion Principle is satisfied (giving me a physical solution)?

I think what you are asking is "What form does $\Psi$ need to take so that the exchange interaction exists when the electronic Hamiltonian given acts upon it?". (Noting that the exchange interaction is due to the fact that electrons are indistinguishable and antisymmetric with respect to their positions).

From what I understand the Hamiltonian you have given acts upon electrons and so does not include the exchange interactions. As far as I'm aware the analytical form of the electronic Hamiltonian that includes exchange is not known. If it were, then density functional theory would not have so many approximate exchange-correlation functionals.

I will try to show the problem by presenting the form of the Schrodinger Equation in a slightly different way than is presented in Szabo and Ostlund's book.

Firstly, it is understood that $N$ in the Hamiltonian you give in the question represents electrons - not spin orbitals and we know that the electron density $\rho$ at $\textbf{r}$ is represented: \begin{equation} \label{eq1} \rho(\textbf{r}) = |\Psi(\textbf{r})|^2 = \langle \Psi|\textbf{r}\rangle\langle \textbf{r}|\Psi \rangle = \Psi^*(\textbf{r})\Psi(\textbf{r}) \end{equation}
The Hamiltonian you give acts upon the electron density by using spin orbitals to represent it.

Note, a physical electron density must therefore arise from a wavefunction that is anti-symmetric.

The Hamiltonian you have given can be seen to act upon the wavefunction as follows:
\begin{align}
E = &\int \Psi^*(\textbf{r}_1)(-\frac{1}{2}\nabla^2-\sum_{A=1}^{M}\frac{Z_A}{|\textbf{r}_1-\textbf{r}_A|}) \Psi(\textbf{r}_1) d\textbf{r}_1 \\ &+\frac{1}{2}\int\int \Psi^*(\textbf{r}_1) \Psi^*(\textbf{r}_2) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi(\textbf{r}_1)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2
\end{align}
Note that the second term - concerning two-electron integrals shows the electron density at $\textbf{r}_1$ interacting with the electron density of $\textbf{r}_2$; this is clear when rearranging the final term (knowing that the $r^{-1}_{12}$ operator is a Hermitian operator) and substituting in the first equation:
\begin{align}
\langle 12|12\rangle &=\int\int \Psi^*(\textbf{r}_1) \Psi(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \Psi^*(\textbf{r}_2)\Psi(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2 \\ &= \int\int \rho(\textbf{r}_1) (\frac{1}{|\textbf{r}_1-\textbf{r}_2|}) \rho(\textbf{r}_2)d\textbf{r}_1d\textbf{r}_2
\end{align}
Now, it is well known that electrons are indistinguishable, (i.e. they exist with a given probability density over all space and an electron cannot be labelled) such that: $$ |\Psi(1,2)|^2 = |\Psi(2,1)|^2 $$ Where $1=\textbf{r}_1$ and $2=\textbf{r}_2$.

It is also known that, as fermions (half-integal spin particles) they are anti-symmetric:
 $$ \Psi(1,2) = -\Psi(2,1) $$
Applying this antisymmetry to the two electron integrals we know that there has to be an antisymmetric electron interaction due to the interchanging of electronic positions. Therefore, the two electron integrals can be written: $$ \langle 12||12\rangle = \langle 12|12 \rangle -\langle 12|21 \rangle $$ $$ \langle 21||21\rangle = \langle 21|21 \rangle -\langle 21|12 \rangle $$ both the negative terms are the exchange interactions associated with the interchanging positions. Which are not included in the electronic Hamiltonian given in the question and their integral forms cannot be rearranged as shown before into terms of the electron density at a single given position.

This is why the Fock Operator is used instead of the electronic Hamiltonian!

I hope this all makes sense! I have probably made a mistake somewhere! So please let me know!

If you would like to know more about calculating the exchange from the electron density for known wavefunctions I recommend the paper: Pendás, A. M.; Blanco, M. A.; Francisco, E. Two-electron integrations in the quantum theory of atoms in molecules. J. Chem. Phys. 2004, 120 (10), 4581–4592. DOI: 10.1063/1.1645788.

For normailised wavefunctions, the exchange is given:
\begin{equation}
K = - \int \int r_{12}^{-1} \vert \rho_{1}(\textbf{r}_{1};\textbf{r}_{2}) \vert^{2}d\textbf{r}_{1}d\textbf{r}_{2}
\end{equation}

Where:
$$\rho_{1}(\textbf{r}_{1};\textbf{r}_{1}^{\prime}) = N_{\text{elec}} \int \Psi_{\text{elec}} (\textbf{r}_{1},...,\textbf{r}_{N_{\text{elec}}})\Psi_{\text{elec}}^{*}(\textbf{r}_{1}^{\prime},...,\textbf{r}_{N_{\text{elec}}}) d\textbf{r}_{2}...d\textbf{r}_{N_{\text{elec}}}$$


How do you know if electron is deexciting or exciting itself? I got a question which goes like this:

  According to Bohr's theory, the electronic energy of H-atom in the n th Bohr's orbit is given by $$E_n=\frac{-21.7*10^{-12}}{n^2}J$$ Calculate the longest wavelength of electron from the third Bohr's orbit of the He+ ion.

For longest wavelength transition, we need lowest energy transition. Therefore the electron can de-excite to $n=2$ or excite to $n=4$. I do not need the answer to the question, but I need to know how to know if the electron is exciting or de-exciting.

From what I know, I think it would be exciting to $n=4$ since the energy is lesser than de-excitation energy.

The 'longest wavelength of electron' does not make any sense. I assume that the question asks for the longest wavelength transition when a photon is absorbed or emitted (as fluorescence). One way to answer this is to look up the Balmer formula for hydrogenic like atoms and choose two principle quantum numbers for the transition you want. Alternatively you take the difference in energy between two levels using the formula given. As you start with n=3 the closest is either going to be n=4 (absorption) or n=2(emission). I leave it up to you to do the calculation but the answer must be positive as it corresponds to a photon's wavelength :) ( Note that the question gives the energy in Joules and you will want the wavelength in nanometres).


Changes of Internal Energy un open system My professor says the internal energy of a system can be calculated as $U=n T c_v$. Now I always thought that we could only calculate a change in internal energy $\Delta U = n c_v \Delta T$.

Following this idea, he says that a recipient that contains gas and loses a certain number of moles at constant temperautre undergoes a change in internal energy of $\Delta U = T c_v \Delta n$.

I've never seen any of this in any book. I believe he is assuming that the internal energy at 0 Kelvin is zero, which I believe is not true. However, in the case of the recipient losing gas,I wouldn't know how to calculate $\Delta U$.

Any thoughts? thanks!

Suppose you assign an reference state for internal energy such that the internal energy per mole u of the gas is zero at $T=T_r$.  This reference state is going to be held fixed for the material both entering or leaving the open system, as well as the material within the system at any time.  Then, in the initial state of a constant-temperature system, $U_{init}=n_{init}C_v(T-T_r)$ and, in the final state, $U_{final}=n_{final}C_v(T-T_r)$.  So, $$\Delta U=\Delta nC_v(T-T_r)$$Typically, in the overall energy balance on the system, including entering and exiting streams, $T_r$ will cancel out (just as it would for a closed system).

Please note, however, that if chemical reactions are occurring, the equation for   the internal energy and the specification of the reference state has to be extended to take this into account (using the concept of "internal energy of formation").


How to decide between different possible substitution/elimination mechanisms under given conditions? I'm trying to figure out the major product. At the moment I'm leaning towards an $\mathrm{S_N2}$ reaction, because there's a polar aprotic solvent, and sort of unhindered electrophile. However to me it could just as easily be an $\mathrm{S_N1}$ reaction (electrophile does have some hinderence and there's a weak nucleophile and weak base). $\mathrm{S_N1}$ would attack the tertiary carbocation and $\mathrm{S_N2}$ would go for the primary.

So how do I know if it's $\mathrm{S_N1}$, $\mathrm{S_N2}$, or even $\mathrm{E1}$/$\mathrm{E2}$?

I guess the main question is I don't know if the electrophile is hindered "enough" so we can say that an sn2 reaction is definitely not happening.

Generally, your substance can react in one of the following four ways that you have already identified:

$\mathrm{S_N1}$ solvolysis on the tertiary chlorine resulting in 4-chloro-2-methylbutan-2-ol;[1]
$\mathrm{S_N2}$ solvolysis on the primary chlorine resulting in 3-chloro-3-methylbutan-1-ol;
$\mathrm{E1}$ elimination from the tertiary chlorine resulting in 4-chloro-2-methylbut-1-ene or 1-chloro-3-methylbut-2-ene;[1]
$\mathrm{E2}$ elimination from the primary chlorine resulting in 3-chloro-3-methylbut-1-ene.

It is highly unlikely that one of the $\mathrm{X1}$ mechanisms happen on the primary chlorine due to the very badly stabilised intermediate primary carbocation we would need, highly unlikely for an $\mathrm{S_N2}$ reaction to proceed on the tertiary chlorine due to steric hindrance (especially since a primary one is available) and moderately unlikely (but not impossible) for $\mathrm{E2}$ to happen on the tertiary chlorine.

We thus need to consider:

How good is the leaving group (influences everything strongly except for $\mathrm{E2}$)?[2]
How strong is a base possibly present in solution (influences which elimination reaction can happen and how well)?[3]
How good is our nucleophile (influences the probability of $\mathrm{S_N2}$ most)?[4]

Starting at the base. This is a case of rule 6:[6] There is no base present in the mixture! (Yes, technically water can act as a base, but you need a strong acid for it to happen. Water’s $\mathrm{p}K_\mathrm{b}$ is $15.7$ if I recall correctly.) This definitely rules out $\mathrm{E2}$ which requires a strong base, and mildly rules out $\mathrm{E1}$ which requires a weak base.

Let’s check out potential nucleophiles next. Well, one of our nucleophiles is water, also known as very lousy, and another possible one is acetone, also known as even lousier. There is hardly any chance that either of them will nucleophilicly attack the primary chlorine site ruling out $\mathrm{S_N2}$.

Next, what about our leaving group? Well, we have chlorine which is also a pretty lousy leaving group. It means that $\mathrm{S_N2}$ will be discouraged even more and the generated chloride will probably hang around close enough to the carbocation required for the $\mathrm{X1}$ reaction paths to disallow $\mathrm{E1}$.

Finally, there are solvent effects, too. Here, the solvent is polar protic; a combination generally preferring the $\mathrm{X1}$ pathways.[5]

All things considered, this leaves $\mathrm{S_N1}$ as the only mildly probable pathway remaining. And thus I would strongly assume the main product to be 4-chloro-2-methylbutan-2-ol.

Of course, one should always include reaction times and temperatures (and yields) to any reaction-predicting exercise. Raising the temperature should increase the likelyhood of $\mathrm{E1}$ over $\mathrm{S_N1}$ because of the entropic factor (two particles become three in $\mathrm{E1}$ resulting in a rise in entropy).[7] However, in exercises, this is usually indicated by the presence or absence of a Δ above or below the reaction arrow. Since we have no Δ, we can assume room temperature.

[1] C. A. Grob, A. Waldner, Helv. Chim. Acta 1979, 62, 1854.
[2] https://en.wikipedia.org/wiki/SN2_reaction#Leaving_group
[3] https://en.wikipedia.org/wiki/Elimination_reaction#E2_mechanism
[4] https://en.wikipedia.org/wiki/SN2_reaction#Nucleophile
[5] https://en.wikipedia.org/wiki/SN1_reaction#Solvent_effects
[6] Rule 6: There is no rule six!
[7] https://en.wikipedia.org/wiki/Elimination_reaction#E2_and_E1_elimination_final_notes


A unique demo for singlet oxygen? I have long known that mixing bleach and peroxide produces oxygen gas:

$$\ce{NaOCl + H2O2 -> O2 + H2O + NaCl}$$

However I only recently learned that the specific form of oxygen produced by the reaction is singlet oxygen:

$$\ce{NaOCl + H2O2 -> O2(a^1\Delta_{g}) + H2O + NaCl}$$

Is there a simple chemical demonstration that can show the difference between singlet oxygen and triplet (ground state) oxygen?  Will singlet $\ce{O2}$ relight a splint "faster", or decolorize a dye faster when bubbled through solution, or otherwise be distinguishable from triplet oxygen in a home lab?

Admittedly, I'm having a professional photochemical bias when it comes to the generation of singlet oxygen!

A rather safe and simple approach - even in home experiments - is the use of air, sunlight and a sensitizer (rose bengal, methylene blue, or even chlorophyll from natural resources, such as Spirulina).

Here, the sensitizer is irradiated and excited to the first singlet state, undegoes intersystem crossing (ISC) and transfers the energy to ground state triplet oxygen, upon which singlet oxygen is formed:

\begin{align}
\ce{Sens &->[h\nu]\ ^1Sens^{\ast}}\\
\ce{^1Sens^{\ast} &->[ISC]\ ^3Sens^{\ast}}\\
\ce{^3Sens^{\ast} +\ ^3O2&->\ Sens +\ ^1O2}\\
\end{align}

A prominent reaction of $\ce{^1O2}$ is the concerted, diastereoselective addition to 1,3-dienes in Diels-Alder fashion.

A classical example for this reaction is the synthesis of the natural peroxide ascaridole (2) from $\alpha$-terpinene (1). G. O. Schenck, the former director of the Max Planck institute for Radiation Research, performed the reaction in batteries of flasks in his garden and published the results in 1944 (Naturwissenschaften 1944, 32 (14-26), 157–157). Ascaridole is a natural anthelmintic (= against parasitic worms), hence the name.

Maybe not really suitable for home experiments due to the chromatographic workup envolved but worth to be mentioned is the Schenck ene reaction of allylic alcohols:

The reaction products can be converted to 1,2,4-trioxanes by peroxyacetalization, as shown by Griesbeck and coworkers (Beilstein J. Org. Chem. 2010, 6 (61))

Note that the 1,2,4-trioxane moiety is the key element in the natural anti-malaria drug *artemisinin.


What are the products in this reaction? I came across this question

I thought that the answer should be option (a)but the answer is given option (d)
I thought that other products might be a result of the following mechanism

But this cannot be the case as a positive charge still remains on the carbon atom. Can you please help me figure out where I am going wrong?
UPDATE:
Or is this a possible mechanism
But I am still unclear about how the double bond will reappear between 3 position and the carbon that carried Chlorine

List item

All the reactions eliminate chloride (a) by the obvious direct attack, the other products b and c arise from methoxide attack on the 3 and 5 positions of the furan and chloride loss via an exomethylene intermediate. The system is somewhat similar to a polyallyl choride where attack can be direct or at the 3 or 5 position.


Why does potassium dichromate always reacts in an acidic medium? On the other hand, potassium manganate reacts in both acidic and alkaline media.

In an alkaline medium, chromium(VI) exists as potassium chromate instead of potassium dichromate:

$$\ce{Cr2O7^2- + 2OH- <=>> 2CrO4^2- + H2O}$$

Wikipedia has more information about this equilibrium.


Effect of +I on the alpha hydrogen in aldol reaction We have $\ce{(CH3)2CHCHO}$. Is the alpha-hydrogen acidic enough to react with a base? There is +I effect of two methyl groups, however there the carbanion is getting stabilised through resonance. And, as we know that inductive is a weak effect, so will it proceed through aldol reaction or will Cannizzaro reaction occur? Moreover, if we are considering  the inductive effect to be significant,  won't it reduce the electrophilicity of carbon of the carbonyl group, hence reducing the chances of Cannizzaro reaction? What could be the products if the reaction is done separately by using conc. base in one time and dil. base for other?

Is the alpha hydrogen acidic enough to react with base?

The answer is yes. As you said +I effect of the methyl group stabilizes the carbocation. It is weakly acidic but not weak enough to not be taken by any base.

Will it proceed through aldol reaction or will Cannizzaro reaction occur?

It will proceed through Cannizzaro mechanism even though it has an alpha hydrogen. That is because if you take it through aldol, the attacking nucleophile in the second step is $\ce{-C(CH3)2(CHO)}$ which is a very bulky nucleophile and so it isn't able to attack on $\ce{CH(CH3)2(CHO)}$. Think, if this reaction undergoes than the product will have $\ce{CH3}$, $\ce{CH3}$, $\ce{CHO}$ and $\ce{-C(OH)(CH(CH3)2)}$ groups all on one carbon. The steric crowding will be so immense that the reaction becomes infeasible. That is why it undergoes Cannizzaro reaction which has $\ce{H-}$ as the attacking nucleophile (in the second step).
I think your other questions won’t arise if you think this way.


Is there a simple way to predict the molecular pi orbitals in conjugated pi systems? Particularly for longer alkyl chains. Or is predicting these molecular orbitals from quantum mechanics too complex to be able to be translated easily as a geometric pattern? If Frost Circles exist, I just figured something could for this as well. The only pattern I've been able to identify is that each molecular orbital is symmetrical, starts with no nodes and goes up by one with each higher energy orbital but for systems with many carbon atoms, this isn't a specific enough set of rules.

This should work for any even number of conjugated carbons (and with slight modification, for cations/radicals/anions of uneven numbers of carbons). You can follow a simple procedure to generate all the $\pi$ orbitals.

Generate the lowest orbital with no nodes
Generate next orbital by creating a node through the middle bond
For all subsequent orbitals, generate by taking the previous orbital, placing nodes to the left and right of the original node, and then removing the original node.
End when all bonds are separated by nodes

We can write these as binary strings where a 1 is one orientation and 0 is the other. So for example, you could write the lowest orbital of N=10 chain as 0000000000, the next as 0000011111, the next as 0000110000, and so on.

The Huckel method mentioned in the comments is a more formal way of obtaining the orbitals by solving for the eigenvalues and eigenvectors of an adjacency matrix. The adjacency matrix is formed by numbering the carbon chain and forming an NxN matrix with the rows and columns numbered like the carbons. In each row, put a 1 if the carbon of the row and column are adjacent and put an x if they are the same. A simple online calculator with a few examples and instructions for this can found here (this only gives the eigenvalues, you would probably have to download or code your own version to give the eigenvectors). You only really need these eigenvectors to get an equation for the orbitals, but for your case you can do that just as well with the simple procedure by simply writing the MOs as linear combinations of the AOs.


Difference between spin-orbit coupling and the Russell-Saunders Effect? The Russell-Saunders effect is the same thing as 'spin-orbit interaction, correct?
The reason I am asking is because I was reviewing the Wikipedia page on 'spin-orbit interaction' and it does not mention Russell-Saunders at all...

I'm not aware of the Russell–Saunders effect, but the Russell–Saunders coupling scheme is definitely a thing. As you noted, the Wikipedia page on "spin-orbit interaction" doesn't talk about it, but a different Wikipedia page does, and basically tells you the same thing as I will.
The answer is... yes and no.
The word "coupling" refers to the coupling of several sources of angular momentum, namely the spin component and orbital component. Now, the issue is that in an atom you typically have many electrons, and every electron has its own orbital angular momentum $\vec{l}$ and spin angular momentum $\vec{s}$,* so you have many, many sources of angular momentum.
The challenge is to bring all of these together in a way that allows us to describe the electronic state of an atom using its angular momentum properties (for example, this is what a term symbol does). There are two approaches towards coupling all of these angular momenta together:

(Russell–Saunders or LS-coupling) Couple all the individual $\vec{l}$'s together to form one gigantic orbital angular momentum $\vec{L}$, and couple all the individual $\vec{s}$'s together to form one gigantic spin angular momentum $\vec{S}$. Then couple these two together to form the total angular momentum $\vec{J}$.
If you have studied term symbols before it is probably using the Russell–Saunders scheme, where you calculate $L$, $S$, and $J$, then write the term symbol $^{2S+1}L_J$.

(jj-coupling) For each individual electron, couple $\vec{l}$ and $\vec{s}$ together to form the total angular momentum $\vec{j}$ for that one particular electron. Then bring all the electrons' total angular momentum together to form $\vec{J}$.
Note that we haven't ever mentioned $L$ and $S$ here, so the term symbols under this coupling scheme are different. Instead, you'd label the term symbols with the individual values of $j$ for each electron. For an example see e.g. Atkins Molecular Quantum Mechanics.

Now, which you use depends on whether electron-electron repulsions or the spin-orbit coupling is a "bigger" term. If spin-orbit coupling is very significant, then it means that the spin and orbital angular momenta on their own (i.e. $\vec{L}$ and $\vec{S}$) are not very useful quantities,† since the interaction between them is large. In this scenario, jj-coupling is a more appropriate way of describing the effects of spin-orbit coupling on the electronic state.
On the other hand, if the spin-orbit coupling is relatively small, then $\vec{L}$ and $\vec{S}$ are useful quantities which are still applicable to the atom's electronic state, so the Russell–Saunders scheme is appropriate.
And of course, sometimes we get stuck in the middle ground where neither scheme is fully appropriate.
TL;DR The Russell–Saunders and jj schemes are both methods which can be used to describe the effects of spin–orbit coupling, but they are not the same thing as spin–orbit coupling.‡

* Well, sort of, anyway: the electrons are indistinguishable, so it's more accurate to say that the $n$ electrons in the atom have $n$ orbital angular momenta $\{\vec{l}_1, \vec{l}_2, \cdots, \vec{l}_n\}$ and $n$ spin angular momenta $\{\vec{s}_1, \vec{s}_2, \cdots, \vec{s}_n\}$.
† To be precise, $L$ and $S$ are not "good quantum numbers" because the operators $\hat{L}$ and $\hat{S}$ don't (approximately) commute with the total Hamiltonian $\hat{H}_0 + \hat{H}_\text{so}$, where $\hat{H}_\text{so}$ is the spin-orbit coupling Hamiltonian and $\hat{H}_0$ is the rest of the Hamiltonian (which does commute with $\hat{L}$ and $\hat{S}$).
‡ If you read the previous footnote, then the spin–orbit coupling itself is represented by the Hamiltonian $\hat{H}_\text{so}$. The two coupling schemes can be thought of as ways of dealing with this term as a perturbation to $\hat{H}_0$. In Russell–Saunders the perturbation is small, and consequently, the "good" quantum numbers are similar to those of $\hat{H}_0$. In jj the perturbation is large.


Does hydrogen peroxide - urea become more active when illuminated? Dentists' tooth whitening systems and consumer kits use gels containing carbamide peroxide, also known as hydrogen peroxide - urea.

Dentists use a light activator: (image from Wikipedia).

Consumer kits typically include 35% carbamide peroxide gel. Some of them use an LED light too, while others don't.

Wikipedia seems to be in two minds about it, stating

  Recent research has shown that the use of a light activator does not improve bleaching, has no measurable effect and most likely to increase the temperature of the associated tissues, resulting in damage

while also talking about which light is "best":

  The ideal source of energy should be high energy to excite the peroxide molecules without overheating the pulp of the tooth

Is a light source relevant here, or is it merely a marketing gimmick?

I hate to answer my own question, but after some digging, I believe that the use of a light source is in fact a gimmick, at any concentration of the reagent, contradicting the only other answer so far (But I'm happy to be corrected, if I'm wrong)

This meta-study https://www.ncbi.nlm.nih.gov/pubmed/22525016 concluded that at high concentrations of the reagent (typical of in-office use), the light is useless; and there was insufficient or no data for lower concentrations (as of 2012):

  Light increases the risk of tooth sensitivity during in-office
  bleaching, and light may not improve the bleaching effect when high
  concentrations of HP (25-35%) are employed. Therefore, dentists should
  use the light-activated system with great caution or avoid its use
  altogether. Further rigorous studies are, however, needed to explore
  the advantages of this light-activated system when lower
  concentrations of HP (15-20%) are used.

A slightly more recent study came to the same conclusion https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4178330/ :

  The in-office bleaching treatment of vital teeth did not show
  improvement with the use of light activator sources for the purpose of
  accelerating the process of the bleaching gel and achieving better
  results.

I'm inclined to think that if light were useful at all, they'd see this at higher concentrations of HP too, but I haven't seen any recent studies that address the lower concentrations specifically.


Does Photoelectric effect occur in a micro-oven while heating food? Does the photoelectric effect occur while heating food on a metal plate in a microwave oven? If not, then why doesn't it occur?

No, the photoelectric effect does not occur in a microwave oven.

The workfunction for a metal is the minimum photon energy required to eject an electron from the metal via the photoelectric effect.  Workfunctions are typically in the range of several electron volts (eV).  Potassium, for example, has a workfunction around 2 eV, see here for a list..

The energy in a microwave photon can range from 0.00001 to 0.001 eV; much to low to eject an electron from a metal.

For more background on the photoelectric effect see this earlier SE Chem Q&A


Is it more energy efficient to desalinate seawater by flash boiling or by flash freezing? In Saudi Arabia I think there are two types of desalination using flash boiling
from electric power production and also from flash freezing using solar absorption
refrigerant freezing.  One requires fossil fuel the other solar energy.

The cheapest way is probably to use the sun, as it is done in Tunisia. In Tunisia, they dig long channels on the beach along the sea. The channels must be long and narrow, and they are filled by sea water. Now the channel is covered by an inclined glass roof. The sun heats the water through the roof, and the steam  is slowly evaporated and condensed on the roof. It makes mist that streams and flow in a drainpipe placed at the lower part of the glass, just above but nearly on a level with the sea. It produce pure water for free.


Does a unit cell have to contain a whole number of atoms? I was given this problem recently to compute the number of ions in a unit cell of ferrous oxide $\ce{FeO}$. The data given was:

Side length $a = \pu{5 Å}$
Density $d = \pu{4 g/cc}$

Using the formula $\displaystyle d = \frac {ZM}{N_0 a^3}$, I obtained $Z = 4.18$. However, the answer had rounded $Z$ off to $4.$
Could a unit cell have  a fractional number of atoms? Intuitively, it doesn't make sense for a cell to have $0.18$ of an atom. If, by experimental analysis, we obtain a fractional number of atoms in a unit cell, do we round them down as is done in this problem?
P.S. Another thought would be that such odd numbers are obtained due to stoichiometric defects.

Crystallographic background

A unit cell cannot contain parts of atoms from a formula unit, otherwise it would not be a geometric unit of repeatability.
The simplest condition arising from translational symmetry is that the total number of atoms present in the unit cell must be either equal to or be a multiple of the number of atoms in the chemical formula, which is an equivalent to the integer number of formula units per unit cell: $Z \in \mathbb{N}.$

Note, however, that the asymmetric unit (crystallographically independent region) contains the $Z/n$th part of the atoms of the formula unit, where $n$ is the order of the crystallographic point group.
Thus, the asymmetric region may contain an integer number of formula units, one formula unit or its fractional part, meaning that some atoms located on symmetry elements, are "divided" across several asymmetric regions.

Data precision

It appears you are using the correct formula for the crystallographic density $ρ$ (I strongly encourage not to use $d$ as a symbol for density as in crystallography $d$ is reserved for spacing between planes):

$$ρ = \frac{MZ}{N_\mathrm{A}V}$$

where $M$ is molar mass, $N_\mathrm{A}$ is Avogadro's number and $V$ is the unit cell volume.
Although you can use the exact value for $N_\mathrm{A} = \pu{6.02214076E23 mol-1}$ and a precise tabulated value $M(\ce{FeO}) = \pu{71.844 g mol-1},$ note that you are given both $a$ and $d$ (experimental data) to the single significant figure, which prohibits your answer from being more precise than these parameters, and you must present a rounded number $Z = 4$ and not $Z = 4.18,$ which doesn't make either crystallographic or mathematical sense.

In real life situation there always will be a systematic difference between the measured and crystallographic densities due to the presence of defects, and the deviation of $Z$ from the integer value is often due to statistically filling of the proper point group by the atoms of a different kind.