What do the numbers in the preferred IUPAC name for ascorbic acid mean? If I look up the IUPAC name of ascorbic acid, I find this: (5R)-[(1S)-1,2-Dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one.

I am used to numbering of molecules in a way like the 3,4-dihydroxyfuran part in this molecule name. But now what I want to know is: what do the 5R, 1S and 5H parts signify?

For the 5H I came up with this: furan has two double bonds, so the H might signify that this is not the case any more, because if there is an extra H at the fifth position, this would be impossible. This can be seen in the IUPAC name of 2-furanone: 5H-furan-2-one. I don't know though if this assumption is correct, also because in 5H-furan-2-one the 5H is before the furan part and in 3,4-dihydroxyfuran-2(5H)-one it is after (shouldn't it be 2(3,4-dihydroxy-5H)-furanone anyway?)

Fact is that I can't find an explanation for the 5R, nor the 1S (as I see no sulfur). Could anyone explain me these weird molecule numberings?

The numbering of (R)-5-[(S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one has four parts:

1) Numbering of the furan ring

2) Numbering of the ethyl side chain

3) Stereodescriptors R and S

4) Italicized element symbols

The italic element symbol H denotes indicated or added hydrogen.


Is this product description wrong and a simple battery and not a fuel cell? I won this "toy" at a science fair...
Now I have to make a short presentation about it in school.
On the package description they describe its energy source as "fuel cell"...
It consists out of a metal plate(magnesium), a black metal plate with a black coating and a paper-cotton-like isolator which is between those plates.
Through putting salt water on the isolator, there's a voltage difference of 1.76V between the black and the magnesium plate.
After a couple of seconds on load the voltage stays constant at 1.31V.
My theory is that it's only a simple battery, because there's no continious source of fuel which should be required for naming it fuel cell (according to wikipedia).

Is my thought right?

My idea about the black coat is, that it is a rubberd graphite or "coal" cathode, rubberd because coal or graphite would fall apart quickly. It is described as air-cathode in the package description. My idea is that the black coal or graphite coat is used to provide a larger amount of air to the reaction...

The magnesium plate gets damaged through this reaction, the black thing not.

Which material is the black coat?

and what is the full reaction?

a picture of the components

Your analysis seems correct, i.e. it is only a fuel cell if the magnesium is being oxidized by the air. You can prove if that is the case, and do so as part of your classroom demonstration, by excluding any air, and therefore oxygen, from the fuel cell.

Use distilled water and fairly pure $\ce{NaCl}$ (rather than table salt) to make the salt solution.
Boil the salt water for a minute or so to remove air bubbles (some oxygen from air dissolves into water).
Put the cell with electrolyte in a plastic bag sealed around the wires and a narrow tube such as a plastic coffee stirrer or soda straw that you'll use later to admit air, carefully squeezing out air bubbles. Seal both bag and tube with plasticine modeling clay or soft wax.
Observe the open circuit voltage of the cell.
Then run it for a while with a load (e.g. a 75 mA 1.5 V lamp) or a 100 ohm resistor, which will draw 1.3 mA), depending on the current output of the cell. You would expect some voltage drop over time, either due to cell polarization or due to any residual oxygen being used up.
To prove that this is a true fuel cell oxidizing the magnesium using air, bubble some air into the plastic bag through the (unplugged) tube. *If the voltage goes back up, then drops down until yet more air is added, you've shown that oxygen is being used and it's a true fuel cell.

Please accept my compliments for your thoughtful question; do post your findings here!


Calculate the number of moles of phosphorus in 15.95g of tetraphosphorus decaoxide I've got this question and I cannot figure out how to get the correct answer.

  Calculate the number of moles of Phosphorus in $15.95\:\mathrm{g}$ of tetraphosphorus decaoxide. ($\ce{P4O10}$)

I tried doing it like this:

Molar Mass of (element) = Number of atoms * Relative Atomic Mass = ### grams/mole

Mass of Compound in Grams = $15.95\:\mathrm{g}$

$M(\ce{P4}) = 4(30.97376) = 123.89504 \:\mathrm{g/mol}$
$M(\ce{O10}) = 10(15.9994) = 159.994 \:\mathrm{g/mol}$

$M(\ce{P4O10}) = 123.89504\:\mathrm{g/mol} + 159.994\:\mathrm{g/mol} = 283.88904 \:\mathrm{g/mol}$

$\ce{P}: (123.89504 \:\mathrm{g/mol})/(283.88904 \:\mathrm{g/mol}) = 0.4364207 \cdot (100\% = 43.64\% \ce{P})$

$15.95\:\mathrm{g} \cdot 0.4364207 = 6.9565\:\mathrm{g} \ce{P}$

$\ce{P} = 6.9565\:\mathrm{g}/ (123.89504\:\mathrm{g/mol}) = 0.056148 \:\mathrm{mol}$

But the answer in the back of the book is $0.2247 \:\mathrm{mol}$

You can go about it more directly as follows.  You have $(15.95/283.889)=\pu{0.0562 moles}$ of $\ce{P4O10}$.  From the molecular formula you see that for every mole of $\ce{P4O10}$ you have 4 moles of phosphorous.  Therefor you have $4 \times 0.0562 = \pu{0.2247 moles}$ of phosphorous.

Your error involved converting moles of $\ce{P4}$ to moles of $\ce{P}$.  If you multiply your answer by 4 you get the correct answer.


Can a network covalent solid be amorphous? All the definitions I'm seeing give examples of crystalline solids like diamond or boron nitride. But I'm wondering if it also includes amorphous macromolecules like vulcanized rubber. I'm thinking "network" probably == "ordered array" but not sure if that's correct.

The most obvious example is amorphous carbon. As mentioned in a comment, another good example is glass (although, whether it's truly "solid" past the glass transition temperature).

Really, a "network" simply means "solid with covalent bonding" and of course it's easy to imagine many possible amorphous network solids.

Vulcanized rubber might qualify too, but it would depend on whether there were many non-interlinked macromolecules. Usually macromolecules would be considered molecular solids due to the high number of non-bonded packing interactions. If the amount of cross-linking is high, classifying as a network solid makes sense.


How can FeS2 be the formula of iron sulphide and CaC2 be the formula of calcium carbide? I know that the valency of iron can either be 2 or 3 and so the formulae of the possible sulphides of iron should be $\ce{FeS}$ and $\ce{Fe2S3}$. But I have recently seen the formula $\ce{FeS2}$ used for iron sulphide. How is this possible.

Similarly the formula for Calcium carbide is supposed to be $\ce{Ca2C}$ and not $\ce{CaC2}$. Why do we write $\ce{CaC2}$ for it?

What you call iron sulphide, in my opinion, is more appropriately referred to as or iron disulphide.

If one were to assign oxidation states to each atom, an appropriate description would be  $\ce{Fe^2+}$ and $\ce{S_2^2−}$. This formalism recognizes that the sulfur atoms in pyrite occur in pairs with clear $\ce{S–S}$ bonds. These disulphide units can be viewed as derived from hydrogen disulfide, i.e $\ce{H2S2}$ (similar to hydrogen peroxide)

Take a look at the crystal structure (of pyrite):

In the center of the cell a $\ce{S_2^2−}$ pair is seen in yellow. The $\ce{S}$ atoms have bonds with three $\ce{Fe}$ and one other $\ce{S}$ atom.

Contrast this with, $\ce{FeS}$, i.e Iron (II) Sulphide where the counter anion is what you would typically expect $\ce{S^2-}$. This is reflected in its crystal structure, which is the nickel arsenide structure:

The bonding situation is completely different in these two compounds and this is reflected in their chemical formulas.

Similarly, for calcium carbide, the crystal structure is a tetragonally distorted $\ce{NaCl}$ lattice, comprised of discrete $\ce{Ca^2+}$
and discrete $\ce{C_2^-}$ ions.

Also, bear in mind what @Jon Custer said in his comment:

  Because the 'rules' on valences are not firm rules. On the $\ce{Fe-S}$ binary phase diagram, $\ce{FeS2}$ exists as a line compound

Also, I'll briefly mention what he probably meant by a line compound:

Vertical lines such as the one seen here for $\ce{FeS2}$ indicate inter metallic compounds, which have precise chemical composition.

References

This article for the crystal structure of calcium carbide: http://pubs.acs.org.ejgw.nul.nagoya-u.ac.jp/doi/abs/10.1021/j100010a022
The crystal structures of pyrite and Iron (II) sulphide both come from their respective wikipedia pages here and here
The phase diagram comes from: http://www.minsocam.org/msa/collectors_corner/arc/scn1.htm (which further credits it to Ehlers, 1972, after Kullerud, 1967 The Interpretation of Geological Phase Diagrams, Fig. 217, p. 232)


Name of corresponding anion of a hemiacetal If a hemiacetal is deprotonated at its hydroxyl group, what is the correct name for the resulting anion (or salt)?

I could not find anything, and assume that these compounds are probably not stable (just like most hemiacetals themselves), but I'd still be interested what's the right nomenclature here - something with -ate, -alkoxide, or the like?

Hemiacetals can actually be named as ‘hemiacetal’ using functional class nomenclature, e.g. pentanal ethyl hemiacetal. However, the preferred IUPAC names are generated using the usual systematic substitutive nomenclature, i.e. as substituted hydroxy compounds, e.g. 1-ethoxypentan-1-ol. The corresponding rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

P-66.6.5.2 Hemiacetals and hemiketals
Compounds with the general structure $\ce{RR'C(OH)(O-R'')}$ have the class name ‘hemiacetals’. They are named substitutively as ‘alkoxy’, ‘alkyloxy’, ‘aryloxy’, etc. derivatives of an appropriate hydroxy parent compound, such as an alcohol; these names are the preferred IUPAC names. Other names are formed by functional class nomenclature using the class name ‘hemiacetal’; similarly derivatives of ketones are denoted by the class name ‘hemiketal’.

According to P-72.2.2.2.2, an anion formed by subtracting $\ce{H+}$ from a hydroxy compound that can be expressed by a suffix ‘ol’ is systematically named by using compound suffix ‘olate’, formed by addition of the ending ‘ate’ to the suffix ‘ol’.

P-72.2.2.2.2 Anions derived from hydroxy compounds
An anion formed by subtracting a hydron from the chalcogen atom of a hydroxy characteristic group, or a chalcogen analogue, that can be expressed by a suffix such as ‘ol’, ‘thiol’, ‘-peroxol’, etc., is preferably named by using suffixes ‘alate’, ‘thiolate’, ‘peroxolate’, etc., formed by addition of the ending ‘ate’ to the suffixes ‘ol’, ‘thiol’, ‘peroxol’, etc. The multiplying prefixes ‘bis’, ‘tris’, etc. are used before these suffixes, to avoid any ambiguity.
(…)

Disappointingly, the current IUPAC recommendations still retain traditional names as preferred IUPAC names for anions of several simple alcohols (such as ‘ethoxide’ instead of the systematic name ‘ethanolate’).
The anion of my example 1-ethoxypentan-1-ol, however, is not affected by such exceptions; the systematic name 1-ethoxypentan-1-olate is also the preferred IUPAC name.


Is Br sp³-hybridised in HBr? $\ce{HBr}$ formation can be explained by simple orbital overlap as well as by $\mathrm{sp^3}$ hybridisation of $\ce{Br}.$ Can't we use dipole moment to find if the lone pairs on $\ce{Br}$ are mutually perpendicular or $\mathrm{sp^3}$-hybridised?

From a theoretical viewpoint we would expect the Br to be sp3 hybridized, because in the HBr molecule the Br atom has 4 pairs of electrons around it (the 3 lone pairs plus the bonding pair), and symmetry (plus the repulsion of electrons for each other) tells us to expect that 4 pairs of electrons arrange themselves in a tetrahedron, and as you know we explain a tetrahedral arrangement of electron pairs around an atom by arguing the s and p atomic orbitals form sp3 hybrid orbitals.

It's the same valence shell electron pair repulsion (VSEPR) logic that leads us  to expect the 4 bonding pairs around the C in CH4 should be arranged tetrahedrally, and, therefore, the C atom is sp3 hybridized.  (We would also expect that a more careful and precise determination would give us some deviation from this prediction for HBr, since unlike CH4 the 4 electron pairs around the Br in HBr are not equivalent.)

But you are interested in the question of whether there could be some experimental verification of this expectation.  I don't think the experimental dipole moment would work, because it's too crude a picture of the electron distribution, it just tells us whether the electrons spend extra time near the H or near the Br, and we already know it will be the Br because of its much stronger effective nuclear charge for valence electrons.  What you want to know is something like whether the valence electrons spend significantly extra time on the "far" side of the Br, away from the H-Br bond, as opposed to the "near" side, so we can conclude they're in sp3 hybrid orbitals that point away from the H-Br bond.  Offhand I can't think of an experiment that would unambiguously test this.

You could certainly calculate the electron density in the HBr molecule, with a good quantum chemistry package, and for a molecule this small the results should be quite reliable.  That would tell you whether it makes more sense to consider the Br lone pairs as in sp3 hybrid orbitals or atomic p orbitals.


Are there square planar complexes with sp2d hybridization? We were taught (Under the section 'Valence Bond Theory') seven types of geometries a transition metal complex may assume and its corresponding hybridization states,

Linear - $\ce{sp}$
Trigonal planar - $\ce{sp^2}$
Tetrahedral - $\ce{sp^3}$
Square planar - $\ce{dsp^2}$   (Inner d-orbital involved)
Trigonal Bi-pyramidal - $\ce{dsp^3}$  (Inner d-orbital involved)
Square Pyramidal - $\ce{sp^3d}$   (Outer d-orbital involved)
Octahedral - $\ce{d^2sp^3}$   (Inner d-orbitals involved)

We were told that there are quite a few instances (for Octahedral complexes), where the outer, vacant d-orbital takes part in hybridization, so the hybridization state would thus become: $\ce{sp^3d^2}$

Now my question is:
Are there Square Planar complexes in which the outer, vacant d-orbitals take part in hybridization (i.e- are there square planar complexes with sp2d hybridization) ? If so, could someone provide a few examples.

0.1) Hybridization is to be used with caution in inorganic chemistry above school level. It is proven not working for one-electron properties at the very least.

0.2) Depending on the details, you may or may not be taught about hypervalent compounds using d-orbitals of outer shell. While this concept fell out of favor, it still is taught here and there.

1) ignoring 0.*, $\ce{PnX5}$ family where $\ce{Pn=P,As,Sb}$ and X is a halogen (typically $\ce{F}$ or $\ce{Cl}$) adopts trigonal-bipiramidal shape and was viewed as an example of $sp^3d$ hybridisation. Square planar compounds for p-elements are much rarer, but $\ce{XeF4}$ adopt such structure.

2) A rare anion $\ce{[Ni(CN)5]^{3-}}$ may adopt such structure, specifically in $\ce{ [Cr(NH3)6][Ni(CN)5]\cdot 2 H2O}$ Actually, it is often said that square planar complexes may coordinate weakly an additional ion to form a square pyramid and this is why they are typically much more reactive, than octahedral complexes.


Which software is suitable to calculate the root mean square deviation between two protein structures? I need to compute the root mean square deviation (RMSD) between two protein structures: the X-ray determined protein structure and the MOPAC optimized structure. Which software do you suggest?

All of these are freely available packages:

Visual Molecular Dynamics - good visualization tool that can also do quick-and-dirty calculations from a graphical interface. The scripting interface has a learning curve, but yes, it can be quite powerful.

Avogadro - not as powerful as VMD, but can be useful for certain things.

Rasmol - Similar to VMD, although I've never used it.

LAMMPS, Gromacs, or NAMD - If one of these is installed on a computer you have access to, each of them can do this sort of computation and will allow you to do more advanced things if needed (they are actually molecular dynamics simulators). Each of these packages requires some familiarity with running code from the command-line and will have a learning curve.

Alternatively, you could always write your own code. RMSD is a pretty straightforward computation, and it's always nice to have more control over your analysis!


How do auxochrome groups work? Functional groups like $\ce{OH}$, $\ce{NH2}$, attached to unsaturated compounds (chromophores) are called auxochromes. These groups tend to shift the absorption wavelength to the infrared region, which is called a bathochromic shift.

Why do these groups shift the light absorption wavelength?
Why exactly to the infrared region; that is, why is the absorption red-shifted or lowered in energy?

Generally speaking, the typical auxochromic groups possess (at least) one pair of non-bonded $n$-electrons and -- if taken alone -- do not absorb in the UV.  If attached, for example to a benzene ring, their pair(s) of $n$-electron interact(s) with the $\pi$-electrons, and the delocalization of the $\pi$-system is extended.

Now the catch:

Bathchromic auxochromic groups (like the mentioned $\ce{-OH}$ and $\ce{-NH2}$) increase the electron density in the $\pi$-system. This is typically caused by a strong $+M$-effect. As they lower the excited $\pi^*$-level, a longer wavelength is needed for the excitation along $\pi{} \rightarrow {} \pi^*$, hence the redshift.
In contrast to the former, substitution equally may alter the $n \rightarrow{} \pi^*$ transition, inducing the opposite effect, moving the absorption towards higher absorption (blue) and a hypsochromic shift.  One example is the to compare the absorption of indigo yielding "jeans blue" with 6,6'-dibromoindigo, main constitutent of tyrian purple. A second, that heteroatoms next to a $\ce{C=O}$ carbonyl group induce a hypsochromic shift like in the case of acetaldehyde example here in comparison to acetic acid here

Note the use of the terms of "bathochromic shift" and "hypsochromic shift" along solvatochromism.


Diastereoselectivity in synthesis problem (4 stereocenters) Why is the particular diastereomer the major product of the sequence starting from racemic enone? I know that hydroboration is syn, therefore the arrangement of ring junction H atom and OH group. Why is one of the methyl groups anti and one syn with respect to the H atom and OH group?

If we get rid of the two methyl groups and just consider what the 3D conformation of the molecule is likely to be (I don't have a pen on me right now, but the model below was minimised using very low level molecular mechanics).

What we can hopefully see is that there is very much a convex face and a concave face of this molecule.

In the absence of all other factors, the hydroboration should be faster from the convex face (the lower face as drawn).

This would be simple facial selectivity, as is often observed on ring systems.

However, we have additional chiral centres that are going to influence the reaction (diastereoselective reactions can be matched/mismatched such that the methyl groups present might help us or hinder us.

The problem in this question is that it isn't specified whether the methyl groups are up or down etc. So presumable what we're actually seeing is a resolution (i.e. we're only forming a small amount of product, but its optically pure because certain combinations just don't react at a sufficient rate).    [I did try and find the paper on Reaxys but had no luck, there is a chance he just made the question up].

So if we now consider the methyl group next to the acetal functionality, it would help us if it were pointing up (as I've drawn it, you could flip the rings to get it pointing down as shown in your product above) to leave the convex face free.

If it were pointing down (as I've drawn it..) then the approach of H-B would be hindered, making attack from convex/concave both unfavourable relative to the approach in which the methyl group is orientated into the concave face.


Why aren't solid pure transition metals colourful? I understand that the compounds of transition metal are colourful because of the appropriate energy gap between different d orbitals. But why aren't they also colourful in their solid pure state? For example, why isn't Fe colourful whereas, its compounds are?

Colour is a property of electronic transitions: the transitions in metals are different to those in their compounds
In many discrete compounds of transition metals the colour arises because there are accessible electronic transitions between molecular orbitals in the molecules (usually involving metal d-orbitals) matching the energy of wavelengths of visible light. These transitions are often fairly narrow so giving a wide range of possible colours.
Pure metals are not like that. To simplify a lot, the definition of a metal is having a continuous band of electronic states (not a simple molecular orbital) for the electrons involved in conduction of electricity (the conduction band). This band involves the whole bulk substance, not just isolated atoms of the metal. So the transitions that might involve colour (d-d transitions in a molecule) are not present in the same way: instead transitions occur inside the conduction band or between orbitals and the conduction band. Either way, the narrow transitions giving colour in compounds are now much broader because  a wide range of continuous electron energies exist in the conduction band.
This is a gross simplification, but it is good enough to explain the key observation if you don't want to get into the theory behind the electronic structure of metals.


What type of gloves should be used while diluting sulfuric and hydrochloric acids? What type of gloves should we use on hands while diluting sulfuric and hydrochloric acid?

Can we use these gloves in chemistry lab to handle other acids?

There are numerous sources online comparing polymer materials resistivity in hazardous environments, but most of them agree that in your case ordinary nitrile gloves would protect in case of incidental contact, and butyl rubber/neoprene/PVC gloves are more suitable if a prolonged exposure is expected.

Nitrile gloves are disposable and cheap, and always should be changed when contaminated with chemicals as they only offer basic protection.


Does the decay of Carbon-14 effect health in biological systems? Please note that this is not fundamentally a Biology question. It relates to the properties of molecules containing Carbon and Nitrogen.
Carbon atoms are everywhere in living systems from sugars to proteins etc.
Carbon-14 is present in living systems in a certain percentage compared to Carbon-12.
This Carbon-14 is decaying to Nitrogen-14 even when the system is alive.
Does this effect the health and functioning of the living system at all ?
I mean take a glucose molecule $\ce{C6H12O6}$, suddenly one of the Carbon atoms randomly decays to a Nitrogen atom. Does this molecule continue to function as a glucose molecule ? This question extends to every molecule in the body that contains Carbon atoms.

I mean take a glucose molecule C6H12O6, suddenly one of the Carbon atoms randomly decays to a Nitrogen atom. Does this molecule continue to function as a glucose molecule ? This question extends to every molecule in the body that contains Carbon atoms.

To answer this part of the question, obviously a glucose molecule which has had a carbon-14 atom decay isn't a glucose molecule anymore. What the glucose molecule turns into is anybody's guess, but no doubt there would be a wide range of molecular fragments from the multiple decays of glucose molecules. Some of the molecular fragments would be highly reactive and cause unusual reactions within our bodies.
Worse, the cabon-14 emits a beta particle when it decays. It would be sort of like a bullet ripping through the body. Thus the beta particle would disrupt the bonding in other molecules, not just in the original glucose molecule.


What is the physical state of the gas adsorbed on a solid adsorbent? I am studying surface chemistry in high school I have this doubt and I am unable to find a answer your help is appreciated
 I searched for the answer on google I asked my chemistry teacher he said the question is beyond high school's curriculum and I even asked some of my friends but all in vain.

Well, it usually depends on the temperature: if temperature is too low, molecules cannot move so freely so they are localized on top of the solid without much movement.

However, in general cases the adsorbed gas behaves like a liquid in two dimensions. Near the surface, the gas molecules are 'trapped' in the force field of the solid, so they don't escape so easily. Then the freedom of movement occurs only in two dimensions (the molecules travel on the surface).

I made some time ago a movie calculating (using molecular dynamics) movement nitrogen molecules adsorbed on a graphene surface at 77 K. This can give you a clear image of what happens on the surface. As you can see, the molecules can travel on top of the graphene in two dimensions (we can call it a 2D phase). This movie is the case when still a monolayer of gas does not complete. There can be cases with less molecules, say, less pressure of nitrogen in the gas phase; or more molecules.

The physical states can be observed more clearly in phase diagrams of the adsorbed surfaces. Typical examples shows a plot of the adsorbed phase transitions in a P vs. 1/T plot. This topic is extensively covered in the nice monograph on Physical Adsorption by Bruch, Cole and Zaremba, for example.


Is graphene chemically reactive? Different internet sources say different things.

If possible, can someone explain why it is (or isn't) chemically reactive?

Graphene in its pristine form, a single sheet of pure sp2 hybridized benzenes, is (relatively speaking) one of the least chemically reactive materials known because of several reasons: one, its enormous size (we keep making larger and larger single "molecules", now thanks to IBM up to 4 inches!). Secondly, its physical hindrance: you can only do bottom/top down chemistry, granted this isn't that big of an issue but it does greatly reduce using large compounds. Last is it's stableness: this thing is crazy stable because of its gigantic p-butts and other effects.

The long story short is that you don't have any functional groups or easy points of entry. The way to think about it is "what does graphene look like and what makes those other molecules reactive?". The answer is...benzene and really any other hydrocarbon with no functional groups. The difference there is their size. But the same rules apply! You can still oxidize graphene, at the edges for example, much like you would benzene. And in the past years, we've seen even more radical methods, the most common being Hummer's Oxidizing, and some other clever ones like a Claisen Rearrangement directly onto the surface of graphene.

Hope that helps!

EDIT: after Greg's comment

Don't apologize! This'll get the discussion going, I was hoping to see more answers other than mine. I think you're right in terms of absolute stability but my statement is a phenomenon harder to describe than mere size. From experience, we noted that on very large graphene flakes, our functionalization of them would distort the sheets, but not break them, creating lots of hindrances and thereby decreasing its reactivity; it basically went from sheet shape to a jumbled up ball. However, on smaller (much smaller) sized flakes, the individual molecules (though hindered and flexed in a similar way) would, at the same concentrations of reactants, get functionalized more, ratio of functional groups to carbons on the sheet.

This is an indirect effect of the massive size of the graphene sheets. So I guess you're right in that size isn't exactly a stability factor but rather, again, it's all about hinderance. However, I wanted to point it out because the massive size of such a flat molecule is quite spectacular and does yield some interesting behavior we don't normally think of.


formaldehyde strong odor neutralization One of our customers used a very high dose of formaldehyde 37% solution (60 liter formaldehyde  in 40 liter water) for disinfection in  his chicken farm (saloon). Now he says the odor of formaldehyde is very strong and can`t start breeding  season.  How can I help her?

Can he  neutralize  formaldehyde odor with some chemical treatment? Maybe he can use sodium hypochlorite (bleach). Do you think chlorine can oxidize formaldehyde?

Ammonia water, AKA ammonium hydroxide ($\ce{NH4OH}$), reacts with formaldehyde to form hexamine, a comparatively innocuous solid (though it is flammable when dry).

Safer might be Neutralization of Formaldehyde Gas by Ammonium Bicarbonate and Ammonium Carbonate, which would greatly reduce the flammability of the product.


Does single-layered Aluminium Phosphide exist? I'm currently working with the band structure of graphene and graphene-like compounds.

α-Boron Nitride (hexagonal, single layer) has many interesting properties. I was wondering if the next period analogue, a hexagonal single-layered Aluminium Phosphide existed. Wikipedia lists this compound, but it seems to only be found with a multilayered sphalerite structure.

Does an allotrope of AlP exist (in a stable state at normal conditions) where the structure is similar to graphene or silicene? If not, why?

I suspect that it does not.  As far as I know, all trivalent phosphorous is highly pyramidal--or at least not planar. This would rule out any graphene-like structure.

The reasoning that I have seen is based on hybridization. In the second row (third period) the 3s and 3p atomic orbitals are well separated, implying that s and p can not mix and that the outer valence is unhybridized. If bonding is with unhybridized p's, then bonding would occur at right angles.

(Kutzelnigg, W. Angew. Chem., Int. Ed. Engl. 1984, 23, 272.)

(Brown, E. Organometallics 2000, 19, 2208-2214.)


Reaction of alc. KOH with alkyl halide The following question was given in FIITJEE study material:

At first, I thought it should be elimination. But, soon I realized that there is no heating, plus a primary alkyl halide is used. So, "no elimination".
Then, I thought it would proceed through $\ce{S_N2}$, and hence (B) should be correct, as it involves concerted mech.. But, again I was wrong. Why is it so?

The correct answer given is:

 (D)

The solvent used is a polar protic solvent which heavily solvates the OH- radical thereby decreasing it's nucleophilicity as the solvent forms a cage around the radical anion .Even though the basic strength of the anion is also decreased by solvation still the decrease in nucleophilicity is dominant as it is more difficult for a large solvated anion to attack at a carbon atom than to sneakily abstract an acidic hydrogen and hence the enolate is formed which then , proceeding through the SN NGP (neighbouring group participation) mechanism eliminates the Cl ion forming a cyclopropane.
Hope that this helps.


Is it possible for the remaining solute of a saturated solution to form a single even phase from top to bottom (similar to that of a colloid)? If a solute that dissolves in a liquid solvent has reached high enough levels to cause saturation within the solution, is it possible for the remaining solute particles to form an evenly distributed phase from top to bottom similar to that of a colloid (where settling would not occur or take very long to occur)? I realize behavior varies depending on the solute and solvent in question, I am just curious if this is possible. Thanks so much!

The density of the solute relative to that of the solvent will primarily determine whether the excess solute will float, sink or tend to disperse throughout the solution.  Additionally, the particle size and porosity will play a role in the rate of settling or floating, as will the viscosity of the solvent, but it is the density that determines whether the particles will float or sink.

But, the answer to your question is no. Even if the density of the undissolved solute is essentially the same as that of the solvent, and the undissolved solute is very fine, then you could have a uniform looking solution, but it would still be a two-phase solution, with a solid phase suspended in a liquid phase.


Free chemical database to search for physical properties? I am looking for an Open-access tool that can search "backwards", eg. give me all the structures corresponding to a given heat of formation, critical pressure, etc. It would be even better if the database canbe automatically searched by API or by downloading it. Currently my best go is ChemExper, where I can search for some of the properties, but it is challenging to make it automatic. Is there anything similar in the field?

As I've mentioned on other related questions, this is very much an open topic. There aren't many open-access databases in chemistry, and far fewer that have APIs.

As suggested in the other answer, there are multiple ways to query / scrape Wikipedia.
ChemSpider offers some property search features.
Reaxys offers some property searching, including spectra features. It's not open access.


How quickly does sulfur hexafluoride mix with the atmosphere? A common science demonstration is to float a "boat" on sulfur hexafluoride.
How long would the sulfur hexafluoride stay in the container (in the experiment linked above) if the container were to be left uncovered and undisturbed at room temperature?

Also, I'm wondering what would happen if you scaled the experiment up to the size of a lake, ignoring the time it would take to fill said lake and the impossibility of obtaining that much sulfur hexafluoride.  Let's also assume that there is no wind to blow it away. Could you have a large permanent pool of sulfur hexafluoride or would it eventually dissipate and mix with the atmosphere?

Would this change if a different dense (non-reactive) gas were used, like xenon?

Up front: this depends on what you mean by "stay in". It would diffuse out of the container and, given the surroundings are closed, obtain a small equilibrium concentration.

Not that I could provide an exact answer to your question (it is rather a physics question), but this paper (paywalled) deals with the diffusion of the $\ce{N2-SF6}$ pair. The binary diffusion coefficient is reported to be 0.0945 cm2/s. In their experiments they used a container of d = 0.5 m, l = 1 m, introduced a small amount (2-5%) of $\ce{SF6}$ at the bottom of the vessel and measured the time it took to obtain the equilibrium concentration of the mixture. For this setup, this was ~15 h, with a reported time constant of 4-5 hours (I believe this is the time it takes to reach a value of [local $\ce{SF6}$ concentration]/[equilibrium concentration] of 0.5).

They used a lengthy formula derived from a one-dimensional model for the comparison of calculated and measured concentrations and in my opinion the measured data agreed quite well with the calculated ones.


Can't we reduce climate change by converting CO2 & CH4? Straight off I want to say, I never took chemistry classes, so my understanding is very limited. My question is considering the equations below, why is it so hard to reduce greenhouse gases? Is it simply a matter of conversion requiring too much energy?

Can't we remove carbon dioxide & methane from the atmosphere by converting them into water and carbon?

$$\ce{CO2 + CH4 -> 2H2O + 2C}$$

Or by converting them into water and propane?

$$\ce{CO2 + 5CH4 -> 2H2O + 2C3H8}$$

There are a few problems with what you're proposing:

The reaction requires energy.
Where do get the methane from?
What will you do with the carbon you're generating?

The energy problem can be solved by using surplus energy from renewable source like solar and wind, but, as airhuff also mentiond in a comment, there is much more CO2 in the atmosphere than methane. Also capturing CO2 from the air isn't simple,  and finally why would you use a valuable fuel like methane to create carbon or propane?

In any case I think a much more useful reaction is
$$\ce{CO2 + 2H2O -> CH4 + 2O2}$$

Water is split into oxygen and hydrogen using electrolysis. The hydrogen reacts with carbon-dioxide to form methane.
There are already small plants that do this, but AFAIK the idea hasn't really caught on yet. This may have something to do with the cost and/or the availability of carbon dioxide or surplus energy.

More recently scientists did something even cooler; they captured CO2 directly from the atmosphere and turned it into methanol (CH3OH) using Ruthenium as a catalyst. A drawback is that the reaction works at high temperatures (about 150 degrees C).


Balancing redox equations in acidic media: to use H+ or H3O+? Is it preferable to balance redox equations with  $\ce{H+}$ or $\ce{H3O+}$? I think the latter, because first, the bare hydrogen proton doesn't exist in water solution. So why write down something we know not to exist?

I ask because in this problem, if we balance using both species, $\ce{H^+}$ and $\ce{H3O+}$, we achieve two different equations. Mass is conserved in both equations as well as charge; all the differs is one coefficient.

So are the two equations equivalent in that they are still describing the same reaction? I think so, because sometimes I see textbook authors write these two equations for $K_\mathrm{w}$:

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

$$\ce{2H2O <=> H3O+ + OH-}$$

However, if we were actually going to run an experiment and needed exact quantities, we would probably prefer the second reaction both above and below because the coefficients in these reactions better represent the chemistry of the system right?

For example, which of the following is "better"?

$\ce{ClO_{3}^- + 6I^- + 6H^+ \leftrightharpoons 3I_2 + Cl^- + 3H_2O}$

$\ce{ClO_{3}^- + 6I^- + 6H_3^+O \leftrightharpoons 3I_2 + Cl^- + 9H_2O}$

This is a question of "style" so there is no correct answer.  In fact, just as $\ce{H^+}$ doesn't really exist in solution, neither does $\ce{H3O+}$ exist in solution.  In reality higher hydrates of $\ce{H^+}$ are what exist in solution. See the first paragraph in this Wikipedia article.  Again, both $\ce{H^+}$ and $\ce{H3O+}$ are just shorthand notations for reality, choose the style that you like best.


Which has one monochlorination possibility? Which would have one monochlorination possibility?

2,2-dimethylpropane:

or 2-methylpropane:

It seems I'm missing something because I thought both would only have one monochlorination possibility. (unless I am allowed to substitute the additional H on the 2-methylpropane with a Cl?)

Replacing Hydrogen is the only way how chlorination (or halogenation) works.

So in 2,2-dimethylpropane the central carbon (carbon-number 2) is quaternary, meaning all four of its valencies are satisfied by carbon atoms, which implies that there are no hydrogen atoms which can be substituted by chlorine (or any halogen), whereas all the remaining 4 carbon atoms are identical, so substitution on any one of those will result in the same compound (1-chloro-2,2-dimethylpropane).

Whereas in 2-methylpropane(commonly known as isobutane) the carbon-number 2 is only tertiary meaning only 3 of its valencies are satisfied by carbon-atoms and it has one hydrogen atom on it. So on substitution of this hydrogen atom by chlorine the product is 2-chloro-2-methylpropane and on substitution of any one of the terminal (last) carbon atoms would all result in 1-chloro-2-methylpropane. That is two different products.


Determining the Energy Density of a Battery given the anode and cathode everyone, I'm using nanohub's introduction to the material science of rechargeable batteries course to do some self-studying until I can go back into the lab. On one of the practice problems, it asks to calculate the energy density of a given battery system. For the first problem, it asks this for a lithium metal anode, LiCoO2 cathode system. We are asked to assume an electrode thickness of 100 micrometers and a porosity of 35%. I assumed no porosity contributions from other components such as binders. When I calculate the value I don't get something close to the answer given in the solution which is 453.6 J/kg.

I have tried calculating this and the other problems a few different ways with different units but I am not getting the correct answer. The formula I am using is

Energy density = (potential difference between anode and cathode) * cathode charge (capacity)* electrode thickness (this was given in the PS)*(1-porosity(this was also given)) below is an image of what the given information and potential answers looks like

Cathode and anode will have different specific charge capacities. These capacities can be made equal by tailoring the thickness of the electrodes. The question is asking you to make sure that the charge capacity ratio is equal to one, which means you will need to find what anode thickness you must have for the capacities match (they give you cathode thickness of 100 microns)
This is the equation to find the energy density of a battery: $$E=VC/m$$
$E$ is energy density in Wh/kg, $V$ is nominal voltage in volts, $C$ is capacity in Ah and $m$ is mass in kg. Why are you multiplying this equation with thickness and porosity?
You were given some electrode geometric parameters and you will have to find the others. Remember, you are looking for the electrodes masses so that you can compute energy density (hint: start by looking into densities of your materials).
Keep in mind that not all electrodes will be porous (for example for Li metal you can assume porosity is 0%). Also, you will have to report results in $J/kg$, so make sure you keep track of your units and convert if necessary.


How can residual entropies be reconciled with the Third Law? The third law of thermodynamics states that the absolute entropy of a pure crystalline substance is $0$ at $0\,\mathrm{K}$. In the analysis of chemical reactions this absolute scale is sued to calculate the entropy change.  However, it is also noted that some substances like $\ce{CO2}$ have residual entropy at $0\,\mathrm{K}$ because their structure is not perfectly crystalline.

Now my question is, given all of the above why is it that in thermodynamics tables such as in the book "Physical Chemistry" by Bawendi the entropy of $\ce{CO2}$ is listed as $0$ at $0\,\mathrm{K}$ when it should be some finite (albeit small) number? The same goes for other substances like $\ce{H2O}$ whose entropy is listed as $0$ at $0\,\mathrm{K}$ when it should be a finite number.

I understand that different tables use different reference scales and the entropy of $\ce{CO2}$ could be arbitrarily taken as $0$ at $0\,\mathrm{K}$. However the problem is then if the tables give the entropy of a substance relative to the entropy of the same substance at $0\,\mathrm{K}$, then the reference scale for each substance is different. This is fine but in all the example problems in the textbook mentioned above (and others) they use the entropy values listed in the table for each substance without accounting for the different reference states.

For example, if I have the reaction

$$\ce{A + B -> AB}\,,$$

the entropy change is

$$\Delta S ~=~S_{\ce{AB}} - S_{\ce{A}} - S_{\ce{B}}\,,$$where $S_{\ce{A}}$ is the entropy of $\ce{A}$ and so on.

Now in order for this to work the entropies must be taken relative to the same reference state (like when we use heat of formation to calculate enthalpies). However in the Bawendi book they use the values listed in the tables for each substance, which are again relative to the entropy at $0\,\mathrm{K}$ of that particular substance. This would all be fine if the entropy at $0\,\mathrm{K}$ was indeed $0$ for all substances but it is not. Could someone please explain why a lot of books do this? or is there something that I am missing?

First of all, entropy is an absolute value; you don't use different reference scales and reference states for it. Internal energy is another story: it never appears in the equations as an absolute value, but only as a difference of starting and ending values, hence you may define a scale for it and place zero on that scale more or less arbitrarily. Ditto for enthalpy and Gibbs energy. But entropy is not like that.

Now to the point. The only thing that has zero entropy at $0\,\mathrm{K}$ is an ideal crystal, which is also the equilibrium state (whatever that might mean) of crystalline compounds at that temperature. Getting to equilibrium in fluids might be anything from slow to very fast; in solids it is usually rather slow, and in solids near absolute zero it is so slow that the Sun might well become a red giant before the last defects leave your real crystal. See, thermodynamics is all about eternity. Wait until then, and you'll see the Third law at work.

That's what we mean when we are talking of residual entropy.

   I know about the possible degeneracy of the ground state, but this post is already getting too long.

Does the said law have no bearing on earthly things, then? Why, it does. Many crystals just get good enough while growing at reasonable temperature. They have very few defects, hence their entropy is already pretty low and gets lower as we cool them down. On the other hand, many other crystals are not that good.

So why would the book list entropy of all compounds at $0\,\mathrm{K}$ as zero? Well, getting to absolute zero is infinitely hard, and getting close to it is pretty hard too. Maybe the authors just put all those zeros there in a well-intentioned attempt to remind the readers of the Third law? Too bad for those who by that point have heard of residual entropy.

Then again, I don't see why would $\ce{CO2}$ have much of residual entropy in the first place. It is a good rigid molecule with nice polar bonds and without nasty things like nuclear spin. Perhaps its entropy near the absolute zero is really $0.0$ within the given precision. Water ice is another story, with all its hydrogen disorder. But wait, isn't it transforming into a hydrogen-ordered form at low temperatures? So maybe the zeros are true, after all?


Common Ion Effect explanations From my textbook, it is stated that:
e.g. If we add a basic salt $\ce{NaF}$, it will fully dissociate to give $\ce{F-}$. Then if we add $\ce{HF}$, since it is a weak acid, the high concentration of $\ce{F-}$ already present in solution shifts the equilibrium to the left, hence fewer $\ce{H+}$  forms than expected so higher $\mathrm{pH}$.
But can I explain like this: after dissociation of $\ce{NaF}$, the formed $\ce{F-}$ reacts with water to give $\ce{OH-}$ as the following equation:
$$\ce{F- + H2O <=> HF + OH-}$$. Due to the presence of $\ce{OH-}$, upon addition of $\ce{HF}$, the $\mathrm{pH}$ will be higher than expected. Are there any flaws in this explanation?

Both your paragraphs are valid, as all present components are balanced by 3, mutually linked equilibrium reactions, with the respective equilibrium constants:
$\ce{HF + H2O <=> F- + H3O+}$, $K_\mathrm{a}$
$\ce{F- + H2O <=> HF + OH-}$, $K_\mathrm{b}$
$\ce{2 H2O <=> H2O+ + OH-}$, $K_\mathrm{w}$
with the relation $K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w}$
$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[F-]}}{\ce{[HF]}}}$


Halogen Oxyacid Trend (acid strength) Why is $\ce{HClO4}$ more acidic than $\ce{HBrO4}$? But $\ce{HCl}$ is less acidic than $\ce{HBr}$? What determines the acidity? Is it the concentration of $\ce{H+}$? If so, how do these trends develop along the halogen group?

In the determination of the $\mathrm{pK_a}$ values of really strong acids, water as a medium is irrelevant at a certain point.

Ivo Leito and coworkers have examined the acidity of various compounds in 1,2-dichloroethane (DCE).

In this medium the $\mathrm{pK_a}$ values for $\ce{HCl}$, $\ce{HNO3}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HBF4}$, an $\ce{HClO4}$ have been determined as -0.4, -1.7, -4.9, -7.7, -10.3 and -13.0.

In this survey, 1,1,2,3,3-pentacyanopropene (1), was found to be even more acidic with a $\mathrm{pK_a}$ of -15.0!

EDIT
It is important to note that these values are likely to be correct in their relative order. But the values are based on picric acid, for which the $\mathrm{pK_a}$ was arbitrarily set to zero. This initially escaped my attention!

Thanks to Martin for pointing out this important detail in his comment below!

Being an oddball, the rather unstable perbromic acid ($\ce{HBrO4}$) was not in the list!

It was synthesized by Evan Appelman (even on a larger scale) by bubbling fluorine gas through an alkaline solution of 1M $\ce{NaBrO3}$ and 5M $\ce{NaOH}$ until the solution becomes acid.

Amazingly, the author further states in a footnote:

  The reaction is not smooth, and small explosions may take place in the vapor above the solution.


Distilling small volume I have 50 ml of a compound I would like to purify. Since the impurities presents have close boiling points (b.p.), I have to use fractional distillation. Am I going to lose significant quantity of my product volume because of loss related to the distillation itself?

If so, what can I do to minimize the loss?

I thought about adding a higher b.p. solvent to the distillation flask so I can distill all the compound without leaving some in the flask. What do think about this method?

Are there any other strategies?

In order to both achieve sufficient separation and minimize loss during fractional distillation, you want

a large inner column surface for optimal contact of the vapour phase (upstream) and the liquid phase (downstream)
a rather small inner column volume

Often, these destillation are performed using a Spaltrohr distillation apparatus. Here, the vertical column consists of two concentrical, threaded glas tubes with a gap of 1-2 mm  in between.


Water conditioning with a descaler? Products known as “descalers” are said by the manufacturers (such as this one) to use electrical impulses to transform calcium bicarbonate (which builds up in pipes and water-using appliances, causing premature failure, and also leaves calcium residue on glassware, clothes, and so on) into calcium carbonate (in a form known as aragonite), which does not adhere -- so the scale problem is solved.

Truth or snake oil?

So, you only need a descaler/softener if you have what is known as hard water, which is basically just water with a lot of minerals such as calcium and magnesium. You can do a test that shows how many GPGs (grains per gallon) your water has. According to http://home.howstuffworks.com/question99.htm,

  If your water tests at 1 GPG (17.1 mg/L) or less, then you have soft water. Water around 1-3.5 GPG (17.1-60 mg/L) occupies a gray zone between soft and slightly hard water and 3.5-7 GPG (60-120 mg/L) is moderately hard. Hard water is around 7-10.5 GPG (120 - 180 mg/L), and very hard water is above that.

So if it does turn out that you have hard water, then you look into descalers and softeners. The reason for this is that the calcium and magnesium come out of the water as scale, which builds up on the insides of pipes, water heaters, etc., to the point where they can become completely clogged.

A descaler is used to remove this buildup. They do this because they are acids and release hydrogen ions when dissolved in water. These ions, in turn, eat away at the scale by pulling at them with the charge. A softener is used to clean the water as it comes, disposing of the minerals safely. Water softeners are mechanical appliances that are put into your home's water system. They exchange the minerals for something else, usually sodium, in what's called an ion exchange. A diagram of a water softener is shown below:

This website provides more information about the exact workings of a water softener.

Hope this helps!


Are non-exothermic explosions possible? Explosives like picric acid and trinitrotoluene are unstable due to the close packing of so many $\ce{-NO2}$ groups onto a benzene ring. These highly strained bonds require only a relatively small amount of energy to cause the molecule to rip itself apart, forming a lot of more stable gas molecules, and releasing a lot of heat in the process.

My question is: is it thermodynamically feasible to have a reaction that rapidly evolves a lot of gas without being extremely exothermic? Would this still be considered an explosion?

It doesn't seem to me that this could be possible, but I never cease to be amazed at how little I know about chemistry. If it is feasible, and the gas molecules are more stable, then where does the energy go? If the gas molecules aren't more stable, then what's the driving force?

Ultimately, explosions are just a type of spontaneous reaction, and even endothermic reactions can be spontaneous, so long as entropy increases enough over the course of the reaction to compensate it. Near room temperature, however, the entropic factor is usually small compared to the enthalpic factor. This is why there are comparatively few examples of reactions which produce strong cooling upon mixing reagents.

That said, none of this strictly precludes slightly-exothermic or endothermic explosions, to the point that the concept has been legitimately investigated. As I mentioned in the comments, there is even a specific term for these types of explosions; entropic explosions.

For a spontaneous reaction to be explosive, there is one property it must have above all others: the reaction must happen very quickly. Now we are dealing with kinetics, which is much harder to analyse theoretically. It is possible, however, to point out another factor that makes endothermic explosions difficult.

By far, most reactions speed up when temperature is increased. In a conventional, exothermic explosive, this establishes a positive feedback effect, where heat from the reaction causes it to react faster, which produces more heat, which causes even faster reaction, and so on, facilitating a runaway reaction and subsequent detonation. If the exotherm is weak, then there is a larger chance that the explosion will fizzle out before enough heat is produce to get things going, or the reaction might be too slow to explode and instead steadily combusts. For an endothermic explosive, very much the opposite happens, and a negative feedback is established; once started, the reaction would decrease the temperature of the explosive, hindering propagation.

So, while there are no arguments which completely forbid slightly-exothermic or endothermic explosions, it seems to be a rather unlikely scenario in practice. Wikipedia only lists one possible contender for an entropic explosion, which is triacetone triperoxide. The original article making the claim is mostly theoretical, and there seem to be some rebuttals to it, so even that is not certain. Of course, we may be in the middle of a giant selection bias effect, because almost all research into explosives is focused on getting more energetic materials, not less. Maybe there are examples waiting to be found, but no one has had the interest in doing so!

As a sidenote, explosions actually don't necessarily produce gas. There are some rare examples, such as cuprous acetylide ($\ce{Cu2C2}$) and silver acetylide ($\ce{Ag2C2}$) where all the products are solids. Also, there are examples of non-chemical explosions which are only slightly exothermic or even endothermic. As MaxW mentions, a pressure explosion from a liquefied gas will be endothermic, as energy is consumed to evaporate the liquid. Another interesting example of an explosion which is probably only slightly exothermic is a Prince Rupert's drop being set off.


Is ammonia an acid or a base? Is $\ce{NH_3}$ an acid or a base? N has a higher electro-negativity (3.04) than hydrogen (2.20), therefore nitrogen attracts the bond's electrons easily. The reaction should produce $\ce{H^+}$ because the electrons of hydrogen are attracted by nitrogen.

According to my Chemistry book, my assumption seems to be wrong because $\ce{NH_3}$ is considered a base. Why am I wrong?

The difference isn't that stark. However, you have to bear in mind the Configuration of electrons on the central nitrogen. We often will draw ammonia with the electron pair atop the nitrogen. That lone pair will readily accept a proton, making it a base.

To put it another way, ammonia is far more ready to accept a proton than to release one thanks to the lone pair.

Of course, it also depends on the environment it's in. But in general, you'll usually see ammonia (or just nitrogens in compounds) referred to as basic


Measuring bioanode potential against Ag/AgCl reference electrode I'm trying to make a bioanode for use in a microbial fuel cell.  I have a piece of carbon cloth in an acetate solution with river sediment.  I'd like to know the potential of the anode, so I connected it to an Ag/AgCl reference electrode also in the sediment/acetate solution (note that ref electrode has 3M KCl, about 0.2V relative to SHE).  Between the reference electrode and anode, I have a multimeter connected with the positive going to the Ag/AgCl reference electrode and negative going to the carbon cloth anode.

The multimeter reads +0.2V.  Does this mean my anode potential is 0.4V (0.2V from multimeter reading + 0.2V for reference electrode correction)?  OR, does this mean that my anode potential is 0V (-0.2V multimeter reading + 0.2V for reference electrode correction). I'm not clear if the positive side of the multimeter and negative side are connected correctly to the electrodes.  If they are connected incorrectly, then the voltage sign should be flipped.

Thank you!

Coming to your experiment, there is a nice description along with sufficient details which shows you how to make connections.

Microbial Fuel Cells & Bacterial Power
https://agsci.oregonstate.edu/sites/agscid7/files/bioenergy/fuel_cell_interactive.pdf

Coming to the confusion: Electrochemists do not measure electrode potentials by using a multimeter as shown in general chemistry textbooks. The numbers shown on a multimeter display only qualitatively showing us that there is a potential difference between these two electrodes. Secondly, carbon cloth is the least reproducible electrodes in terms of its potential (heavily dependent on the history of how it was made). You won't find its values in standard tables. Therefore we cannot compare your experimental values with standard conditions (in the tables, the values are calculated under controlled conditions) because your system is nowhere close to standard conditions. It is like comparing apples and oranges.

An interesting part for you is, how to decide which one is cathode and which one is anode of your cell? Connect the electrodes in such a way that you get a positive potential difference on the multimeter. If you happen to get it, you have made the right connections i.e., negative terminal of your cell is connected to the negative terminal of your multimeter and so on.


How to determine the number of stereoisomers of a di-substituted allene, exhibiting both axial and central chirality? What are the diastereomers and enantiomeric pairs of the compound below where A are chiral groups (central chirality)? There should be 3 diastereomers and 3 enantiomeric pairs (last page).

I previously asked a question about tetra-substituted allenes.

These are the three diastereomers with their enantiomers on the right side. The top and middle diastereomers are connected via axial chirality (R and S are used to mark the configuration of A groups).


Regioselectivity of alpha halogenation of ketones Why does alpha halogenation of a ketone in acidic conditions happen at the more substituted alpha carbon atom, and in basic conditions at the less substituted alpha carbon atom?

The acidic and basic catalyzed alpha-halogenation of a carbonyl proceed through an enol and an enolate intermediate respectively.  The difference in regioselectivity arises because of the different process used to arrive at the enol/enolate in the acidic and basic cases.

In the acidic case, 1) the carbonyl oxygen is protonated and then 2) a proton is removed from the alpha carbon to form the enol.  See here for a good picture of this mechanism.  In this process, the reaction is an equilibrium so thermodynamic control will lead to formation of the most stable enol.  Therefore, a proton will be removed from the more highly substituted alpha-carbon to produce the more stable double bond (the one that is more highly substituted).

When the reaction is run under basic conditions, the process is no longer a thermodynamically controlled equilibrium, but rather proceeds under kinetic control.  The same link used above also shows the mechanism of the base-catalyzed alpha-halogenation.  The rate determining step is removal of an alpha-proton by the base.  The lowest energy proton to remove is the one on the least substituted carbon. Remember that a primary carbanion is more stable than a secondary carbanion which is more stable than a tertiary carbanion, so removing a primary proton will produce the most stable carbanion-enolate.  Hence, in the base catalyzed case the proton will be removed from the least susbstituted alpha carbon so to produce the most stable carbanion-enolate intermediate.

To summarize the acid-catalyzed halogenation proceeds under thermodynamic control to produce the most highly substituted (most stable double bond) enol.  Whereas the base-catalyzed halogenation is kinetically controlled and produces the least substituted (most stable carbanion) enolate anion.


Which calcium salt is good enough to make calcium benzoate? This question is inspired from the following questions:

Reaction of citric acid and calcium chloride
Precipitation of citric acid by CaCl2 and Ca(OH)2?

The answer to the 1st question says that calcium citrate is highly unlikely to form from calcium chloride and citric acid because of its inability to be present in deprotonated form and will not precipitate in acidic medium. 2nd question also addresses a similar kind of problem. But what about calcium benzoate?

Which calcium salt can succesfully precipitate calcium benzoate from benzoic acid? Would calcium hydroxide work for that? I am pretty sure calcium chloride won't work because it is a good dehydrating agent and would rip off water molecules to form something else. Calcium benzoate is also somewhat more soluble in water than calcium citrate. So, what should be the necessary measure to successfully precipitate calcium benzoate?

Thanks to @Shoubhik who pointed out that calcium hydroxide is good enough to make calcium benzoate. To support this claim, I found a patent which indeed describes that calcium hydroxide was used to make calcium benzoate (See comments). Here is the summary of the patent:

  The invention relates to a processing method for synthesizing calcium
  benzoate at low temperature. The method comprises the following steps:
  heating and stirring 65-85kg of industrial-grade benzoic acid, 28-32kg
  of calcium hydroxide, 1-3kg of catalysts comprising triethanol
  distearate, ethylene distearamide and ethylene bislauramide and
  300-500kg of clear water in a reaction kettle at 333-363 K to react
  for 2-4 hours; and then adding 5-15kg of separant, stirring for half
  an hour, filtering and drying. The method provided by the invention
  can be used for directly synthesizing calcium benzoate, thereby
  lowering the equipment investment, reducing the working procedures,
  lowering the labor intensity of workers, increasing the output,
  enhancing the quality, lowering the cost and preventing the
  environmental pollution.


Can I add citric acid to KCl for iron removal in a water softener? I recently installed a water softener and having a fair amount of clear iron in our water, I've been using Morton Rust Remover pellets which are basically NaCl with citric acid added as an additive.  I'm guessing the iron removal function has something to do with the citric acid since it's the only notable difference between this salt and regular water softener salt.

At some point I'd like to stop using NaCl and switch the KCl since it seems to be more environmentally friendly.  However, I really notice a difference using the iron removal pellets and I've been unable to find an equivalent product for sale which uses potassium.

Can I simply mix some citric acid in with the potassium in the water softener to achieve the same result?  Or do the citric acid and the sodium have some special relationship which helps with the iron removal?

Your ion-exchange column removes $\ce{Fe^{3+}}$ ions from your water by exchanging them for $\ce{Na^+}$.

When you back-flush you use concentrated solutions of $\ce{Na^+}$ ions and citric acid to remove the iron out of your ion-exchange column. The flush water is waste of course which must be disposed.

The exact chemistry mechanism of all of this depends on a number of factors:

NaCl has a solubility of 36 g/L at 20 C which is 0.62 molar but
KCl has a solubility of 34.2 g/L at 20 C which is 0.46 molar.
I'd expect potassium citrate to be less soluble than sodium citrate but
I didn't find any data to support this notion in a quick search.
I'd expect that you column has a better attraction to $\ce{Na^+}$ ions
than to $\ce{K^+}$ ions.
If you have some sort of automatic monitor then it may not
function properly with $\ce{K^+}$ ions. I'd guess that the monitor
just looks to determine if the water is yellow, so I'd guess that
$\ce{K^+}$ ions would work fine in this regard.
Not sure what pH you'd want. I'd guess maybe half trialkali citrate with
half dialkali hydrogen citrate to keep the back-flush solution
slightly acidic.

All in all my guess would be that potassium/potassium citrate flush solution would work. I'd expect the column to need flushing more often. Maybe doing a flush once in a while with the sodium/citric acid mixture would give the ion-exchange column a good cleaning.

The other consideration here is cost. Since the sodium/citric acid mixture is the common back-flushing salt, I'd expect a potassium/citric mixture to be more expensive.


Entropies of Solvation for Chlorate and Nitrate Anions Can anyone find any data on the entropies of solvation for these two anions in water solution? I can't find any.

Look at the graph on p. 65, it puts the entropy of solvation (298 K) for chlorate and nitrate at about 25 and 21 kJ/mol respectively.  This reference suggests that the entropy term for nitrate is 116 J/(K-mol), multiplying that by 298 K yields 34.6 kJ/mol.


Why is phenyl weakly activating? The susceptibility of a benzene ring to electrophilic attack depends on the type and number of groups bonded to the ring.

Activating groups donate/release electrons and increase the electron density in the benzene ring and so render it more vulnerable to electrophilic attack.

Deactivating groups do the opposite, withdrawing electrons and reducing electron density in the ring.

The phenyl group is said to be weakly activating, suggesting that it donates electrons to a benzene ring bonded to it.

One possible explanation I can think of is that phenyl group donates electrons through resonance with the other ring and it therefore increases electron density to a greater extent relative to hydrogen, which is the reference point for activating ability.

In the following diagram, which carbocation is more stable, A or B?

The answer is A because the positive charge on the positively charged carbon can be delocalized into the attached benzene ring (I've drawn one of the possible resonance structures to illustrate this).

The same thinking can be applied to your question. Look at resonance structures C and D (we can draw many more) for the attack of an electrophile on a benzene ring.  If there is a second benzene attached and it is attached ortho or para to where the electrophile attacks, then the positive charge can be delocalized into the second benzene ring.

This additional delocalization will stabilize the system. Hence, the second benzene is an ortho-para activating substituent.  However, since this stabilization involves electrons involved in pi bonds, the stabilization (activating effect) will be weaker than when a substituent with a "more readily available" lone pair of electrons (amino, hydroxyl, etc.) is involved.


Gibbs energy and Entropy In Gibbs energy equation:

$$\Delta G=\Delta H-T\,\Delta S$$

At constant pressure,

$$Q=\Delta H\tag{1}$$

and we know that $T \Delta S=Q$ (reversible)
so finally

$$T\,\Delta S=Q\tag 2$$

Putting values of Equation $\text{(1)}$ and $\text{(2)}$ in Gibbs energy equation ,
we should always get $\Delta G=0$ for a reversible reaction. ..

Why isn't it so?

I don't know how to search it on Google; if anyone could give me a link or otherwise explain it?

Reason 1:

while $T\,\mathrm dS=\mathrm dQ$ for a reversible process, this does not mean $T\Delta S = Q$

Reason 2:

The term "reversible reaction" is unrelated to "reversible process".

Lame Reason 3:

Because your "always" statement does not repeat the "constant pressure" requirement assumed for $Q=\Delta H$


Partial Coverage During Zinc Plating Bath I am experiencing partial coverage of a part during a DIY home zinc plating operation and was hoping someone here may be able to identify the cause of this.

I am using a solution of household vinegar and epsom salt to zinc plate some pieces during a motorcycle restoration but I am having difficulty getting the zinc to deposit on an area of a specific part.

On accompanying picture of a footpeg mount it is evident where the plating has not adhered.

I have done multiple runs and cleaned the part thoroughly between each run and even oriented it in different directions to the zinc anode and even connected my cathode connection point directly to the area not accepting any zinc all to no avail and was wondering if anyone has any ideas on what may be happening here and how I may resolve the issue. (My guess is there is an electrical or magnet field being generated by the "sides" of the part and preventing plating in this area as all other areas not visible have zinc adhered from the first run)

It is hard to imagine this would be due to an electric field or a magnetic field.

I would say this has to be a different material or coating in the region where you are having bad coverage.

Would it be possible to sandpaper this piece to get rid of any potential coating? Even leftover glue from some sticky on top would cause an issue.

You could also try to electrochemically clean the surface immediately prior to the plating. Applying a highly cathodic potential should lead to some electrolysis that would evolve hydrogen gas which would knock any impurities off the surface.

Either method could roughen the surface somewhat, I don't know enough about motorcycles to say if this would be a dealbreaker.


Metal-Ligand-Complex and bond directions: What defines coordinate system of the complex ion? I have a question concerning metal ligand complexes.

In every chemistry text book I have read which treats metal ligand complexes, following situation is discussed:
When 6 ligands (in octahedral coordination) approaches a metal ion with d-orbitals, 2 of the three orbitals are energetically raised because these orbitals have their lobs in direction of the ligands.

But why is this always true?
Why does always point the lobes of the z² and x²-y² orbital towards the ligands coming from "above" and "below"? Isn't it possible, that the z-axis of orbital system of the metal is a little bit inclined with respect to the octahedral ligand system surrounding the middle metal ion?

The only thing I can think of, is that this is because of energetic reasons. That in total, it's energetically more favourable when 3 of the 5 d-orbitals are arranged, so that electron occupying these orbitals have the lowest possible energies.

But up to now, I haven't found anything discussing this question.

Thanks in advance,
Jürgen

The answer lies in symmetry.  When placed in an octahedral field, the $d_{x^2-y^2}$ and $d_{z^2}$ orbitals are directed at the surrounding ligands because they have $e_g$ symmetry.  The remaining three orbitals have $t_{2g}$ symmetry and are directed between the ligands.  This explanation can be found in an undergraduate inorganic chemistry textbook such as that by Miessler, Fischer and Tarr.  (In the 5th edition, that's section 10.2.1 - Crystal Field Theory).


Dangers of Colbalt Chloride Paper Burning I am a science technician and foolishly decided to heat a small amount of cobalt chloride paper on a glass petri dish over a Bunsen. Unfortunately the petri dish broke and the paper hit the metal gauze under and burnt. I immediately put the rest of the paper in water and opened the window.

The cobalt chloride paper was white (not pink) and approximately 3 x 1 cm in size. I am very concerned that I have released very toxic material and would welcome any comments.

Not a good story to share ...

It is not surprising that a rapid heating of a Petri dish provided by a Bunsen burner cracks the former.  By application of the heat, the glass wants to expand, to dilate; hence you swirl test tubes over a Bunsen, provided your lab still has access to natural gas.
A more gentle heating of the glass, to dry the paper impregnated by $\ce{CoCl2}$ were gradual heating on a heating plate.  By this you equally would obtain the colourless, waterfree form, in a more safe manner (no open flames, no ignition).
Yes, $\ce{CoCl2}$ is considered toxic, and dangerous for the environment.  It is suspected carcinogen. You set yourself unnecessarily in danger, not only because of the flames, but that the dry salt, too: finely dispersed in air, it is then easily inhaled.  You won't die on the spot, however, do not repeat this. And ChemSE is not a subsitute for a physician.
Depending where you did this, disposal of heavy metal, and aqueous solutions of heavy metals (like $\ce{CoCl2}$) is regulated by law, and hence the rinse with water was probably fine to extinguish the fire, but the waste water generated is not normal household waste.  Especially if it were on a larger scale.
For future reference: look up safety informations of chemicals you intend to work with, chemical suppliers include them in MSDS sheets. In chemistry lab classes, students are requested to get familiar with these information prior to performing the experiments.


Obtaining atomic form factor from electron density? I know that atomic form factors for x-ray scattering are obtained by Fourier transform of the electronic charge density.

I am looking for a citable elementary source (book, review, article) which covers the systematic and explicit derivation of the relationship between electronic charge density and the atomic form factor, to which people unfamiliar with the connection can be referred to for better context.

Here is a possible source: Modern X-ray analysis on single crystals: a practical guide by Peter Luger, De Gruyter, Inc. 2014.

The relevant excerpt describing how to estimate the density of an atom j, and with that, the scattering factor (really scattering function) is below.

So atomic scattering factors are derived by taking the Fourier transform of the computed electron density of a stationary atom (the density has spherical symmetry, so the scattering factor is a function that only depends on $s = \sin \theta / \lambda$).


What does the osmotic pressure do in an electrolytic solution? I have learnt that in an electrochemical cell, the osmotic pressure in the solution hinders the flow of ions from the electrode to the solution, rather some ions may accumulate on the electrode because of this osmotic pressure. As far as I know, osmotic pressure is a measurement of the tendency of the solvent molecule to flow from one solution to another.

My question is, how does this pressure cause the ions in the solution to gather at the electode?

Actually the ions cause what is called osmotic pressure, not the pressure cause the ions to flow. As positive charge accumulates on the electrode, it has a tendency to attract extra negative charge towards itself, as alike attract each other. It starts behaving as the cathode and this phenomenon starts to develop what is called osmotic pressure.


How to study Sodium hypochlorite stability with materials first of all I'm not not sure to be in the right stack exchange site, but I'm a material engineering student and this year I have a project: choose a material for a container filled with Sodium hypochlorite, my university has given us CES Edupack, that contains a lot of infomartion about mechanical and physical properties, but nothing about material stability.
Since we haven't done any lecture about materials stability/decomposition, I'm here to ask if there is someone that could enlight me about how to proceed to evaluate the stability of a material.The only thing that I came across while thinking about it is to avoid materials(polymers) that has got active groups with Sodium hypochlorite, is it a good approch?
EDIT
Maybe my question is a little dispersive so I make it clear here: my main concert is more general: how do I study the stability of sodium hypochlorite inside the container? What should I check to be sure that the container doesn't "melt"?

Sodium Hypochlorite  is soluble in water and reacts with all organic solvents. Sodium hypochlorite is extensively used in house-holds as a bleaching agent and it is a chemical compound with formula $\ce{NaClO}$.

Household sodium hypochlorite bleach was introduced to Americans in 1909 and sold in steel containers, then in glass bottles. In the early 1960s, the introduction of the plastic jug brought a cheaper, lighter, and nonbreakable packaging alternative. It reduced transportation costs and protected the safety of workers involved in its shipping and handling. Additionally, the thick plastic did not permit ultraviolet light to reach the bleach, which improved its chemical stability and effectiveness. In recent years, how-ever, plastic containers have become an environmental concern because of the time it takes the material to decompose in a landfill. Many companies that depend on plastic packaging, including bleach manufacturers, have begun to reduce the amount of plastic in their packaging or to use recycled plastics. In the early 1990s, Clorox introduced post-consumer resins (PCR) in its packaging. The newer bottles are a blend of virgin high-density polyethylene (HDPE) and 25% recycled plastic, primarily from clear milk jug-type bottles.

Read more: http://www.madehow.com/Volume-2/Bleach.html


Chemical compatibility I'm an electrical engineer trying to understand a failure in a plastic part that I'm working with.

I am using a plastic called polysulfone- and the data sheet has a lot of words ending in "amine" that are listed as not-being compatible with said plastic.

I want to bond this plastic with an epoxy, but the very-vague datasheet states that the hardener contains 80% "Modified Cycloalipathic Amines".

With this admittedly limited information, should I look for a different adhesive?

The process temp is $5$ to $85~^\circ\mathrm{F}$ ($-15$ to $29~^\circ\mathrm{C}$).

I suggest taking the experimental route and exposing another test piece of the plastic you are working with to the hardener. Keep an eye out for any softening, swelling, change of color and any other indicators of a reaction occuring.


How is the alpha carbon defined in alkyl halides? The α carbon refers to the first carbon that attaches to a functional group. According to this information, the alpha carbon in 2-bromobutane should have been the carbon next to the one with bromine. But in my book it says that the alpha carbon in that compound is the one with bromine atom. How is this possible?

The α carbon is the one directly attached to the functional group. There is no contradiction to your book’s claim that the α carbon of 2-bromobutane is the one directly bound to bromine.

This is due to the different definitions of functional groups and what they include. The keto functional group is defined as $\ce{-C(=O)-{}}$, i.e. it includes the carbon atom doubly bonded to oxygen. Halogen functional groups, for example the bromo functional group are defined as only the heteroatom, so in this case $\ce{-Br}$. Note that because the carbon is not included in the definition of this functional group, it is instead the α carbon.

This means that reducing the ester functionality of a β-halo ester will result in a γ-halo alcohol even though no group migrated.


Is there a chemical reaction which could give off enough heat underwater to cook something? The mer-people demand cooked food. They don't want to go on land to make it.

Is there a chemical reaction they could use which would create enough heat to cook their food underwater in a safe manner?

Interesting prompt!
There are plenty of reactions which could be performed underwater to release a large amount of energy. In fact, many substances react vigorously with water itself. You may have heard about how alkali metals such as sodium are not compatible with water, and there's a good reason for it!
Explosive reactions aren't particularly suited to cooking food, though, and a sustained flame would be much more useful. Interestingly enough, there was legitimate research performed on making underwater flames, for a very specific purpose: the Olympic torch!
For the 2000 Summer Olympics in Australia, during the classic Olympic torch relay before the start of the games, the torch was actually carried ablaze underwater for three minutes at the Great Barrier Reef. According to Wikipedia, the usual fuel for the burning torch is a mixture of liquefied propane/butane, which burns using oxygen gas in the air as an oxidizer (more details of a torch's construction can be found here). There is very little dissolved oxygen in ocean water, however, so the fuel/oxidizer mixture had to be changed. From the BBC article:

The special flare was developed after nine months by a team of chemists and engineers at Melbourne pyrotechnics company Pains Wessex.
The company's managing director Charles Tegner said designing the torch was a challenge - not only to produce a flare to burn underwater at such a depth, but to burn like the Olympic torch flame as well.
"It had to be clearly visible," he said. "Such flares don't normally exist."
The flare's chemical composition, pressed into a steel tube, produces sufficient oxygen and nitrogen to maintain a very hot flame.
The flare burns so fiercely at more than 2,000 degrees Celsius that this creates enough pressure to keep the water from entering the tube.
Its intensity is produced by a mix of oxygen-generating chemicals as well as the combustible element magnesium in a finely powdered form

I haven't found the detailed composition of the fuel/oxidizer mixture, but this is already a very good hint. Magnesium is an alkali earth metal which burns vigorously with a bright white flame in many conditions, be it in pure oxygen, air, pure nitrogen, or even pure carbon dioxide, and it will also react with water when hot producing hydrogen gas, much like sodium (this is why reactive metal fires are so hazardous; don't even think about using a $\ce{CO2}$ or water extinguisher on them, that just feeds the fire!). In this case, the pyrotechnicians decided to mix finely pulverized magnesium metal with a solid oxidizer, very likely something such as potassium chlorate, $\ce{KClO3}$, which decomposes under heat releasing oxygen gas. The flame was lit in air until the reaction between the magnesium and solid oxidizer started going, and then it was safe to submerge the torch while the reactants lasted. The stunt has in fact been repeated since.
To add a few more options to this answer, there is a general class of reactions with water called hydrolysis reactions, some of which are quite energetic. Here, the starting material tears apart water molecules directly and uses the fragments to modify its structure to a more stable form. For example, this is what happens when phosphorous pentoxide is allowed to come into contact with water. This avidity for water makes phosphorous pentoxide a useful material in a laboratory, as it is capable of scavenging trace amounts of water from a liquid or gas. However, these types of reactions already tend to release much less energy per gram of reactant, compared to combustion reactions.
Lastly, I'll cite one more thing. Even without any chemical reaction, the dissolution of some solids in water is capable of releasing some heat. This happens when starting material has a weaker attraction to itself than to water molecules. This property can be exploited in disposable heat packs. These types of processes usually release even less heat per gram of starting material than combustion or hydrolysis, but at least the process is conceivably easily reversed, provided energy is given to dry the material again.
Therefore, it is possible to get a lot of heat from reactions underwater, though it is quite a more elaborate process than starting a fire in air.
Edit: Now that I saw the original question at Worldbuilding.SE, I realize part of it wondered about how cooking would be done in earlier stages of technological development. In this case, I'm hard-pressed to think of a chemical reaction which happens using materials found easily underwater and releases a lot of energy. The difficulty is great enough that it has actually been suggested previously as a general argument for why technologically advanced or even sentient life would be unlikely to develop underwater, as any hot thermal processing of food or materials would be severely hindered. Except for incredible strokes of luck, hydrothermal vents on the sea floor sound like the most reliable heat source which could be accessed by a primitive civilization.


Is cyclohexane saturated or unsaturated? I have this confusion that cyclohexane is of form $\ce{C_{n}H_{2n}}$ and so it must be an unsaturated compound, but also it does not have a double or triple bond so it must be saturated.

Cyclo comes from the Greek word 'kyklos' which means circle.

The cyclohexane molecule is in the form of a circle therefore it can have the formula $\ce{C_nH_{2n}}$ and still be saturated.


How is Langmuir adsorption isotherm different from Freundlich's? I see no difference between the two. Both say that the degree of adsorption reaches a limiting value at high pressures and at very low pressures degree of adsorption is directly proportional to the pressure.
Where does the difference lie?

There is a fundamental difference between both models.

Langmuir's model was a theoretical construct, while the Freundlich isotherm is empirical.

In the Langmuir model, it is assumed that at maximum coverage, there is only a monomolecular layer on the surface. This means that there is no stacking of adsorbed molecules. The Freundlich isotherm does not have this restriction.


How to prepare cyclohexanone without starting with cyclohexanol? What's an easy way to prepare cyclohexanone if I dont have direct access to cyclohexanol? I'm willing to synthesize cyclohexanol from more common materials, then make cyclohexanone from that. But most of the instructions for one cyclohexano- say to start with the other cyclohexano-.

I'm looking to start with common materials and synthesize cyclohexanone or cyclohexanol. If i have cyclohexanol, i can use bleach and acetic acid to make cyclohexanone.

Since I dont have access to lab deliveries at the current moment, what are the most common chemicals to start with to make this happen? AKA, stuff I can order legally online without a lab cert.

Do you have access to phenol? If you have, then use that to make cyclohexanol (hydrogenation of phenol using nickel catalyst):

Then treat cyclohexanol with copper(II) and chromium(III) oxides to convert it to corresponding ketone (Dehydrogenation):

Benzene could work as a starting material but the reaction might be costly and hectic as the catalyst is not a household item$\ce{^{[2]}}$. Secondly, it is used as a starting material in the synthesis of nylon and the reaction intermediate is a mixture of cyclohexanol and cyclohexanone(mixed oil)$\ce{^{[3]}}$ which add to your task of separating them. Better convert it to phenol$\ce{^{[4]}}$.

Notes and References

http://www.leck.co.uk/demo/eci3/polymers/polyamides.html (above reaction synthesis source). It is a part of polyamide synthesis
https://link.springer.com/article/10.1023/B:CATL.0000011080.03035.eb
http://www.greener-industry.org.uk/pages/nylon/4nylonpopup.htm
https://www.quora.com/What-is-the-method-to-convert-benzene-to-phenol
https://russianpatents.com/patent/220/2205819.html


Totally confused about the location of "s,p,d,f" Orbitals inside the atom Here is the other crazy question... I know that electrons revolve around nucleus in their orbits.But my question is "how the orbitals are located in the orbits? Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits?? Totally confused....

I know that electrons revolve around nucleus in their orbits.

You should "unknow" that, because that is the failed Bohr Model of the atom.

The are no fixed orbits for electrons in atoms.

The Bohr Model has been superceded by the Schrodinger model.

The atom is described in terms of probabilty density: for each point in space the probability that an electron will be at that point (when the atom is in a given state such as 1s, 2s, 2p, etc).

  my question is "how the orbitals are located in the orbits?

There are no orbits

  Do they really posses Spherical, dumb bell, double dumb bell shapes inside the orbits??

The shapes you seen in drawings that are spherical, dumb bell, etc. are meant to show regions of higher electron probabilty density.  For example, someone might draw a sphere on the basis that there is a 90% probablity of the electron in that orbital being within the sphere at a given moment.  However, in reality, for example in the 1s state, the electron could be anywhere in space. Other states, such as 2s, have nodes, with the electrons possibly being anywhere except at the nodes.

So think of the shapes as showing where the electrons in that orbital are often, but remember that they can really be anywhere excpet at the nodes.


Is a pyrylium ring aromatic? I am trying to figure out which rings in callistephin are aromatic:

I know that the benzene rings definitely are aromatic, but I'm not sure about the ring with the positively charged oxygen (highlighted in red).

How does the Hückel $4n+2$ rule apply to this ring?

Pyrylium is aromatic. Following Hückel's rule, pyrylium is uninterrupted (no $\mathrm{sp^3}$ hybridization), all atoms have p-orbitals and there is 6 π electrons. the 6 π electrons indicate that Hückel's rule $(4n+2)$ is agreeable for pyrylium, $n = 1$.


Why the atomic (metallic) radius of gallium is less than that of aluminium? It is a well known fact that on moving down the group of p-block elements, the atomic radius is expected to increase. However, gallium is an exception. The atomic radius i.e., the metallic radius of gallium ($135~\mathrm{pm}$) is less than that of aluminium ($143~\mathrm{pm}$).

My book provides the following reason:

  It is due to the presence of additional 10 d-electrons in gallium which offer poor screening effect for the outer electrons from the increased nuclear charge.

But if that is the case, why the effect is not that same in case of the heavier members?

Moreover, there must be something else which governs this exceptional property of gallium. Maybe the structure, but I am not sure. Can someone elaborate?

First, you have to look at the definition of metallic radius, which is the half distance between two atoms in a lattice. It has a significant dependence on crystal structure.

  Tanget relevant to the question and other answers:  Gallium has an orthorhombic crystal structure (CN = 6) whereas as aluminum has a face-centered cubic crystal structure (CN = 12).  This difference in coordination requires a goldschmidt correction to compare values as if gallium were 12-coordinated which makes the gallium's corrected metallic radius actually even smaller compared to aluminum at about $\mathrm{130pm}$.

Gallium exists as a diatomic solid in the lattice, which causes the atoms to be closer together on the  a axis because they are covalently bonded. Gallium's covalent radius is about $\mathrm{122pm}$ which is very much smaller than the metallic radii of either element. This value between the bonded atoms lowers the average distance between atoms giving gallium a lower metallic radius.

If you look at the Van der Waals radii of the elements, (more representative of a monoatomic gaseous atom) you will see that aluminum is actually smaller though not by a lot, giving some credence to the d-electrons explanations, but not a complete explanation in regard to the metallic radius.


HCl rota-vibrational lab: ab initio simulation of rotation constants I'm trying to set up the simulation portion of the classic undergraduate lab studying the rotavibrational spectrum of HCl and DCl. In terms of simulation (using Gaussian 16), the equilibrium geometry and vibrational energy calculations are trivial. Including anharmonicity is also simple through the use of the freq=anharm keyword. I'm running into trouble with the inclusion of vibrational-rotational parameters, which the program is supposed to support, but I can't find appropriate output information. I'm either missing the output data because I don't know what I'm looking for, or something is wrong in the calculation. Specifically, I'd like the students to get ab initio values for $D_e$, $\widetilde{\nu}_0$, $B_e$, and $\alpha_e$ in the following equation ($m$ is a substitution for $-J''$ or $J''+1$ for the $P$ and $R$ branches, respectively):
$$\widetilde{\nu}(m) = \widetilde{\nu}_0 + (2 B_e - 2 \alpha_e) m - \alpha_e m^2 - 4 D_e m^3$$
I'm currently able to find $D_e$ and $\widetilde{\nu}_0$.

The only clue I can find in the calculation is the following lines:

WARNING: Anharmonic treatment of linear tops is experimental.
         Moreover, an hybrid treatment is used to simulate spectra:
         - Energy: equations including degenerate modes are used.
         - Intensity: summation done on N' modes, considering only one mode
           per couple of degenerate modes. No variational correction done.

    ==================================================
                  Coriolis Couplings
    ==================================================
[No Coriolis Coupling for diatomic molecules

Later, in the entry for Vibro-Rot alpha Matrix (where I'd expect to see entries for $\alpha_e$), I get:

Vibro-Rot alpha Matrix (in cm^-1)
---------------------------------
            A(z)        B(x)        C(y)
Q(    1)    -0.00000        NaN        NaN

I'm assuming the message for Coriolis Couplings is connected to the zero-valued alpha matrix. Is there a way to pull out these parameters in a Gaussian simulation? If not, is there another ab initio package that will perform these calculations? I've tried (without luck) to look through the NWChem documentation.

The input might be useful to reproduce the calculation. Unfortunately, the output is larger than SE.com will allow. Let me know if you require additional lines from it (or a full copy).

Input file:

%nproc=4
%mem=400MB
# MP2 def2tzvp
# opt freq=(anharm,vibrot,noraman) scf=tight freq=ReadIsotopes

Vibrational analysis for D35Cl

0   1
 Cl 0.0 0.0 0.0
 H  0.0 0.0 1.27

298.15  1.0
35
2

Update:

After more reading through the Gaussian references, I've found the rotational constants (Ba(x), Ca(y)) under the following section:

     ==================================================
          Vibrational Energies at Anharmonic Level
     ==================================================

 Units: Vibrational energies and rotational constants in cm^-1.
 NOTE: Transition energies are given with respect to the ground state.

 Reference Data
 --------------
                              E(harm)  E(anharm)     Ba(x)      Ca(y)
 Equilibrium Geometry                             10.643785  10.643785
 Ground State                 1525.714  1515.023        NaN        NaN

 Fundamental Bands
 -----------------
     Mode(n)      Status      E(harm)  E(anharm)     Ba(x)      Ca(y)
        1(1)      active      3051.429  2951.452        NaN        NaN

 Overtones
 ---------
     Mode(n)                  E(harm)  E(anharm)     Ba(x)      Ca(y)
        1(2)                  6102.858  5802.928        NaN        NaN

 WARNING: Anharmonic transition moments for symmetric and linear tops
          are not yet fully implemented.

However, I still can't find an entry for $\alpha_e$, the coupling constant between vibrational and rotational states. Is Gaussian capable of putting out this constant? If so, am I looking for it under the wrong name?

Disclaimer:

I never used Gaussian, so I am just guessing about the program.

Coriolis coupling:

Coriolis coupling couples bending modes with rotation and there is no bending in diatomic molecules.

There is no Coriolis coupling in Gaussian for HCl, because there can't be one. The relevant output could also be NaN, it is simply not defined.

Alpha "matrix":

For diatomic molecules you have two degenerate rotations and one stretching mode. Hence your alpha "matrix" is actually well represented by a scalar and used as such in your equations. For this reason one would expect only one alpha value.

That there is written -0.000 and not 0.000 is an indication for a truncated negative number. In addition other values are properly set to NaN. I think that your one reported alpha value is simply close to zero. You can try a verbose output to verify this.

Solution:

If you want to get a non zero alpha
I would recommend a molecule with a shallower potential hyper surface, i.e. a lower force constant. Or you could use this example to discuss floating point arithmetics and truncation with your students.


Is it possible to burn things in a halogen atmosphere? Fires can burn in oxygen. But, can they burn in halogen atmospheres (fluorine and chlorine only)? I am wondering this because combustion reactions involve oxygen and carbon and other elements. But, can they burn in fluorine and/or chlorine? If they do, what color will the flame of paper and wood be in fluorine and chlorine atmospheres respectively?

They certainly can. Put some water in the bottom of a beaker, fill the beaker with chlorine gas, and drop in bits of calcium carbide. The carbide will produce bubbles of acetylene, which will spontaneously burst into flame when they hit the chlorine.

Flame colors depend on many things, even in oxygen. In general, flames in a chlorine atmosphere will probably look slightly more greenish than flames in air or oxygen, simply because the gas itself is green. Fluorine isn't as strongly colored, and flames from combustion in fluorine will be VERY VERY BRIGHT.