Reduction of oximes with lithium aluminium hydride I know that $\ce{LiAlH4}$ is good a reducing  agent and it donates hydride ions but then I am not able to understand the mechanism of this reaction.

I attempted this way

I am not understanding what to do with the extra charge on nitrogen as there is no acidic medium so that I can donate $\ce{H+}$ to it or can remove ($-\ \ce{OH}$) by any protonation.

Comprehensive Organic Synthesis, 2nd ed. writes (emphasis mine)1

  The reduction of oximes or oxime ethers and esters to primary amines is a useful synthetic transformation that is used as a substitute for direct reductive amination to convert ketones or aldehydes to primary amines. The reaction involves reduction of C=N bond and reductive cleavage of N–O bond. Thus, the transformation requires the use of strong hydride reagents or reagent
  combinations.

I would guess that after the addition of hydride to the $\ce{C=N}$ bond (incidentally, you should not draw it as free hydride) the next step is a substitution at nitrogen. $\ce{X-Y}$ $\sigma^*$ orbitals ($\ce{X},\ce{Y}$ being heteroatoms) tend to be low in energy and rather electrophilic, so this is probably no different.

If it helps you to feel better about the negative charges, you should probably draw the nitrogen as being coordinated to a lithium and the oxygen as being coordinated to an $\ce{AlH3}$ fragment, or some permutation thereof. The close association of the heteroatoms with metal fragments also encourages "hydride" to attack. Again it's not really hydride attacking but rather the $\ce{Al-H}$ $\sigma$ bond.

1 Abdel-Magid, A. F. Reduction of C=N to CH–NH by Metal Hydrides. In Comprehensive Organic Synthesis, 2nd ed.; Knochel, P., Molander, G. A., Eds.; DOI: 10.1016/B978-0-08-097742-3.00802-8.


Liquid for cooling beverages I am looking for any kind of fluid which has a freezing point below the temperature of a regular freezer  (about -20 C). My plan is to keep this liquid in my freezer for fast cooling of beverages, so the liquid needs to be harmless to drink in case some is left on the can.
Suggestions? Or maybe the liquid will cool down the beverage too fast making it impossible to get a proper even drinking temperature without freezing the beverage?

If brine is either unpalatable or has too high a freezing point for you, other options include a solution of glycerol in water. Glycerol is perfectly safe to ingest in modest quantities -- it's used as a food additive and a medicine -- although it tastes a little weird (a bit like extra-sour sweet and sour sauce).

Although it's mostly been replaced by ethylene glycol now, glycerol used to see a lot of use as an antifreeze, and is still used in domestic applications like ice cream makers. Apparently it's best used in a 60-70% solution, when it can get a freezing point below -40°C.

The paper linked below puts the optimum point at -46.5°C with two-thirds glycerol, one-third water by weight.

"Freezing Points of Glycerol and Its Aqueous Solutions", Leonard B. Lane, Ind. Eng. Chem., 1925, 17 (9), pp 924–924, DOI: 10.1021/ie50189a017


Magnetic moment from given CFSE value Spin magnetic moment of a octahedral complex having CFSE = $– 0.8∆$ and surrounded by weak field ligands can be:
A. $\sqrt{15} \space \mu_B$
B. $\sqrt{8} \space \mu_B$
C. (A) & (B) both
D. None of the above

The CFSE value given in the above question corresponds to only $d^2$ configuration for weak field ligands. Hence, according to me, the value of spin magnetic moment should be only $\sqrt{8} \space \mu_B$.
For the answer to be $\sqrt{15} \space \mu_B$, the configuration would have to be $d^7$ requiring the CFSE value given in the question to have an additional 2P(twice of pairing energy) but that is not the case.
Hence answer should be option (B) but the source of the question says that the answer is option (C).
I am unable to understand the consideration of both $d^2$ and $d^7$ configurations.
Can anyone please explain?

Even for weak ligands, the electrons will eventually have to pair up, once the number becomes more than $5$. So, $\mathrm{d^7}$ is indeed possible with a weak field ligand. Consider $\ce{[CoCl6]^{4-}}$, for example.


How to merge PDB-files to build trimer from monomers Problem

Using WinCoot or SWISS PDB viewer, how would one merge monomeric structures that should be in dimer formation?

What I have done:

I have downloaded the PDB-file which shows the monomer with a bound drug (Infliximab Fab fragments), but it should be a trimetric structure as whole -although you only get the monomer when you download the file (TNFa) .
I have opened the file in both coot and SWISS-PDB viewer, and opened three copies of the monomer, but they just overlap and I am not able to combine them to the proper structure

I also know that it is possible to download the "biological assembly" PDB-file that has the entire trimetric structure, but the point here is to be able to do that on my own (which should be easy for those who know it?).

I am very new to all of this modeling, so there might be something basic I don’t understand. But I though that you can somehow "merge" the monomers by using some "layer" function? I am really unsure.

The PDB-file only contains the asymmetric unit, and no information on a potential biological relevant multimer. So you will need to get information about the state in solution experimentally.

That being said, one can often deduce from the size of the contact surface whether you have a potential quarternary structure. You can do this by hand, but the PISA-server is quite helpful there, and it will also output the multimer PDB-file:

http://www.ebi.ac.uk/pdbe/pisa/

If you just want to apply symmetry operators to your asymmetric unit in order to create symmetry mates, you can do so in

Pymol:
- click Action/generate/symmetry mates/within X Å
- from the command line:
        symexp prefix, selection, cutoff
        , e.g. symexp sym,1GVF,(1GVF),5
   see also http://pymolwiki.org/index.php/Symexp

Coot:
- if you just want to look at the symmetry mates, activate View/Cell&Symmetry with an appropriate radius and/or select "Symmetry by molecule"
- if you would like to actually create the symmetry mates, try Extensions/Modelling/New Molecule from Symmetry Op  - however you will have to specify the SymOp manually.

You will need to have the CRYST1-record in your PDB-file with the correct space group.


Why Cu+ is unstable in aqueous medium? This was one of the question that was given in the school exam.

But what I am expecting is that $\ce{Cu+}$  attains a $d^{10}$ configuration by losing one electron from s-subshell. Since it has fully filled d-orbital, it should be stable. But it is found that it is unstable and tends to disproportionate in aqueous medium or with moist air to form $\ce{Cu^2+}$ where  $\ce{Cu^2+}$ has incompletely filled d-orbital ($\ce{d^9}$ configuration) . For example:

$$\ce{Cu2O(s) + H2SO4(aq) -> Cu(s) + CuSO4(aq) + H2O(l)}$$

$$\ce{Cu2O(s) + 2H+(aq) -> Cu(s) + Cu^2+(aq) + H2O(l)}$$

Oxidation number changes: $\ce{2Cu(I) -> Cu(0) + Cu(II)}$

(source)

Why is this so?

You are quite correct in that it appears at first sight that $\ce{Cu+}$ should be more stable than $\ce{Cu^2+}$, but in aqueous media it isn’t.

Stability in aqueous conditions depends on the hydration energy of the ions when they bond to the water molecules (an exothermic process). The $\ce{Cu^2+}$ ion has a greater charge density than the $\ce{Cu+}$ ion and so forms much stronger bonds releasing more energy.

The extra energy needed for the second ionisation of the copper is more than compensated for by the hydration, so much so that the $\ce{Cu+}$ ion loses an electron to become $\ce{Cu^2+}$ which can then release this hydration energy.  A nearby $\ce{Cu+}$ ion is the most facile reduction target for the removed electron, which is why $\ce{Cu(s)}$ is also formed.

Hence, $\ce{Cu^2+}$ is more stable than $\ce{Cu+}$ in aqueous medium.


How can I formulate a rate equation using a solid catalyst? When formulating a rate a equation two experiments must take place: one in which the reactant concentrations are varied, and one in which the catalyst concentrations are varied.

If I am given a sample of a solid catalyst, such as $MnO_2$, how can I change its concentration? Because I can't dilute a solid and increase the volume, which is how I'd do it for a catalyst in solution.

You will perform the reaction inside a container or solution of some particular volume. This means that you can change the catalyst concentration inside that volume by adding more or less catalyst in different experiments. So you don't increase the volume of the solid (i.e. change its density), you change the mass of solid inside the reaction volume.


2-step reaction mechanism for the decomposition of NO2Br I need to propose a 2-step reaction mechanism for the following reaction:
$$\ce{2NO2Br -> 2NO2 +Br}$$
And prove the mechanism is consistent with: $$v=k[\ce{NO2Br}]^2$$
For the mechanism to be consistent with the rate equation, the first equation would have to be the limiting one and also be in this form:
$$\ce{2NO_2Br -> Products}$$
I've tried using:
$$\ce{2NO2Br -> N2O4 +Br2}$$
$$\ce{N2O4 -> 2NO2}$$
But  the problem with this mechanism is that $\ce{N2O4}$ is not unstable enough to be consumed immediately.
Any help with this problem?
Would also appreciate some advice on proposing reaction mechanisms. Thank you.

It is more probable  like
$$\begin{align}
\ce{NO2Br &-> NO2 + Br} \\
\ce{NO2Br + Br &-> NO2 + Br2} \\
\ce{2 Br &->  Br2} \\
\end{align}$$

The last reaction is a minor one in case concentration of $\ce{Br}$ is low.

The reaction rate order can be concentration dependent and need not be the integer.

In fact, it is rather mathematical parameter, related to solution of differential equations for a complex reaction system.

If the 2nd reaction is fast enough, the overall reaction rate is given by the slow rate of generation  of $\ce{Br}$, which fast reacts to form $\ce{Br2}$

If the 2nd reaction is slow enough,
it's rate $$k_{\rm 2}\cdot [\ce{NO2Br}][\ce{Br}]$$ can be written as $$k_{\rm 2a}\cdot [\ce{NO2Br}]^2$$

The exact solution is to solve system of differential equations for the rates of the concentration changes.

$$\frac{\mathrm{d}[\ce{Br}]}{\mathrm{d}t}=k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}] - k_3 [\ce{Br}]^2$$

For the dynamic equilibrium of the steady concentration of $\ce{Br}$:

$$\begin{align}
0&=-k_1.[\ce{NO2Br}] + k_2.[\ce{NO2Br}][\ce{Br}] + k_3 [\ce{Br}]^2 \\
[\ce{Br}]&=[ -k_2.[\ce{NO2Br}]+\sqrt((k_2.[\ce{NO2Br}])^2+4.k_3.k_1.[\ce{NO2Br}])]/(2.k_3) \\
\frac{\mathrm{d}[\ce{NO2Br}]}{\mathrm{d}t}&=-k_1.[\ce{NO2Br}] - k_2.[\ce{NO2Br}][\ce{Br}]\\
\end{align}$$


Is picric acid a very strong acid? Picric acid has nitro groups in para and ortho positions, so it has great possibility to go for resonance, thus it is supposed to be a very good acid.
But a question arises! The $\ce{O}$ of $\ce{-OH}$ has intramolecular $\ce{H}$-bonding with ortho-nitro groups present on both sides. So will the firstly mentioned theory work well or is the answer something else?

Yes it is very strong, albeit not compared with Inorganic acids like HCl...

As far as organic acids are concerned, it is really one of the strongest ones... as its pKa is 0.3 [ more acidic than acetic, benzoic, formic, oxalic, trichloro-acetic acids...]

Possibility of intramolecular H-bonding, though exists, is offset reasonably as the negative charge of phenolic oxygen is very nicely delocalized over the 3 nitro groups. [In fact, if we compare the pKa's of phenol (9.98) with o-, m-, p- nitro phenols (7.23, 9.88, 7.15 respectively); we can see that H-bonding is not decreasing the acidic nature "upto greater extent"]

It is always possible that due to bulkier -NO2 groups, the repulsions betweens the LP of O atoms in -OH might be leading to the H of -OH group to be oriented far off from the range of effective H-bonding


Why some compounds form different crystal structure at different temperature? Lets take an example :-

  Sodium superoxide has three different crystal structure, the marcasite
  structure at nearby 81 K, the pyrite structure between 196 K and 223 K
  and calcium carbide structure at room temperature.

There are many more compounds which have different structure at different temperature. If temperature is increased, bonds breaks, atoms scatters and decomposes into a different product (compound or element) . But this does not happen. Rather bond breaks, atoms rearranges to form a different structure of same compound.  Why does this happen?

You are just entering into the field of Polymorphism. I couldnt find any source for Why polymorphism occurs but you can think about it as Energetically favorable structure at a particular temperature.

Related - Why polymorphs form?


Tap water pH adjustment using citric acid I've lowered tap water pH from 7 to 6 using citric acid. After couple of hours pH raised again to 7.

What is the reason for this raise? Citric acid overcomes waters buffering  feature then raise the $\ce{H+}$ concentration in water. Why pH raise again?

The pH was measured using a pH meter.

Maybe citric acid have had a negligible effect on the $\mathrm{pH}$. Everybody knows that in contact with air, pure water is carbonated by $\ce{CO2}$ from the atmosphere. And the $\mathrm{pH}$ may go down to $5.5$. Later on, the water may loose its $\ce{CO2}$, and the $\mathrm{pH}$ goes back to $7$.


Etching PCB - dealing with anhydrous ferric chloride I'm looking into PCB etching,which requires ferric chloride (There are other etchants, but they're pretty hard to come by where I live and too expensive if I can find them).
The issue is that where I live, only anhydrous ferric chloride is sold (Which is stupid. They don't even label it properly as anhydrous - got exposed to HCl fumes for 2-3 hours, thankfully the doctor found nothing abnormal with me and I got sent off with some anti allergy meds).
I'd like to know what equipment I'll need to deal with this stuff safely. I always use nitrile gloves with these, but in particular what kind of respiratory protection will I need? I plan on just having 4 PC fans blow these fumes outside the window. If I must, I will invest in respiratory protection (It's very expensive where I live, the NIOSH approved ones at least). I'm assuming a 3M 6002 will be enough.
Please keep in mind I will be using this stuff at home. If you have any suggestions regarding alternatives, do mention them. If I can find it, I'd rather use that than this.
If this isn't the proper place to ask these questions, please direct me to a place more suitable for these questions.

$\ce{FeCl3}$ ( and similarly $\ce{AlCl3}$ ) has very different properties, comparing its anhydrous and hydrate form. They cannot be easily transformed one to the other. Heated hydrate decompose to the oxychloride or oxide eventually. Anhydrate put in water undergoes violant hydrolysis.
I suggest obtaining the hydrate, or preparing it yourself, e.g. reacting of iron, $\ce{HCl}$ and $\ce{H2O2}$ ( this mixture of the 2 reagens was an alternative Cu etching solution, when $\ce{FeCl3}$ hydrate was not easily available in the communist Czechoslovakia, but is more aggressive then $\ce{FeCl3(aq)}$ ).
Another option is to let react the anhydrate with limited water outdoors and that adding concentrated HCl. For the dissolving itself it would be the best to put in directly to the acid:water 1:1, but it would develop dangerous fumes.
Avoid mixing concentrated stuff, it would turn into a chlorine chemical weapon. Start with 10% peroxide and 1:1 acid, calculate the ratio. Tune it later at will. Keep the etching solution rather mild than strong, to simulate etching performance of FeCl3. My experience with etching $\ce{Cu}$ circuit boards was that too strong solution was often etching metal under the protection layer. Be aware of proper after treatment, to prevent future corrosion.
Usual gloves, coat and eventually shield. Well vented place is needed, at least for first experiments, until you know the behaviour ( I mean rather passively - like outdoor - and just temporary for initial experiments. when it is settled, it is not much needed. ).


How to create different fluorescent colors using household items This is a request:
I wanted to create fluorescent liquids that have different colors.

I'm going to dye my flowers in different fluorescent colors and I'd need a recipe for fluorescent colors that aren't too toxic.

Please bare in mind that the injection of the liquid can only happen by spraying or watering with the liquid.

The liquid needs to be able to handle being Waterdown by tap water.

I'd recommend using household items to make this more helpful to the average day user. Also any color will do but, the most wanted Colors are:

List item

Blue
Green
Red
Yellow

If you dismantle highlighter pens, you can find the felt material inside. This material contains fluorescent chemicals, such as pyranine. This felt like material is usually encased in a plastic film, remove the film and soak the felt in water.

If you use too much water then it may become too dilute and you may start to encounter an issue with the ink/water mixture sticking to the flower.

Are you using a black light to view the fluorescent colour?

it might be worth trying to cut the stems and keep the flowers in the ink containing water. Then as they absorb the water they will absorb the fluorescent ink as well and may take on some of the colour (and viewable under black light).


What is meant by work done "expanding against atmosphere"? In the book, Chemistry : A molecular approach, I came across this line:

Over here what does the author mean by work done by expanding against atmosphere?

Imagine someone was pushing you against a wall. If you spend some energy overcoming that push, wouldn't you say that you did some work (expenditure of energy) against that pressure?

With a similar line of reasoning, the atmosphere is exerting a pressure of one atmosphere trying to compress your gas down. That's a huge pressure (nearly $10^5$ pascals!) and the gas in question has expanded against it. The wording makes sense.

In general, whenever gases are produced as a result of some reaction (in this case combustion), they are initially at a higher pressure than atmospheric pressure. This isn't an equilibrium situation. So, they expand, and lose their pressure until it finally becomes equal to that of the atmosphere.


'Chemical' First-order Phase Change First order phase changes occur when one local minima of the Gibbs Free Energy becomes deeper than another. Thus at 1 atm and 99 °C, the Gibbs Free Energy of liquid water is less than the Gibbs Free Energy of steam at those conditions (though both states form local minima). As the temperature increases, the relative depths of the two minima change: at 100 °C they are the same depth, and above 100 °C the vapour-phase minimum is lower.

I'm wondering if similar first-order changes can occur in chemically reacting systems? It seems to me that equilibrium constants always change continuously with temperature: are there any situations in which this isn't the case, and there exist discrete jumps in chemical compositions? And if these jumps in the deepest minima of the Gibbs Free Energy never occur in chemically reacting systems, do we know why not? Of course, situations in which a chemical reaction is accompanied by a phase change don't count!

Well, let's see. As you know, the equilibrium constant is determined by the reaction's $\Delta_rG:\;K=e^{-\frac{\Delta G}{RT}}$, so a discrete jump in the former means a discrete jump in the latter. What is that $\Delta_rG$? That's the sum of $\Delta_fG$ of products minus that of starting compounds. Now what is the $\Delta G$ of formation of a compound? A continuous function, unless there is a phase transition, which by your problem definition is not the case. Well, a linear combination of a few continuous functions is still a continuous function.


Imposing particle number subspace on second quantized Hamiltonians Here is my question:
Second-quantized Hamiltonians act on any state, regardless of how many electrons the state has. They are also block diagonal on different particle-number subspaces, as they commute with the electron number operator.

Now, suppose that one writes down a molecular Hamiltonian in the second-quantized formalism. In order to get the ground state energy, does one need to project the Hamiltonian on the correct electron subspace and diagonalize?

Or is the bare diagonalization of the Hamiltonian (without projecting) sufficient to pick the subspace with the right number of electrons?

$\newcommand{\Ket}[1]{\left|#1\right>}$
$\newcommand{\Bra}[1]{\left<#1\right|}$
$\newcommand{\BraKet}[2]{{\left<#1}\left|#2\right>}$

It's an older question, but one worth answering!

For the uninitiated, second quantization is a bookkeeping technique that describes many-particle systems as excited states of a field, usually taken to be the reference field or vacuum state. I found the most clear description of it in atomic_rabbit's answer to this reddit post. The second-quantized operator is useful because this form allows it to be applied to an indeterminately sized system (the number of particles is not explicitly stated in the second-quantized operator).

Now, onto the question:

The "first-quantized" (electronic) Hamiltonian has the form (in atomic units):

$\hat{H} = \sum_i\frac{1}{2}\nabla^2_i - \sum_i\sum_A\frac{Z_A}{r_{iA}}+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}} = \sum_i\hat{h}(i)+\frac{1}{2}\sum_i\sum_j\frac{1}{r_{ij}}$

which, upon second quantization, goes to:

$\hat{H} = \sum_{p,q}\Bra{p}\hat{h}\Ket{q}a^+_pa_q + \frac{1}{4}\sum_{p,q,r,s}\Bra{pq}\Ket{rs}a^+_pa^+_qa_sa_r$

In the case of "first quantization", we construct the matrix form of the operator by projecting it onto whatever basis functions we're using. We then find the eigenvalues of the operator by diagonalizing its matrix form. Your question is an excellent one, because in second quantization, this projection seems to be already done within the operator, and so we should be able to diagonalize the operator outright. Right?

The catch is that the second-quantized operator still has the creation and annihilation operators that have not yet been projected onto any basis. This is equivalent to us not knowing whether there is a particle in the p$^{th}$ state. And what are the states? Those are properties of the field, which we must find out by projecting the second-quantized operator onto said field.


Do both H (enthalpy) and ∆H, or ∆H alone have a physical significance? My chemistry teacher said that H (enthalpy) alone has no physical significance it is ∆H or change in enthalpy that has a physical significance. But my book says that both have a significance and H alone represents the total heat content of the body. Thus I am in a confusion, please help me.

Offcourse H and ∆H have physical significance. H represents the enthalpy of a system. Enthalpy refers to the sum of the internal energy of a system plus the product of the system's pressure and volume.

The ∆ symbol is used to represent change. Therefore, ∆H represents the change in enthalpy of a system in the reaction. So your book is completely correct.


Are triglycerides macromolecules? I've previously been taught that triglycerides are macromolecules. However, a current professor of mine gave the definition of macromolecules as "polymers of covalently linked monomers". Since, triglycerides aren't formed from repeating monomers, are they considered macromolecules?

Depending on who you ask you will receive a different concept to define macromolecules. These concepts typically arise from the sub-field of those who made the definition, so a polymer chemist is more inclined to consider repeating monomers of exactly one type since that is what plastics are. At the same time, the branch of macromolecular chemistry usually deals with systems that are much smaller than your average multi-domain protein.

The definition most can agree on (to a certain extent) is that macromolecules must be ‘larger’ than ‘small molecules’ (note the pairs of inverted commas). In this definition, it is perfectly reasonable for lipids to be grouped as macromolecules as they are much larger than most secondary metabolites. However, it is true that they are not made up of subunits linearly connected (whether repeating or not) so one may not consider them biopolymers.


Dissolving copper minerals in acid I have copper minerals in rock that has been pulverized. These copper minerals are pure (native copper), chalcocite, chrysacolla, malachite, tenorite, cuprite and azurite. Does anyone know which acids will dissolve which minerals? I need this information as a lab has proposed to use $\ce{HF}$, $\ce{HNO3}$ and $\ce{HClO4}$ to perform an assay and I'm not sure it will work 100%.

If I remember correctly, we (an analytical laboratory) would generally get all copper minerals into solution using a nitric/perchloric digestion, depending on the sample matrix.  Unless the sample matrix as a whole were highly silicious in nature an additional HF digestion was not required. As commented above though, $\ce{HNO3/HF}$ may work for you also.


Indirectly Measuring Diffusion of NO2 into Water I am working on a data science project with the central hypothesis is that air-born pollutants can become water-born through diffusion. The example I am using as a proof of concept is $\ce{NO2}$. The problem I am having is that I do not understand the chemistry of how it interacts with $\ce{H2O}$ outside a laboratory setting that well.

$\ce{NO2}$ reacts with $\ce{H2O}$ to form $\ce{HNO3}$, $\ce{HNO2}$, $\ce{H+}$, and $\ce{NO3-}$. In addition, $\ce{HNO2}$ also reacts with water to form $\ce{NO}$, $\ce{H+}$, and $\ce{NO3-}$.

Do these reaction happen so fast that measuring dissolved $\ce{NO2}$ or $\ce{HNO2}$ in a sample an unreliable measurement?

If so, what is the best measurement I can use? $\ce{H+}$ and $\ce{NO3-}$ are both part of a lot of other very complicated biological pathways and have even more possible sources, so there is no way they can be relied upon. That leaves $\ce{HNO3}$, $\ce{HNO2}$, or $\ce{NO}$ as possibilities. Which of those would be the best to use as an indirect measurement?

Thanks for any help you can offer in advance. I am decent enough at chemistry in a lab, but application to real world systems is way beyond my abilities.

NO2 is a free stable radical, better express as $\ce{.NO2}$, and being a radical, does not directly react with H2O, or H+ or OH-.

The reaction mechanism likely proceed via a self-reaction creating nitrogen tetroxide:

$\ce{.NO2 + .NO2 <=> N2O4 }$

In a diluting medium of air, depending on the concentration of nitrogen dioxide, this reaction is kinetically expected to be slow.

When it does occur, it can be followed by:

$\ce{N2O4 +  H2O -> HNO2 + HNO3 }$

Source: See Reactions (9) and (10) on Page 28 of this paper.

Now, one can ignore the intermediate reaction mechanics, for simplicity, however, this does present a problem for some, as yourselves, as clearly the rate determining step is not even cited.

Also, if you review the paper, there are other reaction equilibriums that you may have to take into consideration. As a sidebar, what the paper does not mention is that aqueous nitrite is a photo-sensitive producing products, so you may wish to control the amount of lab light exposure as well.


Why does the composition of Ocean Water list sodium and chloride separately? This is the site I am using.

It is also found here, but this uses mg/L instead of percents.

It lists Sodium at 30.8%, and Chloride at 55.3%. How can this be? Shouldn't the two be listed as Sodium Chloride or salt? Because as far as I know, there is no way to obtain chloride on its own.

First of all, of course you could provide NaCl content as a weight or molar percentage, and I can demonstrate if necessary.

However, this doesn't really reflect the reality: ocean water is a solution of many salts. Chloride does not necessarily only come from NaCl, but can also come from KCl, MgCl$_2$, and so on. All the ions are solved individually, and if you start evaporating the water, you will get a mix of salts.

So there's really no point in showing NaCl percentage.

Of course, you're right in that it is impossible to obtain pure chloride ions. But the chlorides you will obtain will not necessarily be those of sodium.


How to calculate he equilibrium constant change for acid-base reactions in a different solvent? So we know that $K_\mathrm w$ is $10^{-14}$. But what if I am working with non-aqueous systems? Suppose I were to add LDA to liquid ammonia, or dissolve HCl in concentrated sulfuric acid. How would I calculate the percentage of dissociation? I have some inkling of how to do this but I don't really see discussions online and want to make sure my procedure is correct. Furthermore, I don't know of what good keywords to find the self-dissociation constant of ammonia or sulfuric acid are and this is really frustrating.

To emphasize: I am looking for solutions for non-aqueous systems.

For all solvents that can dissociate into ions, a dissociation constant can indeed be measured. Expect them to be unpredictable in each direction.

Furthermore, the $\mathrm pK_\mathrm a$ values you know are typically those measured in water, although some organic chemists rely more heavily on the DMSO tables e.g. the one by Evans. Just a quick look at the first column shows you that there is no way you can predict the $\mathrm pK_\mathrm a$ value in one solvent from the one in a different solvent:

$$\begin{array}{lcc}\hline
\text{compound} & \mathrm pK_\mathrm a(\ce{H2O}) & \mathrm pK_\mathrm a(\ce{DMSO}) \\ \hline
\ce{HF} & 3.17 & 15\phantom{.0}\\
\ce{HCN} & 9.4\phantom{0} & 12.9\\ \hline\end{array}$$

While $\ce{HF}$ is the stronger acid in water, $\ce{HCN}$ is the stronger acid in $\ce{DMSO}$.

Couple this with a different dissociation constant means that you can calculate nothing ab initio.


What is the correct way to illustrate a bond to something irrelevant in a structural formula? You can probably tell I'm not really a chemist (hi from physics!), but I've suddenly found myself having to draw some structural formulae, hence the stupid question.

I've drawn a diagram of the four DNA bases:

I don't care what they're attached to, and I don't fancy drawing an entire DNA molecule, but I'd like to show where they bond to the rest of the nucleotide. I've used \cdots ($\cdots$) currently, but that doesn't feel right. I considered R, but I'm not sure if an entire DNA molecule really counts as an R group. I saw a wavy bond once, but it feels like those are confined to stereochemistry.

What's the convention in cases like this? (And what do I Google to find it next time I want to know?)

According to the Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), in the structure diagram of the parent structure, variable substituents should be indicated by an unambiguous label.

  This label should be chosen so that it cannot easily be confused with other valid element symbols or fragment abbreviations. Traditionally, the letter “R” is widely used, followed by a superscripted number (R1, R2, etc.) or prime(s) (R', R", etc.) if necessary to distinguish different groups. It is not acceptable to use a letter followed by a subscripted number, since subscripted numbers are more commonly used to indicate repeat counts, and confusion between the two is inevitable. Other common symbols include X, Z, and G. (…)

The label may be used with or without an explicit list of allowed substituents (e.g. in tabular form).

The wavy line is preferred in structure diagrams of the variable substituents, in cases where the external connection points must be indicated for the variable substituents.

Therefore, if you consider nucleobases with variable substituents, you may use labels such as “R”. However, if you consider the nucleobases to be the variable substituents of another (larger) parent structure, you may use a wavy line.


Naming an ensemble of compounds As part of intern work I am running simulations on a group of compounds, but as it is I have little in the way of O-Chem experience and I wanted to be able to at least get some direction on naming these compounds (maybe some examples).

I think I've got the names of the functional groups down right, and I've recognized a few of the reaction types, but after that what do I do?

What are the other 2 reaction types?

How should I take the names of the products/reactants and the functional group names and stick em together.

What about when the functional group listed is bigger than the main group. (example R8 and ring opening)?

I should have written F1-F8 but R1-R8 represent functional groups if I'm not mistakened.
Here are my best guesses

R1:alkane

R2:alkene

R3:alkyne

R4:amine

R5:alcohol

R6:ketone

R7:amide

R8:benzene

Pericyclic ring opening, cyclobutene to butadiene (Woodward-Hoffmann rules).
Claisen rearrangement, vinyl ether to ketone (re bullvalene).
Diels-Alder addition.
Sn2 displacement; AgF will move that along (not a soft Lewis base solvent).
Alkene to expoxide is not done with water.  Direct catalytic oxidation, base elimination from the cyanohydrin (first adding HOX), peracids.
Primary alcohol to aldehyde is Jones reagent (in water), Collins reagent (in anhydrous pyrdine, and much better. There is no product in the sludge), Swern oxidation (makes you unpopular downwind), fancier stuff.
Markovnikov addition of HCl to an alkene.

The Woodward-Hoffmann rules tell you what conditions are required (thermal vs. photochemical) and the associated stereochemistry (suprafacial vs. antarafacial) re conservation of orbital symmetry.


Does electron mass decrease when it changes its orbit? I have studied  this in my chapter atomic structure that when an electron changes its orbit from lower energy to higher energy state , it does not state in my book that it moves there but that it disappears.It just kind of reaches that orbit without crossing the path.We don’t know whether it moves there or not.
So , I also read about the fundamental properties
of electron in which it said that when an electron moves or is in motion , it’s mass changes.
If we see , We know that electron only absorbs certain wavelength of light. If we can find that when it changes its orbit , if it’s mass decreased or not during that interval.Is it that we can find that whether  electron moves and changes orbit or it just disappears kind of.
Yes , I am only talking electron in an hydrogen atom here.
Please do give your answer in reference to other atoms if it is helpful.

I'd like to copy the answer by John Rennie to a similar question at Physics.SE, since it's much better than current answers here IMHO. Although it mostly speaks about binding energy, the same principle applies to the quanta of excitation energy.

The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron together then they come to $1.67353272 \times 10^{-27}$ kg. The difference is about 13.6eV, which is the ionisation energy of hydrogen (though note that the experimental error in the masses isn't much less than the difference so this is only approximate).
This shouldn't surprise you because you have to add energy (in the form of a 13.6eV photon) to dissociate a hydrogen atom into a free proton and electron, and this increases the mass in accordance with Einstein's famous equation $E = mc^2$. So this is a direct example of the sort of mass increase you describe.
However you can't say this is an increase of mass of the electron or the proton. It's an increase in mass of the combined system. The invariant masses of the electron and proton are constants and not affected by whether they're in atoms or roaming freely. The change in mass is coming from a change in the binding energy of the system.


Restricted open-shell versus unrestricted methods for open shell systems Open-Shell systems (radicals and so on) can be modeled by single-determinant quantum chemical methods using unrestricted (u) or restricted open-shell (ro) methods (I am aware that single-determinant approaches are not always reasonable for open-shell systems, but let's ignore that for a second and suppose we have to use a single-determinant method).

Unrestricted calculations may suffer from spin-contamination. Restricted open-shell calculations remedy this, they are unaffected by spin-contamination. However, restricted open-shell methods are much more computationally expansive. Also, Koopmans' theorem cannot be rigorously applied in the case of ro calculations.

My question is:

Apart from the before mentioned issues, are there any disadvantages of restricted-open-shell computations in comparison to unrestricted calculations? Are there any situations where an unrestricted calculation would be more appropriate than a restricted-open-shell calculation?

[Methods of interest are - for example - MP2, DFT and HF.]

There is already a somewhat related discussion: U- or RO-method for Singlet-Triplet Gap?. However, my question is much broader, it includes for instance: Are there cases when it is absolutely not ok to use ro methods, and you have to use unrestricted methods?

To be more precise, it's not so much that restricted open-shell calculations are unaffected by spin-contamination as that you are forcing them to give back the multiplicity you want.

Broadly speaking, closed-shell methods, restricted open-shell methods and constrained unrestricted open-shell methods all give correct spin multiplicities, but that's because all of them assume from the beginning which is the correct multiplicity and force it into the calculation - there is always a restriction in place. For closed-shell methods, it is that the orbitals occupied by both spin populations are the same; for restricted open-shell methods, the same applies to all but the excess spin electrons from the larger spin population. But more recent developments point to restricting multiplicity after the fact - by adjusting orbital population to minimise spin contamination. There are a couple of methods to do that, but they start from an unrestricted calculation (much faster, as you mention), and then force the population in a way that minimises spin contamination - either by forcing orthogonality constraints between the $\alpha$ and $\beta$ determinants, or by parametrising spin contamination in a way that allows it to be fitted to the expected total multiplicity by constraining occupation. This retains the computational advantage and simpler mathematical implementation of unrestricted methods but makes sure that the multiplicity of the model system reflects the multiplicity of the modelled system with generally very small additional computational effort.

I'd suggest that this is going to be an active area of research that will see increased use in the next few years.


Why doesn't natural gas combustion release only one-sixth as much carbon dioxide as coal burning? Burning hydrogen releases a little more energy than burning carbon, correct?

And methane is CH4, but coal is mostly C, correct?

So shouldn't burning pure carbon release five times the CO2 (Assuming combusting carbon releases the same energy as hydrogen), or even six times per unit energy 'released'/'created' by the power plant (since hydrogen is more energy-dense per unit mass than carbon)?

When you do these calculations be sure to keep track of your units and the actual reactions you are considering.

Burning methane is not the same as burning the same amount of carbon plus twice the amount of dihydrogen. You need to check the specific heats of combustion and you need to make sure you keep track of the units used to measure the heat produced in the reaction (per mol, per kg or per mol of carbon dioxide?).

The heat of combustion (assuming all products are at STP) for carbon is 393 kJ/mol and for methane it is 889 kJ/mol. Both reactions produce the same amount of carbon dioxide if combustion is complete. So burning methane produces about 2.3 times as much heat per mole of carbon dioxide as burning pure carbon (anthracite coal will be close to this).


How to properly denote imaginary magnitudes of vibrational frequencies Vibrational frequency computations provide the magnitudes of vibrational frequencies for each vibrational mode of a molecule ($3N-6$ degrees of freedom for non-linear molecules and $3N-5$ for linear molecules).  Sometimes these magnitudes can be non-real (i.e. imaginary).  This is common when characterizing a transition state which contain one, and exactly one, imaginary mode of vibration.

Some popular software packages will print these magnitudes as a negative number.  This syntax has unfortunately propagated itself into common, everyday practice even if technically incorrect.

My question is, what is the best way to denote these imaginary magnitudes, especially in the context of a peer-reviewed journal?  Should we stick to using the minus sign convention (e.g. -15) with a note explicitly defining the meaning of the $-$ sign,  or should they simply be listed with an $i$  (e.g. 15$i$)?

(I've just been told to NEVER use the minus sign convention as this will incur the wrath of the reviewers.  Please feel free to verify.)

The suffix i to the positive wavenumber to denote an imaginary frequency is seemingly accepted by reviewers, see Photochemical transformations of 5-methyltetrazole. Matrix isolation FTIR and DFT studies as one example I could find in a very quick search.


Unexpected result (sodium hypochlorite & steel wool) So I got myself a large Erlenmeyer flask, some steel wool and the bleach from the bathroom, mixing them under safe conditions in small quantities to see the sparky reaction I'd heard about. Trouble is, I didn't know the bleach we had was non-chlorine!

EDIT: My silly mind assumed that no reaction meant I had non-chlorine bleach, but staring me in the face is hypoCHLORITE. My apologies, I'd not meant to mislead. The brand was Shaw's at the time, or "Signature Selections" which is their store brand. I got no sparks and no visible fumes nor immediate discoloration.

Disappointed, I left the mixture alone and uhm.. forgot about it.

Going back to check on it, I found the steel wool dissolved into a powder at the bottom, and the rest of the fluid was purple.

Curious, I used a giant coffee filter and strained the rusty cake-mix residue out of the rest of it, which was now a lighter purple.

I didn't know what I'd made and thought it best to simply leave it be for a while again. I kept it in the basement, visible when the door was open and kept an eye on it. Slowly the purple went away and the fluid decreased leaving a white powder and bits of clear-quartz like appearing crystals. After several months the crystals were larger, and the white powder has climbed the entire container and spread down the sides. It's now been about 18 months since the initial experiment, and the white powder (which is similar to efflorescence in appearance) has climbed about halfway down the outside of the container, and the inside crystals have grown slightly.

What in the world have I made?

After the edits by the original post, the OP clarified that they used a hypochlorite (not an oxygen bleach as stated earlier) and left a piece of steel wool in it. Due to corrosive nature of chloride/hypochlorite, the steel wool crumbled to iron (III) oxides. When you gently heat rust in highly alkaline medium in the presence of hypochlorite ion, it forms a purplish sodium ferrate. This has iron in the (+6) oxidation state.  Over time the purple color disappeared because ferrate slowly decomposes back to lower iron oxidation states in water solution, releasing its excess oxygen.  Ferrates are more stable in alkaline solution than in neutral or acidic ones, so absorption of carbon dioxide over time, making the solution less alkaline, could have promoted this decomposition.

This decomposition of ferrates by acids, carbonic acids, and acidic salts was known as early as 1890s as noted in a review by Moser in Journal für Praktische Chemie 1897, 56, 425 (see relevant comments). Ferrates were known in 1700s.

Also note that hydrogen peroxide and ferrate decompose each other immediately, just like permanganate ion in aqueous solution.

The quartz like crystals are nothing but sodium chloride with perhaps some sodium carbonate.

Image from: https://www.youtube.com/watch?v=5u8LOfTrO6w


What should be the name of the given compound? Following compound is named as

(1) 6-Mercaptocyclohex-4-ene-1,3-diol
(2) 1-Mercaptocyclohex-2-ene-4,6-diol
(3) 1-Mercaptocyclohex-5-ene-2,4-diol
(4) 4-Mercaptocyclohex-2-ene-1,5-diol

As per my knowledge if I start numbering the compound from the alcohol then to mercapto group, then alkene group.
So, according to me, the numbering should be from bottom alcohol as number 1 and making the way to the top alcohol as number 3. Then mercapto will come at number 4 and alkene at 5.
So as per me the name should be 4-mercaptocyclohex-5-ene-1,3-diol, but according to the answer key the answer should be (1). Is there any rule which I am missing out on?

The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

For the compound that is given in the question, the principal characteristic group is $\ce{-OH}$. It is expressed as suffix “ol”. Thus, low locants are assigned first to the $\ce{-OH}$ groups according to Rule (c), which yields “…1,3-diol”.
Next, a low locant is given to the double bond according to Rule (e), which gives “4-ene-1,3-diol”.
This step also establishes the numbering of the other atoms of the ring so that there is no longer a choice for the last substituent. Nevertheless, strictly following the rules, a low locant can be assigned to the remaining substituent according to Rule (f).
Thus, the complete name is 6-sulfanylcyclohex-4-ene-1,3-diol. Note that the prefix “mercapto” is no longer recommended.


How do I know if concentration of Ammonia affected my absorbency measurements? This is for determination of copper in a penny Lab. After making standard solutions we made a solution with a penny, dissolving it in nitric acid and adding ammonium hydroxide to form a more absorbtive copper complex. We didn't control the amount of ammonium hydroxide used in our samples.
How can I test whether or not it has an effect, specifically?

The concentration of ammonia will matter, but only up to a certain point.

When you dissolve copper in nitric acid, you form copper nitrate, which is very soluble, but as you say, not very absorptive.

$$\ce{Cu(s) +2HNO3(aq) -> Cu(NO3)2(aq)+H2(g)}$$

One of the problems with copper nitrate is that it dissociates readily in water.

$$\ce{Cu(NO3)2(aq) -> Cu^{2+}(aq) + 2NO3^{-}(aq)}$$

The copper(II) ions then coordinate with varying amounts of water molecules. Each balue of $\ce{x}$ in the equation below is a slightly different aqua complex with slightly different optical properties.

$$\ce{Cu^{2+}(aq) +xH2O-> Cu(H2O)_{x}^{2+}(aq)}$$

The addition of ammonia changes the situation. Ammonia binds strongly to copper to preferentially form the tetraamminecopper(II) complex, technically $\ce{Cu(NH3)4(H2O)2^{2+}}$, which is highly absorptive. Since this formation is highly favored, you only need to add enough ammonia to provide at least 4 stoichiometric equivalents.

If the ammonia solution you added was labeled "Ammonium Hydroxide" (as you say in your question), then you know how much ammonia you added. Your instructor just might not have told you how to do it.

Ammonium hydroxide exists only by the ionization of ammonia in water (which is very small except at high concentrations). Labeling a solution as "Ammonium Hydroxide" has come to mean a "saturated solution of ammonia in water", which is 35% ammonia by mass (approx. 18 molar). Now you can calculate the amount of ammonia that you added (as long as you know the volume of solution that you added).

To determine the effect of amount of ammonia, you can do the following:

  Dilute the original ammonium hydroxide solution. For example, dilute 25 mL of the solution to 50 mL with water to get an ammonia solution of about 9 molar. Then repeat the penny experiment and add exactly the same volume of the new ammonia solution. This guarantees that the overall copper concentration remains constant even when the ammonia concentration changes. Then measure the absorbance. Then dilute the ammonia solution again (either dilute the new solution or dilute the original solution by a different amount). Rinse and repeat.


Comparison of alkali and alkaline earth melting points - MO theory If you compare the melting points of alkali and alkaline earth metals, you see that the latter have higher melting points. However, if you look at MO theory, then alkaline earth metals should have quite a considerable proportion of the antibonding part of the bands occupied (even if there is an overlap with the p-band). So, why should the bond in alkaline earth metals be stronger?

(I realise that according to the electron sea model you just say that alkaline earth metals contribute two electrons, not just one, wich solves the problem. MO theory should also be applicable, so what am I overlooking?)

From "The Nature of the Interatomic Forces in Metals" by Linus Pauling

  This qualitative description of the interactions
  in the metal is compatible with quantum mechanical treatments which have been given the
  of
  problem, and it leads to an understanding
  such properties as the ratio of about 1. 5 of
  crystal energy of alkali metals to bond energy of
  their diatomic molecules (the increase being the
  contribution of the resonance energy), and the
  increase in interatomic distance by about 15
  percent from the diatomic molecule to the
  crystal.
  The alkaline earth metals, by assuming the
  configuration nsnp, are able to form twice as
  many bonds as the alkalis. Similarly the suc-
  ceeding elements in the periodic table can form
  bonds in increasing number.

You are right on the fact that alkaline earth metals should not bond if we only considered their highest atomic energy levels. The dissociation energy for diatomic clusters of alkaline earth metals is low.

However, the disassociation energy gets higher for tetra-atomic as shown in this computational study: http://www.sciencedirect.com/science/article/pii/0039602885902328

Now, the reason why alkaline earth metals have higher melting points to be due to the p orbitals. The previous study relates p hybridization with bond strength. And so does this study: http://scitation.aip.org/content/aip/journal/jcp/77/8/10.1063/1.444313

Both journals state that p orbitals are involved in bonding. And the more involved the p orbitals in bonding, the stronger the bond. This trend is shown for alkaline earth metals. Beryllium, which has the highest melting point, also has more p orbital character. Also, take a look at the density of states of alkali metals and alkaline earth metals:
http://www.colorado.edu/engineering/MCEN/MCEN5024/

You can see that not all p orbitals are entirely above the s orbitals.

So, to answer your question. The reason why alkaline earth metals have higher melting points is due to p orbitals. As Pauli states in his review, the np orbitals "hybridize" with the s orbitals. This allows for more bonds to form.

I think that p orbitals participate in bonding in alkali metals. But I couldn't find information that compares p orbital "hybridization" of alkali and alkaline earth metals directly.


Why is there a discrepancy among sources on the atomic radius of some elements? My book says the atomic radius of gallium is less than that of aluminium, but I found out different atomic sizes on different sites.

For example, this says gallium is 136 pm and aluminium is 118 pm, but Wikipedia says Ga is 130 pm, Al is 125 pm.

Both of them give different atomic radii but are consistent in ordering: both sites say Ga is larger than Al. What is correct?

My book says atomic size of Cr, Mn, Fe is about 117 pm and Zn is 125 pm; these again differ from the values give in the above websites. Which source provides correct values of the atomic radii?

Atomic radii are not really well-defined. There is a great deal of uncertainty associated with them, mainly due to the somewhat philosophical question ‘where does an atom end?’ Orbitals, in theory, extend into infinity for all (mathematical) intents and purposes, yet the probability of locating an electron in Far, Far Away is zero by all sensible rounding methods. However, the decrease is exponential, so you need to define a cutoff value if you want to know where an orbital ends to draw the beautiful pictures you probably have seen before.

This only gets worse when considering entire atoms rather than orbitals. We typically consider atoms as spherical (hence radius). But for every element except hydrogen, helium, lithium and beryllium (possibly including boron, too) there is at least one additional non-spherical orbital occupied. Can we still truly define an atom as a sphere? And even if we can, consider the case of transition metals; should we be talking about some cutoff value for the outermost s orbital or about one for the lower shell’s d subshell, whose electrons are considered valence electrons?

Now in theory, you could ignore all those philosophical discussions and just take the solid-state structure of an element, measure the shortest distance between two nuclei, divide it by 2 and call that the atomic radius. Well, that is what is done. However, many elements have different structures that they adopt or can adopt at room temperature and standard pressure. For other, especially non-metals, the shortest interatomic distance is not an atom-atom contact but an interatom bond. In a select few, the interatomic bond can have different bond order values; sixtytrees noted carbon whose most stable phase is graphite (bond order 1.5) but which also features a diamond allotrope with a bond order of 1.

The covalence problem can be overcome with a set of correction factors that introduce further uncertainties and are added differently by different sources.

All these issues mean that there is not one ‘true’ way to determining atomic radii; and since the methods differ, so will the results. It is only natural for some elements to be larger than others in one scale but smaller in another.

The whole problem gets infinitely worse with ionic radii. Ionic compounds exist as crystals with anions and cations — but where to the cations end and the anions begin? How do we deal with the fact that no bond is truly ionic and no bond between two different elements truly covalent?

Even if you determine e.g. the ionic radii of $\ce{Na+}$ and $\ce{Cl-}$ by various methods, (e.g. from comparing sodium fluoride, sodium carbonate, magnesium chloride, ammonium chloride …) and then add the two values together to arrive at an interionic distance in the $\ce{NaCl}$ crystal, it is very likely that your result is off.


Resonance with different size orbitals The equilibrium for this reaction is on the left side.

Why is this? I thought since the sulphur in molecule X has bigger p-orbitals than the oxygen in molecule Z, resonance would be a less important factor for stabilization, meaning Z is more stable, and the equilibrium should go to the right.
What part of the concept am I misunderstanding?

The fundamental question is: which is more acidic, 2-pyrrolidone or 2-pyrrolidinethione?

If you took a 50-50 mixture of the two and added enough alkali to neutralize half of the mix, what would be the composition of the solution? The mismatch in p orbitals between C and S leads to more C=N bonding, rather than C=S, with a charge separation, leading to more negative charge on sulfur in Y and more positive charge on N, so this N will be more acidic than the N in W. The charge separation by oxygen, leading to more double bond character between positions 1 and 2 in 2-pyrrolidone is an effect also, but apparently not as large as the p-p mismatch between C and S.

The Lewis diagrams don't show clearly enough the actual charge distribution in the anions.

And I thank Jan for pointing out that I completely flipped my actual explanation when I thought I understood what two effects were happening. Although this is not a trick question, I think it needs more context to be able to be of predictive value. Maybe just a way to get you thinking: what could be going on here, and where else might it explain.


Solving detergent powder in a solution I used a detergent powder mixed with water to clean my espresso machine. Some micro particles of powder are stuck into machine's inner tank. So when I pull the hot water out of the machine's nozzle, it contains micro powders and it tastes powdery.
What solution can I use instead of water to wash out all those sticky powders?
A harmless solution to break bond between powders and solve the powder into itself.

I think using acetone will be helpful. Acetone will evaporate fastly and it is polar and can form hydrogen bonds to the detergent powder like water. Though you need to use much acetone to wash all detergent powders and after that, wait for some minutes for acetone to vaporize from your machine.


Interhalogen with alkene Due to electronegativity differences, the iodine will be electrophilic and the chloride nucleophilic. Correct me if I am wrong here.
My question is one of regioselectivity: will it proceed through an iodinium ion (like a bromonium ion which is observed when Br2 is added) where the iodine and Me groups are on opposite faces. Then, chloride attack at the most substituted carbon? Or would it be a more akin to a planar tertiary carbocation, so attack at either face almost equally probable? **chlorine should be "back" in first structure

Iodine monochloride is usually highly specific for anti addition to alkenes. While bromonium ions form more readily than do iodonium ions, the latter do form. Additionally, the large size of the iodine atom further reduces the chance of syn addition. To answer your question in particular, it is unlikely that a planar carbocation is formed, and the anti-addtion would be expected to be the main product.

The more typical issue with iodine monochloride is that while the chloride is typically added to the Markovnikov carbon, the regioselectivity is sometimes relatively mediocre. For example, the addition of ICl to propene gives 2-chloro-1-iodopropane and 1-chloro-2-iodopropane in a 69:31 ratio; the yields of the expected regioisomer are better when styrene is used, giving approximately 95% of the expected 1-chloro-2-iodo-1-phenylethane. (J. Chem. Soc., 1931, 2742) This is likely due to the greater stability of the benzylic stabilization of the cation.


Pulsed NMR change in magnetisation vector When a second, radio frequency magnetic field $\ce{B1}$ is applied perpendicularly to $\ce{B0}$, why does the magnetisation vector M rotate in a direction perpendicular to both $\ce{B1}$ and $\ce{B0}$?

I understand how electric charges move in magnetic fields, and I understand the classical electromagnetism magnetic moment, but I can't figure out how $\ce{B1}$ move M in this direction.

Any help would be appreciated.

Image from http://chemistry.mit.edu/sites/default/files/DCIF-IntroNMRpart1-theory-o07.pdf

In the nmr experiment the sample is placed into an intense but static magnetic field. By the (nuclear) Zeeman effect the spin energy levels that are degenerate in the absence of a magnetic field become split in energy and thus more spins reside in the lower energy levels that in higher one. The number of any level is given by the Boltzmann distribution at the prevailing temperature. The total spin angular momentum of the sample is called the magnetization. To make a measurement a magnetic field from a radio frequency source is applied at right angles to the large, static field. This tips the magnetization into the x-y plane making measurements possible. I try to explain how and what happens below.

The spins have a magnetic moment $\bar m$, which is a vector quantity meaning it has direction and magnitude, and as a result the static field causes magnetic induction $\bar B$ in the sample. This is also a vector. The magnetic induction (confusingly but commonly also called the magnetic field) is induced in the sample by the the large static magnetic field often labelled $H$.

If all the individual spin moments are added up these form the magnetization $ \bar M$. This is a magnetic field and is also a vector and so has both magnitude and direction.    All nuclear spin magnetic moments experiences a torque ( a rotational motion generating angular momentum ) in a magnetic field and is given by classical electromagnetism as $\bar m \times \bar B$ where $\times $ means this is a vector cross product; $\bar M \times \bar B=|M\|B|\sin(\theta)$ where $\theta$ is the angle between the vectors.  The change in angular momentum produced is given by $\displaystyle \frac{d\bar L}{dt}=\bar m\times B$. As the angular momentum is given by $\bar m=\gamma \bar L $,   where $\gamma$ is the magnetogyric ratio (these are known and for H is $27.75\cdot 10^{7}$ rad/T/s), then

$$ \frac{d\bar M}{dt}=\gamma \bar M\times B \tag{1}$$

This equation can be broken down into components in x, y and z directions. The z direction is conventionally taken as the direction along which $\bar B$ lies, and this has no x or y components. The three equations produced are

$$\frac{d M_x}{dt}=\gamma m_xB;\quad \frac{d M_y}{dt}=-\gamma m_yB; \quad \frac{d M_z}{dt}=0  \tag{2}$$

If there is no external perturbation then these equations show that the magnetization remains along the z axis and as we place the detector along the x or y axis then no signal can be observed. This is  because the magnetic field of the magnetization has to 'cut the coils' of the detector as so induce a signal that can be measured. It clearly cannot do this if it has magnitude only along the z axis.

The equations (2) can be solved if some initial conditions at time zero are chosen and physically important solutions are found only when the x and y components are not initially zero. In that case the solutions show that then magnetization in x and and y are proportional to $\sin(\omega_0t)$ and $\cos(\omega_0t)$ respectively where $\omega_0= \gamma B_0$ where $B_0$ is the z component of $\bar B$. The Larmor frequency is $\nu_0=\omega_0/(2\pi)$ and is the frequency measured in an nmr experiment.

As these solutions require that the magnetization has to be in the x-y plane this  condition has to be satisfied and is done so by exciting the sample with rf radiation directed along the x (or y) axis. This radiation has a magnetic  field that is weak compared to $\bar B$ and of magnitude $B_1$; its frequency we will call $\omega_r$. Classically the magnetic field applies a torque to the magnetization and rotates it away from the z axis.  In a molecular (quantum) explanation, if the rf energy is resonant with the energy gap, formed by the large static field $B$, then spins are excited from the lower to upper level and this changes both the size and direction of the magnetization $\bar M$.

In a classical  explanation the  polarised rf field $B_1$ is considered to be composed to two oppositely rotating fields. One moves in the same sense as the magnetization precesses, and so interacts with it, the other moves in the opposite sense and has little interaction and can be ignored. The experiment is now such that the spins experience both a static magnetic field $\bar B \equiv B_0$ in the z direction and a rotating field $B_1$ in the x-y plane. Clearly this is now a very complex problem but can be simplified if one assumes that the axes x and y rotate about z at the frequency of the $B_1$ magnetic field. This is called the 'rotating frame'. The result is that the equations similar to (1) are produced but with $B_{x,y,z} $ replaced by $B_{eff(x,y,z)}$ where $\displaystyle B_{eff}^2=(B_0-\frac{\omega_r}{\gamma})^2+B_1^2$.


Internal (Z-Matrix) to Cartesian coordinates conversion Is there any way to convert from Internal to Cartesian coordinates using simple formulas?

I have studied and implemented: INTERNAL TO CARTESIAN by Jack D. Kunitz.

But somehow the x, y and z values are not correct. I am not implementing the transformation back to the initial frame as it says, because I do not understand that step. Is that the reason for not being able to get correct answers or is this not a good tutorial or solution to follow.

Sure, it's just trigonometry. There are a variety of open source implementations.

For example:

InternalToCartesian from Open Babel
ChemKit
QCL Python

There are dozens in a search on GitHub if you don't like C++ or Python.

There's also a web app from Shodor for z-matrix conversion.


Is heating oil identical to fuel oil? I don't quite understand the categorization of middle distillate petroleum products.

Is heating oil the same thing as fuel oil which is the same thing as diesel, or not? Can someone explain the differences (if any) between the common middle distillates?

Home heating oil and motor diesel are very similar in terms of density and chain length (basically same stuff) with minor differences in sulfur content (at different eras in different countries) and with colorant for tax purposes.  [Similar to comment above.]

In general the middle distillates would have a bit more sulfur in them if just based on distillation cut.  In order to meet lower sulfur requirements for diesel, additional processing is needed in the refinery to get the sulfur out.  As the US (and other countries) started putting in stricter and stricter sulfur requirements for diesel (for pollution reasons), it has had the impact of driving diesel prices up to be more than gasoline--traditionally it was cheaper.

[Note also that, at least in the US, it is possible to buy untaxed, undyed diesel for use in farm machinery.]

Note that "RFO" or residual fuel oil is a very different beast from home heating oil.  It is much heavier and higher boiling point fraction in the column.  It also typically has MUCH more sulfur in it (from concentration, the sulfur containing compounds tend to be heavier).  RFO is used in large onshore power plants (declining in the US but still some remain and many overseas) and also in ship power plants ("bunker fuel").  Ship use is both in steam boilers and in "marine diesels".  Larger shipboard diesels have evolved to become larger and larger, competing with steam turbines, and they are designed specially so that they can use a more crappy fuel (don't need motor diesel). RFO is also often burned (for heat exchangers and/or cogen) in island refineries (that lack access to natural gas).

Note that shipboard use of RFO is expected to decline after 2020 because of new restrictions on open ocean fuels (the "IMO law").  RFO burning is already prohibited in close proximity to North American and European coastal waters. The issue is sulfur content.  Ships can continue to burn the RFO if they install onboard sulfur scrubbers, but current indications are most will just use a lower sulfur fuel.  The easiest way to do this is to use something close to a motor diesel (rather than desulfurize RFO itself).  Will have a little more sulfur than most land diesels, but still somewhat desulfurized...and just because it is a middle distillate "cut" inherently has much less sulfur than RFO (which collects the nasties when distilling).  Also there are still questions about enforcement (i.e. cheating) or even legality of regulating open ocean operations.  But overall impact will be a big drop in prices for RFO as market goes away and more pressure on middle distillate prices.  [Ships can also shift to LNG but this is even trickier than scrubbers because, in addition to engine modifications, the infrastructure to bunker LNG does not exist in most ports.  But perhaps over time, we will see migration to LNG, a zero sulfur fuel.]

A decent reference is here:  https://www.amazon.com/Petroleum-Refining-Nontechnical-Language-William/dp/1593701586 It is a few years old so won't have the IMO discussion but will give you a good overview on the whole situation and is VERY friendly in terms of plain language.  I bought it before going into the industry and worked through the whole book in a few days...totally helps with some initial grounding in things.  After that you can fill in the blanks with whatever you are doing (marketing, buying, refining) by just asking for specs.

See also: https://www.eia.gov/tools/glossary/index.php?id=petroleum EIA glossary--you may also find some other info if you scratch around that website...a lot of pages and FAQs are a little buried.


Internal and external pressure during a gas expansion There is a solid substance in a rigid cylinder with a piston placed on top such that the distance between the piston and the bottom of the cylinder is essentially zero, and then the substance decomposes and forms gas that expands and pushes the piston up against a constant external pressure to a maximum point. It then falls slowly as heat from the reaction is lost to the surroundings. Is it true that at the maximum point, the internal pressure is equal to the external pressure?

If the frictionless massless piston is not moving, then internal and external pressures must be equal.  If the piston has mass, the internal pressure must be greater when it is stationary, to support the weight.  If there is static friction, all bets are off.

Suppose the piston were a right cone rather than a right cylinder.  Would the excess surface area change the outcome, all other things being equal?


What about sodium triphosphate in drinking water? Recently, at my place, a sort of filter for hard water was installed. They told me that it contains sodium triphosphate which is dissolved into all the water that enter home, i.e. both cold and hot water to prevent calcium carbonate to accumulate into the plumbing system, dishwasher, etc.... From what I can understand, there is no way this device is considering the flow rate in order to decide how much sodium goes into the water. It just happens.
According to what I googled, sodium triphosphate is used as an additive in many foods and, even though is not considered harmful, high levels may lead to health risks. Also, I have found some links (not in english) stating that such devices should be installed in the hot water plumbings since sodium triphosphate and calcium carbonate interact only at certain temperatures. According to these sources, drinking cold water is not harmful since no chemical reaction took place (but still sodium is present in the water, I believe) and, also, once I use the same cold water for cooking it gets hot and a reaction may take place.
I am quite puzzled and I don't know what to do with this filter. I can move it to hot water only, or I can keep it where it is (both cold and hot) if it's not harmful but I couldn't find any clear information. Is there something that may help?
EDIT 1:
this is the exact product I am using.
EDIT 2:
its chemical formula is Na5 P3 O10
EDIT 3:
this seems to be the device that was installed and this is one of the components that is taking care of the dosing (that is equal to 4-5ppm of sodium triphosphate, apparently)

It appears to be safe, provided that you are putting it into one of the delivery systems that it’s made for which I found through the link you provided.
Here is one of the delivery systems:
https://www.pineco.com/eng/products/dosing/polyphosphate-dosers/spillo-polyphosphates-doser-dm
Per the specs, it delivers 3ppm, which means that you’re only introducing 1ml per 333L of water, which is approximately what the average person uses per day.
Contributing to phosphate load shouldn’t be a concern as someone (with “normal” diet &renal function) has a concentration of 10-300ppm of phosphorus in their urine!
This company also sells water filters for regionally specific water issues like high arsenic, and water softeners with different ion exchange resins based on the ions present in the water. They appear to have a solid understanding of drinking water chemistry, but it’s your domestic water, not mine, so caveat emptor


Why is compressed air flammable? I have a can of compressed air. The type you use to clean the insides of computer. It's "100% ozone friendly" and "pure compressed gas". So I assume its everyday regular breathable air. I used it to try blow out a candle. It had precisely the opposite effect.

Why?

Most probably some hydrocarbon like butane = big flame. A quick way to check is if the can makes a sound when you shake it. If it does then it's most likely hydrocarbon. It would have been less confusing if it said "compressed gas".


How do chemical bonds increase molecular refractivity? I see that the molecular refractivity of a compound consists of the refractivity indexes of its composite atoms and chemical bonds, and that the refractive index $n$ of a substance increases with its polarizability $\alpha$, as described with the Lorentz-Lorenz equation:

$$
{n^2-1\over n^2+2} = {4\pi N \alpha \over 3},
$$

where $N$ is the number of molecules per unit volume.

I think that the formation of a covalent bond restricts the electron cloud of its composite atoms, delaying its release of an electromagnetic wave after being stimulated by a wave of the same frequency, and decreasing the wave length of the resulting created by the interference between the two electromagnetic waves affected and unaffected by the electron cloud. In the same way, I would explain the comparatively high refractivity index of a double bond between the same atoms as a stronger restriction of the shared electron cloud, delaying the release of an electromagnetic wave later than the covalent bond.

So I want to know, is there anything fundamentally mistaken with my interpretation?

The polarisability is the ability of any atom to have its electron 'cloud' displaced by an electric field. Polarisability is defined as the ratio of the induced dipole, formed by electron displacement, divided by the electric filed strength. It has units proportional to volume usually; values are in units $4\pi\epsilon_0\times 10^{-30} \mathrm{m^3}$. Larger atoms/molecules  naturally have larger values, in general, than smaller ones, e.g. $\ce{CH4}\; 2.6,\; \ce{CCl4}\; 10.5$. In a general sense also the number of electrons and how tightly they are held affects the magnitude of the polarisability.

In the case of light it is its oscillating electric field that affects the atoms in the material, a solvent medium for example. In doing so the speed of light in the medium is reduced and the ratio of this to that in vacuum is the refractive index. The energy of the photon remains the same inside as it is outside the material (assuming no absorption or scattering ) so the frequency remains the same, but inside the material the wavelength changes because $c/\lambda =\nu$ is constant and c changes. In forming the refractive index, the tails of electronic absorption bands  are important and are usually far into the ultraviolet.

In terms of working out the refractive index of some arbitrary molecule usually a semi-empirical method is used with the very many measured values available; values for different bonds can be extracted and parameterised. From these values then the refractive index of unknown compounds can be calculated. I found the pages www.pirika.com which has some explanation of how this is done.

In bringing atoms together to form molecules the atom/bond polarisability is used to form the refractive index and  should be additive provided that in forming the bond a strong absorption band is not brought into the visible part of the spectrum.  This is just what happens when a double bond or aromatic is formed so in these cases some extra empirical adjustment is needed.


Why is Frost's Circle an extension of Huckel's 4n+2 rule? What physics leads to the emergent property that the circle can predict relative MOs? The Frost Circle is a useful mnemonic to determine the relative energies for the molecular orbitals of an aromatic ring. One vertex of the polygon of the cycle is drawn pointing downwards and the entire polygon is inscribed in a circle. At this point, the location of each vertex corresponds to the MO relative energies. What is the physical explanation as to why a Frost Circle can accurately be used to predict the relative energies of molecular orbitals in aromatic compounds? What quantum mechanical phenomena lead to this emergent property? I'm looking for a connection between an equation and the circle I suppose.

The MO energies are obtained from Hückel theory: see Pi molecular orbitals of polyenes (and the Internet in general, there are many excellent expositions) for more details.

Regarding the Frost circles, they do not actually have any meaning: they are simply visual representations of the energies obtained from Hückel theory for cyclic polyenes. The last line of porphyrin's answer is relevant here:

  In the case of circular polyenes the solutions are $E_j=\alpha + 2\beta \cos(2\pi j/n)$.

where $n$ is the number of p-orbitals (e.g. for benzene $n = 6$) and $j$ is an integer satisfying $1 \leq j \leq n$. (The convention $0 \leq j \leq n-1$ is also sometimes seen, but it is mathematically equivalent.)

I am not a mathematician so some of the notation may be less than ideal, but the idea is as follows. Note that if you draw a circle with radius $r$, then the coordinates of any point on the circle $P$ may be described as $(r\sin\theta, -r\cos\theta)$ where $\theta$ is the (clockwise) angle that $\overrightarrow{OP}$ makes with the negative $y$-axis $\overrightarrow{OY}$. For example, here is a circle with $r = 2$. The angle $\theta$ here is equal to $\pi/4$, so the coordinates of the point $P$ are given by $(2\sin(\pi/4), -2\cos(\pi/4)) = (\sqrt{2}, -\sqrt{2})$.

Now we simply need to choose a circle with radius $r = -2\beta$ (note that $\beta < 0$, hence we need the negative sign) and select the six points on the circle where $\theta = 2\pi j/n$. The $y$-coordinates of these six points will then be the six MO energies. I have illustrated this for $n = 6$ and $\beta = 1$. We need to pick the points with

$$\theta = \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi$$

and these are $A$, $B$, $C$, $D$, $E$, and $F$ in order. Note that $\angle YOA = \pi/3$, $\angle YOB = 2\pi/3$, and so on. And note that whatever the value of $n$, there will always be the case where $j = n$ and hence $\theta = 2\pi$: that corresponds to the vertex at the bottom of the circle.


The Propinquity Effect I am currently in the process of studying enzyme catalysis and am struggling to get to grips with this concept. As far as I understand, the propinquity effect allows substrates to bind with enzymes in the conformation which would facilitate the reaction, thus increasing the reaction rate. Could someone expand on the why and/or how enzymes are able to do this? What specific reactions benefit from this the most?

how: an enzyme is able to effectively make a reaction intramolecular by binding the reactants in the correct conformation in close proximity within the active site this can speed up the rate of a reaction dramatically, intramolecular rates avoid diffusion control and have prearranged conformations which is far more efficient than typical intermolecular reactions,
why: enzymes are outrageous, evolution has pushed them, through trial and error, towards optimal configurations that, almost magically, provide ideal conditions for a reaction to occur


Directing effect in Intramolecular Friedel Crafts Alkylation This reaction should cause ring closure at either the $o-$ or $p-$ position relative to the methyl substituent on the ring (as methyl is an $o,p-$directing group), giving the following products:

The question is: Which one of these is the major product? Is substitution preferred at the ortho position or para position, and why?

A few more examples of these types of questions:

This would give which of the following?

Electron giving groups like $-OCH_3$ (by resonance) and $-CH_3$ (by hyperconjugation) are Ortho-Para directing.

But, in cases where there is no increased stability by H-bonding or other factors at ortho position, the para substituted product will give major yield.

This can simply be explained by stating that there is less steric hindernace on the para position compared to the ortho position. Which is further emphasized when you take this example:

So, the answer is para-substituted in both cases.

References

Aromatic reactivity


How a titration curve is affected when a poorly soluble salt is formed? Suppose that I have $\pu{100 mL}$ solution of strong acid, $\ce{HA}$, at $\pu{0.10 mol L^{-1}}$.
This solution is titrated with a strong base solution, $\ce{BOH}$, at $\pu{0.10 mol L^{-1}}$.
Suppose yet that a poorly/sparingly soluble salt, $\ce{BA}$ $(K_\ce{sp}=\pu{2.30×10^{-3}})$, is formed during titration.

The reactions involved are:

ionization of strong acid: $$\ce{HA + H2O -> H3O+ + A-}$$
dissociation of strong base: $$\ce{BOH -> B+ + OH-}$$
self-ionization of water: $$\ce{2 H2O <=> H3O+ + OH-}\qquad K_{\ce{w}}=\ce{[H3O+][OH-]}$$
Equilibrium of poorly soluble salt: $$\ce{BA <=> B+ + A-}\qquad K_{\ce{sp}}=\ce{[B+][A-]}$$

Charge balance of the solution is given by the equation: $$\ce{[H3O+] + [B+] = [OH-] + [A-]}$$

Once that $\ce{B+}$ and $\ce{A-}$ are spectator ions, are come from strong substances, and disregarding the ionic strength effect, I think that titration curve don't be affected. I'm right?

If not, how to write a mass balance expression to take account into the precipitation of $\ce{BA}$?

How a titration curve is affected when a poorly soluble salt is formed?

If by "I think that titration curve don't be affected" you mean that the graph of pH vs volume of strong base will not be affected by the precipitations, then you are right. This is because the species that form the precipitate are not part of any other equilibrium reaction.

  how to write a mass balance expression to take account into the precipitation of BA?

In general terms, the mass balance (really the amount balance) can be expressed in concentrations if the system is homogeneous (all in solution). For example, it could look like this:

$$[E] + [F] + [G] = \mathrm{constant}$$

If there is a solid species $H$ that you have to consider, use an expression based on amount of substance:

$$n_E + n_F + n_G + n_H= \mathrm{another\ constant}$$

If you want to, you can express parts of this as concentration again:

$$([E] + [F] + [G]) \cdot V_\mathrm{solution} + n_H= \mathrm{another\ constant}$$


Safe drying agents for drying machine When I was milling NaOH for some purpose, I noticed that the powder is incredibly hygroscopic. Because I have problems with drying certain chemicals safely (without extensive heating), I thought this might be a safer solution:

But I'm not so sure about that safety. Can NaOH exit into the air and pollute the compound in the second beaker? If it can, which compounds are safe and strongly hygroscopic?

Could I use this technique to increase sulfuric acid concentration? What about acetic acid?

The first part of your schematic basically describes a dessicant dryer or deliquescent dryer. These are common industrial air drying processes that produce very low dew point properties. Generally air (or other inert gas) is passed through a column of desiccant, under increased pressure.

Desiccant dryers use common desiccants that include activated alumina, molecular sieves and silica gel. In all cases, there is a maximum amount of water that can be adsorbed/absorbed before the desiccant needs to be regenerated; typically regeneration is done by heating the desiccant column, although other methods, such as purging can be used. Industrial desiccant dryers have the heater element installed as part of the construction, although any bench chemist can probably tell you about heating red silica gel beads from desiccators in the oven until they turn blue.

Deliquescent dryers are similar, but involve a highly hygroscopic agent that slowly dissolves, or deliquesces, as it absorbs more water. Usually the agent cannot be regenerated, and the waste liquid is drained and discarded. Fresh drying agent then needs to be added. This is most likely the scenario for NaOH.

So, this would need to be a major consideration for your proposed system; how to safely and effectively dry your NaOH. This is a very caustic material to have to try to continually regenerate or replace.

Yes, I would be concerned that you would transfer NaOH from the first chamber. To check this, you could easily test the gas outlet with some litmus.

Can you use dry air to remove excess water from other materials (such as acids)? In theory, this may be possible and will depend on the relative hygroscopicity of the drying agent and the material to be dried; you need to be sure that the air you are supplying your wet material with is dry enough to receive water and not simply supply more. Purge drying (passing dry air over a desiccant) is another form of regenerating industrial desiccants, but is not as effective as heating. Sulfuric acid is itself used as a desiccating agent as it is very hygroscopic. I don't know the relative hygroscopicity of NaOH and H2SO4/CH3COOH, but there is no way I would ever consider passing air from one container directly into the other chamber. This is a very hazardous setup, and is fraught with much danger. Heating an acid solution in a suitably ventilated environment under controlled conditions is still, to my knowledge, the most effective way to increase the concentration of the solution.

You should also consider the energy efficiencies of trying to 'dry' a sulfuric acid solution by passing dry air over or through the solution versus heating it. The amount of time and materials required to produce sufficient dry air to effectively concentrate a sulfuric acid solution is likely to be far in excess than the time/material investment required to simply heating it.


Determination of (R) and (S) configurations from a Haworth projection Short version of question : How do I know which groups are equivalent to "out of and into the page (like in the Fisher projection)" for the Haworth projection (so that I can send the lowest priority group to the "back")?

Details:

I am comfortable enough finding R and S configuration of Fischer projections, but am unsure which groups are coming in or going out of the plane of the page in a Haworth display.

For example:

Figure 1

Taking carbon $3$ for example, the priorities are as shown below in blue:

Figure 2

By definition, horizontal groups are out of the page (i.e. towards us) and vertical groups are into the page, the lowest priority group (hydrogen) is out of the page. If hydrogen was into the page, carbon $3$ would have a clockwise R configuration, but as hydrogen is drawn out of the page, carbon 3 has the opposite, hence an  anticlockwise S configuration.

If I now took carbon 3 in the Haworth for the structure above, I am assuming it is still an S configuration as the groups are still in the same position around the carbon 3 atom, relative to the other groups.

Update: After seeing the following image (annotated with numbered carbons from this source) I am not so sure about that anymore as carbon 4 has different configurations in the Fischer vs the Chair and Haworth.

Figure 3

But how would I determine the R and S configuration of carbon 3 without going back to the Fischer projection?

Mainly, how do I know which groups are equivalent to "out of and into the page (like in the Fisher projection)" for the Haworth projection (so that I can send the lowest priority group to the "back")?

I would think carbon 4 is "into the page" as it is further away from our point of view (which matches the Fischer projection, as it is drawn vertically there).

Based off this, carbon 2 would also be into the page (as its the same as Carbon 4 in the Fischer projection). Hence, the $\ce{OH}$ and $\ce{H}$ would be out of the page.

The priorities would be as follow:

Figure 4

If $\ce{H}$ was into the page, as draw it would be R, but as $\ce{H}$ is into the page, its the opposite and hence carbon 3 is an S

So I arrive at the same answer, but mainly due to comparing the Haworth off of the Fischer projection. How would I judge what was in and out of the page without the Fischer projection and only the Haworth?

In a Haworth projection, the hexagonic shape of the ring already implies the direction of two bonds. The two bonds on the ring are always ‘bent ring-inwards’ when viewed from the outside. Thus, the two perpendicular bonds (typically vertical) must ‘point away’ from the ring: in a forward direction for the front carbons of a ring and in a backward direction for the rear side of the ring.

However, I notice that the question uses (R) and (S) a lot and attempts to argue a lot with it, so a note must be dropped on those stereodescriptors, too. It is important to realise that the (R) and (S) stereodescriptors can change even if the conformation of the carbon in question did not invert or otherwise change conformation. This is due to the way priorities are determined in the CIP system. Let me exemplify this.

First, remember that α-D-glucopyranose, β-D-glucopyranose and the aldehyde form of D-glucose are three different compounds with different physical properties. The α and β forms are diastereomers, more specifically epimers, more specifically anomers of each other and e.g. display different optical rotation. The chain-form of glucose is different altogether. I will construct all further discussion of this answer in a way that it does not matter whether we are dealing with α or β-D-glucopyranose, so you may always assume a mixture.

Now let’s consider the absolute configuration of carbon 3 in both the chain and pyranose forms.

Pyranose

Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbon atoms of the (b) and (c) priorities; initially undistinguishable. Thus, we must go down one level.
Both carbons are attached to the atoms (in order of priority) $\ce{O, C, H}$. Again, this is a tie; we must go down one level.
Both oxygens are attached to a hydrogen; this is identical and the end of the sub-chain. We must go back to the next, lower priority of the former atom.
The two carbon atoms differ: one is the acetal group which expands to $\ce{O, O, H}$. The other is connected to $\ce{O, C, H}$ (carbon 5). Thus, we have resolved the issue.
Going towards lower carbon numbers is the higher priority.

Therefore, carbon $\ce{C{3}}$ has an (S) configuration.
Chain form

Again, oxygen obviously has highest priority.
Equally again, hydrogen obviously has lowest priority.
The two carbons are initially equal.
Both are again attached to $\ce{O(H), C, H}$.
The two subsequent carbons differ again. This time, one is the aldehyde group which expands to $\ce{O, (O), H}$, the other is $\ce{C{5}}$ which expands to $\ce{O, C, H}$. Since a ghost oxygen atom has a higher priority, going up the Fisher projection has the higher priority.

We arrive at the same (S) conclusion as before.

Now, we will construct the same discussion for carbon $\ce{C{4}}$.

Pyranose

Obviously, the oxygen atom has highest priority.
Obviously again, hydrogen has lowest priority.
We have two carbons, initially indistinguishable again. We go down a level.
Both carbons ($\ce{C{3}}$ and $\ce{C{5}}$) expand to $\ce{O, C, H}$. We must go down a level.
One of the two oxygens is connected to a carbon atom. Thus, $\ce{O(C), C, H}$ wins over $\ce{O(H), C, H}$.
Going towards $\ce{C{5}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (S) configured.
Chain form

Bla, oxygen.
Bla, hydrogen.
Blabla, two carbons. Again, we descend a level to $\ce{C{3}}$ and $\ce{C{5}}$.
Both carbons expand to $\ce{O, C, H}$. We descend a level.
Both oxygens are attached to a hydrogen and nothing else. They are indistinguishable.
The two carbons we are now examining are $\ce{C{2}}$ and $\ce{C{6}}$. The former expands to $\ce{O, C, H}$, the latter to $\ce{O, H, H}$. Therefore, Going towards $\ce{C{3}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (R) configured.

Note that while the configuration of $\ce{C{3}}$ did not change, $\ce{C{4}}$’s absolute configuration changed from (R) to (S) upon formation of the pyranose ring. Thus, you cannot rely on absolute configurations to verify a correct Fischer to Haworth transformation for carbohydrates!


Instructions for handling toxic liquids I would like to know what precautions should be taken when handling liquids like trichloroethylene, tetrachloroethylene, dichlormethane, and mercury. Of these, which is most toxic?

I have seen people handle mercury in rooms without fume hoods. I don't have access to a fume hood, so is it adequate protection if I wear a suitable mask when handling mercury?

Also, I am interested in the conditions in which mercury vapor is formed. Why is it formed when its gaseous temperature is high?

Each liquid produces some vapours. When you leave wet clothes, they eventually dry. Same here. It is true, that vapour pressure for mercury at room temperature is extremely low, but since mercury accumulates and has quite high toxicity, it is still a trouble. Same for other liquids.

Mercury must be kept in sealed container and/or under layer of water. Any works with it must occur in large pan, so no tiniest drop of mercury escaped. Escaped drops must be collected using thin copper plate, washed in concentrated nitric acid. All solutions of mercury compounds and remaining mercury should be utilized according to local procedures (there is an agency for it at my city). Metallic mercury may be transformed into salts using $\ce{FeCl3}$ concentrated solution. While not removing the mercury, this transforms it into compounds that does not produce vapours. However, in case of mercury leak a special agency must be informed to perform all necessary procedures. Consult local regulations on the cases.

Dichloromethane has regrettably low boiling point. It must be kept away from heat sources and in case of high temperatures it must be cooled or kept in sealed container. It evaporates swiftly on open air, carrying away much heat.

All the liquids should be used under fume hood. In case one is unavailable, an acceptable alternative is open air as long as there is a gentle wind. The chlorinated hidrocarbons mentioned has strong odor, strong enough to make handling them without good ventilation impossible. None of the liquids should contact skin. It is not immediately fatal, but still. In case the contact occurred, the chloroalkanes will evaporate quite quickly, though washing skin with soap and water is highly encouraged. Again, one-time breathing-in of the vapours is not fatal with this compounds, but repeated breathing-in guarantees health problems.

Good thing: none of the liquids is flammable, so flammability is not a trouble here. Unfortunately, heating them at presence of air is still highly unrecommended as well as continuous exposure to sunlight: it may produce phosgene.


Mass percent composition How would my sample be affected if the sample was not totally dry? What would happen to the mass percent composition? I believe that it will increase but someone argues that due to ratios, it will decrease.

The IUPAC Compendium of Chemical Terminology defines the mass fraction $w$ as the

  mass of a constituent divided by the total mass of all constituents in the mixture

$$w_i = \frac{m_i}{\sum\limits_{i=1}^{N}m_i} \tag{1}$$

This means that $w_i$ can adopt values between 0 ( = 0 %) when the compound in question is not contained in the sample), and 1 ( = 100 %) when the sample only consists of the compound.

If your sample isn't totally dry, the mass of water adds to the total mass of the sample.

As a result, the denominator in $(1)$ will become larger and the mass fraction $w_i$ for any component (except the water) will decrease.