How to systematically find the reducible representation of a molecule? (eg: XeF4) I am attempting to use the reduction formula to find the irreducible representation of $\ce{XeF4}$ to determine its IR stretching vibrations. I know the point group of $\ce{XeF4}$ is $D_{4h}$ and have the character table but I am having a hard time understanding how to get the characters for the reducible representation. I was taught to do it by considering moved/unmoved vectors emanating from the point of the molecule giving values of 1, 0 and -1 for each vector after the symmetry operations have transformed them.
Using this method I to find 2$E_u$ irreps, but I found in a paper (http://sces.phys.utk.edu/~moreo/mm08/penchoff.pdf) that $\ce{XeF4}$ also has other irreps including an Au irrep involved in IR activity. How do I find the irreps that I missed? I noticed the paper uses different reducible representations ($\Gamma_{xyz}$, $\Gamma_{unmoved}$ and $\Gamma_{Tot}$) and I don't understand why.

A way to do it is to put 3 axes ($x$, $y$ and $z$) on each atom (to represent the 3N degrees of freedom) and determine the characters for each axis.

After that, if you need the representations for the 3N-6 (3N-5 for a linear molecule) vibrational modes, you need to remove the representations for the 3 translational modes (what they call $\Gamma_{xyz}$), as well as the representations for the 3 (2 for a linear molecule) rotational modes. Those are usually stated in the character table for the space group, but you can also determine them by checking the characters for a vector.

Example with $\ce{H2O}$

Character table for point group $C_2v$

C2v | E  | C2 (z) | v(xz) | v(yz) |  lin  |    quad
-------------------------------------------------------
A1  | +1 |     +1 |    +1 |    +1 |   z   | x2, y2, z2
A2  | +1 |     +1 |    -1 |    -1 |   Rz  |     xy
B1  | +1 |     -1 |    +1 |    -1 | x, Ry |     xz
B2  | +1 |     -1 |    -1 |    +1 | y, Rx |     yz

Character table for all 3N = 9 axes

C2v | E  | C2 (z) | v(xz) | v(yz)
----------------------------------
Ox  | +1 |     -1 |    +1 |    -1
Oy  | +1 |     -1 |    -1 |    +1
Oz  | +1 |     +1 |    +1 |    +1
H1x | +1 |      0 |     0 |    -1
H1y | +1 |      0 |     0 |    +1
H1z | +1 |      0 |     0 |    +1
H2x | +1 |      0 |     0 |    -1
H2y | +1 |      0 |     0 |    +1
H2z | +1 |      0 |     0 |    +1
----------------------------------
Γtot| +9 |     -1 |    +1 |    +3

Reduction of $\Gamma_{tot}$ to irreducible representations

Using the reduction formula $n_i = \frac{1}{h}\sum_R \chi_{\Gamma_{tot}}(R) \chi_i(R)$ for each irreducible representation $i$ of the point group, we get :

A1 : $\frac{1}{4} \times (9 + (-1) + 1 + 3) = 3$
A2 : $\frac{1}{4} \times (9 + (-1) + (-1) + (-3)) = 1$
B1 : $\frac{1}{4} \times (9 + 1 + 1 + (-3)) = 2$
B2 : $\frac{1}{4} \times (9 + 1 + (-1) + 3) = 3$

Therefore, $\Gamma_{tot}$ = 3 A1 + A2 + 2 B1 + 3 B2.

We can check that there are the 3N = 9 representations that are associated with the 3N degrees of freedom of the molecule.

For degenerate representations ($\chi(E) > 1$), the representations correspond to $\chi(E)$ modes.

Isolation of the vibrational representation

Determine $\Gamma_{trans}$ and $\Gamma_{rot}$ (from the character table or by hand).

$\Gamma_{trans}$ corresponds to the linear terms in the table, $\Gamma_{rot}$ corresponds to the rotational linear terms in the table.

For $C_2v$ :

$\Gamma_{trans}$ = A1 + B1 + B2
$\Gamma_{rot}$ = A2 + B1 + B2

You can check that there are 3 modes for $\Gamma_{trans}$ and 3 (2 for a linear molecule) modes for $\Gamma_{rot}$

For degenerate representations, you will see things like $x + y$ in the character table. This means that this representation corresponds to both x and y and should be included only once

Finally, $\Gamma_{vib}$ = $\Gamma_{tot} - \Gamma_{trans} - \Gamma_{rot}$ = 2 A1 + B2.

Activity of the vibrational modes in IR and Raman

IR activity is determined by the dipole moment and a mode will be active if the irreducible representation contains a linear part (x, y or z).

Raman activity is determined by the polarisability and a mode will be active if the irreducible representation contains a quadratic part ($x^2$, $xy$, ...).

For $C_2v$, all the representations correspond to active modes in Raman and all except A2 correspond to active modes in IR.

Therefore, the 3 vibrational modes of $\ce{H2O}$ (2 A1 and B2) are active in both Raman and IR.

Two modes have the same irreducible representation: A1. That does not mean that they have the same wavenumber. For $\ce{H2O}$, there are 3
vibrational modes with distinct wavenumbers.

For degenerate representations, there will be $\chi(E)$ modes with the same wavenumber.

In groups that have an inversion center (like $D_4h$), linear and quadratic terms are mutually exclusive. Therefore, an irreducible representation will always correspond to a mode active only in IR or Raman, not both.


Is this the right conclusion for a thought experiment? In this thought experiment, let’s consider the surrounding to be air that is composed of entirely water vapor (no other species like $\ce{O2}$ and $\ce{N2}$ are present). The surrounding is infinitely large. The system in question is a pan containing a thin layer of liquid water. The liquid water is brought to its boiling point by uniform heating. This pan is open (not covered in any way) and is exposed to the surrounding water vapor.

By definition, boiling occurs when the vapor pressure of the liquid is equal to the external pressure. So in this case, at boiling point, the vapor pressure of the liquid $\ce{H2O}$ is equal to the external pressure of gaseous $\ce{H2O}$. Thus, the system is in a dynamic equilibrium, and the rate of vaporization is equal to the rate of condensation.

This implies that the amount of liquid water will not change over time. In particular, it will not all “boil off” as steam. Is this the right conclusion? It certainly seems a little absurd.

This conclusion assumes that the enthalpy of the water remains entirely constant, which is not the case. Given that the water is being uniformly heated, but no end to the heating was specified, the enthalpy of the water will continue to increase, and thus force more and more of the liquid water into vapor form. Additionally, as you specified a very large surrounding, any water that enters the gaseous form would not largely impact the air pressure of the environment, so there would be few things acting to maintain some kind of equilibrium here.


Solution of the Roothaan Equations of H2 by Symmetry Arguments FWIW my background is in physics and maths, but I am just starting a chemistry PhD (the last time I took a chemistry class was high school). I have only some background in representation theory, and only an abstract taste, as it was in a maths class (bonus: suggest a reasonable representation/group theory reference with applications to chemistry, preferably a concise treatment).
In Szabo and Ostlund's Modern Quantum Chemistry, section 3.5.2 they solve the ${\rm H}_2$ molecule (in Hartree-Fock theory) using a minimal STO-3G basis, $\phi_1$ and $\phi_2$ (centred on each proton). I believe I understand how to set up the Roothaan equations, guessing the density matrix, and solving self-consistently. However, they go on to say "The canonical molecular orbitals will form a representation of the point group of the molecule." Um, okay. Why? And what is meant by canonical here? They go on to say they can be labelled by their symmetry $\sigma_u, \sigma_g, \pi_u, \pi_g$, etc. (I know these terms from wikipedia, and it is discussed briefly in another chapter) I don't fully understand why they (can) do this.
Now, I understand we have two molecular orbitals so it makes sense that they choose only the 2 lowest energies. However, from the set of options above, they state without proof that the $\sigma_g$ is the lowest energy, followed by $\sigma_u$ (I also don't really understand where the wave function comes from). How do we conclude this? Is it because the hydrogen atoms are 1.4 Bohr radii away, hence close and hence we expect bonding (not antibonding?), which I understand to be represented by $\sigma_g$? If this is correct, then it's still not rigorous so I would appreciate any more explanation.

After speaking to a coworker and reading some more I think I have a better understanding.
The symmetry of the wave function needs to follow the symmetry of the molecule (unless we are studying a broken symmetry solution, e.g. charge transfer, but that doesn't apply in this case). We are using a minimal basis, and hence we have only two orbitals. From atomic theory, these are s functions. Hence, they cannot form $\pi$ orbitals, and we are left with $\sigma$ orbitals. Between $\sigma_g$ and $\sigma_u$, $\sigma_g$ has the lower energy because it has no nodes, whereas $\sigma_u$ contains a node. From QM we know that this generally implies greater energy, but there's a nice explanation on this here (at the end he also specifically discusses a homonuclear diatomic molecule).
The specifics of how to get the symmetry representation for the molecule and match those with wave functions is still not entirely clear to me, but for this I will read a book on group theory and its chemical applications. It is, however, clear that $\phi_1+\phi_2$ and $\phi_1-\phi_2$ respect $\sigma_g$ and $\sigma_u$ symmetry respectively.


Does partial charge violate the law of quantization of charge? Well, we know that charge is quantized. Quoting wikipedia below,

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

And we know that there is partial charge. Quoting wikipedia below,

A partial charge is a non-integer charge value when measured in elementary charge units.

Don't the above two things (and both being from the same respectable source) contradict with each other?
If yes, then which one is correct? And if both are true, then in which conditions do they follow?

Both are.

Any isolated object has integral charge, as any elementary particle has integral charge. However, as we know, elementary particles are not truly located in one point.

When we talk about stationary states in physics, we can assume, that particles inhabit some volume with special density function indicating how much of the particle in question is in any given unit of volume. When we consider isolated molecule, we can assume that it is in a stationary state. This means, that electrons may be viewed as electron clouds, and different parts of said cloud may be assigned to different atoms in the molecule. In case if total amount of electron density, assigned to specific atom is non-integral, we gain a partial charge.

Whenever atomic partial charges are estimated, there is a significant arbitrariness in the process, as there is no unquestionable way to define, which parts of electron density assigned to any given atom, as there is not strict definition of atom borders (atomic radii are estimated using equilibrium positions in some circumstances, but in truth electron density is non-zero even far beyond that radii). Still, the concept is useful enough.


Instability of gem dihydroxy compounds Organic compounds containing two hydroxy groups on one carbon (gem dihydroxy) are unstable and tend to lose water molecule. However, Boron, sulfur and phosphorus with two OH groups are stable. What is the reason?

The instability of gem-diols is due the fact that hydogen bonding between the 2 OH- groups increases the repulsion causing a strain on the C-O-H bonds.
In case of compounds of boron, a greater bond angle (due to sp2 hybridization of boron) prevents intramolecular hydrogen bonding.
Compounds of sulphur, phosphorous always contain a double bonded oxygen along with the OH- groups on the central atom. This oxygen atom forms hydrogen bonds with OH- groups preventing hydrogen bonding between the OH- groups, thus preventing any strain.


What classifies as a solvent/base in SN1, SN2, E1, and E2 reactions? I need to clarify some terminology for an intro to organic chem class. We are studying substitution and elimination reactions and it has not been made clear what exactly these terms describe. For instance, in the bottom example of this image I have seen CH3OH on both the top and bottom of the arrow. Does this imply that it is the nucleophile/base when it is on top, and a solvent when it is on the bottom?

Okay in the question i am only seeing CH30H that simply means that methanol is acting both as a base and a solvent.

Since methanol is acting both as a solvent and a reagent, it will undergo E1 reaction as you will know that it is a polar protic solvent and if the solvent was like **water(polar) or DMSO **  then it would follow SN1 and SN2 respectively and *CH30 (-) * would act as the nucleophile.  Thus solvents play an important role in understanding the type of reaction it would undergo.  So it is really important to understand the solvent. As per as i know there are no such conventions..( That  solvent is written below and reagent/base at the top).  If only one reagent  is given it is acting as a both solvent and a base. I would like to tell another thing that sometime we are mistaken a reagent to be a solvent :
DMSO is often used as a oxidizing agent in presence of phosphoric acid.  We mostly confuse it and hold phosphoric acid as the oxidizing agent.


What is difference between T1g and T2g? I know that the “$\mathrm{g}$” stands for symmetry of octahedral shape, but how are labels “1” and “2” given in $\mathrm{A_{1g}}$ and $\mathrm{A_{2g}}$ representations too…

Is it related to group theory? If so, then please elaborate.

Well, not quite. Here's the quick summary on subscripts in Mulliken notation for point group symmetry.

“g” stands for "gerade" and refers to the symmetry around an inversion center. (That is, the representation retains symmetry around the inversion center.) Yes, octahedral ($O_\mathrm{h}$) molecules have an inversion center, but so do others.

“u” stands for "ungerade" and refers to anti-symmetric around an inversion center.

“1” or “2” refer to symmetry (1) or anti-symmetry (2) around a secondary $C_2$ axis (e.g., in a $D_x$ point group like benzene).

The $'$ and $''$ assignments (if needed) are used for symmetry around the $\sigma_\mathrm{h}$ operation.

And, while you didn't ask, for “A” versus “B”, these are one-dimensional representations that differ with respect to the "main" rotational axis. “A” representations are symmetric “+1” and “B” are anti-symmetric “−1” for that axis.

You only need enough subscripts to uniquely identify a representation. For example, here's the character table for $D_{4h}$:

$$\begin{array}{c|cccccccccc|cc} \hline
D_\mathrm{4h} & E & 2C_4 & C_2 & 2C_2' & 2C_2'' & i & 2S_4 & \sigma_\mathrm{h} & 2\sigma_\mathrm{v} & 2\sigma_\mathrm{d} & & \\ \hline
\mathrm{A_{1g}} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_{2g}} & 1 & 1 & 1 & -1 & -1 & 1 & 1 & 1 & -1 & -1 & R_z & \\
\mathrm{B_{1g}} & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & 1 & -1 & & x^2-y^2 \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & 1 & -1 & 1 & -1 & 1 & & xy \\
\mathrm{E_g} & 2 & 0 & -2 & 0 & 0 & 2 & 0 & -2 & 0 & 0 & (R_x,R_y) & (xz,yz) \\
\mathrm{A_{1u}} & 1 & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & & \\
\mathrm{A_{2u}} & 1 & 1 & 1 & -1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{1u}} & 1 & -1 & 1 & 1 & -1 & -1 & 1 & -1 & -1 & 1 & & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & & \\
\mathrm{E_u} & 2 & 0 & -2 & 0 & 0 & -2 & 0 & 2 & 0 & 0 & (x,y) & \\ \hline
\end{array}$$

Note that all the “g” representations have +1 or +2 for the character of the “$i$” symmetry operation. We don't need “1” or “2” to distinguish the $\mathrm{E_g},$ but we do for the different $\mathrm{A}_{x\mathrm{g}}$ representations. For those, we go to the $C'_2$ axis, which separates $\mathrm{A_{1g}}$ from $\mathrm{A_{2g}}$. These are unique and we don't need to assign $'$ or $''$ for this point group.

(If you look up the $D_\mathrm{3h}$ character table, you find $'$ and $''$ needed.)


I want to make NaOH from NaHCO3, what is the easiest way to do it without electrolysis? I want to make NaOH from NaHCO3.
I can't use electrolysis as I lack the proper equipement.

You could heat it to turn bicarbonate into carbonate, and let it stand with Ca(OH2) and water mix. That's essentialy how potash lye was made for millenia


Are redox half-reactions "real"? Are the half-reactions used in analyzing redox reactions a real thing or are they a conceptual shortcut?

Is there a way to, however briefly, run a half-reaction by itself without the corresponding other half?  For example, reduce metal atoms to metal ions without the corresponding oxidation half-reaction?

Electrolysis/Galvanic cell is basically redox reaction split in half-reactions, running on different electrodes. For example, hydrogen fuel cell has two electrodes. On one hydrogen donates electrons to electrode:

$H_2 = 2H^+ + 2e^-$

on another, oxygen drains electrons from electrode

$O_2 + 4 e^- + 4 H^+ = 2 H_2O $

The electrons travel from one electron to another via external electricity consumer. If electrons are forced to travel by external force, the reaction may be reversed. This ideas are utilized in electric chemical batteries, accumulators, fuel cells, industrial electrolyzers and so on.

This way, half-reactions are separated between different electrode. Running half-reaction by itself without other half elsewhere is problematic at best, because of electroneitrality violation. You'll need extreme voltage source and even then it will stop quickly.


Is there any scenario in which the size of a molecule increases due to an increase in temperature? Take an ice cube for example. Heat is applied in a closed container until it is vaporized completely. Will the molecule's size be larger (on average)? Is there a substance that you know of that has odd behaviour such as changing molecular size due to temperature change? Obviously there would be more pressure in the case of $H_2O$. Have you heard of any substances that are exceptions to this rule? I would love to hear about it if you have.

If this question hasn't quite boggled your mind or you are having trouble thinking of an exception, let's say that an average diamond at about room temperature (something with entropy, ie: not a perfect lattice) and we say that this is a single macro-molecule because every carbon atom present in the diamond shares a bond with another carbon in the diamond.

Pretend we heat said diamond up a bit or let it sit for awhile. Are covalent bonds being broken and reformed as time goes on? If we heated it up enough without breaking the lattice completely, would the diamond itself, the molecule as a whole, increase in size at all? Or would the bonds just be broken more often and reformed, and the size change would be so negligible that it could hardly be considered an increase in volume?

First, let me remark to “the size of a molecule” is not particularly well-defined. I assume in the following that you are interested about bond lengths (which are averages over time of distances between bonded atoms), as seems to be the case in your example. In the simplest case, you can consider an isolated diatomic molecule (think: gas phase N2), and we suppose we're not going to temperature high enough to ionize it or otherwise break its bond. Because the interatomic potential is asymmetric:

you can see that, as the temperature increases and the system is allowed to explore more of the potential energy curve (climb higher up the well), the average bond length will increase and not stay constant. So, for a single bond in an isolated molecule, it is expected that the bond lenghtens with temperature.

Now, one case where the overall “size expansion” can be clearly studied is that of crystalline materials. In solids, the evolution of volume as a function of temperature is named thermal expansion, and is characterized by the parameter of the same name, defined as: $$\large \alpha = \frac{1}{V} \frac{\partial V}{\partial T}$$ Because bonds grow longer with increasing temperature, most materials display a positive thermal expansion, i.e. they expand with temperature. This is, for example, the case for diamond, with $\alpha \approx +10^{-6} \text K^{-1}$.

However, some materials display the oppositive behavior, and shrink upon temperature increase. This phenomenon is called negative thermal expansion (NTE), and is rather uncommon.


What is the actual reason for geometrical isomerism? My teacher explained it as restricted rotation between the double bonds. But I think there is more to it. After all, why is it considered to be a type of isomerism itself? Whether an alkene is cis or trans only depends on how you make the structure on paper. So what is the basic idea behind it?

Geometric isomerism occurs when two structures with the same connectivity are not interconvertible.

Cis-Trans isomerism is common and easy to recognize kind of geometric isomerism. The carbon-carbon double truly has limited rotation. The double bond in the alkene functional group consists of a $\sigma$-bond located within the plane of the molecule and a $\pi$-bond located perpendicular to the plane of the molecule. In the image below (created by David Pilz for Wikipedia), you can see the $\pi$-bond in a perpendicular plane.

The $\pi$-bond is made from $p$-orbitals, so it cannot have the same symmetry as the $\sigma$-bond. You can see the formation of the $\pi$-bond from $p$-orbitals in this picture by Wikipedia user JoJan. In order for the $\pi$-bond to exist the $p$-orbitals must be aligned.

In order for the molecule to rotate from $cis$ to $trans$, the bond axis would need to rotate so that the $p$-orbitals are no longer aligned and can no longer form the $\pi$-bond.  This slide show from James Condon at Roane State(PDF) explains the same concept. Rotation around the double bond requires enough energy input to break the $\pi$-bond, at least $250 \ \text{kJ/mol}$, which is a lot for room temperature.

The clearest demonstration that geometric isomers are clearly different molecules is to look at some examples with drastically different properties. It is important to remember that the experimental observation that these compounds with very similar structures have very different properties predates the explanation that the double bond has limited rotation.

The two isomers of 1,2-dichloroethene have different physical properties due both the difference in polarity and the difference in shape:

  $cis$-1,2-dichloroethane

Dipole moment = $1.9 \ \text{D}$
Boiling point = $60.2 \ ^\circ\text{C}$
Melting point = $-81.47 \ ^\circ\text{C}$
Density = $1.28 \ \text{g/cm}^3$

  $trans$-1,2-dichloroethane

Dipole moment = $0.0 \ \text{D}$
Boiling point = $48.5 \ ^\circ\text{C}$
Melting point = $-49.44 \ ^\circ\text{C}$
Density = $1.26 \ \text{g/cm}^3$

  However, my favorites are maleic acid and fumaric acid. Notice the extreme difference in water solubility as well as the difference in acidity. These molecules behave very different chemically.

Maleic Acid $cis$

Melting point = $135 \ ^\circ\text{C}$ (decomposes)
Density = $1.59 \ \text{g/cm}^3$
Solubility in water $788 \ \text{g/L}$
Acidity $\text{p}K_{a_1} = 1.9$, $\text{p}K_{a_2} = 6.07$

Fumaric Acid $trans$

Melting point = $287 \ ^\circ\text{C}$ (does not decompose)
Density = $1.635 \ \text{g/cm}^3$
Solubility in water $6.3 \ \text{g/L}$
Acidity $\text{p}K_{a_1} = 3.03$, $\text{p}K_{a_2} = 4.44$


Is there a simple way to get the circular dichroism of a molecule from its structure? Are there any heuristics to get the relative absorbtion of left and right circularly polarized light by a molecule from its molecular structure? Is it even possible to predict which polarization is selectively absorbed?

Is there a simple way to get the circular dichroism of a molecule from
  it's structure?

Short Answer:  Yes, for many molecules if you know the molecule's structure, then you can predict the shape of the optical rotary dispersion (ORD) or circular dichroism (CD) curve.  Conversely, and perhaps more importantly, if you know the shape of the ORD or CD curve, then you can predict the molecule's structure, in particular, the absolute configuration at chiral centers in the molecule. There are many empirical techniques, such as the "Octant" rule.  These methods are based on numerous experimental observations that have been codified into sets of rules for different classes of compounds.  These empirical rules can be used to determine the absolute configuration of cyclic ketones, conjugated dienes, alpha, beta-unsaturated ketones, twisted biphenyls as well as other systems.

Long Answer: When polarized light is passed through a chiral sample and the light is absorbed, there are two physical phenomenon that can be measured.  One is called the circular birefringence and it measures the difference in the refractive indices ($\ce{n_{R}, n_{L}}$) of the sample when a beam of right and left circularly polarized light is passed through the sample.  When this difference is plotted against wavelength an ORD curve results.  The other parameter is called circular dichroism and it measures the difference in the molar extinction coefficients ($\ce{\epsilon_{R}, \epsilon_{L}}$) when right and left circularly polarized light passes through the sample.  When this difference is plotted against wavelength a CD curve results.  The dependence of these two phenomenon on wavelength is termed the Cotton effect.

The following figure shows the uv-vis absorption (blue line), ORD curve (green line) and CD curve (red line) for the two enantiomers of camphor sulfonic acid.  Note how the ORD and CD curves for one enantiomer are mirror images of the same curves for the other enantiomer.  It is the carbonyl chromophore's $\ce{n->\pi^{\ast}}$ absorption that is responsible for the curves seen in this image.  In ORD spectra there are typically 2 maxima, one positive and one negative.  The positive maxima is a "peak", while the negative maxima is termed a "trough". If, when moving from longer to shorter wavelength, a peak is encountered first, then the ORD curve is said to be positive (+).  Conversely, if a trough is encountered first, then the ORD curve is negative (-).

image source

Chromophores can be divided into two groups, symmetric or inherently dissymmetric.  A carbonyl, nitro and olefin are examples of groups that, by themselves, are symmetric.  They may be perturbed by asymmetry on other atoms around them, but alone, by themselves, they possess no inherent asymmetry.  On the other hand, groups such as helicenes, twisted biphenyls and unsaturated ketones possess a twist or handedness; such molecules are inherently dissymmetric.  The intensity of an ORD or CD curve from a molecule containing an inherently dissymmetric chromophore is usually much greater than the peak intensity from a molecule containing a symmetric chromophore.

With that information as background, we can now apply these concepts to the examination of real molecules.  One of the most studied classes of molecules where these concepts have been applied are the substituted cyclohexanones.  In the early 60's  Djerassi, Moffitt, Woodward, Moscowitz and Klyne (some of the greats!) developed an empirical set of rules to predict the absolute configuration of chiral centers in substituted cyclohexanones.  The rule became known as the "Octant" rule and came about because of the work they were doing with complex natural products that often contained a substituted cyclohexanone ring.  They would run a reaction on a pure enantiomer of a cyclohexanone containing natural product and wanted a method to quickly determine if a chiral carbon in the molecule retained its absolute configuration or if the configuration was inverted in the reaction.  After comparing ORD curves from countless cyclohexanone containing natural products, they were able to compile their observations into the Octant rule.

As its name implies, the Octant rule divides the region around a cyclohexanone carbonyl into 8 regions or octants.  As the picture below (left part of figure) illustrates, the octants are generated by bisecting the carbonyl group with 3 planes.  Two of the planes contain the carbonyl group.  One divides the carbonyl into top and bottom regions and the other divides the carbonyl into right and left regions.  The third plane actually passes through the middle of the carbonyl and is perpendicular to the 2 other planes.  This third plane divides the molecule into front and back regions.  There are a few molecules that do have substituents in the front region, but most molecules do not, so we can usually simplify our analysis by only considering the 4 rear octants.  Each octant is also assigned a plus or minus symbol telling us whether we would expect a substituent in the octant to contribute in a positive or negative manner to the ORD or CD curves.

image source

The figure on the right shows how a cyclohexanone would be placed in the octant diagram and how its projection appears as a rectangle when sighted down the carbonyl group.

Let's examine the case of 3-methyl cyclohexanone.  First let’s draw the possible conformers of both S- and R-3-methylcyclohexanone

Now let’s take the 2 possible conformations for R-3-methylcyclohexanone (axial and equatorial methyl group) and project them onto the octant diagram.  We see that the equatorial conformation has a positive contribution to the Cotton effect, while the axial conformer has a negative contribution.  At equilibrium, the molecule exists predominantly in the conformation with the methyl group equatorial, therefore we expect the positive Cotton effect from this conformer to outweigh the negative contribution to the Cotton effect from the axial conformer.  Therefore, we expect a positive ORD and CD curve for R-3-methylcyclohexanone.  Similar analysis leads to the expectation of a negative ORD and CD curve for the S enantiomer.  Indeed, experimentally it is found that the ORD curve for R-3-methylcyclohexanone is positive and negative for S-3-methylcyclohexanone.  The full name for the R enantiomer is R-(+)-3-methylcyclohexanone

image source

More detail on the 3-cyclohexanone example, as well a nice, concise review of ORD can be found here.

Books have been written on this subject so there is much that I have left out.  For those who would like to explore further here are a few phrases that, in addition to the previous link, can help you get started:  Cotton effect, ORD biological applications, and magnetic circular dichroism


What method do I use to synthesize propyl-2-(2-methylpropyl)-ether and why? I need to know which method to use to synthesize propyl-2-(2-methylpropyl)-ether.

I considered electrophilic addition of alcohols to alkenes, but I'm not really sure how the method works, and anything I found on it was hard to understand. From my understanding I need to "add" electrophilically activated alkenes to alcohol.

I don't really know what that means. I was thinking of using t-butanol and propylene, but the whole part of electrophilically activated alkenes threw me off. Any advice or help is greatly appreciated!

There are two basic choices here if you are trying to do this on a lab scale. The alcohol plus alkene with catalytic acid is not an approach I would consider unless I needed a lot of this. t-BuOH plus acid is a recipe for isobutylene. Sodium or potassium t-butoxide is commercially available and reacted with an excess of iPr-I should give a decent yield of the ether.

The other approach is to use t-Butyl 2,2,2-trichloroacetimidate compound details here - a general reagent for the formation of t-Butyl ethers originally reported by Armstrong et al, Tet. Letts 29 2483 (1988)


How to determine if two calibration curves are parallel, within 5% I have complex matrices and I need to make sure the matrix does not affect the measurement of my analyte. To do that, I need to prepare samples of my analyte in a solvent and of the matrix reinforced with my analyte. Both curves are going to have 5 concentration points, each done 3 times and, for sake of argument, let's assume both have comparable levels of precision. Then, I linearize these two calibration curves and check if they are parallel, which is where my question lies.

How do I prove that both curves are parallel (meaning no matrix effects are present), within a margin of 5%?

I have found the following method:

Get the y-data of the matrix data and apply them to the calibration function of the solvent and get an "apparent concentration" of the matrix.
Perform a linear regression of the concentration values of the solvent with the apparent concentrations of the matrix.

Then, I have an intercept a and a slope b. A value of b different from 1 indicates a concentration dependent bias. I have both standard deviations of the fit, stdev_a, stdev_b, so I have to determine the confidence interval (95%) of b and check if their values are withing these confidence intervals.

I have found two ways to do that in Excel and I am unsure which one is correct. For example, the slope:

Do I use the conf = CONFIDENCE.T(0.05, stdev_b, N-4), get b+-conf, and check if 1 is within this interval b+-conf?
Do I use t = T.INV(0.95,N-4), multiply that by the stdev_b, get the range b+-t*stdev_b and check if 1 is within this range?

I have used N-4 for the degrees of freedom, since it's two linear regressions, each consuming one degree of freedom, correct? In this case, N-4 == 11.

In summary:

Which function do I use to calculate the confidence intervals for this example in Excel, and
Is this method of checking parallelism correct?
Is there an easier/simpler or more correct way of doing this parallelism check?

Define the angular coefficients of the two resulting lines and evaluate the parallelism between them. Perform the statistical evaluation of the data.
For the parallelism test between the curves, considering the equations of the lines to be compared as: $y_1 = a_1 + b_1x + \varepsilon$ and $y_2 = a_2 + b_2x + \varepsilon$, the hypothesis for comparison of the lines slopes is given by:
$$H_0 : b_1 = b_2$$
$$H_1 : b_1 \neq b_2$$
When the null hypothesis ($H_0$) is true, the regression lines become: $$y_1 = a_1 + b_3x + \varepsilon$$
$$y_2 = a_2 + b_3x + \varepsilon$$
Considering a significance level of $5\ \%$, the null hypothesis will be true when the test returns a P-Value above $0.05$.

More information in the link:
http://www.real-statistics.com/regression/hypothesis-testing-significance-regression-line-slope/comparing-slopes-two-independent-samples/


Can 2‐chloropropane and acetyl chloride give the iodoform test? Can $\ce{CH3-CHCl-CH3}$ and $\ce{CH3COCl}$ give the iodoform test?

I thought $\ce{CH3COCl}$ having the keto-methyl group would show the iodoform test, while $\ce{CH3-CHCl-CH3}$ will not since it's not a secondary alcohol but an alkyl halide.

As mentioned by user55119, ethanoyl chloride will not give iodoform test, because acid derivatives will show nucleophilic substitution reactions. Resulting will be acetate ion, which is no more a good nucleophile. For a more detailed answer check out here.

2-chloropropane will show iodoform test, since through bimolecular substitution reaction $(\mathrm{S_N2}),$ chlorine will be replaced by $\ce{-OH}$ group. Propan-2-ol so formed will be oxidised to acetone, which will show iodoform test.

Remember that only $\ce{CH3COR}$, (where $\ce{R}$ can be alkyl or aryl part) or which can be converted to it, give haloform tests, until steric hindrance prevent even some of these compounds to react.


Where does the energy of formation of covalent bonds come from? Consider two $\ce{H}$ atoms. Since the proton in one attracts the electron in another, they attract each other, and form a covalent bond ($\ce{H-H}$). Bond forming requires energy (436 kJ/mol for an $\ce{H-H}$ bond) – where does that energy come from?
If I put two $\ce{H}$ atoms in a container, would they form an $\ce{H-H}$ bond by themselves, or would I need to add energy to the system in order for them to bond?

Your view of bonding is a little simplified and confused. For the details it is worth reading the great answer referred to in some comments already: Fundamental forces behind covalent bonding .

I'll stick with a simpler view that focusses just on the energy involved and not the mechanism of chemical bonding. Firstly, you are wrong that boding two hydrogen atoms needs an input of energy. It actually releases energy. This is a thermodynamic view of the reaction. This tells us whether there is more or less energy in the product of the reaction being described than in the chemicals being combined to create the product.

But a lot of chemical reactions are thermodynamically favourable (that is the reaction releases energy) don't happen easily (this is probably not true for atomic hydrogen which should react very easily to form H2). But there are plenty of other reactions that release energy according to thermodynamics but that don't happen easily. For example, elemental carbon plus oxygen will release energy if converted to CO2. But neither the diamond in jewellery or the coal in your fireplace (which is basically impure carbon) reacts easily to give carbon dioxide despite being in an environment with a lot of Oxygen gas in it. Here the problem is that room temperature oxygen molecules and carbon just don't have enough energy to kick off the process that creates the product. This is where we have to look at the kinetic view of the reaction rather than just the thermodynamics. Although the overall reaction is favourable according to thermodynamics, we need to add energy (firelighters for coal fires, a blowtorch or diamond rings) to the mixture to give the atoms in the reactants enough energy to overcome the kinetic barriers to the reaction. Think of having to climb a tall mountain to get to a lower valley on the other side. Or think of giving oxygen molecules enough energy to break carbon atoms apart from their normal solid structure.

Once coal or diamond are warm enough, the reaction will happen spontaneously and you have either destroyed your engagement ring or you have a nice coal fire. In this case you can feel the energy released from the reaction as a pleasant warming of your room.

Back to the question. The energy involved in a reaction is an accounting of the amount of energy used in the bonding of the products versus the reactants. The energy released in reactions such as burning carbon goes into the environment as heat. Other reactions happen spontaneously (try dropping a small lump of sodium into water). Some reactions require a tiny amount of encouragement (oxygen and hydrogen in certain concentrations will react explosively to form water when encouraged with the tiniest spark).

We can drive some reactions the other way using inputs of energy. For example we can electrolyse water into a mixture of hydrogen and oxygen. This requires us to add energy to the system (about 286kJ/mol) because the products are in a higher energy state then the reactants.

If you just focus on the thermodynamics of a reaction the numbers (kcal or kJ/mol) tell you how much the energy difference between the products and reactants is. But many reactions that should release energy require some input to kick off the reaction. This is confusing, but chemistry wouldn't be interesting and we wouldn't exist if thermodynamics were the only thing driving reactions.

Also energy is not created or destroyed here, just moved around. You can put energy in to a system to drive a reaction or the environment gets energy back from reactions that release it.

So, some reactions require energy to drive them because the products have higher energy, some release energy because the products have lower energy but many release energy but, confusingly,  require some energy input to kick off the actual reaction process.


Ring compounds of non-carbon atoms Are there compounds that are rings of non-carbon atoms, say a ring of six oxygen atoms, or 5-6 nitrogen atoms with attached hydrogens?
Or are these too unstable to exist for long (if at all), like long oxygen chains ($\ce{HO_{n}H}$)?

Pentazole exists - an $\ce{N5H}$ ring. It is stabilised by aromaticity, with 6 $\pi$-electrons in a cyclic, planar system.

Sulphur forms many allotropes which are rings but these do not contain hydrogen. $\ce{S8}$ and $\ce{S7}$ are the most common.

Phosphorus also forms cyclic allotropes such as white phosphorous, $\ce{P4}$.

DavePhD also mentioned in his answer that a silicon analogue of benzene has been synthesised recently but it is not $\ce{Si6H6}$ because it has other groups attached to some of the silicon atoms.


Which has a greater dipole: chloroform (CHCl3) or trichlorofluoromethane (CCl3F)? Does chloroform have a greater dipole because the $\ce{C-H}$ dipole is weaker than $\ce{C-Cl}$ dipole thereby making the overall net dipole greater in chloroform, as opposed to trichlorofluoromethane where the $\ce{C-F}$ bond dipole being more similar to the $\ce{C-Cl}$ dipole makes the molecule more stable with a smaller net dipole?

Imagine a $\ce{CHCl3}$ molecule and a $\ce{CFCl3}$ molecule fixed with the carbon atom of each molecule at the origin of a three-dimensional coordinate system. Select the two coordinate systems such that the hydrogen atom of the $\ce{CHCl3}$ molecule and the fluorine atom of the $\ce{CFCl3}$ molecule both lie on the $z$-axis above the origin, and let $u_z$ be the $z$ unit vector. If the four single covalent bonds of each molecule are symmetrical (here, tetrahedral), then it follows that the net vector sum of any four vectors that are equal in magnitude and are directed in the four directions from the origin to each non-carbon atom will be the zero vector. In both molecules, the net vector dipole moment arising from the three $\ce{C-Cl}$ bonds will equal the dipole moment of a single $\ce{C-Cl}$ bond with the chlorine atom in the direction of $-u_z$ from the carbon atom.

Assuming arguendo that the dipole moment of each bond is proportional to the difference in electronegativity of the two atoms between which that bond lies (on the Pauling Electronegativity Scale, hydrogen rates a $2.1$, carbon rates a $2.5$, chlorine rates a $3.0$, and fluorine rates a $4.0$), then to find the dipole moment of the $\ce{CHCl3}$ molecule, we add the vector dipole moment of the $\ce{C-H}$ bond ($[(2.1 - 2.5) \cdot u_z]$, or $-0.4 u_z$) to the vector dipole moment of the $\ce{C-Cl}$ bond ($[(3.0 - 2.5) \cdot (-u_z)]$, or $-0.5 u_z$), obtaining $[(-0.4 u_z) + (-0.5 u_z)]$ or $-0.9 u_z$.

To find the dipole moment of the $\ce{CFCl3}$ molecule, we add the vector dipole moment of the $\ce{C-F}$ bond ($[(4.0 - 2.5) \cdot u_z]$, or $1.5 u_z$) to the vector dipole moment of the $\ce{C-Cl}$ bond ($[(3.0 - 2.5) \cdot (-u_z)]$, or $-0.5 u_z$), obtaining $[(1.5 u_z) + (-0.5 u_z)]$ or $1.0 u_z$.

One might thus conclude that the dipole moment of $\ce{CFCl3}$ might be some ten percent larger in magnitude (as proportional to 1.0) than is that of $\ce{CHCl3}$ (as proportional to 0.9).


Stereochemistry of 1,4 disubstituted cyclohexane How are these cyclohexanes identical?

Take the molecule on the left and ideally flip it by rotating 180 degrees along the vertical axis of the diagram. This inverts the cyclohexane without having to break or modify the molecule in any way


Difference between polarizability and dipole moment? In my head, any molecule that is very polarisable, has a dipole moment. Is this true? I am very confused and can't seem to find anything that helps my thought, and I can't seem to distinguish between the two without both intertwining.

First, the quick definitions. A dipole moment ($\mu$) implies a permanent separation of charge across a distance. For example, HF has a positive charge near the hydrogen atom and a negative charge near the fluorine:

Instead, polarizability $\alpha$ indicates the degree to which electrons (and thus charge) re-arrange in an applied electric field. Indeed, when a polarizable molecule experiences any type of electrostatics, there is an induced dipole moment:

$ \mu_{induced} = \alpha \vec{E} $

Now this induced dipole moment (from the polarizability) is completely distinct from the permanent dipole moment. For example, benzene is non-polar (no net dipole moment) but highly polarizable. In the HF case, I could find some large applied field $\vec{E}$ that pushes electrons from the F to H.

Here are my mental analogies:

A dipole moment is like a hill or tilted plane - water (charge) moves "downhill" and it's always there.
Polarizability is like hitting the water with a cannonball. I can drive water from one side of a bathtub to the other!

The two properties can be related, but it's possible to find highly polarizable molecules with no dipole moment - particularly if they are symmetric. Here's a plot with ~80,000 molecules from PQR - note that all combinations occur.


Predicting color of precipitate Is there any way of predicting whether a precipitate will be colored? I know that for solutions, transition metal ions with unfilled d-orbitals will have color because of d-orbital splitting in the complex ions. However, this isn't applicable to precipitates because those are just ionic solids, not complex ions.

For example, $\ce{PbI2}$ is a yellow solid, and $\ce{Al(OH)3}$ is white. Is there a way to predict that $\ce{PbI2}$ will be colored while $\ce{Al(OH)3}$ will not be, or is it something we have to memorize to know?

There are a few wrong assumptions in your question that I would like to clear up.

Yes, d-orbital splitting is part of the reason for colour. However, $\ce{[FeF6]^3-}$ displays a beautiful d-orbital splitting which could display colour from the energy difference — but doesn’t. There is more to colour than d-orbital splitting; the most important terms are spin-forbidden and Laporte rules.
You somehow got the impression that ionic solids have nothing to do with coordination complexes. Please remove this idea from your brain; it is wrong. Ionic solids are equally coordination complexes as present in solution only that the environment and sometimes the composition of the complex is different. E.g, $\ce{CuSO4(aq)}$ is essentially a solvated sulphate ion plus $\ce{[Cu(H2O)6]^2+}$. When precipitated or crystallised as ‘$\ce{CuSO4 . 5H2O}$’, what you actually got was $\ce{[Cu(H2O)4(\mu{-}SO4)2].H2O}$ — clearly a coordination compound.

To answer your question, it depends on the nature of the compound you are precipitating — the colour can derive from cation, anion or their combination. For example, aluminium salts will be colourless unless the anion dictates otherwise. Likewise for zinc salts. For a very large number of salts where the cation is responsible for the colour, these will be the same as in solution. But sometimes ($\ce{PbI2, Fe4[Fe(CN)6]3}$ and others) you just have to know because only this specific combination will give this colour.


How does temperature affect the concentration of flavonoids in orange juice? I am in high school, and I would like to do an experiment on this for a project.

I've done quite a bit of research into the different methods I could use (see 2.4.2 here) but I still cannot find enough information to formulate a hypothesis. It would be useful iff you could point me in the right direction. So, how does heat affect how much flavonoids are present in commercially available orange juice? (without pulp)

Orange juice is often subjected to temperature variation during processing, pasteurization, storage and refrigeration at 4$^{\circ}$C. According to some research, there appears to be a dramatic decrease in the concentration of certain flavonoids at storage temperatures especially above 20$^{\circ}$C.

A variety of flavonoids found in orange juice includes didymin, eriocitrin, hespertin, narirutin, poncirin, apigenin and neoeriocitrin.

Effective analytical methods used to identify and quantify individual flavonoids include gas chromatography mass spectroscopy (GCMS) and high-performance liquid chromatography (HPLC).

Whilst chromatography is the most effective method of identifying specific flavonoids, you may be interested to know that ultra-violet spectrophotometry has been used to measure total flavonoid content before & after treatment.


Why, in an applied magnetic field, is electron with spin parallel to the magnetic field higher? My lecturer today said that when an applied magnetic field, $B_0$, is applied, the electron with spin parallel to $B_0$ is higher in energy than the electron with spin antiparallel to $B_0$. Similarly, in protons, it is the opposite; parallel spin is lower in energy and antiparallel spin is higher in energy. Why is that?

In addition to the other answer, the energy  of a magnetic moment in a magnetic field is $E=-\bar{\bf{\mu}}.\bar{\bf{B}}$ where the dot indicates the dot product between the two vectors $\bar{\bf{\mu}}$ and $\bar{\bf{B}}$, thus the energy is lowest when the magnetic dipole is parallel to the field $B_0$; $E=-\mu B_0$. The sign of $\mu$ is opposite for nuclear and  electron spin angular momentum.The magnetic moment of a nuclear spin is defined as $\mu=+\gamma I$, where $I$ is the nuclear angular momentum and if the field is along only the $z$ direction  $\mu_z=\gamma I_z=m_z\gamma \hbar$ (Joule/Tesla) where $\gamma$ is the magnetogyric ratio, (unit radians/Tesla/sec) the components are $m_z$, the projection, magnetic or azimuthal spin quantum number on the $z$ axis, the axis parallel to the magnetic field $B_0$. $I_z$ is the $z$ component of the nuclear spin angular momentum. (The magnetogyric ratio is $\gamma = g\mu _0/\hbar$ where $\mu_0$ is the nuclear magnetron and $g$ the nuclear spin factor, different for each type of nucleus). The energy is then $E=-m_z\gamma\hbar B_0 $. For a proton $m_z=\pm 1/2$.

The spin magnetic moment of the electron is defined in a similar way to that for nuclear spin but with a negative sign; $\mu_e=-\gamma _e I_e$ where $I_e$ is the electron spin angular momentum and $\gamma _e$ the electron magnetogyric ratio. (  $\gamma _e=g_e\mu_B/\hbar$,  $g_e$ is the electron $g$ factor, $g_e =2.0023..$, and $\mu _B$ the Bohr magnetron. Some authors give a negative $g_e$, in this case the negative sign is removed from $\mu _e$).
(The nuclear magnetogyric ratio can be positive or negative, e.g. $\ce{26.75.10^7 rad T^{-1} s^{-1}}$ for protons. For a few types of nuclei e.g. $\ce{^{15}N, ^{29}Si}$ $\gamma$ is negative. The magnetogyric ratio for the electron is much $\ce{larger 1.76 10^11 rad T^{-1} s^{-1}}$.)


What is to be called as solvent? My book says:

  In a solution consisting camphor and nitrogen gas, camphor (solid) is solute whereas nitrogen gas is solvent.

Doesn't this contradict the statements that:
(1) Generally the component present in large proportion is the solvent.  I mean camphor will be in larger proportion as it is a solid. We can diffuse much nitrogen gas.
(2) Solvent determines the physical state in which solution exists. If nitrogen is solvent, then how can the solution be in gaseous state? Or if the statement is true, does that mean that the solution exists in gaseous state with camphor particles of collidial size?

Regarding part 1

Camphor will readily sublime and so, in proportion to the nitrogen gas, it is the minor constituent

For part 2,

  If nitrogen is solvent, then how can the solution be in gaseous state?

if the nitrogen is the solvent then of course it will be in the gaseous state as nitrogen is a gas.


Structure of Niobium and Tantalum carbonyls $\ce{V(CO)6}$ is well known to be an unusual 17 valence electron, paramagnetic octahedral complex.

Do the corresponding tantalum and niobium species form simple dimers or more complicated structures ?

From the abstract of J. Chem. Theory Comput., 2011,
7, 2112-2125 (DOI) is seems that niobium forms a dinuclear $\ce{Nb2(CO)12}$ complex, in which two $\ce{Nb(CO)6}$ units are connected by a 340 pm $\ce{Nb-Nb}$ bond.


What mechanisms might make a substance travel different distances on the same TLC plate? When doing TLCs to monitor reactions, it is considered good common practice to not only spot the reactants and the reaction mixture (likely after some mini-workup), but also to include a ‘cross-spot’: every other substance (mixture) spotted somewhere on the plate is spotted together on one spot, commonly labelled something like X.

When asking why during my bachelor's thesis, the PhD student supervising me said something that roughly translates to:

  You can't be sure that the components don't run differently in the mixture than by themselves.

But in a perfect world, they should. Aside from the obvious case of having a polar reaction solvent that didn't entirely evaporate and skewes the elution mixture on parts of the TLC, are there other reasons why the components of a mixture should have different Rf values when spotted separately?

are there other reasons why the components of a mixture should have different Rf values when spotted separately?

One might speculate that two polar reactants form a less polar "adduct". This should however result in an additional spot, rather than in altered $\mathrm{R_f}$ values for the individual compounds.

But this is speculation. Admittedly, I have never used the cross spot method. Neither have I ever observed that spots of the reactants had different $\mathrm{R_f}$ values in the reaction mixture.

On a side note: What always worked for me was:

Don't apply spots but bands and overlap them.
"Sharpen" the bands on the base line by developping the TLC in methanol or another polar solvent first.

Together, these two tricks allow to destinguish between two compounds with very similar  $\mathrm{R_f}$ values.


Which electronic effects are responsible for the reduction of nucleophilicity of a hydroxy group orthogonal to an ester? I am facing a similar synthetic problem as Boekman Jr. et al noted in their synthesis of Tetronolide.[1] I will quote their’s rather than disclosing my actual synthetic problem, but note that the key element — an α-hydroxy-γ-lactone is identical.

The authors’ intended convergent synthesis involves coupling the two compounds 1 and 2 shown in scheme 1 in a ‘tandem ketene-trapping/[4+2] cycloaddition’, which requires the free hydroxy group in 1 to nucleophilicly attack the ketene.

Scheme 1: Tandem reaction attempted by Boekman Jr. et al; would include a nucleophilic attack of 1’s hydroxide onto 2’s ketene motif.

However, as indicated this reaction did not produce 3. The authors note:

  However, despite considerable experimentation, all attempts to couple lactone [1] with [2] to afford β-ketoester lactone [3] failed. We surmised that the failure stemmed from impaired nucleophilicity of the tertiary OH group owing to a stereoelectronic effect resulting from the enforced alignment of the lactone carbonyl π-orbitals and the adjacent CO bond. (Compound numbers adapted to fit this scheme.)

Instead, the authors used alcohol 4, which readily reacted with the masked ketene 2 to give the ester 5 as shown in scheme 2.

Scheme 2: Successful tandem reaction attempted by Boekman Jr. et al; included a nucleophilic attack of 4’s hydroxide onto 2’s ketene motif.

My experimental findings have confirmed a similar effect for my compound. However, I have a bit of trouble discerning which electronic effects exactly cause this unreactivity. My first assumption is that the conformation of the five-membered ring forces the hydroxy group into an angle of approximately $90^\circ$ with respect to the carbonyl-π system. This should allow mixing of σ and π orbitals of whichever kind.

My second assumption was tracing the problem back to the anomeric effect. In that effect, a nonbonding electron pair interacts with an adjacent σ* orbital, lowering its own energy but reducing the σ bond strength. The required electrons could be supplied by the π system, resulting in the effect being π and σ*. However, I do not quite see yet, how this kind of mixing would reduce the nucleophilicity of the oxygen atom.

Thus, the questions are:

Which electronic interactions exactly are responsible for this effect?
Has this been the topic of a research paper?

Reference:

R. K. Boekman Jr., P. Shao, S. T. Wrobleski, D. J. Boehmler, G. R. Heintzelman, A. J. Barbosa, J. Am. Chem. Soc. 2006, 128, 10572. DOI: 10.1021/ja0581346.

If we focus on the 2-oxo-3s-hydroxy-tetrahydrofuran the electronic  effects present are the hyperconjugation and the dipole minimization. "Both effects contribute to the preferred (Z)-conformation of esters over the (E)-conformation. In the (Z) conformation, the lone pair of electrons in the alpha oxygen can donate into the neighboring σ* C-O orbital. In addition, the dipole is minimized in the (Z)-conformation and maximized in the (E)-conformation. But this electronic effects occur between the carbonyl carbon and the oxygen of the ring but the hydroxyl group." [a]

In contrast, I calculated the PES of hydroxy group in the following two systems:

2-oxo-3s-hydroxy-tetrahydrofuran [S1]
(1S,2S)-isopropyl 1-hydroxy-2-((vinyloxy)methyl)cyclohexanecarboxylate

The latter one is:

The oxygen of iprO-(C=O)- group was pointed away from [S2] and towards to [S3] cyclohexane. The PES is: (It was calculated with am1 due to computational resources, it was done with my laptop.)

It can be observed that the intramolecular hydrogen bond (IHB) in [S3] is more stable than in the other systems. The order of intramolecular hydrogen bond preference is [S3]>[S2]>[S1]. In other words, the IHB formed with the oxygen SP$^3$ of iprO-(C=O)- group is preferred over the carbonyl oxygen. [b] This can be seen on the rotational barrier between both minima. This conformational preference allows the lone pair of hydroxyl oxygen to be more nucleophilic than in the other systems. As it can be seen in the following figures:

[S1]

[S2]

[S3]

Related to the electronic effects present in this IHB, it can be complemented that "this is a consequence of interacting the lone pair on the accepting oxygen with an empty hydride antibond of the donating O-H atoms, $n_O$ → $\sigma^*OH$. Occupancy of the antibonding orbital raises the energy of the O-H species while stabilizing the oxygen." As Albrecht et. al. concluded in her paper.[c]

Bibliography

[a] Cuevas, Eusebio Juaristi, Gabriel (1995). The anomeric effect. Boca Raton: CRC Press. ISBN 0-8493-8941-0.
[b] Rincon et al. J Solution Chem (2011) 40: 656.
[c] Albrecht et al. J. Phys. Chem. A. (2012) 116: 3946


How do we turn PTFE powder into a paste we can dilute in a solvent We want to create a binder we can use in our filter elements that is PTFE based, so we need to somehow create a paste, we can thin down with a solvent that will still hold the PTFE in suspension, long enough to dip our virgin glass microfibre elements in and get an even distribution through the element structure.
So far we have tried a DMP and DIBK mix to make a paste, and then diluted in MEK but we cannot get the PTFE to remain in suspension long enough.

You might try making a solution, rather than suspension. PTFE is very resistant to most solvents, but can form a 3% solution in perfluorinated hydrocarbon solvents and, if I remember correctly (I could not find it on the web), in $\ce{Na}$ dissolved in liquid $\ce{NH3}$.


Why hot water is more effective than cold water in cleaning Today while washing curtains, my mother dipped them in hot water(containing detergent) for an hour and then washed them.When i asked her that why she done that,her answer was simple(it will make the cleaning easier and effective). It seems to me that there is a scientific phenomena is lying behind it,but i don't know what it is.

The question i m asking is why hot water is more effective than cold water in cleaning (cloths,curtains,bedsheets etc).

I shall be thankful to have your answers.

Many physical and chemical reactions are affected by temperature. Usually higher temperature causes the reaction to occur faster, and lower temperature will slow it down.
Water solubility (the ability of substances to go into solution in water, i.e. bond with water molecules) is a reaction that is affected by temperature for many substances. Changes in temperature may increase or decrease the solubility of a certain substance in water, depending on the substance properties.

Since high temperature will cause water molecules to move around relatively faster, it will also cause the hydrogen bonds between them to form and break at a higher rate (that is why hot water vaporizes more quickly than cold water). In addition, the high temperature increases the water's energy, and this energy may be used to break bonds between dirt and curtains for which colder water will not have enough energy.
This means that the detergent/water solution will be able to dissolve (form bonds with) the dirt faster and better, because the bonds between the detergent and water and the bonds between the detergent and dirt will form faster and will be strong enough to prevent the dirt from reforming the bonds with the curtains. As mentioned earlier, these bonds will also break faster, but the presence of a detergent helps to stabilize the solution and keep the dirt dissolved in the water. Once the dirt is dissolved in the water, the curtains are clean, and the dirty water can be easily discarded.

In summary, hot water helps the cleaning process (i.e. moving the dirt from the curtains to the water) because it can dissolve the dirt faster and break strong bonds (up to a limit, of course) of dirt with the curtains.

It is worth mentioning that between the detergent and the hot water, the detergent contributes more to the dissolution of dirt in the water than does hot water alone (that is the same reason it is better to wash your hands with soap and water than with water alone). However, for optimal results, using both detergent and hot water will clean the curtains better (and faster).

Here are some images I found online that may help clarify this point:


How to get a protein's dihedral angles ordered by variance? The setting: I want to simulate protein docking and let some dihedral angles vary, but in order to keep it low-dimensional I have to select those which are most likely to change.

What I've thought of so far:

The easiest solution would be some tool/database that simply returns a list of dihedral angles ordered by their variance (something like http://www.ifm.liu.se/compchem/msi/doc/life/cerius46/qsar/theory_descriptors.html#515786, but that only returns the number of "meaningful" rotamers)
Take a set of different conformations and use principal component analysis on the phase space consisting of the dihedral angles.
Take Ramachandran plots and try to combine all those conditional variances (if φ is …, then ψ has some amount of freedom) into a more "holistic" statement.

The problems I encountered:

I couldn't find such a list.
Searching rcsb.org, I only found about ~100 conformations (for albumin), but if I'm not mistaken the format isn't the exactly the same, so it'd be kinda hard to compare. Besides, ~100 samples won't yield good results for >100 dimensions (=angles).
The only approach for which data seems to be available, but it just looks like a real pain.

If I understand you, I can't imagine how to solve the your problem with this approach. Also I don't get the physical meaning of the proposed solution.

If you have to deal with a molecule with more than 100 rotatable bonds, you are just in problems.
You could find the rotatable bonds with some tools (see for example OpenBabel). But the problem will just start.

At first, you can not just try one by one the $N$ angles ($\{\theta _i \}$) (by changing the angle value $n$ times and trying each of them), because for exploring all the conformational space you will just have $n^N$ points. And you just can't know a priori which are the most important angles (in case (I think not) that effectively exist the most important angles).

Also note that not every point ($\theta _1, \theta _2, \cdots,\theta _N)$ is acceptable, just because parts of the molecule can superpose with other parts.

  "Take a set of different conformations and use principal component analysis on the phase space consisting of the dihedral angles."

I think that this is not possible. The values of the rotatable angles will be just chosen by you.
Even more, suppose that you have two rotatable bonds very close each other, in a linear part of the molecule. If not a large superposition ocurrs, you will have no diference in the efect of them. But the effect of them will be very correlated. So, you can get useful info. with PCA.

A possible approach

As I know very few of biology, I would make a research about if there are prefered configuration, that are much more common than others (because of biological facts).

If you are forced to work with the entire configurational space, the computational approach that comes to my mind is:

Use very fast method for energy computations that allow you set the connectivity of the atoms. This will prevent changes of the covalent bonds due to superpositions.
Use the Monte Carlo method, this should explore the configurational space giving you a reasonable idea of this space.

Final note

I find extrange doing docking with a large protein and ¿without a ligand? If there is a ligand for binding to the protein that we are talking about, I won't take too much care of the protein. The scoring functions or energies are few accurate, I find hard to believe that just one found configuration would represent accurately the problem (considering the large degeneracies involved)


Water cooling by evaporation For many years of drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from $\pu{35 ^\circ C}$ to $\pu{20 ^\circ C}$ by the evaporation of $\pu{60 g}$ of water?
Heat of vaporisation of water in this temperature range is $\pu{2.4 kJ g-1}$. Specific heat of water is $\pu{4.18 J g-1 K-1}$.

I'm pretty confuse by what the question is asking for. This is my attempt for the evaporation of $\pu{60 g}$ of water:
$$q = \pu{60 g} \cdot \pu{2.4 kJ g-1} = \pu{144 J}$$
I then subbed this energy unit ($\pu{144 J}$) to the heat change formula to find the mass of water. Why am I wrong?

It helps to write down the formulas, before you plug in the numbers.

Heat you need to extract to cool the water:

$$Q=m_1 c_p\Delta T$$

Heat needed to evaporate water:

$$Q=m_2L_v$$

So:

$$m_1 c_p\Delta T=m_2L_v$$

So:

$$m_1=\frac{m_2L_v}{c_p \Delta T}$$

Now plug in the numbers.


According to KMT, is the velocity of an ideal gas always sqrt(3RT/M)? It is a common textbook question to treat the root mean square velocities of an ideal gas as given by the following equation:
$$v_\text{rms}=\sqrt{\frac{3RT}{M}}$$
I was wondering about the validity of this equation, particularly as it applies to gases that are not monatomic. For example, in the following question from Zumdahl et al., it says:

Calculate the root mean square velocities of $\ce{CH4}$ and $\ce{N2}$ molecules at
$\pu{273 K}$ and $\pu{546 K}$.

The answer key gives $\pu{652 m/s}$ for methane at $\pu{273 K}$, $\pu{493 m/s}$ for nitrogen at $\pu{273 K}$, $\pu{922 m/s}$ for methane at $\pu{546 K}$, and  $\pu{697 m/s}$ for nitrogen at $\pu{546 K}$. These were all calculated with the equation $$v_\text{rms}=\sqrt{\frac{3RT}{M}}$$
Why is this equation valid for all ideal gases? Shouldn't it be $$v_\text{rms}=\sqrt{\frac{6RT}{M}} \text{ for methane}$$ (because there are $3$ rotational degrees of freedom in addition to $3$ translational degrees of freedom), and
$$v_\text{rms}=\sqrt{\frac{5RT}{M}} \text{ for nitrogen}$$ (because there are $2$ rotational degrees of freedom)?
If my thoughts are not true, then why is it commonly reported that the average molar specific heat of an ideal monatomic gas (at constant volume) is $3/2RT$, that of an ideal diatomic gas is $5/2RT$, and that of a nonlinear gas is $3RT$?

$E=\frac 12mv^2 \implies  v=\sqrt{\frac{2E}{m}} $ is valid for translational kinetic energy and the speed of the centre of mass.
Vibrational or rotational energy does not count. An object may vibrate or rotate even if it's centre of mass has zero speed.
As each available degree of freedom has the mean energy $E=\frac 12kT$,
and as there are 3 independent translational degrees of freedom,
the mean translational kinetic energy is $E_\mathrm{transl}=\frac 32kT \le E_\mathrm{kin}$
and the respective
$v_\mathrm{RMS}=\sqrt{\frac{2E_{transl}}{m}}=
\sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$
even for ideal gas consisting of  multiatom molecules.


Is there a universal minimum energy state? I'm imagining this: if everything was separated (even on the nuclear level) and then the parts were recombined so that they were in the lowest possible energy state, do we have any way of knowing what that might be? Particularly, would it be something simple like a helium atom or could it be something very complex and unusual? And I mean that you might grab a few parts and make a helium atom or grab a whole bunch of parts and make whatever else.

I guess one problem is we need a way of totaling how much the energy is decreasing per some amount of something else. Similar to how nuclear binding energy is often expressed on a "per nucleon" basis.

I'm interested to see what people think.

This is an interesting and actually surprisingly hard question, which reaches the edge of our understanding of physics, so really a physicist would be more equipped to answer this.

I'm not sure if there is rigorous mathematical/physical proof that energy must have a minimum. Energy is a continuous, real-valued, state function, which as far as I know makes it mathematically very well-behaved, so it's unlikely that it can display weird behaviour. The extreme value theorem suggests that taking any finite collection of states into account, at least one of them must be a global energy minimum. However, I'm not sure if the Universe has a finite amount of states it can assume (probably not, if it is infinite in size). All that said, even if strictly speaking there is no global minimum, it's likely there is some asymptotically low value. So what is it?

What you're trying to imagine, essentially, is a Universe where there is no kinetics, only thermodynamics. Equivalently, you're thinking about the state of the Universe at $\mathrm{t→+ \infty}$. All finite kinetic barriers are surpassed and all finite distances are traversed. With this much time, a lot of unintuitive events take place. Two very interesting sources to read are "Time without end: Physics and biology in an open universe",  (1979) Rev. Mod. Phys., 51, 447, and "A dying universe: the long-term fate and evolution of astrophysical objects", (1997) Rev. Mod. Phys., 69, 337. There are many processes, varying from almost certain to extremely speculative, which can happen in overlapping time scales, so it is hard to determine exactly which route the Universe (or a toy model of a universe) might take. I'll present several events which may happen as the Universe marches on towards $\mathrm{t→+ \infty}$, and thus which reach closer to the minimum energy state of the Universe, but be aware that there are many uncertainties, especially near the deep end.

Starting with chemistry, allowing systems to evolve for very long periods would have some interesting consequences. Only the most stable compounds could survive, so for example, all nitrogen atoms would eventually recombine to form gaseous $\ce{N2}$. Other very stable molecules would be $\ce{H2O}$, $\ce{CO}$, and $\ce{H2}$. The Universe would turn into a mixture of small molecules. This would take at least something of the order of $10^{65}$ years. For comparison, the Universe is currently $1.38\times 10^{10}$ years old.

But why stop there? Chemical energy is only one of many factors to be taken into account in the total energy, and really it's one of the smallest. For example, nuclear energy scales completely dwarf chemical energy scales, so forget about optimising molecular abundances to minimise energy, because optimising atomic abundances will reach much lower energies. What would really happen is that, given tremendous amounts of time, light/heavy nuclei would overcome the barrier to fusion/fission, even at very low temperature and pressure. As you mention, the ultimate effect would be that all of the atoms in the Universe would fuse/fission into $\ce{^{62}Ni}$, the nucleus with the highest nuclear binding energy per nucleon. This would take on the order of $10^{1500}$ years.

This still isn't the end, though. Given a ridiculous amount of time, much of the nickel could still coalesce and compress into degenerate neutron matter (like neutron stars), releasing even more energy. The kinetic barrier for this is truly colossal, though, so it would take some $10^{10^{76}}$ years.

It's possible to keep going. Sufficiently large chunks of matter could compress further into black holes, releasing even more energy. Depending on the mass of the lightest possible black hole, this process may take "only" $10^{10^{26}}$ years, or up to a scale comparable to neutron matter formation. If the lightest possible black hole has a non-zero mass, then objects lighter than that threshold may be stable indefinitely. The black holes that did form would evaporate almost instantly relative to this time scale, producing an abundance of subatomic particles which would eventually decay into photons and neutrinos.

There's another speculative process which may severely undercut the previous possibilities, however. Currently many types of string theory posit the existence of proton decay. If it does happen, it could happen as quickly as $10^{40}$ years. This would be the end of Chemistry in the Universe; all atoms would eventually decay into photons, neutrinos, electrons and positrons. Black holes would survive this, but they would evaporate by $10^{100}$ years at the latest, quickly resulting in a Universe sparsely filled only with photons and neutrinos.

Last but definitely not least, it may be that even completely empty vacuum is not the minimum energy state of the Universe. It's conceivable that the vacuum we know is less stable than another type of vacuum, separated by an enormous kinetic barrier. If this is the case, then given enough time there could be a vacuum metastability event, a tremendously energetic process in which the Universe as we know it would cease to exist, being replaced by another with entirely different physics.


Identifying Copper Purity At Home Are there any ways to figure out if the copper foil and mesh I have are pure, as-close-to 100% copper as possible (and as advertised)?

As a home tool, but not necessarily a definitive solution, I would suggest the dissolution of your copper metal/alloy in a galvanic cell where the Cu is the anode and as a noble cathode, use a source of graphite.

This approach for the dissolution has been favorably argued in commercial patents in a 1999 U.S. Patent 6,692,632. To quote in parts:

  The method is preferably used in synthesis, but can also be useful in the destruction of pollutants etc. In certain embodiments of this nature, the use of external current applied to the system is not necessary, and a simple galvanic cell can be set up...

On copper particularly, for the home chemist, I have demonstrated a so-called Bleach Battery approach as an electrosynthesis path discussed in 'Subject: Experimental Variants of Bleach Battery to Obtain Metal Oxides (like CuO). However, I would replace the cathode, as in the current case as Nickel could be present as an impurity also in your copper metal sample, with very noble graphite in its path to CuO and other metal oxides that may become evident (or be chemically identified).

As the copper can be alloyed with metals like Zinc, Tin, Aluminum, Nickel, Silicon,..., with the first three creating respective oxides from the dissolving a small piece of the copper metal/alloy, potentially revealing distinct metal oxides as, for example, CuO is black, ZnO is white, SnO is blue-black (as produced by heating the tin(II) oxide hydrate, SnO·xH2O
where x<1), and Al2O3 varies from white to a clear transparent gel. The metals Nickel and Silicon are expected in their metallic state.

I say, convert a bit of it into metal oxides and proceed from there!


Fraction of cube Volume empty The unit cell in a body centered cubic lattice is given. Each sphere has a radius r and the cube has a side a. What fraction of the total cube volume is empty?

Assuming the $a$ is large enough to accommodate a BCC lattice with spheres of radius $r$, we may define

$V_s = \frac{4}{3}\pi r^3$ is the volume of the sphere

$V_c = a^3$ is the volume of the cube.

in a BCC lattice, there are 8 corner points, and one central. Because the lattice points on the corners are defined with respect to the center of each of the spheres, we know that each of the corner spheres will contribute $\frac{1}{8}$ of their total volume to the occupied volume of the unit cell. In addition, the sphere at the center will contribute its entire volume to the occupied volume. We may then write the total volume occupied in the unit cell as

$V_{occ} = 8\left(\frac{1}{8}\right)V_s + V_s =2V_s$

We know that the total volume of the unit cell is that of the cube, meaning that the unoccupied (empty) volume of the unit cell may be written

$V_{unocc} = V_c - V_{occ} = a^3 - \frac{8}{3}\pi r^3$

Wherefore we may write the empty fraction as

$\frac{V_{unocc}}{V_c} = \frac{a^3 - \frac{8}{3}\pi r^3}{a^3} = 1 - \frac{8\pi r^3}{3a^3}$


What is this equation for? I have been given a formula on the topic of alloys and temperature, however am unsure specifically what this equation calculates. The equation is as follows:

$xC_{1} + (1 - x) C_{2} = C_{3}$

I am assuming that the C's will be given (and possibly represent Specific Heat?) - however exactly what is this equation to represent? What does solving for $x$ achieve?

I know that the letters used may vary, however our teacher has provided no more information due to sickness. I have provided all the information I know.

I fear that without additional information it will be difficult to provide you with a definitive answer. The $x$ usually represents the mole fraction and in the context of a 2-component alloy this seems plausible. The $C$s might indeed be specific heat constants, i.e. $C_{3}$ would be the specific heat of the alloy and $C_{1}$ and $C_{2}$ would be the specific heats of the components making up the alloy, whereby component 1 would have a mole fraction of $x$  and component 2 would have a mole fraction of $(1-x)$. For a mixture of ideal gases such a formula would be correct but I doubt that it will work well with alloys in the real world. But if this came up in the context of a school lesson then the approximation might be ok.


Will (3R,4S,5R)-1,3,4,5,6-pentahydroxyhexan-2-one tautomerise in nitric acid? The following question was sent to me .The question is " Which of the following is  A ?"
The given answer was a,b,c

Background
If both ends of an aldose chain are oxidized to carboxylic acids the product is called an aldaric acid.

source : https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/carbhyd.htm#carb2
My thoughts

Options a and c are direct reactions.
Option b gives 2 and  it can only rotated in the plane it is drawn (2D) by 180 degrees (http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-7.html) to give 1.
How ever option d in acidic medium could tautomerise to give 3d , that on oxidation again gives 1.

Therefore answer should be a,b,c,d.
My answer hinges on tautomerisation of ketone is acidic medium. Would this compound (3R,4S,5R)-1,3,4,5,6-pentahydroxyhexan-2-one tautomerise in nitric acid ?
Any help in this regard is very much appreciated. Thankyou.
reference :
1 http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch07/ch7-7.html

Well I am just a high school student, so I may be wrong, and I beg pardon in advance, but I would like to share my notions about this topic.

@ChakravarthyKalyan I would like to ask you to review your last step in the conversion of 3d to 1. 3d on oxidation should yield this product (in the image)--which is by my knowledge not same as 1.

So even your theory does not yield (d) as the correct option, I assume. Also I doubt the double tautomerism of the compound in option (d) in acidic medium, though your mechanisms were convincing. This closely resembles Lobry-Debryn-van-Eckenstein rearrangement that occurs only in presence of alkaline medium.

As per my knowledge, ketoses in presence of strong oxidizing agents like conc. HNO3, undergo oxidative cleavage forming 2 carboxylic acids(acc. to Popoff's rule), or as @user55119 had pointed out to me, that alpha ketoacids can undergo oxidative decarboxylation and form CO2 and an acid(though its mechanism is quite unclear to me as well).


Aromatic ether to phenol reagents I have been asked to identify suitable reagents for the following transformation:

(Apologies it's back to front - the question is a map of reactions and I took a snapshot).

I really have no idea about this one. Don't even know where to begin, sorry. For the purposes of the question I only need reagents but I would appreciate a mechanism if you know it. Thanks.

The reaction is demethylation of methoxybenzene. You can do it by using an acid which has a nucleophilic counter ion.

It is not actually obvious why the iodine nucleophile attacks the methyl group and not inside benzene ring. The other option would be:

This reaction doesn't happen because otherwise the iodine ion would appear inside the benzene ring and resulting product would be in a very strange conformation (high energy).
More information:
 Clayden, Jonathan; Greeves, Nick; Warren, Stuart; Wothers, Peter (2001). Organic Chemistry (1st ed.). Oxford University Press. p. 434, 589.

As it was correctly stated in the other answer there are also other reagents which can be used to cleave the methyl ether. $\ce{BBr3}$ is a Lewis acid which also can be used in this reaction. A relatively new paper describes a computationally calculated bimolecular mechanism for this reaction.


How to synthesize tetrabutyl titanate from titanium dioxide powder? How do you synthesize tetrabutyl titanate $(\ce{[CH3(CH2)3O]4Ti})$ from titanium dioxide ($\ce{TiO2}$) powder? As I know, we can get $\ce{TiO2}$ from tetrabutyl titanate, but how can we do the opposite?

Treat with a chlorination reagent the titania to form titanium tetrachloride. You could do this with coke and chlorine in an electric arc furnace. Then distill the titanium tetrachloride. Next react it with a mixture of sodium butoxide and butanol at a low temperture.

Next vacuum distill the product to obtain pure titanium butoxide


Reaction speed of Na2CO3 with vinegar vs that produced by NaHCO3 with CH3COOH I'm working on a project requiring the easy production of a gas not corrosive to "normal" household materials (e.g. carbon dioxide). I settled on vinegar (acetic acid) and baking soda or washing soda, as it is very easily available to most people. Looking at the chemical formula for both sodium carbonate and sodium bicarbonate, I can assume that roughly the same amount of gas will be produced by the reaction of both with vinegar.

However, is there any difference at all in reaction speed? I don't care about the end product as long as it can be washed down the drain.

Sodium carbonate, $\ce{Na2CO3}$, will not create gas when mixed with an equimolar  portion of acetic acid (i.e. $1\  mol\ \ce{Na2CO3}:1\ mol\ \ce{CH3COOH}$). The reaction would proceed as follows: $$\ce{Na2CO3} + \ce{CH3COOH} \ce{->} \ce{NaHCO3} + \ce{CH3COONa}.$$ Therefore, you would need another equimolar portion of acetic acid to protonate the bicarbonate ion, according to this reaction: $$ \ce{NaHCO3} + \ce{CH3COOH} \ce{->} \ce{CH3COONa} + \ce{H2O} + \ce{CO2_{(g)}}.$$ This should illustrate that using sodium carbonate simply causes you to use an extra portion of acetic acid to make the same amount of gas.

Thus, sodium bicarbonate and acetic acid will produce gas in less overall time than sodium carbonate and acetic acid. When starting from sodium carbonate, the reaction will proceed through both reactions above. When starting with sodium bicarbonate, only the second reaction must proceed. Technically, the rate of gas evolution will be the same once gas formation begins (assuming equal starting molarities). However, the sodium carbonate reaction wouldn't produce any gas until the first reaction had completely finished, taking more time overall.


E-glass degradation in water At the moment, a question puzzling me is why and how glass fibers degrade in water environment? As the degradation of the glass fibers occurs, its reduces the strength of them. As far as I know, the degradation occurs slower in seawater (containing ions) than in distilled water, which in my opinion indicates that process is related to dissociation of glass material.

How does this degradation occur and why is it different for seawater and distilled water? Do I understand correctly, that E-glass is slowly partly dissociating into ions in water environment? Or is it ion-exchange or some other mechanism?

Simply the glass degrades because it dissolves in the water.

First the how. There are two factors which impact this process which is the dissolution of sodium and the dissolution of silica. The dissolution of sodium occurs first as evidenced by a plot of pH vs time for a glass powder in water and proceed via electrophilic attack:
$$\ce{Si-O-Na_{(glass)} + H2O_{(l)} \longrightarrow Si-O-H_{(glass)} + OH-_{(aq)} + Na+_{(aq)} }$$

The second step is the dissolution of the silica which proceeds via nucleophilic attack:
$$\ce{(-O-)2Si(-O-H)2_{(glass)} + H2O_{(l)} <-> Si(OH)4_{(aq)} }$$

The solubility of the silica is, of course, low, but measurable and significant. This process at high temperature increases the solubility of the silica (hydrothermal) and is how natural quartz crystals were formed (hydrothermal vents).

Second you say that degradation occurs slower in seawater than distilled water and that, in fact, is correct. At first, one might suspect that the dissolved sodium reduces the dissolution due to a common ion effect or reduced chemical potential but is actually not the case. The operative word here is seawater.  Seawater already has a large amount of dissolved silica in it from the rocks and sand that lines coastlines and ocean beds such that it is saturated. Since the second reaction is reversible the increased concentration prevents the silica from being leached appreciably faster than it can be redeposited thus reducing the corrosion of the fiber


What is the net ionic equation of sodium hydroxide when it dissolves in water? What is the net ionic equation of sodium hydroxide when it dissolves in water?

For the net ionic equation I got
$$\ce{NaOH(s) + H2O(l) -> NaH+(aq) + OH- (aq) + H2O(l)}$$
but it was wrong.
Then I tried:
$$\ce{NaOH(aq) -> Na+(aq) + OH- (aq)}$$
and it was wrong as well.
Can someone please explain?

When you write down the net ionic equation, you have to consider that the educt and the products are in different phases, a solid and a liquid phase. Undissolved $\ce{NaOH}$ is a solid, denoted by the index $\ce{(s)}$, while the dissolved $\ce{Na+}$ and $\ce{OH-}$ ions are solvated in the aqueous phase (index $\ce{(aq)}$ in the equation):

$$\ce{NaOH(s) ~->~ Na+(aq) + OH^{-}(aq)}$$


What are the enthalpy and entropy contributions to carboxylic acids dissociating in water? Carboxylic acids such as acetic acid do not dissociate much in the gas phase. In aqueous solution, however, they act as acids (proton donors).

$$\ce{CH3COOH(aq) <=> H+(aq) + CH3COO-(aq)}$$

Are reaction entropy and reaction enthalpy positive or negative, and what are there magnitudes?

In the gas phase, we have clear expectations:

$$\ce{CH3COOH(g) <=> H+(g) + CH3COO-(g)}$$

The reaction enthalpy will be positive (we need to break a bond) while the reaction entropy will be positive (more gas particles). As the reaction does not go forward in the gas phase, the enthalpic contribution must be of larger magnitude.

So another way of stating the question is to ask what the effect of solvation is on the enthalpy and entropy of reaction.

Electrostatic screening associated with solvation is obviously the key factor enabling dissociation in aqueous solution when compared to the gas phase. Without such dielectric screening, dissociation would be too costly ($\approx \pu{1 MJ/mol}$ or $\approx\pu{14 eV}$,  roughly the energy required to ionize a hydrogen atom) to occur to a significant extent at RT. The dielectric constant of water reduces the energy of the gas phase electrostatic interaction by a factor of $\approx 80$. This is an enthalpic effect. This energy is further reduced by water oxygen bonding to $\ce{H^+}$ to form $\ce{H3O+}$ (proton to hydronium).

Differences between dissociation constants of different carboxylic acid groups originate from enthalpy or entropy changes due to anion formation (since the formation of protium is something shared in common by all acids, its solvation cannot explain differences in behaviour of different acids).  The effects on the solvent are often described in terms of "structure making or breaking". As a rule there is an entropic penalty ($\Delta S^\circ<0$) with dissociation that can be attributed to solvent rearrangement around the ensuing charged species. The penalty is seen in thermodynamic data for many weak acids.  Trends are otherwise not quite as clearcut. Beyond the entropic penalty, other factors (enthalpic or entropic) may play a role in encouraging or discouraging dissociation, for instance charge delocalization (inductive, mesomeric effects) or intramolecular rearrangements.


Ionization energy of S²⁻ vs S Which of $\ce{S^2-}$ and $\ce{S}$ has smaller ionization energy?

On one hand, ionization energy of $\ce{S^2-}$ can be predicted to be smaller than that of $\ce{S}$ because it is "easier" to remove and electron from an electron-rich ion, which already experiences a lot of electron-electron repulsion and reduced $Z_\mathrm{eff}$.

On the other hand, $\ce{S^2-}$ has a stable noble gas electron configuration, and removing an electron would get rid of the noble gas configuration.

I could not find relevant data for the second electron affinity of sulfur, so I am unable to check the reasoning.

Start by considering that electron affinities and ionization energies are related. Thus $\mathrm{EA}$ of $\ce{S-}$ is equal to $-\mathrm{IE}$ of $\ce{S^2-}$. (Note the convention used here is to report the value of the signed energy change associated with the process (not the absolute value), see the Wikipedia for an explanation if necessary).

A good place to continue is reading chem.libretexts.org:

  the first electron affinity can be greater than or equal to zero or negative, depending on the electron configuration of the atom. In contrast, the second electron affinity is always positive because the increased electron–electron repulsions in a dianion are far greater than the attraction of the nucleus for the extra electrons.[...] Similarly, the formation of all common dianions (such as $\ce{S^2-}$) or trianions (such as $\ce{P^3−}$) is energetically unfavorable in the gas phase.

This means that the electron affinity of $\ce{S-}$ is positive. Since $\mathrm{IE}(\ce{S^2-}) = -\mathrm{EA}(\ce{S-})$, this means that  $\mathrm{IE}(\ce{S^2-}) < 0$.

Note in comparison that the first ionization energy of sulfur is $\pu{999.6 kJ/mol}$.

Therefore $\mathrm{IE}(\ce{S^2-}) < \mathrm{IE}(\ce{S})$, with $\mathrm{IE}(\ce{S^2-})$ exothermic.


Formation of azobenzene from nitrosobenzene and N-hydroxyaniline I'm trying to create a reasonable reaction mechanism for the reaction of nitrosobenzene and N-hydroxyaniline, which happens as part of the reduction of nitrobenzene to azobenzene using magnesium as reductor and methanol as solvent. Ron was very helpful to give me a general outline of this entire mechanism:

  The key step is formation of nitrosobenzene (this is just a reasonable proposal).

  Some of the nitrosobenzene can be further reduced to aniline and
  N-phenylhydroxylamine, both of which are nucleophiles and can react
  with the nitroso N=O bond (analogous to a carbonyl). Elimination of
  water in both reactions produces azobenzene and azoxybenzene
  respectively. The azoxybenzene can be further reduced to azobenzene.

I went on to try and create a reasonable mechanism for the reactions of N-hydroxyaniline with nitrosobenzene and aniline with nitrosobenzene (both in the presence of magensium with methanol as solvent):

I wonder whether this is a correct reasonable reaction mechanism. I tried to move radicals as much as possible from positively charged/highly electronegative atoms (e.g. the nitrogen in the nitro functional group / oxygen) to the less electronegative atoms (e.g. carbon). My reasoning was as follows (I have not yet learned about in radical chemistry, but I give it a try): radicals can be removed in two ways: 1) the radical electron leaves the molecule, which results in a positive charge. This is less likely to happen with a more electronegative atom like oxygen in comparison to for example carbon, so that the radical is more stable on oxygen. 2) The radical reacts with an other molecule and obtains an extra electron forming a new covalent bond (in this case the electron is donated by methanols C-H bond), resulting in a neutral molecule with no radicals. Since oxygen is more electronegative than carbon it is more "greedy" to fill the singly occupied orbital, making the radical more stable on carbon.
Furthermore the second way to lose the radical is much more likely than the first, since it leaves a neutral molecule instead of a charged one.

At last, I was told that reactions involving metal surfaces (like magnesium in this case) are often not well understood. Why is that? Because there are too many pathways?

Below is my proposed mechanism. You'll see that we go through some of the same key intermediates, so I think you are on the right track. In your second step, it is much more likely that the proton is removed by acid base reaction instead of through the electron transfer process you've shown. Since the initial intermediate (my second structure) is zwitterionic, we can just say that it will neutralize itself. If the reaction conditions are not acidic (from what I see, it's just methanol and magnesium), then there will not be an appreciable amount of the protonated hydroxylamine. The mechanism doesn't require it, as nitrogen is strong enough to "kick out" hydroxide. The result is essentially an azoxybenzene, so we just need to figure out how to lose the other hydroxyl. As the cationic form, it will want to accept electrons, such as from the magnesium as a free electron. Back to a neutral (radical) compound, the other hydroxyl can be kicked out, giving a radical cation on nitrogen. Filling in a second electron from magnesium results in the azobenzene product.