Why does hydrogen have a lower ionization energy than fluorine? I found here that the ionization energy of hydrogen is $\pu{1312kJ/mol}$ and for fluorine, it is $\pu{1681kJ/mol}$. Now clearly, from the data, we can see that hydrogen has a lower ionization energy when compared to fluorine.

I did not expect this because I saw here that fluorine is twice as big as hydrogen (I am referring to the empirical values of the atomic radii). This means that it would be significantly easier to remove an electron from a fluorine atom than from a hydrogen atom. That's why I expected hydrogen to have a higher ionization energy. But I found that it's the other way round. Why is it so? What is wrong with my reasoning? Is this an exception?

Fluorine is bigger, but there are $9$ protons in the nucleus, compared to $1$ in case of hydrogen. So, the outermost electron in fluorine experiences more attraction than what hydrogen's single electron experiences, in spite of larger size of fluorine.

However, the outermost electron in fluorine doesn't experience attraction from all the $9$ protons (nuclear charge). Due to improper shielding, the effective nuclear charge experienced by one $2p$ electron of fluorine becomes -
$\mathrm{ [9 - {(6 \times 0.35) + (2 \times0.85)}]= 5.2}$.

In comparison, the single electron in hydrogen experiences effective nuclear charge of $1.0$

Note: For hydrogen, there is no other electron to shield nuclear charge. So, effective nuclear charge = actual nuclear charge.

Source: To know how to calculate effective nuclear charge visit this Wikihow link.


Suitability of liquid plastic resin for plumbing application Which synthetic resins are suitable for fabrication of plumbing components e.g pipe sections?

I'm trying to find an easy way to mold/house/encase a custom designed data logger.
The water temperature ranges between 10 to 50 degrees C.

EDIT: I'm planning this for both indoor & out door use. Under direct sunlight.

Epoxy should work OK and you have different options that are easy to get.


How to find the molar mass of an unknown metal carbonate through a gas evolution reaction? As part of a pre-lab exercise in Chemistry, we went through the necessary calculations to identify an unknown metal in a metal carbonate. The method was gas evolution by means of hydrochloric acid (see equation below).

$$\ce{HCl + M2CO3 -> MCl + CO2 + H2O}$$

The unknown metal was Alkali, and we had three possibilities: Lithium, Sodium and Potassium as $\ce{M}$.

Using data given to us, we first converted grams of released $\ce{CO2}$ to moles. Then here was the part I was confused on-

To find the molar mass of the metal carbonate, we took the mass of the sample used (the metal carbonate) and divided it by moles of $\ce{CO2}$ released, which was just calculated. The rest here on out was a little algebra. What is the reasoning behind dividing the mass of the sample by released $\ce{CO2}$? What makes it "legal"? Yes, in the balanced equation, the ratio of $\ce{M2CO3}$ to $\ce{CO2}$ was $1:1$.

$$\ce{HCl + M2CO3 -> MCl + CO2 + H2O}$$

Since you already figured out, that there is a one-to-one ratio of carbonate to carbon dioxide, you can simply write
$$n(\ce{CO2}) = n(\ce{M2CO3}).$$

With the formula for the Molar mass,
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{M2CO3})},$$
you can simply substitute one for the other, hence
$$M(\ce{M2CO3}) = \frac{m(\ce{M2CO3})}{n(\ce{CO2})}.$$


Does the arrangement of atoms around a chiral carbon matter? In other words, are these two structures equivalent?


Does the methyl rotation in propene have a symmetric or asymmetric potential energy surface? Depending on how one imagines the methyl rotation of propene to occur, it seems like it could exhibit either a symmetric or an asymmetric potential energy surface.  If the $\ce{H2C=CH\! -}$ portion were held fixed, the rotating methyl would seem to be symmetric due to the local $\mathrm C_3$ axis:

On the other hand, if the methyl group were held fixed, the rotating $\ce{C=C}$ portion would seem to be highly asymmetric:

Obviously this rotation can't be both symmetric and asymmetric. So which is it, and why?

Intuitively, given the logical contradiction of the above, my inclination is to assume that the rotor is asymmetrical. My best guess as to a more rigorous argument for it being asymmetrical is that it's not valid to imagine any part of the molecule as being "held fixed", since it's not actually anchored to anything.  Thus, both of the above images are probably subtly incorrect.

These rotations are the exact same physical processes for a molecule because we cannot fix either end in place. In order for rotation to mean anything, we must choose an origin. Mathematically, I must be able to draw a vector from all of the nuclei to some arbitrary point, but be consistent about this arbitrary point. Or, I must determine positions of nuclei by some measurement relative to other nuclei.

Imagine I pick the carbon of the methyl rotor as the origin. If you were standing at this origin and this rotation you describe were happening, you would see the so-called "asymmetric rotation". If, however, we put the origin on C1, we would see the "symmetric rotation." We could also define internal coordinates so that we simply changed a dihedral angle and observed the same rotation. In this case it's almost like both motions are happening. Clearly only one physical process is happening, so we might be picturing this wrong in our heads.

I believe the problem you're encountering is that if you had a macroscopic object shaped like this, you could imagine grabbing either end and you would have two separate rotations which we could choose to call symmetric and asymmetric if we so desired (though I'm not convinced of this naming). We know the motions really are different because we are spinning around different total masses. On a large scale where we hold half of this molecule in place, it would seem easier to do the symmetric rotation than the asymmetric rotation. Stated another way, for an equivalent impulse to each of the possible rotors, the initial rotational frequency of the methyl rotor would be larger than that of the $\ce{C=C}$ rotor. Note that we would still be resisting torque in the hand with which we are holding the object and the size of this torque would change depending on the rotation.

Molecules, however, cannot be thought of as fixed in this kind of way, and the resistance to rotation does not come from friction as it probably would for a macroscopic object. The "resistance" to rotation for the molecule comes from how much energy it takes to rearrange electrons for a change in nuclear positions. You can measure these nuclear positions relative to each other using whatever method you want, three of which we already discussed, and this will not change the minimum energy of the electrons for the set of nuclear positions. This minimum energy path (dare I call it a gradient?) is the path along which the rotation would happen. Whichever actual rotation you picture in your head does not matter at all. Because the molecule is free during this rotation, you can get to any of the many rotations by imagining yourself on a different atom, but this will not change the electronic energy because the relative nuclear coordinates will remain unchanged when you change the origin. In the language of one of your comments, because no part of the molecule is fixed, the inertial properties of the rotation are determined by the PES, so whatever symmetry you're thinking of is the same as the symmetry of the PES.

I think I addressed the potential pitfall here. It is indeed quite weird to imagine what the "actual" internal rotation of the molecule looks like, and maybe someone has a better intuition for what that means, but I think describing the actual motion is fairly arbitrary.


How to determine order of nucleophilicity among commonly used nucleophiles? How to determine order of nucleophilicity among commonly used nucleophiles? For example, in Finklestein reaction, $\ce{NaI}$ (reagent) - taken in acetone (solvent) - displaces $\ce{SH-}$ ion from alkyl group via SN2 mechanism to form an alkyl iodide. But in a separate question we may be asked to write the product formed when $\ce{NaSH}$ reacts with an alkyl iodide in the presence of acetone. Since I do not know which of the two is a stronger nucleophile, I will displace $\ce{I-}$ and form $\ce{alkyl-SH}$. But definitely only one of the reactions shall occur.

So, my question is that is there an order of strength of nucleophilicity (like a metal reactivity series)? I am preparing for JEE Advanced 2015 (college entrance exam in India), so I am expecting answers within its scope.

The relative strengths of nucleophiles can be correlated with three structural features:

A negatively charged nucleophile is always a more reactive nucleophile than its conjugate
acid. Thus $\ce{HO^-}$ is a better nucleophile than $\ce{H_2O}$ and $\ce{RO^-}$ is better than $\ce{ROH}$.
In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basicities. Oxygen compounds, for example, show the following order of reactivity:
$$\ce{RO^- > HO^- >> RCO_2^- > ROH > H2O}$$
This is also their order of basicity. An alkoxide ion ($\ce{RO-}$) is a slightly stronger base than a hydroxide ion ($\ce{HO^-}$), a hydroxide ion is a much stronger base than a carboxylate ion ($\ce{RCO2^-}$), and so on.
When the nucleophilic atoms are different, nucleophilicities may not parallel
basicities. For example, in protic solvents $\ce{HS, CN}$, and $\ce{I}$ are all weaker bases than $\ce{HO^-}$, yet they are stronger nucleophiles than $\ce{HO^-}$.
$$\ce{HS^- > CN^- >  I^- > HO^-}$$

Nucleophilicity versus Basicity: While nucleophilicity and basicity are related, they are not measured in the same way. Basicity, as expressed by $\mathrm{p}K_\mathrm{a}$, is measured by the position of an equilibrium involving an electron pair donor (base), a proton, the conjugate acid, and the conjugate base. Nucleophilicity is measured by relative rates of reaction, by how rapidly an electron pair donor reacts at an atom (usually carbon) bearing a leaving group.

For example, the hydroxide ion $\ce{OH^-}$ is a stronger base than a cyanide ion $\ce{CN^-}$; at equilibrium it has the greater affinity for a proton ($\mathrm{p}K_\mathrm{a}(\ce{H_2O})= 16$, while $\mathrm{p}K_\mathrm{a}(\ce{HCN})= 10$). Nevertheless, cyanide ion is a stronger nucleophile; it reacts more rapidly with a carbon bearing a leaving group than does hydroxide ion.


Is concentration polarisation an observable phenomenon over time in a galvanic cell? If polarisation means: "any mechanical side-effect by which isolating barriers develop at the interface between electrode and electrolyte."

And if concentration polarisation is "changes in the electrolyte concentration due to the passage of current through the electrode/solution interface. Here polarization is understood as the shift of the Electrochemical potential difference across the cell from its equilibrium value. The cause of the changes in concentration (emergence of concentration gradients in the solution adjacent to the electrode surface) is the difference in the rate of electrochemical reaction at the electrode and the rate of ion migration in the solution from/to the surface."

Thus, should concentration polarisation cause a decrease in the performance/efficiency of a galvanic cell over time and would this be shown with a decrease of voltage/current through the cell? Or does voltage and current remain the same throughout a galvanic cell reaction?

There are two different factors that cause a battery to have a lower energy output.

First regardless of the "state" of the battery, there is the open cell voltage and the obtained voltage under load. State meaning here how much of the battery is used up. For the open cell voltage the voltage is measured with a high impedance voltmeter so that essentially no current (meaning a very very small current) is being drawn. Next the battery is put under some load typical for its use and the voltage is measured again while under the load. This second load voltage measurement will always be lower that the first measurement for the open cell voltage. This voltage drop is caused by the diffusion limits of the chemicals in the battery to and from the electrodes.

This behavior can be essentially modeled by assuming that the battery has an internal resistance. So as the current draw increases the voltage drop internal to the battery increases and the output voltage drops.

Now the second problem is related to the Nernst equation. Let's assume that the battery is used up in 10% increments in a series of tests. At the start of each test the voltage and current is measured. Of course during using each 10% of the battery the reactants in the vicinity of the electrodes are used up and both the voltage and current out for a given load drops.

Now if the battery is left to equilateral for some time then the reactants redistribute and the battery regains some power. However the battery will not recovery to the level at the start of the previous load test. Essentially the bulk concentration of the reactants is dropping by about 10% after each test. So even after a recovery period both the voltage and current that the battery can supply for a constant load drops.

Now with all of this being said there are many types of batteries. Some types produce a more constant power output than others.


Phenol or ethanol as a nucleophile for ether formation? If we react phenol with $\ce{C2H5I}$ in $\ce{C2H5O- Na+}$ (excess), $\ce{C2H5OH}$ (anhydrous) solutions .

In this reaction why is diethyl ether not formed and the actual product is phenyl ethyl ether?

You are adding excess sodium ethanolate so there is the possibility of deprotonation of phenol: $$\ce{C6H5OH + C2H5O- <=> C6H5O- + C2H5OH}$$

If we look at the $\text{p}K_\text{a}$ of ethanol and phenol we have 16 and 9.95 respectively and so the equilibrium above will strongly favour the right hand side. Therefore, the phenoxide ion will dominate as a nucleophile in the mixture, and the major product will be ethyl phenyl ether.


Reaction of enols with electrophiles In a reaction between an enol and an electrophile which will attack the electrophile, the oxygen of the $\ce{-OH}$ group or the pi electrons of the the double bond?

This is my attempt at a solution: Since, the oxygen atom has two lone pair of electrons it should be more electron rich than the double bond and therefore it should attack the eletrophile, Am i correct?

The functional group in question is a vinyl alcohol, and it is a mistake to consider these groups as isolated. The oxygen is in conjugation with the vinyl group, meaning that resonance favors the entire system attacking. There is no way to leverage conjugation as an energetic driving force if the oxygen attacks by itself.

This is almost the same as your other question:

Hydrolysis of 3,4-dihydropyran


Photoelectric effect with 0 Kinetic Energy? Does the photoelectric effect take place if the frequency of the photon is exactly equal to the threshold frequency of the metal? I know the electron is going to be free, but it will have 0 kinetic energy; it will be stationary. If so, can voltage be still created if the metal surface is connected to a positive terminal side?

If you irradiate a metal with the exact amount of energy to produce a free electron with KE = 0, it will just stay free, in the "sea of electrons" of the metal, while the nucleus, with its positive charge, also sits in the same sea - near "other" electrons, each of which has KE due to temperature. Then, after a while, an electron drops in around that nucleus, giving a bit of (thermal) shock. The observable effect will be that irradiation with this wavelength warms the metal as much as possible without exciting any electrons to the "outside" world.

Now if you attach a negative electrode to the metal and a positive electrode close to the surface, you can push/pull that electron (or one other from the sea of electrons) off - the application of a potential gives KE to a free electron. Actually, you don't even have to irradiate: if you apply enough potential, you can pull electrons out of a metal. If you heat the metal, the electrons will come out more easily.

This is the basis of the Edison effect, which makes vacuum diodes possible: a heated cathode "boils off" some electrons which would stay near the wire, but if you have a nearby anode (an electrode made positive by an outside battery), electron current will flow easily from the heated cathode to the (cold) anode. But not from the cold electrode to the hot electrode. So you can rectify AC current to pulsed DC.


Combustion of thiazole I am learning Chemistry from a book Heterocyclic compounds and I have bumped onto a following image with given tasks:

What is the balanced reaction?
What is the RMM of thiazole?
What is the fuel-air mass ratio?
What is the oxygen depletion?
What is the yield of CO2?
What is the yield of H2O?
What is the yield of N2?
For ideal combustion, what is the yield of carbon-monoxide?

Now, from this I can see that aformentioned product is Thiazole( checked on wiki ) and that products of combustion are carbon dioxide, water, hydrogen sulphide (H2S) and nitrogen. Now how can I solve this matter having in mind following data that air is consisted of 21% O2 and 79% of N2.

First I have to ask about the premise of your question- why are you teaching yourself chemistry from a book on heterocyclic compounds?  The rest of your question suggests you've skipped the basics and jumped to fairly advanced stuff.  There are many good introductory chemistry that will walk you through problems similar to this one that will be a far easier starting point than heterocyclic compounds.

Assuming your general question is "how do I do calculations on oxidation reactions?", you fill find whole chapters and even books on the topic.  I would highly recommend starting with something like "What is the balanced reaction of combusting propane (C3H8) in pure oxygen?"  You will need this kind of basis to be able to work up problems like the one you ask about with many possible products and spectator species.

This will get you started: http://en.wikipedia.org/wiki/Combustion#Chemical_equations with answer to the first question.  In general though, pick up a used introduction to chemistry book up off Amazon for $10 and start with those walk throughs.

Hope this helps answer your real question and please let me know if I misunderstood and you had a more specific question about this particular problem.

EDIT: For the yield questions, you can use the the balanced equation and molecular weights to come up with a yield.

You'll have something like:
$$yield=stoichiometricRatio*\frac{MW_{product}}{MW_{thiazole}}$$

$$\frac{g_{product}}{g_{thiozole}}=\frac{mol_{product}}{mol_{thiozole}}*\frac{g_{product}}{mol_{product}}*\frac{mol_{thiazole}}{g_{thiazole}}$$

If they give you a mass (or mass flow rate) of thiazole, you could further convert this to a mass (or mass flow rate) of products.


How is adenosine synthesized in biology? I understand the constituent parts of adenosine (adenine and ribose) and that it can be broken down from ATP anabolically. Can adenosine also be formed catabolically? If so, what is the mechanism?

What's the mechanism for combining ribose and adenine in the cell?

A general explanation or citation would work. No need for someone to go out of their way to draw pictures. Thank you very much!

Adenosine is synthesised from Inosine Monophosphate (IMP) as the nucleotide form (adenosine monophosphate).
IMP does not accumulate in the cell but is rapidly converted to AMP and GMP. AMP, which differs from IMP only in the replacement of its 6-keto group by an amino is synthesized in a two-reaction pathway.

  In 1948, John Buchanan obtained the first clues as to how nucleic
  acids and their components are synthesized by describing the synthesis
  of purine ribonucleotides.

Buchanan investigated how this process occurs de novo by feeding a variety of isotopically labelled compounds to pigeons and chemically determining the positions of the labelled atoms in their excreted uric acid (a purine).

The biosynthetic origins of purine ring atoms. Note that C4, C5, and N7 come from a single glycine molecule but each of the other atoms is derived from an independent precursor.

The actual pathway by which these precursors are incorporated into the purine ring, was elucidated in subsequent investigations performed largely by Buchanan and by G. Robert Greenberg. These investigations showed that the initially synthesized purine derivative is inosine monophosphate (IMP),

(Here I am not going to explain the biosynthesis of IMP but focus on the former)

IMP is converted to AMP (or GMP) in separate two-reaction pathways:

In the first reaction, aspartate’s amino group is linked to IMP in a reaction driven by the hydrolysis of GTP to GDP + Pi to yield adenylosuccinate.
In the second reaction, adenylosuccinate lyase eliminates fumarate from adenylosuccinate to form AMP (adenosine monophosphate).

It is important to note that free adenine (and guanine) can be reconverted to corresponding ribonucleotides through salvage pathways requiring the enzyme Adenine phosphoribosyltransferase (APRT) (mediates AMP formation)

References

Voet and Voet Biochemistry: Section 28-1. Synthesis of Purine Ribonucleotides


Why isn't a polar protic solvent used in the Finkelstein reaction? Usually, in a nucleophilic substitution reaction, a stronger nucleophile replaces a weaker nucleophile.

In any polar protic solvent, the order of nucleophilicity is $\ce{F-}<\ce{Cl-}<\ce{Br-}<\ce{I-}$ (due to stronger solvation of the smaller anions).

While in any polar aprotic solvent, the order of nucleophilicity is $\ce{F-}>\ce{Cl-}>\ce{Br-}>\ce{I-}$.

Why does the Finkelstein reaction use acetone in spite of the fact that it is a polar aprotic solvent? Wouldn't using a polar protic solvent be better since it would make $\ce{I-}$ the strongest nucleophile of the group.

The key point is driving the equilibrium where you want it. You could do it by relative reaction rates. The Finkelstein reaction, however, uses a better method: precipitation of one of the products. Acetone is specifically chosen because sodium chloride, bromide and fluoride are insoluble in it while sodium iodide is soluble. So the equation becomes:

$$\ce{R-X + Na+ ({acet.}) + I- ({acet.}) -> R-I + NaX v}\tag{1}$$

Nothing beats the equilibrium-driving force of precipitation.


Priority order of attack of grignard reagents on functional groups Recently I was solving carbonyl compounds and I came across this. Is there some priority order, or logic behind why alcohol reacts first with Grignard reagent and then ketone. Whereas ester could've reacted with 2 moles of RMgX but didn't.

For the compound given in the question, the priority order of Grignard reagent attack follows,
$\ce{R-OH > R-CO-R' > R-COOR'}$
Reason:

Acid-base neutralisation are very fast.

Carbonyl carbon is more nucleophilic than carboxylic carbon, as the latter one is stabilized by resonance.

Therefore, first alcohol is attacked and hence P is (a), then the carbonyl part is attacked and hence Q is (b).
Finally, one more mole of RMgX gets used up on the carboxyl part, hence R is not (d).


Develop slow smoke for mini windtunnel I want to create a small wind tunnel for testing of aerodynamics for Radio controlled units (Cars, planes and helicopters).

In real wind-tunnels water mists are used but water isn't a good thing for the electronics.

I have considered the following:

Smoke bombs, but the smoke develops too fast.
Burning stuff like laves or coal, but it creates odor, and i don't want my RC stuff to smell.
Incense, same as above.

My chemistry level is a College equivalent degree of chemistry that has been unused for 15 years.

And the materials to make this should be somewhat harmless, and legally available.

Any suggestions on how to do this would be highly appreciated.

The DLR (German Aerospace Center) has developed a couple of DIY instructions for school education. Among these is a PDF in German for the construction of a wind tunnel.

For the smoke, they suggest to use either

incense (the sticks or cones that you usually find in asia shops or esoteric shops) or a
fog machine, like the ones used for entertainment (dance clubs, etc.)

In the latter case, a mixture of 1,2-propanediol and water is heated and pressed through a nozzle at elevated temperature. Given that properly insulated machines, build for safe use at home, are available from DIY electronics shops for roughly 30 € up, I wouldn't start to build one myself.


Identifying lone pairs when looking for formal charges Why are there no formal charges on the 1st compound? I am specifically referring to the nitrogen atom, which I would think should exhibit a +2 formal charge? However, in David Klein's organic chemistry textbook it is said that the 1st compound shows no formal charge anywhere. The second compound, of course, exhibits a +1 formal charge on the nitrogen atom and -1 formal charge on the oxygen atom. This would make more sense if I were to take into account the lone pair for Nitrogen in compound 1 (and the 2 lone pairs for oxygen), but then why would they hide the lone pairs in compound 1 and exhibit them in compound 2? Please, correct my incorrect thinking.

Drawing common electron pairs like they appear in most organic molecules is only then relevant, when

they play a keyrole in the reaction process
they should explain a special molecule structure
they are of an unusual quantity and thus resulting in a (intermolecular) formal charge

Otherwise especially molecules with a lot of atoms with unpaired electrons (e. g. saccharides) can get confusing. And in addition: they take more time to draw ;)
Of course when learning organic chemistry its good to think of them a lot, because of their high reaction potential.

You can be a 100 % sure that when they would not be there in compound 1 that it would be highlighted somehow, because that would result in a very unstable state.


Mechanism of formaldehyde / phenol condensation The production of bakelite using this method, but with an acid as the catalyst is simple enough; a protonation of the carbonyl oxygen, followed by an electrophilic substitution to the ring. The subsequent cross-linkage formation is the same for both cases, but I can't figure out how the initial reaction
steps would work with a base as the catalyst. Google searches and the texts I have yield nothing.
How would the mechanism proceed with a base as the catalyst here?

@Harry Holmes: Here is a simplified example of how the formation of Bakelite may occur under base-catalyzed conditions to incorporate formaldehyde as -CH2- at the ortho and para positions of phenol. There are a myriad of permutations as to the order of condensations. I have selected one that illustrates the Michael-like addition that seems to have eluded you.
Phenoxide 1 adds to formaldehyde at the para position in what amounts to the first step in an aldol addition. The addition is likely adding to the ortho position of phenoxide as well. Species 3 may condense with phenoxide at this point but I have chosen to incorporate two more equivalents of formaldehyde at the ortho positions of 3 to form 4. Now elimination of hydroxide from any of the three positions of 4 leads to a Michael acceptor, such as 5, which reacts with phenoxide or any substituted phenoxide having an open ortho or para position. Structure 6 simply illustrates the mechanism for incorporating formaldehyde. The o-methylenedienone 5 is the product of an aldol condensation wherein phenoxide is the enolate.


What is the accuracy of scf=qc when convergence criteria is lowered? I am trying to run single point calculations with Gaussian 09 for a Mulliken population analysis on large graphene/graphite (940-1411 atoms) sheets/clusters.

I was having difficulty with memory issues (single CPU license) running DFTBA for single point calculations. I have stepped my convergence (beginning with conver=1, and increasing by reading in my previous checkpoint files).  I found my results oscillating and failing.

I decided to try scf=(qc,conver=5,maxcycle=500).  This worked, which concerned me.

It not only worked, but it worked much faster than my failed sp calculations.  I know Gaussian's old default convergence for single point calculations was N=4.  As my calculation was done in less than two hours compared to the days it was taking for DIIS, I have become concerned at the accuracy of the QC using conver=5.  I cannot find anything indicating my approach is bad, but also I cannot find anything syaing my approach should yield reasonable results.

Can anyone please tell me if this approach is reasonably accurate?

Something that I have discovered is that when Gaussian (or other good ab initio software) fails to converge, there is something sketchy about what one is trying to calculate. Sometimes DFT calculations just aren't appropriate, especially in cases that require multiconfigurational SCF for even a zero-th order description.

In this case, given the sheet-like character, I suspect the problem is that the molecular orbitals are so close together, and so similar, that there are serious linear dependency problems that lead to all sorts of havoc in the linear algebra routines and convergence accelerators.

If you are trying to simulate an infinite sheet, then I suggest that you try plane wave codes like those found in solid state chemistry (Gaussian's PBC code, NWChem, abinit, etc.)  These sorts of calculations have gotten easier in recent years, but sometimes they require serious horsepower.

There really does seem to be a problem with SCF convergence, and this could be because you are approaching some kind of band structure limit. I have always used QC as an absolute last resort when I must do something wacky.

(One other thing, you have "capped" your peripheral benzenoid rings with hydrogens?)


How to measure the water content of opaque liquids? Water content of (semi) transparent liquids could be approximated via absorption spectra. What method(s) can be used to measure the water content of an opaque liquid such as milk for example?

I would add Karl Fischer titration.

Of course for your milk example, you would have to dilute your sample quite a lot in a dry (or known-dryness) solvent before titration. Also, the strong redox solution might interact in complicated ways with other materials in your sample (proteins, etc?).


If salt crystals make a cube shape, what is this 'X' marking? I am an amateur, with very little professional chemistry knowledge. I was doing a demonstration / experiment with my 6 year old child.

We took very warm normal tap water, and dissolved into it a large pile of normal table salt. I didn't measure anything particularly, because I was just showing the basic concept of 'dissolving'. That's all. My kid was amazed to see the white pile of salt disappear visibly, but upon tasting the water understood what had happened.

Next we left the bowl up on a high shelf for 2 weeks, to let the water evaporate. This morning we took a look, and were rewarded by a dry bowl full of beautiful perfectly square salt crystals, many of them between 4-6mm in width.

That the crystals were perfect squares did not surprise me, as I have read that sodium and chlorine atoms form a cube shape at an atomic level.

But I could not answer the question about what the 'X' pattern marking was on each square. I have tried to attach photographs to show what I mean. The 'normal' picture is blurry, but if you look you can faintly see an 'X' on each crystal.

The other 2 pictures are each at 20x, using the Natural History Museum Pocket Microscope.

My question is, what explains those 'X' shapes? If possible I need to translate the explanation into 6-year-old talk :-)

When salt (NaCl) crystals are grown under normal conditions, the edges of the cubes usually grow faster than the faces (because the edges have more contact with the saturated salt solution than the faces). So the crystals grow with a square-pyramidal indentation on each face. The "X" you observe is your view of the edges of that inverted pyramid. On the other hand, if the crystals are grown really slowly under controlled conditions, this does not occur and you get a more or less perfect cube.


Entropy Changes in Electrolytic/Galvanic Cell? One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$
If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.
In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.
What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.
I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

[OP] One of the most fundamental equations in chemical thermodynamics states: $$ \Delta_rH_m^⦵ = \Delta_rG_m^⦵ + T\Delta_rS_m^⦵ $$

This is just the definition of the Gibbs energy. By itself, it does not give you any insight into chemical reactions.

[OP] If we look at this equation in context of net chemical reaction in electrolytic or galvanic cell, it is usually interpreted as follows: Enthalpy of reaction denotes total amount of energy at constant temperature and pressure which needs to be supplied (electrolytic cell) or which is released (galvanic cell) during a reaction, standard Gibbs energy of reaction denotes what amount needs to be supplied or which is released in form of electrical energy (electric potential difference between electrodes), the last term including standard entropy of reaction denotes what amount of heat is exchanged with surroundings during a process.

This is not the usual interpretation. Instead $ \Delta_rG_m^⦵ $ is the maximal work you can get out of the reaction (or, if the reaction moves away from equilibrium as written, the minimal work you have to put in to run it in that direction). $ \Delta_rH_m^⦵ $ is the sum of heat and non-PV work exchanged with the system. It could be negative or positive irrespective of whether this is an electrolytic or galvanic cell. Finally, the entropy of reaction is the same as for any reaction - roughly how much the dispersion of energy and particles changes within the system when the reaction proceeds.

[OP] In electrolytic cell, as entropy of reaction is mostly positive, last term is usually interpreted as heat which comes from surroundings and as such it increases entropy of the system. As heat comes to the system, it helps us during electrolysis since we don't need to put in the whole enthalpy of reaction in form of electrical energy, but only Gibbs energy. In galvanic cell it is the other way around.

You can't make any generalizations about the heat transfer when there is also work exchanged between system and surrounding. In many thermodynamic arguments, you talk about processes that happen in the absence of non-PV work, and in a reversible manner (i.e. the entropy of the universe is near-constant, which is the same as saying it is a near-equilibrium process).

[OP] What I don't understand is the interpretation of that last term ,which includes entropy of reaction, which I found on hyperphysics page. According to hyperphysics, this term denotes heat exchanged with surroundings and they say that entropy change in the system is due that heat exchanged.

Without a specific link, it is hard to understand what statement on the hyperphysics site you are referring to, and it what context it was made. In general, the entropy change of the system reflects changes in the system.
EDIT: The remainder was written after OP provided the link.

[Hyperphysics] Since the electrolysis process results in an increase in entropy, the environment "helps" the process by contributing the amount TΔS. The utility of the Gibbs free energy is that it tells you what amount of energy in other forms must be supplied to get the process to proceed.

If there is no work and the reaction is run near equilibrium, the environment shows a change in entropy of $-\frac{\Delta H}{T}$. TΔS is not directly related to the change of entropy in the surrounding. The reason that $\Delta H$ is related to the entropy change of the surrounding is that for a boring surrounding (i.e. one that only shows changes in temperature, and very little at that) the change in entropy is proportional to the heat transferred.

I would say that entropy change is not due to heat exchanged, but due to the fact that during chemical reactions entropy changes because products and reactants have different entropies since they are different compounds with different structure and aggregate state. Entropy of reaction denotes such entropic changes and not entropy changes due to heat exchanged. What are your thoughts?

That is correct. Entropy is a state function, so if the entropy of the system changes, it is because the system changes, no matter what happens in the surrounding (you can't break the second law, though). The hyperphysics quote does not say that the entropy change of the system is associated with heat exchange. Their (implicit) argument is the second law:
$$T \Delta S_\mathrm{system} + T \Delta S_\mathrm{surrounding} > 0 $$
So if we have a positive change in entropy of the system like in the example, we can get away with cooling down the surrounding a bit (negative change in entropy of the surrounding). Typically, though, we don't want a super-slow reaction, so you would apply a higher voltage than necessary, and the heat transfer would go in the other direction (as you can tell by the heat typically dissipated from a fast battery charger).
Applying this to the first law, some of the energy input for the reaction can be in the form of heat transfered from the surrounding, it does not have to be all work in this case. (In other cases you have to put in more work, not for the first law energy balance but to satisfy the second law.) For a graphic illustration of the different cases, see https://chemistry.stackexchange.com/a/112958/72973.


Nicotine concentration in e-liquids E-liquids are blends of nicotine, propylene glycol, vegetable glycerin, and aroma.
I need to develop a method for determining the concentration of nicotine in a vial.
(to verify if the information listed on the bottle are correct.)

Does anyone have any idea how to proceed?

Nicotine concentration is a serious concern, as the material is a serious poison (dose matters).  Liquid Chromatography/Tandem Mass Spectrometry is a little pricey.

http://oxfordchemserve.com/e-cigarette-refill-nicotine-quantitative-test-kit/
  Specific test kit "designed with the non-chemist in mind"
http://oxfordchemserve.com/wp-content/uploads/products_img/Oxford_Nicotine_Test.pdf
   Explanation - it is a pH titration of nicotine base.
http://www.eliquidtest.com/
   Another titration kit.  Google for the product spectrum.

Saliva or urine nicotine test kits detect nicotine persistent metabolite cotinine and are therefore irrelevant.

http://www.medpagetoday.com/Pulmonology/Smoking/44238
Warning about E-liquid refills.
http://www.veppocig.com/product_images/uploaded_images/eliquid-nicotine-strength-guide-2.png
Claimed dosage guide.  Caveat emptor.
http://www.learn.eversmoke.com/nicotine-strength.html
A second dosage opinion.

Glycerin can pyrolyze to acrolein.  This is not a good thing to inhale.  NEVER inhale butter flavor.  Butanedione destroys lungs: "popcorn lung," bronchiolitis obliterans, obliterative bronchiolitis, constrictive bronchiolitis are life-threatenting, cumulative, and irreversible.


Why is the the rate of a reaction measured in concentration? Why is the rate of a reaction measured using concentration against time, rather than mol against time?

Simply, because if you used an absolute amount (moles, for instance), by changing the quantity of reagents used, you couldn't use the existing data (or you would still need the volumes, to calculate the concentrations, in order to scale up/down the existing data, and fit them with your reaction).

Since the kinetics doesn't change with volumes, but with concentrations (for instance, if you use the same concentrations of reagents in 100 ml, or in 1 liter, the reaction, at least in theory, doesn't change) it's much easier and useful to express it in terms of concentrations.


Similarities between Co and Ni Elements $\ce{Co}$ and $\ce{Ni}$ share many similar/same properties and usually if one of them is an example of some group (e.g., ferromagnetism, same possible oxidation states, packing fraction and coordination number in lattice, melting and boiling point,  similar properties of their alloys, etc.), the other is too.

I understand that it may be due to similar sizes and atomic numbers, but many other elements have these properties in common. Can you give any other example of such a pair and why only few of these exist, or why this relationship between  $\ce{Co}$ and $\ce{Ni}$ special?

Cobalt and Iron ions are the only ones to produce colored complexes when mixed with thiocyanate ion $\ce{SCN^-}$. If you mix up a dilute acidic solution of Iron(III) Chloride $\ce{FeCl_3}$ and a similar solution of ammonium thiocyanate $\ce{NH_4CNS}$, a extremely dark red solution is obtained, due to a complex containing the ions $\ce{Fe^{3+}}$ and $\ce{SCN^-}$. If you replace $\ce{FeCl_3}$ by a solution of Cobalt Chloride $\ce{CoCl_2}$, you obtain a blue color due to a complex containing $\ce{Co^{2+}}$ and $\ce{SCN^-}$. No other metallic ions are known to produce such colored complexes with thiocyanate ions, at least among the usual ions studied in chemical analysis courses. The blue Cobalt complex has also the property of being soluble in pentanol, where it can be extracted if the initial solution is colored.

Any way it should be mentioned that :

1) The reaction with Iron is extremely sensitive. The red color may still be seen if the concentration of the complex is 1 - 2 ppm. The blue complex made with Cobalt ions is only visible at much higher concentrations.

2) If this reaction is made with a Cobalt solution containing Iron as an impurity, the solution will turn red, as the blue color of Cobalt will be hidden. In this situation, a couple of milligrams of Potassium fluoride $\ce{KF}$ should be added to the solution. This will produce a colorless complex of $\ce{FeF_6^{3-}}$ and the blue color of the cobalt complex is restored.


Numbering bonds in cyclic compounds Let's say I have a double bond between carbons 6 and 1 on a 6 carbon cycle (the numbering is imposed because of the positions of the suffix functional groups), is the bond numbered 1 or 6?

6 seems logical because it is not ambiguous (clearly if it starts on 6 it ends on 1), but usually you must minimize numbers (however numbering the bond 1 means there's ambiguity between it being between 1 and 2 and between 1 and 6).

I've tried finding the official rule but haven't managed to.

however numbering the bond 1 means there's ambiguity between it being between 1 and 2 and between 1 and 6

It might appear to you all of a sudden that there's an ambiguity, but actually there's none. Look at these two molecules carefully:

Note that the numbering for the molecule on the left will increase in the anticlockwise direction, while the numbering for the molecule on the right will increase in the clockwise direction. The pi bond in each case is named as a -1-en (and it should, because both the molecules are identical)

You seem to be assuming that the numbering for the molecule on the left also applies on the molecule on the right. But, that is not the case, as shown above.

The disubstituted cyclohexene ring complicates the scenario, but only slightly. Consider the case when functional groups of equal priority are present:

Now, recognize that - even before assigning a name - that both the molecules are different. Thus, their names will also be different. The molecule on the left begins numbering from the top $-\ce{COOH}$ and increases anticlockwise. The molecule on the right begins numbering from the lower $-\ce{COOH}$ and increases clockwise. The first pi bond is assigned a name "-1-en" while the other is assigned "-3-en".

If a lower priority functional group is present instead, like so:

the numbering will increase along the direction of the pi bond, beginning from the top $-\ce{COOH}$. Hence, their names will be 3-hydroxycyclohex-1-ene-1-carboxylic acid and 5-hydroxycyclohex-1-ene-1-carboxylic acid respectively.

A possible way to achieve a "-6-en" is to take equal priority suffix groups on positions 1, 2 and 4, and then place a pi bond between positions 1 and 6, like so:

It is named "cyclohex-6-ene-1,2,4-tricarboxylic acid" giving the suffix functional groups a lowest locant set of "1,2,4", consequently naming the alkene as a "-6-en". Actually, though, this is not 100% correct. A slightly technical detail is missing, which can be found in Loong's answer.


How do things glow in the dark? What makes something glow in the dark? The only explanation that I can come up with for what makes things glow in the dark is that there is probably a chemical reaction slowly releasing the light that was previously absorbed. Am I correct?

How does this phenomenon work?

The spontaneous emission of light by a substance not resulting from heat is luminescence (a type of cold-body radiation).  $\ce{^1}$

A few types of luminescence are:

Bioluminescence: Made by living creatures such as fireflies, glow-worms, and many marine creatures.
Chemoluminescence: made by a chemical reaction. Glow sticks work this way.
Electroluminescence: made by passing electricity through something like in a light emitting diode (LED).
Photoluminescence: made by shining light at "luminous" (phosphorescent) paints.
Röntgenoluminescence: made by shining X-rays at things. (The curious name comes from Wilhelm Röntgen (1845–1923), the discoverer of X-rays.)
Sonoluminescence: made by passing energetic sound waves through liquids.
Thermoluminescence: made when photons are emitted from hot materials.
Riboluminescence: made by rubbing, scratching, or physically deforming crystals.
Phosphorescence: photoluminescence as a result of triplet-singlet electronic relaxation (typical lifetime: milliseconds to hours)

A chemical absorbs and then re-emits a photon of light, where the energy of the photons matches the available energy states and allowed transitions of the chemical takes place. In phosphorescence, the electron that absorbs the photon undergoes intersystem crossing $\ce{^2}$  into a triplet state.

The excited in triplet state returns to the lower energy singlet state. Such transitions, are kinetically unfavored and progress at a slow rate. Most phosphorescent compounds are fast emitters with triplet lifetimes in milliseconds. If triplet lifetimes are minutes or even hours, very slow degradation occurs.$\ce{^3}$ $\ce{^4}$ In phosphorescent materials, there's a delay between them absorbing energy and giving out light.These substances are so-called "glow-in-the-dark" materials.

To make glow-in-the-dark toys manufacturers mix zinc sulfide or strontium aluminate with plastic giving in glow-in-the-dark toys.$\ce{^5}$ $\ce{^6}$

Fluorescence

Fluorescent materials produce light instantly, when the excited atoms return to normal, in as little as a hundred thousandth of a second.$\ce{^4}$ $\ce{^7}$

Chemoluminescence

In the case of chemiluminescence, two chemicals interact which giving energy as a byproduct in the form of light( heatless glow).

When glow sticks are bent, an inner glass tube is broken releasing hydrogen peroxide. Hydrogen peroxide reacts with diphenyl oxalate giving 1,2dioxetanedione, which is unstable. It decomposes to carbon dioxide releasing energy. The energy is absorbed by electrons in dye molecule, that falls down to ground state releasing energy in the form of light.$\ce{^8}$ $\ce{^9}$

Chemiluminescence can happen naturally as well. In such instances — like fireflies — it is bioluminescence.

Some creatures, such as fireflies and jellyfish, contain chemicals within them that cause them to glow. Some of these creatures glow for protection, camouflage or to attract mates.

Ex 1.Bobtail squid $\ce{^12}$ $\ce{^13}$

Ex.2 Mycena lampadis ( mushrooms )$\ce{^14}$

The principal chemical reaction in bioluminescence involves some light-emitting molecule and an enzyme, generally called the luciferin and the luciferase.$\ce{^10}$ $\ce{^11}$ $\ce{^15}$

Electrochemiluminescence $\ce{^16}$is a special form of chemiluminescence in which the light-emitting chemiluminescent reaction is preceded by an electrochemical reaction.$\ce{^17}$

References

1 https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Map%3A_Principles_of_Instrumental_Analysis_(Skoog_et_al.)/15%3A_Molecular_Luminescence_Spectrometry

2 https://www.shsu.edu/chm_tgc/chemilumdir/JABLONSKI.html

3 https://en.wikipedia.org/wiki/Intersystem_crossing

4 https://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/lumin1.htm

5 http://www.phys.ubbcluj.ro/~dana.maniu/OS/BS_7.pdf

6 https://science.howstuffworks.com/innovation/science-questions/question388.htm

6 https://www.wonderopolis.org/wonder/how-do-things-glow-in-the-dark

[7] https://en.wikipedia.org/wiki/Fluorescence#Lifetime

[8] https://www.compoundchem.com/2014/10/14/glowsticks/

[9] https://en.wikipedia.org/wiki/Glow_stick#Fluorophores_used

[10] https://en.wikipedia.org/wiki/Bioluminescence#Chemical_mechanism

[11] https://www.nationalgeographic.org/encyclopedia/bioluminescence/

[12] https://www.nationalgeographic.org/encyclopedia/bioluminescence/

[13] https://en.wikipedia.org/wiki/File:Sepiola_atlantica.jpg

[14]https://www.timeforkids.com/g34/bioluminescence-2/

[15] http://photobiology.info/Branchini2.html

[16] https://www.sciencedirect.com/topics/chemistry/electrochemiluminescence

[17] https://www.sciencedirect.com/science/article/pii/S0008622317304943


Why does NaCl act as a catalyst for the reaction of CuSO4 + H2O + Al? I am analyzing a reaction between water sulfate copper and aluminum. A chunk of $\ce{Al}$, e.g., aluminum foil. I realize that a layer of $\ce{Al2O3}$ forms on the surface of the $\ce{Al}$, passivizing it, and I realize that the $\ce{NaCl}$ (or rather just the $\ce{Cl^-}$) cleans the aluminum oxide off the surface of the chunk of aluminum, but I don't understand why. I'd appreciate greatly if somebody could offer me an explanation.

Furthermore, (I didn't exactly include this in the topic of my question,) why is it that the reaction looks like this:

$$\ce{6H2O + 3CuSO4 + 5Al \rightarrow 3H2 + Al2(SO4)3 + 3Al(OH)2 + 3Cu}$$

(or something along the lines of that) rather than:

$$\ce{3CuSO4 + 2Al -> Al2(SO4)3 + 3Cu}$$ ?

Something along these lines has to be the case, because the reaction yields gas..

Aluminum resists corrosion in neutral or slightly off-neutral water because of the very insoluble Al2O3 film on the metal. If you break this film, it will corrode the bare metal and reform.

But if you scratch the Al and attach a cathode (a less active metal), you have a galvanic cell, and H2 can be evolved from this cathode as Al dissolves. The simplest example of this is when you put Al in H2O in contact with a drop of mercury. Scratch the Al thru the Hg; the Hg amalgamates with the Al (attaching itself as cathode), allowing H2 to be evolved as Al dissolves elsewhere.

The experiment with CuSO4 is similar: scratch the Al; some Cu will be deposited and act as a cathode and the rest of the Al will eventually dissolve in the H2O. If you are too lazy to scratch the Al (I'm being facetious!), you can add a little Cl- ion to the H2O, which will begin corrosion of the Al, deposition of Cu, galvanic cell production, dissolution of all the Al.

There is an electrochemical example posed as a test question: would it be better to make a ship out of aluminum with copper rivets or out of copper with aluminum rivets? An aluminum boat? Nah! It would corrode in seawater. Copper would survive longer. But NO! The copper boat would have a huge cathode and tiny aluminum anodes (the rivets), which would corrode rapidly and the copper plates would fall apart. On the other hand, an aluminum boat would not last forever in seawater, but the tiny cathodes would limit corrosion current, and the large aluminum anodes would have corrosion distributed over the whole ship, so it would survive longer than the other way. Of course, any intelligent person would use copper rivets on copper plates, and aluminum rivets on aluminum, but it's just to make a point.


Does the speed of an ion affect the current that can flow through a solution? Typically, when a cathode and anode are inserted into a solution that allows current to flow, the ions are moving with a certain kinetic energy.

If the ions were to gain kinetic energy (through transfer of thermal energy for example), will more current be able to flow, and will the potential difference increase?

My assumption for this is that increased ion kinetic energy will speed up the rate of ions being attracted to each electrode, thus causing more charge to arrive in a given period of time. As $I = Q/t$ where $I$ = current, $Q$ = charge and $t$ = time, would this mean an increased current?

Edit: I've performed a basic experiment in which I got a Tupperware container, stuck in some multimeter probes, and tested the difference in potential (voltage) for $\ce{NaCl}$ (table salt) in solution when cold and hot, then for water cold and hot. I appreciate there are many issues with reliability, but the general results were:

Water cold = 0.082 V
Water hot = 0.000 V (actually went into the negative then stabilised at 0)
NaCl cold = 0.065 V
NaCl hot = 0.130 V

Now, my real experiment involved $\ce{AgNO3}$ being titrated into $\ce{KCl}$ (mixed with deionised water and a little $\ce{HNO3}$ to acidify) – assuming the temperature remained constant, what effect could increasing the ions' velocity have on conductivity, without heating it up (increasing kinetic energy) to alter viscosity?

There is a strong dependence on temperature in conductivity of ionic solutions; see http://www.eutechinst.com/techtips/tech-tips44.htm, for example. However, thermal energy, i.e. molecular motion, is random, and does not favor current flow in the forward (or backward) direction. There are a number of factors that temperature influences:

Viscosity usually changes  inversely with temperature, so ions move more freely at higher temperature.
Ionic solutions have other species than OH- and H+; for example OH3+ [water + hydroxyl radical] is the hydronium ion; see http://en.wikipedia.org/wiki/Hydronium. The amount of each species varies with temperature, as shown in that reference. See also http://en.wikipedia.org/wiki/Self-ionization_of_water. Depending on mass, hydrogen bonding etc. different species have different conductivity.


Does negative sign of s-orbital exist? I know that the signs '+' and '-' are written in case of lobes of p and d atomic orbitals, which denote the sign of the wave function of that orbital. But s-orbital is also a wave. So, it should also have negative regions. However, most books show the s-orbital with a '+' sign only. What is the significance of this?
Again, the 2s, 3s, 4s, ... etc orbitals have some nodes. So does these nodes indicate the seperation of '+' phase and '-' phase? The 1s orbital does not have any node.So where is the '-' part of the 1s orbital ?

The sign of the wave function does not have any physical meaning since for every solution $\Psi$ of the Schrödinger equation $-\Psi$ is also a solution. In fact there are infinitely many equivalent solutions that only differ in their overall phase.

So the (meaningless) answer to your question is yes.

It should be noted that $\Psi \Psi^*$, that can be physically interpreted as a probability distribution for finding a particle at a certain point, is always a non-negative real number.


Why is the rate inversely proportional to the square root of temperature in Grahams Law of Diffusion? In my book it is given that:

  The general form of the Grahams Law of Diffusion can be stated as follows when one or all of the parameters are varied:
  $$\text{rate} \propto \frac{PA}{\sqrt{TM}},$$
  where $P$ - pressure, $A$ - area of the hole, $T$ - temperature, $M$ - Molecular weight.

How can rate be inversely proportional to the square root of temperature? I feel that on increasing temperature it must increase, but how can the rate decrease?
Is the formula in my book correct or incorrect?

It helps to rewrite
$$\text{rate} \propto \frac{PA}{\sqrt{TM}}$$
by assuming the ideal gas law holds, as follows:
$$\text{rate} \propto \frac{RA}{V_m}\sqrt{\frac{T}{M}}$$
where $V_m$ is the molar volume or inverse of molar particle density.

Written this way it is clear that if the particles occupy the same volume, increasing their temperature increases the rate at which they effuse through the apperture.


Why is $^3P_1$ an atomic term symbol for carbon? I'm using these rules, $^{2S+1}L_J$ with $J=L+S$ and it seems to imply that $^3P_1$ means $S=1$, $L=1$, and $J=1$ but clearly $1+1 \ne 1$ so according to me this shouldn't be an atomic term symbol but it is. Why?

with $J=L+S$

Nope. In general, it can be shown that the permitted values of the $J$ quantum number for the total angular momentum that arise from two sources characterized by quantum numbers $J_1$ and $J_2$ are given by
$$
    J = J_1 + J_2, J_1 + J_2 - 1, \dotsc, |J_1 - J_2| \, .
$$
In this particular case, $J_1 = L$ and $J_2 = S$, so $J$ ranges from $L + S$ to $|L - S|$ in integral steps and not necessarily equal to $L + S$.


Can o-phenanthroline be used as a complexing agent for lead(II)? I’ve been trying to find a method to perform a spectrophotometric analysis to determine lead concentration (on the ppm level). This would require some chelator capable of forming a coloured complex with lead in the aqueous phase.

Fe(II) ion can bind to o-phenanthroline to form an orange colour (with deeper hues based on the present concentration of iron irons).  I was wondering if lead(II) can too complex in the same way.

Note: The source of lead is $\ce{Pb(NO3)2}.$ This means some nitrate ions will be present in solution after dissociation of this chemical in water.

Yes, there are numerous known lead(II) o-phenanthroline complexes with thiocyanate, perchlorate and bulkier organic ligands.
As for nitrate, Bytheway et al. [1] reported synthesis (from lead(II) nitrate in methanol) and crystal structure of bis((1,10-phenanthroline)-(nitrato-O,O'))-lead(II):

Figure 1. crystal structure of $\ce{[(phen)2Pb(NO3)2]}$ (CCDC: 1246661, CSD: RAQMAX).

Unfortunately, there is no UV-Vis/NIR spectra available.
Harrowfield et al. [2, p. 1030] reported that the crystals are colorless, which likely eliminates visible region of detection.

References

Bytheway, I.; Engelhardt, L.; Harrowfield, J.; Kepert, D.; Miyamae, H.; Patrick, J.; Skelton, B.; Soudi, A.; White, A. Lewis-Base Adducts of Lead(II) Compounds. XIII. Synthetic, Structural and Theoretical Studies of Some 2:1 Adducts of 1,10-Phenanthroline With Lead(II) Oxoanion Salts. Aust. J. Chem. 1996, 49 (10), 1099. DOI: 10.1071/CH9961099.
Harrowfield, J.; Miyamae, H.; Skelton, B.; Soudi, A.; White, A. Lewis-Base Adducts of Lead(II) Compounds. VII. Synthetic and Structural Studies of Some 1:1 Adducts of Linear Polyamines With Lead(II) Nitrate. Aust. J. Chem. 1996, 49 (10), 1029. DOI: 10.1071/CH9961029.


Which gas has the greatest volume? I was given $4$  gases, which gases had the same temperature ,pressure and mass and was told to determine the gas with the greatest volume.

The gases were $\ce{CH4}$, $\ce{NO}$, $\ce{C2H2}$ and $\ce{CO2}$.

So I assumed the mass was $1$ gram and used the formula $n=\frac{\text{mass}}{\text{molar mass}}$  and found that $\ce{NO}$ had the greatest volume but the correct answer given to me was $\ce{CH4}$.

I cannot find out where I went wrong?

First of all, if in a certain condition it is said that mass (or any other parameter) is constant, you cannot just assume ${1gm}$ (or any other number). Mass being constant means $\Delta m =0$, not that it is ${1gm}$. You have no way of knowing what the mass is (unless stated in the question, of course). The correct approach is to assume the mass to be $'m'\space{units}$. Now, this mass $'m'$ can assume any value, according to the condition.

Moving on, you have made a mistake while comparing molar mass and volume.

The ideal gas equation gives us the relation,

$$PV=\frac{\text{Mass}}{\text{Molar Mass}}RT$$

For your question,

$$V=\frac{1}{\text{Molar Mass}}.\frac{(\text{Mass})RT}{P}$$

$$V\propto \frac{1}{\text{Molar Mass}}$$

Recheck your calculations.


Is there geometrical isomerism in case of 2-methylcyclopentanone? How can 2-methylcyclopentanone show geometrical isomerism?
A ring and a double bond are in different planes, so even after having restricted rotation there will be no change in areal distance.

If you look at the structure of 2-methylcyclopentanone, you'll notice that it is asymmetric about the C-2 atom

This gives rise to two isomers; the R isomer on the left and the S isomer on the right

These are definitely optical isomers, more specifically enantiomers of each other. Whether these are 'geometric isomers' or not is a matter of how you define 'geometric' isomerism; the IUPAC Gold Book strongly discourages the usage of this term, instead preferring to use 'cis-trans isomerism'. If it is defined by the latter, then this compound does not display cis-trans isomerism, as there are no groups for the methyl substituent to be trans to.


Why not simulate every particle properties and interactions? Today we have huge computational power (which is even significantly larger with supercomputers). I know that computational chemistry is used sometimes to predict particle properties. As I read on Wikipedia:

  Present algorithms in computational chemistry can routinely calculate the properties of molecules that contain up to about 40 electrons with sufficient accuracy.

If that's so, why bother to try to find chemical interactions and properties experimentally, at least up to 40 electrons? For example, every year new drugs are being discovered. Wouldn't be it easier at least to find new chemical compounds, if not their properties, simply by computer simulation? What are the constraints and where do they come from? (I know that such constraints exist, but I'd like to know why).

Forty electrons is tiny. Even if we limit ourselves to just the valence electons, cyclohexane already has 36 electrons. Anything drug-like has way more electrons that 40. For example, viagra has 178 valence electrons, and that's not necessarily a "large" drug. (Compare with vancomycin, for example.)

Even if you're dealing with things inorganic compounds, where the total number of atoms in the formula unit is small, the properties of the material don't come from a single formula unit, but come from the interaction of a large number of atoms. -- That's an example of a more general principle. The important properties for most materials you use (including drugs) don't come the molecule in isolation, but come from the interactions of the molecule with other molecules, either of the same chemical or of different chemicals. To be accurate, all of those interactions need a system with much more than 40 electrons.

The 40 electron limit comes from the implict assumption here that you're talking about quantum mechanical calculations. QM calculations are rather computationally expensive, as you have to account for all the interactions of all the electrons with each other at all positions in their delocalized superposition. There's various tricks (like DFT) which make the calculations for large numbers of electrons easier, but note that "easier" doesn't mean "easy". Even with DFT and other approaches, large systems take a lot of computer time to calculate accurately.

There are other approaches which don't suffer from the same limit as QM does, but they are able to make their gains in efficiency because they make approximations. For example, molecular mechanics approaches are able to simulate systems in the hundreds of thousands of atoms region. But they're able to do so because they don't actually calculate the position of electrons. Instead they treat the system "classically", and experimentally fit interaction potentials which approximate the underlying quantum effects. (For example, they don't exactly calculate the bond stretching potential, but instead approximate it as a harmonic one. That's "close enough" to the true bond stretching potential for the range of bond lengths typically seen in such simulations, but not 100% quantum mechanically accurate.)

There's many groups and companies which do use molecular mechanics and other similar approaches to inform their drug and material development process. The issue is that because the energetic potentials being used are only approximate the results from the simulation are also only approximate. Depending on what you're trying to simulate, the results of the simulation may or may not be accurate. As such, these simulations are treated mostly as a first step, to find potential leads/hypotheses, and then the scientists actually have to go into the lab and test the results to confirm.


What determines the strength of reducing and oxidizing agents? The best reducing agents are located at the bottom left of the periodic table (low electronegativity) and the best oxidizing agents are located at the top right of the periodic table (high electronegativity), excluding noble gases.

Is electronegativity the factor that determines the strength of reducing agents and oxidizing agents?

Because compounds can be oxidizing agents like potassium permangenate (KMnO4) and reducing agents LiH4, what makes a compound an oxidizing or reducing agent is oxidation & reduction tables. As oxidation is the giving off of electrons and reduction is the acceptance of electrons, like plus vs minus relationship, if you have an oxidation table you can turn it into a reduction table by putting the flipping the table, changing the signs and reversing the equations. Anyway, reduction tables are more standard, where the strongest oxidizing agents have the most positive/largest standard reduction potentials and the strongest reducing agents have the most negative/smallest standard reduction potentials.

The standard reduction potentials are determined with a volt-meter connecting two cells together as electrons pass through the salt-bridge.

This site explains it  rather nicely

But there are some patterns such that

$\ce{2M(s) + 2H2O(l) -> 2M+(aq) + OH^{-}(aq)}$ $\ce{ M = Li, Na, K, Rb, Cs}$

Cesium reacts more violently (in an explosive reaction which occurs as hydrogen gas is ignited from the heat of the strongly exothermic reaction) than all the metals above it because it is a stronger reducing agent reducing agent being oxidized itself more than the metals above it because it has the the lowest ionization energy due to shielding as the many shells of electrons around the nucleus of Cesium reduce the pull of the positive charged nucleus on the electrons due to the electrons in the electron shells repelling the valence electrons further from the nucleus by like charge repulsion. Ionization energy is simply the measurement of the heat energy required to cause an atom to lose an electron in the gas phase.

This makes sense, but there are anomalies to consider not going by the electrochemical series from experiments and just trying to rationalize off of stuff like ionization energies, electronegativities, etc that I started off my answer with. The reason that you might find anomalies is because the reduction potentials in voltages of elements are calculated in aqueous solution, while ionization energies are calculated in the gas phase, though many oxidation-reduction reactions occur in liquid solutions, one potential difference being the failure to consider enthalpy of solvation when considering ionization energies. Furthermore, Pauling's electronegativites can be calculated from running some equations through elements considering the physics of ionization energies, so ionization energies and electronegativites give you a similar perspective.


Ni-Mh battery makes fizzy sounds while and after charging - why and do I need t replace it? I have some Ni-Mh rechargeable batteries (size AA/Mignon 1.2V, 2100 mAh) and just started to charge them using an automatic charger with 450 mA output.

Now one of those batteries surprisingly started to make silent fizzy sounds, like if you're frying something in a pan. It started within seconds after putting it in the charger and keeps going for minutes after removing it from the charger again. There is nothing visible from the outside though, I don't see any bubbles or leaking fluids.

What exactly is the chemical process here that causes this sound (probably something is producing gas) and why does it happen in this one battery and not in the others I have of the same type?
Is that battery still usable (safely?) or do I need to replace it?

Update: As of now, the battery in question is replaced and returned to a recycling station. I'm still interested in the chemical process behind this noise though.

It's not possible to give a proper answer to (a) without examining the battery but it is likely due to a compromised casing.  The noise is likely hydrogen gas leaking out as it's produced. Nickel metal hydride (NiMH) batteries use a metal hydride (MH) for energy storage.  When this Hydride is exposed to moisture in air it reacts to form hydrogen gas: $$\ce{MH_x + xH2O \rightarrow M(OH)_x + x H2 }$$

For (b) I would say that your battery is probably not usable—safely or otherwise—anymore and may present a fire hazard. You should replace the battery as rechargeables don't cost near as much as the device they can leak inside of and destroy or the cost of fire damage to a house.


Can silicon dioxide melt at 20 °C? Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$?

I have searched the web for $\ce{SiO2}$ phase diagram, but it seems to me that almost all the graph I can find have the temperature axis where the minimum temperature is in the hundreds degrees Celsius.
Is it theoretically possible that with an high enough pressure I can melt $\ce{SiO2}$ in a temperature range like $\pu{-30 ^\circ C}$ to $\pu{50 ^\circ C}$? Or is it something theoretically simply impossible?

Question: Can $\ce{SiO2}$ melt at $\pu{20 ^\circ C}$?

According to experimental and calculated data values, my answer is no. See the phase diagram of pure silica based on the experimental and calculated data given in Ref.1:

Reference 1 states that:

An internally consistent data set on the thermodynamic properties of the silica polymorphs stable up to $\pu{15 GPa}$ (α‐quartz, β‐quartz, tridymite, cristobalite, coesite, and stishovite) and the liquid phase is presented. The data set was produced through a computer‐based assessment of the properties in which the available thermochemical (calorimetric), physical (bulk modulus and thermal expansion), and solid‐state and melting transition data (including some newly determined data on the high‐pressure polymorphs cosite and stishovite) were considered. The data set can be used to calculate phase relations at pressures of $\pu{0.1 MPa}$ to $\pu{15 GPa}$ and temperatures of $300$ to $\pu{3200 K}$. The calculated phase diagram using these data agrees quite well with the phase equilibrium determinations except for the high‐temperature part of the coesitestishovite boundary. The properties of the liquid phase obtained are also in good agreement with the available data.

Keep in mind that $\pu{0.1 MPa} \approx \pu{1 atm}$. According to the phase diagram, cristobalite form is in liquid form at temperature grater than $\pu{1750 K}$ in pressure range of $\pu{0.1-0.7 MPa}$. The extrapolation of that boundary would show it would never cross $\pu{20 ^\circ C}$ point on $x$-axis without going to negative pressure. Therefore, it is impossible to have any form of $\ce{SiO2}$ in reduced or high pressusituations at room temperature or below.
References:

V. Swamy Surendra, K. Saxena, Bo Sundman, J. Zhang, “A thermodynamic assessment of silica phase diagram,” Journal of Geophysical Research 1994, 99(B6), 11787-11794 (https://doi.org/10.1029/93JB02968).


Dispersion forces - what determines which molecule will have its electron cloud repelled? Here is an image I found which illustrates dispersion forces:

One helium has its electron cloud repelled by the other helium's electron cloud, which exposes the nucleus and gives an induced dipole which holds the atoms together weakly. But when two helium atoms approach each other, what determines which atom will have its electron cloud repelled? I would think that both electron clouds would equally repel each other. Also I am guessing that this effect will get stronger as temperature is lowered (since helium is a liquid at very low temperatures), but I am not sure why.

You are correct that the electron clouds would repel each other equally.

The induction as shown in your second picture would require another outside entity, polarizing one of the He atoms, which would then affect the other.

At an arbitrarily show distance, we might describe a molecular orbital formation between the two helium nuclei, as shown here:

(source: mhhe.com)

In this case, without the external perubation, the orbital shape (and hence the electron clouds) is (are) quite symmetrically distorted.

However, it becomes nonesense to describe a system of two atoms.  In liquid helium there are manifold other charge fields distorting these nice shapes, and you may get some asymmetry as shown in your picture above.


What factors promote honey's crystallisation? Honey is a complex mixture containing, probably, more than hundred compounds.
I understand that the % of those components will affect the crystallisation.
However, I am wondering what factors may speed up the crystallisation, apart from heat?

The decisive factor is the plant origin of honey. The major components ( reportedly by some beekeeper books of my passed away father ) have ratio glucose : fructose about 1:1 to 1:1.7. Glucose reportedly crystallizes more readily, triggering fructose crystallization as well. So honey closer to 1:1 ratio has stronger tendency to crystalize.(**)
Aside of already mentioned factor of temperature, another factor is obviously the ripeness of honey. Lower water content in more matured honey means higher crystallization probability.
Seeding by already crystalline honey helps to trigger or speed up crystalization.
If crystallization is already happening, or if it is intended and desired, then stirring speeds up crystallization.
Seeding and stirring leads to many small crystals, what is applied during production of the pasted honey. This form is favourite one among people. It often dissolves faster then liquid honey during drink sweatening. It is also very convenient for bread spreading, as its flow is almost frozen.
Slow spontaneous static crystallization leads to big crystals, forming mechanically tough product, requiring big force to separate a chunk of honey(*). The crystallization rate, the size of crystalsfraction of residual liquid phase and physical properties of the product seem to be composition dependent as well.
If such honey is wanted to be liquid again, it is recommended to carefully heat it up in $\pu{40 ^{\circ}C}$ warm water bath. High temperatures affect the present probiotics and enzymes. Note that honey is along fermented milk the major source of natural probiotics.

(*) My father stored honey in 4 L flasks for home-made pickles. When honey was crystallized, we often digged it up by table spoons or forks. But unless they were very strong, like heavy stainless steel ones from Soviet Union, they had tendency to bend.
(**) Infamous one is larch tree honey. It has so strong crystallization tendency it often crystalizes in the honeycomb. It is not being able to be directly centrifuged out without honeycomb destruction.


Is (2R,3S)-2,3-dichloropentane optically inactive? Two confusing thing about (2R,3S)-2,3-dichloropentane:

The 2 stereocenters have opposite absolute configuration.
But there is no plane of symmetry.

So, is (2R,3S)-2,3-dichloropentane optically inactive?

(This is actually a follow-up question of:
Monochlorination of 2-chloropentane and possible enantiomer products)

To have an optically inactive meso compound with two chiral centers:

1)  the chiral centers must have opposite configuration

2)  the chiral centers must have the same set of fragments bonded to them

Among other things the 2 carbon has a methyl group bonded to it versus the 3 carbon having an ethyl group instead.  You do not satisfy (2), so you do not have the symmetry required for a meso compound, and the molecule is still chiral and optically active.


Mechanism for reaction of Tollens' reagent with aldehydes Upon treatment with Tollens' reagent (ammoniacal silver(I) nitrate), aldehydes are oxidised to carboxylic acid, and silver(I) is reduced to silver metal.

I am trying to find a mechanism for the this reaction online, but the only thing I can find is the balanced equation. Can someone propose or help me find the mechanism?

Here are the two half reactions:

$$\begin{align}
\ce{[Ag(NH3)2]+ + e- &-> Ag^0 + 2NH3} \\
\ce{RCHO + 3OH- &-> RCO2- + 2H2O + 2e-}
\end{align}$$

which together yield the overall reaction

$$\ce{2[Ag(NH3)2]+ + RCHO + 3OH- -> 2Ag^0 + RCO2- + 4NH3 + 2H2O}$$

Here is a diagram of the reaction mechanism. The carbonyl group is oxidized in the process and the $\ce{Ag^+}$ is reduced. The resultant oxidized aldehyde (now a radical cation) reacts with hydroxide to form a tetrahedral intermediate. A gem-diol like intermediate is formed via a hydrogen shift, which then continues on to the final carboxylate anion.


Allotropy and polymorphism Polymorphism is defined as the ability of a solid material to exist in more than one form or crystal structure, whereas allotropy is defined as the property of some chemical elements to exist in two or more different forms, in the same physical state, known as allotropes of these elements. (From wikipedia).

In the case of crystal solids, it seems clear that allotropy is a particular case of polymorphism... but this is always true? That is, both terms, polymorphism and allotropy refer always to crystal structures? Polymorphism definition mentions "crystal structures" but "crystal" does not appears in the definition of allotropy...

Thanks in advance.

UPDATE
Thank you for your answer Then, in some cases allotropy is more general than polymorphism. For instance, dioxygen and ozone are allotropes but not polymorphisms. But... my question arises from my TRUE/FALSE set of questions which I have to answer, and one of them is:

  In a non-crystal material the concept of allotropy or polymorphism does not exist.

What would you answer to that? My conclusion is that the question is not clear... For polymorphism is true whereas for allotropy it is not.

In my own work, I've never used the term polymorph for anything but molecular crystals in a highly but differently ordered solid state.

  In the case of crystal solids, it seems clear that allotropy is a particular case of polymorphism.

Yes, allotropy is the polymorphism of elements in the same state. This does not mean that each solid allotrope has to be highly ordered: amorphous forms of carbon, phosphorus or selenium are considered as allotropes too.

The definition of allotropy isn't confined to the solid state: ozone is an allotrope of dioxygen.


If propane with specific parameters were to combust in open air, what is actual volume of CO2 produced Lets say we have a sample of propane gas having a volume of 2.70 L at standard room temperature and 1.75 atm. This gas is exposed to our air. The equation that governs this reaction is:

$mC_3H_8 + xO_2 + 3.76xN_2 \rightarrow yCO_2 + zH_2O + dN_2 + eO_2$

Here $m,x,y,z,d$ and $e$ are coefficients. I want to find out what the volume of carbon dioxide given this information. The trouble I'm having is finding out what pressure the air is and I strongly suspect that I would use the sea level or 1 atm and the volume of nitrogen and oxygen with which propane reacts with.

Rephrased homework question? Anyway, you are going about this the wrong way, thinking much too complicated.

If you combust m moles of propane, how many moles y of CO2 do you get? Now remember that moles and liters of ideal gas are totally proportional, and you have the answer.


Vapor pressure of immiscible liquids I have read that vapour pressure of a mixture of two immiscible liquids is
$$P=P^0_A + P^0_B$$ where A and B are immiscible liquids.

If they are immiscible, they would form separate layers. Thus since the more dense liquid would be below the less dense liquid, it would have no interface with the gas phase. Given this lack of an solid-gas interface, should the pressure due to the lower layer be zero?

This is not just some vapor pressure. This is the equilibrium vapor pressure. Thermodynamics is all about equilibrium, you know. And equilibrium, roughly speaking, is what takes place in a closed container after a billion years.

Immiscible as they are, the liquids still have some solubility in each other (maybe extremely low, but anyway). Over the course of centuries, molecules of the lower liquid will slowly creep one by one through the upper layer and vaporize. Eventually they will saturate the gas phase and create the same vapor pressure as if the upper liquid were not there.

Stirring the whole mixture, as Klaus suggested, is quite likely to speed things up a bit.


Does all processes that consume energy from outside result in negative entropy change? In any spontaneous process entropy increases.In biological systems disordered elements are converted into highly ordered tissues etc(-ve entropy change) which make up living organisms. But, it requires energy from the sun(ultimately). So,does that mean all processes that consume energy from outside result in negative entropy change? Endothermic reactions consume energy in the form of heat from outside so, is all endothermic reactions causing the entropy to decrease?

Entropy increases in a spontaneous evolution of an isolated system… other than that, there is no general law on the evolution of entropy.

  does that mean all processes that consume energy from outside result in negative entropy change?

No. When you heat a solid, it increases its entropy.

  endothermic reactions causing the entropy to decrease?

Endothermic reactions which cause entropy to decrease have both ∆H positive and ∆S negative, so they will have a positive ∆G, i.e. their are not thermodynamically favorable.


Distance requirement for hydrogen bonding to occur I am trying to understand the distance requirement for hydrogen bonding to occur (more specifically in the context of protein folding). There needs to be a donor and acceptor atom for bonding to occur. What is the range of distances between these atoms where one can say that a hydrogen bonding is likely to occur when the polypeptide is in an aqueous environment? Is it element pair specific?

You will probably find all this information in biochemistry textbook. To give you an idea in proteins, typically; alpha helices H..O distance is 2.06 Angstrom, and N..O, 2.99. In beta sheets (parallel) H--O 1.97, N--O 2.29 and in anti parallel sheets, H--O 1.96 and N--OI 2.91. These H bonds are not linear, e.g C=O..H-N typically 150 to 160 degree for COH or NHO angle.

In Watson-Crick base pairing in DNA, (C-G pairs) the linear H-bond distances  are  $\mathrm{N_4-(H)-O_6}$ 2.91, $\mathrm{N_3-(H)-N_1}$ 2.95, and $\mathrm{O_2-(H)-N_2}$ 2.86.


Why does depurination happen at a higher pH (or under less harsh conditions) than depyrimidation? I've always wondered why the N-glycosidic bond, which connects the adenine or guanine purine bases to their sugars, was the easiest to cleave via acid hydrolysis, or why purines have the lowest ionization potentials?  What is it about the rings / resonance / etc. of purine vs. pyrimidine bases in nucleic acid polymers that makes this so?  For example, according to (Lindahl, Nature 362, pp. 709–715 (1993), http://www.nature.com/nature/journal/v362/n6422/abs/362709a0.html), at least in the context of deoxyribonucleic acid (DNA) at pH ~ 7 (probably the rate is much slower in RNA under the same conditions due to the 2'-OH) the rate of depyrimidation is $\approx 20$ fold slower than the rate of depurination.  Note that the actual base rate of depurination is $\approx 4 \times 10^{-9} \space \mathrm s^{-1}$ at 70 °C and pH ~ 7.4 (Lindahl & Nyberg, Biochemistry 11(19), pp. 3610–3618 (1972), http://pubs.acs.org/doi/abs/10.1021/bi00769a018), implying a base rate for depyrimidation of $\approx 2^{-10} \space \mathrm s^{-1}$ under the same conditions.

If I had to guess, I'd say that the reason has to do with purine bases (or the local context of the N9-glycosidic bond on purine bases) being less electrophilic, which is why I mention generally lower ionization energies for purines vs. pyrimidines.

Quoting from: http://www.nrcresearchpress.com/doi/abs/10.1139/o74-139#.U32-3vmSyDZ

"Contrary to previous suggestions, pyrimidine as well as purine deoxynucleosides are subject to base cleavage, the relative ease of thermolysis being 2′-deoxyguanosine > 2′-deoxyadenosine > 2′-deoxycytidine > 2′-deoxyuridine > thymidine."

Note that this actually seems to match the relative (experimentally measured) ionization energies of the bases --- Guanine (7.77 eV) < Adenine (8.26 eV) < Cytosine (8.68 eV) < Thymine (8.87 eV): Orlov, V. M., Smirnov, A. N., Varshavsky, Y. M. Ionization potentials and electron-donor ability of nucleic acid babes and their analogues. Tetra Lett. 1976;17(48):4377–4378.

If my guess regarding the lower electrophilicity of purine bases is true, it's not clear to me, however, why this is the case?

Update / Thought:

If we look at the four bases (where uracil is just missing the C5 methyl group on thymine) - (http://img.sparknotes.com/figures/7/749a4182b7527e44d289a612e420f40c/dna_bases.gif) - we can notice that the N9 nitrogen connecting purines to their sugars is only ever adjacent to carbon atoms with single- or double-bonds to nitrogen.  However, the N1 nitrogen connecting pyrimidines to their sugars is adjacent to a C2 carbon double-bonded with an oxygen atom.  Given that oxygen is more electronegative than nitrogen (oxygen has an electronegativity of ~3.44 vs. nitrogen's electronegativity of ~3.04) could this be a hint as to why depyrimidation is much less common than depurination?  Are there pyrimidine base analogues with a less electronegative group attached to the C2 carbon, and do depyrimidation rates here match or exceed depurination rates?

This could, of course, be completely off.  For example, maybe N7 protonation on purines presents a route to depurination, since guanine having a greater depurination rate than adenine (or lower ionization energy than adenine) is consistent with a decrease in the N7 nitrogen's basicity.  Maybe pyrimidines are much less prone to destabilizing protonation.  Etc.

*

My objective here: I'd like to know the "most significant" factor for (and hopefully most general reason for) the greater observed rates of depurination vs. depyrimidation.  For example, perhaps the aforementioned oxygen on pyrimidines decreases depyrimidation rates by a factor of two or so (this isn't meant to be an actual guess), but what explains the $\approx 20$ fold difference noted by (Lindahl, 1993)?

Really interesting question!
Mechanism of acid catalyzed hydrolysis of purine nucleosides is known to proceed through unimolecular departure of base moiety(DOI: 10.1021/ja00709a055).

It seems that someone else has also asked the same question: "Why reaction rate with pyrimidines is so much slower?" -and published paper of the reaction mechanism (DOI:10.1039/P29940000309).

In the paper it is suggested that pyrimidines follow different reaction mechanims. This would explain huge differences in reaction rates.

  In other words,
  the structure-reactivity correlations lend additional support to
  the previous suggestion, according to which the acid-catalysed
  hydrolysis of 4-oxopyrimidine nucleosides follows an openchain
  mechanism

Open chain mechanims:


How can functional groups be bounded? When I read about the different kinds of functional groups, they usually have formulas like $\ce{-NH2} $, $\ce{-COOH}$, $\ce {-OH}$, etc.

What does the bar represent? Any bond? Or can it only be a covalent one?

Functional groups are a unifying idea in organic chemistry. There are tens of millions of different organic chemicals but many of their key properties are determined by the functional groups they contain. For example, amines (compounds with the $\ce{-NH2}$ group) are basic and often smelly; carboxylic acids (containing the $\ce{-COOH}$ group) are acidic.

The bar in the symbol represents the bond to the rest of the molecule. Sometimes chemists write this slightly differently by including a specific symbol for the rest of the molecule like this: R-OH where R is the rest of the alcohol. So a chemist might talk about a series of similar alcohols by defining R as $\ce{CH3}$ (methanol) $\ce{CH3CH2}$  (ethanol) or $\ce{(CH3)2CH}$ (isopropanol).

This simplifying terminology allows chemists to talk about large groups of related compounds without having to be too verbose and writing out all the details of every structure.

This is an organic chemistry concept so the functional group attaches covalently to a carbon. Some groups can also occur in inorganic compounds but the terminology would be different. For example, Sodium Hydroxide contains an isolated OH but this isn't an alcohol and wouldn't be written or described using terminology related to functional groups.


Calcium ion and sulfate did not form precipitate I was in the chemistry lab earlier this week. We were doing a qualitative experiment (no concentration given for any solution).

I added an unknown solution that contains $\ce{Ba^2+},$ $\ce{Ca^2+},$ and $\ce{Mg^2+}$ with $\ce{(NH4)2SO4},$ I got barium precipitate and added $\ce{(NH4)2C2O4}$ and another precipitate appeared. I was told the second precipitate is $\ce{Ca^2+}.$

But I recall from solubility mnemonic rule, I should be able to see precipitate from both $\ce{Ca^2+}$ and $\ce{Ba^2+}$ after adding ammonium sulfate. Why didn't I? What are the mechanics behind this? How did calcium just sit there as a spectator ion?

While calcium sulfate is usually termed insoluble, it is not the ‘sitting at the bottom like a rock’ type insoluble; rather, it is the ‘there’s no practical way for me to get the two ions into the same solution without precipitation, but I’m still able to identify both ions in solution’ type insoluble. Those sentences don’t really help, so let’s look at numbers.
\begin{array}{lcr}
\hline
\text{salt} & \text{mass solubility} & \text{molar solubility} \\ \hline
\ce{BaSO4} & \pu{2.45e-3 g/l} & \pu{1.05e-5 mol/l}\\
\ce{CaSO4} & \pu{2.1 g/l} & \pu{1.54e-2 mol/l}\\
\ce{CaC2O4} & \pu{6.7e-4 g/l} & \pu{5.23e-6 mol/l}\\
\hline
\end{array}
To calculate your actual experiment, you would have to transform these values into $K_\text{sp}$ values and then calculate how much would precipitate at each given step. Without wanting to do that tedious calculation, you can still arrive at the general conclusion:

addition of sulphate precipitates practically all barium ions

addition of sulphate precipitates a large part of calcium ions
but: a non-neglegible amount of $\ce{CaSO4}$ remains dissolved

addition of oxalate precipitates practically all the remaining calcium.

In fact, the solubilities of $\ce{BaSO4, SrSO4}$ and $\ce{CaSO4}$ are such that a saturated solution of $\ce{SrSO4}$ will always give precipitate when barium is added but never with calcium or strontium; while a saturated solution of $\ce{CaSO4}$ will give a precipitate if either strontium or barium is added but never with calcium. This can be used to identify the heaviest alkaline earth metal in an unknown sample.


Acid-catalyzed hydration of alkenes So I'm basically doing an experiment where I measure the effect of pH on the yield of alcohol within the acid catalyzed hydration of ethene. Essentially, all other details aside, I get a mixture of ethanol and ethyl chloride solutions at the end of my experiment. I want to find the total amount of either. I want to use a simple method to determine the amount of each in the solution. I want to avoid using fractional distillation, if possible. I need help to do this.

If you are certain that there are no other halogenated byproducts or reagents remaining in your solution, and that any levels of bromine and iodine would be negligible compared to chlorine (fluorine would not be detected), then one of the total halogen analysis (TX) methods would be a good choice.

If this is an ongoing analysis that you need to set up, I suggest the Carius halogen method. Wikipedia gives a good overview here:

  The Carius halogen method in analytical chemistry is a method for the
  quantitative determination of halogens in chemical substances. A
  known mass of an organic compound is heated with fuming nitric acid in
  the presence of silver nitrate contained in a hard glass tube known as
  carius tube, in a furnace. Carbon and hydrogen present in the compound
  are oxidised to carbon dioxide and water. The halogen present forms
  the corresponding silver halide (AgX). It is filtered, washed, dried
  and weighed.

Alternatively, depending on the equipment available to you, you could also do an oxygen bomb combustion of the sample with a bicarbonate absorbant, followed by either titration of the resulting chloride or ion chromatographic (IC) analysis. In the case of the IC finish, you can usually just use the IC eluent as the adsorbant.

If you are only going to do this analysis once, or only rarely, then I would strongly suggest a commercial TX anaysis using Standard Methods - For the Examination of Water and Wastewater, Method 9020. Method 9020B is for the analysis of total organic halogens (TOX) in water, which is not appropriate for your analysis as a whole because it includes a step for extracting trace amounts of organics from water onto activated carbon. But, then a TX analysis is performed on the activated carbon, which is what you would want to do for your compound rather than for the activated carbon. This is typically about an $80 analysis.

Regardless of the TX technique used, you would of course then assume that all of the halogen came from the ethyl chloride and calculate the ethyl chloride concentration from there.


Suggest possible structures for the compound PF3(Ph)(NMe2) using the NMR data use these data to suggest the possible structure for the compound PF3(Ph)(NMe2):

19F Spectrum, two signals:
a doublet of doublets, J=56 and 819Hz, -39.5ppm, relative intensity 2
a doublet of triplets J=959 and 56Hz, -68.3ppm, relative intensity 1

31P Spectrum, one signal:
a doublet of triplets, J=959 and 819Hz, -53.0ppm

part b,
    suggest an experiment or measurement that might be helpful to discern possible isomers of the compound

My answer so far:
There are two types of structure that a Phosphorus with five bonds could be, trigonal bipyramidal and square based pyramid. I think that because there are two equivalent fluorines that this must be the trigonal bipyramidal structure as it had the two axial positions that would be equivalent.

I agree that the trigonal bipyramid structure is what we are dealing with, so let me focus on that structure.

The nmr data shows that 2 of the fluorines are equivalent and one is different. This results in 3 possible isomers as shown below.

Both $\ce{^31P}$ and $\ce{^19F}$ are spin 1/2 nuclei, so each of these isomers would produce phosphorous and fluorine nmr spectra with splitting patterns and integration ratios consistent with the reported data. That is, the phosphorous will split each different fluorine into a doublet, the two equivalent fluorines will split the singular  fluorine's doublet lines further into triplets.  The one non-equivalent fluorine will split the other two equivalent fluorine doublet lines into doublets.  Similar analysis of the phosphorous signal leads to the expectation of a doublet (from the singular fluorine) of triplets (from the two equivalent fluorines).

Two additional pieces of information:

Fluorine prefers the axial position in pentacoordinate phosphorous
(see here)
Because the orbitals involved in axial and equatorial bonding in pentacoordinate phosphorous compounds are different, the axial and equatorial coupling constants will be different.  Generally, $\ce{J_{PF(eq)}}$ > $\ce{J_{PF(ax)}}$ because there is increased s-orbital density in the equatorial bond (see here, for example)

The larger reported P-F coupling constant (959 Hz) is associated with the intensity = 1 fluorine signal.  This suggests that structure I, with one equatorial fluorine (with the larger $\ce{P-F}$ coupling constant) and 2 axial fluorines (with the smaller coupling constant and as preferred electronically - point 1 above) is the structure being observed.

Part B Experiments:

The trigonal bipyramid structure could be confirmed by heating the sample while in the nmr spectrometer.  As Berry pseudorotation commences the 2 fluorine signals should coalesce and determination of a barrier height around 35-40 kcal/m would be expected.

Also, compounds II and III must have similar energies and both would be expected to be present at equilibrium.  If we heated our sample so that pseudorotation took place, and then upon cooling we saw an additional set of signals, that would be consistent with II or III being our starting compound.  No new signals upon cooling would support structure I.