How to go from, Iron(II) Sulfate (anhydrous, FeSO4) to, Ferrous Sulfate Heptahydrate (FeSO4.7H2O) In the lab I need some ferrous sulfate heptahydrate ($\ce{FeSO4.7H2O}$). We have some old anhydrous $\ce{FeSO4}$ from several years ago.

It seems that at normal humidity, $\ce{FeSO4}$ will hydrate to the heptahydrate on its own after a few days at most.

Is this correct? Can I "hydrate" my $\ce{FeSO4}$ just by leaving it out overnight?

Thank you.

According to this Wikipedia article:

  Iron(II) sulfate (British English: iron(II) sulphate) or ferrous
  sulfate denotes a range of salts with the formula $\ce{FeSO4·xH2O}$.
  These compounds exist most commonly as the heptahydrate (x = 7) but
  are known for several values of x.

While the anhydrous form appears as white crystals, the article above continues:

  Iron(II) sulfate can be found in various states of hydration, and
  several of these forms exist in nature.
  $\ce{FeSO4·H2O}$ (mineral: Szomolnokite, relatively rare)
  $\ce{FeSO4·4H2O}$ (mineral: Rozenite, white, relatively common, may be  dehydratation product of melanterite)
  $\ce{FeSO4·5H2O}$ (mineral: Siderotil, relatively rare)
  $\ce{FeSO4·6H2O}$ (mineral: Ferrohexahydrite, relatively rare)
  $\ce{FeSO4·7H2O}$ (mineral: Melanterite, blue-green, relatively common)

This all suggests that simply letting the anhydrous form sit in air overnight is not a reasonable procedure for the quantitative production of the heptahydrate form.

One qualitative hint that you could get regarding the degree of hydration of your old anhydrous material is the color.  While both the anhydrous and quadrahydrate are white, the heptahydrate is a bluish green color.

You could also measure the melting point of the old material; it should be $\mathrm{60-64^oC}$ if it has hydrated all the way to the heptahydrate.

If you really want to try to convert the old anhydrous material to the heptahydrate, you could expose it to a humidified environment overnight.  Then observer any color changes and measure the melting point. You could even weigh a sample of it then heat to ~$\mathrm{110^oC}$ for a few hours then re-weigh it and calculate the degree of hydration.

That's all a bit crude, but you could try it. I recommend just purchasing some of the heptahydrate compound if possible.


Hartree Fock and DFT - Exchange term this might be a stupid question but I am really confused at the moment. I'm currently preparing for an exam in computational chemistry. In the lecture I noted always down that the exchange term (Pauli-Principle), which arises as part of the two-electron functions is not determined in Hartree-Fock and added to the cusp-problem, when two electrons come close together. Those two then form the correlation energy, which is determined in post-Hartree-Fock methods.

Then in DFT we have several attempts to solve the exchange part, and I took a note back then, that basically in DFT or in the Kohn-Sham approach, as we do not know the DFT-functional, you create something, that looks like a Fock-operator, with the difference, that it also contains the exchange part.

Now I looked at some tasks and one question was why you do not simply take the HF expression for the exchange part and insert it into the DFT equations. I was a bit shocked what this expression might be and it turns out, when I look into other scripts I cannot find any note that in modern HF the exchange part would be neglected in any way.

So for simple HF (not post-HF) calculations, is the exchange term calculated as well and the only thing you miss would the the correlation?

Most of the original question has been solved by discussion in the comments, but a pertinent question remains:

  [W]hy you do not simply take the HF expression for the exchange part and insert it into the DFT equations [?]

The answer is that it has been tried and does not work well. In a sense, the DFT quantities "exchange energy" ($x$) and "correlation energy" ($c$) do not quite match the HF quantities "exchange energy" and "correlation energy".$^1$ In fact, even within DFT, the quantities do not match between functionals, as can be seen in some $x$- and $c$-functionals working together well and some not so well. In particular, those that have been fitted in combination with each other often only work in that combination. This suggests to me an illusion of a clear-cut way to differentiate the two in the DFT framework. It is also true that in some functionals, the $x$-term appears to account for some static correlation (which, broadly speaking, often becomes important when a single Lewis structure does not describe the molecule well: transition states, resonance structures). Entirely replacing it with Fock exchange forgoes this advantage.

Some programs, such as NWChem, and to some extent Gaussian, allow you to combine functionals and Fock exchange (often with arbitrary weights) to conduct your own experiment. A small benchmark set for main-group thermochemistry would probably give you an indication.

I wish I could you references for this, but this has been related to me verbally by my advisor back in the day and I no longer have access to most of the literature. If anybody has such references, I would be happy to see them included.

$^1$ The notation is a bit sloppy, but we're not striving for rigor in this answer.


Lewis dot structure of beryllium fluoride I'm so confused because I'm trying to draw $\ce{BeF2}$ … and the way it's drawn (something like $\ce{F-Be-F}$) just doesn't make sense to me because Be can only hold 2 valence electrons. But this structure shows it having 4.

You are correct. $\ce{BeF2}$ is octet deficient.

To solve this problem, they would form polymers by themselves, but that would be out of syllabus.

(I cannot find the polymer of $\ce{BeF2}$, so here is the polymer of $\ce{BeCl2}$:)


At which temperature does a concentrated aqueous solution of sodium chloride begin to boil? We know, solutes dissolved in water increases the boiling point of water. This question is an MCQ. The options are 96,99,100,104 degree Celsius. The correct answer is 96. Why? If the boiling point INCREASES after adding salt then why not 104?

The boiling temperature always increases when adding salt into water. It looks as if salt was attracting water molecules and preventing them from quitting the liquid. So that it is necessary to overheat the solution to let water molecules get into the vapor phase. I have often made this measurements in class and always found about $\pu{104°C}$.


Are melting points for mixtures of solids ALWAYS below the melting points of the individual components? In every chemistry lab we are taught that mixtures of compounds are going to have melting points lower than either of the components pure melting points.

Is this always true? Or can there be ratios of compounds that would generate a melting point above one pure compound and below the other?

For example, lets take acetanilide (MP: 114 C) and mix it with sodium acetate salt (MP: 324 C) in varying ratios. What would the melting points be for these?

1.) 98% acetanilide, 2% sodium acetate (slightly below 114?)

2.) 75% acetanilide, 25% sodium acetate (slightly above 114?)

3.) 50% of each (significantly above 114?)

4.) 25% acetanilide , 75% sodium actate (below 324?)

5.) 2% acetanilide, 98% sodium acetate (slightly below 324?)

NOTE: I chose these two compounds completely arbitrarily. Perhaps there is literature that has examined melting points for varying ratios of two particular compounds?

Follow up question:

Does there exist a mixture and/or ratio of solids that intermolecularly bond so strongly to each other that the melting point of that mixture is higher than the melting point of the individual components? What about a specific azeotrope with a higher boiling point than the liquids that make it up?

Is this always true?

No. As a general rule of thumb, if two compounds are miscible in each other, that is they exhibit solid solution, then the melting point of a (chemical, not mechanical) mixture will be intermediate. If the two compounds are not miscible and it's only possible to make a mechanical mixture and not a chemical mixture, then you will have a lower melting point.

Let's try to see this with several examples of phase diagrams, taken from FACT.

  ...mixtures of compounds are going to have melting points lower than either of the components pure melting points.

This is called eutectic relationship, for example sodium chloride and sodium sulfate. There is no intermediate compound, and they have a eutectic relationship, where the melting point of any mixture is lower than the melting point of both compounds. Note that if you are not in the pure liquid field, you will have a pure compound together with a liquid that is a mixture.

Silver and gold are a good example for a case when there is complete miscibility. All intermediate compounds can exist, therefore the melting point of the mixture changes continuously with the composition:

The double line is there because there is a very thin temperature range in which a liquid coexists with the solid of a different composition. See the solid solution page for more about that.

These were two simple cases. More complex cases arise when you have discrete compounds rather than a complete solution solution. A rather good example is cerium and nickel:

This system consists of several eutectic and peritectic points.


Why do metals tend to lose electrons, as opposed to maintaining electric neutrality? Metals tend to lose electrons to obtain the stable noble gas configuration of 8 valence electrons.

Why do they want to obtain this configuration, and how does the strength of their "desire" to obtain this configuration compare with the "desire" to maintain neutral charge.  If the answer depends on the chemical, I'm happy for you to provide some examples.

Thanks.

Firstly, atoms "want" to achieve the noble gas configuration of 8 valence electrons because it is the most stable form. All that means is that it doesn't tend to react under normal conditions that we experience on Earth, therefore it will stay in that configuration for quite a while and are less likely to react. There is a more complex quantum physical answer for that but you'll have to go elsewhere for than.

The main force that keeps electrons in atoms is the electrical attraction between the electrons and the protons in the nucleus and so, if it is more energetically favourable to lose that electron in order to form a bond, then that is what will happen.

Focusing on the Alkali metals as an example, as you move down the group, they get more and more reactive. This is because of two main reasons that are a result of the electrons being further away from the nucleus:

Because they're further away, the attraction between the protons and the outer most electron is less
Secondly, taking Rubidium as an example, it has 37 electrons and 37 protons. From the perspective of the outer-most electron, there are 36 electrons repelling it, and 37 protons attracting it, therefore acting as a net charge of 1. However, if you take into account the first point, the repulsion of the closer electrons is stronger than the attraction of the protons so it could even be less than one


Do bare protons exist, even transiently, in aqueous solution? Generally $\ce{H^+}$ ions in water remain in the form of hydronium ions ($\ce{H3O+}$), and not as "free" aqueous protons, as far as I've been taught.

My question: is this always the case, even when considering dynamics at extremely short time scales? By this I am not meaning interconversion with the Zundel ($\ce{H5O2+}$) or other states. I mean to say, does bare $\ce{H+}(\mathrm{aq})$ ever exist  on, say, a femtosecond time scale?

According to Myths about the proton. The nature of H+ in condensed media Accounts of Chemical Research, vol. 46, pp 2567–2575:

  But to put this into
  perspective, an electron-free proton has an estimated acidity $10^{56}$(!) times greater than 100%
  H2SO4, the threshold of superacidity.$^2$ Adding a bare proton to a molecule is a real event in the
  gas phase but only a solvated hydrogen ion can be transferred in condensed phases.


How does activated carbon bind the adsorbate? I read some articles and journals about activated carbon and found a problem about how the adsorbate could bind to the activated carbon. Some said it is because of intramolecular forces (van der Waals) and some said it is because of chemisorption. If chemisorption plays a role, what would the bonding be?
And if the bonding is just caused by intramolecular forces, what is the active site of the activated carbon? An answer or illustration for better understanding is greatly appreciated.

Activated carbon works by enlarging the surface area of elemental carbon. Activated carbon can reach surface areas of several hundred square meters to several thousand square meters per gram. Also elemental carbon is fairly non reactive, non toxic and cheap.

As seen in the picture, activated carbon forms a micro structure with incredibly small pores, so that many molecules can be adsorbed (the forces involved decrease with a factor $1/r^{6}$ with distance). The adsorption itself is an application of the London dispersion force (one of the Van der Waals forces).

I don't believe chemisorption has anything to do with this.


Does acid and base react directly, or do they react first with water? If an acid $\ce{AH}$ reacts with a base $\ce{B}$ in aqueous solution, in textbook they write

  $$\ce{AH + B -> A- + BH+}\label{rxn:1}\tag{1}$$

But first I add acid to water and then base, so why we don't write the equations like this:

When I add acid to water:
$$\ce{AH + H2O -> A- + H3O+}$$
When I add base to the acidic solution:
$$
\begin{align}
\ce{B + H2O &-> BH+ + OH-}\\
\ce{B + H3O+ &-> BH+ + H2O}
\end{align}
$$

So the equation $\eqref{rxn:1}$ isn't true because the base doesn't react directly with the acid, but with $\ce{H3O+}$ formed by the acid reaction with water. Is what I have written true?

With strong bases and acids, only $\ce{OH-}$ and $\ce{H3O+}$ react, right, because $\ce{AH}$ and $\ce{B}$ (practically) don't exist in aqueous solution.

For weak acids and bases, that's not true, but still there the water can "transport" protons much faster between AH and B than direct exchange after diffusion. As was noted below, that's called the Grotthuss mechanism.

So yes, you are right, but the formula (1) is a valid simplification which is, for most uses, as good.


How to determine the order of acidity of the following dimethyl nitrophenols This is how I went about it, although I got a wrong answer:

3,5-Dimethyl-4-nitrophenol (2) shows the nitro group to be sterically hindered, due to which it should go out of plane and this will be an example of the steric inhibition of resonance. Hence, there will only be a weak withdrawing inductive effect, and no resonance. Thus, this should be the least acidic.
4,5-Dimethyl-2-nitrophenol (3) shows a hydrogen bonding between the nitro and the hydroxyl group, due to which there will be a lower tendency to release the proton. Hence, even this should be less acidic.
Between 2,6-dimethyl-3-nitrophenol (1) and 2,4-dimethyl-5-nitrophenol (4), (1) should be weakly acidic due to the hyperconjugation destabilisation caused by the two methyl groups, and hence the answer should be (4), with lower hyperconjugation, lower induction from the methyl groups, and higher induction from the nitro group (no resonance at the meta position).

However, the answer is given to be (2), which I ruled out first.

Can someone please explain what is happening, and where I've gone wrong?

The right answer should be 3 as the $\ce{NO2}$ group has a negative mesomeric effect, which is only possible with 3. In 2, due to steric inhibition of resonance, it exhibits only -I effect, and although 3 contains H-bonding, it slightly affects acidity.


What is the IUPAC name of this organic compound? Variety is the spice of life. I have made an organic compound myself. Can you please tell me the name?
I thought that the correct answer is -
3,3-dibromo-2,4-dichloro-1,4-difluoro-2,4-diiodo-1-methylbutan-1-ol, because the haloalkane group is to be given in the alphabetical order and the priority of the alcohol group is higher than the haloalkanes.
Is it correct? What will be the derivation of the name if the alcohol group is replaced with hydrogen?

As of Loong's suggestion, I'd put the answer removing it from comment section: The name should be 4,4-dibromo-3,5-dichloro-2,5-difluoro-3,5-diiodo-pentan-2-ol as Maurice pointed out. However, keep in mind that there are 3 chiral centers within the molecule. But, the indicated name is okay since stereochemistry of the molecule is not shown.
If $\ce{OH}$ group is replaced by $\ce{H}$ atom, the name of new compound is 2,2-dibromo-1,3-dichloro-1,4-difluoro-1,3-diiodopentane. Again, there are still 3 chiral centers within the molecule.


Is it more economical to liquify natural gas to ship overseas or convert it to propane (C3H8) and ship it overseas? Problem statement: Companies want to frack oil wells and then export natural gas overseas using transport via high pressure pipeline followed by liquefaction at a terminal port.

If the methane component in natural gas ($\ce{CH4}$) can be converted to propane $\ce{C3H8}$, that would leave $\ce{H2}$ which can be combined with nitrogen to create ammonia-based fertilizer.

If the process is done at the well, propane could also be reverse-fed to wells in liquid form to use in place of water for fracking, after which the propane could be drawn off by lowering the pressure (such that it is then in the gas phase) and recompressed to liquid at $160 \ \mathrm{psi}$. This reduce water contamination and propane would not be lost.

In the liquid phase at $160 \ \mathrm{psi}$, propane has a volume $\frac{1}{250}$ as it does in the gas phase, and consequently propane would require a pipeline much smaller than natural gas for same volume sent.

On the receiving end, propane has a higher value for its ease of transport and storage as well as larger consumer base.

Gas to liquid technology is known, but propane isn't a good choice of end product

The idea of converting natural gas (which is mostly methane) into liquids is widely used not least because the earth has, inconveniently, located the best natural gas fields some way from the places that most want to use natural gas.

Right now (in 2016), though, it seems it is still not desperately uneconomical to convert the methane into a liquid by refrigeration and ship it to customers in the largest known thermos flasks (LNG tankers). Both the ships for transporting liquefied natural gas (LNG) and the plants to create it are very expensive (tens of billions of dollars for the minimal economic scale). There are something like 410 of these ships and there seem to be dozens of new units being built every year (see this report on the industry for more detail: warning pdf).

But there are some examples of plants that convert the local gas supply directly to more transposable liquids. Shell, for example, have developed a process that converts gas directly into clean liquid fuels (the initial plants produce very clean diesel, but other fuels are possible). See this description of the Shell process. Some other firms are operating similar processes.

So it seems that, right now, LNG is the best economical option. But if you are going to use a technology to go directly to something more transportable than methane, it is better to go the whole hog to a widely used fuel (diesel and petrol are both basically longer chain hydrocarbons than propane).

This may, of course, change.


ligand pi system irreducible representation for trigonal bipyramidal transition metal complex I am considering a D$_{3h}$ trigonal bipyramidal transition metal complex MX$_5$.  I have found that the irreducible representation for the ligand s and $\sigma$ systems is 2A$_1'$+A$_2''$+E$'$.  I have
\begin{array}{|c|c|c|c|c|c|}
\hline
&E& 2C_3 & 3C_2 & \sigma_h & 2S_3&3\sigma_v\\ \hline
 \Gamma_s&5& 2& 1&3&0&3\\ \hline
 \Gamma_\sigma&5& 2& 1&3&0&3\\ \hline
 \Gamma_{\pi+\pi'}&10&0&-2&0&0&0\\ \hline
\end{array}
However, I think that my $\Gamma_{\pi+\pi'}$ is incorrect since I get fractional values for coefficients for the irreducible representations when I try to reduce it to irreducible representations.  But from sketching the trigonal pyramidal complex, I cannot see how my $\chi(R)$ are wrong for the pi system representation, $\Gamma_{\pi+\pi'}$.  Any help would be appreciated.

I believe you have the wrong character for the $\hat{C_3}$ rotation. None of the p orbitals in the equatorial positions remain in place so they all contribute zero. The axial p orbitals are rotated by $120^°$ which means they each contribute $\cos(120^°)=-\frac{1}{2}$. Since there are two of them at each of the two axial position, $\chi(\hat{C_3})=-2$. Using this new value for the character (assuming all the others are right, which they seemed to be), I obtained: $a_2',a_2'',2e',2e''$


hydroxide ion in water -- How does the dynamic change? It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions. Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion. If I am correct at this point, how would the dynamics be different in these cases?

It's well known that $\ce{H+}$ ions in water always form $\ce{H3O+}$ ions.

This is not quite right. Because of the hydrogen bonding a wide variety of $\ce{H3O+(H2O)$_n$}$ ions are in dynamic equilibrium, where $n$ can take values as large as few tens. It has been shown that $\ce{H3O+}$ can donate three hydrogen bonds, so that it's first solvation shell of 3 water molecules, and the hydrogen bonds between $\ce{H3O+}$ and $\ce{H2O}$ are quite strong. This means that the $\ce{H3O+}$ cation can actually be considered as $\ce{H9O4+}$ in solution. For more details, see here.

  Likewise, $\ce{OH-}$ should exist always as $\ce{H3O2-}$ ion.

Reasoning by analogy is a very dangerous road. As mentioned by sheogorath there is no possibility for a coordinate covalent bond as in the case of $\ce{H3O+}$. But the hydrogen bonding is possible and the hydroxide ion is indeed strongly hydrated as well as $\ce{H3O+}$. The details of the hydration are not well known yet, but recent studies indicate $\ce{OH- (H2O)4}$ as the prominent complex in an aqueous solution. For details, see here.


Difference between voltaic cell in chemistry and battery in physics In chemistry's voltaic cell, the anode (negative electrode) of the battery supplies electrons to the cathode (positive electrode) of the battery.

In physics, the electrons flow from negative terminal (cathode) of the battery to positive terminal (anode) of the battery. I don't nderstands this because in thermionic emission, electrons flow from cathode to anode too through an electric source ( which i believe is a voltaic cell). I dont understand this. Please help. Thank you.

I think you've confused the flow of electrons with the flow of current. Before it was discovered that electricity is the result of the flow of negatively charged electrons it was proposed that electricity result from the flow of a positive charge. We called this current and this convention has remained to this day.

So current is effectively the flow of positive charge not the flow of electrons, which are negatively charged. So current flows from the cathode to the anode, but electrons flow from the anode to the cathode. This is true in both physics and chemistry.


Ring Formation and Alkenes I was consulting Wade and I came across this problem and solution:

While I agree with the solution manual, I can't help but wonder if my proposed mechanism is correct.

I had the water attack the hydrogens attached to the alkene. I knew that alkenes are slightly electron-withdrawing and this makes the hydrogens an appealing target for a base.

Once the hydrogens were cleaved, I had the remaining pair of electrons attack the carbocation. Is this a valid pathway? Valid as in plausible? Would you accept this mechanism on a test?

The main problem I see with this mechanism is that yield is exceedingly low as sp2 carbons are not especially acidic, and water is a poor base. On the other hand the original problem does specify that this product is minor. But then, it appears that I chose the more minor of two minor pathways.

Greg E.'s comments sum up the flaws with your version of the mechanism. While yours contains the same steps, it combines the nucleophilic attach and the proton transfer in a single step. Unless you have experimental evidence for a bimolecular step, the unimolecular step is much more likely. In this case, the unimolecular step is a 5-endo-dig cyclization, which is particularly fast. The alkene nucleophile is already present when the carbocation forms. The rate of the deprotonation is diffusion controlled, as we have to wait for a water molecule to find it.

Also, your logic relating the minority of the product to the unlikeliness of the mechanism is inherently flawed. A minor product does not occur at a lower proportion because the mechanism contains less valid steps. Instead, the distribution of products is related to the relative rates of different mechanisms and the relative energies of the products. That product is a minor product because some other product either forms faster or is more stable or both.

If this was on my exam, and I like the problem so I might use something like it, I would dock you a few points for combining steps that are not likely concerted into a concerted step. For example, if I decided this mechanism is worth 10 points, then you might receive 7 or 8.

  Final note: Remember that experimental evidence trumps all else in science.

Regardless of how intuitively and aesthetically unusual your mechanism appears, if experiments support it, then it becomes a more valid mechanism. As an example of some of the non-intuitive things that are in "accepted" mechanisms, check out halonium ions and sigmatropic rearrangements, especially the one in the Baeyer-Villager Oxidation. Evidence in support of your mechanism might include a failure to observe or trap that final carbocation, a demonstration that the rate depends on the amount of water (since final deprotonation steps tend to be very fast), and others.


Molecular Orbital Hybridization vs Delocalization for positive charge stabilization While studying Molecular Orbital Theory, I ran into a comparison of basicity between pyridine and piperidine. The latter was concluded to be more basic because of the sp3 hybridization of the positively charged nitrogen opposed to the sp2 hybridization of the pyridine nitrogen. The p orbitals are higher in energy, hence, the higher the p influence on the final hybrid orbital the bigger it will be in energy. Higher energy orbitals result in higher stability for positive charges.

On another moment, when reading about carbocation stability, it was stated that tertiary carbocations are more stable due to the delocalization of the positive charge.

Both statements regarding the positive charge stabilization makes sense, but here's where I started getting confused:

If positive charge delocalization plays a big role on carbocation stability, wouldn't the π-system of the pyridine help delocalize the positive charge through resonance? Wouldn't that compensate the effect of the orbitals lower energy (due to higher s orbital character)?
Considering the answer to the above question was that the orbital hybridization actually plays a bigger factor in positive charge stabilization than delocalization, then why alkynes like methylacetylene are attacked in the secondary carbon (which are sp hybridized in the carbocation form?) and not in the sp3 hybridized carbon? (Consider R = CH3 in the image below)

Could someone give me a light on those matters? I assume I might be making a confusion regarding the carbocation hybridization of methylacetylene, would the secondary carbon have a sp geometry when it's positively charged?

Basicity is the ability to donate a pair of electrons.
In the case of pyridine, the lone pair is in a sp2 orbital. This orbital does not participate in any resonance, and is smaller than the sp3. This means the electrons will be closer to the nucleus due to stronger attraction, making the pyridine a weaker base, since it's harder for it to donate these electrons.

Tertiary carbocations are more stable, that is correct, but you should look closely on what that means. A tertiary carbon is carbon atom that is to 3 other alkyl groups , not just any other atom. Alkyl groups are weak electron-donating groups, therefore stabilizing the carbocation, because they increase the electron density in the electron-poor center.
That is why according to Markovnikov's rule, the electrophile, in this case the proton, will bond to the less substituted carbon - in order to create a more stable secondary carbocation, which the nucelophile (the chloride) will attack.


Which is most likely the acidic proton in ascorbyl palmitate? The structure of ascrobyl palmitate contains a vinyl group, so I assume the red proton would be:

The other H (hydrogen) will be more acidic. The ascorbate anion is stabilized by electron delocalization, as I have show below in terms of resonance between two canonical forms.


Applying Henderson–Hasselbalch equation to amino acids: which pKa to use to calculate Z-/HZ ratio? I was wondering which $\mathrm{p}K_\mathrm{a}$ to use when calculating the ratio of $\ce{HZ}$ to $\ce{Z-}$ of amino acids, the Henderson–Hasselbalch formula used:

$$\mathrm{pH} = \mathrm{p}K_\text{a} +\log\frac{[\ce{Z-}]}{[\ce{ZH}]}$$

Because most amino acids have two $\mathrm{p}K_\mathrm{a}$'s, one for the amino group and one for the carboxyl group (and sometimes a third $\mathrm{p}K_\mathrm{a}$ for the side chain). For example, leucine:

$$\begin{array}{cc}
\hline
\text{Residue} & \mathrm{p}K_\mathrm{a} \\
\hline
\ce{NH3} & 9.74 \\
\ce{COOH} & 2.33 \\
\hline
\end{array}$$

Let's say I want to know the ratio of $\ce{HZ}$ to $\ce{Z-}$ for a $\mathrm{pH}$ of $1.00$ and $11.00$, which $\mathrm{p}K_\mathrm{a}$ should be used in the Henderson–Hasselbalch equation and why?

There are two approaches.

The two groups are treated as independent. You would calculate the ratio of protonated and deprotonated group one using the pKa for that group. Then, you would repeat that for the second group. Independent treatment would be a good approach, for example, to estimate the average charge of the hundreds of side chains in a typical protein that have a pKa near physiological pH.
The two pKa values are obtained experimentally from a titration.$$\ce{AH2 <=> AH- + H+ <=> A^2- + 2 H+}$$ Here, the $\ce{AH-}$ species could be ambiguous if the pKa values are close to each other (group 1 deprotonated and group 2 protonated or vice versa). Or, for a case like hydronium, water and hydroxide where the same atom can accept two protons, $\ce{AH-}$ would represent a single defined species. You can get the ratios of the concentrations from the pH and the respective pKa ($\ce{AH2}$ vs $\ce{AH-}$ via $K_1$, and $\ce{AH2}$ vs. $\ce{AH-}$ via $K_2$), and turn that into fractions of the total for each species. You can use that for a weighted average to calculate the average charge, for example.

  Let's say I want to know the ratio of HZ to Z− for a pH of 1.00 and 11.00, which pKa should be used in the Henderson–Hasselbalch equation and why?

This depends what HZ and Z- refer to. With two ionizable groups, there are really four species, $\ce{HZYH, HZY^-, ^-ZYH and ^-ZY-}$. If the pKa values are far apart, one of the mixed species would be very unlikely in aqueous solution (in your example, protonated carboxylate with deprotonated amino group). At pH 1.00, the amine would be 100% protonated, and the carboxylic acid would be mostly protonated (comparing its pKa of 2.33 to a pH of 1.00). At pH 11, the carboxylic acid would be 100% deprotonated, and the amine would be mostly deprotonated (comparing its pKa of 9.74 to a pH of 11.00).


Why is fabrication of ordered graphite substrates difficult? While reading this paper (suggested by @Janice here), I came across the following line:

  Surface reconstruction is known to be extremely important for Si, the material of  20th century electronics. However, surface reconstruction has been much less studied for graphite. One of the reasons might be graphite's limited use in electronics because fabrication of ordered graphite substrates is difficult.

(emphasis mine)

I've always believed that graphite isn't much used in semiconductor tech since it isn't a metalloid and conducts in an entirely different manner, probably making doping not work. This reason is new to me, but quite interesting.

I can sort of guess why "fabrication of ordered graphite substrates is difficult" — I'm not too sure though.

At any rate, I'd like to know more about it. Exactly why is such fabrication difficult?

Graphite is a semimetal, and no more mambojumbo than with metals, but it doesn't do anything with the problem.
Silicon/silica can be melted, purified, doped, can single crystals grown, can oxidize the surface, making the silicon semiconductor an silica insulator. By "can" I mean that engineeers know well tested methods, optimized for half a century.

Graphite cannot be reasonably melted, cannot grown to single crystals, cannot doped with similar techniques, highly anisotropic, so cutting maybe an issue, and when you make thin layers, it is hard to oxidize in a controlled way to make an insulator out of it. The resulted oxides are semiconductors, and there is a large variety of local structures that can appear with different electronic properties. By cannot I mean that no one really though about these problems until the last decade, so there might be equivalent solutions with e.g graphene, nanotube or other nanocarbon structures.


Avoiding snow melting with salt I've been quite stunned during the last Winter Olympics in Sochi by the following fact.

Temperatures were quite high, sometimes 10° Celsius, and the organizers had to do something to prevent the snow from melting. I heard on the TV the commentators saying that they were discharging salt on the tracks.

However where I live (France), we have occasionnally snow in the winter, and to keep the roads clear for the vehicles or the sidewalks clear for the pedestrians, city services also discharge salt !

How is it possible to have these two different effects with the (seemingly) same component ?

This article and this article (posted by @Jun-Goo Kwak in the comments above) say that the salt was used to lower the melting point of the snow on the surface, which would cause it melt. It would then re-freeze (I am assuming either at night or due to being in contact with snow beneath it), forming an icy layer on the surface. I don't know anything about Alpine skiing, but it seems that an icy surface is better than a slushy surface.

The ice would also form an insulating layer on the surface, which might help keep more snow from melting. Ice is a better conductor of heat than snow is (snow contains a lot of air, which is a terrible thermal conductor), so it is unlikely that it would act as an effective conductive insulator. However, it would shield the sub-surface snow from convective heat transfer caused by the movement of air. This is how igloos and snow caves work.

There could also be an effect due to the phase transition of melting - when ice melts, it is an endothermic transition, which means it absorbs heat. Some of that heat comes from the air, but some of it will also come from the snow directly below the surface. The result could be that the snow melts more slowly than it would if it were not covered by a layer of ice. The reverse of this effect is the reason that spraying an orange crop with water right before a hard freeze can save the crop (pg. 3) - as the water freezes, it releases heat (freezing is exothermic) and some of that heat goes into the orange trees.

It is hard to guess which effect is most important (insulation vs. endothermic heat transfer), but I would say that both should be considered.

So, to answer your question, in both cases the idea is to melt snow and/or ice, but in the case of roads, we want the ice to melt immediately and flow away, so we use lots of salt. In the case of Alpine ski events, we want a little snow to melt and then re-freeze, so we use a little salt.


How to determine the iron content in spinach with a spectrometer? I used a spectrometer to find the absorbency of 5 solutions with different iron concentrations.

\begin{array}{rrrrr}
  \ce{Fe (III)} / \pu{mol L-1} &
  \text{Absorbance} &
  V(\ce{HCl}) &
  V(\ce{FeCl3}) &
  V(\ce{KSCN}) \\
  \hline
  0.00005 & 0.194 & \pu{9 mL} & \pu{1 mL} & \pu{10 mL} \\
  0.0001  & 0.424 & \pu{8 mL} & \pu{2 mL} & \pu{10 mL} \\
  0.00015 & 0.674 & \pu{7 mL} & \pu{3 mL} & \pu{10 mL} \\
  0.0002  & 0.89  & \pu{6 mL} & \pu{4 mL} & \pu{10 mL} \\
  0.00025 & 1.113 & \pu{5 mL} & \pu{5 mL} & \pu{10 mL}
\end{array}

With this data I've constructed a graph and a line of best fit.
\begin{align}
y &= 4608x - 0.0322 &
R^2 &= 0.9994
\end{align}

I took $\pu{5.10g}$ of spinach leafs and burned them, then added $\pu{10.00 mL}$ of $\ce{HCl}$ and $\pu{10 mL}$ of $\ce{KSCN}$ then filtered the contents (so ashes weren't in the beaker).

The absorbency of this spinach extract was $0.015$.

My question is: How do I calculate how much iron is in this sample? I need to determine the mass of iron per $\pu{100 g}$ of spinach.

Using the line of best fit question, solve for $V(\ce{Fe(III)}) = x$ concentration:
\begin{align}
0.015 &= 4608x - 0.0322\\
x &= 0.0000102431
\end{align}

Convert $\pu{mol/L}$ to $\pu{g/L}$ to $\pu{mg/L}$:

\begin{align}
0.0000102431 \times 55.845 &= 0.00057202591\\
\pu{0.00057202591 g} &= \pu{0.572 mg}
\end{align}

But this is clearly wrong because it's not taking account the amount of spinach I started with ($\pu{5.10 g}$). Anytime I try to include the amount, either by multiplying the density my answer is way off.

Is it reasonable that I'm getting a number above $\pu{10 mg}$? The iron amount reported by the USDA says that there is $\pu{2.7 mg}$ of iron per $\pu{100 g}$ of spinach.

Can anyone guide me in the correct direction?

Here's how I approached the problem.

Plot the Data

Below is the plot I quickly hacked together, and a fit of the form $g(x) = a\cdot x$. (Note that I do want the line to go through (0,0) because the absorption at zero concentration should be zero. And it's a pity you didn't check that with a measurement... One never knows whether the calibration was successful.)

The parameter found was $a=4.43236~\mathrm{L\, mmol^{-1}} = 4.43236~\mathrm{mM^{-1}}$.

Calculate the Concentration of Fe Ions Present

You have measured an absorbance of 0.015 with a corresponding concentration of $c_\text{meas} = 3.38\cdot 10^{-3} ~ \mathrm{mmol\, L^{-1}}$.

The mass of iron in the solution is given by
$$ m_\ce{Fe} = c_\text{meas} \cdot M_\ce{Fe} \cdot V = 0.00378~\mathrm{mg} $$
with the volume $V=0.02~\mathrm{L}$ and molar mass $M_\text{Fe} = 55.845~\mathrm{g\, mol^{-1}}$.

Calculate Mass Percentage

The mass of iron in one gram of spinach is easily calculated:
$$ w_\text{Fe,spinach} = \frac{0.00378~\mathrm{mg}}{5.10~\mathrm{g}} = 7.4 \cdot 10^{-4} ~\mathrm{mg\, g^{-1}} $$

(I could just calculate away the units, but I was too lazy to do it here.)

The mass in $100~\mathrm{g}$ of spinach follows accordingly:
$$ m_\text{Fe,100g} = 7.4 \cdot 10^{-4}~\mathrm{mg\, g^{-1}} \cdot 100~\mathrm{g} = 0.0740~\mathrm{mg} $$


Boiling point elevation and liquid solutions $\pu{1 L}$ of an aqueous solution of urea having density $\pu{1.06 g mL-1}$ is found to have elevation in boiling point $\Delta T_\mathrm{b} = \pu{0.5 °C}.$ If the temperature of this solution is increased to $\pu{101.5 °C},$ then calculate the amount of water which must have vaporized by doing so. Ebullioscopic constant of water is $K_\mathrm{b} = \pu{0.5 K kg mol-1}.$

I tried solving this question using the Clausius-Clapeyron equation and the basic ideal gas equation. I calculated the change in the number of (gaseous) moles of water that would occur by raising the solution's temperature to $\pu{101.5 °C},$ and then multiplied it by its molar mass, hoping to get the mass of water thus evaporated.

Am I doing something incorrectly? My answer does not seem to match the answer given in my book at all. If yes, please help me out here.

Use the ebullioscopic equation (the first equation) in this Wikipedia article,

$$\Delta T = K_\mathrm{b} m$$

First solve for $m_\mathrm{init}$, the initial molality of urea. Second figure out at what molality $m_\mathrm{fin}$ the boiling point is elevated by $\Delta T = \pu{1.5 °C}.$ Since the amount of urea is constant, the % change in the mass of water is then given by

$$\% \left(\frac{\Delta w_{H2O}}{w_{H2O,\mathrm{init}}}\right)= \frac{m_\mathrm{init}^{-1}-m_\mathrm{fin}^{-1}}{m_\mathrm{init}^{-1}}$$

where $m$ is the molality of urea, $w$ mass solvent.


Proteins and their solubility I have 2 statements , and i am not sure, whether they are true or false.

Proteins like albumin and haemoglobin are insoluble in water

(in my view , almost all proteins are soluble- so this is a wrong  statement)

Proteins which are soluble in water have complex 3D structure and they are globular molecules. - (in my view I think this is also wrong)

Please explain these statements.

All proteins have complex structures. There are three broad classes: globular, membrane and scleroproteins. Globular proteins are generally water soluble while membrane and scleroproteins are not.

Albumins and haemoglobin are globular and water soluble.


How to compare the Heat of Combustion qualitatively? I have been doing practice problems in General Organic Chemistry. I got some questions that asked me to compare the Heat of Combustion of some molecules. I couldn't get any of those questions and the logic behind their answers. I tried googling but it didn't help. The search results were not direct. So, I'd like to know the factors on which the heat of hydrogenation depends. I am not looking for an answer that describes it quantitatively but rather qualitatively.

Here are some example questions:

  Which one of the following has the smallest heat of combustion?

The answer is (c). But why is that?

  Rank the following substances in order of decreasing heat of combustion (maximum $\rightarrow$ minimum)

The answer is 2>4>1>3. But again, why is that?

I knew that the heat of combustion depends on the number of carbon atom. But in these two  examples the number of carbon atoms are same. Is there anything I need to watch out for?

For isomers differences in the heat of combustion are due to differences in the internal energy of the educts. If intermolecular interaction (lattice energy etc.) can be neglected it reflects differences in the internal energy of the molecules.

In your first example C has the smallest heat of combustion since it has the lowest internal Van der Waals strain.

In the second example the order is given by the ring strain, and the Van der Waals strain between the two methyl groups at the cyclobutane ring.


Why does the acetyl group drop the pH? I have a molecule that is 8% of the R groups are replaced with Acetyl groups.  When I dissolve it in water it drops the pH.  I am trying to understand why.

Because the electrons in the double bond C=O help to stabilize the CH3 group after it has released an H+ (it is similar in this way to what happens with carboxylic acids), the Acetyl groups are acid. Hence the pH drop.


Find product of the oxidation of 2-methylnaphthalene with chromium trioxide This is a question from an organic chemistry book.
I did not understand the solution given in that book.
Here's the solution given in that book,

I know that CrO3 oxidizes methyl group of toluene, but here
Why are alpha positions more reactive and alpha positions of which functional group have they mentioned?

Fun fact: this synthesis is/was the commercial method used by BASF to make the product, menadione, which is often found as a supplement in animal feeds (patent DE092952709). Apparently the use of stoichiometric chromium is worth it.

Why are alpha positions more reactive? There are two potential questions here, I'll try to answer both.

1) Why does methylnaphthalene react differently from toluene?
The naphthalene ring is generally more reactive than benzene because reaction intermediates do not fully destroy aromaticity (see below image, left)

2) Why is the alpha position more reactive than the methyl group?
This question of reactivity is tricky because other oxidants and reaction conditions will indeed oxidize 2-methylnaphthalene into 2-naphthoic acid. In general, electron-rich naphthalenes are able to oxidize into naphthoquinones, even just by exposure to $\ce{O2}$. If we assume (big assumption!!!) that these reactions to go through radical mechanisms, then we are either oxidizing the naphthalene pi system (left) or abstracting an H* (right). There's various arguments to be made about which pathway is more stable based on the extent of conjugation, and obviously that will depend on the reagent and solvents in use as well.

Alpha positions of which functional group have they mentioned?

The alpha nomenclature they use here is actually referring to the two different positions in naphthalene:


Calculating the concentration of product in a solution based on the peak area of HPLC Please help me resolving the following:

A sample solution, $\pu{0.24mg/mL}$ Compound A, is exposed to heat to form Compound B. From HPLC data report I have peak area for Compound A and Compound B.  The  STD  concentration is 0.3 Compound A. The concentration of Compound A in sample solution is $\pu{0.20mg/mL}$ Compound A calculated using STD Compound A information. Which is the correct way to calculate concentration of Compound B?

My advisor in the project I'm participating insist I must calculate the concentration of Compound B since is a degradation of Compound A  in term of the amount lost of Compound A is the amount gained by Compound B (she call it a indirect method for calculating the concentration). I'm at lost! Originally I performed the calculation using a STD Compound A at the LOQ level of the impurity:
$$\text{[Compound B]} = \frac{\text{Area Compound B} \times \text{[STD Compound A]}}{\text{Area STD Compound A}}$$
The new approach is calculate the [compound A] using a STD at the level of the sample and then indirectly calculate the [Compound B].

Your advisor is making a pretty dangerous assumption (although sometimes pharma people do it). The assumption is that the impurity is structurally similar to the main compound and its UV-Vis spectrum is almost similar i.e., their extinction coefficients are the same.
The idea in your formula is that the calibration curve for A will be same as B. Well that might be a first order approximation when the impurity is known to be structurally similar.
If B is a known compound, you should buy a separate standard, and make a calibration curve.


Reactivity and transport safety of substituted methyl benzoates This was a practice question from an MCAT book whose answer I didn't really understand:

The normal reactivity of methyl benzoate is affected by the presence of certain substituents. Which of the following substituents will decrease methyl benzoate reactivity making it safer for transport?
a) $\ce{NO2}$
b) Hydrogen
c) $\ce{Br}$
d) $\ce{CH3}$

They give the answer as (a) because "in order to decrease the reactivity of  benzene molecule, an electron withdrawing deactivating substituent must be used." However, I thought the reactive part of the molecule would be the carboxylic acid, and so I would've expected (d) to decrease the reactivity of the molecule because it's electron donating.
I suppose if the question meant the reactivity of the ring itself, then the answer is (a) since electron withdrawing substituents on a benzene ring lower its reactivity.
So my question is: How would one reasonably infer that the reactive part of this molecule is the benzene ring itself instead of the carboxylic acid?

I’ve already addressed this in a comment, but now I went the lengths to check out the material safety data sheets of these compounds from Sigma Aldrich. My suspicions have proven correct: methyl benzoate is not a dangerous compound to transport so any recommendations to reduce its reactivity by adding a substituent must be moot.

In short, section 14 of the MSDSs as I access them contains transport information; and this entry for the compounds methyl benzoate (CAS 93-58-3), methyl 3-nitrobenzoate (CAS 618-95-1), methyl 2-methylbenzoate (CAS 89-71-4) and methyl 2-bromobenzoate (CAS 610-94-6) contains nothing but ‘not dangerous goods’ and the required iterations and variations of that theme to the different categories there.

Therefore, the question must be rejected as totally pointless: the starting compound is not considered a dangerous good for transportation and neither is any of the three possible substituted ones (one generic placeholder for each; I’m not going to check an additional six MSDS sheets).

Naturally, the only answer you can remove by deduction is b, because a hydrogen ‘substituent’ does not change the properties of the benzoate residue since hydrogen is already present there five times.

Considering we have found out that the compound is not dangerous to transport, how can we approach this question? Well, it’s hard. Concerning the reactivity of electrophilic aromatic substitutions — which is low anyway because of the deactivating ester substituent — of course an electron-withdrawing substituent would help; the nitro group suggests itself. On the other hand, concerning the reactivity of the ester group’s carbonyl function, only electron-donating substituents would help; the strongest being the inductive effect of the methyl group. So we are left with two possibilities and it is unclear which one we should follow.

Of course, neither the reaction with electrophiles nor the reaction with nucleophiles is inherently more dangerous than the other. We have again reached a vicious circle in which we cannot proceed at the risk of taking the wrong path.

Therefore, the best advice I can give is to ignore the question in the book. If it is given in an exam, challenge the exam for there is no reasonable answer to it.


How to determine which reagent to use H2; lindlar catalyst vs Na;NH3(l)? I do understand that H2; Lindlar will give a cis double bond and Na;NH3(l) will give trans double bond. However,  I am having difficulty understand how to determine which  reagent will work in this question between H2; lindlar catalyst and Na;NH3(l). The answer is given as H2;Lindlar. Second step is to add Br2 to the alkene. Then a strong base such as NaNH2 will deprotonate OH and the compound will undergo a Sn2 reaction to form the ring. The part I am having difficulty is  choosing between H2; lindlar catalyst vs Na;NH3(l).

Thanks for any help.

This is a really interesting question. I'm not sure I agree with the end product, but I will get to that in a second. There does appear to be a piece missing from this equation, which is a second step (something has to be responsible for the new bromine in the product).

First thing is to identify how to get to the final product from an intermediate. You've correctly ruled out $\ce{NaNH2}$ which is just a strong base. It's frequently used to deprotonate a terminal alkyne (which you do not have), and in this example, all it would do would be to remove the hydrogen on the alcohol.

You've also ruled out $\ce{H2, Pt}$. This is a standard hydrogenation and will reduce a triple bond fully to the alkane. If that happens, you won't have the functional groups to do anything like a ring closure.

That gives two possibilities, hydrogenation with Lindlar's catalyst or a sodium ammonia reduction. Both of these reactions reduce the alkyne to an alkene (and then stop). The difference is that concerted hydrogenation will add hydrogen in a syn fashion to afford the cis-alkene, whereas the sodium/ammonia reduction is not concerted and will give the more stable trans-alkene.

The cyclization happens via addition of bromine. This is a standard reaction: halohydrin formation. Except that the intramolecular opening via the attached alcohol will be fast, so the alcohol attacks instead of water to provide an ether. Assuming this is the mechanism, you have to think about how the relative stereochemistry of the alkene is carried forward.

The key intermediate is the bromonium ion (the three member ring with a bromine). The intermediate retains the stereochemistry of the alkene. You might need a model to identify this, but the correct bromonium ion that gives the indicated product has the hydrogens on the same side. That means that they were originally syn across the double bond, so the double bond was cis. Then Lindlar hydrogenation was the appropriate reaction for the first step.

Here's my issue with this question: I don't actually think you'll get this product. The technical point is that the opening can proceed via a 5-exo-tet attack or a 6-endo-tet attack. Generally, in these types of reactions, the exo attack is favored for stereoelectronic reasons. So my money is on obtaining the 5-member ring instead of the 6-member ring.


Which acid to choose when designing a world with acid seas? I'm currently writing a narrative and hoping to include a world which is home to a sea of some form of 'nasty' substance (doesn't have to be acid).

The varieties of inhabitants of this world have evolved mostly on the lands and eventually built their cities right up to the shorelines of this acidic sea. (In some cases I may even have cities that float and migrate across it).

I'm imagining some kind of scientific or industrial endeavor they perform whereby they've created ships that can withstand the chemicals (at least in short term) by lining their hulls with a type of ablative shielding.

Also desirable (although not essential) would be for the composition to not be immediately fatal. A carbon life form (or would they likely be silicate?) could fall in and maybe swim for a few minutes and leave with only some semi minor chemical burns. Yet prolonged exposure would most definitely be fatal.

I'm also wondering if the inhabitants could harness this chemical in some imaginative way such as for propulsion or agricultural/industrial means?

My question is, what kind of chemical/cocktail could I use to fit these kinds of parameters?

Coca Cola or a mixture of Coca Cola and Pepsi Cola would probably be a good choice.

It's not immediately fatal for carbon life forms that fall into it. Only prolonged exposure will cause digestion and damage to bones and teeth.

Due to the phosphoric acid ships with iron hulls would withstand corrosion due to passivation through the formation of iron phosphate (also a molluscicide to protect from shipworms).

In agriculture phosphate is also a potent fertilizer.

For propulsion purposes, Cola together with Mentos is a well-known hybrid rocket propellant.


If LPG gas burners can reach temperatures above 1700 °C, then how do HCA and PAH not develop in extreme amounts during cooking? According to What is the temperature of heat generated from LPG gas?, the temperature of (1) LPG gas is above 1700 degree Celsius.

Many of us are familiar with pan-frying meat (steak, pan-frying the chicken breast etc).

Next, consider this: when you're (2) cooking meat at high temperatures ("above 200 °C) then it'll start forming heterocyclic amines (HCAs) and polycyclic aromatic hydrocarbons (PAHs), according to healthline.com and precisionnutrition.com.

Note: The way they're formed is when fat drips onto the flame or the utensil, which is at very high temperature, and then reacts with the high heat to form HCAs and PAHs.

Lastly, there is research that (3) corelates HCAs and PAHs with causing cancer in our bodies.

Sources: They're are all mentioned in the two articles that I linked above.

Question: If you combine these 3 facts, then we must be cooking a huge amount of HCAs + PAHs. It's a wonder that we aren't dead yet! How is it that we haven't developed cancer by eating meat cooked over a pan on an LPG gas (which I believe is quite common)? Or are we all consuming HCAs / PAHs slowly over time and will develop cancer? Or are one of the 3 facts above wrong?

It is the temperature of the pan that matters not the flame

The flame temperature is irrelevant if you are cooking in a vessel. the only temperature you need to worry about is the temperature of the surface of the pan (or–even more importantly–the temperature of the meat). The surface of the pan will rarely get above around 220 °C if you are monitoring it. An unattended pan can get hotter, but, if you let it get too hot, your food will rapidly burn and will not be edible. Conversely, if the food is palatable, you probably haven't heated it enough to get lots of nasty HCAs and PAHs.

But the research that caused your worry is also frequently overstated. Yes, large quantities of PAHs or HCAs may be nasty, but the amounts in food–even food cooked on an open flame barbecue–are very small and there have been no convincing studies showing a notable effect on health. There are some studies showing  a very small effect of meat on health (diseases such as bowel cancer have been linked to some meats but the studies are statistically weak and the effects are very small. Moreover these studies link to meat not cooking products but would probably have spotted any effects based on nasty cooking by-products). It is worth remembering that people have been cooking on open flames since we invented fire. If that were really dangerous, there would be strong evidence of harm and/or primitive man would have developed good defences to avoid the harm (as milk-drinkers evolved lactose tolerance in adults because their diets consisted of a lot of dairy products).

There have been many food scares based on observations of known nasties in cooked food. Acrolein and Nitites, for example. None of these have been shown to have any notable measurable effect on people in the concentrations present in the diet.


Formation of water vapour during combustion When we burn fuel, water vapour is produced because hydrogen is burnt as well. But since the vapour is in gaseous state already, it shouldn't steal any heat from fuel gases, should it?

Water vapor produced during the burning of hydrogen-containing fuels should not steal any heat from the fuel, unless they condense on solid surfaces. In such cases, it may require a small amount of heat to evaporate them again. However, in my experiences, there is a stage during burning that there is a lot of water vapor on the solid object, and as the object is warmed up, there is no more vapor condensed.


Separating a powdered mixture of Zinc Oxide and Silver Oxide I have some powders that contain silver, $\ce{Ag2O, ZnO}$ and maybe some $\ce{AgO}$
Can I wash powders with $\ce{NaOH}$ to remove $\ce{ZnO}$ (as Sodium zincate) without any silver or silver oxides loss?

Your second suggest that you also want to save the silver oxides apart from zinc oxide. Sodium hydroxide won't help because it also reacts with silver oxides. So, you need a weak acid to selectively react with zinc oxide but neither silver oxide or silver. Hydrobromic acid will work because it reacts with zinc oxide to form zinc bromide which is highly soluble in water. leaving silver oxide and silver untouched. Silver bromide is also expected to form but hydrobromic acid is not strong enough to do that. Instead, if an alkali bromide salt was used, silver bromide would definitely form. From wikipedia article of zinc bromide:

  $\ce{ZnBr2 · 2H2O}$ is prepared by treating zinc oxide or zinc metal
  with hydrobromic acid.

  $$\ce{ZnO + 2 HBr + H2O → ZnBr2 · 2H2O}$$

  The dihydrate can be dehydrated by passing it over hot $\ce{CO2}$


Why is biphenyl not planar? I've seen in a lot of books that biphenyl is not planar. I don't know why that is because both the carbon are $\mathrm{sp^2}$ hybridized so it should be planar. Can someone explain it to me?

Biphynyls are a class of compounds with two phenyl rings attached by $\ce{C-C'}$ carbons in each phenyl ring (See Figure $\bf{\text{A}}$):

The phenyl rings can rotate around this $\ce{C-C'}$ bond (acts as an axis). Keep in mind that any molecule capable of rotating one part of it around fixed axis (a bond) is never planner. As a rule, if the rotational energy barrier is less than an activation energy barrier of $16$ to $\pu{19 kcal mol-1}$ at room temperature, the part of the molecule constantly rotate around the said axis. The experimental barriers for rotation about the $\ce{C-C'}$ bond of parent biphenyl (no substitution) are $\pu{6.0 \pm 2.1 kcal mol-1}$ at $\theta = 0^\circ$ and $\pu{6.5 \pm 2.1 kcal mol-1}$ at $\theta = 90^\circ$ with minimum energy barrier at $\theta \approx 44^\circ$ (Ref.1 & Ref.2; See Figure $\bf{\text{B}}$ for $\theta$). This values are less than required energy barrier to prevent rotation ($16$ to $\pu{19 kcal mol-1}$) at room temperature, thus, the biphenyl is rotate around $\ce{C-C'}$ axis (See Figure $\bf{\text{B}}$).

In addition, the rotation about $\ce{C-C’}$ can be restricted if each of the phenyl rings of a biphenyl has at least one same or different ortho substituent (e.g., 2,2'-dicarboxylic acid in $\bf{\text{C}}$). However, if these substituents are not large enough, it won’t prevent the rotation about $\ce{C-C’}$ and relevant conformers would not be able to isolate at room temperature (e.g., 2,2'-dicarboxylic acid in $\bf{\text{C}}$). If there are two different ortho substituents on each of two phenyl rings of a biphenyl (e.g., 6,6’-difluro-2,2'-dicarboxylic acid in $\bf{\text{C}}$), the rotation about $\ce{C-C’}$ would be not possible at lower temperatures since thermal (kinetic) energy is not sufficient to provide the necessary activation energy for racemization. As a result, the twisted $\theta = 90^\circ$ dihedral angle conformer becomes chiral. In order to inter-convert such conformers with their mirror image structures, a rotation through the higher energy coplanar form must be made. How easy this inter-conversion would be depends on the size of the ortho substituents. Unlike 2,2'-dicarboxylic acid, the two additional substituents in 6,6’-difluro-2,2'-dicarboxylic acid have a higher activation energy for racemization, and hence, can be resolved if care is taken to avoid heating them (Note: As a rule, an activation energy barrier of $16$ to $\pu{19 kcal mol-1}$ is required to prevent spontaneous room temperature racemization of substituted biphenyls). Since fluorine is smaller than a nitro group, 6,6’-difluro-2,2'-dicarboxylic acid racemizes more rapidly on heating than does the nitro compound (6,6’-dinitro-2,2'-dicarboxylic acid in $\bf{\text{C}}$). In fact, two conformers of 6,6’-dinitro-2,2'-dicarboxylic acid can be isolated at room temperature. Those conformational isomers that are isolable due to high energy barriers are called atropisomers.

For example, those conformational isomers have been studied comprehensively and computationally (Ref.3):

  Using B3LYP/6-311+G* and other methods, twist angles and torsional energies were obtained for biphenyl (BP), 2-halogen BPs, 2,2'-dihalogen BPs, and 3,3'-dihalogen BPs, the halogens ranging from $\ce{F}$ to $\ce{I}$. The results were compared with available gas phase and X-ray data. 2,2'-difluoro BP has a rotational double minimum, at $57.9^\circ$ and $128.9^\circ$, whereas the other 2,2'-dihalogen BPs have a single minimum at dihedral angles ranging from $84.9^\circ$ to $94.8^\circ$. All 3,3'-dihalogen BPs have a double minimum at about $45^\circ$ and $135^\circ$. Optimized twist angles and energy barriers were also calculated for 2,2'-dimethyl BP and for perfluoro as well as perchloro BP. Most structures are accounted for by steric effects. For 2,2'-dihalogen BPs, however, attractive forces also appear to play a role, as evidenced by the dihedral angle of 2,2'-dichloro BP lying well below $90^\circ$.

References:

Mercedes Rubio, Manuela Merchán, Enrique Ortí, “The internal rotational barrier of biphenyl studied with multiconfigurational second-order perturbation theory (CASPT2),” Theoretica chimica acta 1995, 91(1), 17–29 (https://doi.org/10.1007/BF01113859).
Otto Bastiansen, Svein Samdal, "Structure and barrier of internal rotation of biphenyl derivatives in the gaseous state: Part 4. Barrier of internal rotation in biphenyl, perdeuterated biphenyl and seven non-ortho-substituted halogen derivatives," J. Mol. Struct. 1985, 128(1-3), 115-125 (https://doi.org/10.1016/0022-2860(85)85044-4).
Friedrich Grein, “Twist Angles and Rotational Energy Barriers of Biphenyl and Substituted Biphenyls,” J. Phys. Chem. A 2002, 106(15), 3823–3827 (https://doi.org/10.1021/jp0122124).


How to create the total reaction out of the two steps behind denitrification? I can see that denitrification consists of processes involving mainly two steps:
Step 1 (nitrate to nitrite):
$$ \ce{3NO3- + CH3OH -> 3NO2- + CO2 + 2H2O} $$
Step 2 (nitrite to molecular nitrogen):
$ \ce{2NO2- + CH3OH -> N2 + CO2 + H2O + 2OH-} $$
Total reaction is supposed to be:

$$ \ce{6NO3- + 5CH3OH -> 3N2 + 5CO2 + 7H2O + 6 OH-} $$

Now I'm a bit rusty.
As far as I can remember, the right hand side on the first step of the reaction will be the starting point for the left hand side of step 2 in the overall reaction process.
In other words, what is present on the right hand side of step 1 and on the left hand side of step 2 can be subtracted from each other (which is logical). Also, the start of the total reaction would be represented by the left hand side of step 1 and the end of the total reaction would be represented by the right hand side of step 2.
The coefficients 5 and 7 seem very strange, but they balance everything so that makes sense. However the balancing of the Os seem to be the challenging part.

The attempt yields:
$$ \ce{3 NO3- + CH3OH -> N2 + CO2 + H2O + 2OH-} $$
Balancing N:
$$ \ce{6NO3- + CH3OH -> 3N2 + CO2 + H2O + 2OH-} $$
C and H atoms are already balanced.
However balancing O is more complicated because changing any of the Os affects the balance of the others (like C, H). Are there any recommendations for easing out that process or methods that can be applied in general?

After balancing carbon and nitrogen, I'd recommend you go for balancing the charge in the equation:

$$ \ce{6NO3- + CH3OH -> 3N2 + CO2 + H2O + 6OH-} $$

Now, we've to balance oxygen and hydrogen without disturbing the already balanced species. It would be interesting to note that here the stoichiometric coefficients of methanol and carbon dioxide should be the same or the number of carbon atoms wouldn't be equal on both sides of the equation.

So, let's assume the stoichiometric coefficient for both these two species to be $x$. Also, let's assume the stoichiometric coefficient of water on the right hand side to be $y$. The equation now looks like this:

$$ \ce{6NO3- + xCH3OH -> 3N2 + xCO2 + yH2O + 6OH-} $$

We now balance the number of oxygen atoms:

$$ 18 + x = 2x + y + 6 $$

And the number of hydrogen atoms:

$$ 4x = 2y + 6 $$

Solving these two equations, you get $x=5$ and $y=7$. And therefore your overall equation finally is:

$$ \ce{6NO3- + 5CH3OH -> 3N2 + 5CO2 + 7H2O + 6OH-} $$


Intermolecular Forces of Attraction Imagine a bottle 85% filled with water and the rest is just air, and the cap is closed. Here are now my questions.

Since the cap is closed completely, the air bubble is formed. Is that air bubble compressed? In other words, since air has the least force of attraction between its molecules, they are more free than the molecules of water. Now since water is creating a pressure against the air bubble, does it mean that the molecules in the bubble are now closely packed than usual?

Is it possible for the air bubble, if compressed a lot, to mix with water?

If the air bubble is able to hold itself even in water, why do we say that air has lesser force of attraction that water?

Assuming that by air bubble you're referring to the pocket of air now sealed in the bottle above the water surface:
Immediately after sealing the bottle, the air pocket is not at all compressed: it's at the same pressure as the surrounding environment (the air outside the bottle).

However, because you have a closed container, some water will evaporate, until an equilibrium is reached between the water vapor in the air pocket and the water in the liquid phase. This occurs when the chemical potential of the water in the two phases are equal.

Due to that vaporization, the total pressure in the air pocket (which now contains a higher amount of water vapor than the surrounding atmosphere outside the bottle) will be larger than the surroundings. This is a prominent effect when it's hot outside (have you ever left a sealed water bottle in a hot car?).

While this effect is usually small, as gases mix very well, it is non-zero and so yes, the air pocket is slightly compressed due to presence of water vapor.

As for whether the air pocket can mix with water: Assuming our container can tolerate such pressures and we were to externally compress it: liquids are much less compressible than gases, and so the air would (given high enough pressure) condensate, mixing into the liquid phase (but you wouldn't achieve this by hand!)

To address your final point: let's say we invert the bottle, so the air pocket travels as a bubble through the liquid phase to re-emerge at the other end of the bottle. One could ask, "why doesn't the air bubble fragment into little bubbles as it progresses upwards?"

To be fair, you do see a few bubbles splinter off, but for the most part, the bubble is intact. The air bubble stays largely intact due to the viscosity of water (the intermolecular attractions between water molecules is larger than the entropic benefits of mixing with the air bubble). So, to answer your question, the fact that the air bubble doesn't mix is not due to its intermolecular attraction, but rather it's due to water's intermolecular attraction.


How does an electron identify whether to absorb a photon of a certain wavelength or not in absorption spectrum? How would an electron identify the specific photons of a certain wavelength which are to be absorbed in a Hydrogen atom while creating an absorption spectrum? Well, I was thinking of it like this-
Way-01: The electron will absorb any photon coming towards it regardless of its wavelength and jump. Then if the wavelength is the exact amount required to jump from one shell to another, then the electron simply jumps to the higher shell. Otherwise, it cannot find a shell and since electrons cannot exist in between shells, they simply re-emit that photon and we perceive the incident as if the photon weren't absorbed at all.
Way-02: The explanation given above is false and electrons choose the photons via some other mechanism
Now, my question is- is the "Way-01" correct? Or do electrons identify the photons via some other way? If so, then what is it?
Note: I'm only a high-schooler, so any relatively simple qualitative explanation will be greatly appreciated.

[OP..] How would an electron identify the specific photons of a certain wavelength which are to be absorbed in a Hydrogen atom while creating an absorption spectrum?

As discussed in the comments, this is an observed phenomenon. We can try to describe this observation in a way that we can make predictions, but that does not answer the questions "how?".

[OP..] Otherwise, it cannot find a shell and since electrons cannot exist in between shells, they simply re-emit that photon and we perceive the incident as if the photon weren't absorbed at all.

There are processes that do something similar as the one you describe, but they involve observable events. When an atom scatters light, it scatters light of the same wavelength as the incoming light, but in all directions. There are also processes where the light is absorbed and then emitted again after a while (sometimes at different wavelength). What you suggest is an explanation for observing no interaction between the electron and the photon, and explaining it with an interaction (absorption and instant re-emission at the same wavelength in the same direction) that we can't observe.
You could have explanation models of the same flavor for many things that don't happen. For example, on a day it does not rain, I could say drop form in the clouds but instantly are destroyed again, so we can't observe them ever.

Way-02: The explanation given above is false and electrons choose the photons via some other mechanism

As you can read in the comments, using the term "choose" for something electrons do is not so helpful. We know under what conditions the photon might get absorbed (not all of them are), but to go beyond that and explain the "how" or even "why" is difficult to impossible (and not taught in e.g. the 4-year college chemistry curriculum, as far as I know).


How can hybrid orbitals be equivalent? How can hybrid orbitals be equivalent? Take, for example, $\mathrm{sp^3}$ hybridization. The wave functions of each hybrid orbitals are given by:

\begin{align}
\psi_1 &= \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\
\psi_2 &= \psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\
\psi_3 &= \psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\
\psi_4 &= \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}\\
\end{align}
No one of them has same linear combination. So, how can they be equivalent? I am not getting the sense.

The standard set of orbitals (s, p, d, f, ...) form an orthogonal basis set, such that you can describe any continuous function as a linear combination of them.  Now, this also means that you can take some subset of them, and make a new set of orthogonal functions, by appropriate normalized linear combinations, that will contain the full symmetry of the subset chosen. The $sp^{3}$ hybridization is just such a set of combinations of $s$ and $p$ orbitals, chosen to make a set of equivalent orbitals emphasizing a mixed symmetry.

Note that the standard orbitals are not the only orthogonal basis set that exists - it just happens to be the one that nicely gives solutions to the Schrodinger equation separating out $n$ and $l$. One could use spherical Bessel functions, or any number of other orthogonal basis sets, but it would not be as clean of a solution just because of the form of the differential equation under consideration.


What gas are the bubbles that evolve in the redox reaction of copper(II) chloride and aluminum? When aluminum (foil) is placed in an aqueous solution of copper(II) chloride, the aluminum atoms lose electrons to the copper(II) ions and replace them to form aluminum chloride, and to free up metallic copper.

The metallic copper has a tendancy to float on the surface of the solution, being buoyed by gas bubbles. The bubbles appear to evolve from the interface of the aluminum foil with the solution. What is the source and nature of the gas bubbles?

I have two hypotheses:

The bubbles are coming from dissolved $\ce{CO2}$ that is coming out of solution due to an increase in temperature close to the site of reactions.

The bubbles are $\ce{O2}$ bubbles coming from the layer of aluminum oxide on the surface of the aluminum foil.

If this is not the case, what happens to the portion of the portion of the foil that was in oxide form prior to the reaction.  How would the aluminum oxide react with copper(II) chloride?

The gas bubbles are very likely just air that was dissolved in the water previously and now gets released. I would expect most of the bubbles to actually be nitrogen, followed by oxygen, argon and then some carbon dioxide — that is assuming you used deionised water which should not have significant amounts of carbonate and hydrogencarbonate dissolved.

The copper particles merely supply suitable nucleation sites for the gas.

The theory on aluminium oxide being oxidised to pure oxygen is not correct. Oxygen is a very oxidative element meaning that you require strong oxidants to oxidise oxide anions back to oxygen. With ‘strong’, we’re talking the oxidative strength of potassium permanganate. Aluminium’s oxidative powers are nowhere near that strong.

Copper(II) chloride and aluminium oxide do not react with each other; only metallic aluminium beneath the passivisation layer is able to react with copper(II) ions.


Wohler synthesis: magnesium cyanate + ammonium chloride? Wohler famously produced urea by reacting silver cyanate with ammonium chloride,

AgNCO + NH$_4$Cl $\rightarrow$ (NH$_2$)$_2$CO + AgCl

Naive question: Could this work instead with magnesium cyanate Mg(OCN)$_2$ ?

(I ask because my undergrad chem lab has HCl, Ammonia, and magnesium strips, and I'd like to make some urea, if possible. Not sure how to make magnesium cyanate, but one step at a time...)

The reason Wohler used silver cyanate is because silver chloride precipitates out, leaving ammonium cyanate, which (in a separate step) was thermally converted to urea. If you used magnesium cyanate, there would be no precipitate since magnesium chloride is very soluble in water. However, if you evaporated the water and heated the mixture, you probably would end up making some urea. But it would not be pure.


Fischer indole synthesis: significance of choice of acid catalyst Lately, as a matter of interest, I have been studying sigmatropic rearrangements. I came across the Fischer indole synthesis, whose Wikipedia article states:

The Fischer indole synthesis is a chemical reaction that produces the aromatic heterocycle indole from a (substituted) phenylhydrazine and an aldehyde or ketone under acidic conditions. ... The choice of acid catalyst is very important. Brønsted acids such as $\ce{HCl}$, $\ce{H2SO4}$, polyphosphoric acid and p-toluenesulfonic acid have been used successfully. Lewis acids such as boron trifluoride, zinc chloride, iron chloride, and aluminium chloride are also useful catalysts for this reaction.

The significance of the choice of acid catalyst to be employed in the reaction is beyond my scope of knowledge. Browsing available articles, it would seem that the [3,3]-sigmatropic rearrangement is stereoselective. In addition, if I am not mistaken, regioselectivity of the Fischer indole synthesis could be increased depending on which acid catalyst was used, such as methanesulfonic acid. However, I was still unable to understand the overall importance of using an appropriate acid catalyst.

I'd like to take this opportunity to give a nutshell review of the Fischer indole synthesis for benifit of future readers. This highly useful synthetic tool, which was named after the famous chemist, Emil Fischer who discovered it in 1883 (Ref.1 & 2)), is considered to be one of the best methods for preparing indoles. Hence, needless to say, it is the most widely used and versatile method for indole synthesis. The method simply converts converts ayrlhydrazones ($\bf{1}$) of an aldehyde or a ketone into corresponding indole ($\bf{8}$) in the presence of an acid catalyst (Ref.3 & 4). The overall reaction is depicted in the top box in following image:

The proposed mechanism of The Fischer indole synthesis is also included for the convenience. As indicated, ayrlhydrazones ($\bf{1}$) are formed from a condensation reaction between an arylhydrazine and an aldehyde or ketone (reference to 2,4-dinitrophenylhydrazine test for aldehydes and ketones). In the presence of acid (Brønsted or Lewis), imine $\bf{1}$ is in equilibrium with $\bf{2}$ (protonated imine), $\bf{3}$ (enamine form), and $\bf{4}$ (protonated inamine). Once formed, $\bf{4}$ undergoes [3,3] sigmatropic rearrangement (ene-reaction; could visualize as azo-Cope rearrangement), forming a new $\ce{C-C}$ bond to give $\bf{5}$ (this is a key step of the mechanism). Imine $\bf{5}$ gains back aromaticity to form $\bf{6}$, which is the most stable isomer between two. The anilinic $\ce{NH2}$ of $\bf{6}$ attacks electron deficient imine group in the side chain ($\ce{C}$ of $\ce{CNH2+}$ group) intramolecularly to give the corresponding dihydroindole derivative, $\bf{7}$ (which is also another key step), which eventually release $\ce{NH4+}$, $\ce{NH3 + H+ <=> NH4+}$, to give stable indole derivative,  $\bf{8}$. Reference 3 has given excellent review on the mechanism and hence, for benefit of interested reader, the PDF has been attached.
The key feature to produce corresponding indole is need of a $\ce{CH2}$ group as noted (as indicated by a green circle; $\alpha$ to the original ketone or aldehyde). If this $\alpha$-$\ce{C}$ is $2^\circ$, then $\bf{7}$ would still release $\ce{NH3}$ to give corresponding cyclic imine derivative (Ref.5). For example, compound $\bf{10}$ was formed ( Ref.5) when $2^\circ$ $\alpha$-$\ce{C}$ is attacked selectively at $\bf{3}$ to $\bf{4}$ conversion (thermodynamically stable enamine):

However, except for the situation to make $\bf{10}$, and the intention is making indole derivative of interest, the stereo selectivity at sigma-tropic rearrangement to give $\bf{5}$ and thereafter would not matter since at the end, all of stereocenters involved would be diminished (see the scheme above). Therefore, the choice of acid to get selectivity in general synthesis of indoles is not as important as it was told elsewhere. In reality, a variety of catalysts (Table I in Ref.3) have been used to effect the indolization of arylhydrazones. According to the Ref.3 (an excellent review of the Fischer indole synthesis):

Generally speaking, it may be said that acids and many compounds capable of complex formation with organic ligands may act as catalysts.

In addition to the variety of catalysts listed in Table I of Ref.3, there are wide variety of other acid catalysts including organic acids such as tartaric acid (e.g., Ref.5) and p-toluenesulfonic acid (e.g., Ref.6). The bottom line is the indole formation by this method is independent of acid catalysis (e.g., Ref.7), although it is markedly accelerated by it.
It is also worth noting that this reaction has proven very useful for the pharmaceutical synthesis of antimigrane drugs (e.g., Ref.8 & 9).

References:

Emil Fischer, Friedrich Jourdan, "Ueber die Hydrazine der Brenztraubensäure," Berichte der deutschen chemischen Gesellschaft 1883, 16(2), 2241-2245 (https://doi.org/10.1002/cber.188301602141).
Emil Fischer,  Otto Hess, "Synthese von Indolderivaten," Berichte der deutschen chemischen Gesellschaft 1884, 17(1), 559-568 (https://doi.org/10.1002/cber.188401701155).
B. Robinson, “The Fischer Indole Synthesis,” Chem. Rev. 1963, 63(4), 373–401 (https://doi.org/10.1021/cr60224a003)(PDF).
B. Robinson, “Studies on the Fischer indole synthesis,” Chem. Rev. 1969, 69(2), 227–250 (https://doi.org/10.1021/cr60258a004).
Sangram Gore, Sundarababu Baskaran, Burkhard König, “Fischer Indole Synthesis in Low Melting Mixtures,” Org. Lett. 2012, 14(17), 4568–4571 (https://doi.org/10.1021/ol302034r).
Seble Wagaw, Bryant H. Yang, Stephen L. Buchwald, "A Palladium-Catalyzed Strategy for the Preparation of Indoles:  A Novel Entry into the Fischer Indole Synthesis," J. Am. Chem. Soc. 1998, 120(26), 6621-6622 (https://doi.org/10.1021/ja981045r).
Okiko Miyata, Yasuo Kimura, Kanami Muroya, Hajime Hiramatsu, Takeaki Naito, "Thermal cyclization of N-trifluoroacetyl enehydrazines under mild conditions: A novel entry into the fischer indole synthesis," Tetrahedron Lett. 1999, 40(18), 3601-3604 (https://doi.org/10.1016/S0040-4039(99)00583-3).
Paul R. Brodfuehrer, Bang-Chi Chen, Thomas R. Sattelberg, Patrick R. Smith, Jayachandra P. Reddy, Derron R. Stark, Sandra L. Quinlan, J. Gregory Reid, John K. Thottathil, Shaopeng Wang, “An Efficient Fischer Indole Synthesis of Avitriptan, a Potent 5-HT$_\mathrm{1D}$ Receptor Agonist,” J. Org. Chem. 1997, 62(26), 9192–9202 (https://doi.org/10.1021/jo971368q).
Majid M. Heravi, Sahar Rohani, Vahideh Zadsirjan, Nazli Zahedi, “Fischer indole synthesis applied to the total synthesis of natural products,” RSC Adv. 2017, 7, 52852-52887 (DOI: 10.1039/C7RA10716A).


Significant figures in pH calculations: does pH 0.30 count as 2 or 3 significant figures? Just say I wanted to give a pH calculation to three significant figures.
Eg: the $\mathrm{pH}$ of a $0.500 M$ solution of $\ce{HCl}$ is $-\log(0.5) = 0.3010299957...$

Would 3 s.f. be $0.301$ or $0.30$?

I know normally when zero comes before a number, it isn't considered a significant number (i.e $0.00034$ and $0.34$ both have two significant figures). But since pH is a scale, can I treat $0.30$ as three significant figures?

Short Answer:

  The number of significant figures in the mantissa of a value expressed in scientific notation equals the number of significant figures to the right of the decimal in the logged value. So in your example, assuming the likely error is the last digit, I'd use two.

Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 2 significant figures, $\pm$ error in the last.

Additionally, just because pH is a scale doesn't mean you can treat $0.30$ as a number with three significant figures, it still has two.

The general principles are given below.
Concepts like this are best illustrated with an example.

Consider, a random number $0.0000273$ Let us write this in scientific notation for the sake of clarity, $0.0000273 = 2.73×10^{–5}$ We have three significant figures.

Now take the log on both sides

$\log(0.0000273)= \log(2.73×10^{–5}) = \log(2.73) + \log(10^{–5})$

Now let us examine the two terms separately,

First, $\log(2.73) = 0.436$: the answer has three significant figures, reflecting the possible error in the last digit of $2.73$

$\log(10^{–5}) = -5$  This is an exact number and has infinite significant figures, i.e there is no error.

Going back, $\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000...$

In this situation the rules for significant digits for sums, dictate that we can only add to the same decimal place as the value with the least significance. In this case, that is the third decimal place from 0.436, so

$\log(0.0000273)= \log(2.73) + \log(10^{–5}) = 0.436 -5.0000... = -4.564$

$\log(0.0000\overline{273}) = -4.\overline{564}$  (the overline indicates the 3 significant figures)

End of example. Now back to your question, you have $[\ce{HCl}]=0.500$ assuming, the error is in the last digit, I'd recommend using 3 digits after the decimal, $\pm$ error in the last. So, $0.301 \pm 0.00\epsilon$


Estimating stability and mean life of anomalous tautomers A pair of bases in the DNA, say, A-T, have a tautomerized form A*-T* (resulting from switching the sides of both protons along the hydrogen bridges).

I have studied how, by means of DFT computations, one can compute the energy difference between the normal and the anomalous forms. But, another point of interest is the stability of the tautomer.

I have read several times that those anomalous tautomers "decay" to the normal A-T form very quickly (much quicker than the time it lasts a round of DNA replication), which is interesting because it implies that tautomerization is not likely to be a source of spontaneous mutations.

What I would like to learn is how can one compute this expected life-time of the anomalous tautomer

I have an idea (which I don't know if it could be rigth or if maybe is completely worthless), which involves making a Born-Oppenheimer separation for the two protons (''after'' the DFT computation of the electronic ground state) and then study the non-adiabatic transitions between the different vibronic states of the couple of protons. I am happy to provide details, if someone whises so.

But perhaps "my method" is too messy, or naive (or both) and / or there are other standar approaches to compute approximately the expected mean life of anomalous tautomers such as the mentioned A*-T*.

I would be very grateful if someone could apport some insigth and/or references into how are these expected life-times calculated typically in computational chemistry!

TLDR: With a quantum/semi-classical transition state theory.
I order to determine the lifetime of the tautomer, you need to know the rate constant for the decay from the tautomer to the conventional dimer.  This is a kinetics question rather than a thermodynamic question so you won't be able to answer this by comparing single point DFT energies.  The rate will also vary with temperature.
A complicating factor is that the nuclei moving are protons in hydrogen bonds.  Hydrogen bonds are frequently double well potentials with two minima, one near each end of the bond - in this case the tautomer and conventional dimer.  In these case the physical distance between the minima is small and the proton may tunnel through the potential barrier between the two minima - hence nuclear quantum effects are likely to be important, at least at low temperatures.
Methods that can calculate nuclear quantum effects can be very computational expensive and may require you to work in a reduced dimensional represention - only considering the two protons quantum mechanically and potentially simplifying the surrounding structure.
Two potential methods that could calculate a rate taking account of nuclear quantum effects are  semiclassical transtition state theory, or ring-polymer molecular dynamics which has been applied within pockets of larger systems such as proteins.


Does double replacement reaction only happen to aqueous reactants? I was taught that double displacement reactions occur when reactants are aqueous. However, I came across this reaction

$$\ce{ZnO(s) + H2S (g) -> ZnS(s) + H2O (g)}$$

This is the reaction that is used to remove hydrogen sulfide that is naturally found in methane gas. This reaction indeed looks like a double replacement reaction. However, the reactants are not aqueous.

Additionally, this reaction happens at the temperature of $\pu{700 K}$. Does that mean if enough energy is given, double replacement reactions can occur without an aqueous medium? How so?

Source 1, Source 2

Generally, reactions occur with liquids and gases (which encompasses solutions). This is because there is the fundamental problem of mass transport. When reactants react, they leave product in the wake of the reaction, which isolates the reactants from each other. For fully solid reactants, it means that more reactants have trouble finding each other. When reactions take place in liquid or a gas, there is the prospect of diffusion, which allows more reactant to find each other.

In the reaction you point out, hydrogen sulfide is a gas so it can easily diffuse to find more zinc oxide. The product zinc oxide might not form a perfect coating (i.e., it has holes in it) that allow the hydrogen sulfide to diffuse through this product to find more zinc oxide.

Other examples: when you burn a piece of wood, this is a solid reacting with a gas. But the combustion reaction is actually between vaporized wood and oxygen. The presence of a gaseous fuel is what allows good mixing between it and oxygen to produce a flame.

Finally, an example from a chemistry demonstration: solid barium hydroxide and solid ammonium chloride are mixed together and shaken. They reaction is quite endothermic and can freeze water on the outside of the reaction vessel. The trick here is that the barium hydroxide is usually the octahydrate. The hydrate provides enough water to create small amounts of solution for the reaction.


What is the effect of oxygen in welding? Welders use a shielding gas to avoid oxygen interact with the electrode. I supposed is might be because of ferric oxide formation as the electrode is steel-made. But fusion temperature of ferric oxide is quite the same than iron so I'm not sure what is the reason for the shielding gas (usually argon). I have gone through some works but found nothing about it.

Any idea what is the effect of oxygen? Is there any reaction involved which plays a role against a good welding or just physical issues?

According to https://www.bernardwelds.com/mig-welding-shielding-gas-basics-p152080 the purpose of shield gas in steel welding is:

  ...to prevent exposure of the molten weld pool to oxygen, nitrogen and hydrogen contained in the air atmosphere. The reaction of these elements with the weld pool can create a variety of problems, including porosity (holes within the weld bead) and excessive spatter.

There is also a picture of a weld that either didn't use a shield gas or didn't have enough. It's pretty obvious that such a weld isn't going to be very strong.

Edit: From this Wikipedia article explaining the purpose of inert gas in welding:

  Their purpose is to protect the weld area from oxygen, and water vapour.

And in what appears to be a bit of a contradiction:

  Oxygen is used in small amounts as an addition to other gases; typically as 2–5% addition to argon. It enhances arc stability and reduces the surface tension of the molten metal, increasing wetting of the solid metal. It is used for spray transfer welding of mild carbon steels, low alloy and stainless steels... Oxygen causes oxidation of the weld, so it is not suitable for welding aluminium, magnesium, copper, and some exotic metals.

When you consider that atmospheric oxygen is almost 21%, the two to five percent used in inert gas welding seems pretty small. However, though I can find no definitive reference, I'd have to say that the prevention of oxides of the metal(s) being joined is a consideration only when extremely reactive metals are being welded. Instead, it seem that the prevention of gaseous materials (probably mostly water vapor) within the weld seem to be of the most importance. Also, considering the temperatures involved in welding, the creation of oxides of nitrogen from atmospheric nitrogen and oxygen would have to be a concern.


Why do lone pairs of electrons have a higher repulsive force? I was just wondering why lone pairs of electrons are more repulsive than bonded pairs of electrons.

I have read that lone pairs have orbitals which are shorter and more round, compared to bonded pairs. I assume, therefore, that this difference in repulsive force could be due to the increased surface area that these orbitals provide, which would cause more repulsive force to be directed towards electrons.

Of course this is just a presumption that I made, which is probably incorrect.

Bonding pairs are stabilized between two atoms. Since there's no atom on the other end of a lone pair, it spreads out more than would be if it were in a bond. Lone pairs therefore repel more because the charge density is spread out more, that is, it's bigger.

  I have read that lone pairs have orbitals which are shorter and more round

This is technically true in the sense that you'd want to have higher $s$ character in a lone pair orbital. But you're comparing possible ways to hybridize orbitals for lone pairs, as opposed to comparing lone pair atomic orbitals with bonding orbitals.


Character table and symmetry operations It's been a while since I've dealt with these, but do character tables include all possible symmetry operations?
I am looking at the $D_\mathrm{\infty h}$ point group table and it lists $C_\infty$, $\sigma_\mathrm{v}$, $S_\infty$, $i$, and $C_2$. But doesn't it also have a $\sigma_\mathrm{h}$? Why is this not included?
Am I missing something here?

$S_{\infty}$ is equivalent to $\sigma_\mathrm{h}$. $S_{\infty}$ means rotate 0° then reflected by $\sigma_\mathrm{h}$, which is just same as taking $\sigma_\mathrm{h}$ itself.