Freezing point of solution It's stated in my textbook that at the freezing point, the vapour pressure of the solid phase equals the vapour pressure of the liquid phase. In a mixture, since the vapour pressure of the solution is lower than the vapour pressure of pure solvent, the temperature has to be lowered in order for the vapour pressures to equalize and hence freezing to take place.

According to what I understand after reading my textbook, when a solution freezes, it's the solvent which converts into solid and leaves behind liquid phase consisting of both the solvent and solute. Thus, how can the pure solid solvent and liquid phase mixture have the same vapour pressure at the same temperature, since the vapour pressure of pure solvent at any temperature is greater than the vapour pressure of mixture?

The figure shows the relationship between the solid and liquid solvent and a solution.

Instead of freezing at the freezing point of the solvent $T_1$ a solution freezes at $T_2$, where the vapour pressure of the solution equals that of the pure solid solvent.

Using the Clausius-Clapeyron equation we can show that the solution curve will always intersect the solid solvent curve. The vapour pressure of the solution follows the line $\displaystyle \frac{d\ln(p)}{dT}=\frac{\Delta H_{vap}}{RT}$ and the pure solid solvent $\displaystyle \frac{d\ln(p_{ss})}{dT}=\frac{\Delta H_{sub}}{RT}$ where $\Delta H_{sub}$ is the molar heat of sublimation. This in turn is given by $\Delta H_{sub}=\Delta H_{vap}+\Delta H_{fus}$ where 'fus' means heat of fusion. Thus $\Delta H_{sub} \gt \Delta H_{vap}$ which in turn means that the slope of the solid solvent curve is always greater than that of the solution $\displaystyle \frac{d\ln(p_{ss})}{dT} \gt \frac{d\ln(p)}{dT}$.  As the freezing point of the pure solvent is higher than that of the solution the vapour pressure curve of the solid solvent and solution must always intersect.


Why do molecules have to have a change in dipole moment in order for them to be IR active? My chemistry textbook keeps saying that in order for a molecule to be able to absorb infrared radiation, it has to have a change in dipole moment when the bond vibrates.

I don't understand why that is. Some clarification would be appreciated.

There is an intuitive reason why a vibration is IR-active only if it involves a change in the dipole moment.

Recall that the typical wavelength of IR radiation ($\sim 10\mu\mathrm{m}$) is much larger than the typical size of a molecule ($\sim 1\mathrm{nm}$). Hence, to a very good approximation, the (time-varying) electric field of the IR radiation is spatially uniform within a molecule.

Now observe that under a spatially uniform electric field, all positive charges, regardless of their positions, are pushed to a common direction, and all negative charges are pushed to exactly the opposite direction. In this case, can the change in the total dipole moment, defined as $\Delta \vec{\mu} = \sum_{i} q_{i} \Delta \vec{r}_{i}$, be equal to zero?

Clearly, the answer is "No." The displacements $\{\Delta \vec{r}_{i}\}$ are parallel between like charges and antiparallel between opposite charges. Therefore, $\{q_{i}\Delta\vec{r}_{i}\}$ are all parallel to one another, and $\vec{\Delta\mu}$ cannot be equal to zero.

So far, I have argued that a motion induced by a spatially uniform electric field always involves a change in the dipole moment. A corollary is that any motion that does not involve a change in the dipole moment cannot be induced by a spatially uniform electric field (of IR radiation).

As a simple example, imagine applying a spatially uniform electric field to a $\ce{CO_{2}}$ molecule such that the direction of the field is parallel to the length of the molecule. Can this field ever induce a symmetric stretch, in which the two oxygen atoms move in opposite directions? Certainly no. Both oxygen atoms are negatively charged by exactly the same amount. Under a spatially uniform electric field, they are pushed to the same direction by the same magnitude, and there is no room for a symmetric stretch.


Are all mixtures linear? I'm writing a computer program to help a factory produce bottles of mixtures. A mixture is usually alcohol, water and some other stuff, like menthol. But could also be water and NaOH and some oils (think schampoo).
My basic understanding is that if one bottle is 500 ml and contains:
50% water
45% alcohol
5% oils
I need:
250 ml water
225 ml alcohol
25 ml oils
Then if I want to make 1000 bottles I just multiply this with 1000:
250000 ml water
225000 ml alcohol
25000 ml oils
However this is with the assumption that all mixtures have linear dependencies.
So my questions is:

Is all (possible) mixtures linear?
Are all reactions linear?
Are there a difference between mixtures and solutions here?

I suspect this is a very easy question for someone who know chemistry, but unfortunately my understanding is basic and my search skills too bad to find this out.

Mixtures are never linear. Mixing $50$ mL water + $50$ mL ethanol gives a $96$ mL mixture. The masses are additive, but not the volumes. I agree the discrepancy is not huge. If you can tolerate a $10$% error, you may consider that the volumes are additive. Anyway take care of the fact that, usually percentage are indicated in mass, not in volume.


About pH of an aqueous solution of SO2 Probably we can have an aqueous solution of $\ce{SO2}$ by dissolving it in water, because we would have an equilibrium between $\ce{SO2(g)}$ and $\ce{SO2(aq)}$:

$$\ce{SO2(g) <=> SO2(aq)}$$

How can I compare pH of an aqueous solution of $\ce{SO2}$ with that of pure water qualitatively?
What happens to the pH of an aqueous solution of $\ce{SO2}$ if I aerate it by bubbling air through it?

Upon adding $\ce{SO2}$ gas to water, it shows its acidic oxide character by hydrolyzing to sulfurous acid:

$$\ce{SO2(g) + H2O(l) <=> H2SO3(aq)}$$

$\ce{H2SO3}$ is a weak diprotic acid in water, with the following acid dissociation equilibria:

$$\ce{H2SO3(aq) <=> H+(aq) + HSO3^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1}=1.4\times 10^{-2}}$$
$$\ce{HSO3{}^{-}(aq) <=> H+(aq) + SO3{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=6.3\times 10^{-8}}$$

Thus, addition of $\ce{SO2}$ gas to water will produce an acidic solution.

What would happen if air were bubbled into the solution? Air is a mixture of mostly nitrogen and oxygen. It so happens that $\ce{SO2}$ reacts with oxygen to form $\ce{SO3}$:

$$\ce{2SO2(g) + O2(g) -> 2SO3(g)}$$

Here, the sulfur atom in $\ce{SO2}$ is oxidized (oxidation number goes from +4 to +6) while the oxygen atoms in $\ce{O2}$ are reduced (oxidation number goes from 0 to -2). Now, $\ce{SO3}$ also happens to be an acidic oxide, and it hydrolyzes in water to form sulphuric acid:

$$\ce{SO3(g) + H2O(l) <=> H2SO4(aq)}$$

This time, $\ce{H2SO4}$ is a strong diprotic acid:

$$\ce{H2SO4(aq) <=>> H+(aq) + HSO4^{-}(aq)}\ \ \ \ \ \mathrm{K_{a1} \approx 10^6}$$
$$\ce{HSO4{}^{-}(aq) <=> H+(aq) + SO4{}^{2-}(aq)}\ \ \ \ \ \mathrm{K_{a2}=1\times 10^{-2}}$$

Assuming that not much $\ce{SO2}$ is lost while bubbling air through the sulphurous acid solution, then you're effectively replacing a weak acid with a strong acid, which would cause the pH to decrease.

As a final remark, if the sulphurous acid solution were bubbled with pure nitrogen, then there would be no oxidation of $\ce{SO2}$ into $\ce{SO3}$, and after a while enough $\ce{SO2}$ will have been dragged away for the solution to become a less concentrated sulphurous acid, which would therefore have a higher pH.


What does peroxide value exactly measure? I understand that peroxide value gives a measure of the extent to which an oil has undergone oxidation, however, I looked and couldn't find any source which explains the chemistry behind PV value. More specifically, how is pv value related with oxidation?

Wikipedia states that "the peroxide value is determined by measuring the amount of iodine which is formed by the reaction of peroxides (formed in fat or oil) with iodide ion." But what is the reaction of the peroxides? And how is it related with oxidation?

Thank you

The test measures the amount of hydroperoxide by a redox reaction with iodide:

$\ce{2ROOH + 2H+ +2I- -> 2ROH + I2 + H2O}$

In the course of the reaction, the hydroperoxide is reduced, while iodide is oxidized to iodine. In principle, this is a variation of the iodine clock reaction of hydrogen peroxide.


How is kinetic energy accounted for when describing diatomic bonds with a Morse potential? The Morse potential can be used to describe the potential energy of the a diatomic bond:

This curve suggests that the lowest state $v = 0$ is the most stable, since it has the lowest potential energy. However, the energy of a system comprises both potential as well as kinetic energy, and stability should depend on the total energy, not just the potential energy. How does kinetic energy come into play here? Is it being neglected for some reason, or is this description of a diatomic molecule insufficient?

The kinetic energy is not ignored. Each quantized state (labeled by the vibrational quantum number) has a total energy given by the expectation value of the Hamiltonian operator for the wavefunction. The contribution of the kinetic energy can be computed as the expectation value of the QM kinetic energy operator, or as the difference between the expectation values of the total energy and potential energy.
Because the math is simpler I will limit the discussion to the harmonic oscillator (ie quadratic potential). For the QM harmonic oscillator the expectation value of the potential energy turns out to be
$$\left< E_\text{pot} \right> =\frac12 \left(v+\frac12\right)\hbar \omega=\frac12 E_\text{tot}$$
The expectation value of the kinetic energy is then
$$\left< E_\text{KE}\right>=E_\text{tot}-\left< E_\text{pot} \right>= \left< E_\text{pot} \right>$$
So it turns out that the kinetic energy and potential values have identical expectation values, consistent with the prediction of the virial theorem for a quadratic potential. This  is all outlined very nicely for instance in Atkins' physical chemistry textbook.
It is difficult to tell what energy levels are displayed on the diagram but they might very well correspond to the total energy given by $ E_\text{tot}  = \left(v+\frac12\right)\hbar \omega$.


Does symmetry lower the energy of molecules? Why does nature "prefer" symmetry? Why is symmetry energy-lowering? I keep hearing statements along the lines of this tossed around by chemists. Here are some examples:

https://physics.stackexchange.com/questions/135671/nature-favours-symmetry

https://books.google.com/books?id=WkRy8b0F2HwC&pg=PA153&dq=nature+favors+symmetry&hl=en&sa=X&ei=V__fVPvpGsuxggSptIG4AQ&ved=0CDAQ6AEwAw

IBrF2 - van der Waals repulsions - and number of unique bond angles (The more symmetric conformation of the molecule is probably preferred).

So:

1) Are any of the above statements true? If so, can you provide example(s)?

2) I think the statement is false at least to some extent. Consider cyclobutadiene ... it's not a square; it's a rectangle.

Symmetry is preferred if it lowers energy, but it does not always.

My high school chemistry teacher had a good rule of thumb. A chemical system will [almost] always tend towards the lowest energy configuration. (I insert "almost," because there are some cases where this isn't true - usually because of entropy.)

So yes, many things are symmetric and this usually lowers the energy.

You raise a good counter-example. Cyclobutadiene is not square. A square cyclobutadiene would be a di-radical, so it undergoes symmetry breaking (i.e., a Jahn-Teller distortion) and becomes a rectangle.

We see here that the rectangular distortion has lower total energy - one orbital goes down in energy and becomes doubly-occupied. The other goes up in energy and is now empty.

So what about benzene? Why is it symmetric, vs. a double-single-double.. bond alternation?

Well, in this case, the symmetric (aromatic) version is lower in energy.
(Figure, Chemwiki)


Is this a plausible reaction mechanism? (H2SO4 cyclization after Friedel-Crafts) in my undergraduate chemistry lab we synthesized the substituted benzene on the left in a Friedel-Crafts reaction.
As a next step we're supposed to reflux for 1.5h at 90°C with concentrated sulfuric acid and obtain the product on the right. A bicyclical product.

We have to come up with a plausible mechanism for the ring closure reaction and I am unsure that what I came up with makes sense or not.

Thank you!

edit: this is the FC reaction we did in the first step:

The Steps for this reaction are as follows:
To a Solution of AlCl3 in abs. DCM the following reagents were added:
-3-chloropropionic chloride.
-p-xylene.
The reaction mixture was stirred at rt for 1h.
Then added to a mixture of 1:1 conc. HCl and ice water.
Then extracted with DCM in a separatory funnel.
Washed twice with a sat. NaCl solution.
The organic phases were dried with MgSO4.
The DCM was then removed in the rotary evaporator.
A yellow-orange Solution was left at the end.
It was left to crystallize over night.

Now this is where the second part happens, the H2SO4-cyclisation reaction.
.......
Some comments suggested that there might be a hydrolyzed intermediate with a displaced chloride. Here is my take on how that would look like:

I would propose the mechanism illustrated below, which would seem plausible in concentrated sulfuric acid:


Can Molisch test be used to separate monosaccharides from polysaccharides? Do the results of Molisch test on monosaccharides have any differences with when it's done to disaccharides or polysaccharides?

No. Mono and poly saccharides both give a positive test.


Density of plutonium oxide I am trying to find the mass density $\rho$ of plutonium dioxide, $\ce{PuO2}$. I could calculate it, but the mass density of oxygen found in the literature is the density of gas oxygen.

So I was wondering where can one find mass densities for compounds that come from experimental measurements?

A quick Google search shows the density to be $11.5~\mathrm{g~cm^{-3}}$.

You might have had trouble finding it since it is actually called plutonium (IV) oxide, as plutonium is an f-block metal (specifically an actinide) displaying multiple  oxidation states (commonly III, IV, V, and VI, while VII is rare).


What is the pH of a solution of HCOONa and HNO3? Calculate the pH of an aqueous solution obtained by mixing
  20 ml of $\ce{HCOONa}$ 0,1 M and 5 ml of $\ce{HNO_3}$ 0.2 M.
  ($K_\mathrm a$ of $\ce{HCOOH}$ $= 10^{-4}$)

My try:

The number of the moles involved are: $n_{\ce{HCOONa}}=0.002~\mathrm{mol}$, $n_{\ce{HNO3}} = 0.001~\mathrm{mol}$.
We have the reaction:

$$\ce{HCOONa + HNO3 -> HCOOH + NaNO3}$$

Where 0.001 moles of $\ce{HNO3}$ si is consumed by reacting with 0.001 moles of $\ce{HCOONa}$ giving 0.001 moles of $\ce{HCOOH}$. Because remains $0.002\ \mathrm{mol}-0.001\ \mathrm{mol}=0.001~\mathrm{mol}$ of $\ce{HCOONa}$, we'll have a buffer solution with the acid $\ce{HCOOH}$.

It'll start the reaction:

$$\ce{HCOOH + H2O <=> HCOO- + H3O+}$$

Where the initial concentration of $\ce{HCOO-}$ is equal to $\ce{HCOONa}$ and is $\frac{0.001~\mathrm{mol}}{(0.02+0.005)~\mathrm l} = 0.04~\mathrm{mol~l^{-1}}$. So we have,

$$[\ce{H3O+}]=K_\mathrm a \frac{[\ce{HCOO-}]}{[\ce{HCOOH}]} \Rightarrow \mathrm{pH} = \mathrm pK_\mathrm a - \log_{10}{\frac{[\ce{HCOO-}]}{[\ce{HCOOH}]}} = \mathrm pK_\mathrm a = 4$$

Is it correct?

Yes. You could also put the molar ratio into the equation for calculating pH (then you didn't need to calculate the concentrations).

And those are not initial concentrations, but equilibrium concentrations.


Why is chloridric acid used in salt bridges? I've been planning an experiment on how to make galvanic cells and during my research I came across an article in which chloridric acid was used, along with a soluble salt and agar-agar,  in salt bridges.
Since then, I've been wondering why chloridric acid was added to agar-agar.

Thanks in advence.

Any ionic solution can function as a salt bridge. Having it acid based helps with the fast conduction paths of H+ in water.

The critical question is the leak rate. Any salt bridge will leak some amount of the ions into the solution. Good ones leak less, bad ones leak more.

So if you cannot tolerate any acidity in your solution, don't use an acid. Similarly, if you cannot tolerate any chloride in your solution, this would be a bad choice.


Producing carbon nanotubes from graphite I am searching for the carbon nanotubes for very low cost. I have one idea, which is to use graphite (such as that found in pencils.) Can it be converted to carbon nanotubes? If so, what would be the procedure for this?

Graphite does not contain CNTs, a simple extraction therefore is not possible.
Graphite seemingly can be converted to CNTs by a "simple" oxidative method published by DW Lee and JW Seo on arxiv.org in 2010.
The authors claim to have used commercially available graphite powder (45 µm, Aldrich) of high purity (>99.995%), but did not provide any information on the yields.
Pencils typically contain a mixture of graphite powder and various amounts of clay as a binder.


Why do we use phenolphthalein as an acid-base indicator? It seems that no resource I come across addresses the fact that the pH range for phenolphthalein's color change is around 8.2-10. In other words, it's still colorless at 8.2. Why then, is it claimed that the indicator changes colors when the solution changes from an acid to a base? Wouldn't you want an indicator that changes colors at about pH 7?

If you analyse a titration curve you will immediately notice the pH shooting up near equivalence point. For this reason, there is very little difference to the amount of base added for pH 7 and pH 8 (due to the logarithmic nature of pH). So, it's not entirely accurate but certainly good enough (for strong acid - strong base and weak acid - strong base), if you want something better use an electronic pH tester


Are hydrogen atoms in acetone molecule acidic? My chemistry book says that acetone is aprotic, but I think that, since the conjugate base of acetone has possible  resonating structures, the hydrogen atom must be free to move out as $\ce{H+}$ (as the resulting structure is more stable than the previous one). Where am I going wrong?

You’re not going wrong anywhere. Acetone can indeed be deprotonated. The resulting structure is called an enolate.

$$\ce{H3C-C(=O)-CH3 <<=> H3C-C(-O^-)=CH2}$$

When compared with truly acidic compounds such as $\ce{HCl}$ and even $\ce{NH4+}$, acetone is a very, very weak acid, though. That does not mean it should be considered aprotic. Given a strong enough base — e.g. $\ce{HMDS}$ ($\ce{(Me3Si)2N-}$) — it can be deprotonated. Weaker bases will allow for partial (but probably hardly verifiable) deprotonations.


What is "dot" sign in •NO? What is "dot" sign in •NO? I know that it is radical nitric oxide, but I don't know if it is necessary to put the "dot". Is there any difference between •NO and NO?

Usually the dot is put there to emphasize that the nitric oxide is a free radical that includes an unpaired electron.  This is especially notable by comparison with $\ce{NO^+}$, which does not have the unpaired electron.

Note that the nomenclature $\ce{·NO}$ should not be rendered as showing the unpaired electron on nitrogen.  The unpaired electron is actually incorporated into the molecular orbital structure of the nitric oxide molecule, covering both atoms.


Relative humidity above various mixtures of table salt and water I heard that if I saturate some water with NaCl and put it in a closed box, the relative humidity near the water surface should be around 75 % at room temperature.

I am using this to calibrate some electronic sensors measuring relative humidity.

If a saturated solution gives 75 % RH, how can i know which RH more diluted mixtures should result in?

--

My setup is a small 1l plastic container where in I put a mixture of table salt (iodized) and tap water (about 150 ml). The sensor is hanging a few cm above the water surface and measures for a day or so.

The solutions I use are:

Fully saturated water (Plenty of undissolved salt crystals)
75 % saturated water (3 parts saturated, 1 part pure water)
66 % saturated water (2 parts saturated, 1 part pure water)
50 % saturated water (1 part saturated, 1 part pure water)

I am also monitoring the temperature in the container, and it is usually swinging a few degrees below 20 °C.

Your information on relative humidity (RH) above saturated $\ce{NaCl}$ is valid within about 0.5% over the range of 0 °C to 80 °C. For other standard humidity solutions, such as $\ce{MgCl2}$ or $\ce{K2CO3}$ the U. S. National Institute of Standards & Technology provides a convenient reference.

However, though it is possible to adjust RH by varying concentration, there are, "special  problems...  because  their  concentrations  must  be  measured  and controlled." If you need only a few accurate RH points, it would be simpler to use saturated solutions of some alternate salts.

BTW, it may take some time for RH to equilibriate because of stratification. It would help to keep the solution on a slow magnetic stirrer, moving not only the liquid but causing eddies in the air above it.


Binding energy vs. ionization energy I am studying binding energy in the context of photoelectric spectroscopy, and I came across a graph of relative numbers of electrons vs binding energy (in units of megajoules per mole). The element in question is lithium. My chemistry resource (Khan academy, to be precise) noted that the binding energy for the furthest right peak corresponds to the binding energy for the electron in the 2s subshell, and that this is also equal to the first ionization energy. However, they go on to say that the the binding energy of the second peak is not equal to the second ionization energy, since the electrons are held more tightly by the nucleus once the outermost electron is removed. I actually have two questions about this. First, why are they more tightly held once the outermost electron is removed? Is the electric field of the protons somehow getting weakened by the presence of the outer electron? I don't think this would be the case, based on my understanding of electric fields. So what is the explanation?
My other question is why the binding energy is different from the ionization energy? Is it because the electrons are removed all at once versus them being removed one-by-one? Or is there some nuance to this situation I am not seeing?
I found a question on this website that seems to be asking a similar thing, but no one answered it. Here's the link to it anyway.
Chemistry: What's the difference between "binding energy" and "ionization energy"?
Thanks in advance for the help.

First some basics....
Lithium has three electrons.

$\ce{Li}$ electronic configuration: $1s^22s^1$
$\ce{Li -> Li+ + e-\quad\quad\mathrm{Ionization\ Energy= 0.520\ MJ/mol}}$
$\ce{Li+}$ electronic configuration: $1s^2$
$\ce{Li+ -> Li^2+ + e-\quad\quad\mathrm{Ionization\ Energy= 7.298\ MJ/mol}}$
$\ce{Li^2+}$ electronic configuration: $1s^1$
$\ce{Li^2+ -> Li^3+ + e-\quad\quad\mathrm{Ionization\ Energy=11.815\ MJ/mol}}$
$\ce{Li^3+}$ electronic configuration: $1s^0$, i.e. no electrons
Ionization energies from Wikipedia

Thus the different ionization reactions remove the electrons from the outer most to the inner most electrons.
Below is an another image of the idealized spectra of lithium that has the binding energies stated. (from https://www.andrew.cmu.edu/course/09-105/S98Slides9.html)

$\mathrm{Binding\ Energy= 0.52\ MJ/mol}$ corresponds to $\ce{Li -> Li+ + e-}$
or $1s^2 2s^1\ce{->} 1s^2$, hence the first ionization energy of a Li atom.
However $\mathrm{Binding\ Energy= 6.26\ MJ/mol}$ corresponds to $\ce{Li -> Li+ + e-}$
or $1s^2 2s^1\ce{->} 1s^12s^1$, which is not the first or second ionization energy of a Li atom. Rather it is the binding energy of the 1s orbital in a neutral lithium atom.
Hence the photoelectron spectra is called the binding energy spectra because that is what it measures. Note that the sweep of the spectra would not have to cover the ionization energy. For example the spectra could have been taken from 5 MJ/mol to 100 MJ/mol.


How to calculate which ionic compound has the highest melting point? Which of the following compounds has the highest melting point: $\ce{PCl5}$, $\ce{H2O}$, $\ce{NaCl}$, $\ce{SrCl2}$, $\ce{CaF2}$?

In general, ionic compounds will have higher boiling points than those of dipole-dipole interaction compounds.

So, the compound above that will have the highest melting point will certainly be an ionic compound, either $\ce{NaCl}$, $\ce{CaF2}$ or $\ce{SrCl2}$.

I thought that I should calculate the molar weight but that won't work, because the correct answer is $\ce{CaF2}$, and $\ce{SrCl2}$ has a greater molar weight.

Between ionic compounds, how can the one with the highest boiling point be predicted?

The strength of the ionic bond depends on Coulomb's law for the force acting between two charged particles where larger force translates to a stronger ionic bond. The equation is

$$F = \frac{-k\cdot q_1 \cdot q_2}{r^2}$$

$k$ is a constant; $q_1$ and $q_2$ are the charges on the ions and $r$ is the distance between the ions.

So the larger the charge the greater the force, and the smaller the radius the smaller the distance between the ions (smaller $r$ in the equation) again leading to a stronger force.

So, first you should calculate the charge on each ion in a molecule. As can see, the charge on strontium ($\ce{Sr^2+}$) and calcium ($\ce{Ca^2+}$) ions is more than the charge on sodium ions ($\ce{Na+}$). Also, calcium ions are smaller than strontium ions. Fluorine ions are also smaller than Chlorine ions. So $\ce{CaF2}$ will have higher bond strength which means it will have higher boiling point since stronger bonds need more energy to break.


Why are isotopes an issue in reading mass spectra? It is my understanding that the isotope distribution for carbon and hydrogen, for example, are skewed heavily toward one species, i.e., one isotope accounts for ~99% of the abundance of the element in nature. If this is true (I question my ignorance on this point and others), then, the other isotope peaks should be negligible in the mass spectra or at least fall below the threshold for processing and remain unprocessed peaks.

I am asking this question within the realm of biological compound mass spectra. I read a post here that states that $\ce{Br}$ has near 1:1 abundance for two isotopes. It would explain the need to consider isotopes for those biological compounds with elements such as $\ce{Cl}$ which has ~3:1 ratio for its two most abundant isotopes. So, in light of the fact that $\ce{C, H, N,}$ and $\ce{O}$ make up 96.5% of a biological organism's weight, are isotopes an issue only when elements such as $\ce{Cl}$ or $\ce{Br}$ are constituents of the compound?

Here is the "periodic table for biomolecules" (leaving out hydrogen, please ignore silicon):

The pie chart shows the isotope distribution. Of the elements C, N O, P, S and H, sulfur has the highest abundance minor isotope (around 4%). On the other end of the spectrum,phosphorous is a mono-isotopic element, with just one stable isotope. There is an interactive version at https://www.isotopesmatter.com/applets/IPTEI/IPTEI.html.

So for small metabolites, there will be little signal due to minor isotopes. However, in a macromolecule with hundred carbons, the 13-carbon isotope (around 1% abundance) does become relevant. The signal for the mass ion will approximately follow a Poisson distribution, with zero or one 13-carbon isotope in the molecule as most probable:

Source: https://homepage.divms.uiowa.edu/~mbognar/applets/pois.html

For smaller numbers of atoms in a molecule, the multinomial distribution is appropriate, as explained here: https://chemistry.stackexchange.com/a/87669/72973


Why does cyclopropane react with bromine? In my exam, I was asked why cyclopropane could decolourise bromine water (indicating that it reacted with the bromine).

All I could guess was that it is related to the high angle strain in cyclopropane, as the C–C–C bond angle is $60^\circ$ instead of the required $109.5^\circ$. No book I have read mentions this reaction. What is the product formed, and why does it occur?

The following ring opening reaction will occour:

You are quite right about the angle strain. Because orbital interactions are not optimal in this geometry. Consider p-orbitals, then a natural bond angle would be $\theta\in [90^\circ; 180^\circ]$. A mixing of s- and p-type orbitals allows a wide range of angles $\theta\in (90^\circ,\dots, 180^\circ)$.

In cyclopropane $\ce{C3H6}$ - which you can also describe as trimethylene $\ce{(CH2)3}$ -  bonds have to be bent to overlap at all. A possible way of describing the bonding situation is regarding each $\ce{CH2}$ entity as $\mathrm{sp^2}$ hybridised. Two of these orbitals are used for $\ce{C-H}$ bonds (not shown) and one forms an inner two-electron-three-centre σ bond (left). This leaves p-orbitals to form some kind of degenerate π-like orbitals (middle, right).

This very general approach can be derived from a Walsh diagram. Schwarz et.al. {@academia.edu} and Hoffmann {@roaldhoffmann.com} described bonding quite similar and it is in quite good agreement with a calculation (BP86/cc-PVTZ, $D_\mathrm{3h}$) I have done. From this I have prepared a chart of all occupied molecular orbitals formed from valence orbitals and the LUMO. Here is a preview. Each orbital is viewed from three different angles:

Especially the symmetrical orbital 8 resembles very well the schematics. A quite rigorous approach for this theory can also be found here.

It is noteworthy - as mentioned by ron - that there is no notable increase in electron density in the centre of the ring. This may be due to the fact that there are much more orbitals having nodes in the centre than there are without.

Now bromine is known to be easily polarised $\ce{{}^{\delta+}Br-Br^{\delta-}}$ and may intercept at any point of the ring causing a bond break and relaxation to a less strained structure. It will most likely attack at the the $\pi$ type orbitals since bromine is an electrophile. The mechanism is analogous to the addition of bromine to ethene, which is nicely described at chemguide.co.uk. The essential part is the attack of the bromine at the HOMO(s).

The ring opening reaction can be reversed by adding sodium.

However, when there are bromine radicals present (UV light) then substitution will occur:
\begin{aligned}\ce{
Br2 &->[\ce{h\nu}] 2Br.\\
&+(CH2)3 -> (CH2)2(CHBr) + HBr
}\end{aligned}


Spin quantum number of electron Why is the spin quantum number of an electron +1\2 or -1\2 in an atom? Is this an experimental value or theoretical value?

It is both experimental and theoretical. Electrons have angular momentum, and sometimes we want to discuss whether it lines up with another direction (ie typically the direction an electromagnetic field, since electrons are charged their angular momentum makes them magnetism and that interacts with electromagnetic field).

How should we talk about the direction of the angular momentum? We can say "up" and "down" or "forwards" and "backwards", and sometimes we do that. But if we want to use it in calculations, it is easier to give them numbers. So what numbers should we give? They are opposite directions so really it makes sense to use opposite numbers. So what opposite numbers should we use? We can use any, so why not use a gap of "1" between them? So what opposite numbers have a gap of 1 between them? 1/2 and -1/2.

There are some other detailed reasons why 1/2 is a convenient number, but it is in my view fair to say we just want to represent two opposite directions in calculations, and the numbers 1/2 and -1/2 suit this well.


How can one reclaim isotopic cadmium from a solution of acid, i.e. H2SO4 or HCl etc.? I have dissolved isotopic cadmium-114 with $\ce{H2SO4}$ and or $\ce{HCl}$.  Now that I have the cadmium-114 in solution, precipitating out $\ce{CdSO4}$ and or $\ce{CdCl2}$, how can I reduce these simply to leave the pure cadmium metal isotope?  I have dissolved $\ce{^{114}Cd}$ with acid in very complex electronic tubes and I wish to reduce the precipitated compound(s) to pure metal for reuse.

$\ce{^{114}Cd}$ or is running about $ 1,500 per gram.  So extremely useful to reclaim it. Can anyone tell me a simple way to do this?

Thank you everyone for the discussion.  Sorry, I have been away for a few days, so was unable to follow up.
Reading the ideas, it is a toss up on a procedure.  Scraping the Cd off of an Aluminum electrode might work, but Al contamination would be very bad for the laser oscillation.  It might work for awhile, but would die from contaminants in short order.
The Al oxide, though only a molecular layer, is extremely tough and can withstand a great deal of heat and abuse.  I make Aluminum edge-wound voice coils for my speaker company and we depend on the oxide layer to act as an insulator from winding to winding.  Our coils can withstand 600 degrees and will melt before they fail.
Stripping the oxide gives you only a few seconds before it reforms in the normal atmosphere.  Any other thoughts?
Perhaps reduction of the salt by heating?
Thanks
Dale


What is the difference between distilled water and rainwater? I am studying different means of collecting water for consumption. From resources available on Internet, I understand that distilled water is not advised for regular consumption because of the lack of minerals and salts, while rainwater can be consumed if not overloaded with unwanted chemicals.

What confuses me is that I know water to be pouring down from clouds, which is water vapor turning into droplets through condensation. To me, this looks to be exactly the same as the process of distillation, i.e. evaporation and condensation of water.

So is my assumption about rainwater formation wrong? What is the difference between rainwater and distilled water? What causes this difference to happen?

Rain water passes through the atmosphere and picks up whatever is present there, including particulates such as dust, trace amounts of atmospheric gases, nitrates and nitrites, and so forth.

Rain water is relatively low in mineral content.  In practice, people drink rainwater from streams and wells, where it has had ample opportunity to pick up minerals from soil and rock.

https://www.google.com/search?q=rain+water+composition


Are there any (simple) molecules with very different absorption and emission dipole directions? When a single molecule absorbs or emits light, it does so perpendicular to the direction of the respective transition dipole. In principle, the directions of the dipoles for absorption and emission can be different, due to different electronic configurations of the ground and excited states of the molecule, but in practice the difference is very small. Photons are thus preferentially reemitted into the same plane in which absorption took place.

A large angle between these two dipoles would probably require a significant change in the molecular structure when the molecule is excited. I have heard of some fluorescent proteins which exhibit large angles between absorption and emission directions, but I do not know any details. (Edit: I found this interesting paper, which reports negative fluorescence anisotropy in YFP and a large angle between the absorption and emission dipoles.)

I would be interested in specific compounds which have a large angle between the two transition dipoles. Particularly, are there any simple compounds (less than ~100 atoms) that show such a behavior?

I have read about twisted intra-molecular charge transfer (TICT), in which a segment of the molecule rotates by ~90 degrees upon excitation. I could imagine that in such a case the emission might take place in a plane perpendicular to the original plane of absorption, but I haven't found any confirmation of this either.

Update: to illustrate, I have created a small sketch that outlines what I have in mind. The molecule I drew is just for illustration and does not necessarily show that behavior; I also don't know how exactly such a restructuring process actually takes place (the one I drew would correspond to a rotation of the rings on the right around the single bond). I further assumed that the dipole directions coincide roughly with the positions of the two rings, which might not actually be the case.

That said, the process I have in mind looks as follows:

A photon with a polarization vector parallel to the dipole direction arrives (its direction of propagation is necessarily perpendicular to the dipole direction)
The photon is absorbed and the molecule is left in an excited state
In the excited state some internal rearrangement of the molecular structure takes place, such that the dipole direction changes
A fluorescence photon is reemitted perpendicular to the new dipole axis. The photon has been effectively "redirected" (and shifted to higher wavelengths)

In part of the naphthalene molecule you draw you indicate a transition moment. This is the direction for absorption into the lowest excited state $S_1$. As naphthalene is highly symmetric the fluorescence is also emitted from a transition moment aligned in the same direction. The second excited state $S_2$ has a transition dipole set perpendicularly to the long axis and so to the $S_1$ direction.  Again excitation and emission would be from similarly aligned transition dipoles if emission was from $S_2$. However, in solution $S_2$ decays very rapidly ( a few picoseconds) to $S_1$ thus the observed absorption and emission dipoles would be at right angles to one another.

The wavefunction (molecule orbital) in an excited state is different to that in the ground state since an electron is excited to an anti-bonding orbital. This means that bond lengths and angles can change a little. If the molecule has polarisable substituents that are asymmetrical placed then the excited state dipole may differ in direction from that of the ground state. However, in molecules such as rose bengal, rhodamine 6G and many dicarbocyanines, which are not strictly symmetric ($C_1$ point group), the absorption and emission dipoles are parallel as determined by experiment.  (Presumably this is because substituents are not sufficiently close in energy to affect the electronic state in a significant way)

How can we know what the angle between absorption and emission is? The probability of absorption is proportional to the cosine squared of the angle $\theta$ between the electric field direction of the light and the transition moment direction, $\approx (\bar e \cdot \bar\mu)^2= e^2\mu^2\cos^2(\theta)$. So we can describe the molecules are being photo-selected, the probability of absorption being greatest when $\theta=0$, i.e. absorption and excitation directions are parallel.

If we use polarised light to excite the molecules, and assume that it is polarised along the $z$ direction, then the  emission intensity observed through a polariser parallel to the $z$ (or vertical) direction is $I_v\approx (\bar e_z \cdot \bar\mu_z)^2$  and observed at right angles to this ($x$ or horizontal direction)  $I_h\approx (\bar e_z \cdot \bar\mu_x)^2$ where $\bar e_z$ is a unit vector aligned along the electric field and  $\bar \mu_z$ the unit vector component of the fluorescence dipole along the $z$ and $\bar \mu_x$ along the  $x$ direction. The amount of polarisation of the fluorescence is usually measured via the  anisotropy which is defined as

$$r=\frac{I_v-I_h}{I_v+2I_h}$$

where the denominator is the total emission intensity.

In fluid solution molecules undergo rotational diffusion and if the molecule is regarded as a sphere, in fact not such a bad approximation in many cases then, the anisotropy becomes $\displaystyle r(t)=r_0e^{-6Dt}$ where $D$ is the rotational diffusion constant. The anisotropy can be measured by recording the fluorescence decay such as by time-correlated single photon counting. Taking the decay back to time zero produces $r_0$, this is the anisotropy that would be observed if the molecules were in rigid solution and effectively measures the angle between absorption and emission dipoles.

Now this relates directly to your question. The value of $r_0$ can be also be calculated as well as being measured and, again, considering the angle between absorption and emission dipoles $r_0=(3\cos^2(\alpha)-1)/5$ where $\alpha$ is the angle between absorption and emission dipoles. When the angle is zero then $r_0=0.4$ which is easily measured and when at right angles $r_0=-0.2$ so this limits the range of anisotropy.

(references. Text. M. Daune. Molecular Biophysics publ. OUP 1999. Some original literature; T Tao, Biopolymers v8,p609, 1969, Y. Yguerabide Meth. Enzymology, v26, p498, 172, Chuang & Eisenthal, J. Chem. Phys. v57, p5094, 1972)


Varying potential diffrence between two graphite strips in NaCl solution I have two identical graphite strips of dimensions 3.5 cm by 1 cm. Both are 3/4 submerged in aqueous $\ce{NaCl}$ solution (0.34 mol/dm$^3$, 2% wt) separated by a distance of 7 cm. I have attached crocodile clips to each strip and a voltmeter is used to record the potential difference between the two graphite strips.

I have found that:

A. There is a potential difference that ranges between 1 mV and 100 mV; and

B. This potential difference can change quite dramatically within that range over time periods of an hour.

Does anyone know of any chemical or physical reason for these potential differences?

What type of voltmeter are you using? to verify the answer to my question, you need to determine 1) the precision and 2) the detection limit of the instrument .

The precision is a measure of the error in a typical measurement. The smaller the standard deviation in repeat measurements, the higher the precision. To get a guess at the precision, prepare a system that should generate a known potential difference in the same range as your current measurements (like 100 mV). If the variation in that value is similar to what you are seeing, then you are recording the noise.

The detection limit is the lowest value your instrument can reliably measure. Generally, you cannot reliably measure values smaller than the twice the precision of your instrument. If you determine a standard deviation of 50 mV when determining precision, then 100 mV is your the detection limit. If this is true, you are recording the noise.

The precision and detection limit of an instrument depend on quality (price), but also on purpose. A voltmeter designed to be used by an electrician will have high precision in the range 1-200 V, but low precision below 1 V (because that range is not as important). A sensitive voltmeter used for electrochemical measurements may have high precision between 1 uV and 1 V, and behave miserably outside that range.


Which species occupy the equatorial positions in VSEPR's trigonal bipyramidal and octahedral geometry Which species (electronegative, double and triple bonded atoms, ions, lone pair etc.) occupy the equitorial position in the trigonal bipyramidal and octahedral geometry when we find the shape using VSEPR theory?

Which species are given more priority and why?

Trigonal planar is a molecular structure that requires five ‘substituents’ around a central atom (counting lone pairs as ‘substituents’). Assuming main-group compounds, that means, we have already left the realm of easily derived molecules and need to resort to 4-electron-3-centre bonds to properly explain the molecule and be in line with the octet rule.

This very good answer by Ron does a good job of explaining what the central atom’s electronic structure should be considered to be and what the 4-electron-3-centre bond is. In a nutshell, assume the central atom to be $\mathrm{sp^2}$ hybridised with an unhybridised $\mathrm{p}$ orbital. This additional $\mathrm{p}$ orbital then interacts with two electronegative substituents forming a 4-electron-3-centre bonds; this bond can be rationalised in a Lewis formalism by looking at these two resonance structures:

$$\ce{X-\overset{+}{A}\bond{...}\overset{-}{X} <-> \overset{-}{X}\bond{...}\overset{+}{A}-X}$$

The bond order of a 4-electron-3-centre bond is approximately ½, in good agreement with a single bond and a non-bond forming equivalent resonance structures for each $\ce{X\bond{...}A}$ fragment. Each outer atom has a partial negative charge while the central atom has a partial positive one.

This already means a few things:

a lone pair can never be in axial position as this position is formed by a $\mathrm{p}$ orbital. If a lone pair were in axial position, no second ligand can be there and the structure is actually better described as trigonal planar.
the more electronegative a substituent is, the more likely it is to occupy the axial positions as these positions require stabilising a partial negative charge. Hence, fluorine is most frequently found here.

This is also somewhat of an example for Bent’s rule — the more electropositive a substituent is, the higher the $\mathrm{s}$ orbital participation for the orbital forming the bond to said substituent. Conversely, the more electronegative the substitutent is, the higher the $\mathrm{p}$ content. Naturally, a pure $\mathrm{p}$ orbital is on one end of that spectrum, clearly preferring electronegative substituents.


Resonance in Propenal I do not understand why the 3rd canonical form of propenal has the oxygen atom donating an electon.

Oxygen in electronagetive so instead of taking electrons why is it giving?

Maybe this why it is most unstable but how is it even formed ??

What you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals.

In the case of propenal the $\pi$ MOs result from the overlap of four $p$ orbitals on the four atoms in the chain. The two occupied $\pi$ MOs show how most of the electron density is between C1 and C2, and C3 and the oxygen but there is also significant electron density between C2 and C3. This reflects the different positions of the double bonds shown in the resonance contributors of propenal.

http://www.chemtube3d.com/orbitalsacrolein.htm


What is C2H5-O-CH=CH2? Someone painted an original graffiti:

source

I wonder what kind of reaction was depicted; what's that product substance?

The graffiti reads:
$\ce{C2H2 + C2H5OH -> C_2H_5-O-CH=CH_2}$

Does it have any more or less common uses?

The particular reaction is of acetylene ($\ce{C2H2}$) and ethanol ($\ce{C2H5OH}$) to yield ethyl vinyl ether (IUPAC name ethoxyethene). It's one example of Reppe chemistry, a fairly diverse set of reactions, often operating on alkenes and alkynes (though certainly not exclusively). The specific reaction in the question is a vinylization reaction, though it was Reppe's carbonylation reactions which were probably of greater practical (read: economic) import historically, as they were famously used in the Monsanto process and in production of various acrylates and other acrylic acid derivatives.

The vinylization reactions could still be synthetically useful, as they provide an extremely simple alternative route to vinyl ethers without having to proceed via elimination, substitution, coupling, olefination, or substrate-specific oxidation reactions.

I'm not aware of any specific uses for ethyl vinyl ether, however it could certainly be used as a building block in any number of organic syntheses.


Greatest Energy Output Reaction to Atmosphere Air First of all i'm no chemist, this is not homework and if you need further clarification please ask. I'll be happy to provide!

Imagine that you:

have 1 kg of any given chemical X
A car battery
A semi-controlled combustion chamber (chamber temp is initially 20ºC, pressure is 1atm, the chamber is either vacuum or filled with atmosphere "air"...)
As much atmosphere "air" as you want, initially at 20ºC...
A not so potent air-pressurizer... you might be able to get 3atm out of it...

The goal here is to get the most energy out of reaction as possible. So the question is: What chemical is X?

Further details:

X doesn't have to be consumed all at once as in a explosion reaction... Damage/Destruction is not the goal
If you can combine X with something in the air, and then use it to fuel the reaction, increasing the energy output, great!

this is a way too generalized question. you certainly need experimental data or extensive screening to make sure you are on the top of the volcano plot. have you heard the term adiabatic flame temperature? The ratio of inert to gasoline/fuel is a determining factor to get high combustion energy. For example the adiabatic flame temperature for methane is higher for a mixture of methane and air with 5-15 mol% methane. 5-15 mol% methane can ignite without any external source of fire.

Please see the attached link to compare fuels according to their energy density:  http://en.wikipedia.org/wiki/Energy_density

And for your question you can't have two variables to optimized with one equation, either you have to specify the chemical X or you have to specify what is the output energy of this process. If you know the chemical X with it's thermodynamic property than using basic heat balance equation you can calculate the output energy.


How to now if the OH group on the C5 carbon of a cyclic sugar is up or down in Fischer Projection I know when using fischer projection I have to look if it's a D or L configuration. if it's D: up = right and down = left and if it's L: up = left and down = right. I think the concept is clear to me, however I dont't know if the OH group on the C5 carbon atom is pointing up or down. For example:

When I have to draw those cyclic molecules to fischer projection I never know if the OH group which will be shown in the fischer projection (not in the clyic structure because it's a hemi-acetal then) is pointing up or down. What is the rule for this?

I want to get this:

But I don't know how to know the side of the circled OH group

The OH group you've circled is always on the right hand side for D-sugars and on the left for L-sugars as its position is in the very definition of D/L sugars.


What chemical equation proves that KHCO3 can extinguish a fire? In a dry powder fire extinguisher, there is a chemical known as $\ce{KHCO3}$ and I was wondering how $\ce{KHCO3}$ works in a chemical equation to extinguish the fire.

According to this Wikipedia entry:

  Decomposition of the bicarbonate occurs between 100 and 120 °C (212
  and 248 °F):

  $$\ce{2 KHCO3 → K2CO3 + CO2 + H2O}, \Delta H>0$$

So, you produce a nonflammable powder and two nonflammable gases, and you absorb heat from the fire in an endothermic reaction.

The picture here is more complex than just a chemical equation however. The $\ce{KHCO3}$ also acts as an energy-absorbing material and provides a solid surface on which free radicals can be destroyed.


Qualitative Comparisons of Dielectric Constants of Organic Solvents This question is inspired by the discussion around this earlier question.
It is relatively simple to do a qualitative comparison of the polarities of different molecules (for example by analyzing the presence of polar bonds alongside the geometry of the molecule).
Now, ethanol and acetic acid have almost identical dipole moments ($\mu$=1.68 for acetic acid, $\mu$ = 1.69 for ethanol) but dramatically different dielectric constants ($\mathcal{E}$ = 6.15 for acetic acid, $\mathcal{E}$ = 24.5).
Is there a similar qualitative analysis that can be done for dielectric constant of a solvent?

Yes, but it is much more complicated. The molecule does not have to have a dipole, in which case the (static) dielectric constant (relative permittivity) depends on the molecule's polarisability and if it does have a dipole it depends on both factors. The equation is $\displaystyle \epsilon_r=\frac{1+2b}{1-b}$ where $b=\left(\alpha+\mu^2/(k_BT)\right)N/(3\epsilon_0)$ and $N=N_A\rho/M$ where $\alpha$ is polarisability, $\mu$ dipole, $\rho$ density,$M$ molar mass, $\epsilon_0$ permittivity of free space.  See Atkins, & Friedman, 'Molecular Quantum Mechanics'. The dielectric constant also depends on frequency of the applied electric field, but the value quoted is normally the static value measured by comparing the capacity of a capacitor with and without the solvent. The name relative permittivity arises from the potential a charge produces. In a vacuum at a distance $r$ this is $\displaystyle \phi=\frac{q}{4\pi\epsilon_0 r}$, in a dielectric medium of permittivity $\epsilon$ this becomes $\displaystyle \phi'=\frac{q}{4\pi\epsilon r}$ and $\epsilon_r=\epsilon/\epsilon_0$.

As an aside you can see why a polar molecule is generally insoluble in a non-polar solvent. If the permittivity is small (say hexane) the electric potential of a charge is higher at any given $r$ than for a solvent with a large $\epsilon$, such as acetonitrile. This means that polar molecules are less soluble in low dielectric solvents than they are in high dielectric ones.


Why do non-polar covalent compounds dissolve in non-polar covalent compounds only? My textbook says that electrovalent and polar covalent compounds dissolve in electrovalent compounds and polar covalent compounds but not in non-polar covalent compounds. I understand the reason for this. It is because the oppositely charged moeities of the two compounds attract each other and the compounds thus dissociate and ionize into ions thus resulting in dissolution.

Now, it also says that non-polar covalent compounds dissolve in non-polar covalent compounds but not in electrovalent and polar covalent compounds. I do not quite understand the reason behind this.

I would like to know what dissolution is at the molecular level and why non-polar covalent compounds dissolve in non-polar covalent compounds only?

There is the oft quoted adage in chemistry, which goes something like:

  like dissolves like

The reason for this can be understood on the basis of polar/ionic compounds have some degree of charge separation. In polar solvents, the partial charges on on the solvent molecules can interact with the partial charges on the solute molecules. These interactions lower the overall energy of the system, and thus dissolution takes place. This part you get, so it's fine.

First let's address why non-polar solutes don't dissolve in polar solvents. Consider the simple case, of a long chain hydrocarbon (solute) in water (solvent, present in larger excess). Typically, water molecules will form (and break) hydrogen bonds with each other. At any given time, a water molecule would like to maximise the number of hydrogen bonds it can form, simply because it is energetically favourable.

Remember that the water molecule, cannot form hydrogen bonds with the hydrophobic molecule; it can only do so with other water molecules.

So when a hydrophobic is dropped in an aqueous medium, hydrogen bonds between water molecules will be broken to make room for the hydrophobe; however, water molecules do not react with hydrophobe. This is bad energetically. So, water molecules would now try to adopt an arrangement that maximises the hydrogen bonding network; to do so they will arrange themselves in a "cage" like structure surrounding the hydrocarbon droplet. While, this allows the water to make more hydrogen bonds, it leads to overral "ordering" and thus decreases entropy. Again, bad. Thus to minimise this hydrocarbon droplets will clump together, and will be exluded by the solvent--this is why oil and water don't mix. (Read more here)

So what happens when non-polar compounds are tossed into non-polar solvents. True, there aren't any dipole-dipole, ion-dipole, etcetera kinds of interaction that you have with polar molecules, but you still have something called London dispersion, and non-polar entities often interact via these weak interactions.


The distance that CO2 molecules travel in the air before they collide with each other Well the title covers most.
But further at 410 ppm, 1 atm and 23 °C.

I believe they call it the mean free path of CO2-CO2 collisions. So not just the mean free path of CO2 in air colliding with every other molecule substances.

I have zero workexperience with gases so I'm having troubles with this one.
I'm Dutch (sadly) so excuse me in advance if my English isn't correct.

You need to calculate the probability of one (CO2) molecule being the nearest neighbour of another.

Let $w(r)$ be the probability that the nearest neighbour occurs between distance  $r$ and $r+dr$ (on a spherical surface). This must be the probability than no molecules exist closer to the victim molecule,  which occurs at a distance $r$, and that this molecule exists in the shell $r$ to $r+dr$. Thus,

$$ w(r)=\left(  1-\int_0^r w(r)dr \right)4\pi r^2 n$$

where $n$ is the average number of molecules / unit volume. (1 mol/m3 $\equiv 10^3N_A/10^{27} = 0.6023 $ molecules / nm$^3$).

Solving this gives

$$ w(r)=4\pi r^2n\exp(-4\pi r^3n/3)$$

This function is zero at zero goes through a maximum and is  small at large. distances, see figure.

If you want a single number then the average distance between molecules is

$$\langle r \rangle =\int_0^\infty rw(r)dr =\Gamma (4/3)/(4\pi n/3)^{1/3} =0.55396n^{-1/3}$$

where $n$ is in number of molecules/volume then $n^{-1/3}$ is in units of length.

( $\Gamma(4/3)$ is the gamma function, sort of equivalent to a factorial but for a non-integer number)

The figure below shows $w(r)$ the probability of a near neighbour at distance $r$ for some concentrations.

If what you want is the total path taken accounting for all collisions then you would have to start with one the of the $w(r)$ distributions corresponding to your concentration and do a Monte Carlo computer calculation to work out the total path, not easy and v long calculation probably. There will be a statistical mechanics way of calculating this I'm sure but I'm not aware of it at present.


What are the IUPAC names for singly- and doubly-protonated piperazine? If ammonia, $\ce{NH3}$, is protonated, it becomes ammonium: $\ce{NH4+}$.

If piperidine is protonated, it becomes piperidinium:

If piperazine is protonated once, what does it become? By analogy, I would assume piperazinium:

If the resulting monocation is protonated again, what does it then become?  The best I've come up with are piperazine dication and piperazin-dium, but that latter seems really awkward:

The relevant rule in the current (corrected) version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is:

P-73.1.1.2 General rule for systematically naming cationic centers in parent hydrides
A cation derived formally by adding one or more hydrons to any position of a neutral parent hydride (listed in Chapter P-2), or whose degree of hydrogenation has been modified (see P-31) is named by replacing the final letter ‘e’ of the parent hydride name, if any, by the suffix ‘ium’, preceded by multiplying prefixes ‘di’, ‘tri’, etc. to denote the multiplicity of identical cationic centers. (…)

Therefore, the preferred IUPAC names are piperazin-1-ium and piperazine-1,4-diium, respectively.


P-phenylenediamine Production Process How p-phenylenediamine (PPD) is produced? PPD is used as raw material for Kevlar (Para amid Fiber).

React benzene with chlorine in the presence of $\ce{FeCl3}$ to yield chlorobenzene.

According to Ullmann's Encyclopedia of Industrial Chemistry:

Most monochlorobenzene is now produced from benzene and chlorine in continuously operated plants. Depending on the ratio of benzene to chlorine chosen, one can achieve either a low rate of benzene conversion and little dichlorobenzene
formation, or almost complete conversion of the benzene with a higher degree of dichlorobenzene formation. [...] The composition of a chlorination mixture containing the highest possible proportion of monochlorobenzene has been given as
4–5% unreacted benzene, 73% monochlorobenzene, and 22–23% dichlorobenzenes.

Nitrate chlorobenzene to obtain a mixture of 2-nitrochlorobenzene and 4-nitrochlorobenzene. Separate the isomers.

In Ullmann's Encyclopedia of Industrial Chemistry, we find:

Nitration of chlorobenzene with mixed acid (30/56/14) typically gives an isomer
mix in 98 % yield consisting of 34–36% 2-chloronitrobenzene, 63–65 % 4-chloronitrobenzene, and only ca. 1% 3-chloronitrobenzene.

"Mixed acid" refers to a mixture of nitric acid, sulfuric acid and water.

React 4-nitrochlorobenzene with ammonia (Nucleophilic aromatic substitution) to obtain 4-nitroaniline.
According to Ullmann's Encyclopedia of Industrial Chemistry, the reaction conditions as as follows:

4-Nitroaniline is produced in an agitated titanium reactor in 99.3 % yield by reacting 4-chloro nitrobenzene with a tenfold excess of aqueous ammonia at 175 °C and 4.2 MPa (42 bar) for 10 h.

Ullmann's Encyclopedia of Industrial Chemistry cites a patent by MONSANTO (DE1768518) as the relevant reference.

Reduce the nitro group with iron.


How did Mendeleev know elements from compounds or mixtures? I read that Mendeleev initially thought didymium was an element, but it was actually a mixture.
How did he know the rest of the chemicals in the periodic table were elements and not compounds, especially since back then most elements would have been incredibly difficult to isolate and get a mono-atomic version?
It seems like an impossible task in the 19th century.

Please do not underestimate the scientists of 19th century. They were as creative, intelligent and perhaps more genuinely dedicated to science than the scientists of the 21st century. Spectroscopy was the tool of the trade to identify and verify that a given substance is not a mixture.
The original reference which established that Didymium was a mixture is in German [1] (English title: "On the decomposition of didymium in its elements").
I don't know enough German to read it conveniently, but it is clear that the authors did fractional crystallization of a lot of salts, made measurements absorption spectra, did emission spectroscopy and finally led to a beautiful result. The original paper has plenty of spectra! From [1, p. 490]:

Da sonach die exacte Zerlegung des Didyms in mehrere Elemente realisirt ist, so schlage ich vor, die Bezeichnung Didym nunmehr ganz zu streichen und beantrage, für das erste Element, entsprechendder Grünfärbung seiner Salze und seiner Abstammung die Benennung:
Praseodym mit dem Zeichen $\ce{Pr}$
und für das zweite, als das „neue Didym“, die Benennung:
Neodym mit dem Zeichen $\ce{Nd}.$
Aus den Atomgewichtsbestimmungen, die in allen Einzelnheiten nach Bunsen ausgeführt wurden, ergaben sich als vorläufige  Werthe für Praseodym = 143·6, für Neodym = 140·8, wobei dem entsprechenden Oxyd die allgemeine Formel $\ce{M2O3}$ zukommt.

English translation with assistance of DeepL:

Since, therefore, the exact division of the didymium into several elements has been carried out, I propose that the name 'didymium' be deleted altogether and that the first element be given a name corresponding to the green colour of its salts and its origin:
Praseodymium with the sign $\ce{Pr}$
and for the second, as the "new didymium", the naming:
Neodymium with the character $\ce{Nd}.$
From the atomic weight determinations, which were carried out in all individual units according to Bunsen, it resulted as provisional values for praseodymium = 143·6, for neodymium = 140·8, whereby the corresponding oxide is assigned the general formula $\ce{M2O3}.$

Anyway, historically, it is not Mendeleev who had to decide whether the elements were actually elements. You may ask the right question, how did the element discoverer decided that it was an element in Mendeleev's time? The answer is that each element has a story to tell and that is why Isaac Asimov wrote a beautiful book called "The Search for the Elements". Go ahead and get hold of it.
Reference

v. Welsbach, C. A. Die Zerlegung des Didyms in seine Elemente. Monatshefte für Chemie 1885, 6 (1), 477–491. DOI: 10.1007/BF01554643.


How is Cu(OH)2 a base if it can't dissolve in water? As the title says, how is $\ce{Cu(OH)2}$ a base if it can't dissolve in water?

I mean if it can't dissolve in water then it won't produce any $\ce{OH-}$ ions.

And if it won't produce any $\ce{OH-}$ ions then how can it be called a base?

Same question for $\ce{Fe(OH)2}$ and every other base which is not an alkali.

I feel like this question is quite obvious and don't know how to describe it more. So please help.

Copper (II) hydroxide has some small solubility in water, determined by its solubility product constant.  More importantly, it will dissolve freely and act as a base if a suitable acid, such as acetic acid, is is available to be neutralized.  See also the comment from @f".


Mechanistic Explanation of Circular Dichroism I'm wondering if someone can give me a physically motivated reason for why chiral molecules interact with left and right circularly polarized light (LCP) (RCP) differently. Particularly, a quantum mechanical view of what's happening here.

I understand why it is that a molecule has to be chiral in order for it to manifest circular dichroism (my use of terminology may not be perfect here so I apologize). That is, both the electric and magnetic dipole operators must be non-zero and there's an argument here on the basis of point groups.

Nothing about that, however, is particularly illuminating when it comes to understanding why I should expect a molecule to interact differently with LCP and RCP light.

I don't know what else I can even say. A good answer would address the different behavior of both the light and the molecule in these different situations, as I imagine that's a necessary piece to the puzzle.

Also, if one has some references (a review or textbook chapter etc.) that would be appreciated.

Some part of a mechanical macroscopic explanation, acoustical activity where mechanical transverse oscillation direction is being rotated, is described in

Frenzel, T., Köpfler, J., Jung, E. et al. Ultrasound experiments on acoustical activity in chiral mechanical metamaterials. Nat Commun 10, 3384 (2019).

The notion “activity” refers the rotation of the linear polarization axis of a transversely polarized mechanical wave. The rotation angle is proportional to the propagation distance and does not depend on the orientation of the incident linear polarization. It has nanometer precision under time-harmonic excitation at ultrasound frequencies in the range from 20 to 180 kHz. Polarization rotations are as large as 22° per unit cell.
Add a net neutral electrical charge distribution to the components of the metamaterial where the negative charge distribution has a different elastic or inertial nature than the positive charge distribution to get a circularly polarized electromagnetic wave.
A quantum mechanical explanation was requested. Just add stochastic bias to the mechanical input so that only inputs of at least a minimum value get statistical significance. This makes the setup equivalent to quantum mechanics.


Difference between lab-grade and food-grade purity? What is the difference in purity betwen lab-grade and food-grade when talking about potentially consumable compounds?

As an example let's take ascorbic acid powder: if it's marked as lab-grade, does that mean it is also fit for consumption?

In particular, I'm wondering whether lab-grade supersedes (and/or improves) a food-grade labeling, or if they are two totally different criteria? Can I say lab-grade > food-grade?

If I can't say lab-grade > food-grade, then what is the actual difference between the two?

So-called food-grade stuff is meant to be consumed by humans. It's important that every single compound of a foodstuff is food-grade or at least GRAS. In the European Union the law giver has defined the state when a compound is good for consumption as "save".

Lab-grade chemicals often contain toxic materials, so when something is called lab-grade it is not save per definition.

For example: $\ce{NaOH}$ is widely used in preparation of foodstuff as well as in the lab. Now it's a fact that lab-grade (standard) $\ce{NaOH}$ (Sigma Aldrich) contains a certain amount of heavy metals. On the other hand, food-grade $\ce{NaOH}$ will not contain any heavy metals (usually food-grade $\ce{NaOH}$ will have a purity of more than $99\%$ while standard $\ce{NaOH}$ pellets are sold with purities less than $97\%$). So using lab-grade $\ce{NaOH}$ in foodstuff is a crime (at least in the EU).

So my advice is: Do not use lab-grade materials for any food-related applications.


Ring closure via intramolecular nucleophilic substitution or intermolecular dimerization: which is favoured? What reaction does the following enolate undergo?

Both are nucleophilic substitution reactions. Ordinarily, the intramolecular product should be favoured, but I know that the transition state has a geometrical constraint in that the incoming nucleophile must approach the C-Br bond from the opposite side. Would that favour the intermolecular reaction instead?

I went through March's Advanced Organic Chemistry and found this:

And this from Clayden's Organic Chemistry:

That's why I am not able to decide if the reaction is intermolecular or intramolecular.

I'd also like to add that the intramolecular cyclisation that you proposed is a 5-exo-tet reaction according to Baldwin's classification.

The ring being formed is 5-membered.
The bond being broken (the $\ce{C-Br}$ bond) is outside the ring, or exocyclic.
The carbon being attacked is a $\mathrm{sp^3}$-hybridised carbon with tetrahedal geometry.

If you look up Baldwin's rules, you'll see that 5-exo-tet reactions are perfectly fine and allowed. Evidently the compound you drew can twist itself a little bit to accommodate the linear transition state. (I am using "linear" loosely to refer to the arrangement of the incoming nucleophile, the carbon, and the leaving group.) Therefore, in this case, the intramolecular reaction will predominate, as is usual.

The disfavoured reactions that you cited from those textbooks are 6-endo-tet reactions. In such compounds, having a linear transition state is an issue as it would lead to too much ring strain.

As an ending note: Baldwin's rules are nothing more than an empirical observation. You might ask why 5-exo-tet reactions are fine, but not 6-endo-tet: where is the threshold or boundary which makes something favoured or disfavoured? The answer is, they did the experiments and found out that that's just how it is. No more, no less.


Equilibrium composition of methane–water system under given conditions Calculate the equilibrium composition of a reactive system consisting of an equimolecular mixture of $\ce{CH4}$ and $\ce{H2O}$ at $\pu{900 K}$ and $\pu{1 atm},$ considered to be an ideal mixture of ideal gases. Under these conditions, the equilibrium constants of the two possible reactions are
$$
\begin{align}
\ce{CH4 + H2O &<=> CO + 3 H2} &\quad K_{p,1} &= 1.307 \tag{R1}\\
\ce{CO + H2O &<=> CO2 + H2}&\quad K_{p,2} &= 2.210 \tag{R2}
\end{align}
$$

I have tried to solve it by expressing each $K_{p,i}$ independently, considering there are five constituents in this ideal mixture and that $\ce{CH4}$ and $\ce{H2O}$ are the starting compounds, with $n^0 = 1.$ I associated a different degree of advance, $\xi_i,$ to each of the reactions and solved the two-equation system, but it seems that this is not the proper solution because the reactions are not independent.
Can anybody figure out a solution?

Let
$$moles\ CH_4=1-e_1$$
$$moles\ H_2O=1-(e_1+e_2)$$
$$moles\ CO=e_1-e_2$$
$$moles\ CO_2=e_2$$
$$moles\ H_2=3e_1+e_2$$
where the $e_1$ is the number of moles of $CH_4$ destroyed and $e_2$ the number of moles of $CO_2$ that are produced.
So the equilibrium equations are going to read:$$\frac{(e_1-e_2)(3e_1+e_2)^3}{(1-e_1-e_2)(1-e_1)(2+2e_1)^2}=K_1$$and $$\frac{e_2(3e_1+e_2)}{(e_1-e_2)(1-e_1-e_2)}=K_2$$


How does fat affect the precipitation of proteins? I am doing an experiment using milk, heavy cream, and vinegar. We are trying to precipitate proteins from milk and heavy cream.

When I did the experiment with milk, it was very quick reaction and easily to isolate the casein polymers with vinegar added to the heated milk. This does not occur when the same process is used with heavy cream. I have tried to find out how this process fails with the heavy cream online and in my book there is a decent portion of the chapter dedicated to fats, but it doesn't mention why polymer creation can not occur in this experiment.

Can some one please help me with the reason the heavy cream doesn't curdle?

Casein is the main protein in milk. When you perform the polymerization experiment, the result is an "aggregation" of said protein, which gives the product a solid consistence.

Products such as butter and cream are made with the fats from milk (not the proteins!), hence they are almost casein-free. Of course, being them almost without proteins, the polymerization doesn't occurr.

If you look up the nutrition facts of milk and heavy cream, you see that proteins constitute the 20% of calories of milk, and only the 2% of heavy cream.


What are radicals in chemistry? I read somewhere that radicals are atoms, molecules or ions that have unpaired electrons, but how do ions have unpaired electrons when an ion has all its electrons paired either by loosing or gaining electrons to attain the nearest noble gas configuration?

but how do ions have unpaired electrons when an ion has all its electrons paired either by loosing or gaining electrons to attain the nearest noble gas configuration?

(Quote from your question, emphasis mine)

Your fallacy is the word I emphasised. Surely it is often taught like that—especially in introductory chemistry—that when ions are generated the noble gas configuration (vulgo: octet or dublet in the case of He configuration) is reached. That is not the whole truth, though.

Especially when you form a polycharged anion such as oxide $\ce{O^2-}$, you can (and should!) consider the electrons being added (or removed) stepwise. Thus, going from an oxygen atom you would initially generate an oxide radical anion $\ce{O^.-}$ before arriving at the classical oxide anion $\ce{O^2-}$. If you look at it energetically, the first electron addition is sometimes (but not always!) exothermic but the second one is always endothermic, so it makes energetic sense to separate like this.

These monoatomic radical ions are not commonly encountered as they typically have very short half-lifes. They often rapidly react to either give the full noble gas ion or the original, neutral atom. However, the vast majority of known ions are molecular or polyatomic ions such as phosphate. Within these, one very common species of radical anion forms rather stable compounds: the superoxide anion $\ce{O2^.-}$.

Structure-wise, it is similar to the peroxide anion $\ce{O2^2-}$ which can be written as $\ce{^-O-O-}$ except with one electron removed. However, it is usually generated  starting from dioxygen ($\ce{O2}$) with one electron added, which turns dioxygen’s double bond ($\ce{O=O}$ in Lewis formality) into a single bond. when the heavy alkaline metals (potassium, rubidium or caesium) are burnt in air, the corresponding superoxide is the product:

$$\ce{K + O2 -> KO2}\tag{1}$$

If lithium or sodium are reacted identically, the product is the oxide or the peroxide, respectivesly:

$$\begin{align}
\ce{4Li + O2 &-> 2Li2O}\tag{2}\\
\ce{2Na + O2 &-> Na2O2}\tag{3}\end{align}$$


Coordination geometry: why is of Cu(acac)2 square planar? Why is $\ce{Cu(acac)2}$ square planar, whereas $\ce{Al(acac)3}$ and $\ce{Fe(acac)3}$ form octahedral complexes? I see how a similar argument to $\ce{Pt}$ complexes would apply, i.e. with $\mathrm{d^9}$, there is only one electron that goes in the very destabilised $2\mathrm{b_{1g}}$ (corresponding to $\mathrm{d}_{x^2-y^2}$) in the square-planar orbital splitting, whereas the rest is preferentially stabilised. However, there are many examples of Cu(II) forming octahedral compounds, in which acac would still bridge only $90^\circ$. Is this something to do with the acac ligands being (pseudo-)aromatic?

Thanks!

With bidentate ligands of the form $\ce{LL'}$, it is difficult to force triple coordination to a copper(II) centre. I don't have data for acac offhand, but I have some data for en binding to Cu. (Source: some past exam question somewhere.)

$$\begin{array}{c|ccc}
n & 1 & 2 & 3 \\ \hline
\log K_n & 10.55 & 9.05 & -1.0
\end{array}$$

The exact numbers don't matter; only the trend does. (Which is why I'm lazy to find a better source.) The binding of the third ligand, which would necessitate coordination at the axial position, is very weak. To get to a $\ce{[Cu(acac)3]}$ complex, you would have to either 1) adopt a perfect octahedral environment and sacrifice the electronic stabilisation from the Jahn-Teller distortion; or 2) stretch two of the acac ligands to a longer length than they would like to be.

As for why axial elongation is favoured over compression, it is perhaps due to electronic reasons: under elongation, the $\mathrm{d}_{z^2}$ orbital is doubly occupied and the $\mathrm{d}_{x^2-y^2}$ orbital is singly occupied. On the other hand, under compression, $\mathrm{d}_{z^2}$ orbital is singly occupied.

Now, there is a second-order mixing which allows the $\mathrm{4s}$ orbital to mix with the $\mathrm{d}_{z^2}$, since they possess the same symmetry ($A_\mathrm{1g}$) under $D_\mathrm{4h}$ symmetry; in the process the $\mathrm{d}_{z^2}$ orbital is stabilised, and the $\mathrm{4s}$ destabilised. The destabilisation of the $\mathrm{4s}$ orbital doesn't matter, since it's empty. However, the stabilisation of the $\mathrm{d}_{z^2}$ orbital is, of course, more profound when there are more electrons in the $\mathrm{d}_{z^2}$ orbital, which corresponds to elongation.

Here's a diagram from my notes.


How to determine the edge length of the aluminium unit cell? I'm having trouble understanding why I can't get the correct answer from this homework problem:

  Solid aluminum forms a face-centered cubic unit cell.  Aluminum has a density of $\pu{2.70 g/cm^3}$. Determine the edge length of the $\ce{Al(s)}$ unit cell in $\pu{cm}$.

Usually if I enter the wrong answer the it will show all the calculations to determine the correct answer. However this one does not show the calculations.

Here is the work I've done:

$\ce{Al }= \pu{26.98 g/mol}$
$\text{Mass} = \frac{26.98}{6.022\times10^{23}} = \pu{4.480E-23 g//atom}$
$\text{Mass of one atom} = \text{mass of one unit cell}$
$\text{Mass of one unit cell} = \pu{4.480E-23 g//unit cell}$
$D = \frac{M}{V}$, thus $V = \frac{M}{D}$
$V = \frac{\pu{4.480E-23}}{2.70} = \pu{1.66E-23 cm^3//unit cell}$
For a cube $V = l^3$, thus $l = V^{1/3}$
$\text{Unit cell edge length} = V^{1/3} = {\pu{1.66E-23 cm^3}}^{1/3} = \pu{1.18E-8 cm}$

The correct answer that the book gives is $\pu{4.049E-8 cm}$.

The only example I have to go off of in my book is for a simple cubic unit cell. Since $\ce{Al}$ is a face-centered cubic unit cell and has 4 particles per unit cell I tried multiplying my final answer by $4$, but that gives me $4.72$. I also tried multiplying the unit cell mass by $4$ which gave me a final answer of $1.87$. I just can't seem to get $4.049$. Any help understanding this would be greatly appreciated.

Correct approach is to do two things:

1)  Put four atomic masses into one unit cell.

2)  Properly extract the cube root of the volume you obtain.  That volume should now be about $6.6×10^{-23}\text{cm}^3$ from part 1.  To take the cube root properly, render the exponent on 10 as a multiple of 3, thus  $6.6×10^{-23}$ = $66×10^{-24}$ where $10^{-24}$, not $10^{-23}$, is $(10^{-8})^3$.

Then your numbers should work out.


Separation of alkylated 1,2,4-triazole in solution I was looking into heterocyclic compounds and their alkylation (i.e. the regioselectivity of the alkylation of the compound).

People are able to give pretty exact numbers, for an experiment done in solution. Take, for example, 1,2,4-triazole, the alkylation can occur in one of two positions and papers state that the yield is about 9:1 in the 1 to 4 position.

How would one go about the separation of the solution to validate these claims? Or is there a method where this can be quantified in solution?

Joule and Mills Heterocyclic Chemistry [1] confirms that 1,2,4-triazoles alkylate with ~9:1 selectivity for N1 over N3, citing a paper in Tetrahedron Letters  from an industrial group at Merck as the source of the data[2].

The ratios reported were based on 1H-NMR of the crude product.[#] In this case, all of the compounds formed are known compounds, and as such it was easy to compare the data for 1-alkyl vs 3 alkyl. If the compounds formed were completely novel, it may be harder to tell which is which, though more complex NMR methods such as H-N or C-N correlation are often used in the pharmaceutical industry where this issue of regioselectivity often arises.

As an aside, the paper points out that the N1 and N3 alkylated products are easily separable, in this case by distillation, but in normal cases, silica gel chromatography (or HPLC if needed) can usually be used to separate the different regioisomers, with sufficient effort!

[1]: Mills, K.; Joule, J. A. Hetrocyclic Chemistry; Wiley:Chichester- Sussex, 2010

[2]: Bulger, P. G.; Cottrell, I. F.; Cowden, C. J.; Davies, A. J., Dolling, U-H. Tetrahrdron Lett., 2000, 41, 1297

[#]: The reaction was worked up, but not purified prior to working out the ratios. This is good practice, as it assures that you haven’t accidentally columned away some of the minor product, as is often the case


Taking a procedure for ICP and adapting it to use with AAS I was reading a paper regarding metal detection and they gave their procedure for an acid digest of a soil sample which was then analysed by ICP-OES.
They used nitric acid, hydrogen peroxide and sulfuric acid.
I want to do a similar analysis, but I only have AAS available to use.
Can I simply dilute the acid digest in a volumetric flask and analyse with the AAS?
Or are there other steps to consider?
Are the choices of acids of concern? Can they cause interference in the flame ionisation?

In short - it depends. For most metals no problem in using the same digestion protocol for ICP (OES or MS) or AAS. Are you using flame or graphite furnace?  If you are using flame AAS -just prepare a blank digest (same ratio of acids but no sample) use that as you blank.  If you are looking to quantify Na, K or Ca there can be significant background issues from soil samples - most trace metals are not a problem - but you need a good blank to qualify.  The LOD (limit of detection) will vary depending upon the matrix S/N (signal to noise).
Hope that helps.  If using graphite AAS you are typically limited as to the amount (volume wise) you can add to  the furnace.  You can try multiple dry downs before atomization.  Again a good plank is paramount.


Calculating valence of oxides Learning about Oxides. Basically when oxygen is combined with a metal.

$$\ce{FeO}$$

This is called "Iron Oxide (II)" according to my book.

Apparently, the II represents the valence. But how come? Look:

$\ce{Fe}$ has the configuration $1s^22s^22p^63s^23p^6\color{red}{4s^2}3d^6$. Since the outermost shell has $2$ electrons, the valence is $2$.

Now $\ce{O}$ has the configuration $1s^2\color{red}{2s^22p^4}$. Its outermost shell has $6$ electrons so oxygen has a valence of $6$ right?

Then how come $\ce{FeO}$ has valence $2$? Shouldn't the valence of $\ce{FeO}$ be like, the combination of the two valences? So $6 - 2 = 4$? (I read somewhere you had to subtract it).

Not only that. I am starting to think that my definition of valence is completely messed up. I just checked in here and apparently Oxygen has $-2$ valence? Why? And Iron has +2, +3, (+4), (+6)? What does all that even mean?

Valency is the combining power of an atom; so, when we talk of iron having a valency of 2, it can combine either with two atoms of valency of one; or with one atom of valency of two.  Oxygen has a valency of two and so the formula for iron (II) oxide is FeO.

In the history of chemistry valency pre-dates the idea of electron configuration which looks like the source of your confusion.

Hope that is clearer.

Edit: you mentioned that oxygen has a valency of minus 2; that's not quite right.  It has a valency of 2 but it can form an oxidation state of minus 2.  You need to separate out the ideas of valency and oxidation state.  Although the ideas are related, they are distinct.


Why is the electronegativity of indium greater than that of thallium? Why is the electronegativity of indium greater than that of thallium?

One possibility which I had assumed was the poor shielding effect by the d and f orbitals which leads to an increase in the effective nuclear charge and hence the anomaly in the 13th group. But with thallium having a f orbital (which indium doesn't) which has a much poorer shielding ability than 'd', why does indium have more electronegativity?

Going down any group of the Periodic Table the general trend is that electronegativity decreases. The relative electronegativities of In and Tl are simply a reflection of this fact. That's the end of the story; you don't need to bring in anything more complicated.

You are correct that the 4f electrons are poorly shielding, and thus the 6p elements (Tl, Pb, Bi, Po, At, Rn) would likely be more electronegative than otherwise expected. But there's no reason why this should make them so electronegative so as to reverse the general trend. None of these elements are more electronegative than their 5p counterparts.


Arrhenius equation derivation I couldn't find the derivation of Arrhenius equation. So, How can we derive the Arrhenius equation:
$k = A \mathrm e^{-E_\mathrm a/(R T)}$?

The equation is rooted in experimental observations. It is not derivable from any physical formalism, and it is not the best way to describe many reactions. Similar equations can be derived from Transition State Theory. It is important to note that the similarities are limited to the functional form.


Converting Chlorine Dioxide to Free Residual Chlorine Our water treatment facility used Chlorine Dioxide to disinfect the produced drinking water. The analysis results indicate concentrations of 0.2 mg/l of chlorine dioxide at the source. Can we analys residual chlorine in the distribution network? We dont have water quality standards for chlorine dioxide, only residual chlorine. Can I calculate the concentration of chlorine in the water from the concentration of chlorine dioxide?

Some chemistry, first add H2O2 and NaOH:

$\ce{2 ClO2 + H2O2 + 2 NaOH -> 2 NaClO2 + O2 (g) + 2 H2O }$

Source: https://chemiday.com/en/reaction/3-1-0-1114
Continue adding H2O2, as:

$\ce{2 NaClO2 + 2 H2O2 ->  2 NaClO + 2 O2 (g) + 2 H2O }$
$\ce{2 NaClO + 2 H2O2 -> 2 NaCl + 2 O2 (g) + 2 H2O }$

which implies a net reaction of:

$\ce{2 ClO2 + 5 H2O2 + 2 NaOH -> 2 NaCl + 5 O2 (g) + 6 H2O }$

So, slowly add H2O2 until the bubbling ceases. Collect the volume of oxygen and convert to moles. Divide by 2.5 to determine moles of ClO2 and your chlorine therefrom.
Here is a related reference, Titrimetric and photometric methods for determination of hypochlorite in commercial bleaches, which cites the use of hydrogen peroxide and acidic permanganate.


Why does the -OH group in β-naphthol direct the incoming diazonium salt towards the ortho position? In the following reactions, the major product W is

I was doing this question which asks me the product of a diazotisation coupling reaction between diazonium chloride and β-naphthol. I know that the diazonium salt will act as an electrophile and compounds like phenol will direct it towards the para position, but I am not able to understand why the diazonium salt attaches itself to the ortho position of the -OH group in this compound.
The salt can also attach itself to the other benzene ring opposite to the -OH group. Also there are two ortho positions with respect to the -OH group. Can somebody please explain this to me?
This is the answer to the question:

It's sort of a process of elimination.  The electrophilic ion could potentially attack anywhere, but:

The hydroxyl group is activating, so the electrophile will prefer that ring.  You are left with three open positions.
One of those three is meta to the hydroxyl group and so less favored, as activating substituents are generally ortho/para directing.  Hydroxyl follows that rule.
Finally -- electrophilic substitution in a naphthalene system is kinetically favored next to the other ring.  See here for example.

We have one position that's on the same ring as hydroxyl, ortho or para to that hydroxyl group, and next to the other ring.  That is the position shown in the book.


Equilibrium with an E2 step E2 mechanisms are known to be irreversible in general, because the forward step takes 2 species and produces three species, and a trimolecular reverse step is unlikely, purely on entropic grounds. (As mentioned in this post.)
Now does that mean that any reaction that is reversible cannot have an E2 step, on the grounds of microscopic reversibility?
Particularly, I want to ask about this hypothetical mechanism, of an isomerisation:

I am not sure how to draw its mechanism. Clearly, some elimination reaction is taking place: it can be E1, E2 or E1cB. Because this is an alkene compound, E1 seems unlikely, only E2 and E1cB are possible. $\ce{MeS-}$ seems like a bad base for both E1cB and E2, so I can't say which one is better. Can I remove E2 from the options purely because the reaction is reversible?
[Note: This is an exercise question which asks me to draw the best possible mechanism if the intermediate detected was the one in the picture. It does not matter if the actual reaction happens in that way or not, the question is hypothetical]

I think your thought process is correct, but you can push it further.
The way I would solve this problem is to think about how you could go from the alkyne to either alkene.
Then, via microscopic reversibility, you would argue that the reverse mechanism--the elimination--is very similar.

 The opposite of conjugate addition + protonation would be most similar to E1cb.


Reaction between manganese(IV) oxide and sulfuric acid $$\ce{2MnO2 + 2H2SO4 -> 2MnSO4 + O2 ^ + 2H2O}$$

How does this reaction occur? We get the oxygen moiety of the manganese dioxide suddenly oxidized to gaseous oxygen, while the manganese gets reduced from +4 to +2.

The sulfuric acid is apparently unperturbed: its proton $\ce{H+}$ and its $\ce{SO4^{2-}}$ parts remain the same in terms of their oxidation states. What role does the sulfuric acid play here at all?

I am afraid I cannot provide the exact mechanism. Nevertheless, here is what I know.

The reaction is most certainly more complex than the equation

$$\ce{2 MnO2_{(s)} + 2 H2SO4_{(aq)} -> 2 MnSO4_{(aq?)} + 3 O2_{(g)} + H2O_{(l,g)}}$$

suggests. One might say it is even misleading, but it is definitely not the whole story.

A purple smoke

After adding concentrated sulfuric acid to manganese(IV) oxide, a purplish smoke manifests. This suggests to the species $\ce{MnO4-}$. Yet, what is the balancing particle? Is it $\ce{Mn^2+}$, or rather $\ce{H}$? I have no clue.
$$\ce{2 Mn_{(aq)}^2+ + 5 MnO2_{(s)} + 6 H3O+_{(aq)} -> 2 HMnO4_{(aq)} + 5 Mn^2+_{(aq)} + 8 H2O_{(l,g)}}$$

A green to brown mixture

The initial mixture is green. This points to either $\ce{MnO4^2-}$ or $\ce{Mn2O7}$, even $\ce{MnO4^3-}$ could be it. Again, which one is it? Maybe it is all of the above? Upon standing, we observe a brown precipitate. The manganese(IV) oxide appears to reform … Speculation (more in the next chapter):

$$\ce{MnO4_{(aq)}^2- + H2SO4_{(aq)} -> H2MnO4_{(aq)} + SO4_{(aq)}^2-}\\\ce{3 H2MnO4_{(aq)} ->  2 HMnO4_{(aq)} + MnO2_{(s)} + 2 H2O_{(l,g)}}$$

$$\ce{3 MnO4^2- + 2 H2O_{(l)} -> 2 MnO4- + MnO2_{(s)} + OH_{(aq)}-}$$

Oxygen, where's that from?

I did the classic match experiment; it did indeed burn more vigorously near the reaction mixture.

$$\ce{HMnO4_{(aq)} -> MnO2_{(s)} + 3 O2_{(g)} + 2 H2O_{(l,g)}}$$
$$\ce{2 MnO4^2- -> MnO4^3- + MnO2_{(s)} + O2_{(g)}}$$
$$\ce{2 Mn2O7_{(l)} -> 4 MnO2_{(s)} + 3 O2_{(g)}}$$

I think you get the idea

Sorry I could not be of more assistance. The mixture is indeed complex with multiple processes happening in parallel. (This is actually the case with every reaction.) It should at least point you in some direction.

Extra

I was a bit loosey–goosey with the ionic equations; you should not do that ;). In my opinion, it made the process more clear.
One might also wish to consider $\ce{Mn2MnO4}$ or $\ce{Mn3O4}$ which, I believe, are also brown.


Modeling 3-center-2-electron bonds in Avogadro Is it possible to model 3-center-2-electron bonds in Avogadro?

It seems to allow users to add however many bonds they would like to an atom, but I can't seem to find documentation on whether this is valid.  When modeling $\ce{B2H6}$, for example, I can draw a $\ce{B-H-B}$ bond, but does Avogadro understand that this is a bridge bond?

For example, below is an $\ce{(AlH3)5}$ cluster I modeled in Avogadro. Two pairs of bridge bonds appear merged after UFF optimization, which doesn't seem right. This is what led me to wonder if modeling the bridge bonds the way I did was incorrect.

First of all, I am not an Avogadro developer, just an occasional user. My view of this matter is the following:

As 3D molecular drawing/editing software, Avogradro, does not need to "understand" B-H-B bonds. If it allows you to draw them in a sensible way, it is already fulfilling its function as a drawing/editing program.
Avogadro is also an interface to a number of codes, either internal or external, that allow you to perform tasks such as geometry optimisations and conformational sampling. Now, there becomes relevant whether or not those methods can "understand" B-H-B bonding. Ab initio and other quantum mechanical methods should afford a reasonable treatment of such molecules, provided that an appropriate level of theory is used, whereas for semi-empirical and molecular mechanics methods it will depend on whether or not a given parametrisation model or a given the force field (version), respectively, have been parametrised to take such situations into account.

In summary, if you are able to draw and edit the structure in Avogadro, the real question is whether or not you are going to do something else with it within Avogadro and whether or not your method of choice is appropriate to treat your molecular systems.


Catalytic iodine in synthesis of phosphonium salts From Clayden et al.

  However, iodine is expensive and a way round that problem is to use a
  catalytic amount of iodide. The next phosphonium salt is formed slowly
  from benzyl bromide but the addition of a small amount of LiI speeds
  up the reaction considerably.

  The iodide reacts as a better nucleophile than Ph3P and then as a
  better leaving group than Br–. Each iodide ion goes round and round
  many times as a nucleophilic catalyst.

In a private conversation with ron I asked him how this catalyzing effect worked, he said that question would be more suited here and I agreed with him. It seems a bit strange at first that a strong nucleophile can also be a good leaving group (he really wants that nucleus, right, why would he leave?).

This is the explanation I came up with (probably wrong, but it is always better to first think self): Iodine is a stronger nucleophile here than bromine because of its higher energy HOMO, which lies closer to the energy of the $\ce{C-X}$ $\sigma^{*}$ (here $\ce{X}$ = $\ce{Br}$) LUMO. So it will react more rapidly than bromine, resulting in an equilibrium.

Now iodine is also a good leaving group, firstly because the $\ce{C-I}$ bond is higher is energy than the $\ce{C-Br}$ bond (less stable bond). And secondly (quoting Ron from another answer. Well explained, no need to rewrite it again): the larger the atom, the more polarizable it is. The more polarizable an atom the more stable is its corresponding anion because we have our charge spread out over a larger volume which is a stabilizing feature. So $\ce{I^{-}}$ will be more stable, making it easier for iodine to leave.

My thought was that iodine can easily knock of bromide but not remain in that position for very long (therefore the equilibrium). Now, if $\ce{PPh_3}$ happens to meet the benzyl iodide it is more eager to react with it than with the benzyl bromide (lower energy $\ce{C-X}$ $\sigma^{*}$). The resulting phosphonium salt is either to stable to be attacked by iodine again or precipitates out (or both?), so that benzyl bromide continues to be transformed into benzyl iodide.

Is this correct?

Iodine is a stronger nucleophile here than bromine

I agree with your HOMO-LUMO reasoning, but this doesn't really matter too much.  As long as iodide is a strong enough nucleophile to displace bromide at least some of the time, then the reaction we want to happen, will happen.  Of course the more nucleophilic iodide is (here's where your HOMO-LUMO argument comes into play), the faster the desired reaction will happen.

I pretty much agree with the rest of what you say Jori, bit I might say it just a bit differently.

I'm not sure how triphenyl phosphine (TPP) is hybridized, I'd guess that it is unhybridized just like phosphine ($\ce{PH3}$).  If it is unhybridized than our lone pair on phosphorous is in an s-orbital.  Since the s-orbital has little "directionality", this lone pair on phosphorous will be a weak nucleophile.  On the other hand, if TPP is hybridized (like triphenyl amine), then although the lone pair is now in an orbital with high p-character (and therefore very directional) it will be extensively delocalized into the 3 aromatic rings (just like triphenyl amine which is a very weak base for just that reason), which again means that the lone pair isn't very available and consequently TPP will be a weak nucleophile.  So now matter how TPP is hybridized, we expect it to be a weak nucleophile.

Iodide is, as you note, a much better leaving group then bromide.  In fact, iodide is such a good leaving group that it can be readily displaced by even a poor nucleophile like TPP. Iodide is also a good nucleophile and can displace bromide.

The reaction is run with just a small (catalytic) amount of iodide.  As soon as iodide replaces a bromide (nucleophilic displacement), the iodide, being such a good leaving group, will now be quickly displaced by the weakly nucleophilic TPP.  During this last step, iodide is regenerated and the cycle starts again.