Hessian Matrix and Physical directions: Potential Energy Surface How do the eigenvectors of the Hessian Matrix for potential energy surface correspond directly to the physical directions of translation, rotation and vibration?

I see that they might correspond to some physical directions, but it seems fishy that they directly correspond to these very neat directions.

I think you have the causality backwards here. It's not as if we have the vibrations to start with and they happen to be matched by the eigenvectors of the Hessian. We choose to look at these particular vibrations precisely because they are the eigenvectors of the Hessian.

Consider the example of water. We know that it has it has $9$ degrees of freedom, because each of its $3$ atoms are described by $3$ Cartesian coordinates. However, if we are interested in the energy of a configuration, we know that some of these degrees of freedom don't matter. For example, if we shift the x coordinate of each atom by the same amount, we haven't changed the molecule at all, only its location. This leads us to seek a set of coordinates that describe the potential energy surface without considering parts that don't matter. For water, we can remove $6$ as shifting along or rotating about any axis doesn't change the energy of the molecule.

This leaves us with $3$ remaining degrees of freedom, internal motions of the molecule that do affect its potential energy. What do these look like? I could come up with all sorts of valid variations on this. Off the bat, I might think that a good representation would be individual stretches of each $\ce{O-H}$. While there isn't anything strictly wrong with this representation, it does have the issue that motions along these coordinates are coupled, meaning changes along a given mode affect some or all of the others.

What can I do to avoid to avoid this coupling? Well, within the harmonic approximation, all the coupling between modes is described by off-diagonal elements of the Hessian matrix. So, if I can find a representation where the Hessian is diagonal, I will have a nice set of uncoupled internal motions. This is exactly what we get from the eigenvectors. Now, instead of two coupled $\ce{O-H}$ bond stretches, the normal modes give two decoupled anti/symm vibrations of both $\ce{O-H}$ bond.

The normal modes aren't any more physically correct than my previous example, but they are easier to analyze because they are decoupled. In this way, the "neatness" of the normal modes is really a fundamental mathematical property rather than a fundamental physical one.


Would adding water to a fish tank with sand lower the level of sand? Let's say I have a typical 10 gallon tank with dimensions 10" x 20" x 12", and it's filled halfway (5" in height) with dry sand for hermit crabs.

If I fill the tank up with water, will the level of sand decrease?

I'm thinking that it doesn't because of this answer -- sand is insoluble.  So the water would just fill the gaps of air first.  If that's not the case though, what are some things to take in to consideration?

Not quite sure where this question belongs, so if it's in the wrong place, let me know -- or mods, feel free to move it.  Thanks!

If your sand is packed in a optimal sphere packing, like a hexagonal packing for example,

nothing would happen to the sand level, because it can't be packed more dense.

If the packing of the dry sand wasn't optimal (which it probably isn't 100%), then the water may help settle it through motion and friction and it would lower a bit. In any case it wouldn't lower so much as that you could easily see it with your eyes.


How to find the valence of transition elements? Let's take Iron for example

As we can see there are 4 desolate electrons in normal state. Can you tell me which electron moves to make the valence 2 and then 3. We have learnt to find the valence only using the configuration formula.

For the 2nd and 3rd rows for 4b-7b groups the highest oxidation state equal to the group number is preferable over all else and for the 1st row for 4b-7b groups the highest oxidation state is equal to the group number.
Unfortunately, for all other information you'll need case-by-case approach (i.e. plainly remember it).


Why does lithium from the nitride whereas the other alkali metal forms the azide? What property of lithium makes them forms the nitride, lithium nitride $\ce{Li3N}$ whereas the other group 1 elements forms the azide, like sodium azide and potassium azide - $\ce{NaN3}$ and $\ce{KN3}$ respectively?

Does lithium azide exist? If so, what is its structure?

The formation of $\ce{Li3N}$ can be explained thermodynamically. The enthalpy of formation of the $\ce{N^3-}$ ion is extremely endothermic (it involves three successive additions of electrons) and with the larger counterions $\ce{Na+}$ or $\ce{K+}$ the lattice energy (which is proportional to $\frac{q_+ q_-}{r_+ + r_-}$) is not sufficiently large to compensate for the large $\Delta_\text{f}H$ of $\ce{N^3-}$. The same argument explains why magnesium also forms a nitride.

According to my very brief research on the Internet, lithium azide does exist but is unstable. It can be prepared via reaction of $\ce{NaN3}$ and $\ce{Li2SO4}$: see Acta Chem. Scand. 1957, 11, 581 for more details.


Why does reduction always accompany oxidation and vice versa? We've all learnt that the reduction always accompanies oxidation and vice versa Why is this so? Why is it not possible that a species releases an electron and the electron just "floats around"? Is this related to the stability of a free electron?

If something is to be reduced then there must be a reducing agent present. Also if something is oxidized then there must be an oxidizing agent present.

You must have a reducing agent or oxidizing agent present in order for a redox reaction to take place. An element can't just dislocate it's electrons. There must be an outside force present in order to do that.


Any popular experiment in chemistry that digital signal processing played a crucial role in? Is there any famous experiment in chemistry where digital signal processing played an important part? I don't mean using a machine that relies on such techniques (they all do) but an experiment where signals were first recorded (e.g. temperature/pressure/voltage/etc. as a function of time) and then analyzed with some sophisticated methods (e.g. time-frequency analysis with wavelets).

Many spectroscopic experiments record analog data, which is subsequently Fourier transformed to give a frequency-domain spectrum. For example, in a nuclear magnetic resonance experiment, the signal emitted by an excited nucleus is picked up as a voltage in an RF coil for 1-2 sec. This time-domain signal is processed then converted to the frequency domain by FT.

As for 'popular', Richard Ernst won the 1991 Nobel in Chemistry for his contributions to this field of research.


Reactivity order of electrophilic addition Find the order of the ease of electrophilic addition of the following:

  $\ce{H3C-O-CH=CH2}$
  $\ce{ F-CH=CH2}$
  $\ce{Cl-CH=CH2}$
  $\ce{NO2-CH=CH2}$

According to me, it should be 1 > 4 > 2 > 3 but in my textbook, it says 1  > 2 > 3 > 4.

If we take $\ce{F-CH=CH2}$ and $\ce{NO2-CH=CH2}$, the result after electrophilic addition would be $\ce{F-CH+-CH2E}$ and $\ce{NO2-CHE-CH2+}$ where E is the electrophile. Since both F and $\ce{NO2}$ are -I groups, further the $\ce{C+}$ carbocation is situated, more stable is the compound. The latter should be more reactive than the former. So why is my textbook going against this logic?

For the electrophilic addition to carbondouble bonds, the $\pi$- electron cloud double bonds first attack the electrophile and then a carbocation is created adjacent to the carbon where the electrophile is added. Then that carbocation is attacked by the nucleophile to form the addition product. $$\ce{R-CH=CH2 + E -Nu -> R-CH^+-CH2-E -> R-CH(Nu)-CH2E}$$The above path is the general electrophillic addition reaction in the case of alkenes. The carbocation position may vary depending upon the $\ce{R}$ group.

When the $\ce{R}$ is $\ce{CH3-O}$, there is a positive mesomeric effect (+M or +R effect) of the group due to which  when the carbocation is formed in the adjacent carbon, it gets  high stabilisation due to the +R effect, which in turn decreases the energy of the transition state of carbocation heavily, and thus the reaction rate becomes very high. That's why, the rate of electrophilic addition is very high in this case is very high, i.e it is the most easy.
When the $\ce{R}$ is $\ce{F}$, there is a significantly stronger +R effect due to $\ce{2p\pi -2p\pi}$ overlap. So, the carbocation gets a slight stabilisation due to this +R effect but there is also a -I effect in case of $\ce{F}$ which is the factor of destabilisation of carbocation. That's why, ease is relatively lower due to significant +R but stronger -I effect.
When the $\ce{R}$ is $\ce{Cl}$, there is a very weak +R effect (vinyl chloride has a weak  effect of +R, which is observed from the lesser dipole moment due to lesser charge separation  than ethyl chloride, where there is no +R), but dominant -I effect. So, in this case carbocation is more destabilised. So, the ease of addition is decreased even further.
When the $\ce{R}$ is $\ce{NO2}$, there is -I effect and also a dominant -R effect or -M effect. So the $\pi$ -electron cloud gets delocalised with the nitro group and thus due to decreased $\pi$-electron cloud density, it becomes difficult for it to even attack the electrophile. So, fractions of molecules forming the carbocation is extremely less, which actually increases the activation energy of the reaction. Thus, the reaction becomes very difficult with nitroethylene.

Consideering the overall effects described above, the ease of electrophilic addition should be,$$\ce{CH3 -O - CH=CH2 > F - CH=CH2 > Cl - CH=CH2 > NO2 - CH=CH2 }$$


Dynamic equilibrium in unsaturated solutions One standard definition of equilibrium in beginner chemistry is that for a reversible reaction, (dynamic) equilibrium has been achieved when the rate of the forward and backward reactions are equal. We also define the equilibrium constant for a given reaction in terms of a ratio involving the concentrations of products and reactants at equilibrium.

Consider adding a small amount of solid $\mathrm{AgCl}$ to a sample of pure water, where the amount added is small enough that the solution does not become saturated. The associated reaction is

$$\mathrm{AgCl}(\text{s}) \rightleftharpoons \mathrm{Ag^{+}}(\text{aq}) + \mathrm{Cl^{-}}(\text{aq}), $$

with $K_{\mathrm{sp}}\approx 1.77\times 10^{-10}$. After the added amount of salt has dissolved, the concentrations of all species are constant since all of the $\mathrm{AgCl}$ has dissolved and if any solid $\mathrm{AgCl}$ is produced by the reverse reaction, it will be shortlived and will dissociate back into ions almost instantly.

Is it not therefore true that the forward and backward reaction rates are equal at this point? By the simple definition of equilibrium given above, we would then conclude that the system has reached equilibrium. On the other hand, the reaction quotient $Q$ is less than the equilibrium constant $K_{\mathrm{sp}}$ since the saturation point has not yet been reached, and therefore the system has not reached equilibrium.

I am seeking an explanation of why this apparent contradiction arises – I expect it may be due to the basic definition of chemical equilibrium simply being inadequate, and that for equilibrium to be 'properly' achieved (in the sense of the equilibrium constant), the forward and backward reaction rates must approach each other smoothly rather than abruptly as they do in this example due to the $\mathrm{AgCl}$ running out before saturation is achieved. Or perhaps at equilibrium, the reaction rates are necessarily equal, but the converse is not true.  You can assume that I'm familiar with thermodynamics and statistical mechanics.

Edit: The equilibrium constant here is calculated subject to the constraint that both a solid and an aqueous phase must coexist as pointed out in Charlie Crown's answer. This is also alluded to in Buck Thorn's answer where they note that there is no equilibrium with the solid below the solubility limit, and in a comment by Karl below their answer where they state that there is no dynamic equilibrium when the concentration of the solid is zero. Hence the apparent contradiction arises here because the static equilibrium achieved before the solubility limit has been reached is not the same as the dynamic equilibrium which $K_{\mathrm{sp}}$ refers to.

Proper definitions of chemical equilibrium will not involve reaction rates whatsoever. Thermodynamics does not care about time. Chemical potential is the work required to form a molecule in solution, irregardless of the time it takes. Statistical Mechanics says chemical potential is the work required to move a molecule from infinitely far away (ideal gas since there are no interactions at infinity) into the solution.

Also, Free Energy is entirely independent of the equilibrium constant, but the equilibrium constant is not independent from Free Energy. At equilibrium, Free Energy will be at a minimum. Throw out definitions that use equilibrium constants. If you so wish, you can calculate the equilibrium constant from the free energy.

\begin{equation}
    K = \exp^{-\frac{\Delta G^0}{RT}}
\end{equation}

Where $\Delta G^0$ is the ideal gas free energy reaction. If you know the ideal gas free energy change of a reaction, you may calculate K. However, K is a useless thing in reality, or at least, leads to more confusion than it is worth.

For chemical equilibrium, this is the definition at global equilibrium

\begin{equation}
   \left( \frac{\partial G}{\partial \xi} \right)_{T,P} = 0
\end{equation}

Where $\xi$ is the extent of reaction. In your case the free energy was minimized by the solid dissolving rather than staying a solid, so your system came to equilibrium given the constraint that there was no more solid that could be dissolved. There was also no constraint saying that there needed to be solid present.

Equilibrium and solubility limit are two different things. All equilibrium points are constrained by definition. At the very least you are constraining T and P. When you are measuring solubility you are actually implementing an additional constraint: find the minimum free energy of the system with the constraint that a pure solid and an aqueous phase must coexist - this necessitates there initially being enough solid to be able to meet this constraint at equilibrium.

Equilibrium reactions do not require time, and I have no time for equilibrium constants :)


No resonance in Anti Aromatic compounds My teacher commented on Anti aromatic compounds as being less stable compounds as it doesn't undergo resonance inspite of having π-bonds/electron pairs in conjugation. I didn't understand why so. Can you help?

As according to Molecular Orbital theory , when anti- aromatic compound is formed, the electron goes to anti bonding, and is unpaired even, which reduces it's stability.,
Or we can say that in anti- aromatic,the plane of electron cloud twists there bonds between carbons, making it unstable, thus resonance is not possible.
Antiaromaticity is a characteristic of a cyclic molecule with a π electron system that has higher energy due to the presence of 4n electrons in it. Unlike aromatic compounds, which follow Hückel's rule ([4n+2] π electrons) and are highly stable, antiaromatic compounds are highly unstable and highly reactive. To avoid the instability of antiaromaticity, molecules may change shape, becoming non-planar and therefore breaking some of the π interactions.
Greater the delocalisation energy(resonance energy) of a compound, more stable it is. The resonance energy can be calculated using Huckel Molecular Orbital Theory(HMOT). Antiaromatic compounds have zero resonance energy hence are unstable.
According to HMOT,
Delocalisation energy=( Pi electron binding energy) - (total energy of isolated double bonds in a classical structure)


Is it better to use a smaller, more accurate measuring cylinder several times or a larger, less accurate one for the same volume? I need to measure out 70 ml of liquid.
I have a 10 ml measuring cylinder with a maximum uncertainty of 0.1 ml and a 100 ml measuring cylinder with a maximum uncertainty of 1 ml.
I've calculated that using the 10 ml cylinder 7 times would give a 0.7 maximum uncertainty, or a 0.7/70=1% percentage uncertainty. Meanwhile, using the 100 ml cylinder, I get a 0.1 ml uncertainty, or a 1/70=1.4% percentage uncertainty. Clearly, the smaller cylinder approach is more accurate (1% < 1.4%)
Meanwhile, the mark scheme says that one must use the 100 ml cylinder, for the 10 ml cylinder would give a total 7*0.1/10=7% percentage uncertainty, compared to the 1.4% from the 100 ml cylinder. (1.4% < 7%)
Isn't adding percentages in this way fallacious? There's no way to accumulate a 4.9 ml error using a 10 ml cylinder 7 times, right? I'd appreciate if someone could clarify which approach is correct and why.

You have three types of errors:

The errors in the accuracy of your measurement.  Assuming you are reading the water levels consistently (at eye level at the bottom of the meniscus), these should be random.
To add random, uncorrelated errors, the standard method is to sum their squares and take the square root.  So, for the 0.1 mL average error for the 10 mL cylinder:
$$\text{average error for seven combined measurements} = \sqrt{7  \times (0.1 \text{ mL})^2} = 0.26 \text{ mL}$$
As you correctly intuited, it wouldn't make sense to simply multiply the 0.1 mL by 7, since this would be saying you have either a +0.1 mL error every time, or a –0.1 mL error every time, but that's not how random errors work—they are just as likely to be positive as negative. In addition, they don't all have a magnitude of 0.1 mL; rather, their magnitudes follow a distribution. Thus essentially what you have is a random walk, with a random distribution of step sizes, where the average step size is 0.1. The square-root-of-the-sum-of-the-squares method gives you the average distance you'd end up from your starting point with 7 of these steps

Error in the calibration of the cylinder.   This would be a systematic error, and it would add to the error of each measurement.  Higher qualities of cylinders (e.g., Class A glassware) have lower errors.

Error in the consistency with which the cylinder delivers the measured amount.  These cylinders are probably calibrated as "TD", which means "to deliver".  That is, it is assumed not all the water will drain out, so the markings on the cylinder are calibrated to account for that.  However, there will be some random error in how closely the amount that drains out corresponds to the assumed amount that drains out.  The only way to determine this error would be to contact the manufacturer (or to test it yourself using a scale).  In addition, if your cylinder is not clean, more water may adhere to the walls, which could cause a systematic error in the amount of water delivered.

Note also the comment by Andrew Morton that there's an additional important practical consideration, which is that if you have to use seven measurements, you could miscount and thus end up being off by  $\pm$10 mL! And this error is much easier to make when measuring with a graduated cylinder than, say, a tablespoon, because each measurement with the cylinder takes focus to get the total volume to 10.0 mL, and that focus can cause you to lose count.  There are of course ways to address this, e.g., making a mark on a piece of paper for each 10.0 mL added.


Is the reaction of hydronium and thisulfate an acid-base reaction? Is this reaction
$$\ce{2H3O^{+}(aq) + S_{2}O_{3}^{2−}{(aq)} -> 3H_{2}O(l) + SO_{2}(g)  + 1/8S8(s)}
$$
an acid-base reaction?

It starts with the protonation of the anion, so that part is acid-base, but then it continues as a redox reaction.


Strong Base combined with Sp2 carbons in Nucleophilic Substitution and Beta-Elimination Reactions When we are given this problem, there is a certain problem. We have a strong base, but hydrogen on the carbon of the phenyl. E2 is impossible. Also, SN2 is not possible since we have a hindered base.

The only product we could get is SN1. My teacher put up the answer in handwritten black in the above picture.

However can SN1 reactions happen when there is a strong base in the solution? Would that strong base be the nucleophile or the solvent tert-butyl alcohol be the nucleophile?

The strong base is capable of being a nucleophile in this sort of situation; deprotonating tert-butyl alcohol will increase its nucleophilicity. In an SN1 reaction the overall rate is only dependent on the loss of a leaving group, so a strong base/poor nucleophile will react via SN1 as long as there are not plausible E1 pathways (e.g. sp3 bonded hydrogen alpha to carbocation).


Why does sp2 hybridization imply adjacent sp2 hybridization? EDIT:  I misquoted my teacher; he said something more along the lines of: in conjugated systems, if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

My teacher said that if an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized.

1) Why is this true?
2) Is this even true?

I'm guessing this has to do with physical limitations. For example, consider a carbonyl oxygen. Oxygen is $\ce{sp^2}$ hybridized. The carbon the oxygen is bound to has already "used up" two of its four bonds. The carbon can only form two more sigma bonds in addition to the one sigma bond it already has in the $\ce{C=O}$ double bond.

But what about "hypervalent" (if there are such beasts) molecules? Here we come to a self-consistency issue. My teacher still believes in hypervalent molecules - i.e. he draws sulfuric acid with six bonds.

So let's consider sulfuric acid in its hypervalent form, with two double bonded $\ce{sp^2}$ oxygens. There are a total of six bonds and four sigma bonds, implying $\ce{sp^3}$ hybridization.

So my revised set of questions is as listed:

1) Is my teacher being inconsistent with himself here?

2) Does $\ce{sp^2}$ hybridization still imply $\ce{sp^2}$ hybridization for linked atoms? Note that sulfuric acid does not disprove my teacher's statement if we take sulfuric acid to have four bonds, not six (i.e. sulfuric acid is properly represented as charge separated due to the high ionic character of its $\ce{S-O}$ bonds. If I remember correctly, on a time-average basis, 1.7 electrons are in oxygen's valence and only 0.3 electrons are in sulfur's valence per bond.

If an atom is $\ce{sp^2}$ hybridized, then every atom it is bound to is also $\ce{sp^2}$ hybridized. Would this be accurate?

Absolutely not, here are a few examples


Reversible and Irreversible adiabatic expansion A $\mathrm{200.0\ m^3}$ balloon at $253.0\ \mathrm K$ ascends to a higher altitude. If the initial pressure inside the balloon is $325$ millibar, and it ascends to a latitude with a pressure of $7.45$ millibar, by what percent does the absolute temperature change:

Assuming the process is reversible and adiabatic
Assuming the process is irreversible and adiabatic

All of the formulae I have on this subject include moles. The question does not give a value of moles. Where do I start?

The change in internal energy $U$ is
$$\Delta U=Q+W$$
where $Q$ is amount of heat transferred to the system and $W$ is work done on the system.
Since the process is adiabatic, no heat is transferred into or out of the system, i.e. $Q=0$ and thus
$$\Delta U=W$$

The reversible expansion is performed continuously at equilibrium by means of infinitesimal changes in pressure $p$.
The corresponding infinitesimal pressure-volume work $\delta W$ is
$$\delta W=-p\,\mathrm dV$$

For an ideal gas, the internal energy $U$ only depends on amount of substance $n$ and temperature $T$. Since $n$ is constant in a closed system, the change in internal energy $U$ is
$$\mathrm dU=C_V\,\mathrm dT$$
where $C_V$ is heat capacity at constant volume.

Equating $\mathrm dU$ and $\delta W$ yields
$$C_V\,\mathrm dT=-p\,\mathrm dV$$
and since $pV=nRT$ for an ideal gas
$$\begin{align}
C_V\,\mathrm dT&=-\frac{nRT}V\,\mathrm dV\\
\frac{C_V}T\,\mathrm dT&=-\frac{nR}V\,\mathrm dV
\end{align}$$

Integration from the initial state ($T_1,V_1$) to the final state ($T_2,V_2$) yields
$$\begin{align}
\int\limits_{T_1}^{T_2}\frac{C_V}T\,\mathrm dT&=\int\limits_{V_1}^{V_2}-\frac{nR}V\,\mathrm dV\\
C_V\int\limits_{T_1}^{T_2}\frac1T\,\mathrm dT&=-nR\int\limits_{V_1}^{V_2}\frac1V\,\mathrm dV\\
C_V\ln\left(\frac{T_2}{T_1}\right)&=-nR\ln\left(\frac{V_2}{V_1}\right)\\
\ln\left(\frac{T_2}{T_1}\right)&=-\frac{nR}{C_V}\ln\left(\frac{V_2}{V_1}\right)\\
\frac{T_2}{T_1}&=\left(\frac{V_2}{V_1}\right)^{-nR/C_V}\\
&=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}
\end{align}$$
where $C_{\mathrm m,V}$ is molar heat capacity at constant volume.

Since $C_{\mathrm m,p}-C_{\mathrm m,V}=R$, the exponent can be rewritten
$$\begin{align}
\frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{R/C_{\mathrm m,V}}\\
&=\left(\frac{V_1}{V_2}\right)^{C_{\mathrm m,p}/C_{\mathrm m,V}-1}\\
&=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
\end{align}$$
where $\gamma$ is the ratio of the heat capacities $\gamma=C_p/C_V=C_{\mathrm m,p}/C_{\mathrm m,V}=c_p/c_V$. For a monoatomic ideal gas, $C_{\mathrm m,p}=\tfrac52R$ and $C_{\mathrm m,V}=\tfrac32R$, and thus $\gamma=\tfrac53$.

Note that, in the given question, the volumes $V_1$ and $V_2$ are unknown. However, the values for the pressures $p_1$ and $p_2$ are given. Since $pV=nRT$ for an ideal gas,
$$\begin{align}
\frac{T_2}{T_1}&=\left(\frac{V_1}{V_2}\right)^{\gamma-1}\\
T_2\cdot V_2^{\gamma-1}&=T_1\cdot V_1^{\gamma-1}\\
\frac{p_2V_2}{nR}\cdot V_2^{\gamma-1}&=\frac{p_1V_1}{nR}\cdot V_1^{\gamma-1}\\
p_2\cdot V_2\cdot V_2^{\gamma-1}&=p_1\cdot V_1\cdot V_1^{\gamma-1}\\
p_2\cdot V_2^\gamma&=p_1\cdot V_1^\gamma\\
\frac{p_2}{p_1}&=\left(\frac{V_1}{V_2}\right)^\gamma
\end{align}$$

In contrast to the reversible expansion, an irreversible expansion is not performed continuously at equilibrium by means of infinitesimal changes in pressure. In the limiting case, the value of the pressure $p$ abruptly changes from $p_1$ and $p_2$. The following expansion proceeds in one step against a constant pressure $p_2$. Hence,
$$\delta W=-p_2\,\mathrm dV$$
and thus
$$\begin{align}
C_V\,\mathrm dT&=-p_2\,\mathrm dV\\
\int\limits_{T_1}^{T_2}C_V\,\mathrm dT&=\int\limits_{V_1}^{V_2}-p_2\,\mathrm dV\\
C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right)
\end{align}$$
Since $pV=nRT$ for an ideal gas,
$$\begin{align}
C_V\left(T_2-T_1\right)&=-p_2\left(V_2-V_1\right)\\
&=-p_2\left(\frac{nRT_2}{p_2}-\frac{nRT_1}{p_1}\right)\\
&=-nRT_2+nRT_1\frac{p_2}{p_1}\\
C_VT_2+nRT_2&=C_VT_1+nRT_1\frac{p_2}{p_1}\\
T_2\left(C_V+nR\right)&=T_1\left(C_V+nR\frac{p_2}{p_1}\right)\\
T_2\left(C_{\mathrm m,V}+R\right)&=T_1\left(C_{\mathrm m,V}+R\frac{p_2}{p_1}\right)\\
\end{align}$$


Can a table salt + vinegar mixture be explosive? I remember having read in a book, Ask Me Anything (Dorling Kindersley), that a combination of copious amounts of table salt and vinegar is explosive.

Now this was a really long while back, and I can't even find my copy of the book anymore. But I'm quite sure there was a mention of salt+vinegar being explosive.

Simply knowing the composition of table salt and vinegar doesn't seem to help in any way. I really don't see any reason for some sodium, chlorine, acetate and hydrogen ions floating around in a bowl to simply react in some godforsaken way and explode. Heck, all salt+vinegar seems like is a good Pringles flavor. I did take into consideration possible impurities in table salt, like other sodium, potassium, calcium and magnesium salts, but even then, I don't see why the explosion would occur at all. I though of iodized salt too, but still, an explosion seems a little far-fetched.

Now the book did mention 'large' quantities of salt and vinegar, maybe there could be something to that?

Is the book correct? If so, what concentration of salt and vinegar have to be used?

I did conduct an internet search, but nothing really turned up.

[ I suspect, the book might've meant baking soda instead of table-salt. Baking soda combined with vinegar in a closed plastic bottle could pass for an 'explosion'. ]

Of Course NaCl is the Salt of strong acid and strong base . It is very stable . Even if you take any amount of NaCl with Vinegar it will not do any explosion .

It may be the baking soda (NaHCO3 ) and Vinegar , As this reaction is a neutralization reaction (Exothermic) and accompanied by the formation of CO2 , If enough concentration is taken it can burst a plastic bottle on applying some pressure .

https://www.youtube.com/watch?v=UclxSfPoapc


What does "sine" mean? I see the suffix "sine" (seen/sin) a lot, adenosine, cytosine, lysine, tyrosine, etc. Most of where I hear it is in amino acid R groups, but it's usually only the prefix that is recognized as significant.

According to Textbook of metabolism and metabolic disorders (1964):

  The ribosides derived from purines have the suffix -osine; those from pyrimidines, the suffix -idine. The corresponding deoxyribosides, with the exception of thymidine, do not have any such simple designation. Occasionally, instead of, for example, guanine-deoxyriboside, the term deoxy-guanosine is used.

Beyond the more common adenosine and guanosine, there are also:

inosine  and

xanthosine


Acid catalysed dehydration of unsymmetric alcohols How to find the major product of acid catalysed dehydration of unsymmetric alcohols. For example, in the dehydration of butan-2-ol using conc. H2SO4, which is the major product? But-1-ene or But-2-ene?
And, in the dehydration of 1-methylcyclohexanol, the major product is 1-methylcyclohexene and not methylenecyclohexane. But how did we work it out??

Zaitsev's rule clearly addresses your query. Or you could look at it this way. The product with more hyper-conjugation is more stable. Hence but-2-ene is more stable than but-1-ene. Always see hyper-conjugation for more stable, major product.


Why are lab coats required in chemical laboratories? Well at first it seems kind of awkward but let me tell you what I was thinking from past few days. Currently I am studying and go to school. In school, whenever we have to go inside a chemistry lab then it’s compulsory for us to wear a lab coat; without one we were not allowed to enter and do experiments. But then one question arises is: Why it is compulsory?

Then I started finding solution for this question. I thought that we wear lab coat just for safety against all those chemicals we use for experiments but then if accidentally someone drops any concentrated acid then I guess that lab coat won’t be able to protect us. So if we wear something solid or something hard which can protect us from those chemical that might be helpful but we only wear a lab coat.

Then I thought that lab coats have so many pockets on them so it might help us to carry a few things but then again why don’t we use something else which is handy rather than a lab coat (still this example is quite useless).

After hours of thinking I am not able to come up with anything.

And also whenever anyone goes inside chemistry lab he/she wears a lab coat which is quite similar to everyone else’s. I.e. everyone wears same type of lab coat (a long, white colored coat) which is quite fascinating.

So my question is really simple: Why do we have to wear a lab coat when we go inside a chemistry lab and how come everyone uses same type of lab coat?

It is protection. It should be worn whenever you work with potentially dangerous chemicals. In German, it is even called (part of the) persönliche Schutzausrüstung, i.e. personal protective gear. Lab coats:

protect you against contamination with chemicals — acids, bases, toxic molecules and salts. They will ‘hit’ your coat and ‘stick’ to it rather than reaching your clothes. Don’t say you are careful. Your labmate may not be or an accident may happen. This is also the reason why labcoats are never allowed where food or drink are consumed.
protect you against spills of liquids. You can quickly take off a lab coat and not be contaminated by the liquid in question. This is basically an extension of the first bullet point.
protect you against fire. This might not be obvious, but lab coats are generally cotton which does not burn well. If you are wearing synthetic fibre below the lab coat, ripping that off could give you valuable seconds for not burning to death.
show other people that you are working with something potentially dangerous. Assuming you are dealing with adults that means they will know better than startle you because they can.

Everything else a lab coat supplies is basically just extension of the protection or meant to help people actually use it. Pockets: If it didn’t have any, people would keep using their trousers’ ones potentially contaminating them. Long sleeves: You arms need protection, too. Being long: so do your legs (at least partially).

And of course, it’s pretty cool if you walk in a lab coat, isn’t it? (That was my feeling at least.)


Group 13 elements reactivity with halogens All the trihalides of the Group 13 elements are known, but $\ce{TlI3}$ is not. Why is this so?

This is due to the Inert Pair Effect. As we move down the 13th group, addition of d orbitals and f orbitals are taking place. Since the d orbitals and f orbitals are away from the nucleus, they have poor sheilding power or capability and thus the s orbitals of the outermost shell are attracted towards the nucleai, making it difficult for s orbital electrons to leave the orbitals and only pairing of the 1 electron in p orbital is possible. Thus +1 valency of Tl is more stable than +3 as it is not energetically favored. This,ofcourse, does not mean that formation TlI3 is not possible. It's just that it requires more energy.


Gauche correction in Bensen group increment I have difficulty understanding the use of gauche correction in Bensen group increment method to calculate the heat of formation. As an example, 4 times gauche correction (0.8 kcal / mol) is considered in the calculation in the hypothetical equilibrium between the following two isomers. How is that figure decided?

Here is how I do it.

Sight down each C-C bond along the backbone of the molecule and draw a Newman projection
Look at all 3 possible conformations in the Newman projection and pick the conformation with the fewest gauche interactions
Count the number of gauche interactions
Repeat with the next C-C bond along the backbone

Here is a drawing of the 3 Newman projections in 3,4-dimethylhexane that contain gauche interactions (there are no gauche interactions along the $\ce{C1-C2}$ or $\ce{C5-C6}$ bonds.  You can see that there are 4 gauche interactions in the molecule.


Does cinnabar react with iodine? Does cinnabar react with bromine and iodine to form elemental sulphur? This reaction was described in "Explorations in the history of science and technology in China" but I have doubts...

I did a cursory search and found some information particularly from books of late 19th century(1,2):
Cinnabar and iodine

On heating, it (cinnabar) is readily decomposed with formation of small scarlet crystals of mercuric iodide.

Cinnabar and bromine

[...] After digesting the cinnabar with excess of bromine liquid (or bromine water containing 3% of bromine or solution of hydrochloric acid containing 13% of bromine) for a few days, the whole is dissolved, and also any metallic mercury present [...] The amount of bromine required is very considerable, 1 part of mercury is obtained and 3.5 parts of hydrobromic acid remains in the solution.

References

Chemical News and Journal of Industrial Science, Volumes 37-38, Chemical news office., 1878
Dictionary of Arts, Manufactures and Mines by Robert Hunt, Andrew Ure, 1878


Purification of primary amines using Schiff base immobilization I want to purify all primary amines containing compounds in a complex mixture (cell-extracted metabolites). I'm unaware of any published methods to do so. So I am thinking of making my own using a reversible Schiff base immobilization.
My input is a 1 mL metabolite extract that contains hundreds of compounds with a total molarity of ~1 mM. Of this ~50% contain a primary amine. Glyoxal agarose typically has 40–50 μmol aldehydes per mL, so capacity is not an issue.
Impurities that I want to get rid of are, among others, organic acids like malate, pyruvate and succinate, sugars like glucose and glucose-phosphate, lipids and all non-volatile salts.
The idea is to use Schiff base chemistry:
$$\ce{R^1CHO + R^2NH2 <=>[$k_1$][$k_2$] R^1CH=NR^2 + H2O}$$
An immobilized aldehyde is used to capture all compounds containing primary amines. A Schiff base is generally stabilized under high pH and destabilized at low pH. It can be further stabilized by dehydration to shift the equilibrium towards the Schiff base side. Therefore, I would add my input at pH = 10 in 95% MeOH, wash under dehydrated conditions e.g. using 100% MeOH, and elude at pH = 2 using 0.1 M aqueous formic acid. Alternatively, adding glycine in the elution buffer to compete off remaining amines.
For immobilized aldehydes I'm thinking about using aldehyde activated beads e.g. glyoxal agarose beads:

My questions are:

Are there alternative ways to purify primary amines from a complex mixture which I'm unaware of?

Did anyone publish a paper that used Schiff base chemistry to purify primary amines?

What would be your odds that this could work? Is it worth pursuing?

Any ideas for improvements? Eluate has to be run on mass-spectrometer, so my requirements are:
a. no non-volatile salts;
b. no solvents with MW above 150 Da.

Let's take your questions one at a time.

Are there any alternative ways to purify primary amines from a complex mixture that I am unaware of?

Your end application is LC-MS.  There are plenty of papers on amine-specific derivatization methods for metabolome analysis. For example, there is one by Zhao et al. [1]. Instead of purifying the primary amine metabolites and isolating the purified mixture as a separate step, they just rely on the analytical instrumentation to be able to identify primary amines from their derivatization pattern.
Alternately, as suggested in the comments by @Waylander, you could acidify the sample to form ammonium ions, and then pass the sample through a cation exchange column, where only positively charged ions will bind.  As you noted, this will not be perfectly selective for primary amines, because species like quaternary ammonium salts etc. will also be retained.
You could combine the derivatization approach with the SCX approach, of course.  If you chose a primary-amine-selective derivatization, you might get a selective method.  At some point, however, the complexity of the method will become a drawback.

Have anybody else, published or unpublished, used Schiff base chemistry to purify primary amines?

Well this question is impossible to answer for the unpublished case.  I couldn't find any published methods in a quick search, which makes me suspect it is not the best approach to your problem.  But I could be wrong of course!

What would be your odds that this could work? Is it worth pursuing?

Unanswerable without data.  I suspect that secondary amines, at least some times, will also react with the aldehydes to form Schiff-base-like iminium cations, and may also be selected by your proposed method.

Any ideas for improvements?

I think derivatization, without separation or isolation of the amines, would be an attractive direction for you to consider.  The paper I linked, its references, and the papers that reference it might be good starting places.
Reference

Zhao, S.; Li, H.; Han, W.; Chan, W.; Li, L. Metabolomic Coverage of Chemical-Group-Submetabolome Analysis: Group Classification and Four-Channel Chemical Isotope Labeling LC-MS. Anal. Chem. 2019, 91 (18), 12108–12115. DOI: 10.1021/acs.analchem.9b03431.


Does the hybridization of sp3d2 type exist? Is the hybridization of $\ce{XeOF4}$ $\mathrm{sp^2}$ rather than $\mathrm{sp^3d^2}$, because the latter type of hybridization does not exist?
Every answer in the Quora thread "How can we find the hybridization of $\ce{XeOF4}$?" concluded that the hybridization of $\ce{XeOF4}$ is $\mathrm{sp^3d^2}$. However, the very last answer posted by Steven Fawl, Professor of Chemistry at Napa Valley College (1985-present) reads: "Where to begin? Let me start here - every answer thus far is wrong! This molecule is not $\mathrm{sp^3d^2}$ hybridized. It is actually $\mathrm{sp^2}$ hybridized and I am going to prove it." The issue that I have with his post is that it only cites a couple of Wikipedia articles as his sources, and it contradicts every chemistry book that I have read by claiming that $\mathrm{sp^3d}$ and $\mathrm{sp^3d^2}$ hybridizations do not exist. The justification he gives for the latter is the following quote from this Wikipedia article:

In 1990, Magnusson published a seminal work definitively excluding the role of $\mathrm{d}$-orbital hybridization in bonding in hypervalent compounds of second-row (period 3) elements, ending a point of contention and confusion. Part of the confusion originates from the fact that $\mathrm{d}$-functions are essential in the basis sets used to describe these compounds (or else unreasonably high energies and distorted geometries result). Also, the contribution of the $\mathrm{d}$-function to the molecular wavefunction is large. These facts were incorrectly interpreted to mean that $\mathrm{d}$-orbitals must be involved in bonding (Ref.1 and Ref.2).
For transition metal centers, the $\mathrm{d}$ and $\mathrm{p}$ orbitals are the primary valence orbitals, which are only weakly supplemented by the $\mathrm{p}$ orbitals (Ref.3). The question of whether the $\mathrm{p}$ orbitals actually participate in bonding has not been definitively resolved, but all studies indicate they play a minor role.

Indeed, the abstract of Ref.2 states that:

No evidence is found to support the traditional notions of $\mathrm{sp^md^n}$ hybridization.

But this article is not publicly available and it may not reflect the current scientific consensus. Making a long story short: Are the authors of Ref.2 and Steven Fawl correct? If so, is the bonding of every hypervalent molecule analogous to $\ce{XeOF4}$ (removing 1, 2 or 3 electrons from the valence shell of the central atom and $\mathrm{sp^n}$ hybridize it)?

References:

Eric Magnusson, "Hypercoordinate molecules of second-row elements: $\mathrm{d}$ functions or $\mathrm{d}$ orbitals?," J. Am. Chem. Soc. 1990, 112(22), 7940–7951 (https://doi.org/10.1021/ja00178a014).
David L. Cooper, Terry P. Cunningham, Joseph Gerratt, Peter B. Karadakov, Mario Raimondi, "Chemical Bonding to Hypercoordinate Second-Row Atoms: $\mathrm{d}$ Orbital Participation versus Democracy," J. Am. Chem. Soc. 1994, 116(10), 4414–4426 (https://doi.org/10.1021/ja00089a033).
Gernot Frenking, Sason Shaik (Eds), eds. "Chapter 7: Chemical bonding in Transition Metal Compounds," In The Chemical Bond: Chemical Bonding Across the Periodic Table; Wiley-VCH: Weinheim, Germany, 2014, pp. 175-218 (ISBN 978-3-527-33315-8).

First, hybridization is a model used in Bond Valence Theory to explain the bond nature between some atoms in some molecules. Actually, the current binding theory is the Molecular Orbital Theory, in which we no longer use hybridisation. But, we study hybridisation because it is a simple and useful tool to explain a lot of compounds.

But, what the professor said in your question, it is not that the sp3d2 hybridization does not exist and cannot be used to explain any molecule. What he is saying is: In hypervalent compounds, there is evidence that the d functions do not participate in the bond.
After reading your question, I have searched a lot about the binding in hypervalent compounds. And what I found was that the nature of bond in hypervalent compounds like XeOF4 has been the subject of a lot of controversy for a long time, because in the represetative atoms, the energy gap between the nsp and nd orbitals is too large, so the d orbitals cannot hold extra electrons. And, nowadays we have evidence through ab initio calculations that the d-functions do not participate in bonding in hypervalent molecules.

However, this is a recent evidence, so, if you have to awnser in a test about the hypervalent molecules, use the d-orbitals to explain them!

I am leaving some links with some articles and books that I read about this subject:

Chemistry of Hypervalent Compounds

The three-center-four-electron (3c-4e) bond nature revisited. An atoms-in-molecules theory (AIM) and ELF study

New bonding concept for Hypervalent molecules, including electron poor and electron odd compounds


Azo coupling with arenediazonium salts - why isn't nitrogen gas lost? Arenediazonium compounds contain a very good leaving group - nitrogen gas - and thus undergo nucleophilic substitution reactions rapidly. With water, for instance, they give phenols.

In the formation of azo dyes from arenediazonium salts, however, the arenediazonium cation acts as an electrophile to carry out a substitution reaction on activated benzene rings like those of phenols and anilines. Why doesn't displacement of $\ce{N2}$ happen here as well? Normally the carbon attached to $\ce{-N2+}$ acts as the electrophilic centre, so phenols and anilines should react at that $\ce{C}$ rather than at the $\ce{-N2+}$ group. Moreover, phenols and anilines have more nucleophilic centres at their $\ce{O}$ and $\ce{N}$ than at the carbon atoms of their benzene ring - so why aren't the final products be diphenyl ether with phenol, and N-phenylaniline with aniline?

A simple explanation would be that the displacement of a diazonium group by oxygen to form a phenol or phenyl ether linkage typically requires heating, indicating that it is a slower process than the very fast room temperature coupling reaction at aromatic carbon to form an azo dye. Why? One explanation is that the diazonium cation and the phenol or amine being coupled are a "soft" acid and base respectively, whereas water is a "hard" base. But a substituted phenol or aniline that could not form an azo dye might well form a phenyl ether with a diazonium salt, as you propose.


Is glycerol/glycerin soluble in ethanol? As titled, I wish to know if glycerol/glycerin is soluble in ethanol (not soluble, slightly soluble, very soluble)

As LDC3 indicated above here it states that "Glycerol is completely soluble in water and alcohol." So, this is correct, as also indicated here.

Now, for the reason this happens it is most likely because of its 3 hydroxyl groups ($-OH$) and the fact that it is a polar molecule.


Why is the zero of standard enthalpy of formation a convention? The standard enthalpy of formation $\Delta H_f^°$ of pure elements is zero by definition. Why is that a convention? It is true that enthalpy is defined unless a constant (like energy and entropy), but enthalpy of formation is actually a variation of enthalpy, so we don't really care.
Also, how could we possibly choose something else than zero if there's no heat transfer?
Note: I'm using General Chemistry (Ralph Petrucci)

It starts with this:  The standard enthalpy of formation of any substance, $\Delta H^\circ_\mathrm{f}$, is (by convention) defined as the enthalpy change for the reaction at 1 bar and a specified temperature (usually $\pu{298.15 K}$), in which the product is 1 mole of that substance, and the reactants are its component elements in their respective standard states.
For instance, $\Delta H^\circ_\mathrm{f}$ for $\ce{H2O_{(l)}}$ is equal to $\Delta H^\circ$ for the following reaction at standard state:
$$\ce{H2_{(g)} + 1/2O2_{(g)}->H2O_{(l)}},$$
since $\ce{H2_{(g)}}$ and $\ce{O2_{(g)}}$ are the respective standard states for hydrogen and oxygen.
Once you've accepted this definition as the convention for determining $\Delta H^\circ_\mathrm{f}$ for any substance, it directly follows that the value of $\Delta H^\circ_\mathrm{f}$ for any element in its standard state must be zero.  For instance,  $\Delta H^\circ_\mathrm{f}$ for $\ce{H2_{(g)}}$ is equal to $\Delta H^\circ$ for the following reaction:
$$\ce{H2_{(g)} -> H2_{(g)}},$$
which is necessarily zero.
To give an analogy: Suppose you define the "altitude of formation", $\Delta z_f$, of any location on earth as the change in altitude necessary to reach that location from sea level.
Hence $\Delta z_f$ for the summit of Mt. Everest is $\Delta z$ for the altitutude change associated with:
$$\ce{sea level -> summit of Everest},$$
which is 29,029 feet.
It necessarily follows, from this convention, that $\Delta z_f$ for any location at sea level is zero, since that would be equal to $\Delta z$ for:
$$\ce{sea level -> sea level}$$.


Run a command at every step of a relaxed scan In Gaussian, I'm trying to do a relaxed scan and for each step obtain the volume. However, when I use the volume keyword, it seems to only output the volume for the initial and final steps. Is there a way to ensure a keyword runs for every step of a scan? A similar question (How to obtain the Raman spectrum along every coordinate of a scan in Gaussian?) was asked earlier this year, but I want to confirm whether Gaussian doesn't allow any other keywords to run during a scan.

Guess what: It is possible (in your specific case).

I was wrong. At least a little bit. Most of what is described below is true for the general purpose, and I will not change that part of the answer.

However, in the specific case where you only want to compute the volume, you are able to do that because I think it is part of the population analysis module. You can actually request a population analysis at every step of the optimisation, but be careful what you wish for (running NBO will take forever). That way the optimisation will be a little slower, but that is the most convenient way. Therefore the solution to your very specific problem is adding the following line to the route section:

pop=always volume

Be aware, that the output will be an even bigger nightmare than before.

Here is a complete input of the $\ce{H3B-NH3}$ I have used previously.

%chk=df-bp86-d3.def2svp.chk
%nproc=4
%mem=16000MB
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
pop=always                         ! Perform a population analysis at every step
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Scan

0 1
  N       0.00000        0.00000       -0.80852
  H       0.92634        0.00000       -1.36025
  H      -0.46317       -0.80223       -1.36025
  H      -0.46317        0.80223       -1.36025
  B       0.00000        0.00000        0.19148
  H       0.52348        0.90670        0.85928
  H      -1.04696        0.00000        0.85928
  H       0.52348       -0.90670        0.85928

B 1 5 S 5 0.2

For comparison (I'm not 100% sure I used the right values):

Energies
Step1 SCF Done:  E(RB-P86) =  -82.7531561505     A.U. after    8 cycles
Step2 SCF Done:  E(RB-P86) =  -83.0511089272     A.U. after    7 cycles
Step3 SCF Done:  E(RB-P86) =  -83.1459883448     A.U. after    7 cycles
Step4 SCF Done:  E(RB-P86) =  -83.1661412429     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1606918387     A.U. after    8 cycles
Step6 SCF Done:  E(RB-P86) =  -83.1485268431     A.U. after    8 cycles

Volumes
Step1 Molar volume =  491.032 bohr**3/mol ( 43.819 cm**3/mol)
Step2 Molar volume =  415.148 bohr**3/mol ( 37.047 cm**3/mol)
Step3 Molar volume =  401.060 bohr**3/mol ( 35.790 cm**3/mol)
Step4 Molar volume =  406.785 bohr**3/mol ( 36.301 cm**3/mol)
Step5 Molar volume =  514.428 bohr**3/mol ( 45.907 cm**3/mol)
Step6 Molar volume =  453.278 bohr**3/mol ( 40.450 cm**3/mol)

The energies differ very slightly only (from the below approach), but the volumes do differ at some points. Now that could be, that I just used the wrong population analysis, or something else. I don't have the time to investigate. (Leave a comment or edit if you find out.)

I've again looked into that, but without some serious hacking via the external keyword interface, I don't think it is possible. I also couldn't find a suitable iOP to control what gets punched to checkpoint file and when, and then use the result from that for a property/analysis run.
The punch keyword is equally disappointing, only outputting the last state.

While the manual suggests that you can access a given step of a scan in the checkpoint file, I found this to be not true. The documentation is fuzzy on that part and it also warns, that not everything is stored. Since it is a binary file, there is also no real way of checking what is contained.
I guess the real reason for the existence of that option is to restart a calculation that went rogue with different parameters from an intermediate calculation.

I was equally unable to find a way to set a given redundant coordinate to a specific value and then freezing it in order to perform a partial optimisation at that point, and then manipulate that coordinate while reading in the guess and perform another calculation.
Even with the new and generally awesome way to define general internal coordinates in Gaussian 16 (GIC) I was unsuccessful.

From all that trying, I came up with the probably easiest (=lazy) way (not computationally most efficient) to do what you want is to perform the scan, extract the (p'optimised) coordinates and run a series of single point calculations.

A slightly less (probably) computationally heavy approach would be to perform an initial scan on a very low level (like pm6, if your molecule is well-behaved), and the do p'optimisations on the resulting steps. Usually pm6 is able to offer reasonable starting geometries, which in many other cases may also help you reduce computational effort. So this is probably easy and quite efficient.
There are various options to extract geometries from a scan, the following question deals exactly with that: Extract all structures of Gaussian 09 molecular dynamics calculation using babel?

There certainly are more efficient ways to do this kind of task, but in the end, most of them require manual labour.

N.B.: I cannot believe it, but the pm6/popt approach is actually a halfway decent application of compound jobs. Although I have to admit I have not tried it yet.
It works as the following example demonstrates. (I used Gaussian 16 Rev. A.03, but it should work with Gaussian 09 Rev D.01, too. Maybe even earlier.)

I set up an initial scan on the pm6 level of theory for $\ce{H3B-NH3}$ and scanned the $\ce{B-N}$ bond:

%chk=pm6.scan.chk
%nproc=2
%mem=8000MB
#P PM6
OPT(MaxCycle=100)
SYMMETRY(loose)
GEOM(ModRedundant)
Volume

title

0 1
 N     0.000000     0.000000    -0.764994
 H     1.019690     0.000000    -1.164250
 H    -0.509845    -0.883078    -1.164250
 H    -0.509845     0.883078    -1.164250
 B    -0.000000     0.000000     0.235006
 H     0.509845     0.883078     0.634262
 H    -1.019691     0.000000     0.634262
 H     0.509845    -0.883078     0.634262

B 1 5 S 5 0.2

I have extracted the pre-optimised geometries with Chemcraft and added a header section and made use of the --Link1-- separator. After the first optimisation I read the MO for a slight speedup. I'm sure some of this can be further automatised, but for the sake of demonstrating, I did it by hand. The commented input:

%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 1

0 1
  N       0.00000        0.00000       -0.80852
  H       0.92634        0.00000       -1.36025
  H      -0.46317       -0.80223       -1.36025
  H      -0.46317        0.80223       -1.36025
  B       0.00000        0.00000        0.19148
  H       0.52348        0.90670        0.85928
  H      -1.04696        0.00000        0.85928
  H       0.52348       -0.90670        0.85928

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 2

0 1
  N       0.00000        0.00000       -0.90852
  H       0.94043        0.00000       -1.36414
  H      -0.47021       -0.81444       -1.36414
  H      -0.47021        0.81444       -1.36414
  B       0.00000        0.00000        0.29148
  H       0.53853        0.93276        0.86317
  H      -1.07706        0.00000        0.86317
  H       0.53853       -0.93276        0.86317

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 3

0 1
  N       0.00000        0.00000       -0.99853
  H       0.95023        0.00000       -1.38525
  H      -0.47511       -0.82292       -1.38525
  H      -0.47512        0.82292       -1.38525
  B       0.00000        0.00000        0.40147
  H       0.55372        0.95907        0.87762
  H      -1.10743        0.00000        0.87762
  H       0.55372       -0.95907        0.87762

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 4

0 1
  N       0.00000        0.00000       -1.07752
  H       0.95590        0.00000       -1.42049
  H      -0.47795       -0.82783       -1.42049
  H      -0.47795        0.82783       -1.42049
  B       0.00000        0.00000        0.52248
  H       0.56746        0.98288        0.89885
  H      -1.13493        0.00000        0.89885
  H       0.56746       -0.98288        0.89885

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 5

0 1
  N       0.00000        0.00000       -1.14730
  H       0.95649        0.00000       -1.46941
  H      -0.47824       -0.82834       -1.46941
  H      -0.47824        0.82834       -1.46941
  B       0.00000        0.00000        0.65270
  H       0.57849        1.00197        0.92763
  H      -1.15698        0.00000        0.92763
  H       0.57849       -1.00197        0.92763

B 1 5 F

--Link1--
%chk=df-bp86-d3.def2svp.rescan.chk
#P BP86/def2SVP/W06                ! Density Functional Theory Calculation
DenFit                             ! Use density fitting
empiricaldispersion=GD3            ! Use Grimme Dispersion
guess(read)                        ! Use the MO from the previous step
opt(MaxCycle=100,Loose,ModRed)     ! Use more optimisation cycles (Loose only for speed)
scf(xqc,MaxConventionalCycle=500)  ! If necessary, resort to quadratic convergence
int(ultrafinegrid)                 ! Larger Grid
scrf(pcm,solvent=water)            ! Use solvent
gfinput gfoldprint iop(6/7=3)      ! For molden
symmetry(loose)                    ! Loosen symmetry requirements
Volume                             ! Report vdW volume

Water H3B-NH3 DF-BP86-D3(PCM)/def2-SVP
Step 6

0 1
  N       0.00000        0.00000       -1.21413
  H       0.95402        0.00000       -1.53258
  H      -0.47701       -0.82621       -1.53258
  H      -0.47701        0.82621       -1.53258
  B       0.00000        0.00000        0.78587
  H       0.58528        1.01374        0.96868
  H      -1.17057        0.00000        0.96868
  H       0.58528       -1.01374        0.96868

B 1 5 F

For comparison:
Energies

Step1 SCF Done:  E(RB-P86) =  -82.7531561507     A.U. after    8 cycles
Step2 SCF Done:  E(RB-P86) =  -83.0511087360     A.U. after    8 cycles
Step3 SCF Done:  E(RB-P86) =  -83.1459860378     A.U. after    7 cycles
Step4 SCF Done:  E(RB-P86) =  -83.1661471840     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1606709617     A.U. after    8 cycles
Step5 SCF Done:  E(RB-P86) =  -83.1484921220     A.U. after    8 cycles

Volumes

Step1 Molar volume =  459.690 bohr**3/mol ( 41.022 cm**3/mol)
Step2 Molar volume =  479.017 bohr**3/mol ( 42.747 cm**3/mol)
Step3 Molar volume =  401.060 bohr**3/mol ( 35.790 cm**3/mol)
Step4 Molar volume =  406.785 bohr**3/mol ( 36.301 cm**3/mol)
Step5 Molar volume =  570.344 bohr**3/mol ( 50.897 cm**3/mol)
Step6 Molar volume =  498.606 bohr**3/mol ( 44.495 cm**3/mol)

Animation


Why can't we oxidise allylic alcohol to aldehyde using pyridinium chlorochromate? I read in my book that we cannot use pyridinium chlorochromate (PCC). Why is that?

There are indeed examples of $\mathrm{PCC}$ oxidation of allylic alcohols to $\alpha, \beta$-unsaturated aldehydes. The first example is from the original Corey-Suggs paper while the other two cases are from the work of Dauben and Michno.

References:

E. J. Corey, J. William Suggs, “Pyridinium chlorochromate. An efficient reagent for oxidation of primary and secondary alcohols to carbonyl compounds,” Tetrahedron Letters 1975, 16(31), 2647-2650 (https://doi.org/10.1016/S0040-4039(00)75204-X).
William G. Dauben, Drake M. Michno, “Direct oxidation of tertiary allylic alcohols. A simple and effective method for alkylative carbonyl transposition,” J. Org. Chem. 1977, 42(4), 682–685 (https://doi.org/10.1021/jo00424a023).


What pressure is created by the electrolysis of water? I am interested to know what is a reasonable maximum working pressure I could expect against the evolution of oxygen and hydrogen when electrolysing a typical water electrolyte.

My aim is to make small quantities of isolated $\ce{H2}$ and $\ce{O2}$ for a hobby hydrogen torch and I am curious how tall I can make the middle branch of the Hofmann Voltameter before the back pressure stops the production of new gas.  Having a predictable gas pressure would allow me to use the collection spaces for gas storage. I do not like the idea of mixing the gases before the torch nozzle like some brave people do so plan to use two collection volumes.

What parameters would be most likely to effect this pressure, electrolyte makeup, temperature, voltage?

Bonus question, is the pressure of $\ce{H2}$ and $\ce{O2}$ expected to be the same in theory and/or practice?

EDIT:
It looks like there is a high pressure electrolysis process out there that operates at many atmospheres of pressure and has been used to pressurise the gas products.  I missed it in my first readings on Wikipedia and further research has confirmed it though information is limited, it seems to be a specialised process usually associated with proton exchange membrane systems I think but I am just hoping to use two metal electrodes.

The maximum gas pressure you can expect in a Hofmann Voltameter is the atmospheric pressure plus the pressure from the water column in the middle cylinder ($1~\pu{atm} / 10~\pu{m}$).

For a reasonably sized apparatus, you will not reach conditions where you had to consider "back pressure" effects that might stop production of hydrogen or oxygen.

Given that the cross section of all three cylinders is the same, the pressure of hydrogen is slightly higher than oxygen pressure since the water column that adds pressure is higher (as can be seen in the Wikipedia article you've linked.)


Viewing a protein or bio structure in Chembio3D in its natural conformation I notice when using SMILES or a IUPAC name for many molecules and generating them in Chembio3D, and then running MM2 > Minimize Energy, the structures produced are most of the time not reflections of any conformation observed physiologically.

I'm wondering if there's a way I can import protein and other macromolecules into Chembio3D, perhaps from a database somewhere, where I can view their precise arrangement either as found in crystallography or in vivo in their usual forms at physiological pH. Most importantly, the proteins would need to be accurate (something I find lacking when generating structures by name in chemdraw/chembio3d).

I appreciate that the proteins and other molecules are always changing conformation and interacting with other molecules, but at least a reasonable prediction of some realistic conformations to view them in 3D would be nice, instead of conformations that might rarely/never exist in a cell. Is this possible? And if so, how would I go about this? What would be the specific steps I need to follow to import these proteins into Chembio3D?

The site you are looking for is the Protein Data Bank.

You can get to it with one of these links.
www.wwpdb.org
www.pdbe.org
www.rcsb.org/pdb
www.pdbj.org


What are some popular science books in chemistry? I have been reading the books The Elements by Theodore Gray and Stuff Matters by Mark Miodownik, both being very popular chemistry science books. Are there some other easy-to-read, popular but scientifically accurate chemistry books available?

The answer to the question should not be "primarily opinion based", since "popular" can be measured.

1) Curiosity: How Science Became Interested in Everything
by Philip Ball
 (https://www.goodreads.com/book/show/13650233-curiosity)

2) Molecules: The Elements and the Architecture of Everything
by Theodore Gray, Nick Mann
(https://www.goodreads.com/book/show/20729580-molecules)


Friedel–Crafts alkylation of benzene with tetrachloromethane There was this question in an exam at our school which asked what the major product will be when excess of benzene is reacted with tetrachloromethane in the presence of aluminium chloride.

All of us got it wrong in the exam, but we realized that it's just Friedel–Crafts alkylation with extra steps. So later we came up with two competing answers: one group claimed it is tetraphenylmethane $(\ce{CPh4})$, while the other claimed it's triphenylchloromethane $(\ce{CClPh3})$ (aka trityl chloride).

Reason in favour of $\ce{CClPh3}$: After the formation of it, the negative inductive effect of the phenyl groups will supposedly be high enough not to allow the release of the last remaining Cl to form a carbocation. Also, there will be high steric hindrance in $\text{CPh}_4$.

Reason in favour of $\ce{CPh4}$: The phenyl groups are bonded to the central carbon through $\sigma$ bonds. So they can just rotate their planes to escape steric hindrance. Also, we do not know if the combined inductive effect of the phenyl groups is high enough not to allow the release of chloride. After all, the chlorine molecule is highly electronegative.

So, we're grappling with this problem and do not know the solution to it. Any help will be highly appreciated.

Fonken1 says that the reaction doesn't occur, but it is not due to steric hindrance or instability of tetraphenylmethane since the molecule is actually thermally stable. Instead, it argues that the formation of tetraphenylmethane due to alkylation of benzene using triphenylchloromethane is reversible in nature. The product readily dissociates into triphenylmethyl carbocation and benzene. Due to this unfavourable equilibrium, only insignificant amounts of tetraphenylmethane is formed.

Ben Norris also mentions that the reaction of benzene with tetrachloromethane is used to produce chlorotriphenylmethane, and provided an Organic Synthesis procedure2 that works on the kilogram scale.

References

Fonken, G. J. Alkylation of Benzene with Triphenylmethyl Chloride. J. Org. Chem. 1963, 28 (7), 1909–1909. DOI: 10.1021/jo01042a510.
Bachmann, W. E. Triphenylchloromethane. Org. Synth. 1943, 23, 100. DOI: 10.15227/orgsyn.023.0100.


What are the correct equilibrium arrows? I was recently subject to some atrocious teaching. One of the gems that found me was the equilibrium arrow usage. I am accustomed to write $\ce{<=>}$. Yet, the entire class had to use ⇆ throughout the semester. Teacher's argument was that each harpoon from $\ce{<=>}$ has a meaning already, the movement of a single electron in a mechanism.

I concur, but I can't think of an situation in which the meaning of a harpoon is unclear because of the context. This is unlike Y being the symbol for both EDTA and yttrium, allowing $\ce{[YY]^{-1}}$ to have a meaning.

I tried to find the correct character, but not even the IUPAC Gold Book (1) (2) has a clear stance on it.

According to IUPAC: Quantities, Units and Symbols in Physical Chemistry (Green Book), the arrows are used differently for composite mechanisms in chemical kinetics.

In case of a simple mechanism that is composed of a forward and a reverse elementary reaction (i.e. reactions that occur at the molecular level in one step), it is conventional to write these in one line using two double-sided arrows $(\rightleftarrows)$.
For example:

$$\begin{align}
\mathrm{A} &\to \mathrm{B + C}\\
\mathrm{B + C} &\to \mathrm{A}\\
\hline
\mathrm{A} &\rightleftarrows \mathrm{B + C}
\end{align}$$

This is distinguished from a net reaction, which is written with two one-sided arrows $(\rightleftharpoons)$.
For example:

$$\begin{align}
\mathrm{A} &\rightleftarrows \mathrm{B + C}\\
\mathrm{B + C} &\rightleftarrows \mathrm{D + E}\\
\hline
\mathrm{A} &\rightleftharpoons \mathrm{D + E}
\end{align}$$

The two-sided arrow $(\leftrightarrow)$, which indicates resonance structures, shall not be used for reactions.

I guess that every elementary reaction that is proceeding in both directions $(\mathrm{A} \rightleftarrows \mathrm{B + C})$ may also be considered a net equilibrium reaction $(\mathrm{A} \rightleftharpoons \mathrm{B + C})$, but not every net equilibrium reaction $(\mathrm{A} \rightleftharpoons \mathrm{D + E})$ is an elementary reaction that occurs at the molecular level in one step $(\mathrm{A} \rightleftarrows \mathrm{D + E})$. Therefore, I assume that one-sided arrows $(\rightleftharpoons)$ may be used for all equilibrium reactions, whereas double-sided arrows $(\rightleftarrows)$ could be wrong when one combines composite mechanisms to obtain net reactions.


Thermodynamics entropy and temperature In thermodynamics, whenever heat is released from the system to the surroundings, temperature of surroundings remains unchanged but entropy increases. But entropy is a function of temperature. How is it possible?

In scenarios like the one you describe, the surroundings are typically modeled as an infinite reservoir, with the capacity to supply or absorb heat with negligible change in temperature.  If you treated it as a large finite reservoir instead, and evaluated its entropy change based on $\Delta S=mc\ln ((T_0+\Delta T)/T_0)$ and its heat absorbed by $Q=mc\Delta T$ (where m is the mass of the reservoir and c is its heat capacity), you would find that, in the mathematical limit of "mc" becoming infinite (at constant Q), the entropy change would approach $\Delta S=Q/T_0$.


Comparing the Hückel and extended Hückel methods I'm very confused about the differences between these methods. From my textbook, it states that the Hückel method only takes into account the π bonding interactions, while the extended Hückel method takes into account all the valence electrons. But what exactly does this do?

To my knowledge, the s, pz, px, and py orbitals are all orthogonal to one another and wouldn't mix anyways. Are shielding effects the only thing accounted for by extended Hückel theory?

In the simple Hückel method (SHM) the basis set is limited to p orbitals. This set is limited in a great extent to pz orbitals which constraints the molecular plane to be the xy plane. Basically, you are limited to planar molecules.
  The inclusion of all valence s and p orbitals in the extended Hückel method (EHM) naturally lifts the spacial constraints and you can work with non-planar molecules.

These are the two first differences between the two methods. The other points are:

SHM:

Orbital energies are limited to same-atom interactions, adjacent-atom interactions while all other interactions are 0.
Fock matrix elements are not actually calculated.
Overlap integrals are limited to 1 or 0.

EHM:

Orbital energies are calculated and vary smoothly with geometry.
Fock matrix elements are actually calculated.
Overlap integrals are actually calculated.

You can look up the derivation and steps for the implementation of these two methods in this book that I used as a reference:

Errol G. Lewars; Computational Chemistry, Introduction to the Theory and Applications of Molecular and Quantum Mechanics, Second Edition;
Springer: 2011. DOI: 10.1007/978-90-481-3862-3


How can isotope labelling be used to determine the mechanism of ether cleavage with hydrogen iodide? When methyl tert-butyl ether (MTBE) is treated with one equivalent of hydrogen iodide, the ether C–O bond is cleaved.

I read that if the oxygen in MTBE is isotopically labelled, it is possible to work out the mechanism, but I am not able to understand the function of $\ce{^{18}O}$.

$$\ce{CH3-^{18}O-C(CH3)3 + HI ->[S_N1/S_N2] products}$$

How does isotopic labelling help determining the mechanism?

If your products are simply an alcohol plus an alkyl iodide, then you really don't need the isotopic labelling. If you have an $\mathrm{S_N1}$ mechanism operating, then iodide ion attacks the tert-butyl cation, and you get the products

$$\ce{Me-O-^tBu + HI ->[S_N1] MeOH + ^tBuI}$$

If you have an $\mathrm{S_N2}$ mechanism, then iodide ion attacks the methyl group, displacing t-butanol:

$$\ce{Me-O-^tBu + HI ->[S_N2] MeI + ^tBuOH}$$

In both cases it should be trivial to take a proton NMR of the crude reaction mixture and determine which set of products are present.

The isotope label is more pertinent if you are forming two alcohols as the products. In that case, if the $\ce{^18O}$ remains attached to the methyl group (for example), it means that the $\ce{^tBu-O}$ bond was cleaved: this indicates an $\mathrm{S_N1}$ mechanism. In other words:

$$\ce{Me-^{18}O-^tBu ->[S_N1] Me^{18}OH + HO-^tBu}$$

Likewise

$$\ce{Me-^{18}O-^tBu ->[S_N2] MeOH + H^{18}O-^tBu}$$

However, you would have to be careful when running this experiment: if you leave it too long, especially under acidic conditions, the labelled oxygen will simply be exchanged with unlabelled oxygen from water by a degenerate nucleophilic substitution reaction, destroying any precious evidence that you had obtained:

$$\ce{R^{18}OH + H3^{16}O+ <=>> R^{16}OH + H3^{18}O+}; \quad \ce{R} = \ce{Me}\text{ or }\ce{^tBu}$$

where the equilibrium position lies to the right because of the larger concentration of $\ce{H3^{16}O+}$. Equilibrium tells us nothing about kinetics, though, so as stated earlier, you will have to stop the reaction (or monitor it using e.g. mass spectrometry, since NMR won't tell you the difference between oxygen isotopes) before this happens.


Eco Friendly Cleaning Solution I would like to find a solution that can kill roof algae, fungus, mold, mildew effectively without using environmentally unfriendly products but don't know enough about chemistry to make a solution myself.

I suggest to stay with commercially available, read-made solutions. These products are thoroughly tested and certified.

They typically contain around 5% of various benzalkonium chlorides, such as N-Myristyl-N-benzyl-N,N-dimethylammonium chloride, as biocides.

UPDATE As far as the mildew is concerned: The orange ladybird (16 white spots) and the 22-spot ladybird apparently feed on mildew. If these two are naturally found in your region, get some. This is as eco-friendly as it can get to keep mildew in the garden close to a minumum.


How to convert glucose to fructose How to convert glucose to fructose? Moreover, can there be direct conversion(not necessarily in a single step), or we have to use indirect means, such as from sucrose, or other higher carbohydrates, and then separating fructose from the products?

After searching on Google about Isomerization of Glucose to Fructose, you can find so many sources that explain the process. I think that this one is easy to understand among them. It explains that

  The isomerization of glucose to fructose is part of the glycolysis cycle that converts glucose to pyruvate. The way this is done is to isomerize the aldehyde (hemiacetal) glucose to the ketone (as a hemiacetal) fructose,and make another phosphate ester. The isomerization takes advantage of the ease of breakage of a C-H bond which involves a carbon next to a carbonyl carbon.

This gives a clear Idea on the mechanism of Glucose Isomerage to fructose. :)

Also see this.


Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane? I tried to draw the frontier molecular orbitals of the following biradical structure:

(A) At the top I've drawn the HOMO and LUMO, because I know that the reaction coordinate belongs to the $\ce{C_{$2v$}}$ point group. These two orbitals of the HOMO (green) lies in the O-C-O plane. In each $p$ orbital I would locate one electron (triplett state is favoured because of Hunds rule). Oxygen has six  valence electrons, so I must bring four electrons in orbitals of each oxygen. I can place two electrons in a $p$ orbital perpendicular to the greenish depicted orbital. Then I would place the other two electrons in a $sp$ hybrid orbital.

(B) In the second possibility (at the right at the bottom) I also used a $sp$ hybridization for the oxygen atom. The two radicals are placed in the $p$ orbital perpendicular to the O-C-O plane. I guess the structure is less favoured than the first possibility (with the two radical in the O-C-O plane) because there are unfavoured filled-filled interactions between pink-pink and green-green while above there is onyl a pink-pink destabilizing interaction.

(C) As a third possibility I've used a $sp^2$ hybridization of the oxygen. The radicals are now perpendicular to the plane. Without exact calculation of the hybrid orbitals I would predict that 4 electrons in $sp^2$ hybrid orbitals plus one electron in a $p$ orbital is lower in energy as case (A) (3 electrons in $p$ orbital and two electrons in a $sp$ hybrid orbital). However the green lobes (both two electrons and hence repulsive interaction) here in the case (C) in the $sp^2$ hybrid orbitals seems to come closer (120°) than the two electrons in the orthogonal $p$ orbitals (also filled-filled) in case (A).

Are these reasons why the HOMO is best described by case (A)>(C)>(B)?

Probably this question has similiarties to the hybridization in a previous question (please note I'm a novice concerning hybridization)
Why allene cannot be described with an allyl system?

I think you are asking what does the HOMO and LUMO of the 1,3-dioxabiradical look like.

Let's start by using cyclopropane as a model compound.  In cyclopropane the $\ce{C-H}$ bonds are approximately  $\ce{sp^{2.46}}$ hybridized and the $\ce{C-C}$ bond is approximately $\ce{sp^{3.74}}$ hybridized (see here).  That's a lot of p-character in the $\ce{C-C}$ sigma bond. Dioxirane is likely similarly hybridized.  As soon as the 1,3-biradical forms (as soon as we break the $\ce{O-O}$ bond), the HOMO and LUMO will look like what is shown in the "initial" drawing below.  If the biradical lives long enough to undergo rotation about the $\ce{C-O}$ bond, then the HOMO and LUMO will look like what is shown in the "rotated" drawing.

These drawings just point out the conformational dependence of the HOMO and LUMO.  Note that the two lone pairs on each oxygen have not entered the discussion.  They are orthogonal to, and have no effect upon, the HOMO and LUMO orbitals.

Imagine bringing two oxygen atoms closer to each other.  If we push them closer and the orbitals containing the odd electrons are pointing towards each other as in the "initial" geometry, we will form a sigma bond and we know what the HOMO and LUMO look like for a sigma bond (just like what is pictured in the "initial" geometry, only the orbitals are closer together).  If we push them closer and the odd electrons are pointing parallel to each other as in the "rotated" geometry, we will form a pi bond and we know what the HOMO and LUMO look like for a pi bond (just like what is pictured in the "rotated" geometry, only the orbitals are closer together). In the 1,3-biradical the carbon atom holds the 2 oxygen atoms together, they are closer than they would be at infinity, but further apart then they would be if there was a bond between them.  This tells us that the HOMO-LUMO energy separation is greater than it would be at infinity (0), but less than if they were closer together.

We could refine our model by noting that, according to Bent's rule, we would expect the orbitals containing the oxygen lone pairs to have a bit more s-character that the carbon $\ce{C-H}$ orbital in cyclopropane.  Therefore, in dioxirane and the corresponding 1,3-dioxabiradical, we would expect both the $\ce{O-O}$ bond and the orbital containing the single electron in the biradical to have even more p-character than $\ce{sp^{3.74}}$.

Again, however, to re-emphasize an earlier point, the lone pairs on oxygen have no bearing on our discussion.  That said, if we would like to speculate on the hybridization of the orbitals containing the lone pairs, we may do so.  Most texts tell us that the oxygen lone pairs in water, ethers and alcohols are in (roughly) $\ce{sp^3}$ orbitals.  However data from photoelectron spectroscopy experiments seems to suggest that one lone pair is in (roughly) a p-orbital while the other lone pair is in an $\ce{sp}$ orbital.


skeletal structure of dimethylamine? HN(CH3)2 I don't understand how to draw this skeleton.
 I know there is a NH2 in an amine structure, but this only has NH?
and dimethyl means there a 2 identical single side chains right? but where on the skeletal structure are they?
I'm totally stuck as to how the skeletal structure is supposed to look.

Please look at the skeletal structure of different types of amines:

I hope it's clear now.


Is carbon–sodium bond mainly ionic or covalent? We all know that carbon mostly forms covalent bonds (almost always) and sodium mostly forms ionic bonds (almost always). But if $\ce{NaOH}$ reacts with ethyne, then $\ce{HC#CNa}$ and $\ce{H2O}$ are formed. Then what type of bond will $\ce{C-Na}$ be: mostly ionic or covalent?

Electronegativity of Carbon is $2.55$ and that of Sodium is $0.93$ (from Wikipedia) so by the Haney-Smith equation (from J.D. Lee's Inorganic chemistry, third edition)
$$\%\text{Ionic Character} = 16 |\Delta\chi| =3.5 (\Delta\chi)^2,$$
where $\Delta \chi$ is the difference in electronegativity of the two atoms. Carrying out the calculation gives us
$$\% \textrm{Ionic Character} = 35.1054$$
So, it shows that the bond $\ce{C-Na}$ has a less ionic character.

In Organic Chemistry by Morrison and Boyd we find

  The carbon-magnesium bond is covalent but highly polar, with carbon pulling the electrons from electropositive magnesium [...].
  But, whatever the metal, it is less electronegative than carbon, and the carbon-metal bond - like the one in the Grignard reagent - is highly polar. Although, the organic group is not full-fledged carbanion [...] it nevertheless has considerable carbanion character.

Note: No bond is completely ionic or covalent, what matters is which character is dominant and hence the bond will behave like that.


Why does denaturation of albumen protein cause pH of solution to increase? I'm interested in the denaturation of proteins by alcohols, specifically by ethanol.
I have devised a simple experiment in which I inject solutions of ethanol (of variable concentrations) into solutions of egg albumen (containing proteins such as conalbumin). I have observed a marked pH change in solution after injection of ethanol. The rate of change of pH is in fact proportional to the concentration of ethanol injected (as expected).
However, I am having trouble explaining the theory behind why ethanol's disruption of tertiary protein structure increases pH. As I understand, ethanol disrupts proteins such as conalbumin by disrupting hydrogen bonds in the tertiary structure of the molecule.
But why does this cause the pH to increase (in effect, the concentration of H+ ions to decrease)?

The main component of egg white, ovalbumin, contains 93 negatively charged residues (Asp + Glu) and 68 positively charged residues (Arg + Lys), as well as 14 His. If you make the solvent less polar by adding ethanol to water, they will tend to form the neutral species (Arg+ and Lys+ will lose a proton, and Glu- and Asp- will pick up a proton).
If you had the protein unfolded already, you would expect that the solution gets more basic if the effect on all charged amino acids is the same because there are more negatively charged ones.
To test whether the observed effect has to do with denaturation of the proteins or with changes in protonation state of side chains already exposed to water, you could measure the pH of amino acids (preferably with a modification of the main chain carboxylic acid and amino functions so only the side chains show acid/base reactions) at different ethanol concentrations.


Why don't periodic acids convert alkenes to 1,2 diols? Periodic acids cleave 1,2 diols by forming an periodate ester as an intermediate. Osmium tetroxide also forms an osmate ester with alkenes and yields syn diols after a hydrolysis step. Given that their structures are so similar, why doesn't periodic acid also succeed in forming diols with alkenes? (These in another step will be further cleaved to ketones). Instead in the Lemieux-Johnson oxidation, osmium tetroxide is taken in the first step - followed by periodic acid.
Osmium tetroxide does not cleave 1,2 diols - though I thought it might be in its being not such a strong oxidising agent as periodate - it may not suffice to cleave a C-C bond. Why doesn't though periodic acid carry out diol formation as well?

Sodium periodate can be used indirectly to form diols from alkene, see:
Emmanuvel, T. M. A. Shaikh, A. Sudalai, Org. Lett., 2005, 7, 5071-5074.
The mechanism for periodate cleavage of diols also strongly suggests the periodate is an electrophile.  This may explain why in general, alkenes are not nucleophilic enough to react under normal conditions without a co-reagent. The osmium tetroxide reaction is favoured as a [3+2] cycloaddition mechanism.


optical rotation of styrene oxide Which optical isomer of Styrene Oxide  would be produced if it is synthesized via epoxidation of the halohydrin. i.e.

Is there a way to deduce this? I assume it would depend on which isomer the halohydrin itself is since that too seems to have an assymetric Carbon atom?

Well, the halohydrin itself is produced from styrene which is optically inactive by the action of hypochlorous acid. Does this enable us to deduce theoritically the optical rotation of the final product i.e. Styrene Oxide?

Styrene and HClO are going to make a racemic mixture, in which case the epoxide will also be racemic, i.e. there should be no optical rotation.

An alternative would be to use the Jacobsen epoxidation, the Shi epoxidation or some other enantioselective epoxidation.


Flame Test: Why does the flame colour of calcium disappear quickly? While performing flame tests for calcium fluoride and strontium fluoride, I observed strontium gives a long-lasting rose-red flame color but the brick red color for calcium quickly disappears. I also found some books mention the flame color for calcium as transient red.
Why does the flame color of calcium is transient?

A flame is a constantly upward "moving" chemical reaction. The velocity is dependent on the flow rate of the gases entering the burner. There is an extensive upward motion (on Earth). In a gravity-less flame, the flame is round, still there is an upward movement. Now another important thing to note is that the Group II flame emission is not due their atoms but their hydroxides, such as CaOH. In an ordinary Bunsen burner flame you can expect "weird" molecules, which cannot exist in an ordinary bottle. The typical flame temperature is not hot enough to cause atomic excitation and completely break calcium compounds into calcium atoms. If this were true you would see a violet flame (422 nm) after introducing calcium. We see reddish color which is always contaminated with sodium's yellow color. The flame color is then dependent on the thermal stability and volatility of the emitting compound being formed in the flame.


Solving for H+ given dissociation of weak acid, c and Ka Can the quadratic equation yield two nontrivial solutions? For example, if a question asks for $[\ce{H+}]$ of a $0.100~\mathrm{M}$ solution of picric acid (monoprotic), given that its $K_\mathrm{a}$ at the system temperature is $4.2 \times 10^{-1},$ following my teacher's instructions I would use the ICE table, which looks like this (for a different example).

$$\begin{array}{cccccccc}
 & \ce{CO (g)} & + & \ce{H2O (g)} & \ce{<=>} & \ce{CO2 (g)} & + & \ce{H2 (g)} \\
\text{Initial} & 0.100~\mathrm{M} & & 0.100~\mathrm{M} & & 0 & & 0 \\
\text{Change} & -x~\mathrm{M} & & -x~\mathrm{M} & & x~\mathrm{M} & & x~\mathrm{M} \\
\text{Equilibrium} & (0.100-x)~\mathrm{M} & & (0.100-x)~\mathrm{M} & & x~\mathrm{M} & & x~\mathrm{M} \\
\end{array}$$

I would then solve for $x$ by using the expression for $K_\mathrm{a}$. However, since $0.100/K_\mathrm{a} < 100$ (lowest initial concentration divided by $K_\mathrm{a}$ is less than 100), $x$ is significant and cannot be omitted from the $0.100-x$. Thus, you need to use the quadratic equation, which yields two solutions for my example above: $x=0.0834, 0.503$. The latter is obviously incorrect because it is greater than the initial concentration of the undissociated compound.

However, my question is: Is it possible that both solutions to a given reaction will be nontrivial, that is, is it possible that both answers appear to work at a glance? (No conflicts with the initial concentration, etc.) This question refers not specifically to the example in question, but rather looks at if it is possible at all (is there a reaction and starting concentration for which both answers would seem correct). In addition, if it is possible, how would one determine which answer is nontrivially impossible?

Whenever I get into a new subject I'm not able to "do" it until I work out some small details that stick with me.  It would be best if the answer had some logical explanation, or even a proof as to why the two answers will never be both possible at a glance, or if the answer shows a way to determine without fail which answer is correct. I don't see an obvious reason why both solutions resulting from the quadratic equation will always have one blatantly incorrect result.

Please ask for clarifications if necessary.

What you have got here is a strong acid ($\mathrm{p}K_\mathrm{a} \approx 0.38$) that is not diluted, whereof the latter tells you that you can safely ignore the autoprotolysis of water.

The pH approximation to choose is therefor:
$$[\ce{H+}] = \frac{[\ce{HA}]_0 K_\mathrm{a}}{[\ce{H+}] + K_\mathrm{a}}$$

or in a quadratic form:
$$[\ce{H+}]^2 + [\ce{H+}] K_\mathrm{a} - [\ce{HA}]_0 K_\mathrm{a} = 0$$

which has two solutions:
$$[\ce{H+}] = -\frac{K_\mathrm{a}}{2} \pm \sqrt{\left(\frac{K_\mathrm{a}}{2}\right)^2 + [\ce{HA}]_0 K_\mathrm{a}}$$

From those two solutions, only one is chemically meaningful, which is the positive combination which yields one of your solutions ... $[\ce{H+}] = 0.0834$.


Not a chemist, needs help with a calculation regarding addition of HCl to sodium formate solution As stated I am not a chemist, but I need help on a question I have.

Assume I have a Formic acid 75% and I add between 5-10 kg to 1.000 liter of water. If I wanted to add sodium formate and HCl 37% to water and achieve a similar result  with regards to pH and "pure" undissociated formic acid how much of each product would I need to add?

I have calculated, probaly wrong, that I would need 8 ml of HCl to 1.000 L to lower pH to around 4 which is close to pKa of FA (3,75), but I do not know if this will give me the same amount of undissociated FA as the original scenario.

For my application idea pH is less important than undissociated (antibacterial) FA.

Hope someone takes up the (over my head) calculation challenge

At $\mathrm{pH}=\mathrm{p}K_\mathrm{a}=3.75$ , there will be 50% of non-dissociated formic acid and 50% of formiate.

You need 50 % of stoichiometric amount of hydrochloric acid to convert 50% of the formiate to the acid.

If we consider the percentage as (w/w) %,  then $\pu{1 kg}$ of 75% formic acid ($M=\pu{46
g/mol}$) contains $\pu{16.3 mol}$ of formic acid.

For usage of $\pu{5 kg}$ to $\pu{1000 L}$, $\mathrm{pH}=\frac 12 \left(\mathrm{p}K_\mathrm{a}-\log(c)\right)\\
=\frac12\cdot \left(3.75-\log{\frac{5\cdot 16.3}{1000}}\right)\\
=\frac12\cdot \left(3.75-\log{0.0815}\right)=2.42$

For $\pu{10 kg}$, it is 2.27.

$\pu{1 kg}$ of the above formic acid  is equivalent to $\pu{ 16.3 mol}$ of sodium formiate, i.e. $\pu{ 1108 g}$

For 50% conversion, you need $\pu{ 8.15 mol}$ of $\ce{HCl}$, i.e. $\pu{ 8.15\cdot 36.5 = 297 g}$ $\ce{HCl}$, what is equivalent $\pu{ 804 g}$ of 37% $\ce{HCl}$


Why does CO2 diffuse through a butyl rubber membrane more readily than air? I am a cyclist, and not a chemist. When I or my cycling buddies get a flat tire out on the road, after removing the offending sharp object (if any) remaining in the tire, we will typically replace the punctured butyl rubber inner tube with a new one, and then inflate the tire using a $\ce{CO2}$ cartridge.

Inflating the tire in this way takes less than a second to bring the pressure up to 120 psi. At home, naturally, we fill our tires with air rather than a $\ce{CO2}$ cartridge, using a floor pump.

When we fill our tire with an air pump the tire remains inflated for several days. But after filling it with $\ce{CO2}$, the tire is severely flat again by the next morning. $\ce{CO2}$ clearly diffuses through the butyl membrane very much more rapidly than air, even though $\ce{CO2}$ is a larger molecule than either $\ce{N2}$ or $\ce{O2}$. Is it the shape of the molecule? Or its chemical properties?

The permeability of a gas through rubber depends mainly on its diffusivity and solubility in rubber.

$\ce{CO2}$ has a significantly higher solubility in rubber than $\ce{O2}$ and $\ce{N2}$, whereas the diffusion coefficients differ not that much.

The result is that carbon dioxide passes ordinary rubber about 5 times faster than oxygen and about 15 times faster than nitrogen[1, 2].


Are there compounds more conductive than Silver (Ag)? Silver is the most electrically conductive element on the Periodic Table. However, I am interested in knowing if there are any compounds (combinations of elements) that would be more conductive than Silver (without Superconductor cooling). The extreme electrical conductivity of specific chemicals has many real world applications, from simply calling someone miles away, to processing enormous amounts of data in servers.

Graphene supposedly has higher conductance.

The [...] resistivity of graphene sheets would be 10−6 Ω⋅cm. This is less than the resistivity of silver, the lowest otherwise known at room temperature.

but it is not a compound - it is a homonuclear molecule.