Difference in the acidity of ethanoic and methanoic acid Peter Sykes attributes the difference in their acidity to differential solvation.

From what I understand, there is not much difference in how well the ions are solvated, since the Delta H is comparable, however, the change in entropy is much greater for ethanoic acid but I don't understand why?

What an interesting way of looking at acidity! How I would look at it would be the more standard way which is to compare the stability of their conjugate bases...

The ethanoate ion has an methyl group, which is electron-donating, donating electron density towards the oxygens, concentrating the negative charge. This makes it a more unstable conjugate base.

In stark contrast, the methanoate ion, does not have the electron donating group and thus, does not have such a high concentration of negative charge. This makes it the more stable conjugate base.

The explanation proposed can also explain the differential solvation of the two carboxylic acids...

With a higher charge density, ethanoic acid would form a more "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is greater.

With a lower charge density, methanoic acid would form a less "ordered" hydration shell as the electrostatic attraction between the water molecules and the negative charge is not as great.

Consequently, the decrease in entropy for the dissociation of ethanoic acid is much greater than that for the dissociation of methanoic acid.

Therefore, the dissociation of ethanoic acid would be less favoured than the dissociation of methanoic acid, based on both considerations of entropy and conjugate base stability, which are actually linked.


What do D and L stand for in the Fischer-Rosanoff convention? In R and S configuration, the R stands for rectus and the S stands for sinister. Similarly, do the D and L descriptors in the Fischer-Rosanoff convention stand for something?

Dextro (right) and levo (left). (I learned it as "laevo", but I don't see that spelling in a quick Google search.)


Computing two-electron integrals with an STO-3G basis set I am trying to implement a restricted Hartree-Fock calculation using an STO-3G basis set, for fun. I managed to perform this calculation where only $\mathrm{1s}$ orbitals are present ($\ce{H2}$ and $\ce{HeH+}$) as explained in Szabo and Ostlund's book. In this book, authors give explicit formulas for overlap, kinetic, nuclear-electron, and electron-electron integrals for $\mathrm{1s}$ orbitals and they work correctly.

In order to generalize my calculation to systems containing $\mathrm{2s}$ and $\mathrm{2p}$ orbitals (for $\ce{H2O}$ and $\ce{N2}$), I used the general formulas I found in Cook's book for the electron-nuclear and electron-electron integrals. In this case, I obtain results that are slightly different from Szabo's book:

$$E_\text{tot}(\ce{H2O}) = -74.4442002765 \text{ a.u.}$$

instead of

$$E_\text{tot}^\text{Szabo}(\ce{H2O}) = -74.963 \text{ a.u.}$$

This is obviously problematic since orbital energies suffer from the same error and this leads to an erroneous ionization potential (0.49289045 a.u. instead of 0.391 a.u., a difference of approximately 63 kcal$\cdot$mol$^{-1}$).

Since I checked my code multiple times and I wrote the two-electron computation code from scratch twice, I was wondering if there is a typo in Cook's book. Is there is a good reference where I can find the (correct) formula to compute two-electron integrals of gaussian functions (in Cartesian coordinates) with arbitrary angular momenta? At the moment I am not looking for a very efficient (recursive) algorithm to perform this task, I only need an exact formula like the one proposed in Cook's book.

Sources:

[1] Szabo and Ostlund, Modern Quantum Chemistry, Dover, 1989.

[2] Cook, Handbook of Computational Chemistry, Oxford University Press, 1998.

Actually there is a mistake in the analtical expression in Cook's Book. On his web page he has a pdf with the corrected verison

http://spider.shef.ac.uk/

Maybe this solves your problem, but I would also recommend to implement the Obara-Saika Scheme or rys-Quadrature since they are really much more efficient. If your are programming in Python, you might have a look at the PyQuante project, which implements all this stuff.
Concerning Obara-Saika you might also read about the Head-Gordon Pople Scheme. It is in principial an adapted version of Obara-Saika which reduces the FLOP count.


Are salt substitutes good alternatives to sodium chloride? I bought a low salt Ketchup from Heinz and it says it uses ‘ALSOSALT’ substitute (potassium chloride). It says it does have 5 mg of sodium. Is potassium chloride (with possible additives) a good alternative to sodium chloride and does it not cause problems for the heart or general health?

Small quantities of potassium are fine. Windsor Salt of Canada makes a 50% blend of sodium chloride and potassium chloride called Half Salt. You could use it like regular salt, although it would be good to decrease the amount you use.

Nerve impairment and muscle function are affected from too much potassium.


translation of a sentence involving antigen/buffer/solution I'm trying to translate a sentence about biochemical testing into English.

So far, I can figure out the following:

Make a 10,000 pg/ml stock solution from[?] antigen into[?] 7.5 % BSA-TSA buffer.

The prepositions are the literal translations of what the foreign language says. However, despite knowing very little about biochemistry, I'm doubtful that either translation is correct in English.

If you were to read this sentence, would you be able to guess how to replace the highlighted words so that the phrase would sound idiomatic in English?

You actually need to prepare a stock solution of 10,000 pg/ml antigen in 7.5 % BSA-TSA buffer.


Is there s-character in the HOMO of dinitrogen? State true or false:
The highest occupied molecular orbital in nitrogen molecule has no s-character.

The answer given is that this false. Why is it so?
The highest occupied molecular orbital in $\ce{N2}$ is the sigma $2\mathrm p_z$ orbital. So it should not have any $\mathrm s$ character and hence the statement should be true.

There is a significant portion of s-p mixing in dinitrogen.
Please see this great answer by Wildcat on Molecular orbital (MO) diagram for $\ce{N2}$ and $\ce{N2^-}$.
For the impatient, I'll quote an image from myself from How to rationalise with MO theory that CO is a two-electron donor through carbon?

I have marked these interactions with the dotted lines. To show these valence MO in all their glory, I've performed a calculation on the RI-BP86/def2-SVP level of theory with Orca.

The analysis (main atomic orbitals) of these orbitals showed, above they are shown in reverse order:
MO  3 (occ.=2.00, E=-1.022230)
  N2 (S on atom N1) [0.416], N16 (S on atom N2) [0.416]

MO  4 (occ.=2.00, E=-0.485050)
  N2 (S on atom N1) [0.435], N16 (S on atom N2) [-0.435],
  N3 (S on atom N1) [0.427], N17 (S on atom N2) [-0.427],
  N15 (S on atom N2) [-0.201], N1 (S on atom N1) [0.201],
  N5 (Px on atom N1) [0.200], N19 (Px on atom N2) [0.200],
  N8 (Px on atom N1) [-0.066], N22 (Px on atom N2) [-0.066],
  N13 (D+2 on atom N1) [-0.015], N27 (D+2 on atom N2) [0.015],
  N10 (D 0 on atom N1) [0.009], N24 (D 0 on atom N2) [-0.009]

MO  5 (occ.=2.00, E=-0.421640)
  N4 (Pz on atom N1) [-0.422], N18 (Pz on atom N2) [-0.422],
  N7 (Pz on atom N1) [0.225], N21 (Pz on atom N2) [0.225],
  N6 (Py on atom N1) [-0.164], N20 (Py on atom N2) [-0.164],
  N9 (Py on atom N1) [0.087], N23 (Py on atom N2) [0.087],
  N11 (D+1 on atom N1) [0.033], N25 (D+1 on atom N2) [-0.033],
  N14 (D-2 on atom N1) [0.013], N28 (D-2 on atom N2) [-0.013]

MO  6 (occ.=2.00, E=-0.421640)
  N6 (Py on atom N1) [-0.422], N20 (Py on atom N2) [-0.422],
  N9 (Py on atom N1) [0.225], N23 (Py on atom N2) [0.225],
  N18 (Pz on atom N2) [0.164], N4 (Pz on atom N1) [0.164],
  N7 (Pz on atom N1) [-0.087], N21 (Pz on atom N2) [-0.087],
  N14 (D-2 on atom N1) [0.033], N28 (D-2 on atom N2) [-0.033],
  N11 (D+1 on atom N1) [-0.013], N25 (D+1 on atom N2) [0.013]

MO  7 (occ.=2.00, E=-0.371870)
  N5 (Px on atom N1) [-0.452], N19 (Px on atom N2) [0.452],
  N3 (S on atom N1) [-0.299], N17 (S on atom N2) [-0.299],
  N8 (Px on atom N1) [0.194], N22 (Px on atom N2) [-0.194],
  N16 (S on atom N2) [-0.134], N2 (S on atom N1) [-0.134],
  N15 (S on atom N2) [-0.074], N1 (S on atom N1) [-0.074],
  N13 (D+2 on atom N1) [0.016], N27 (D+2 on atom N2) [0.016],
  N10 (D 0 on atom N1) [-0.009], N24 (D 0 on atom N2) [-0.009]

MO  8 (occ.=0.00, E=-0.071630)
  N7 (Pz on atom N1) [-0.528], N21 (Pz on atom N2) [0.528],
  N4 (Pz on atom N1) [0.513], N18 (Pz on atom N2) [-0.513],
  N9 (Py on atom N1) [-0.032], N23 (Py on atom N2) [0.032],
  N20 (Py on atom N2) [-0.031], N6 (Py on atom N1) [0.031],
  N11 (D+1 on atom N1) [0.013], N25 (D+1 on atom N2) [0.013],
  N14 (D-2 on atom N1) [0.001], N28 (D-2 on atom N2) [0.001]

MO  9 (occ.=0.00, E=-0.071630)
  N9 (Py on atom N1) [0.528], N23 (Py on atom N2) [-0.528],
  N6 (Py on atom N1) [-0.513], N20 (Py on atom N2) [0.513],
  N7 (Pz on atom N1) [-0.032], N21 (Pz on atom N2) [0.032],
  N18 (Pz on atom N2) [-0.031], N4 (Pz on atom N1) [0.031],
  N14 (D-2 on atom N1) [-0.013], N28 (D-2 on atom N2) [-0.013],
  N11 (D+1 on atom N1) [0.001], N25 (D+1 on atom N2) [0.001]

MO 10 (occ.=0.00, E=0.456580)
  N3 (S on atom N1) [-3.853], N17 (S on atom N2) [3.853],
  N8 (Px on atom N1) [-2.584], N22 (Px on atom N2) [-2.584],
  N16 (S on atom N2) [0.543], N2 (S on atom N1) [-0.543],
  N5 (Px on atom N1) [0.266], N19 (Px on atom N2) [0.266],
  N15 (S on atom N2) [0.162], N1 (S on atom N1) [-0.162],
  N13 (D+2 on atom N1) [0.035], N27 (D+2 on atom N2) [-0.035],
  N10 (D 0 on atom N1) [-0.020], N24 (D 0 on atom N2) [0.020]

The main concern is MO 7, which shows significant s contribution.


How do scientists check which kind of unit cell is present in a crystal? There are many types of unit cells like:

Tetragonal        Monoclinic      Orthorhombic        Rhombohedral
Hexagonal         Triclinic       Cubic

But how do scientists identify the type when a sample of the compound is given to them?

X-ray diffraction. Usually computer will tell you automatically nowadays but in the old days you had to check whether the values you got for the miller planes were possible with the symmetry of different unit cells. They discarded the ones that didnt match until the would find the one unit cell which matched the miller planes from XRD


How do Electrons Cross Nodes in Orbitals? I was wondering in orbitals that have nodes such as the p orbital, how does the electron move from one lobe to the other? I know that people say it is because of the wave/particle nature of electrons but they don't actually explain how they travel across the node without actually passing it.

My teacher says it is because that electrons are able to act like waves and hence like how EM waves can pass through a wall, electrons are able to pass through the node. However when he says that electrons are like a wave, does he mean that it is literally a wave so does an electron have a wavelength, amplitude, frequency, etc? Because I remember reading somewhere which stated that electrons aren't literally like waves. However they can be mathematically described using concepts that apply to waves. Is this true?

From the comments, they state that electrons can be thought as standing waves. After looking at the internet I came across the following image which explained how electrons act as a standing wave:

However this doesn't make sense to me. Don't electrons exist in orbitals not in orbits around the nucleus? If so, isn't this wrong? Also, since electrons move in orbitals, how can there be any end points as it can move anywhere and its movement isn't restricted to a close loop. Hence how can it act as a standing wave?

Electrons (indeed all matter and energy) behave both as wave and as particle, depending how the observation is made. The most obvious example is the single-electron interference pattern.

"In 1974 the Italian physicists Pier Giorgio Merli, Gian Franco Missiroli, and Giulio Pozzi repeated the [interference] experiment using single electrons, showing that each electron interferes with itself as predicted by quantum theory. In 2002, the single-electron version of the experiment was voted "the most beautiful experiment" by readers of Physics World." Wikipedia.


What does *oxidizing* have to do with *hydrogen*? An oxidizing acid is a Brønsted acid that is a strong oxidizing
agent. All Brønsted acids can act as oxidizing agents, because the
acidic proton can be reduced to hydrogen gas.

(source: Wikipedia)
But what does oxidizing have to do with hydrogen? Isn't it about oxygen?
Like how iron becomes iron oxide after prolonged exposure to oxygen in the atmosphere?
So why is hydrogen relevant?

An oxidizing acid is a Brønsted acid that is a strong oxidizing agent.
All Brønsted acids can act as oxidizing agents, because the acidic
proton can be reduced to hydrogen gas."

The whole sentence is written in a very convoluted fashion, and it is partially wrong as well. Someone should improve this Wikipedia section.
All they are saying is that strong acids do react with many metals and oxidize them. In turn their acidic proton is reduced to hydrogen.
For example, metallic Zn reacts with dilute sulfuric acid to form zinc sulfate and hydrogen gas. Zinc (0) has been oxidized to Zn(II). Consequently, Zn(0) also reduced the hydrogen ions to hydrogen gas.
Why this statement is wrong? The reason is that it says ALL Bronsted acids can do this, which is simply an over-generalization. For example, benzoic acid would not produce hydrogen with zinc. It is sparingly soluble in water.
Independent of this point, sulfuric and nitric acid are strong oxidizers in concentrated form. Here it is not behaving like an acid. Sulfuric acid, if concentrated will oxidize copper to Cu(II), and itself get reduced to sulfur dioxide.


Why is it the case that KE = 0.5mv^2 To the best of my knowledge, energy equals work, $\mathrm{E = W}$; work equals force multiplied by distance, $\mathrm{W = Fm}$ ; force equals mass multiplied by acceleration,  $\mathrm{F = MA}$; and acceleration equals distance per second squared, $\mathrm{A = m \setminus s^2}$. However, when I substitute those values in for the dimensions of $ \mathrm{W}$, I can't transform the result of those substitutions into  $ \mathrm{W= \frac{1}{2}Mv^2}$. Perhaps I've done the algebra incorrectly.

Does the result of those substitutions transform into $ \mathrm{W= \frac{1}{2}Mv^2}$?

If it doesn't transform into that equation, why doesn't it do so, and why do we use an equation that isn't compatible with our other equations?

The kinetic energy $E_\text{k}$ equals the work $W$ done on an object:
$$E_\text{k}=W$$
The work $W$ is the result of a force $F$ on a distance $s$. If $F$ is constant, the work $W$ is equal to the force $F$ multiplied by the distance $s$:
$$W=Fs$$
The force $F$ on an object is equal to the mass $m$ of that object multiplied by the acceleration $a$:
$$F=ma$$
For uniform acceleration $a$, the final velocity $v$ is
$$v=at$$
and the distance $s$ is
$$s=\frac{1}{2}vt=\frac{1}{2}at^2$$

Therefore,
$$\begin{align}
E_\text{k}&=W\\
&=Fs\\
&=mas\\
&=ma\cdot\frac{1}{2}vt=ma\cdot\frac{1}{2}at^2=\frac{1}{2}m(at)^2\\
&=\frac{1}{2}mv^2
\end{align}$$


What are the differences between the HMQC and HSQC experiments? This afternoon, I was having some trouble with an HSQC (Heteronuclear Single Quantum Coherence Spectroscopy) spectrum I was trying to acquire, and the spectroscopist recommended I try HMQC (Heteronuclear Multiple Quantum Coherence) spectra, as they are more robust. I did so and got a very nice spectrum in about half the time an equivalent HSQC would have taken, and didn't have to phase it, to boot.

So, what is the downside? Why doesn't everyone use HMQC spectra instead of HSQC? I know they are rather different on a quantum-mechanical level, but as an end-user what, if any, is the difference in the data the two techniques give me, what are the advantages and disadvantages of each, and in what situations might I prefer one over the other?

You get a better resolution in the indirect dimension with an HSQC compared to an HMQC. The HSQC signal is a singlet while the HMQC is usually a multiplet as you see certain couplings on it. So generally you should prefer the HSQC over the HMQC.

One advantage of the HMQC is that it uses significantly fewer pulses than the HSQC. This means that errors in the pulses don't accumulate as much as in the HSQC, you'll lose less signal if your pulses are wrong than in the HSQC.

If you had trouble setting up an HSQC, that could mean that the probe wasn't tuned and matched properly for your sample, or that your pulse lengths were wrong. The proton pulse should be determined on each sample individually; the $\ce{^13C}$ and $\ce{^15N}$ pulses should be set up correctly already. If you're not using a probe with automatic tuning and matching you should also do that manually before the measurement.

There are a lot more HSQC and HMQC variants that are used, e.g. Sofast-HMQC or TROSY sequences that are used for larger molecules like proteins. Another interesting variant you should consider is the multiplicity-edited HSQC, where $\ce{CH2}$ signals will have the opposite sign from $\ce{CH}$ or $\ce{CH3}$ signals.


Does hydrogen bonding strength correspond to Brønsted basicity in a given medium? It does seem that there is a correlation. Both are measures of proton affinity. Heck, isn't hydrogen bonding present in the transition states of most acid base reactions (excluding those involving proton tunneling?) Also, it seems that hydrogen bonding would be subject to the same limitations as Brønsted basicity; i.e. it seems that steric hindrance in some molecule would inhibit both proton transfer and limit hydrogen bonding capability. At least it seems that way to me prima facie.

Also why do we depict acid base reactions as concerted? Is it because the time scale is extremely small for proton transfer reactions, so who cares if it's actually stepwise or concerted?

@Dissenter, you really need to stop asking questions like this.  Once again you point out that the emperor has no clothes!  :)

It turns out that the mechanism of simple proton transfer is an active area of investigation.  There is not a single yes or no answer that would adequately address your questions.  But if this truth were told in the introductory texts who would continue on when the answer to seemingly simple questions leads to such a morass?  Instead, particularly in introductory circumstances, there is a gentlemen's agreement of sorts that we will simply say that the emperor's clothes are the finest we've seen.

Let's look at your two questions in reverse order.

  why do we depict acid base reactions as concerted?

In many introductory texts, and even some that are for advanced courses, proton transfer is presented as a concerted process based largely on Occam's razor (the simplest explanation is often the correct explanation).  However, enough experiments in this area have been run that I think it would be fair to say that a stepwise mechanism is now favored, but by no means mandated for all proton transfers (we'll see why some proton transfers proceed in a concerted fashion below).  The stepwise mechanism proceeds as follows:

step 1, formation of the encounter complex $\ce{A-H}$ and $\ce{B-}$ diffuse through solution until they are close enough to form a hydrogen bond

$$\ce{A-H + B- -> A-H..B-}$$

step 2, the $\ce{A-H}$ bond elongates while the $\ce{B-H}$ bond contracts and a transition state is reached.

$$\ce{A-H..B -> [A..H..B]}$$

The transition state is not necessarily linear or symmetric.  There could also possibly be an intermediate accompanying the transition state.  This depends upon your definition of an intermediate.  How long does a species need to survive in order to be called an intermediate; is it 10 vibrations, 1 vibration or what?  Obviously it is difficult to detect such short-lived molecules, but ultra-fast spectroscopy has made great strides and we may soon know whether an intermediate exists along this multi-step reaction coordinate.

step 3, the transition state elongates in the opposite direction

$$\ce{[A..H..B] -> A..H-B}$$

step 4, the hydrogen bond is broken and the complex diffuses apart

$$\ce{A..H-B -> A- + B-H}$$

  Does hydrogen bonding strength correlate with Brønsted basicity in a
  given medium?

If we throw out cases where geometry interferes with hydrogen bonding (e.g. one example would be where steric factors preclude close approach or proper alignment of the species involved), then it would not be unreasonable to expect some correlation between hydrogen bond strength and the pKa difference between the two species involved.  There is certainly experimental evidence to support this view.  However, it has been found that if we step back and examine a larger range of pKa differences, the correlation breaks down.

Here is a diagram showing how the rate of proton transfers might change as a function of the pKa difference between the two species involved.

If we look at a small portion of the plot, the part at low pKa, we could fairly easily convince ourselves that there is a linear correlation.  But eventually as the pKa becomes large, the rate of proton transfer becomes diffusion controlled and insensitive to any further increase in the pKa.  The same concept applies to hydrogen bonding.  There could be reactions (acid-base proton transfers for example) where there is no activated process, the reaction is diffusion controlled; as a consequence we cannot expect to accurately measure or compare hydrogen bond strengths as a function of Brønsted basicity differences in such a regime.

Summary Answers

  why do we depict acid base reactions as concerted?

To keep it simple in introductory classes.  It turns out that many acid-base proton transfers are not concerted, but rather involve a multi-step process.  Some other proton transfers, for example those involved in diffusion controlled reactions, are probably concerted.  This is still an active area of research.

  Does hydrogen bonding strength correlate with Brønsted basicity in a
  given medium?

The answer is "yes" if we 1) examine only a small range of basicities and 2) look in the range of basicities where the hydrogen bonding is described as an activated process.


Why do lithium and sodium corrode so easily? I want to know what is the phenomenon and explanation behind this corrosion. What is the reaction?

$$\ce{2Li + 2H2O -> 2LiOH + H2}$$
$$\ce{2LiOH + CO2 -> Li2CO3 + H2O}$$
$$\ce{4Li + O2 -> 2Li2O}$$
$$\ce{Li2O + CO2 -> Li2CO3}$$
$$\ce{6Li + N2 -> 2Li3N}$$
Lithium and sodium are reacting with the gases in the air; a destructive physical process.


Why does a molecule with a Thiol group form Hydrogen bonds? I'm doing the GRE Biology practice test, and question #1 asks:

The answer is E. I understand why II and IV are polar, but why is $\ce{R-S-H}$ polar? R is a side chain to the sulfur, sulfur's valence shell is full, and sulfur is highly electronegative. But without knowing what $\ce{R}$ is, how do we know the molecule's total polarity?

Actually it's all four.  See here for I.  The article describes a hydrogen bonding interaction between methane and water.  It's weaker than the "usual" hydrogen bonds but still contributes to forming methane-water clathrates.

The larger lesson here is that hydrogen bonding is not just the electrostatic interaction often described in textbooks.  It has a molecular orbital component that works even with primary bonds of low polarity.


Amending soil pH to change the color of hydrangea flowers I would like to ensure that my hydrangeas are blue. I am told that adding ammonium sulfate or aluminium sulfate will cause the soil to be more acidic. I am also told that adding elemental sulfur will help. Will adding magnesium sulfate to the soil help?

Although it is true that soil $\mathrm{pH}$ plays a role in the color of hydrangeas, the presence of aluminium is actually what determines the flower color. Aluminium is a naturally occurring mineral that is found in most  soils, however, just because a mineral is in the soil doesn’t mean the plant is absorbing this mineral. This is where soil $\mathrm{pH}$ comes into play – plants are able to absorb aluminium when the soil $\mathrm{pH}$ is between 4.5 and 5.8 with the best range being between 5.0 and 5.5. When the $\mathrm{pH}$ moves to 6.0 and above then the plants will not be able to absorb the aluminium despite it being present in the soil.

Adding either sulfur or magnesium sulfate (epsom salt) to the soil will lower the $\mathrm{pH}$, but this will not help with the presence of aluminium in the soil, which is why many aluminium sulfate is so highly recommended.

This is not to say that magnesium sulfate would not be helpful in other ways, however, as magnesium is the central atom of pigment chlorophyll, which will help to improve the foliage quality of your hydrangea.


How can the iron in the reaction of iron (ii) chloride and zinc not go to zero oxidation state? I know that the zinc half equation would be
$\ce{Zn -> Zn^{2+} + 2e-}$.

However I am confused about the half equation for iron. I would think that since iron is completely displaced, its oxidation state goes to zero, so its half equation would be $\ce{Fe^{2+} + 2e- -> Fe}$.

However my teacher said that whenever you have iron, it will always change oxidation states from 2+ to 3+ or vice-versa in a redox reaction, and will not go to zero. For instance, she gave us an example in which there was a reaction between iron(iii) chloride and zinc. I thought that the zinc would completely displace the iron so that it goes from 3+ ion to 0 in the solid iron metal. However my teacher insisted that it goes to 2+, and not to zero.

In the example of iron(ii) chloride, I don't even see how it is possible for it to go to 3+ because then both zinc and iron are oxidised!

Generally, to predict redox reactions, we consider the Standard Reduction Potentials at $25\ ^\circ\mathrm{C}$: The bigger the difference between the potentials of two redox couples, more favorable is the reaction.
$$E^0(\ce{Fe^3+ /Fe^2+})=0.77\ \mathrm{V}$$
$$E^0(\ce{Fe^3+ /Fe})=-0.04\ \mathrm{V}$$
$$E^0(\ce{Fe^2+ /Fe})=-0.44\ \mathrm{V}$$
$$E^0(\ce{Zn^2+ /Zn})=-0.76\ \mathrm{V}$$
So, when $\ce{Fe^3+}$ reacts with $\ce{Zn}$, we get $\ce{Fe^2+}$ (not $\ce{Fe}$) and $\ce{Zn^2+}$.

As for your reaction: when $\ce{Fe^2+}$ reacts with $\ce{Zn}$, we get  $\ce{Fe}$ and $\ce{Zn^2+}$.


How does the Wave-Mechanical model reconcile with the particle definition of an electron? While just starting chemistry, I have a little pit of particle physics knowledge. I was introduced with the wave-mechanical model of an atom and was slightly confused, because from particle physics, I know an electron is a fundamental particle.

How does this reconcile with the idea of electrons as "standing waves"? Or is the standing wave idea something that explains how electrons never really have a definite location? I have had zero years of calculus so please keep calculus equations and jargon to a minimum.

Wave particle duality is a fundamental idea in the quantum mechanics required to understand the behaviour of fundamental particles

One of the key ideas in quantum mechanics is that particles fundamental behaviour cannot be explained if we assume their "real" behaviour is Newtonian (or billiard-ball like). The behaviour of particles like electrons depends on how we measure them and sometimes they appear wave-like sometimes particle like.

This is very important in chemistry as it leads to good explanations of chemical bonding and the shapes of molecules. Electrons in atoms and molecules, for example, were originally thought to behave like a mini solar system where they orbited the nucleus. But this explanation doesn't work well to explain much in chemical bonding or atomic behaviour.

Schrödinger developed a better way to think about their behaviour with his famous wave equation. This describes the shapes of electron clouds around atoms. One way to think of the output of this equation is that it describes clouds of probability describing the possible locations of an electron in an atomic orbital. Those orbitals are the "standing waves" of electrons bound to atoms and they have specific definite shapes (related to spherical harmonics which are the 3D equivalent of the standing waves in a vibrating violin string).

This description of the properties of electrons is the core of our understanding of chemical bonding and the shapes of molecules.

What are electrons? Quantum mechanics doesn't tell us what they really are: it just tells us how they behave when we make certain measurements. And, in chemistry, the measurements that matter mostly give results where the wave-like behaviour is what matters.


How can I generate 250 GPa to create metallic hydrogen? In the preparation of metallic hydrogen, how does one conjure up such immense pressures?

I mean, I read that something like diamond anvils were used to contain the hydrogen, but now, in accordance with Newton's 3rd law, shouldn't the diamond also face the same extreme forces?

I mean a 10 on the Mohs scale is very high, but how does that suffice to withstand the immense force applied back by the compressed gas?

According to The Compressive Strength of a Perfect Diamond 410-480 GPa is possible.  So, yes, the diamond experiences the applied pressure, but it is strong enough to sustain it.

For more recent work, see Implementation of micro-ball nanodiamond anvils for high-pressure studies above 6 Mbar


Ring formation via oxymercuration-demercuration I'm solving past problems from a 2014 Organic Chemistry II course (from a publicly available Harvard summer school Canvas page), when I came across this question:

The answer (which is the reactant) looks like this:

Now I understand how the alkyne would turn into an alkene, but I don't get how a furan would form from what looks like an oxymercuration-demercuration reaction. Any help would be much apprecited.

Observe that there is no nucleophile give in the mercuration step and recall that the three-membered mercury ring intermediate is very reactive. Hence, this is suggestive that the oxygen would attack the intermediate with its lone pair on the more substituted position as follows:

Edit: Oops! Missed a carbon on the last step as correctly pointed out by @Waylander. I have updated the answer with a new image that corrects the mistake.

For reference see  Addition reactions of Alkene's in David Kleins organic chemistry (around page-410 ) and for a quick reference see wiki


What is the OCNH2 group called? what is the product of reaction of Traut's reagent with aliphatic hydroxyl called? I am reacting Traut's reagent with Fmoc protected Doxorubicin and I just need to know the name of the product after Traut's regent reacts with the hydroxyl group. What is $\ce{OCNH2}$ called? I am looking for a name for $\ce{OCNH2}$ circled in red.

The group itself unprotonated is called a carboximidate as noted by the Wiki page.  The acid that the group is derived from is carboximidic acid.  As is common practice with imines, the nomenclature for the corresponding protonated halide is done by removing -es, adding -ium, and the halide name.  If this is done using the acid name, the group will be called a carboximinium chloride.


Gas Stoichiometry & Ideal Gas Law I need a little help with one of the problems below. Any help on the problem below will be appreciated. I don't want to sound like an idiot, I will also provide my work.

  Solid $\ce{NaHCO3}$ decomposes to form water vapor, carbon dioxide and sodium carbonate.
  $$\ce{NaHCO3 -> H2O + CO2 + Na2CO3}$$
  If $2.00\ \mathrm g$ of $\ce{NaHCO3}$ decompose at $500\ \mathrm{^\circ C}$ and $700\ \mathrm{mmHg}$, how many liters of $\ce{CO2}$ should be produced?

$$\mathrm{\frac{750\ \mathrm{Torr}}{760\ \mathrm{Torr}}=0.99\ \mathrm{atm}}$$
$$pV=nRT$$

$$2.00\ \mathrm g\ \ce{NaHCO3} \cdot \frac{1\ \mathrm{mol}\ \ce{NaHCO3}}{84.02\ \mathrm g} \cdot \frac{1\ \mathrm{mol}\ \ce{CO2}}{2\ \mathrm{mol}\ \ce{NaHCO3}} \cdot \frac{22.4\ \mathrm L \ce{CO2}}{1\ \mathrm{mol}\ \ce{CO2}} = 0.287\ \mathrm L\ \ce{CO2}$$

$$0.99\ \mathrm{atm}\cdot0.287\ \mathrm L=n\cdot773\ \mathrm K\cdot0.0821\ \mathrm{L\ atm\ K^{-1}\ mol^{-1}}$$
$$n = 0.004159\ \mathrm{mol}\ \ce{CO2}$$

$$0.004159\ \mathrm{mol}\ \ce{CO2} \cdot\frac{22.4\ \mathrm L}{1\ \mathrm{mol}\ \ce{CO2}} = 0.09317\ \mathrm{L}\ \ce{CO2}$$

I do not like to post an answer for my own question, however I feel the need to clarify my mistakes ton anyone else. So here is the answer:

Calculate the moles of $\ce{NaHCO3}$
Then I need to plug in the moles to $\ce{PV = nRT}$
After finding the volume of $\ce{NaHCO3}$, now I plug it in a dimensional analysis to figure out the volume of $\ce{CO2}$. That is all and the answer would be $\ce{.77\ Liters\ CO2}$


Can Gay-Lussac's law be observed in the reverse direction? Gay-Lussac's law states that if you decrease the temperature of a gas at constant volume, the pressure decreases by a corresponding amount. This is easy to carry out experimentally. But in practice is it possible to perform this the other way round, i.e. decrease the pressure and thus cause the temperature to drop, while keeping volume constant?

It seems to me that in order to reduce the pressure, you would go about lowering the chamber's volume by removing air from it. However, this would also remove a proportional number of moles from the chamber, so the temperature would remain unchanged.

When you decrease the temperature of an ideal gas held at constant volume, what you are doing is transferring energy as heat from the gas into the surroundings. You do this by keeping the surroundings at a targeted final temperature and placing the gas in thermal contact with the surroundings, allowing heat to dissipate. When you reach the final state, you measure the temperature and pressure of the gas and find, in agreement with Gay-Lussac's law, that both T and P have decreased. Then if you ask yourself, which did I lower first, you cannot say. They decreased simultaneously.

The pressure and temperature are coupled (dependent) properties of the ideal gas when the volume is held constant. This is a consequence of the thermodynamic state postulate that for a simple compressible system, the properties of the system are fixed once two independent ones have been specified. The fallacy in the question here is the assumption that one might be able to change the temperature and that a pressure change somehow follows after. The cooling experiment changes both simultaneously. You never actually choose to alter one variable rather than the other.


Hydride Donating Ability We know that some compounds can donate $\ce{H-}$ anion to attain stability. What are the conditions under which it occurs and what should be the extent of stability of the product for this pathway to be viable?

Many compounds are good hydride donors, if it results in carbonyl bond formation or extended conjugation. Sometimes, a compound can donate H- ion if it results in stable pi bond formation.
(e.g. Tertiary butyl carbanion)


Enantiomerism and Diasteriomerism I am really confused between them. I know that an enantiomer is an isomer having optical activity, non super-imposability of image on original configuration, same molecular formula(Correct me if I am wrong). The thing that is causing me trouble is what diffrentiates this from diasteriomerism. It would be appreciated if you provide an example to clarify like a fischer projection

Your definition of enantiomer is correct.

This simple further classification might be helpful.  Let's take 2 molecules with the same molecular formula.  These molecules are either constitutional (or structural) isomers or stereoisomers.  If they have different structures (or connectivity patterns)  and are separated by an energy barrier, then they are constitutional isomers.  If they have the same structure or connectivity pattern, then they are stereoisomers.

Among stereoisomers there are 3 classes that we can place a pair of molecules into.  Take any two stereoisomers and compare them, they are either

the same (2 identical molecules)
mirror images of one another (enantiomers)

If they are not the same or enantiomers, then they are

different, and we call them diastereomers.


Changes in pressure for an equilibrium reaction containing a solid/liquid on one side only For a reaction where there is only one solid/liquid present, but gases present as both reactants and products (are there any examples of this type of reaction?):

$$\ce{A(s) + B(g) <=> C(g)},$$

if pressure is increased, it will favour the formation of the side not containing the solid/liquid (e.g. the forward reaction favoured), so as to decrease pressure of system – since one mole of a gas molecules has a considerably lower volume than a mole of any solid or liquid (far more spread out).

And the opposite if pressure decreased.

E.g. all that is considered is which side the solid/liquid is on, not the moles of gas present on either side. Is this true?

There are many examples of such reactions, for instance:

$$
\begin{align}
\ce{S (s) + O2 (g) &<=> SO2 (g)}\\
\ce{C (s) + 2 H2 (g) &<=> CH4 (g)}\\
\ce{Ni (s) + 4 CO (g) &<=> Ni(CO)4 (g)}
\end{align}
$$

However, your assumption

  all that is considered is which side the solid/liquid is on, not the moles of gas present on either side

is not correct. Since shift of equilibrium according to Le Chatelier’s principle depends on partial pressures, which in turn, rely on molar ratio of gaseous reactants and products, decrease/increase of pressure doesn't depend on which side of the reaction solid/liquid participant is located; instead, it is defined for each system individually based on ratio between gaseous products and reactants.


Why isn't the hydroxyl group of citric acid deprotonated while the carboxyl groups are? My AP chemistry book "Chemistry: the Central Science 12th edition" says that one molecule of citric acid generates 3 H+(aq) ions when dissolved in water. Looking at the structural formula:

There are three carboxyl groups. However, there are four instances of an -O-H link. Why don't all of these break?
My guess is that the carboxyl groups can form resonance groups and thus obey the octet rule.

You are correct in saying that the carboxyl groups will have resonance structures after being deprotonated. Thus, the carboxyl groups are fairly acidic and will deprotonate in water. On the other hand, deprotonation of the hydroxyl group will not lead to any resonance stabilization, such that the hydroxyl group will be much less acidic than the carboxyl groups. In fact, the $\mathrm{p}K_\mathrm{a}$ of alcohols is usually in the range 16-20, which means that the alcoholic hydrogen will not deprotonate in water. It can deprotonate, however, in other solvents.


What does it mean if a functional group has "higher priority" in terms of IUPAC naming of organic compounds? What does "higher priority" mean exactly in the following statement from this wikipedia article (article name: "E-Z notation") on naming organic compounds in IUPAC nomenclature of organic compounds?

  If the two groups of higher priority are on opposite sides of the
  double bond, the bond is assigned the configuration E (from entgegen,
  German: [ɛntˈɡeːɡən], the German word for "opposite").

Whether trying to figure out if a double bond is "E" or "Z", or if a chiral center is "R" or "S", the Cahn–Ingold–Prelog priority rules (see the section "Assignment of priorities") are used to prioritize the groups.  In the case of a double bond there are 2 substituents (a lone pair of electrons can be a substituent) at each end of the double bond.  Priorities (1 and 2) are assigned to the substituents at each end using the Cahn–Ingold–Prelog rules provided in the link above.  The first rule used to set priorities is based on the atomic number of the atoms attached to the double bond.  If there is a tie (the 2 attached atoms at one end of the double bond have the same atomic number), then we move out to the next atoms and compare these.  The process is continued until the tie is broken (eventually the tie must be broken if the molecule being analyzed is really capable of being described as "E" or "Z").  After the priorities have been assigned, if the two priority-1 substituents at opposite ends of the double bond are on the same side of the double bond, then the molecule is the "Z" (zusammen, together) isomer, otherwise it is the "E" (entgegen, opposite) isomer.

Example:

image source


Finding Kp using log Kp I am working through some MIT lecture notes trying to understand adiabatic flame temperature. There is a section in these notes where $K_p$ is calculated for the following reaction at $\pu{1600K}$.
$$\ce{CO + \frac{1}{2}O_2 \rightleftharpoons CO_2}$$
The notes find $K_p$ by using the following values:
@$\pu{1600K}$:
$$\log K_p (\ce{CO}) = 8.234$$
$$\log K_p (\ce{O2}) = 0$$
$$\log K_p (\ce{CO2}) = 12.940$$
$$\log K_p (\text{RXN}) = 12.940-0-8.234=4.706 \implies K_p \approx 5 \times10^4$$
What is the significance of these $\log K_p$ values? I'm confused because I thought that $K_p$ was a value pertaining to a reaction as a whole, not specific reactants. Also, where can I find these values online? My notes pointed me towards https://webbook.nist.gov/chemistry/, but I am having difficulty finding these $\log K_p$ values in that database. I would appreciate any guidance/intuition you may be able to provide!

The $K_p$'s here are those for the standard formation reactions.
If you go to the following link, you will see $\Delta G^o_f$ values for $\ce{CO2(g)}$  and $\ce{CO(g)}$, in $100 \,\mathrm{K}$ increments, up to $1500 \,\mathrm{K}$:
https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjbt56Z9IbvAhXVjp4KHbD-BS0QFjAJegQIAhAD&url=https%3A%2F%2Flink.springer.com%2Fcontent%2Fpdf%2Fbbm%253A978-3-319-41616-8%252F1.pdf&usg=AOvVaw3rJgF0nzpqXtq9ML9qvGaV
Thus, doing a simple linear extrapolation to $1600 \,\mathrm{K}$ gives us the following:
$$\Delta G^o_f,\ce{CO_2(g)} = -396.309 \mathrm{\frac{kJ}{mol}}$$
$$\Delta G^o_f, \ce{CO(g)} =  -252.3298 \mathrm{\frac{kJ}{mol}}$$
Solving for $\ln K_p$ (using $\Delta G^o = -RT \ln K_p$) gives:
$$\ln K_p,\ce{CO_2(g)} = 29.7906$$
$$\ln K_p, \ce{CO(g)} = 18.9677$$
Dividing by $\ln(10)$ to convert from the natural logarithm to the base-10 logarithm gives:
$$\log K_p,\ce{CO_2(g)} = 12.940$$
$$\log K_p, \ce{CO(g)} = 8.238$$
[$\Delta G^o_f$ for $\ce{CO(g)}$ changes more rapidly with temperature than that for $\ce{CO2(g)}$, so my extrapolation for the former may have a larger error, which could explain the small discrepancy vs. the value you were given.]
In addition, if you appropriately combine the formation reactions to give the net reaction (as you would for a Hess's Law problem), you will see that you get the supplied equation for $\log K_p$ for the net reaction (I'll leave this as an exercise for you).


Unified theory of chemical bonds My question is related to the edge between Subatomic Physics and Chemistry, but I decided that here is the best place to discuss it.

As we all know, Theoretical Physics is a well developed field of knowledge, able to describe the very tiny forms of interactions. However, after some researches for some kind of formula to perfectly describe the interactions on the atomic level (not subatomic), more specifically, a way to calculate the strength of chemical bounds, I keep comming across frustrating highschool explanations, such as

  ionic > covalent > H-bond > dipole-dipole > van der Waals

The most precise data that I was able to find comes at page number 20 and 21 of this pdf, that apparently shows some experimental results (I'm guessing).

So here is my question: is there really a unified formula? What if I want to simulate chemical interactions (proton-by-proton, electron-by-electron) on a computer, is it that complex to do, even assuming that I want a simple simulation of about 20 atoms?

Edit: When I reffer to calculating chemical bonds strength, I also include predicting many other chemical properties, as electron affinity or reactivity of the elements, which are all related.

Note: I can accept as answer any pdf that clarifies my question, because I wasn't able to find. Maybe I'm looking it with the wrong key-words.

The short answer: no. We do not know how to solve the Schrodinger equation analytically, even for systems as simple as the He atom.

The much longer answer:

The Schrodinger equation is the basis for almost all theoretical chemistry: chemistry that attempts to predict properties of compounds without an actual experiment. The problem is that the Schrodinger equation is unsolvable for any compound with more than two electrons, and so we must make some approximations to solve it. This leads to methods such as Hartree-Fock and Density Functional Theory.

While these approximations are usually pretty good, they are still only approximations, and sometimes stop working correctly. There is an added problem that the Schrodinger equation is only an approximation, and the proper theory to use to describe light-matter interaction is Quantum Electrodynamics, which is incredibly difficult to solve for a single photon reflecting off of a mirror.

Long story short: there may or may not be a "unified theory" for describing exactly how all bonds behave in every situation, but there are currently methods that are good enough to predict a large portion of the behavior that we observe.


Building a Voltaic Cell Without Metal Ions For an assignment in my Chemistry class we have to build a voltaic cell using materials that are provided to us and that outputs a voltage as close to 0.8v as possible. The list of materials is as follows:

Glass plate
4 filter paper strips

1 ml of the following 1.0M solutions

HCl
NaOH
NaCl
Water

Small metal strips:

Cu
Fe
Pb
Mg
Zn

I originally thought of doing something like this:

This is essentially a small scale version of a 2 beaker voltaic cell with a salt bridge that you see in textbooks. I would attach a multimeter between the Zn and Cu strips and read a voltage.

What I'm confused by however, is that I'm not provided with any solutions of aqueous metal ions, which are required for the half reactions powering this cell to go forward. I know that this has something to do with the HCl and NaOH that I'm provided with, but can't figure out exactly how they factor in.

I've done quite a bit of googling but I'm completely stuck. How would one make a voltaic cell out of the materials I've listed? Thanks for any help in advance.

If you just put a copper strip and a zinc strip into some NaCl solution and try to write the Nernst equation for the whole system, you will notice that the electrochemical potential will be undefined. Since the electrochemical potentials involve the cation concentrations which are all zero by preparation.

What this means is that there is infinite driving force for some metal to get oxidized into the solution until there is some equilibrium. Therefore, as one of the comments to your question indicate, you will be dissolving some of the metals into the solution by the very act of putting them into such an arrangement.


qv and qp relationship Is there any relationship between q at constant volume and q at constant pressure that I could derive with just the reaction equation, and the enthalpy of formation and Cp of the products and reactants?
I am aware that q at constant pressure is enthalpy of formation of products - enthalpy of formation of reactants.

Let us assume that the initial and final temperatures in both the isobaric and isochoric processes are same. We will limit this discussion to ideal gases only. To calculate heat transfered we will use the expression $$q = nC\Delta T$$ Here C is molar heat capacity of the gas. For isobaric process, this will be $$q_p = nC_p\Delta T$$ For the isochoric process $$q_v=nC_v\Delta T$$
$\Delta T$ for both these processes same. Substracting these two equations we get $$q_p - q_v = n(C_p - C_v)\Delta T$$ The value of $C_p - C_v$ is $R$. Therefore $$q_p - q_v = nR\Delta T$$


What happens to the shielding of the proton when an alcohol is oxidized? The full question:

  For H-NMR, the magnetic field experienced by a hydrogen atom can be
  externally applied or internally induced or some combination of the
  two. When the applied and induced fields act in concert, a proton in
  these fields is said to be deshielded. Which of the following
  processes will result in the greater deshielding of the proton? Please
  choose from one of the following options:

  Removing nearby carbonyl groups
  Adding electropositive atoms like silicon
  Oxidation of primary alcohols with $\ce{KMnO4}$
  Reducing terminal alkyne groups

This is what the question is asking me, but I'm confused because if you oxidize a primary alcohol, this gives an aldehyde or a carboxylic acid (depending on reaction conditions), and the proton in question goes into forming water. So if an H disassociates, does this equate to a total deshielding, or how is this explained with respect to proton NMR?
I realize that water does show up on NMR spectra, but I'm unclear what I am being asked. It also obviously depends if this is primary, secondary, or tertiary alcohol.

The answer given is (3) Oxidation of primary alcohols with $\ce{KMnO4}$. I don't understand why.

The alcohol proton becomes an acid proton. Potassium permanganate oxidizes primary alcohols to carboxylic acids. In the process, the hydroxyl proton goes from 2-5 ppm to >10 ppm (generally). Therefore, since it is shifted to a higher ppm, it is being more deshielded.


How can a nonspontaneous reaction occur? $$\frac{-\Delta G}{T}=\Delta S_{universe}$$

(This equation applies under isobaric and isothermal conditions.)

I understand that if $\Delta G$ is positive, the reaction is nonspontaneous, and adding that amount of energy to a system can drive that reaction to occur. However, if you’re increasing the Gibbs free energy of the system, then by that equation, the entropy of the universe would be decreasing, which would violate the Second Law of Thermodynamics. Can anyone explain this?

When you add energy from somewhere, let's say a battery, you are creating a greater amount of entropy there (in the battery or whatever). That's all there is to it. The Second Law is not violated.
The electrolysis process is not non-spontaneous. Yes, it would not occur in the absence of the battery. But the fact that it does occur with the battery there must tell you that overall it is spontaneous. The fact that you are coupling a strongly spontaneous reaction (discharge of the batter) with a non-spontaneous reaction (the electrolysis) means that overall, the electrolytic process is spontaneous.
It's incorrect to treat the electrolysis as if it occurs without the battery.


where is sulphate going here? electrolysis of magnesium sulphate I am electrolyzing a water + magnesium sulphate solution using copper for both electrodes. I am getting a light-blue precipitate forming which I believe to be a mixture of copper hydroxide and magnesium hydroxide, which are both poorly soluble. The solution is remaining clear which tells me that basically no copper sulphate is forming.

My question is, where is the sulphur from the sulphate going?

The mixture is not becoming acidic as if sulphuric acid was forming, and I do not detect any smell of sulphur is if some sulphur oxides were being released as gas.

In electrolysis of aqueous solutions sulfate ions just sit there.  Protic hydrogen (or the metal in the cation if this metal is not highly electropositive like magnesium) is reduced more easily at the cathode, and water is more easily oxidized at the anode.

Sulfate ion (actually, bisulfate ion) can be oxidized electrolytically at the anode if you use sulfuric acid instead of water as the solvent.  See the preparation methods for potassium peroxydisulfate.


Create combinatorial library of biologically plausible chemical compounds based on library of known compounds I recently downloaded all the molecular structures (about 13,000) stored in Pherobase, an open library of insect pheromones. I would now like to create a derived, more extensive combinatorial library of compounds that would likely occur in nature, but right now don't happen to be included in the library (e.g., similar aliphatic compounds but with a different chain length or different positions of methyl branches etc).

I know I can create these semi-manually by specifying certain series using SmiLib, or using ChemAxon's Reactor. I was wondering though if there are any good cheminformatic methods to recognize biosynthetic series, which could, e.g., subsume lists of compounds in the library in  Markush notation, which could then later be expanded. Or what could be other good ways to expand the library to include other biologically plausible related compounds?

Could Chemaxon's Metabolizer perhaps help, maybe after specifying plausible biosynthetic/enzymatic reactions? Any other ideas?

Just looking in Pubchem for structurally related compounds I think wouldn't work as that would result in too many biologically implausible compounds ... (The reason I want to do this is in the context of making a large QSPR library with in-silico predicted retention indices, and later perhaps also predicted EI-spectra).

I think your best bet would be to go to the KEGG Reaction Database (http://www.kegg.jp/kegg/reaction/) which is a comprehensive database of known biochemical (enzyme-catalyzed) reactions, plus various associated tools. It includes a module that predicts a likely biosynthetic pathway for any pair of structures. All your pheromones are natural products and are formed via enzymatic reactions, so you should be able to explore out from your starting set by exhaustively applying the reactions in the KEGG database to each one of them. There may be a module that does this for you  automatically, but it is not documented on the KEGG web site.


What is the temperature of heat generated from LPG gas? I'm working on a project and I require around 300-400 degree Celsius of temperature. As of now I'm taking the normal LPG gas that is available in our house for cooking purposes for my project, but I'm not sure whether it can reach 300-400 degree Celsius or not? I can definitely use multiple burners but how much heat/temperature does 1 burner generate?

Maybe I'm wrong because heat is counted in calorific value, but then how do we know the temperature of that calorific value? Or is there some relation which says that x amount of LPG would give x degree Celsius or maybe x calorific value? But yet again, how can I know the temperature?

LPG is mostly propane and butane both of which have adiabatic flame temperature near 1970 degrees C when burned in air.  See Flame Temperatures of some Common Gases.

So 300-400 degrees C is certainly reachable.


Does the Clemmensen reduction reduce the C=C bond of α,β-unsaturated ketones? The Clemmensen reduction is a chemical reaction which reduce aldehydes and ketones to alkanes.

In my book, it is written that under such conditions, double and triple bonds are not reduced, but my teacher told me of one exception:

  α,β-unsaturated carbonyl compounds undergo reduction of both the olefinic and carbonyl group.

But according to my book, when this compound undergoes Clemmensen reduction, the product is pent-2-ene.

Who is correct?

Both are not quite correct. A quick literature search doesn't turn up any hits for the Clemmensen reduction of this particular aldehyde, but March's Advanced Organic Chemistry, 7th ed. (p 1541) writes about the Clemmensen and Wolff-Kishner reductions:

Neither method is suitable for α,β-unsaturated ketones [...] Under Clemmensen conditions, both groups of these molecules may be reduced or if only one group is reduced, it is the C=C bond.

Large structural rearrangements are common under the vigorous conditions: see, for example, Helv. Chim. Acta 1976, 59, 962. "In all cases but one the formation of cyclopropyl acetates was observed."
It seems that there are some situations in which the ketone is selectively reduced, but this is not general: see Tetrahedron 1986, 42, 6615.


What is the physical state of a single atom? Can a single atom on its own be either a solid, liquid, or a gas? Or is it none of them?

This answer has been written so it hopefully can be understood by people that do not have a degree. If something is not correct, let me know, but be aware that throwing around fancy words will not help anyone understanding this any better. Consider the other posts here concerning liquids and solids, I don't want to be redundant and repeat what has been said already.

The answer is a bit complicated. One would intuitively say no, but that doesn't tell the whole story.

Take the interstellar medium. We have about 1 atom per cm³ in there and it is called a "gas". Now that's what I would call an isolated atom in the gas phase.

An atom is an object that belongs into the realms of quantum mechanics. A solid, liquid or gas is something that belongs into our classical world. If you zoom in close enough, that means if we look at a problem microscopically, words like "solid" or "liquid" no longer make sense. If you now begin to zoom out, the quantum effects vanish. This is sometimes called quantum decoherence and we enter the world of classical physics where solids and and liquids exist again.

So I would argue it all depends on how closely you look at the problem. Isolated atoms in a large volume? A gas if you ask me. But if you have to zoom in to talk about an individual atom within a larger compound consisting of many more atoms, let's say one atom within a piece of iron, you no longer can say that this single atom is "solid".

Think of it like this:

Can a single person have a political system?

Yes, why not. I can proclaim my own nation and be the only person in it. But mostly, a nation is something that many people form together. If you look at one individual person, you are in the realm of psychology (or biology). If you look at an entire state, you are in the realm of (for example) politics. If you zoom in at one individual person, you can no longer say that person has a certain political system. But if you zoom out far enough, you suddenly get a political system. Zooming out of course is a bit more abstract here, but, as I said, this is the realm of psychology/politics ...


Thermoplatics into Thermosetting plastics I must confess, I am not a chemistry student but have always been a fan of it, so I am not sure if this is a blunder.

I have often wondered if Thermoplastics can ever be converted to or made to behave like a Thermosetting plastic. In other words, can a Thermoplastics be made more heat resistant?

Depending on the plastic, it is possible to convert a thermoplastic to a thermoset. The way to achieve this, is by linking the polymer chains in a thermoplastic to one-another and is called 'cross-linking'. Then, the chains cannot 'slide' along-side each other and will therefor become a thermoset. This means the plastic will not melt and start to degrade when the temperature becomes to high for the molecular bonds to hold.
Note that cross linking the polymer chains is a chemical reaction which needs certain functional molecules. When the thermoplastic chains have functional groups left to attach a 'cross linker' to, a bridge between two chains can be made. When the thermoplastic has no functional groups left however, it might never be possible to convert the thermoplastic into a thermoset.

Lastly, I would like to subscribe to the idea that there are no stupid questions (or blunders). Just people stupid enough not to ask them. Good job asking this one :)

Hope this helped!


Why use a heating block? Is a heating block necessary for heating small chemistry samples? I figure it helps a ton (not to mention it stabilizes too) in getting heat to the reaction but is there anything else it does?

You have already mostly answered the question yourself ;)

A heating block, made from a material with a high thermal conductivity, helps to keep (a lot of) samples distributed over a larger area (several vials, Eppendorf tubes, etc.) at exactly the same temperature even if the primary heat source is punctual or just a layer of heating wires at the bottom of the instrument.


Carius method for quantitative analysis of halogens Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Fig.12.17) in a furnace

I want to know :

What is the significance of using $\ce{HNO3}$ ? Can we carry out oxidation with other acids like $\ce{H2SO4}$?
Why $\ce{AgNO3}$ is added to $\ce{HNO3}$ and not to the organic compound? Will it be wrong if I add $\ce{AgNO3}$ directly to the organic compound after it is subjected to fuming nitric acid?
Wouldn't precipitation of compounds other than that of halogens occur since we are carrying the reaction in a closed furnace?

Any help is greatly appreciated.

1) $\ce{HNO3}$ is a more oxidizing acid than $\ce{H2SO4}$.

2) The silver nitrate would not dissolve well in the organic solution.

3) Sulfide is an example of something you might worry about interfering as it forms a very insoluble compound with silver. But silver sulfide and most anything else that might interfere will decompose in the fuming nitric acid.


Why is arsenic toxic to humans? I don't understand why arsenic is toxic and why it does damage to humans. Could it be that it is so similar to phosphorous? If so, would that make antimony toxic as well?

Arsenic is toxic to humans due to the fact that it imitates phosphorous almost identically in its bonding patterns, as both elements are in the same group.

However, why would this have any affect? The most likely reason for this is due to the fact that when the phosphorous is substituted by arsenic in many of a cells vital molecules such as DNA and other phosphorous containing groups, the properties of the new substance is slightly different in terms of its polarity, reactivity, etc. Because the proteins and enzymes within a cell interacts with these phosphate containing groups in such a specific manner, these slight irregularity in chemical nature can have detrimental effects (you can think of the processes of these proteins and enzymes between the phosphate as clockwork, i.e. if the parts are slightly different, it will not work).

I am not as sure about the chemistry of antimony, as I have not researched that topic greatly, but I would guess it being in the same group adds to its toxicity.


Why so differance in alkali halide's crystal forms? I always wonder that though Both $\ce{NaCl}$ and $\ce{CsCl}$ are alkali halides, why they adopt different crystal forms, FCC and BCC respectively? and I have a doubt about $\ce{CsCl}$ structure that Why the $\ce{Cs+}$ or $\ce{Cl-}$ ions are not located at voids/lattice points as like as $\ce{NaCl}$ structure?

Perhaps you have heard of the following terminology: "rock salt", "cesium chloride", "zinc blende" structures? If so, you may recall that in this case, the main factor dictating structure is "size" or atomic radius ratio, rather than group. For example, in the case of the rock salt structure, it is most likely that there is "a cation/anion radius ratio of 0.414 to 0.732". As such, you could consider the radius of cesium in comparison to sodium: significantly larger. Thus, it adopts the cesium chloride configuration, which is preferred when "ions are of roughly the same size ".


TPE fabric vs PVC fabrics? I am currently in charge of purchacing fabrics from asia. My personal policy is to try and stay as enviromentally freindly as possible. Your answer to the following question would be of great help and would prevent me from buying the wrong product in massive quantities.

What is a TPE fabric and how does it compare to PVC fabrics?

This question is impossible to answer without knowing which TPE you mean - TPE is a mixture of thermoplastic elastomers, which is a huge class of compounds.

PVC is polyvinylchloride, which is a specific compound. I found an MSDS here:

http://www.fishersci.com/ecomm/servlet/msdsproxy?productName=AC183320010&productDescription=POLY

My advice is to contact the supplier for each fabric you are considering and to ask them for the MSDS. In that you will see what the acute and long-term exposure effects are for people. "Environmental effects" is a very broad subject, and you are not very likely to find research for any proprietary blends, particulary if they don't come from the US or EU. However, sometimes MSDS list non-proprietary compounds, and you can then check the EPA website for those names to see if there are any reports on broad environmental effects.

If you are also interested in the environmental impact of the entire production chain (which you should be, since end-use is usually the least significant part of that), then you should also ask the suppliers about their manufacturing process, whether they have any "green" certifications, etc. You can look up the patents for each production process (most of the time they will be patented) to see what their raw materials are, and then check the MSDS and EPA reports for those. It is also a good idea to see if you can tour their production facilities, once you know what the potentially dangerous precursors are and how they should be handled. It is easy to temporarily hide bad working conditions, but it is harder to fake treatment and containment facilities. Of course, if there isn't good regulatory oversight, there is a very good chance that the producer will take shortcuts to reduce costs, so keep in mind that what they tell you isn't necessarily going to be the truth.


Why are there more fluoride compounds formed with Xenon? Based on WebElements, of the Noble Gases, $\ce{He}$ and $\ce{Ne}$ do not react with any of the halogens; however:

$\ce{HArF}$ has been detected at low temperatures (thank you to @bon and @Martin in the comments)
$\ce{KrF2}$ can be synthesised at low temperatures.
Xenon can react with fluorine to produce $\ce{XeF2}$, $\ce{XeF4}$ and $\ce{XeF6}$
Radon can react with fluorine to produce $\ce{RnF2}$

I understand that fluorine is the only halogen that seems to react with the Noble Gases (partly due to the related question Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides?.  My question is

Why are there more fluoride compounds formed with Xenon compared to the other Noble Gases?

The reason that we discovered more fluorides of xeon than helium, argon, and krypton is quite obvious and you might already know it. As lighter noble gases have more stable shell configuration, it is more difficult to make any kind of compound, including fluorides.
Also, the higher amount of fluorine atoms in a molecule of the compounds means drawing more electron from the noble gas, which is thermodynamically not preferred. Thus, $\ce{KrF4}$ and $\ce{KrF6}$ are far less stable than $\ce{KrF2}$. The quantum chemical calculation confirmed this (Dixon 2007). The same reference said that they are stable, but have not been synthesized yet, and the synthesis would be very difficult.

For radon it is a different problem. Briefly, there might be more fluoride compounds of radon than xenon, but the synthesis is difficult to be confirmed.
Even though radon is less inert than xenon and its compounds were predicted and calculated to be more chemically stable than xenon compounds (Chernick 1962, Thayer 2005), it is difficult to study. Radon is is rare and radioactive (half-life of 3.8 days). Studying chemical compounds in trace amount is a difficult task. There are claims for discoveries of $\ce{RnF4}$ and $\ce{RnF6}$ (also in the above references), but confirming that you have synthesized new compounds in trace amount is extremely difficult because the evidence is not obvious. $\ce{RnF8}$ was calculated to be unstable, but should be more possible than $\ce{XeF8}$.

References:
1 David A. Dixon, Tsang-Hsiu Wang, Daniel J. Grant, Kirk A. Peterson, Karl O. Christe, Gary J. Schrobilgen, "Heats of formation of krypton fluorides and stability predictions for $\ce{KrF4}$ and KrF6$\ce{KrF46}$ from high level electronic structure calculations," Inorganic Chemistry 2007, 46(23), 10016-10021 (https://doi.org/10.1021/ic701313h).
2 C. L. Chernick, H. H. Claassen, P. R. Fields, H. H. Hyman, J. G. Malm, W. M. Manning, M. S. Matheson, L. A. Quarterman, F. Schreiner, H. H. Selig, I. Sheft, S. Siegel, E. N. Sloth, L. Stein, M. H. Studier, J. L. Weeks, M. H. Zirin, "Fluorine Compounds of Xenon and Radon," Science 1962, 138(3537), 136-138 (DOI: 10.1126/science.138.3537.136).
3 John S. Thayer, "Relativistic Effects and the Chemistry of the Heaviest Main-Group Elements," Journal of Chemical Education 2005, 82(11), 1721-1727 (https://doi.org/10.1021/ed082p1721).


How to determine the acidity or basicity of dissolved salts? One kg of each substance below is dissolved in a bathtub full of water. Will it be acidic, basic, or neutral?
  $\ce{KNO2}$
  $\ce{HF}$
  $\ce{(CH3)3NHBr}$
  $\ce{HO-CH2-CH2-O-CH2-Cl}$

$\ce{KNO2}$, $\ce{K+}$ $\ce{NO2-}$ $\ce{K+}$ is inert and $\ce{NO2-}$ is the base of strong acid, so it should be basic overall.
$\ce{HF}$, $\ce{H+}$ $\ce{F-}$ $\ce{H+}$ is strong acid and $\ce{F-}$ is strong base so it should be neutral overall.
$\ce{(CH3)3NHBr}$, Molecule acts like a $\ce{H3NHBr}$ or $\ce{NH4+}$ $\ce{Br-}$ so $\ce{NH4+}$ is acidic $\ce{Br-}$ is neutral, so overall solution is acidic.
$\ce{HO-CH2-CH2-O-CH2-Cl}$, No idea.

Can someone please explain them, if they're wrong? Thank you.

The $\ce{KNO2}$ is basic, the HF is acid, the $\ce{(CH3)3NHBr}$ being an ammonium salt is acid too, the last one is an alcohol, I would consider it neutral for your purpose. But you are missing one step to get your answer quite right.
Last part, in the pH world, everything turn around two ions H+ (or H3O+) and OH- and the equilibrium between those two. A substance is considered acid when it either give a H+ as disolving or takes an OH-. On the other hand a substance is considered basic when it give an OH- or takes a H+. Like HF is an acid (H+, F-) and NaOH is basic (Na+, OH-). For KNO2, it can take an H+ to go back to HNO2, so it is basic. For organic compounds, alcool groups are usually considered inerte as they don't ionate in normal conditions.
Now, maybe I'm wrong but your question sound like if you put all those in a bathtub what'S the final result? Is it right or not?


Sound of ringing gels I'm searching for an audio file of the ringing of a ringing gel - does anyone know a source?
(I already searched on different video- and sound-platforms and looked up in the literature)

From this question on Physics SE, there is a linked audio track in Soundcloud here, which is what you are looking for. The OP references a jar of hair-stying gel which, upon being tapped with a mallet, produces a "reverberating" sound instead an otherwise expected "short 'tok' sound."
Note the the ringing gels are common ingredients in pomades, shampoos, and other products where rheological properties are engineered for various reasons. See this patent application for a ringing gel, wherein it is stated:

The phenomenon of ringing is a result of the manner in which the dispersed droplets are entrapped within the network structure of a gel. It has been suggested that the network within a gel consists of a rod-like or lamellar structure. Such a network could allow dispersed particles to vibrate within a gel much like a ball bounces off the walls of a box.

The practical application of the gels appears to be mainly in the cosmetics industry.


Can it be said that on our planet we have roughly the same number of electrons as we have protons? My question is quite simple I think. Is it true that we have approximately the same number of electrons as we have protons and the average electric charge of the molecules in our planet is roughly neutral?

Yes

We can see that the charges in the observable world are close to balance by considering what we would observe if they were not.

Big imbalances of charges cause strong electrical fields. Those are measurable and their effects are observable. We can create such imbalances locally (that's what a Van de Graaff generator does) and the effects are spectacular: really big sparks as the strong fields ionise air and cause big electrical discharges. These separations also occur in nature (that's what thunderstorms are). In both cases, the consequence of significant temporary separations of charge are spectacular natural phenomena that act to restore the balance.

It could be true, however, that the earth as a whole had a net charge. We wouldn't notice this while on the surface as the net charge would be relative to interplanetary space. On the other hand any such charge would probably be neutralised by selective interaction with the charged particles in the solar wind which we could observe. Also, if the net charge were substantial, we would observe significant effects on planetary orbits (the electrical field is a lot stronger than the gravitational field.

These observations together suggest that both locally and globally we have approximately the same number of protons as electrons on the earth.


Acetic acid freezing distilation How can acetic acid be distilled by freezing when its freezing point is above the freezing point of water.

Examples on the web show the water portion of the vinegar as ice and the acid as a liquid but Wikipedia shows that freezing point of acetic acid is 16.6°C.

The procedure you are describing sounds like fractional freezing. Acetic acid is frequently purified by this method, hence 100% acetic acid is often called "glacial" acetic acid.

  As an example of how this works, let's consider a mixture of acetic acid and water.

Acetic acid has a freezing point of 16.6 oC. Water has a freezing point of 0  o. They form a eutectic at -26.7 oC (although I do not know the composition). The phase diagram probably looks something like (but is not) the following.

Phase Legend:

A = Liquid solution
B = Solid solution of acetic acid in water PLUS liquid solution
C = Solid solution of water in acetic acid PLUS liquid solution
D = Solid solution of acetic acid in water
E = Solid solution of water in acetic acid
F = Solid solution of acetic acid in water PLUS Solid solution of water in acetic acid

A mixture of acetic acid and water that is approximately 72% acetic acid is cooled until it begins to freeze (solid blue vertical line), which happens around 2 oC. The phase changes from A (liquid solution) to C (Solid solution of water in acetic acid PLUS liquid solution). The solid solution is enriched in acetic acid compared to the original liquid solution. The composition of the remaining supernatant liquid solution is enriched in water compared to the original liquid solution. The composition of the solid solution can be determined by drawing a horizontal (red) line to the phase boundary between C and E and reading down to 91% acetic acid. The composition of the remaining supernatant liquid solution can be determined by drawing a horizontal (green) line to the phase boundary between A and C and reading down to 63% acetic.

The supernatant phase is removed and discarded, and the solid phase is allowed to melt. The process is repeated. Each successive freezing yields a new solid solution that is enriched in acetic acid, until the purity is as high as you want it.

Note, that since water and acetic acid have a eutectic point, if we has started on the left side of the eutectic point, the phases would have been reverse. The solid would be enriched in water and the supernatant would have been enriched in acetic acid. The supernatant would be remove and kept, and the solid would be discarded. The supernatant would then be cooled and fractionally frozen again to further enrich the supernatant in acetic acid until the eutectic point is reached (at which point the composition of the supernatant is equivalent to the composition of the solid phase). If you want to isolate acetic acid from a solution that is mostly water, I would distil the solution until the composition is mostly acetic acid, and then fractionally freeze it.


Why can a sigma bond rotate? My book says

  Because it (the sigma bond) lies along the line joining the two carbon atoms, the sigma bond can rotate about this axis.

But that doesn't make sense to me. So what if the sigma bond lies along the bond axis? How does that matter?

Sigma  bonds  are  defined as  having their  electron  density  along the  bond  axis, while  pi  bonds  have  their  electron  density  above and  below  the  bond  axis. What this mean is that pi bonds cannot rotate the same way as sigma bonds since rotation would break the pi bond interaction. See the picture below for clarification.

If you're wondering why it rotates then the more  freely  that  a  bond  rotates, the  more  favorable the entropy.  In  general, the  more  kinds  of  motions  and  the  more  unconstrained  those motions are, the  more  favorable the  entropy.


Atomic emission spectra: classical vs. quantum mechanics In class, we were first introduced to classical mechanics, and how Bohr used the hydrogen electron to explain the atomic emission spectrum of that element. And it made sense. But then classical mechanics fails to explain other things, so we jump to quantum mechanics. And we learn about Newton's ideas of particles of light, and how Einstein said that light quanta are called photons, and how Louis de Broglie suggested that particles of matter can also behave like waves.

But I'm frustrated because, while the book does explain the history of quantum mechanics and the Heisenberg uncertainty principle, it doesn't explain how quantum mechanics has anything to do with atomic emission spectra. So my question is, how does quantum mechanics explain the atomic emission spectrum of elements better than classical mechanics does?

And how much of classical mechanics fails to account for what quantum mechanics does? So is the idea that an electron moving from one high energy level to a lower energy level in an atom releases energy/light is only true for hydrogen, and false for all the other elements? How can this be true?

First of all, classical mechanics does not explain the emission spectrum at all, since there is no energy quantization out there. Bohr introduced energy quantization in his model in an ad hoc way by requiring the angular momentum to be an integer multiple of a fixed unit $\hbar = h/2\pi$ (which in turn resulted in energy also being quantized). So, Bohr model is not a classical mechanical model, since there is quantization of angular momentum and energy. But it is also not a quantum mechanical one since quantization of angular momentum is not justified, but rather asserted.

I'm saying all this just to correct OP: it was not a fight of classical vs. quantum mechanics; classical mechanics was already knockouted and quantum mechanics didn't yet fully emerged. It was a battle between semiclassical-semiquantum theories and reality which today could be viewed as well as a fight between semiclassical-semiquantum theories and the real quantum theory invented later.

Anyway, there are in general many possible electron transitions for each atom (or molecule) that contribute to its emission spectrum and each transition has a specific energy difference and probability. The Bohr semiclassical model had shortcomings with respect to both these characteristics of electron transitions:

Multi-electron atoms do not have energy levels as predicted by the Bohr model, and consequently, the model predicts wrong positions of the spectral lines. In fact, it fails already for helium atom.
There is no general way of calculating probabilities of transitions, thus, it is not possible to provide relative intensities of spectral lines.


Why does the carbonyl of carboxylic acids get protonated and not the hydroxyl group? Is the O of the C=O group more electronegative or is it for the sake of getting from product A to B that the carbonyl is protonated rather than the hydroxyl?

Because the protonated acid is resonance-stabilized when the carbonyl is protonated, but isn't if the hydroxyl is protonated:

So the first protonated form is much more energetically favourable than the second.

The same resonance in the non-protonated acid means that the lone pair on the alkoxide has reduced basicity, meaning that the carbonyl lone pair is the more basic since it isn't involved in any energy-lowering orbital interactions(/resonances).


Why do frying pans stick less with heat? I know this is simple, compared to most questions asked here, but I notice this several times a week and I'm curious.

I notice if I break an egg and drop it into the frying pan, and the pan is still warming up, the egg sticks to the pan.  But if I wait until the pan is quite hot, and I break an egg and drop it in, the egg barely sticks to the surface at all.

How does the heat make a non-stick surface work better then when it is merely warm or is still warming up?

Heat creates steam bubbles, which hold the egg away from the pan until it is somewhat cooked - see Leidenfrost effect.


Do MO's span a bond or the whole molecule? I'm reading Organic Chemistry by David Klein (2nd Ed), and when discussing the causes of torsional strain, he writes this

Based on recent quantum mechanical calculations, it is now believed that the staggered conformation possesses a favorable interaction between an occupied, bonding MO and an unoccupied, antibonding MO. This interaction lowers the energy of the staggered conformation.

This favorable interaction is only present in the staggered conformation. When the C—C bond is rotated (going from a staggered to an eclipsed conformation), the favorable overlap above is temporarily disrupted, causing an increase in energy.

I've searched online and most sources provide a similar explanation and diagram. My issue is mainly with the diagram: why are the MOs illustrated as though they span only the C—H bonds? I thought that MOs were supposed to span the entire molecule, rather than being localized along/parallel to the internuclear axes of two nuclei. Are these MOs correct? Do MOs cover the entire molecule or do they each exist between two bonding atoms?

The unfortunate answer is: both, and it is rather complicated. It can be noted at this point, however, that between two atoms connected by covalent (or even ionic bond) there are MOs.

Molecular Orbitals, the Wave Function, and the (Electron) Density

In quantum chemistry, we can define:

(spin) MO: a one-electron wave-function. Contains one electron, can be combined to with another spin MO to form what we usually think of as an MO, that is, doubly occupied.
Wave Function (WF): Typically a product of MOs, at least in the simpler approaches (Hartree-Fock [HF], common Density Functional Theory). The product is anti-symmetrized such that it obeys the Pauli principle. These products of MOs can be linearly combined to yield more accurate, complicated and expensive approaches.
Density: The expectation value of a WF when applying the density operator. Out of the three, this is the only true observable, for instance by X-ray diffraction. If you have a open shell molecule, you may be able to observe a spin-density, for instance by Electron Spin Resonance spectroscopy.

What's the point of these definitions?

They illustrate that a MO and the observable density have a few steps in between. Let's assume that someone gives us a set of MOs which yield the correct density of a molecule (as well as other correct observables). This is already a tall order, by the way. It is a mathematical fact that one can "rotate" the MOs, that is, create new MOs from linear combinations of old ones and still yield the correct density. So the set of "correct" MOs is not unique by the reasonable measure that it should give us the correct density.

OK, so there is more than one set of "correct" MOs. So what?

For one molecule/method/basis set combination, many ways of "choosing" your MOs are possible and lead to very different pictures, even when choosing reasonable contour values (say, 85%) for the visualization. An incomplete list is:

canonical MOs: these are eigenfunctions of the Fock operator, which is central to HF, but also of relevance for post-HF methods. They also preserve meaningful orbital energies (which relate to ionization potentials), but they are usually completely delocalized over the whole molecule.
localized MOs: "rotations" of the MOs have been performed to yield orbitals of (hopefully) minimal spatial extent. They are more easily interpreted, but the mathematics involved is iterative, meaning that minor changes in molecular geometry could lead to rather different local MOs. There is also more than one scheme to do this, and the results can be fairly different.
natural MOs: suppose you have performed the very accurate/very expensive calculation of full Configuration Interaction (full CI) for your molecule. The resulting orbitals are then eigenfunctions of the true Hamiltonian (in the given basis set). You will probably never encounter these due to the sheer cost of the calculation - and they may not look all that different from canonical orbitals.

And then there is the small issue of virtual/unoccupied orbitals

They do not enter the density (duh), but are often relevant to discussions of mechanisms, such as initial point of attack or stabilization. In my opinion, there is basically no way to pick a good scheme for them - canonical orbitals sort-of yield electron affinities.

Then all hope is lost. There's no way to choose.

Well, for occupied orbitals, there kinda is. One should compare orbitals (of one computational level) with each other across molecules to get to useful results. The textbooks usually do not give you this opportunity because you have no idea where those orbitals come from.

Back to OP's question: yes, canonical orbitals would extend over basically the whole molecule. You are being shown, in all likelyhood, orbitals that have been localized somehow. They are used as illustrations by the authors and that's usually not a bad idea. Take 'em with a grain of salt.


Normal and heavy water mixing This question seems silly, but this just came in my mind while i was learning about hydrogen.

Means will there be any reaction when $\ce{H2O}$ will be mixed with $\ce{D2O}$?

Yes, a reaction known as "hydrogen exchange" will take place.  When $\ce{H2O}$ and $\ce{D2O}$ (heavy water) are mixed, hydrogen exchange will take place rapidly to form $\ce{HDO}$ as a third component of the mixture.  Depending on the initial amounts of $\ce{H2O}$ and $\ce{D2O}$ added, a statistical mixture of the 3 compounds will result. So if you mixed equal amounts of the $\ce{H2O}$ and $\ce{D2O}$, then you would wind up with something around a 1:2:1 (it won't be exactly 1:2:1 due to the effect of the slightly different stabilities of the 3 compounds - see LDC3's comment) mixture of $\ce{H2O}$, $\ce{HDO}$ and $\ce{D2O}$.


A different type of photosensitive material I'm curious as to whether there is any small, solid or gel-like material that dissolves/liquefies/evaporates under brief exposure to bright artificial light? So, some small element or agent that would form a 'wall' between two liquids but then vaporise when exposed to light (preferably artificial)

I think I found a crystal that changed to oxygen awhile ago via google, but I cannot locate it again so any help would be well appreciated.

There are many types of photoresist that might suit your purpose. Many are commercially available from electronics suppliers; even Amazon carries some. A thin film of photoresist could be used to separate two liquids, and dissolve when exposed to light, if at least one of the liquids is transparent to the wavelength needed.

You might also consider a photo-sensitive valve using a plastic that changes shape in light.


Could milk rust a steel teaspoon? Recently, while cleaning a neighbour's fridge (turned off for a few weeks), I came across a cup (closed with a lid).  Inside the cup was, to my olfactory horror, congealed milk, with a steel (iron) teaspoon nicely wedged into it.

The milk had separated into a blue cheese-like solid, overlain with a semi-translucent liquid.

When I took the teaspoon out of the congealed mess, there was a layer of rust on the teaspoon where it was in the liquid. But, I observed almost no rust on the parts of the teaspoon in the blue-cheese-like solid and that exposed to the air.

The illustration below is a rough schematic of where the rust occurred (the 2 layers observed are labelled A and B, and the air in the top of the container is labelled C and the diagonal line represents the teaspoon:

A = congealed blue-cheese-like solid = virtually no rust on this part of the teaspoon
B = semi translucent liquid = heavily rusted part of the teaspoon
C = air = virtually no rust on this part of the teaspoon

How could the milk, or the liquid separation, cause rust on the teaspoon?

A hunch I have (and I could be way off the mark), is it possible that the corrosion could be due to, at least in some significant way, due to the fact that milk is slightly acidic $^{(1)}$?

$^{(1)}$ According to the International Livestock Research Institute, they state that

  Fresh milk has a pH of 6.7 and is therefore slightly acidic

Related question and answer Silver and milk (or milk products)

Technically, even simple water can cause rust, so nothing surprising here.
However, spoilage of milk most probably produced a lot of organic acids (lactic acid and similar) which can speed up any corrosion process. Lactic fermentation is a natural fermentation process in milk, when bacteria start converting the sugar content of milk to lactic acid. It is not the only possible product, but a very common one. Link: https://en.wikipedia.org/wiki/Lactic_acid_fermentation

There are two important factors that help corrosion: First, acidic environment (low pH) itself generally help corrosion. Second, the organic acids produced in fermentation can often form metal complexes, therefore they help remove any protecting oxide layer from the surface of metal objects as well as shift the  redox equilibrium towards an oxidized complex form.


Zwitterion form of aminoacids What I am wondering, are aspartic acid and glutamic acid at physiological pH present in zwitterion form. In my textbook it says that all aminoacids are at pH 7,4 or 7 in zwitterion form. I wonder how if pKa values of side chains of aspartic acid is 3,65. There are also arginine, lysine, cysteine and tyrosine that  will not have overall neutral charge at pH = 7.

The IUPAC goldbook defines a zwitterion as being overall neutral. Any other definition is thus contrary to the standard definition.


Correct term/word for bias in calculations? I ran an experiment recently and made several approximations in the calculations that led to the results being around 2 orders of magnitude less then the literature values.  I will be re-doing the calculations but in the meantime I need a word to describe what is wrong with this data set (ie what the effect was on the data by making the approximations).

I thought saying the approximations resulted in a "statistical bias", or "analytical bias" might be right, but this is a little different that regular bias because the data is all fine and unbiased, its the approximations/calculations that are making the final results off by ~100x literature values.

I'm not sure if this is the right place to ask this, so if you think it belongs on the statistics stackexchange, please let me know.

If you're interest (or if it matters), the approximation was to ignore epithermal neutron capture in NAA analysis.

I believe the fundamental notion that you're looking for is a "systematic error in the data analysis".

A "statistical bias" would imply some sort of sampling problem which doesn't seem to be the case from your description.

An "analytical bias" would imply a different/better laboratory technique which again doesn't seem to be the case. So although the overall results are biased, you don't need to redo the experiment itself but rather reanalyze the data using the proper calculations.


Lone pair electrons on secondary carbon vs primary carbon I know that a molecule is more stable when a free radical is on a secondary carbon over a primary carbon, but how about when there is a lone pair of elections that can be placed on a secondary or primary? Is there any difference in the stability of the molecule or does that only apply to free radicals?

I am not sure if you are referring to carbanions or carbenes (I suspect the latter), because both contain a carbon with a lone pair, so I will give explanations on both in the following.

Carbanions are destabilized by the exact same effect that stabilizes carbocations and radicals: hyperconjugation. Donation of electron density from an adjacent $\sigma$ bond into the empty/half-filled $p_z$ orbital stabilizes the cation/radical, but not a carbanion center with its negative charge in the fully-filled $sp^3$ orbital lobe. Therefore, a primary carbanion is more stable than a secondary one.

The inverse is true for carbenes: Secondary carbenes are way more stable than primary ones. This is because carbenes are per se a very reactive species, and because of this property they are often used in organic synthesis as intermediates which are generated in situ. To obtain a reasonably stable carbene, the reactivity of the lone pair needs to be somewhat reduced. Two substituents at the carbene carbon are needed which stabilize the carbene either by steric hindrance (i.e. making the lone pair difficult to reach for other reactants) or mesomeric effects (for example, lone pair donors which alleviate electron lack of the "sextet" carbon, delocalization of the carbene lone pair into an aromatic system). The Arduengo carbene (named after its discoverer) is a good example for this. The stability of this compound is such that it can be isolated as a crystalline solid at room temperature. The carbene lone pair can be delocalized into the five-membered ring which then contains 6 $\pi$ electrons (2 double bonds + lone pair of uncharged nitrogen) and is aromatic. Bulky $\mathrm{R}^1$ groups like adamantyl further contribute to stability by shielding the lone pair from potential reactants (source).

Arduengo carbenes are not the only persistent (especially stable) carbenes, and the understanding of their stability is still a subject of ongoing research, because structural elements (bulky substituents, aromaticity) which had been regarded as necessary for stability were often disproved as such by the synthesis of carbenes which did not contain them. The compounds do not need to be cyclic, as shown by bis(diisopropylamino) carbene, and even heteroatoms next to the carbene center do not need to be present, like in the case of bis(diisopropylamino)cyclopropenylidene.


Does "silica colloidal anhydrous" contain aluminium? Some silicas such as sodium aluminosilicate contain aluminium. Is silica colloidal anhydrous one of these substances?

It will probably have some aluminium in it, likely to be less than 0.1%, and most probably less than 0.01%. Not only that, also expect Ti and Zr in those quantities.

This level of impurities does not matter for most practical applications. If you do require exceptionally pure silica, you can buy it for a premium.

A note about “anhydrous”: when exposed to air, powdered silica will absorb moisture. Particularly if it’s amorphous. If you want it to be really anhydrous, dry the thing before using. Above 100 C will remove most of it, but completely removing all traces of water will require firing at ~1000 C in a noble metal crucible such as platinum.


Why does density decrease across the period 4 and 5 transition metals? I know that density decreases across period 4 because the last two elements are liquids and gases which means that their densities are less but why are the densities increasing and then decreasing within the transition metals themselves?

Density is generally proportional to how far from a full octet (or in your case, an 18 electron set of outer-most orbitals) a material is, hence this effect, and its inverse - that earlier elements in a period are less dense. This does not apply only to transition metals, but all elements, as illustrated here. (Note: I am not sure what pressure is in that chart, so I suggest ignoring gas-phase elements, including O, F, and Cl.)

The reason for this is fairly simple. The further from a complete octet an element is, the more energetically favorable orbital overlap becomes. The more favorable orbital overlap becomes, the closer elements will pack at a given temperature. There is no energetic benefit to an element with a full octet (ideal gases) overlapping, so they typically do not, and thus form a gas. There is an intense energy benefit in elements with both half-full octets overlapping.


How do chemists decide on which detergent to use? I found betaine in my face wash. I think it is there because it has a hydrophobic (non-polar) end that can pick up grease, and a hydrophilic (polar) end that means it is attracted to polar solvents, such as water.

My question is why are betaines used over other detergents? What are the factors that contribute to the which detergents are used in face wash (or indeed shampoo/shower gels)?

If it's polar, then it won't pick up grease because lipids (fats and oils) are generally non-polar. I don't know specifically how a detergent is chosen, but soaps are chemically the salts of a fatty acid. When choosing a detergent, you want a molecule with a polar head with a long, nonpolar tail.

Recall that water is polar and polar substances dissolve polar substances, and nonpolar dissolve nonpolar. The detergent in a sense forms a "bridge" between the water and the nonpolar substances (dirt, grease, oil, etc.) which would not dissolve in pure water. The detergent molecules form little "cells" (my choice of this word is not a coincidence - this is also how your cell membranes work) with the polar heads facing out and the nonpolar tails mingling with nonpolar dirt molecules inside. This allows the dirt be washed away by water.

As to why certain detergents are chosen, it is probably a balance struck between safety, effectiveness, cost to mass-produce, possibly environmental considerations associated with production, etc.