How to get force constants in internal coordinates in Gaussian? I have tried quite a few different ways to get Gaussian to print the force constants in internal coordinates. If the size of the molecule is small then it prints both the internal hessian and Cartesian hessian in the log file on using Iop(7/32=5), it outputs only the Cartesian hessian without the Iop command. But if the size of the molecule is large then it stops printing the hessian in the .log file. Though the Cartesian hessian is always printed in the checkpoint file. How do I get the internal hessian? I am also using #p to get Gaussian to write everything in the log file.

You have to use freq=intmodes and use the command

formchk -3 xxxx.chk xxxx.fchk

to convert the binary checkpoint file to a formatted checkpoint file, which consists of text. The -3 requests the version 3 of formatted checkpoint files, which support all features of Gaussian 09 and 16.

Source: Ruixing Wang's answer on Researchgate, where I asked the same question.


Mutually immiscible liquids How many mutually immiscible liquids can be stored in a glass bottle at S.T.P.?

According to

  PATTLE, R. E. (1950). Systems of Mutually Immiscible Liquid Layers. Nature, 165(4188), 203–204.

a system of 6 immiscible phases can be formed (including mercury), but the ammonia layer disappears after a month. They suggest gallium could add another layer, giving:

ammonia solution
isobutyl alcohol
liquid paraffin
potassium carbonate solution
methylene iodide
gallium
mercury

Could silicone oil and fluorocarbons create additional stable layers, and have there been any other systems suggested since 1950? I saw a 2006 reference with allegedly 8 phases.

Better than gallium, could the alloy Galistan be used? It’s a mixture of Gallium, indium and tin. Other liquids like Bromine, Cesium and NaK might have some potential but may just react explosively.


Limitations of Henry's law Consider the diagram

Henry's law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid. The proportionality factor is called Henry's law constant.

Suppose the weights on the piston are so increased such that the pressure of the gas becomes numerically equal to Henry's law constant.
Then Henry's law gives that the mole fraction of the gas in solution is equal to 1.
Mole fraction of solute in solvent being equal to 1 implied that there is absolutely no solvent.
But in the situation described by the diagram, the solvent molecules are always present.
How is this so? Perhaps there is a limitation to Henry's law which has not been mentioned?

(Source: chem.libretexts.org)
In the image, we see that for a "real gas" (indicated by non-ideal solution), Henry's law is applicable for $\ce{\chi} << 1$.
As, $$\ce{p = K_H \cdot \chi}$$ hence the pressure should be low.
Limitations of Henry's law:

The pressure should be low and the temperature should be high i.e the gas should behave like an ideal gas.

The gas should not undergo compound formation with the solvent or association or dissociation in the solvent.


Can 1 mole of H2 react partially with one mole of O2? I am currently studying stoichiometry (high school level) and we came across the following equation in school:

$\ce{2H2 + 1O2 -> 2H2O}$

which is clear to me. We learned in school that reactions in chemistry only happen if there are enough atoms/molecules to react (so every molecule/atom can react), therefore the following reaction won't do anything:

$\ce{1H2 + 1O2 -> nothing happens}$

So now my first question: Why does the following not happen:

$\ce{1H2 + 1O2 -> 1H2O + 1O}$

And, as a follow up question, why does the following not work?

$\ce{H2 + O2 -> H2O + 1/2O2 (Left) }$

What do I do wrong? (maybe I simply misunderstood our teacher)

Are there any explanations for this? I'm especially interested in the answer of my example with $1$ mole of $\ce{H2}$ and $1$ mole of $\ce{O2}$. According to our teacher nothing will happen at all. I would assume that at least all $\ce{H2}$ molecules react with half of the oxygen molecules to $\ce{H2O}$ - leaving half a mole of $\ce{O2}$ left over.
(otherwise this would mean that all molecules "know" each other so they are "aware" that not all molecules/atoms can react)

Can somebody help me? Or point out what I am doing wrong?

As you have said, you are studying stoichiometry at High School Level. From this I can guess, that you have probably not studied the concept of Limiting Reagent yet.
What is Limiting Reagent?

In a chemical reaction, the limiting reagent, also known as the "limiting reactant", is the substance which is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent since the reaction cannot proceed further without it. The other reagents may be present in excess of the quantities required to react with the limiting reagent.

As per your question, $\ce{H2}$ is limiting reagent. Hence it will get totally consumed when reaction happens and 1/2 Moles of $\ce{O2}$ will be left.

1 mole of H2 and 1 mole of O2 won't make 1 mole of H2O and 1/2 mole O2.

1 mole of $\ce{H2}$ and 1 mole of $\ce{O2}$ will make 1 mole of $\ce{H2O}$ and 1/2 mole $\ce{O2}$ will be left.
$\ce{H2 + O2 -> H2O + 1/2O2 (Left) }$
And here is an advice for you. If your teacher is saying that this will not happen at all, tell him to be sure that reactant in reaction are single molecules, not 1 mole.
Read this to be more clear on concept of limiting reagent.


Sulfuric acid symmetry point group On the following page: https://cccbdb.nist.gov/pglist.asp It appears that the point symmetry group of sulphuric acid is $C_2$. I know that $C_2$ means that there is an axis of symmetry of order 2 (180 degree rotation).
But, I am not able to visualize it in the molecule.

In fact, you can see $C_2$ and other symmetry elements for yourself in 3D by opening the CIF [1, COD-2005680] with Mercury (free, available for Windows, Linux, MacOS):

Figure 1.
Unit cell of crystalline sulfuric acid $\ce{H2SO4}$.
Green lines parallel to crystallographic axis $b$ depict $C_2$ rotation axes.

This can be achieved by drawing the unit cell content with Calculate → Packing/Slicing…  and adding symmetry elements of interest via Display → Symmetry Elements….

Figure 2.
Single molecular unit $\ce{H2SO4}$ in two projections with the $C_2$ placed in drawing plane (left) and perpendicular to drawing plane (right).

References

Kemnitz, E.; Werner, C.; Trojanov, S. Reinvestigation of Crystalline Sulfuric Acid and Oxonium Hydrogensulfate. Acta Crystallographica Section C Crystal Structure Communications 1996, 52 (11), 2665–2668. https://doi.org/10.1107/S0108270196006749.


Common ingredient between two disinfectants? I find one of the most powerful associations to be odour, placing me very clearly in a time and place. Recently we were using Dettol in the house and I was reminded very clearly of a another disinfectant my grandmother used to use to clean greyhound kennels called Jeyes Fluid.

Is there a common ingredient between the two of them? If, so what is it; if not why do they smell similar (IMO)?

Dettol's active ingredient is chloroxylenol.
Jeye's fluid's is 4-chloro-m-cresol.

It's still possible that both disinfectants simply have an added scent to make them smell "clean". However the active ingredients above are both phenols and will likely have a similar phenolic smell (think hospitals), though I haven't sampled them both myself.


Charles's Law confusion Charles's Law states that for every 1 degree Celsius increase in temperature of a gas, the volume increases by 1/273 of the original volume (at constant pressure).

What I am wondering is, doesn't gas have an indefinite volume, therefore it is not possible for it to have volume at all? Am I missing something here?

A gas occupies the volume of whatever the container it is in. Take a balloon filled with Helium for example. Now the volume of Helium gas is the volume of the ballon.

If you increase the temperature by 1 °C at constant pressure the volume of the balloon increase by 1/273th of the original volume of the balloon. Hope this might clear things up.


Oxidation state of metal in metal oxo complexes always positive? Do all metals in metal oxo complexes (eg $\ce{Fe}$ in Ferrate, $\ce{Mo}$ in $\ce{MoO4}$,...) have positive oxidation states?

Yes. Oxygen is much more electronegative than all of the transition metals, so it is given a formal -2 charge. Keep in mind, though, that these values are not intrinsic properties of these elements/molecules. Their use is intended to help us rationalize some of the chemistry and properties associated with these materials. As an interesting example, consider oxygen difluoride, $\ce{OF_2}$. Fluorine is the most electronegative element (that forms compounds with other elements) and is given a formal charge of -1. To balance the formal charges then, oxygen must have a +2 charge!


Testing isopropyl alcohol purity I'm curious how to test purity of isopropyl alcohol? I simply got very curious today. I bought rubbing alcohol and it doesn't say what's the purity anywhere. It could be 70%, but it could also be 90%, or 99% or even 99.9%. I'm a curious man so I started googling but couldn't find anything.

If I were a chemist, how would I test the purity?

If the isopropanol mixture was bought commercially, the chances that it contains significant amounts of methanol or ethanol are fairly small, I think.  (I've never seen commercial isopropanol in the US that contains > 3% of these impurities, at least, but the situation may vary in other countries or markets.)
If you can assume that the only other main component is water, the easiest way to estimate purity is simply by measuring the density.  100% pure isopropanol has a density at 20 °C of 0.786 g per mL.  Pure water has a density of 1.00 g/mL.  At intermediate concentrations of isopropanol, the density is in between those values.  If you can accurately measure both (a) the volume and (b) the weight of a portion of isopropanol, you should be able to measure the density.  I know my kitchen scale and kitchen measuring cups would be precise enough for this.
If you can't assume that water is the only other impurity, other methods such as NMR, chromatography, or spectroscopy would probably be required.  I think NMR would probably be the most informative (but perhaps the least accessible to the home chemist.)


What is the vapor pressure of a liquid if it is not in a sealed container(i.e., a river or lake)? Is it just the atmospheric pressure? From what I understand, the boiling point of a liquid is when its surrounding pressure is equal to its vapor pressure. The vapor pressure is the pressure of the gas form of the liquid in a container when the rate of condensation equals the rate of evaporation. If this is the case, what is the vapor pressure of a liquid that is not placed in a container(i.e., a river or lake)? Is it the same as the pressure of the atmosphere? If so, then how does this affect its boiling point?

(Please correct me if what I know is not true).

From what I understand, the boiling point of a liquid is when its surrounding pressure is equal to its vapor pressure.

This is backwards. The boiling point is when the vapor pressure of the liquid is equal to the atmospheric pressure. So given the average pressure of 1 atmosphere at sea level water would boil at $\pu{100 ^\circ C}$. However on some mountain top water might boil at $\pu{95 ^\circ C}$ since the air pressure is lower.

  The vapor pressure is the pressure of the gas form of the liquid in a container when the rate of condensation equals the rate of evaporation.

For a particular temperature, in a sealed container, when the rate of condensation equals the rate of evaporation then the system is at equilibrium. The partial pressure of the gaseous form of the liquid is then the equilibrium vapor pressure of the liquid.

Without being at  equilibrium the partial pressure could be anything.

  If this is the case, what is the vapor pressure of a liquid that is not placed in a container(i.e., a river or lake)? Is it the same as the pressure of the atmosphere? If so, then how does this affect its boiling point?

If I have an open beaker of benzene on an outside picnic table then the system isn't at equilibrium and benzene has no defined vapor pressure. There just isn't enough benzene to ever reach an equilibrium with the whole atmosphere of the earth.

This wouldn't effect the boiling point of the benzene. Assuming that the air pressure is 1 atmosphere, benzene would boil at the same temperature whether if it was being boiled  in the open air or refluxed in a lab.

This bring up another point. If we say that the vapor pressure of water is $\pu{0.0231 atm}$ at $\pu{20 ^\circ C}$, we are referring to "local conditions." In other words, that gravity can be assumed to be  a constant within the system. Obviously if I had a cylinder 200 miles high at $\pu{20 ^\circ C}$, the vapor pressure wouldn't be the same at the top and bottom of the cylinder due to the difference in gravity.


Algorithm used by PubChem to determine the 2D layout assignment for Compound entries? The PubChem database provides 2D structures for entries in its Compounds database:
ftp://ftp.ncbi.nlm.nih.gov/pubchem/Compound/ (ftp link to download examples)

On the page linked below, under the "NCBI-generated records" heading, it's mentioned that the data processing of records for the Compound database "includes 2D coordinate layout assignment":
https://pubchem.ncbi.nlm.nih.gov/docs/subcmpd_summary_page_help.html#DataProcessing

What algorithm does PubChem use to generate these 2D structures?

I'm familiar with similar algorithms used in RDKit (e.g. compute2dCoords defined in rdkit/Code/GraphMol/Depictor/RDDepictor.cpp), but I'm specifically interested in a resource that describes the algorithm that PubChem uses.

Thanks very much!

According to Xemistry, PubChem uses Cactvs for 2D layout coordinate computation.
A friend pointed this out to me. If you scroll to the bottom of the CACTVS Toolkit FAQ (http://www.xemistry.com/docs.htm), it says the following:

Interesting. I did not know that PubChem is built with Cactvs. I heard it was implemented with OpenEye's OEChem library.
Well, that is not wrong, but only part of the story. Both Cactvs and OpenEye software are used, in combination with the in-house NCBI toolkit. Software built with Cactvs is in charge of structure submission pre-parsing, structure input, structure ASN.1 encoding, 2D layout coordinate computation, structure rendering, most property computations, structure unification, the complete structure search system and Web-based structure sketching for queries. The OpenEye toolkit is used for structure standardization, some property computation (names, canonic SMILES), and SD file output. There are applications at NCBI which link to both OpenEye and Xemistry libraries - they can peacefully co-exist.


Difference between quick lime, slaked lime, lime water and soda lime The chemicals "quick lime", "slaked lime", "lime water" and "soda lime" all have "lime" in common. What is the difference between them?

Quick lime is calcium oxide, $\ce{CaO}$, in the solid state.

Slaked lime is calcium hydroxide, $\ce{Ca(OH)2}$, in the solid state.

Lime water is a dilute solution of calcium hydroxide, $\ce{Ca(OH)2}$, in water.

Soda lime is a mixture of sodium hydroxide, $\ce{NaOH}$, (also known as caustic soda) and calcium oxide, $\ce{CaO}$, (also known as quicklime).


How do I create a concentrated solution of singlet oxygen (in the dark)? Without appealing to reactions that require photons / light, how do chemists achieve a very high concentration of continually produced, though quite short lived, singlet oxygen in aqueous solution (i.e. $\ce{{}^1O_2}$ where dioxygen is typically in the triplet $\ce{{}^3O_2}$ state)?  I suppose I'll also ask the same question if we have the freedom to select an organic solvent of choice for the buffer?  By "high concentration" I mean at concentration from $\approx 10~\mathrm{\mu{}M} - 100(+)~\mathrm{\mu{}M}$.

My guess would be that should be a way to do this starting with a solution containing some amount of $\ce{H_2O_2}$, but I'm not quite sure what a reasonable process might be.

Here's the wikipedia link for singlet oxygen ($\ce{{}^3O_2}$)

This reaction should happen at or around room temperature.

The reason I'm specifying that I'd like to do this in the dark, and to have the singlet oxygen continually produced at room temperature (given its short lifetime, which is $\leq 100~\mathrm{\mu{}s}$ under any conditions I've heard of), is because I'm curious if I could easily set up some buffer that lets me dose a sample with singlet oxygen on a lab bench with a pipette.  Given the short lifetime of singlet oxygen, having to use a continuous radiation source presents difficulties that, while surmountable, lead me to ask this question.

Provided the above, I'd like to avoid reactions that vent gases like $\ce{Cl_2}$ and require fume hoods in the sense of "... well, if you value your life ..." vs. "... OSHA says so ...".

Potassium ferricyanide (and other ion complexes, like hemoglobin) will decompose hydrogen peroxide into water and (presumably) singlet oxygen. This process is necessary for the luminol reaction.

$$\ce{2H2O2 \overset{Fe}{->} 2H2O + ^1O2}$$


Is food that burnt in an 18/10 stainless steel pot still edible? I was cooking Christmas tamales in an 18/10 stainless steel pot and I accidentally let the water evaporate and the pot started burning.  I think it was probably burning for around 5 minutes or less.  There were a lot of tamales in there and I'm wondering if they are safe to eat or if I have to throw them out?

Your stainless steel pot is thoroughly thermally stable well above the temperature of your stove, and there are no volatile compounds present that could migrate out of the pot and into your food. It is (was) perfectly safe to consume the tamales cooked in this pot.


Calculation of mass from mmol/L Sorry I understand this looks like a "homework question", (I'm actually a 46yr old dentist looking to see what amount of calcium phosphate should be effective in a proprietary toothpaste!). However, if the actual question below can not be worked out can you let me know what the process of the calculation would be to determine mass of a salt to add to gain a mmol/L concentration?

Original Question:
I wonder if you can help on this? I want to calculate how much calcium phosphate to add to 1g of toothpaste to have between 18mmol/L to 1.5mmol/L of calcium and phosphate on the tooth surface? Also if you are able to work this out can you let me know what assumptions you have made in this calculation?

There are several calcium phosphates.  I'll assume you mean monocalcium posphate, $\ce{Ca(H2PO4)2}$, because the others are not nearly soluble enough in water to allow even a 1.5 mmol/L solution.  (I assume the toothbase's most abundant ingredient is water.)

The formula weight is 234.  (I won't bother with many decimals because your range of molarities, 1.5 to 18 mmol/L, is so wide.)  I got this number from
http://en.wikipedia.org/wiki/Monocalcium_phosphate although it is easy to compute from scratch by adding the atomic weights for 1 Ca atom, 4 H atoms etc.

To calculate how much  of this compound you need given the molarity, ideally you should be looking at a known volume of paste.  Instead you have a known weight, 1g.  If you know the density of the paste, you can compute the volume by dividing 1g by the density.  Or you can just use an approximation.  I'll assume that 1g = 0.9mL (this will be a good approximation).  From now on I will assume volume = 0.9mL.

Multiplying 1.5mmol/L by 0.9mL gives you 1.35 micro moles (1.35$\times 10^{-6}$ moles).  Since 1 mole is 234 grams (formula weight), this means 316 micrograms or 0.315 mg.  The same calculation with 18mmol/L gives 3.79 mg.

So your answer (discarding  unwarranted decimals) is: between 0.31 and 3.8 mg .

I hope if your proprietary formulation is commercially successful you will make a generous contribution to the organization behind stackexchange!


Different elution profiles of carvacrol and thymol I have identified carvacrol and thymol via thin-layer-chromatography and via gas chromatography
On a normal silica plate the thymol has a higher Rf-value then carvacrol. The explanation for this would be the position of the hydroxyl-group and the difference size of the nearby methyl or isopropyl group. I think thymol should be more hydrophobic then carvacrol, explaining the less interaction with the silica gel resulting in a higher Rf-value.

The problem is, that the analysis by gas chromatography with a non-polar column shows a different elution profile. I would expect that thymol, according to the higher hydrophobicity, should have a higher retention time then carvacrol. This isn't the case. Carvacrol (27.08 min) elutes after thymol (26.51 min).
Is there any explanation I'am not aware or is my thinking simple wrong?

For gas chromatography, boiling point is one of the most important determinants of elution time.  For thin-layer chromtagraphy, that isn't true.

The comments from @Molx and @Buttonwood on this question expound on this point quite well.

@Molx:

  Thymol has a lower boiling point (232 °C) than Carvacrol (237.7 °C, both from Wikipedia), I'd bet this is the reason because AFAIK the volatility has a more important effect on the separation than the column polarity in gas chromatography.

@Buttonwood:

  ... gas chromatography's separation is more influenced by volatility (related to boiling point) of the constituents of a mixture than their adsorption to the stationary phase. In addition -- as already stated in the original question -- the column deployed in GC was a non-polar one. Except explicitly stated differently, thin layer chromatography more often uses a polar stationary phase (silica, acid/neutral/basic alumina, for example) where different strength of adsorption is important. [...]


Why is the solution of (NH4)2[Ce(NO3)6] used as an oxidizing agent? In an MCQ, the correct statement regarding the abovementioned compound is that its solution is used as an oxidizing agent.
I do not understand why that is so.

Most lanthanides display stability in their +3 oxidation states. Cerium does the same, which is why it has a very high reduction potential. The reduction potential for the $\ce{Ce^{4+}|Ce^{3+}}$ couple is 1.61 V (for comparison, $\ce{Cl2}$ has a potential of 1.32 V)

Cerium in the form of ceric ammonium nitrate is also quite stable compared to other oxidizing agents and can be stored on the shelf for a longish period of time. It's also quite usable: the bright orange colour is decolourized to a pale yellow on reduction of $\ce{Ce^{4+}}$. This is why it's used as a common oxidizing agent.


Sp5 hybridization in cyclopropane? I have never before heard/read about something as a $sp^5$ hybridization. Today, Henry Rzepa's blog post made me aware of the existance of such a bonding system. That made me search a little bit and I found an entry in a german chemistry forum, where this question was also asked ... they answered it with a mathematical construction$^\ast$:

  Cyclopropane has the following angles:

  $\angle \ce{HCH}=118^\circ~\text{resp.}~\gg 120^\circ$
  $\angle \ce{CCC}~\text{with bent bonds:}~60 + 2 \cdot 21 = 102^\circ$

  The orbitals towards the protons are $sp^2$ because of the $120^\circ$ angles.
   The orbitals towards the carbons originate in the following relation:

  $$1 + a \cos~\alpha = 0$$ ... where $\alpha$ is the bond angle and $a$ is the p-amount in sp$^a$ for the orbitals, which make up the angle.

  This means for the orbitals, which span the 102 degree angle: $$1 + a \cos 102^\circ = 0$$ $$1 + a \cdot (-0.20) = 0$$ $$a = \frac{-1}{-0.20} = 5$$
  $$\Rightarrow \text{sp}^5 \text{-orbitals}$$

  Test:

  In a single sp$^a$ orbital, the s-amount is: $\frac{1}{1+a}$, because $1+a$ equals the sum of all amounts of s and p
  In a single sp$^a$ orbital, the p-amount is: $\frac{a}{1+a}$
  For s:

  In the orbitals that are oriented towards the protons, the s-amount is $\frac{1}{1+2} = \frac{1}{3}$
  In the orbitals that are oriented towards the carbons, the s-amount is $\frac{1}{1+5} = \frac{1}{6}$
  Addition of all s-amounts at a single carbon with all four bond orbitals yield: $\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{6}=1$, which is correct, because there is only one single s-orbital at every carbon atom.

  For p:

  In the orbitals that are oriented towards the protons, the p-amount is $\frac{2}{1+2} = \frac{2}{3}$
  In the orbitals that are oriented towards the carbons, the p-amount is $\frac{5}{1+5}=\frac{5}{6}$
  Addition of all p-amounts at a single carbon with four bond orbitals yield: $\frac{2}{3}+\frac{2}{3}+\frac{5}{6}+\frac{5}{6}$, which is correct, because there are 3 p-orbitals at every carbon atom.

  This means, that the bent bonds with $21^\circ$ from the $\ce{C-C}$-bond are spanned by sp$^5$ orbitals.

So math-magically this seems to make sense, but is there another explanation that might base more on chemical intuition or "real" chemical concepts?

A fast calculation ($\omega$B97X-D/def2-TZVPP) and a subsequent analysis of the isosurface of the Laplacian of the electron density, showed at least the expected "nonlinear", slightly curved bond between the carbon atoms.

Whoever might want to see the electron localization function (ELF), which also shows the bent bonds quite good:

$^\ast$ While I tried to translate it to my best, some errors might have been introduced by this . . . please correct me, where I'm wrong.

Until this post, $\mathrm{sp^1,~sp^2~\text{and}~sp^3}$ meant for me, that the hybrid orbitals would consist from one s orbital and one, two or three p orbitals. I thought of it being so, as this seems to be what the most textbooks suggest, as can be seen in the following image:

One 2s orbital and three 2p orbitals hybridize to four $\mathrm{sp^3}$ orbitals.

This scheme seems to work but it also seems to be a working simplification as can be seen by my question about $\mathrm{sp^5}$ hybrid orbitals, which simply cannot consist from one s orbital and five p orbitals, as there are only three p orbitals in reach.

Then what does this nomenclature actually mean?

As Martin commented, it is more like the following:
$$\mathrm{sp^n \equiv s^\frac{1}{n+1}p^\frac{n}{n+1}}$$ or to say it slightly different, a $\mathrm{sp^n}$-bond has
$\frac{100}{n+1}~\%$ s-amount and $\frac{100\,n}{n+1}~\%$ p-amount.

\begin{array}{lll}
\hline
\text{bond} & \text{s-amount} & \text{p-amount}\\
\hline
\mathrm{sp} & 50~\% & 50~\%\\
\mathrm{sp^2} & 33~\% & 67~\%\\
\mathrm{sp^3} & 25~\% & 75~\%\\
\mathrm{sp^4} & 20~\% & 80~\%\\
\mathrm{sp^5} & 17~\% & 83~\%\\
\ldots & \ldots & \ldots \\
\mathrm{sp^n} & \frac{100}{n+1}~\% & \frac{100\;n}{n+1}~\%\\
\hline
\end{array}


Need help with stoichiometry problem I'm pretty new to chemistry and I've been stuck on it for hours.

Question: $\pu{10g}$ of the hydroxide of a metal on ignition gave $\pu{8g}$ of oxide. The equivalent weight of the metal is:
a) $\pu{136g}$
b) $\pu{40g}$
c) $\pu{56g}$
d) $\pu{28g}$

I used the law of equivalence but the answer I got was $\pu{3g}$. what am I doing wrong?

How I went about it:
10/E=8/16;
E=20;
E=M(equivalent mass of metal)+16+1;
M=3

I found your numbers confusing, but I think the first equation you wrote is not correct. This is how I'd solve it:

Let's say we have $\mathrm{n}$ equivalents of metal in the sample, and let's call the equivalent weight of the metal $\mathrm{e}$. For the hydroxide sample, we have:

$n(e+17)=10$

where $\mathrm{e+17}$ is the molar mass of the hydroxide $\ce{MOH}$.

For the oxide, we have:

$n/2(2e+16)=8$

where $\mathrm{2e+16}$ is the molar mass of the oxide $\ce{M_2O}$. Note that $\mathrm{n}$ becomes $\mathrm{n/2}$ because of stoichiometry:

$\ce{2MOH->M2O + H2O}$

So you have a system of two equations with two unkowns ($\mathrm{n}$ and $\mathrm{e}$), which you can solve normally. For instance, if you divide the equations:

$\cfrac{n(e+17)}{n/2(2e+16)}=\cfrac{10}{8}$

$\cfrac{e+17}{e+8}=\cfrac{10}{8}$

$8e+136=10e+80$

$2e=56$

$e=28$

So the equivalent weight of the metal is 28g.

Makes sense?


What does it mean "A oxidizes on B" I have just managed to understand the difference between oxidation and reduction. So basically if A oxidises B, then B loses electrons, right? But what does this formulation mean: "A is oxidized on B"? What is this "on" suppose to mean? That A is losing electrons to B or vice-versa?

Sorry if this sounds like a very stupid question, but I'm really confused by presence of "on something"

EDIT:
If you search for this

  "Carbonate-based electrolytes commonly used in Li-ion batteries are
  reduced on negative electrodes"

on Google, you will find the first paper and this specific formulation.

And here you can find the paper: http://pubs.acs.org/doi/abs/10.1021/acs.jpclett.5b01727?journalCode=jpclcd

This answer is a summary of the string of very good comments to the OP's original set of short questions.  This is both my take on the questions and an attempt to produce a more readable "net" answer than those contained within the set of individual comments.

Regardless of the intended context, "A is oxidized on B" means that A looses electrons.  It's seems this context likely implies that B is an electrode surface "on" which A is being oxidized.

Regardless of context, if A is oxidized by B, then B is reduced.

If B is an electrode surface, then the electrons from A needn't reduce B itself, but could flow through that part of the cell (the anode).

I hope this (correctly!) answers each of the OP's questions.


Reaction of dimethyl sulfate with Protocatechualdehyde I know that $\ce{(CH3)2SO4}$ reacts with phenols to give methoxybenzene. But the question that came in INChO-2018 was:

  When one equivalent of $\ce{NaOH}$ and one equivalent of dimethyl sulfate reacts with Protocatechualdehyde, K is formed in 87% yield. When carried out with excess of $\ce{NaOH}$ and only one equivalent of dimethyl sulfate, M is formed in 95% yield. Identify K and M.

How would I know that ether formation take place at p- to $\ce{-CHO}$ group or at m- to it?

This is Protocatechualdehyde

When one equivalent of dimethyl sulfate reacts with 3,4-dihydroxybenzaldehyde (protocatechualdehyde), there are two nucleophilic sites in the molecule, namely 3- and 4-hydroxy-groups. If one equivalent of strong base such as $\ce{NaOH}$ is used, it will deprotonate the most acetic group between them, in this case, 4-hydroxy-group. For example, $\mathrm{p}K_\mathrm{a}$ of 4-hydroxybenzaldehyde is $7.61$ while $\mathrm{p}K_\mathrm{a}$ of 3-hydroxybenzaldehyde is $8.98$ at $\pu{25 ^\circ C}$ (Ref.1). Thus, it is safe to assume that product K is 3-hydroxy-4-methoxybenzaldehyde (isovanillin).

However, the regioselectivity of the synthesis (when two nucleophilic sites are present) can be controlled by the number of base equivalents in this kind of alkylation steps (Ref.2). Thus, in excess $\ce{NaOH}$, both phenolic hydroxy groups of 3,4-dihydroxybenzaldehyde would be deprotonated and methylation would occur at 3-phenolate, because it is the most reactive at this point between two (basically, negative charge of 4-phenolate should be largely delocalized within aromatic nucleus with benzaldehyde group than that of 3-phenolate without direct resonance effect from benzaldehyde group).  Therefore, the major product M in this case would be 4-hydroxy-3-methoxybenzaldehyde (vanillin). You can probably compare with methylation of ascorbic acid in different conditions.

Note: Some literature mentioned that when one equivalent of $\ce{NaOH}$ is used, reaction is must faster and tends to use both methyl groups of dimethyl sulfate. Thus, I won't surprised if the given answer for K is 3,4-dimethoxybenzaldehyde.

References:

E. P. Serjeant, Boyd Dempsey, Editors, IUPAC chemical data series no. 23: Ionization Constants of Organic Acids in Aqueous Solution; Pergamon: Oxford, United Kingdom, 1979 (Values extracted from PDF here).
Tobias Schneider, Vladimir Kubyshkin, Nediljko Budisa, "Synthesis of a Photo‐Caged DOPA Derivative by Selective Alkylation of 3,4‐Dihydroxybenzaldehyde," Eur. J. Org. Chem. 2018, (18), 2053-2063 (https://doi.org/10.1002/ejoc.201701749).


What actually happens in strong acid- weak base reactions? I'm going back through my notes from a chemistry class, which I did well in despite not having a very good conceptual understanding, in order to hopefully develop that understanding, but there's something I can't find in the book or online. Why is it that, when solving for the pH of a solution that results from the mixture of a strong acid with a weak base or a weak acid with a strong base, we can treat the acid-base reaction as going to completion before solving any resulting equilibrium problem, or completely ignoring any equilibrium in the pre-mixed solutions.

For a concrete example, when mixing 40.0 mL of 0.50 M ammonia with 25.0 mL of 0.30 M hydrochloric acid, the approach the class would have me use is to first solve the limiting reaction problem as if the reaction goes to completion. Doing so, we see that there are 0.0075 mol of chloride ions, 0.0075 mol of ammonium, and 0.013 mol of ammonia. Then, because we don't have excess hydrogen ions, nor excess hydroxide, nor only neutral ions, we solve the equilibrium problem of ammonium with ammonia.

What I don't understand is how we can ignore the fact that the ammonia solution would have already been in equilibrium when mixed with the hydrochloric acid, and as I understand it should be constantly arriving towards equilibrium as the reaction is taking place.

Alternatively, if there had been excess hydrogen ions, I was told to use the concentration of hydrogen ions from the completed reaction, but that would also be ignoring the fact that the ammonium contributes to the pH (because it again ignores it's equilibrium).

The strategy of major species

You correctly describe the strategy. First, let reactions go to completion (in the direction that makes sense, i.e. weak base and strong acid forms weak acid and spectator ion - never the other way around). Then, check what the major species are and estimate the pH. Finally (and this step is often omitted), see if equilibria involving minor species (such as hydroxide for acidic solutions) need adjustment.

Why does the strategy of major species work?

For the example, the OP already calculated the estimated amount of species: 0.0075 mol of ammonium, and 0.013 mol of ammonia. With a total volume of 65 mL, that comes out as:

$$c_\ce{NH4+} = \pu{0.115M}$$
$$c_\ce{NH3} = \pu{0.20M}$$

From this, we can estimate the pH to be a bit more basic than the $\mathrm{p}K_\mathrm{a}$, i.e. 9.49. Now we can think about the minor species, hydroxide and hydronium. At pH = 9.49, there should be an excess of hydroxide over hydronium, by about 0.00003 M. Dissociation of water does not give an excess, so it has to come from ammonia turning into ammonium:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

However, when this reaction makes hydroxide, it also changes the ratio of ammonium to ammonia, changing the pH. The only reason we don't keep running in circles, adjusting one equilibrium and then the other, is that this adjustment is tiny compared to the concentration of the major species. It amounts to decreasing the ammonia concentration by 0.00003 M and increasing the ammonium concentration by the same amount. The shift in pH is so small that it makes no difference when the pH is written with appropriate number of significant figures.

Alternate strategy

For the calculators we use, this is a good strategy to deal with multiple equilibria (in your case, that of the weak acid/base pair and autodissociation of water).

If you had an analog computer that adjusts the concentrations based on a "pH slider" and reports out whether all the constraints are met, there would be no reason to do this in multiple steps. Instead, you would just push your pH slider from one extreme to the other and stop when all equations are satisfied.

In your case, you could calculate the ratio of ammonia to ammonium from the pH and the $\mathrm{p}K_\mathrm{a}$, and the hydroxide concentration from the autodissociation constant of water. Then, you vary the pH until the charge balance (chloride plus hydroxide has to match hydronium plus ammonium) is achieved. It is a fun exercise to program this into a spread sheet.

What actually happens?

All reactions (any of the acids - water, hydronium or ammonium - reacting with any of the bases - water, ammonia or hydroxide) go on at the same time. The further from equilibrium a reaction is, the faster is the net change. Once all the reactions are at equilibrium, there is no more net change. The details depend on the kinetics but are not relevant to the equilibrium state that is finally reached.


Does 1,3‐dichloroallene show geometrical isomerism? Since the attached groups are present in different planes, $\ce{ClCH=C=CHCl}$ should not show geometrical isomerism, but my book say that it does. Why?

This is a chiral compound; it has two enantiomers.  This source defines "geometrical 8somerism" as isomerism involving different geometrical arrangements, and a pair of nonsuperimposable mirror images would seem to qualify.  But in the body they talk about about cis/trans or E/Z isomers, not chiral compounds.  The latter, which applies here, should be described by more precise terms like "enantiomers".


Are children's sparklers based on a magnesium reaction? We were letting our kids play with sparklers on New Year's Eve, and my friend's son asked his Dad:

  What would happen if we threw the sparkler into the water? Would it keep burning under water.

My friend pondered that for a moment - and said it depended if the reaction was based on magnesium or not.

My question is: Are children's sparklers based on a magnesium reaction?

To answer the question, let us understand sparkler chemistry. A sparkler consists of:

An oxidizer: potassium nitrate or potassium chlorate. They burn off a mixture and are part of oxidation-reduction reaction:
$$
\begin{align}
\ce{\underset{potassium nitrate}{2 KNO3 (s)} &-> \underset{potassium nitrite}{2 KNO2 (s)} + O2 (g)}\\
\ce{\underset{potassium chlorate}{2 KClO3 (s)} &-> \underset{potassium chloride}{2 KCl (s)} + 3 O2 (g)}
\end{align}
$$
Fuel: firework fuel is charcoal and sulfur. Both sulfur and carbon (charcoal) are reducing agents forming sulfur dioxide and carbon dioxide.
Metal powder: aluminum, iron, steel, zinc, magnesium, titanium or ferrotitanium dust or flakes. These metal flakes become incandescent and shine brightly in the presence of oxidizers. If temperatures are high enough, these metal flakes burn. The temperatures of these sparks range from 1800 °F to 3000 °F (1000 °C - 1600 °C)(Ref. I). Many elements heat up and emit at a characteristic wavelength (Ref. II).

A combustible binder: binders hold oxidizers, fuel and metal powders together. Common binders are dextrin (a sugar) dampened by water.
Colorants: aluminium, magnesium and titanium all give near brilliant white sparks. Iron produces orange sparks.

Putting all together: these chemicals are mixed together to form a slurry. The slurry is coated on to wire and allowed to dry. When ignited the metal flakes  aluminum, iron, steel, zinc or magnesium dust or flakes create the bright, shimmering sparks. As to your question, children sparklers are based on metal oxidizing reactions.

References

Ref. I https://www.wired.com/2014/07/the-awesome-physics-in-a-simple-sparkler/
Ref. II https://www.forbes.com/sites/startswithabang/2016/07/01/the-physics-of-fireworks/#3895e0711eee
http://www.compoundchem.com/2014/11/04/sparklers/
http://www.compoundchem.com/2013/12/30/the-chemistry-of-fireworks/
http://www.compoundchem.com/2014/07/02/the-chemistry-of-gunpowder/
http://www.skylighter.com/fireworks/how-to-make/sizzler-sparklers.asp


Hybrization of H in HNO There is a problem that states: Find the molecular geometry of each molecule and the hybridization of each atom in the molecule. The two molecules are HNO and HCN. I found HNO to be sp3 hybridized and HCN to be sp hybridized. However, it says that only N and O are hybridized in the first. While all three are hybridized in the second. Why is hydrogen listed as sp hybridized?

Hybridization refers to the orbitals that each atom is using to make bonds.  In $\ce{HNO}$, the $\ce{N}$ and $\ce{O}$ are $sp^2$ hybridized.  This means they form a sigma bond with an $sp^2$ orbital from each and the pi bond from the interaction of the non-hybridized $p$ orbital on each.  The $\ce{N}$ also has a $sp^2$ orbital overlap with the $1s$ of $\ce{H}$, which forms a sigma bond.  Each of the lone pairs on the $\ce{N}$ and $\ce{O}$ is in an $sp^2$ orbital.

In $\ce{HCN}$ the carbon is $sp$ hybridized, as is the $\ce{N}$.  $\ce{C}$ and $\ce{N}$ form a sigma bond with $sp$ overlap, and the pi bond from the interaction of both of their non-hybridized $p$ orbitals (to form a triple bond).  The lone pair of $\ce{N}$ is in an $sp$ orbital, and the $\ce{C-H}$ bond is from an $sp$ overlap with the $1s$ of $\ce{H}$.

None of these molecules have $sp^3$ hybridization.


How to calculate ion concentrations for a precipitation from the solubility product constant? The equilibrium constant for dissolving the insoluble substance gold (I) iodide, AuI (s), in aqueous
  solution is 1.6 × 10^-23
  at 25ºC. Write the equilibrium constant expression and calculate the concentration
  of Au + (aq) and I (aq) ions in a solution in which 0.345g of AuI(s) is in equilibrium with the aqueous ions.
  The answer is 4x10^-12.

So 1.6x10^-23=products/reactants. The products = x^2, but what does the reactant equal to? I think they got the answer by ignoring the reactants, is this possible?

The concentration (the activity, to be precise) of solids is assumed 1. [Au+][I-]/[AuI]=[Au+][I-] = Ks. Here Ks is the solubility constant. If water originally didn't have any I- or Au+, then the amounts of [Au+]=[I-]=x. Then you have x*x=Ks and x= square root of Ks. The amount of AuI doesn't seem to play any role because they didn't specify the amount of water it is dissolved in.


Strength of Hydrogen Bond and Angle This link has a short paragraph on the very first page, which says: "There are two useful criteria to evaluate the presence and the strength of a H-bond. The first is to look at the distance between the H and the A atom, which should be shorter than the sum of their Van der Waals radii. A second method suggests to evaluate the angle formed by X-H∙∙∙A, centered on the H atom: the closer this angle is to 180°, the stronger the H-bond is"

I wish to understand why the part in bold is true, that is, what is the angle dependence (qualitative or quantitative) of the strength of hydrogen bonds.

To add to the above discussion, in an introductory biology course at university, we were also given the following example -

Well, once again, why is it so?

I did some research, didn't really find much, but came across an interesting paper that may be worth a read if anyone's really interested. Attaching it here!

Thanks in advance!

Empirical data

If you analyze crystal structure data for short intermolecular distances between hydrogen atoms and potential acceptor atoms, you will find a distribution of these angles. The simplest explanation for this distribution is that there is a favorable interaction that is strongest for the most commonly observed angles. The fit between empirical data and computational results is fairly good, e.g. https://www.researchgate.net/publication/242862021_Distribution_of_hydrogen_bond_angles_in_molecular_crystals

Dipole-dipole interactions

If you treat hydrogen bonds as pure dipole-dipole interactions (Q-Y ... H-X), you expect the bond to be directional. After all, you maximize the interaction by close approach of the negative partial charge on the acceptor and the positive partial charge on the donor (i.e. the hydrogen atom in a polar bond). However, for this model the angular dependence should be symmetrical, i.e. there should be a similar decrease in bonds strength when you rotate the donor or the acceptor away from a colinear arrangement. The "rules" for hydrogen bonds say otherwise. It is more important that the angle at the hydrogen atom is maximal (Y ... H-X) than that the angle at the acceptor atom is maximal (Q-Y ... H).

Covalent contribution

There is a covalent contribution to the hydrogen bond that explains why a linear arrangement around the hydrogen atoms is favored. The strongest hydrogen bonds are those where hydrogen bond and covalent bond with hydrogen become interchangable ("symmetric hydrogen bonds"). In his 2002 review, Steiner wrote:

  All hydrogen bonds can be considered as incipient proton transfer reactions, and for strong hydrogen bonds, this reaction can be in a very advanced state.

There is some controversy whether the hydrogen bond has covalent character, and some question how to describe it theoretically (as a 3 center 4 electron bond?). There is, however, strong evidence from NMR that there is indeed a covalent contribution because nuclei couple in a through-bond fashion across hydrogen bonds (e.g. this review).


Biological Consequences of Asteroid Mining—Death by Isotope? It's been documented that NASA hope to capture an asteroid in 2025, and have subsequent aims to mine that asteroid. If if this is successful, we would expect other asteroids to be mined in the future.

A consequence of this is that relative atomic masses of elements mined—those with two or more stable isotopes—will no longer be faithful to our current periodic table. Ruthenium alone, a high-value rare earth, has five stable isotopes alone, ranging from $^{98}\ce{Ru}$ to $^{102}\ce{Ru}$. The relative abundances of these isotopes are bound to differ in other places other than Earth, and so would the weighted average.

Say after mining our ruthenium, we use it in an electronic device and consequently that device is thrown away. For argument's sake, also say that this ruthenium—of an isotopic distribution never before seen—leaches into the environment and later comes into contact with biological systems. Could it be possible that compounds hitherto considered non-toxic on Earth become toxic to biological systems by virtue of a newly realised isotopic discrimination?

As an example, $^{13}\ce{CO2}$  is effectively discriminated against in uptake by plants in comparison to $^{12}\ce{CO2}$  (due to it being a diffusion limited reaction). Ruthenium is purported to be carcinogenic, but the most abundant isotope on Earth of Ruthenium is $^{102}\ce{Ru}$ which could be effectively non-toxic by virtue of the fact that  compounds containing these atoms takes so long to diffuse, they don't diffuse across biological membranes at all.

If a sample of Ruthenium from the moon found its most abundant isotope in $^{98}\ce{Ru}$, then this would be expected to diffuse much faster than its heavier counterparts. Therefore could it be possible that a compound previously thought of as non-toxic could become toxic via this effect. Does this sound plausible?

NB: I've deliberately missed out other routes of exposure such as ingestion to only consider diffusion controlled reactions.

This is an interesting question and you raise a number of points, let's step through them.

  A consequence of this is that relative atomic masses of elements
  mined—those with two or more stable isotopes—will no longer be
  faithful to our current periodic table.

But this is already happening.  $\ce{^235U}$ constitutes 0.72% of uranium found on earth and decays to the stable isotope $\ce{^207Pb}$, which is found in a natural abundance of 22.1%.  Before we get to the asteroid the abundance of various isotopes and their weighted average mass is already shifting.

  98 Ru , then this would be expected to diffuse much faster than its
  heavier counterparts

Diffuse much faster?  First off, we're talking about diffusion, not chemical reaction.  Elements or compounds usually enter the body by ingestion of one sort or another (eating, breathing) rather than diffusion, and these ingestion pathways wouldn't involve isotopic discrimination.  That is, if we breath air that contains isotopes in a certain ratio, then that's the ratio that will initially appear in our lungs.  Nonetheless if diffusion were found to be an issue, my guess is that isotopic discrimination by diffusion would be a very small effect.  For a gas the maximum separation of two isotopes is given by
$$\mathrm{\sqrt{\frac{[MW~of~compound ~with ~isotope~1]}{[MW~of~compound ~with ~isotope~2]}}}$$
In the case of separating $\ce{^235UF6/^238UF6}$ this amounts to a fractionation ratio of 1.0043 after 1 pass.  Ruthenium is lighter, so the effect would be larger, about 1.02 ($\ce{^102Ru/^98Ru}$) after 1 pass, even less if it is in a molecule.

Chemical reaction within our body of ingested isotopic compounds would also show  discrimination due to primary kinetic isotope effects.  The maximum primary kinetic isotope effect is proportional to the reduced mass as follows:
$$k~ \thicksim ~\sqrt{\frac{m_1 + m_2}{m_1 m_2}}$$
Applying this to $\ce{^102Ru}$ and $\ce{^98Ru}$ yields a primary kinetic isotope effect of 1.020, again, even less if the element is incorporated into a compound.

To me, the effects look small and since shifts in natural abundance have been occurring for a long time here on earth, with no one raising a flag, my guess (and that's all it is, a guess) is that it's not something to worry about.  Still, I'd feel better if a NASA toxicologist looked at the question and said "no problem"


Distinguishing Brown Ring Test for Nitrates and Nitrites My question is how will one detect the presence of nitrates by brown ring test when nitrites are also present in the solution?

Brown Ring Test for Nitrite
$\ce{NO_2^- + H+ -> HNO2}$
$\ce{3HNO2 -> H2O + HNO3 + 2NO}$
$\ce{FeSO4 + 6H2O -> [Fe(H2O)6]SO4}$
$\ce{[Fe(H2O)6]SO4 + NO -> [Fe(H2O)5NO]SO4 + H2O}$

Brown Ring Test for Nitrate
$\ce{2NO_3^- + 6Fe^2+ +4H2SO4 -> 6Fe^3+ + 4SO4^2- + 2NO + 4H2O}$
$\ce{FeSO4 + H2O -> [Fe(H2O)6]SO4}$
$\ce{[Fe(H2O)6]SO4 + NO -> [Fe(H2O)5NO]SO4 + H2O}$

Note: I personally think that it may have something to do with preventing oxidation of nitrites to nitrates and the release of NO (nitric oxide). But I am not quite sure how to go about that.

Yes, you are quite right. It has something to do with prevention of oxidation of nitrites. By controlling the pH, the nitrite present is reduced to nitric oxide ($\ce{NO}$) which reacts with ferrous sulfate to form a brown-colored complex ion, $\ce{FeSO4.NO}$. Nitrates do not react in such way and thus not registered in the test. It uses a special agent to prevent color formation or precipitation of common
interfering ions. Here is the relevant information[1]:

  The high range nitrite test is a modification of the classical brown
  ring test for nitrate using ferrous sulfate. By controlling the sample
  pH, the nitrite present is reduced to nitrous oxide, which reacts with
  the indicator to form a greenish-brown color. Nitrates are not
  registered in the test. [...]

  In an acidic medium ferrous sulfate reduces nitrogen in nitrite
  ($\ce{NO2–}$) to form nitrous oxide ($\ce{NO}$). Ferrous ions combine
  with the nitrous oxide to form a brown-colored complex ion, the color
  intensity of which is in direct proportion to the nitrite present in
  the water sample. Color development follows Beer’s Law.

  $$\ce{2Fe^2+ + 4H+ + 2NO2‾ → 2Fe^3+ + 2NO + 2H2O}$$

  $$\ce{NO + FeSO4 → FeSO4·NO}$$

Reference

https://www.hach.com/asset-get.download.jsa?id=7639984196


Opening a shaken up bottle quickly = no fizz? I recently watched this video:
https://www.youtube.com/watch?v=Lqk3lGic3D0

Summary: Opening up a shaken bottle of Carbonated drink extremely quickly results in almost zero bubbles forming in the drink and then the soda is 'flat'

How exactly does this work? The extremely fast sudden opening seems to causes the CO2 to react differently than normally expected?

Thanks.

The trick is simply waiting for the CO2 to be released ahead of time.

For a full and cold soda this trick will not work, the pressure in the head-space will be slightly lower due to the lower temperature and with less total volume of CO2 -which makes the "pop" part of the trick fail- and the CO2 initialization points within the solution will be distributed -which causes the fizzing and overflow of liquid.

With a half filled soda that's been sitting out in a comfortably warm room, the soda is warmer. This means that you are working with a (even more) super saturated solution of CO2 in water and acid. This additional instability combined with a greater head-space for pressurization allows a greater build-up of CO2 in the bottle allowing for the "pop".  If you didn't wait for the bubbles to go away you would still have fizz formation (and even flatter soda). If you did wait for the bubbles to disappear but you opened the bottle slowly, the soda would still be just as flat as with the pop-trick but with a slight hissing sound and limited additional bubble formation.


Validity of Trouton's Rule I would like to know under what circumstances is Trouton's rule obeyed by liquids and why certain systems i.e. substances may deviate from Trouton's rule.

By the way I know what Trouton's rule is and I have a vague idea that sometimes Trouton's rule is not obeyed due to the presence of hydrogen bonding in liquids such as water but I would like to know more about other reasons why certain situations the rule is invalid.

The rule says that for many liquids the entropy of vaporization is almost the same, around ${\rm 85\,J K^{-1}} $.

The (partial) success of the rule is due to the fact that the entropy of a gas is so much larger than that of any liquid. Therefore, the entropy of the initial state (e.g. the liquid) usually makes no big difference.  However, if the liquid presents hydrogen bonding or any other kind of high ordered structure, its entropy will be particularly low and the entropy gain during vaporization will be larger than usual, too.


What happens during Glycolysis? Can I get brief explanation of the reactions that occur under Glycolysis? What do the curved lines represent?

Glycolysis is a process of which converts Glucose(Described as C$_{6}$ in your diagram), to two Pyruvate (2C$_{3}$). Then in downstream of glycolysis, Pyruvates are then converted to AcetylCoA to be used for many things in body. Glycolysis is part of a metabolic pathway, and in total it uses 1 Glucose, 2 ATP, 2 NAD+ to produce 2 Pyruvate, 4 ATP and 2 NADH. So those curve lines represents the Total input and output of energy. Each step is facilitated by different enzymes, so you could search for enzymes in each step for more information about each reaction. Wikipedia has table of each reactions.


How does an MRI work? This is my interpretation of an MRI:
There's an external magnetic field and most of the hydrogen nuclei will align themselves with that field (which is the lower energy state) but some are not aligned (which is the higher energy state). A chemist can then direct a wide range of radio waves with different wavelengths. The specific wavelength that causes the hydrogen nuclei to go from their lower energy aligned state to their higher energy unaligned state can be determined by the wavelength of the energy emitted by the photon when the hydrogen nuclei return to its ground state. And that wavelength determines the environment. Is this accurate?

The simplest way to detect where a spin in a sample is is to place this in a magnetic field that changes in direct proportion to depth. Then the frequency at which spins respond to the r.f. radiation can be interpreted as a depth. Clearly it is going to be far more complex to get 3D images but in essence is the same; the detected signal has to be encoded with a position.


Effects of pressure on equilibrium Which system at equilibrium will not be influenced by a change in pressure?
$$
\begin{align}
\ce{3 O2(g) &<=> 2 O3(g)}\tag{A}\\
\ce{N2(g) + 3 H2(g) &<=> 2 NH3(g)}\tag{B}\\
\ce{2 NO2(g) &<=> N2O4(g)}\tag{C}\\
\ce{H2(g) + I2(g) &<=> 2 HI(g)}\tag{D}\\
\ce{2 W(g) + X(g) &<=> 3 Y(g) + 2 Z(s)}\tag{E}
\end{align}
$$

My attempt: Equilibrium will be influenced if the number of moles on one side of the equation is different than the other side. Thus, I can eliminate option A,B, and C. But how to differentiate between options D and E? They both look right to me, because there are the same amount of moles of gases on both sides, and solids do not play a factor in determining equilibrium.
For reference, the answer given in the book is D.

As Zhe explains in the comments, the answer is D. Even though small changes in pressure are not expected to significantly affect the chemical potential of a solid, it is a function of pressure:

$$\mu(\mathrm{s}) = \mu^\circ(\mathrm{s}) + \int_{p^\circ}^{p}V_\mathrm{m}\,\mathrm dp \qquad T = \mathrm{const}$$
This means that a small change in the chemical potential of the solid also contributes to the total free energy change of the system when the pressure is altered. Usually this response is ignored as $V_\mathrm{m}$ is orders of magnitude greater for gases than for condensed phases.


Convert Free base to 2HCl form I've been using this drug SID 347829124 - PubChem in my experiments from a vendor that supplied it as a $\ce{. 2HCl}$ version, and I switched to another supplier that has a free base instead. The two compounds have different solubility  in water and it would be desirable for me to obtain the $\ce{. 2HCl}$ form to avoid using $20\%$ dimethylformamide to solubilize the free base powder. I thought that using water with a stochiometric quantity of $\ce{HCl}$ would do the trick, but if somebody from the community could confirm this it would be very helpful.

It should work but it is not guaranteed and will depend on application. Drug efficacy is sensitive to the polymorph of the drug and for that reason, a slight change in manufacturing requires a complete redo of any certification process. I assume you are not in the manufacturing business, but it is something to consider and be aware of that your results may vary just by this simple change.


Do metal ions of 4d and 5d series always form low spin complex? I've recently come across a source where it stated that all 4d and 5d metals form low spin complexes irrespective of the strength of ligand.

I'm not sure of the authenticity of the source and couldn't figure out the reason for this phenomenon, thereby enquiring about it.

I feel that the splitting energy is higher than pairing energy in this case, but I am not able to justify it properly.

While the vast majority of the compounds and complexes of the heavier transition metals are low spin (where that concept has any meaning) $\ce{PdF2}$ is one exception, which is unusual not only this way but also having $\ce{Pd^2+}$ in an octahedral environment (see e.g. Wikipedia). Thus your source is not quite correct.

As for the reason why 2nd and 3rd row transition metals are more likely to form low spin complexes than the lighter elements, the reason is given in the answer linked above in the comments. In fact, while the question may be different, the answer is almost a duplicate.


Selective nitro reduction of poly nitro compounds Is there any way through which I could selectively reduce one $\ce{NO2}$ group from a dinitro- or trinitro- compound to an amine.

I was reading Solomons and Frhyle and it says that m-dinitrobenzene can be selectively reduced by using $\ce{H2S}$ in aqueous or alcoholic ammonia. It also says that when it is carried out with 2,4-dinitrotoluene it leads to reduction of the 4-nitro group.

So, is there any way to predict which nitro group will be reduced, and/or any mechanism related to this. I have been frantically searching in books and even an internet article would suffice.

Is there any way to predict which nitro group will be reduced

After having gone through three pages of Google Scholar and reading several research papers looking for an answer, there is definitely no way to predict the product here for every general case. The most detailed paper on the topic of Zinin reduction by far is Porter, H. K. Organic Reactions 2011, 455–481 (link). Even this paper went as far as to say that (p458):

  It is evident that much remains to be learned about the mechanism and that studies using modern techniques and instruments would be desirable.

However, it does give reasonably detailed examples of of selective reduction of substituted dinitrobenzene which are worth mentioning. To begin with:

  The positions of the substituents play an important role in the progress of the reaction. ... In polynitro compounds, the relative positions of the groups determine whether reduction gives monoamino or polyamino derivatives and determine the isomer distribution of the monoamino derivatives.

For example: (p459):

  In substituted dinitro- and trinitro-benzenes, the least hindered nitro group is preferentially reduced.

  In dinitro- and trinitro-phenols and their ethers, a nitro group ortho to hydroxy or alkoxy is preferentially reduced.

This is the most you need to know about this reduction. The other details are more technical, and are attached below are interested readers.

Further detailing:

In the end of the paper, Porter has provided huge tables detailing several (substituted) mono-, di-, tri-nitrobenzene derivatives and their major products on Zinin reduction. So, I studied nearly a hundred different reductions through those tables, and tried to devise other reasonable ways to predict the major product than the two already mentioned above, especially in asymmetrical polynitrobenzene derivatives. I've not listed below mononitrobenzenes since they have only a single nitro group to be reduced. Further, symmetrical polynitrobenzenes (like 2,4,6-trinitromesitylene) have all their nitro groups equivalent, so the product obtained on reduction of either group is same. I've ignored biphenyls and napthalenes in my calculations to keep results simple. Further, exceptional products are not listed here, as they have already been detailed in the paper (p460-p464 under "Side reactions").

The following findings may be used to predict the major product in simple polynitrobenzene derivatives. Keep in mind that exceptional side-reactions do occur, so the prediction may not always be accurate. Yet, a substantially large list of predictable products is obtained as shown below. Examples of compounds are enclosed in parentheses.

The less sterically hindered $\ce{-NO2}$ is reduced in alkyl derivatives of polytnitrobenzene (tables 2 and 3).
Example: 5-t-butyl-2,4,6-trinitro-m-xylene yields 4-t-butyl-3,5-dinitro-2,6-xylidene i.e. a fancy way to say that the $\ce{-NO2}$ group para to the tert-butyl group got reduced.
When a hydroxy or alkoxy group is present, then the ortho nitro group is reduced, even in presence of other halogens on the ring.
(table 4: 2,4-dinitrophenol, 2-chloro-4,6-dinitrophenol, 4-bromo-2,6-dinitrophenol)
If both an alkyl substituent and a hydroxy/alkoxy group are present, and: (table 4)

a single nitro group is ortho to the hydroxy/alkoxy, then that ortho nitro group is reduced
(2,4-dinitroanisole, 2,4-dinitrophenetole, 2,3-dimethoxy-5,6-dinitrotoluene)
no nitro group is ortho/both nitro groups are ortho to the hydroxy/alkoxy group, then the least sterically nitro group hindered is reduced
(2,3-dimethoxy-5,6-dinitrotoluene; 1,2,4-trimethoxy-5,6-dinitrobenzene yields 3,4,6-trimethoxy-2-nitroaniline)

Table 5, 6, 7, 8, 10, 11 do not have any analyseable compounds.
Table 9 lists nitro derivatives of aniline.

preference in reduction is given to the nitro group ortho to the amino group.
(2,4-dinitroaniline, 2,6-dinitroaniline, 3,5-dinitroaniline)
even in presence of other alkoxy or alkyl substituents, the nitro group ortho to the amino group is only reduced.
(4,6-dinitro-o-toluidine, 4,6-dinitro-o-anisidine, 4,6-dinitro-2,5-xylidene)
the ortho nitro group is reduced even in case of N-alkyl anilines, though they are much more bulky.
(N-ethyl-2,4-dinitroaniline, N-methyl-2,4-dinitroaniline)

There are a few cases which I was unable to resolve. The author has not noted them as an exception either. Suggestions are welcome:

Table 4 states that 2,4-dinitroanisole yields 5-nitro-o-anisidine. It is not entirely clear to me why the expected 4-nitro-o-anisidine was not formed. The same yield is observed in 2,4-dinitrophenetole yieldings 5-nitro-o-phenetidine.
Table 7 gives the reduction of 2,4-dinitro-$\alpha$-toluenesulfonic acid as 2-amino-4-nitro-$\alpha$-toluenesulfonic acid, which is not clear why, given the steric constraints at the ortho position.
Table 9 states reduction of 3,5-dinitro-2,4-xylidene gives 2,4-dimethyl-5-nitro-m-phenylenediamine. It is not clear to me why the more sterically hindered nitro group is reduced. It isn't even ortho to the aniline group.

I believe this is a reasonably detailed analysis of the selective reduction of the polynitro compounds.


Graphical representation of reductive powers of NaBH4 and LiAlH4? I don't know whether this is a valid question, I will delete it if it is inappropriate.

I am looking for a graphical representation of the reductive powers of $\ce{NaBH4}$ and $\ce{LiAlH4}$. Such as this

but then including acyl chlorides and the other important organic functional groups.

This is perhaps more than what you are asking for, but the textbook by Clayden et al.1 has a good overview of reducing agents and their substrates:

Reference

(1) Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry, 2nd ed.; Oxford UP: Oxford, U.K., 2012; p 534.


pH of dilute HCl solution What is the pH of a solution of $3.4 \times 10^{-10}$ moles of $\ce{HCl}$ in a volume of $\pu{150.7L}$?

This is dilute so I expect pH to be high.

$$\frac{3.4 \times 10^{-10}\ \mathrm{mol}}{150.7\ \mathrm L} = 2.26 \times 10^{-12} \frac{\mathrm{mol}}{\mathrm L}$$

Then pH is 11.65. Is this correct?

Like you guessed, the pH should be around 7, since the concentration of HCl is very low. Obviously, it can't be 11.65, since it's still an acid. The reason for this discrepancy is that, for low concentrations, you must consider the self-ionization of water in order to get the correct concentration of $\ce{H3O+}$ ions.

Consider the equilibrium:

$$\ce{2H2O <=> H3O+ + OH-}$$
$$K_w = \ce{[H3O+][OH^{-}]} = 1 \times 10^{-14} \tag 1$$

and also the dissociation of the acid:

$$\ce{H2O + HCl -> H3O+ + Cl-}$$

Then
$$c_0 = \ce{[Cl^{-}]} = 2.26 \times 10^{-12} \tag 2 $$

and

$$\ce{[H3O+] = [Cl^{-}] + [OH^{-}]} \Rightarrow \ce{[OH^{-}] = [H3O+] - [Cl^{-}]} \tag 3$$

If we replace $(2)$ in $(3)$ and that in $(1)$, we end up with

$$ K_w = \ce{[H3O+]} (\ce{[H3O+]} -c_0) $$

$$ \ce{[H3O+]} (\ce{[H3O+]} -c_0) - K_w = 0$$

using $ x = \ce{[H3O+]}$

$$x^2 - c_0x - Kw = 0$$

This can be easily solved. Taking only the positive value of $x$:

$$ x = \frac{c_0 + \sqrt{c_o^2 + 4K_w}}{2}$$

Inserting the values of $x$ and $K_w$, the result is:

$$ x = 1.000023 \times 10^{-07} = \ce{[H3O+]}$$

So

$$pH = -\log\ce{[H3O+]} = -\log{1.000011 \times 10^{-07}} = 6.999995 $$

Which is not surprising at all, but is the calculated answer.

For a less diluted solution of $\ce{HCl}$ with concentration of $1 \times 10^{-8} \frac{mol}{L}$, which still can't be calulated directly, the result is $pH = 6.98$.


Where can I find example MS, IR, NMR spectra for teaching molecular spectroscopy? I am teaching a course on analytical chemistry, and I would like to give my students MS, IR and NMR spectra to practice with.

It would be ideal if they could analyze spectra in a more "realistic" way than looking at photocopies of spectra from SDBS. Ideally we'd get some newish (not 30 years old) spectra in an interface where students could, for instance, integrate NMR peaks and zoom into multiplets, automatically get a list of wavenumbers in IR spectra, maybe even deconvolute mass spectra.

I talked to MestReNova, but we can't afford it.

BioRad's SpectraBase might be the best resource I found so far, but it's not as interactive as I'd like it to be.

There is a nice collection at UCLA specifically for teaching but it's 20 years old, mostly static, and pixelated.

I'd appreciate any pointers to suitable collections of molecular spectra, or to free (or cheap) software for spectral analysis.

An elder way would be the via book examples, like «Organic Structures from Spectra» by Field, Sternhell, Kalman, with dozens and dozens of combined exercises in a format like the following:

and currently available in the 5th edition by Wiley (e.g., here). There is a  related project of «Interactive Spectroscopy Problems» by the University of Calgary (here), too.

German VCH published years ago Peter Bigler's training course «NMR-Spectroscopy: Processing Strategies» (e.g., here) including a CD full of 1D and 2D NMR data, which comes close to your aim of interactive sessions to process fids till the plot.  Bruker's WIN-NMR included is old (think Windows 95 era, could be still 16bit), but the experiment-based NMR data should still be processed by current software.

Almost forgotton:  NIST webbook chemistry has for many compounds which may qualify as «easy» IR, UV (condensed phase) and mass spectra, too.  While not aiming to be a teaching set, there are options to download the spectra not only as picture, but equally in the JCAMP-DX format to process, e.g., benzaldehyde.


Nucleation and Phase Equilibrium at the Melting Temperature I used to think that at the melting point, the solid and liquid phases are in equilibrium. That’s because the Gibbs free energy change for the transformation is 0, as is characteristic of equilibria.

However, I recently learnt about nucleation in phase transformations, and I’m very confused. My textbook tells me that below the melting temperature the Gibbs free energy change is not 0 and this energy can go into increasing nuclei surfaces and surmounting the nucleation activation energy barrier. It goes on to say that at the melting temperature, no such energy is liberated (since ∆G = 0) and the nucleation activation energy barrier cannot be achieved. Thus, no nucleation occurs at the melting temperature.

If no nucleation occurs, then how can the solid and liquid phases be in equilibrium? I know about supercooling and that the principle behind it is that it takes time for such an equilibrium to be achieved. However, if nucleation simply doesn’t occur at the melting temperature, then equilibrium will never be achieved...

You probably know that water can be supercooled down to well below -30°C. And it has probably been suggested (if not outright stated) that at those temperatures, the water is in a "metastable" liquid state and that "eventually" it will spontaneously solidify - even if "eventually" may require trillions of years. Since thermodynamics is empirical, you could wonder how that has been empirically observed...but let's not go there. You've been taught that ΔG is a measure of the spontaneity of a process at constant temperature and pressure and at equilibrium. (How a "process" happens when the system is at equilibrium (steady-state) is often ignored. It's fair to say that nothing "happens" at equilibrium, rather when we say "equilibrium process" we mean "a process occurring sufficiently close to equilibrium". But I think I'm getting off topic, again. In considering nucleation, it is vital you distinguish between homogeneous and heterogeneous nucleation. This is the crux of the issue. Water is almost never without contaminates which provide nucleation sites, even if those sites are only on the surface of the container holding the liquid (getting the water this pure is an enormously difficult feat, and how you create a liquid phase without a container I leave to your imagination (hint:rain)). So, when reading about nucleation, you need to distinguish between the situations where there is preexisting centers (contaminants) or whether the water is assumed to be absolutely pure and centers must form spontaneously. Perhaps to answer your question (it would have been most useful if you had copy/pasted the exact text from your book). Once centers (seeds?) exist, the continued growth is pretty much identical whether their origin was homo- or hetero-. Obviously there is no equilibrium between a dust particle (solid) and liquid water. So how can nuclei exist at the MP? Obvious, right? They're hetero-, no formation is required. (Here I avoid the issue of whether FP is (or is not) identical to MP). Wikipedia has a short article on Classical Nucleation Theory which may help. Consider a solid. At its surface, the atoms (or molecules) are different than in the bulk (for most solids, bulk properties aren't reached for top 2 or 3 (or more) layers) but just consider an ideal crystal. Obviously, the surface has fewer bonds and hence has higher energy (is less stable). We call that surface tension (or exactly equivalently, surface energy). The classical theory states it's always positive. This is the energy which needs to be "overcome" to get nuclei, and it's proportional to the surface area. You should note here that our discussion has gone down to the microscopic level, where classical thermodynamic quantities become less and less useful. In other words, the macroscopic classical world thermodynamics needs to be treated carefully here. Statistical thermodynamics is most suitable transition from the classical to the quantum world. These types of processes can be studied using either, but fundamentally the world behaves according to the rules of (statistical) quantum mechanics when you're concerned with atoms depositing on or leaving a surface. The perspective of Classical Thermodynamics is of limited use. (In this case, I think it is probably just as useful, if not more useful, than the quantum mechanical approach. But you're unlikely to get many "fundamental" insights using it.) I'll point out one other apparent flaw in the "activation energy" approach: if the energy barrier is symmetric, then at its peak there should be no preferential direction: the odds of the system moving towards the solid or towards the liquid should be exactly the same (i.e. 50:50), yet the Classical Nucleation Theory requires a fudge factor for this probability - it is an indication that there is more going on here than this approach can handle, that the nucleation activation energy is, at best an approximation. Meaning don't take it too seriously; unless it can be derived from the quantum mechanics, it has no fundamental basis. I don't know much about this area of chemistry. I don't know whether or not it can be derived from the QM, but given that the Classical Nucleation Theory article in wikipedia is speaking about radius as if it is continuous, then I have my doubts. In some systems you begin to see bulk properties at N=100 to N=1000 (N=number of particles), so you'd have to be above this threshold for "classical" behavior to emerge.


Arrangement of elements in the reactivity series I have seen a variety of reactivity series of elements online and also have consulted some books regarding the arrangement of the elements. I understand that the arrangement of elements is based on degree of reactivity of elements with cold water, steam, oxygen, dilute acids etc.
Lithium is less reactive with respect to the criteria mentioned above yet it is placed on the top of more reactive elements in the reactivity series that I have observed.
What is the reason for the position of Lithium above let us say potassium and sodium which are more reactive.
Link to one of the references:
https://kaiserscience.wordpress.com/chemistry/metal-reactivity-series/

The reason that lithium is on the top of the activity series is due to its behavior in aqueous reactions with water and acid. Due to the lithium atom's very small size (only two energy levels), the lithium ion interacts well with water forming ion-dipole interactions.  The lithium atom's small size compared to sodium and potassium is more entropically favorable in the formation of ion-dipole interactions with water and consequently so are reactions involving lithium ion formation favored to to other alkali metals.

The reasoning behind this is that reactions favor a greater state of entropy (disorder) in the universe. With larger cations, more water molecules have to become ordered around the cation to form the appropriate ion-dipole interaction.  However, due to the lithium cation's small size, less water molecules are required to form the ion-dipole interaction, thus less order, being more favorable in terms of entropy.


Does a diatomic gas have one or two vibrational degrees of freedom? Many sources claim that the vibrational degrees of freedom for a diatomic gas is one, but there are also a few which claim it to be two. Which is correct, and why is there even any ambiguity to begin with?

I think this confusion arises from quite different definitions of degrees of freedom. Here are three different definitions that could reasonably arise:

Positional degrees of freedom. This leads to the common $3N$ degrees of freedom based on movement in three directions. Refer to Jan's answer. (Question: Bonds constrain atoms, so why don't they reduce the number of degrees of freedom? Answer: Bonds are not rigid constraints; the atoms don't necessarily remain a fixed distance apart.)
Phase space degrees of freedom. Here we have $6N$ degrees of freedom: $3N$ from the positions of all particles, $3N$ from the velocities of all particles. Positions and velocities are independent, so to specify a system of $N$ particles exactly requires $6$ degrees of freedom per particle. (We also presume no internal structure; as example, for rigid bodies, there are $12$ degrees of freedom: the $6$ mentioned above, $3$ more for specifiying the orientation of the body, $3$ more for describing the speed of rotation of the body.)
Quadratic degrees of freedom. This may seem like a strange distinction to make, but is a key requirement of the equipartition theorem, which is one of the areas where the concept of degrees of freedom is first introduced. These degrees of freedom are the subset of the phase space degrees of freedom whose elements all contribute quadratically to the system energy.

We will clarify these degrees of freedom with the example of a diatomic gas. In this case, there are $6$ positional degrees of freedom, $12$ phase space degrees of freedom, and $7$ quadratic degrees of freedom. Jan's answer describes the $6$ positional degrees of freedom well; specifying the $6$ analogous velocity degrees of freedom gives the $12$ phase space degrees of freedom. Let us enumerate the $7$ quadratic degrees of freedom: $$\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}, q, \dot{q} \quad \Longleftrightarrow \quad \frac{1}{2}m\dot{x}^2, \frac{1}{2}m\dot{y}^2, \frac{1}{2}m\dot{z}^2, \frac{1}{2}I\dot{\theta}^2, \frac{1}{2}I\dot{\phi}^2, \frac{1}{2}kq^2, \frac{1}{2}\mu\dot{q}^2$$ Positional degrees of freedom do not play into this, because there is no quadratic energy contribution associated with them. Instead, we have the six velocity degrees of freedom from the two atoms, $\{\dot{x}_1, \dot{y}_1, \dot{z}_1, \dot{x}_2, \dot{y}_2, \dot{z}_2\}$, which we replace by the conventional center-of-mass coordinates $x, y, z$ and the two angles $\theta, \phi$ that describe the orientation of the molecule, $\{\dot{x}, \dot{y}, \dot{z}, \dot{\theta}, \dot{\phi}\}$. One degree of freedom has "gone missing", because by symmetry there is no (quadratic) energy associated with rotation about the bond axis. The bond also produces two more quadratic degrees of freedom, represented by the vibrational coordinate along the bond, $q$, corresponding to the potential and kinetic energies of vibration.

Hence in one sense there exists only one vibrational degree of freedom; in another sense, two.

Remark. One might argue that the two vibrational degrees of freedom $q, \dot{q}$ are not independent, as they should be related by the equation of motion for a harmonic oscillator; this is important for the statement of the equipartition theorem. This is true for an isolated system, but presumably collisions decorrelate $q$ and $\dot{q}$---I don't have a better answer for this.

At any rate, experimental evidence (heat capacities) does corroborate seven quadratic degrees of freedom for diatomic molecules at high enough temperatures, when vibrational modes are active.


Synthesising Complex Aromatic Compound Synthesize the following compound using only toluene as the carbon source-

I have never done something like this before and I have no clue on how to start.
I am even more confused by the fact that the top two phenyl groups are meta substituted despite having a methyl group which is suppose to be ortho and para directing.

From an answer below it states I need to use Friedel–Crafts acylation. However I am not entirely sure how to do this. This is my guess:

Convert toluene into benzoic acid using $\ce{KMnO4}$ and then react half of it with $\ce{SOCl2}$ to form $\ce{C6H5COCl}$. Then react that with the other half of the benzoic acid followed by Wolff–Kishner reduction to form the below molecule:

After this, I am not sure how to chlorinate it so that I do another acylation.

An alternative approach might be something like this. The tertiary alcohol product with two identical groups suggests that a Grignard reaction with an ester might be a useful step.


Why is 4-hydroxypyridine more acidic than benzoic acid? 4-hydroxypyridine (or 4-pyridone) has a $\mathrm{p}K_\mathrm{a}$ in the ballpark of $3$ to $3.33\:^1$ (in the pdf linked below, the $\mathrm{p}K_\mathrm{a}$ of 1-methyl-4-pyridone has been given to be $3.33$, however I suppose removing a methyl would only cause a slight decrease in the $\mathrm{p}K_\mathrm{a}$ value). This seems an unexpectedly low value to me. 4-pyridone seems more acidic than I expected it to be, since the $\mathrm{p}K_\mathrm{a}$ of other common acids are:$\:^2$
\begin{array}{lr}
\hline
\mathrm{Acid} & \mathrm{p}K_\mathrm{a} \\
\hline
\text{Benzoic acid} & 4.20 \\
\text{Formic acid} & 3.75 \\
\text{Acetic acid} & 4.75\\
\text{Phenol} & 10.0 \\
\text{Carbonic acid} & 6.37\rlap{\:^1} \\ \hline
\end{array}
Now I had expected 4-hydroxypyridine to be more acidic than phenol, because of the negative charge in the former's conjugate base being delocalised onto the electronegative nitrogen, yielding a more stable resonating structure than the latter's corresponding resonating structure. I thought that 4-pyridone's $\mathrm{p}K_\mathrm{a}$ might be somewhere around $\ce{H2CO3}$'s $\mathrm{p}K_\mathrm{a}$, but I was wrong. Also, I wasn't quite sure that it would be acidic enough to have a $\mathrm{p}K_\mathrm{a}$ value any close to other carboxylic acids. But it seems to me that it is more acidic than benzoic acid, formic acid and acetic acid.
Why is this so? I can't see any excetional characteristic in its structure which might impart it such a low $\mathrm{p}K_\mathrm{a}$. This does seem more of an exception to me, as of now, but it would be very helpful if someone clarified and explained such a trend.
References:
$[1]:$ http://gissmo.nmrfam.wisc.edu/static/pKa_compilation_R_Williams.pdf
$[2]:$ https://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/acidity2.htm

The following are two resonance forms of 4-pyridone,

So, when 4-pyridone gets de-protonated, the negative charge on nitrogen gets delocalised on more electronegative atom i.e., oxygen, resulting in more stable resonance structure!
Note that, this explanation is identical to that one which is given when we compare $\mathrm pK_{\mathrm a}$ of benzoic acid and phenol.

Ref: Stability of 4-pyridone vs 4-pyridinol


Which atom goes in the middle of a lewis dot structure? I've read online in multiple sites that the least electronegative atom goes in the middle with the exception of H, which always goes on the outside.

However, in the molecule NaCN, C is in the middle. Could anyone explain to me what is going on with this molecule?

ETA: I'm unclear about your question.  Are you asking why nitrogen isn't the central atom or why sodium isn't?  If it's the former, the explanation is below.  If it is the latter, sodium, then it's because this is an ionic compound, and we build the lewis structures of the cation and anion separately.

First and foremost, carbon (EN=2.5) is less electronegative than nitrogen (EN=3.1), so by your rule it should be in the center.

A more complete explanation of why that is:

Your molecule is sodium cyanide, an ionic compound.  The cyanide group, CN-, carries a negative charge.  If we consider where that charge lies in the ion, it is a bit confusing.  The basic structure of the cyanide is  ${\ce :C:::N:^-}$  .

At first glance we might think that the nitrogen would be the more negative end of the ion, since it is the more electronegative atom.  However, if we consider the formal charges of the atoms in the ion, we see that nitrogen has a formal charge of 0 and carbon a formal charge of -1.  Formal charge is not the same as actual charge, but it does tell us that the carbon will be the negative end of the ion, and that is what we see in its bonding.

Here, the sodium ion, Na+, bonds with the carbon in the cyanide ion.  Even if it weren't a sodium ion, but something like hydrogen cyanide (with covalent bonding) the bond would be formed at the carbon end.  This is because that structure would allow carbon to minimize its electron excess and reach a lower energy state than if the hydrogen stuck to the nitrogen.  Again, formal charges predict the HCN structure over the HNC structure.


Predict the Product(s) of a reaction of tertiary benzylic alkyl halide I could find two products:
The first one where is $\ce{Cl}$ is replaced by $\ce{CN}$.
AND the second one where $\ce{CH3CH2O}$ replaces $\ce{Cl}$.
Did I make a mistake or miss anyone?

$\ce{CN-}$ is a significantly stronger nucleophile than $\ce{EtOH}$ and will therefore primarily displace the chloride. The facts that substitution takes place at a tertiary benzylic carbon, and that a polar protic solvent (ethanol) is used indicate that the reaction follows the $\ce{S_{N}1}$ mechanism (sources: 1, 2, 3). Due to the achiral intermediate (carbocation), the nitrile will be obtained as a racemic mixture, i.e. 2 products.

Elimination is also a possible reaction pathway which needs to be considered, since cyanide is a good nucleophile but a weak base, which favors an E2 reaction unless an excess of $\ce{CN-}$ is present. In this case, the product with the more substituted double bond should be favored.


Convergence issue in Gaussian I have been struggling with geometry optimization of a prostaglandin molecule (20 carbon atoms). I finally got the optimization, all right. Then I did the Frequency job on the optimized structure (I opened the Optimization job .LOG file scrolled up to the last geometry (highest number) and launched the Frequency job from there). Below is my route line from the .gjf file:

# freq=noraman cphf=noread b3lyp/6-31g(d) geom=connectivity

The Frequency job terminated normally and that's what I got in the Frequency job .LOG file:

          Item               Value     Threshold  Converged?
 Maximum Force            0.000005     0.000450     YES
 RMS     Force            0.000001     0.000300     YES
 Maximum Displacement     0.013516     0.001800     NO
 RMS     Displacement     0.003867     0.001200     NO
 Predicted change in Energy=-2.250043D-08

Unfortunately, there is no statement: "Stationary point found", however according to Gaussian manual: "A stationary point is found when the Maximum Force and RMS Force are two orders of magnitude smaller than the thresholds shown, regardless of the values of the displacements."

Is not it just the case? If so, why wouldn't Gaussian kindly state that the stationary point was found?

Is not it just the case?

Looks like it is indeed not the case. I think the phrase OP quoted from the manual, which says that the optimization stops when

  The Maximum Force and RMS Force are two orders of magnitude smaller
  than the thresholds shown, regardless of the values of the
  displacements.

can be interpreted a bit differently, but I guess that algoritmically it is implemented in a pretty simple way: the above mentioned quantities should be at least 100 times smaller that the corresponding thresholds. And this is not the case for the calculation under consideration: Maximum Force value (0.000005) is not 100 times smaller (but only 90) than the threshold (0.000450).


Calculating the concentration of a diluted solution I have this problem:

From a stock solution of initial concentration of $10~\mathrm{mg/l}$, I take $2\,\mathrm{ml}$, and then I add $10\,\mathrm{ml}$ of distilled water.

I am supposed to calculate the final concentration. I have solved the problem in this way.
$$\begin{align}
C_1 \cdot V_1 &= C_2 \cdot V_2 \\
C_1 &= 10~\mathrm{mg/l} \\
V_1 &= 2~\mathrm{ml} \\
V_2 &= (2~\mathrm{ml} + 10~\mathrm{ml})= 12~\mathrm{ml} \\
C_2 &= \frac{C_1 \cdot V_1}{V_2}  \\
&= \frac{(10~\mathrm{mg/l}) \cdot (2~\mathrm{ml})}{12~\mathrm{ml}} \\
&= 1.67~\mathrm{mg/l}
\end{align}$$
Is this correct?

The answer is basically correct (see note at the end of my answer about significant figures) but there is a simpler method, using a dilution factor.

The original volume of your solution was 2 mL, and the final volume was 12 mL, so the dilution factor is simply $\frac{2}{12}$.

The original concentration times the dilution factor gives the final concentration:

$$\pu{10\frac{mg}{L}}\cdot\frac{2}{12} = \pu{1.7\frac{mg}{L}}$$

Note that only 2 significant figures were given in the problem and your answer reported 3.


Why hydrogen bonds are stronger than Van der Waals forces? Why the hydrogen bonds are more intense than Van der Waals forces in the case where the molecules with the Van der Waals forces have a stronger dipolar moment than the molecules with the hydrogen bond?
For example, ethanol (hydrogen bond) has a boiling point of 78 ºC and ethanal 20 ºC (Van der Waals forces) but their dipole moments are respectively 1,69 D and 2,69 D.
I thought the intensity of the hydrogen bond was due to the high polarization of the bond between a small and very electronegative atom like N, O, F, but this is not the case in the previous example, as I understand.

Distance
In vacuum, the calculated (classical) interaction between to aligned dipoles decreases with the square of the distance. If the dipoles can't come very close to each other, the interaction will be weak. If you treat hydrogen bonds as a special case of dipole-dipole interaction, you will find that a N-H ... O=C interaction is much stronger than a C-H ... O=C interaction. For one, the N-H group has a larger dipole moment than the C-H group. For another, the distance between the donor and acceptor are larger for the weak C-H hydrogen bond.
When the dipoles of two ethanol molecules approach, they can come fairly close because the hydrogen on the hydroxyl group lacks inner electrons. For ethanal, in contrast, the carbonyl carbon of one molecule has to come close to the oxygen atom of the other. Both atoms have filled inner shells (and more electrons in the outer shell), so that approach is limited by electronic repulsion.
Covalent character
The hydrogen bond has covalent character, explaining why it is a directional interaction. This covalent character (like all covalent bonds) can not be fully explained using classical (Coulomb interaction) arguments. So in a hydrogen bond, we have an additional attractive term that is lacking in pure dipole-dipole interactions. In extreme cases, you might think of the hydrogen bond as a 4 electron-3 center bond (the 4 electrons are the lone pair of the acceptor and the bonding electrons of the covalent bond of hydrogen with the rest of the molecule; the three atoms are the hydrogen, the electronegative atom it is covalently bound to, and the acceptor atom).
Nomenclature
Some use Van der Waals forces as an umbrella term for (permanent) dipole-dipole, dipole-induced dipole, and temporary dipole-induced dipole interactions. Others use Van der Waals forces (or dispersion forces or London interactions) as a separate and weaker kind of interaction and contrast it with the stronger dipole-dipole interaction.


If you add sodium salt to carbonated water (sparkling mineral water), could you produce sodium bicarbonate? I know that $\ce{CO2}$ can dissociate into bicarbonate ions within $\ce{H2O}$. If Na is added, then sodium bicarbonate is produced.

Could you then create sodium bicarbonate using carbonated mineral water and sodium in the form of table salt or does that have too many minerals aside from $\ce{H2O}$ in it?

Assuming $\ce{Na}$ is used to denote sodium metal and $\ce{NaCl}$ is used to denote table salt, the following reactions in water occur:

\begin{align}
\ce{Na + 2H2O &-> 2Na+ + 2OH- + H2(g)} \tag {1}\\
\ce{NaCl(s) &-> Na+ + Cl-} \tag{2}\\
\ce{CO2_(g) <=> CO2(aq) &<=> H2CO3(aq) <=> H+ + HCO3-} \tag{3}\\
\ce{H+ + OH- &-> H2O} \tag{4}
\end{align}

So (1), (3), and (4) would give you sodium bicarboante if you dried the solution.

(2) and (3) would give you just sodium chloride back if you dried the solution.


Dipole moment of symmetrical molecules While studying dipole moment in organic chemistry, I got stuck at a particular statement written in the book :

  Symmetrical molecules without lone pairs of electrons will have $\mu = 0.$

The main problem was, why would they have $\mu =0$ ?

What type of Symmetrical molecules are possible?

The book gave the example of $\ce{BH3}$. How is this a symmetrical molecule?

Yes, the statement in your book is very vague.  Molecules are often classified by the symmetry elements they contain.  For example, $\ce{BH3}$ contains a $C_3$ axis perpendicular to the plane containing the $\ce{BH3}$ molecule.  "$C_3$" means that if you rotate the molecule around this axis by $360^\circ/3 = 120^\circ$, you'll get a molecule that is indistinguishable from the molecule you started with.  $\ce{BH3}$ also contains 3 $\sigma_\mathrm{v}$ planes of symmetry.  These planes contain the $C_3$ axis and one of the $\ce{B-H}$ bonds.  $\ce{BH3}$ also contains one $\sigma_\mathrm{h}$ symmetry plane that is perpendicular to the $C_3$ axis and contains the $\ce{BH3}$ molecule.  In addition to these symmetry elements, a number of other symmetry elements exist as well.

It turns out that only molecules that belong to symmetry classes (point groups)

$C_n$ (the molecule only contains a $C_n$ axis
$C_{n\mathrm v}$ (the molecule contains a $C_n$ axis and $n$ total $\sigma_\mathrm{v}$ planes)
$C_{\mathrm s}$ (the molecule only has a plane of symmetry that contains
the entire molecule)

can have a permanent dipole moment.  If the molecule belongs to any other point group it cannot have a dipole moment.  Since $\ce{BH3}$ contains a $\sigma_\mathrm{h}$ plane of symmetry ($\ce{BH3}$ belongs to point group $D_{3\mathrm h}$) - it therefore does not belong to groups $C_n$, $C_{n\mathrm v}$ or $C_{\mathrm s}$ - and cannot have a dipole moment.


Name of this glassware? Could you tell me the name of this glassware? I found it in the laboratory with a label that says "container of stoichiometry", but I have not found more information about it, especially about the uses that it could give.

Thank you!

It looks like a vessel for sublimation. The tube going in the top would be filled with cold liquid making a tiny cold finger. The crude mixture is put in the cylindrical cup at the bottom. A vacuum is pulled from the bottom to lower the pressure. The side arm is for pressure release. After the material sublimates and deposits on the cold finger, it can be removed with the pure material on it in order to easily recover the material.


How to calculate the pH of a buffer after dilution You have a beaker containing $\pu{542 mL}$ of a solution of $\pu{0.48 M}$ acetate buffer ($\mathrm{p}K_\mathrm{a} = 4.76$) at $\ce{pH 4.2}$. To this solution, you add $\pu{342 mL}$ of pure water. What is the $\ce{pH}$ of the resulting solution?

The answer is $4.6$. However, I do not know how to arrive at that answer. I know the H-H equation must be used:

$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

Adding water means diluting the buffer.

Use the known concentration, the volume of the buffer and the volume of water added to calculate the new concentrations.


nodes and probability density The square of the wavefunction gives probability density of finding an electron somewhere in the orbital.
The text I'm referring to says that the value of probability density is always higher than zero at any finite distance from the nucleus.
so if probability density cannot be zero, then what about nodes? Arent they regions where probabilty density is zero?

The text is simplifying, as what it says is valid for and only for 1s orbital.
The text author had probably in mind that there is no sharp end of electron density as misinterpretation of orbital shapes may suggest. There is an asymptotic exponential-like ceasing of the wave function values toward infinity.
All higher s orbitals with n>1 have n-1 radial nodes, having zero $|\Psi|^2$ value of wave function at n-1 different radii.
Similarly, 2p orbitals have the node plane passing the center between the two lobes and upper p orbitals have additional node surfaces. Similar for d and f orbitals.