Identifying protons in NMR using the J values When looking over an old organic midterm I came across this question.

As you can see I got half of it right, but honestly I'm not sure how I got that one right or how you would answer this in general.  Looking back at this now it appears that the first signal was split into a doublet and each peak was then split into a triplet, which makes sense for proton c (first split by b and then by d).

The second one seems trickier.  It looks as though the signal was split into a quartet, and then each peak into a triplet.  To me that looks like it should be d - split into the quartet by e and then split into the triplets by c and b.  So why is the answer b?  These J values wouldn't have been supplied for now reason - is there a way I could have used them?  More generally, is it possible (and if so how) to identify the type of proton and the structure of a molecule if you just had access to the NMR and the J values like this?

The proton(s) in question absorbs at 5.95 ppm, the olefinic region - so it cannot be D which is expected to absorb upfield in the aliphatic region.  B is the correct answer.  Proton B would be split by proton C into a doublet with a 15.5 Hz separation (the coupling constant $\ce{J_{BC}}$).  Each line in the doublet would be further split into another doublet with a J=8 Hz spacing by proton A.  So now we have 4 lines, a doublet of doublets.  Each of these four lines will be further split into a triplet with 1.5 Hz spacing by the two (hence the triplet) D protons.

Using the chemical shift of a proton and the observed splitting (e.g. the coupling constants, or J values) can go a long way in helping to determine a structure, particularly in relatively simple molecules.  As molecules get more complicated you can do advanced experiments to help with the structure determination.  One example would be spin decoupling experiments.  For example, if we "know" which signal is due to the D protons, we could irradiate them with energy during our nmr experiment.  If they are coupled to proton B (they are), then the triplet in the B signal should diappear and we would just see our doublet of doublets.  This type of experiment confirms which protons are coupled to which other protons.  There are many other advanced nmr techniques that can be used to help in structure elucidation.


Is there a name for the intersection of the graphs of first and second order reactions? Is there a name for the point where a first order rate of reaction meets a second order one.

This point highlighted above.

As far as I know there isn't a specific term for that point.

If you have a one-step first order reaction

$$\rm A \xrightarrow{k_1} B + C, \quad \text{rate} = k_1[A]$$

that can run backwards as a second order reaction

$$\rm B + C \xrightarrow{k_{-1}} A, \quad \text{rate} = k_{-1}[B][C]$$

and the rates of those reactions are equal, you can call that point "equilibrium".

You can't really plot the rates separately the way you have unless you have a common concentration, e. g. plot a change in concentration $x$ where $\rm [A] = [A]_0 - x$, $\rm [B] = [B]_0 + x$, and $\rm [C] = [C]_0 + x$.


What is the process for purifying KOH? I'm partnered with a jewelry production company whose plating baths require $\ce{KOH}$. We're located in Indonesia, where import fees are absurdly high, and reliable suppliers are rare. A local shop sells $\ce{KOH}$ at \$5/kg, but whatever impurities are present in their mix cause enough problems to spoil the entire bath. (Expensive mistake when doing gold plating.) The only alternative we can find is a Merck redistributor, who sells $\ce{KOH}$ at close to \$100/kg, which is ridiculous.

Is there a relatively simple process we could learn (of the DIY variety) for removing impurities, turning non-lab-grade $\ce{KOH}$ into lab-grade $\ce{KOH}$?

According to Armarego, et al. (Purification of Laboratory Chemicals, 6th edition), they only provide a procedure to purify a solution of $\ce{KOH}$, likely owing to its very high solubility.  Further, the only impurity of note they cite is carbonate (from standing in contact with air), so if this is not your impurity then this procedure will not help.

If this is your impurity, they recommend adding a saturated solution of a barium salt, either $\ce{BaCl}$ or $\ce{Ba(OH)2}$, shaking, and then setting aside to allow the insoluble $\ce{BaCO3}$ to settle.  It sounds like you are working on large scale, so it is important to recall that the procedure just described should be done in several small batches to work best.

Not certain this is what you are looking for, but I hope it's a start.


How can I determine if a chemical will dissolve aluminum oxide, on paper? I am trying to clean some corroded aluminum. In my quest to find out how, I've stumbled across a lot of bewilderment and misinformation, and a lot of different feasible techniques as well.

My question is: How can I determine, on my own, whether something will dissolve aluminum oxide (room temperature) or not? Like, just by looking at chemical formulas.

I'd also need to be able to check to make sure a chemical that dissolves aluminum oxide doesn't also dissolve the good aluminum.

My goal is to be able to figure out not only if a given chemical will dissolve it, but also to come up with an idea of which chemicals that I don't know about may also dissolve it.

As a bonus it'd be nice to determine the temperature at which a reaction would occur if it's a bit higher than room temperature so I can determine on my own if I can apply a reasonable amount of heat to make something work.

I know very little about chemistry except a basic understanding that things react and can be described by equations, and that reactions absorb or release some amount of quantifiable energy.

Aluminum oxide is Al2O3.

Sorry I do not know how to tag this.

Also I'm not sure if "dissolve" is the right word. To be clear I use it in a way that means if I apply the chemical to corroded aluminum, it'll make the corrosion go away into the air, or turn back into aluminum, or at least make it easy to scrub or wipe off.

Normally you would learn about properties of chemicals from general chemistry textbook. Al2O3 is pretty stable and is hard to dissolve in anything at room temperature.

It also depends a lot on what is the physical state of Al2O3. Fresh fine Al2O3 is much more reactive than aged (or coming in big chunks) Al2O3.

Searching in general chemistry literature is the best way to go, but Al2O3 is really robust.


Can one reduce solubility of NaCl by adding CO2 under pressure? Will the salt fall out of solution? One can salt CO2 out of solution by adding salt to carbonated water.  Can one do the reverse?  If one adds more and more CO2 to the water under pressure, can you drive out the salt?

NaCl is vastly more soluble than carbon dioxide, so this method won't work.  You can get some NaCl to precipitate (assuming you have a concentrated solution) by bubbling HCl in, thanks to the common-ion effect, so you could for example get pure NaCl from rock salt this way.  But somehow I don't think that's the situation you have in mind :-)


Trend in atomic radius for noble gases In an exam, we were given the following graph and asked to explain why the slope of the change in r vs. Z changes dramatically at a point along the curve. I understand that the d block causes the change in slope for all of the series, but why is the slope from $\ce{He}$ to $\ce{Ne}$ less than that from $\ce{Ne}$ to $\ce{Ar}$? Why is this trend not present for any of the other series? I initially thought that this might have to do with the introduction of the p-block, but that would not explain the absence of the trend for the other series.

I have a theory for the question.

First lets look at what the graph is telling us about the radius and the atomic number. For the Noble gases, as the atomic number is increasing the radius increases. It is because of the addition of more electron shells (simplification for now).

Helium has a full 1s orbital. Whereas Neon has a configuration of $1s^2,2s^2,2p^6$. Therefore the electronstatic force occurring on the electrons is slightly bigger than the electron on the helium atoms. Whereas Argon has a configuration of $[Ne]3s^23p^6$.

Now my theory. In He, we only had $1s^2$ and in Neon we added the extra shell and we went up til the 2p. Now the shells also act as barriers sometimes, therefore we have something called an effective nuclear charge. Now, since He has 1s2 and we added an extra shell, well the $1s2$ does not act much of a barrier and the nucleus is still able to attract the electrons in the second shell ($2s^2 2p^6$) electrons and the radii does not change drastically. But when you add a full shell on top of that, well then we have a problem. Because then the second shell electrons can act as a barrier in the electrostatic force, and the effective nuclear charge decreases and the repulsion between the electrons in the third shell will be greater, hence a greater atomic radii occurs between Ne and Ar, compared to He and Ne.

Now this trend will not be seen in the other ones. It is because the electron configuration is one thing but is not the only thing that is being shown on the graph. All of the other ones are ions. When something is an ion, for example the positive ones, then there are two things to look at, electron configuration and the charge. Meaning in a positive ion, you have more protons than electrons, meaning that the electrostatic force will much more greater in $Li^+$ than the one in $He$, even though they have the same electron configuration.

Whereas the trend in the noble gases is based on the fact that they all have $e^- = p^+$ and are neutral, whereas the others are not.

I hope I make sense here. But like I said, its just my theory for the question and I can be wrong :)


Volume of composition of air My teacher stated:

  At standard conditions, 1 liter of air at 21% oxygen possesses $\pu{0.21 L}$ of oxygen. Since at STP, 1 mole of gas occupies $\pu{22.4 L}$, simply divide $0.21/22.4$, to arrive at 0.0094 moles of oxygen.

I have a question regarding this. What if I trap $\pu{100 ml}$ of air in a vessel and maintain $\pu{273 K}$ and pressure $\pu{1 atm}$? So, in that case, what does it mean that oxygen has volume of $\pu{21 ml}$? Won’t the volume of oxygen and nitrogen be same inside vessel? Or am I wrong somehow?

If you took the 100 mL sample of air, and took out everything but the oxygen, then the amount of oxygen trapped would fill 21 mL at STP.

One way to think of it is, even though the oxygen would expand to fill the container, its pressure would need to reduce according to Boyle's law: P1V1 = P2V2. This is where the concept of Partial Pressure comes from. But if you then compressed that oxygen so that P = 1 atm (since you specified initial pressure to be 1 atm), the oxygen would take up 21 mL of volume.


Oxygen-based explosives I know that oxidanes are oxygen chains and rings. For example trioxidane is $\ce{HO-O-OH}$. I can see this turning into $\ce{O2}$ and $\ce{H2O}$ like hydrogen peroxide does. Since the double bonds in $\ce{O2}$ are strong do oxygen compounds without nitrogen ever explode?

They do indeed explode. Oxyliquits are probably the most comprehensive example of liquid-oxygen-based primary explosives (which are considered obsolete today). Theoretically liquid oxygen can also be substituted with ozone-oxygen mixture, which in reality has never been utilized due to technical difficulties -- oxyliquits were popular in the beginning of 20th century.


Lanthanoid Contraction Why is the radius of Europium so unusually high and out of the general trend ? Moreover, at different sources I am getting different values of radii. In some (e.g., Wikipedia) the radii are following the regular trend and in some Europium has been shown to be bigger. (Source : NCERT Class 12 Chemistry Part-1)

Well spotted! I dare say, the book appears to have used a covalent radius for Europium, while using atomic radii for the other lanthanides.

See "atomic radius" vs "covalent radius" values: for Europium, Samarium, and Gadolinium.

Also see this article if you don't like my referencing Wikipedia :P


Dissolving collagen in water: Why does the weighing scale reading fall? Every day I take 10g of hydrolysed collagen dissolved in about 200ml of water.

First I put the cup of water on my digital scale (which measures in 0.1g increments) then I add exactly 10g of collagen and stir it with a spoon until it dissolves.

As the collagen dissolves, the total weight always gradually decreases by about 1.5g - 2g in total. If the total weight of the cup, spoon and water is 300g, when I add the collagen, it becomes 310g, then as it dissolves over the next 15 seconds it decreases to about 308g. All I've done is stir it.

I can't see how this is due to evaporation since I can leave the pure water on the scale for 5 minutes beforehand and the weight doesn't change; but as soon as I stir the collagen in it always decreases.

What's happening?

Because collagen does not react with water and you say, no (measurable) evaporation of water is evident, there are only two options as to what is happening:

Matter is converted into energy:
If virtually no evaporation or spillage took place, and your scale is working properly, this is what must have happened. Using the formula of mass-energy equivalence $E=mc^2$ (E = energy, m = mass, c = speed of light in a vacuum) we can calculate the amount of energy that should result from a "loss" of 2 grams:
$$ E = 0.002~kg\cdot(2.998\cdot10^8~m/s)^2 $$
$$ E = 0.002~kg\cdot8.988\cdot10^{16}~m^2/s^2 $$
$$ E = 1.798\cdot10^{14} (kg~m^2)/s^2 = 1.798\cdot10^{14}~J$$
so... that would be a LOT of energy.
Your scale is not working as intended.

I'd go with option 2.

Now on a serious note: We do have a scale (precision of 0.0001 g) in our lab that occasionally has the same problems. Sometimes I would measure an empty reaction flask and it would be fine but some other time I could watch the measured weight drop by the second, "losing" up to 100 mg. The scale was checked by a professional but he could not find anything wrong with it.

My advice is: try another scale, see if the problem persists.


Can someone please explain to me where the flaw is in my proposed synthesis mechanism? For this I have to convert benzene into the product in the box using the reagents listed. The order of reagents that makes sense to me is IDGBEH. However, this is not the correct answer. Somewhere I am missing 2 steps to the synthesis and I am at a complete loss. I understand all of the basic and complex concepts of organic chemistry we've covered in my class thus far, so I'm confused as to how my proposed mechanism has an issue. If it's any help I know the mechanism is supposed to be 8 steps.

In my mind IDGBEH makes complete sense and should give me the product that I want, unless I am missing something or I misunderstood my professor.

The steps for conversion of Benzene to 1,3-dichloro-5-(propan-2-yl)benzene are shown below (Scheme).


AX Spin system ordering I would like to ask a question about the ordering of the $\ce{X}$ spins on the right hand side of an $\ce{AX2}$ energy diagram.

I was given the diagram here below:

and asked to evaluate why this diagram results in a triplet. This is simple, since there are 4 possible energy transitions, but one of them is equivalent to the singlet transition, hence only 3 transitions would appear in the NMR spectrum.

Prior to this though, I was confused because I thought the ordering for the bottom 3 spins was incorrect. Taking $\ce{\alpha}$ as the lower energy spin, I thought the bottom 3 spins should read as:

$$\begin{align}&\beta \beta \\
 \beta \alpha &~~~~ \alpha \beta \\
 &\alpha \alpha \end{align}$$

and the energy levels would be drawn more appropriately.
I attempted but failed to mention this to the lecturer during the presentation.

Am I correct in this instance?

There are a few things that seem to be wrong with the diagram you present, but since your focus is on the relative energies of the $\alpha$ and $\beta$ spin states I'll address only this. If you check a reliable source such as Cavanagh et al.s NMR textbook [ref 1] it is explained that

  the state with $m=+\frac{1}{2}$ is referred to as the $\alpha$ state, and the state with $m=-\frac{1}{2}$ is referred to as the $\beta$ state.  If $\gamma$ [the gyromagnetic ratio] is positive, then the $\alpha$ state has lower energy than the $\beta$ state.

Therefore for protons $\alpha$ is lower energy (since $\gamma>0$). If the nucleus is $\ce{^{15}N}$ ($\gamma<0$), $\beta$ becomes the lower energy state.

So while your concern is valid, it is a relatively minor flaw (in my opinion) with the diagram. Good for you for noticing it.

Reference

John Cavanagh, Nicholas Skelton, Wayne Fairbrother, Mark Rance, Arthur Palmer, III. Protein NMR Spectroscopy, Principles and Practice, 2nd Edition. Academic Press, 2006, ISBN: 9780121644918.


Do proteins bind to carbon? If so how? Can someone explain me whether a protein (like an antibody) would bind to carbon? The carbon arises due to electron beam induced deposition of Polyethylglycol (PEG). The carbon molecules that arise are either due to the PEG that dissociates or due to molecules, that are present in the microscope chamber as contamination, that fall apart. I am thus not sure about what the carbon molecule is specifically. You can look in this experiment to see more of the procedure.

And why? (what bond?)

I can't tell you for sure what is happening, but I can tell you what could be happening.
Physical Adsorption
It is possible that the proteins are not chemically adsorbed, but rather physically adsorbed. The article you cited implies that this is the case.
Many proteins in the presence of nano-scale hydrophobic "features" will partially denature, and the hydrophobic core will physically adsorb (attach via van der Waals forces). This effect is amplified if water is present, since the hydrophilic parts of the protein will be stabilized in water, allowing more of the hydrophobic parts to attach to the carbon feature.
Now you are probably wondering, "Why doesn't it attach to the PEG surface as well, since that is hydrophobic?" There are two possible reasons.
PEG is not as hydrophobic as you think

The PEG chains have oxygen in each monomer, which would reduce the non-polar character of the chains. It is possible that the difference is enough to make adsorption of non-polar amino acids on PEG thermodynamically unfavorable, while on pure-carbon structures it would be favorable.
Shape matters
Another possibility is that the actual shape of the features makes the difference. In order for a protein to denature and "attach" to carbon feature via hydrophobic interactions, it has to first expose the hydrophobic groups in the core. For a flat surface, the amount of initial denaturization (maybe not a real word) would need to be higher than for a "sharp" feature. Sort of like how a needle can slip into your skin without doing much damage. So, the smaller the characteristic length of the features, the more likely the protein would adsorb. There are also protein-protein effects on the surface, which are again reduced by smaller-scale features. In this scenario, you could think of the proteins as kinetically trapped in solution when near a regular PEG surface, while the activation energy for adsorption to a small feature is smaller.
I can think of two ways to test this idea:

Use different proteins on the same type of surface features - the ones with less exposed hydrophobic groups should adsorb less easily
Use different sizes of features - there should be a feature size that gives a maximum adsorption for a given protein.

Chemical Bonding
The other possibility is chemical bonding between the carbon features and the protein. It is unlikely that this will happen if the carbon features are really pure carbon (graphene, for example). However, if they are creating them by ionizing PEG with an electron beam, then it is highly likely that they are functionalized with something. Carbonyl groups would be my guess, but depending on the conditions there could be nitro groups as well. In this case, there is a possibility that hydrogen bonds would form, and if the surface is reactive enough, chemical bonds might form as well.
To test this idea, I would do the same thing as before - only this time, if we are right:

Proteins with more exposed hydrophobic groups should be less likely to adsorb
Feature size should not matter at all

If this turned out to be the case, you could then systematically test polypeptides made up of different amino acids to verify it and determine whether hydrogen bonding or covalent bonding was more likely.


What is the origin of the names P-1 and P-2 of nickel boride catalysts? From Chapter 2 of Catalytic Hydrogenation in Organic Syntheses:

  Highly stereospecific hydrogenations of acetylenes to cis-olefins can be achieved with a nickel (P2) catalyst, prepared by sodium borohydride reduction of nickel acetate in ethanol, and ethylenediamine as a promoter.

The Wikipedia article on nickel boride describes two forms, P-1 and P-2, differing in their ratio of $\ce{Ni2B}$ to $\ce{NaBO2}$.

What is the origin of the names "P1" and "P2"? Does it have to do with whether the catalysts add hydrogen in a syn or anti fashion?

The actual differences between the two forms are well-documented on Wikipedia. As for the names: as far as I can tell, the choice of "1" and "2" seems to be arbitrary. The choice of the letter "P", though, is described by C. A. Brown (who carried out investigations into the synthesis and reactivity of these catalysts).1,2 In footnote 13 of reference 1, Brown writes:

  The designations are derived from Purdue University, where this work was initiated.

So, it is within reason to assume that the P stands for Purdue.

References

Brown, C. A. Catalytic hydrogenation. V. Reaction of sodium borohydride with aqueous nickel salts. P-1 nickel boride, a convenient, highly active nickel hydrogenation catalyst. J. Org. Chem. 1970, 35 (6), 1900–1904. DOI: 10.1021/jo00831a039.
Brown, C. A.; Ahuja, V. K. Catalytic hydrogenation. VI. Reaction of sodium borohydride with nickel salts in ethanol solution. P-2 Nickel, a highly convenient, new, selective hydrogenation catalyst with great sensitivity to substrate structure. J. Org. Chem. 1973, 38 (12), 2226–2230. DOI: 10.1021/jo00952a024.


Reactions between Ca(OH)2 and CO2, and Sr(OH)2 and CO2 So I'm doing an investigation, and it involves using supersaturated solutions of Ca(OH)2 and of Sr(OH)2. I notice that there's always a flaky precipitate formed on top. I need to know about the reactions with carbon dioxide. Is it carbonate that's formed? Or is it hydrogen carbonate? Why?

Your solid is the respective carbonate. Group II bicarbonates can form aqueous solutions, but do not form stable solids, ergo the solid flakes must be carbonate and formed along the lines of:

$$\ce{Ca(OH)2 (aq)+ 2CO2 (aq/g) <=> Ca^2+ + 2(HCO3)- (aq) <=> CaCO3 (s) + CO2 (aq/g)}$$


What compounds are created by mixing Styrofoam and paint thinner? Yesterday I watched several videos that make a waterproof coat by disolving Styrofoam with paint thinner (my guess is that they use acetone). Will this recipe work for an emergency? Is it toxic (I don't want my pets to fell sick)?

Polystyrene will be dissolved in acetone. The long polymer molecules will dissociate from each other and float around in the acetone solvent; but the polymer chains will likely not be broken down into their small styrene monomer units. If you mix acetone and polystyrene ("styrofoam"), you will simply end up with polysterene in acetone. Not significant chemical reaction is expected to take place, only the physical process of dissolution/solvation/dissolving.

I would assume such a solution is toxic and keep it far away from animals, eyes and mouths.


The difference between peptide bonds and the bonds between polypeptides? I was doing some tests for the multiple-choice final we've got ahead. And it was on me to count the peptide bonds in an Insulin hormone with 51 aminoacids arranged in two polypeptides with 30 and 21 aminoacids. (these are not true in reality) the number of bonds were 49, not 50, and that means the bond between two polypeptides doesn't count as a peptide bond. Additionally, I know that polypeptide bonds make the proteins' molecular structure, as it is now. (just look at that shape.) PEPTIDE COVALENT BONDS CAN NEVER cause that kind of 3d orientation in space. So there must be some fundamental difference between those bonds. What is it?
 My research couldn't find any results as simple things have jammed the internet.

The term peptide bond refers to amide bonds. The 21 amino acids in insulin's A-chain are covalently linked by 20 amide bonds. The 30 amino acids in insulin's B-chain are linked by 29 amide bonds. The two polypeptides are covalently linked by disulfide bonds, which are not amide bonds.

The 3D structure of proteins is a complex topic and a field of intense research. It's worth pointing out though that even polypeptides (with no disulfides or other covalent linkages) have a more complex 3D structure than you might expect. I suggest you look into alpha-helices and beta-sheets as an entry into the topic.

Took some info from the Beta Cell Consortium.


My text seems to say that the high(est) energy orbitals are not valence orbitals - everything else says otherwise. What's true? A paragraph in my text reads

  Bonding involves the valence orbitals almost exclusively because these orbitals have the appropriate energies to interact strongly. Examine the electron energy-level diagram of a fluorine atom. The core $1 \mathrm{s}$ orbitals on a fluorine atom are at much lower energies than any of the $n = 2$ orbitals and do not participate effectively in bonding with other atoms. On the other hand, orbitals with $n > 2$ lie at considerably higher energy than the $n = 2$ orbitals. These orbitals can interact, but the resulting orbitals are too unstable to form strong covalent bonds. Therefore, the only orbitals of fluorine that form chemical bonds are the valence orbitals, those with $n = 2$. Typically the highest energy occupied orbitals are the ones used to bond with other atoms.

In contrast this simple paragraph from SparkNotes expresses what I had believed to be the case

  The outermost orbital shell of an atom is called its valence shell, and the electrons in the valence shell are valence electrons. Valence electrons are the highest energy electrons in an atom and are therefore the most reactive.

I had thought that the outermost shell, the valence shell, the most energetic shell, and the shell involved in molecular bonds were all the same thing. However, the quoted paragraph of my text says otherwise. Where have I gone wrong?

The problem is the words used to describe energy. I am not native english speaker so I am not the best person to answer your question but I'll try.

As smaller $n$, smaller energy. Problems arises when the words higher lower are used to compare numeric values (as those of the energy). They must say greater or smaller.

Without accurate words all of us intuitively agree in that:
"2 is higher that 1", and we should say that 2 is greater than 1 (2>1)

When negative numbers are involved it is not so clear. For some authors "-2 is higher than -1" and "-2 is higher than 1", for others "-2 is lower than -1" and "-2 is lower than 1". In the later cases the correct word is "smaller" (-2<-1).

In this context energies are negatives because of the definition of the reference states. So the energy of an $1s$ orbital is smaller than the energy of a $2s$ orbital.

I have some discrepancies about the meaning and usage of orbital energies but I adopted the common usage to avoid extra confusions.


What is the volume of the produced SO2 gas at STP? In a electricity producing plant $3.1\times10^7\ \mathrm{kg}$ coal with $2.4\ \%$ rhombic sulfur is burnt every year. What is the volume of the produced $\ce{SO2}$ gas at STP?

My effort  for solving this answer :

In one day how much sulfur is burnt. Then I wrote the equation

$$\ce{S + O2 -> SO2}$$

With the unitary method I tried but I am confused from where to start, how to start? Please give me a hint how to start the answer.

You have an excess of oxygen, right...?

So the limiting reactant is the sulfur, which is $2.4$% of the coal.

Spoiler:
Step 1:

   So calculate the sulfur content, which is $3.1 \times 10^7$ kilograms multiplied by $2.4 \times 10^{-2}$ which is $7.44 \times 10^5$ kilograms of sulfur.

Step 2:

   The $\ce{SO_2}$ is $50$% sulfur by mass (32 amu of sulfur vs. 64 amu total) so there is $1.488 \times 10^6$ kilograms of $\ce{SO_2}$ produced.

Step 3 (for moles, optional):

   If you want moles, divide by $64$, which is $2.325 \times 10^4$ moles (someone check my math)

Then again, you emphasized STP, so proceed with caution (sulfur doesn't react at $20$ degrees Celsius).


How does a lone pair of a central atom affect the dipole moment? Dipole moment is the degree of polarity, i.e. the seperation of positive and negative charges. But I am not getting the intuition why and how lone pairs affect the polarity and dipole moment. I cannot connect the  separation of positive and negative charges  definition with lone pair.

How does lone pair of a central atom affect the dipole moment?

There is no single answer to your question, let me explain.  Unlike a typical covalent bond where the electrons are shared between two nuclei and the electron density is spread out over the entire bond, in a lone pair the electrons are not shared and the electron density is more localized around the atom that has the lone pair of electrons.  This increased electron density could lead to a more significant contribution from the lone pair electrons to the molecular dipole moment than from electrons spread out more diffusely in a covalent bond.  Next we must understand the directionality of the lone pair of electrons.  Consider the two molecules pictured below, ammonia and phosphine.  The molecules appear to be very similar, they are in the same column in the Periodic Table.

However in ammonia the $\ce{H-N-H}$ angle is around 107 degrees and the molecule is roughly $\ce{sp^3}$ hybridized, the lone pair and the 3 $\ce{N-H}$ bonds roughly pointing towards the corners of a tetrahedron.  You can see that in this case (as shown by the arrows, the "arrowhead" end representing the negative end of a dipole), the lone pair on nitrogen will make a contribution to the molecular dipole moment.  Next, let's examine phosphine.  The $\ce{H-P-H}$ angle is around 90 degrees and the molecule can be viewed as being unhybridized, the lone pair is an $\ce{s}$ orbital and the 3 $\ce{P-H}$ bonds are constructed from phosphorous $\ce{p}$ orbitals.  You can see that in this case, the lone pair on phosphorous, due to its spherical symmetry will not make a contribution to the overall molecular dipole moment.

So in summary, a lone pair of electrons can make a significant contribution to the magnitude of a molecular dipole moment due to the fact that they are more localized than bonding electrons and consequently there is a high electron density.  But, the directionality (or lack thereof) of the lone pair must also be assessed, since a lack of directionality may preclude it from making a significant contribution to the overall molecular dipole moment.


What is the hybridization of oxygen in furan? I fail to understand why the hybridization of oxygen in furan is $\ce{sp^2}$. I think that since there are 2 bond pairs and two lone pairs so it should be $\ce{sp^3}$ hybridized?

Is it to do with the aromaticity of furan?

Furan is aromatic because one of the lone pairs of electrons on the oxygen atom is delocalized into the ring, creating a $4n+2$ Hückel aromatic system.

The steric number is sometimes incorrectly described as the sum of the number of sigma bonds and the number of lone pairs. However, it is more accurately given by the number of sigma bonds plus the number of localised lone pairs.


Alphabetical order in nomenclature Shouldn't 4-isopropyl-2-cyclohexyl-3,5-dimethylheptane be written as 2-cyclohexyl-3,5-dimethyl-4-isopropylheptane, by following alphabetical order?

Or maybe you don't look at the cyclo or the di prefix when you look for alphabetical order.

Both names that are proposed in the question, ‘4-isopropyl-2-cyclohexyl-3,5-dimethylheptane’ and ‘2-cyclohexyl-3,5-dimethyl-4-isopropylheptane’, are unambiguous and describe the correct structure. However, they are not in accordance with IUPAC recommendations.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, cyclohexyl, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. For example, ‘3,5-dimethyl’ is considered to begin with ‘m’. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.

  P-14.5 ALPHANUMERICAL ORDER

  Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved

  Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.

  (…)

  P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order for the example given in the question corresponds to the name 2-cyclohexyl-4-isopropyl-3,5-dimethylheptane since ‘cyclohexyl’ begins with ‘c’, isopropyl begins with ‘i’, and methyl begins with ‘m’.

However, note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred prefix is ‘propan-2-yl’. When using this prefix, the correct alphanumerical order  corresponds to the name 2-cyclohexyl-3,5-dimethyl-4-(propan-2-yl)heptane.

Anyway, to construct a preferred IUPAC name (PIN), the ring is selected as the parent hydride. Therefore, the PIN for the example given in the question is [3,5-dimethyl-4-(propan-2-yl)heptan-2-yl]cyclohexane. (In general nomenclature, the context may favour the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.)


How does one analytically quantify whether or not a compound is a SIRT1 activator? So there has been a lot of noise over whether or not certain molecules activate SIRT1. Resveratrol was thought to be a specific SIRT1 activator, but many have argued against this claim.

Similarly, this has also happened to SRT1720 too.

My question is - why is it so hard to quantify whether or not a molecule is an SIRT1 activator? How would one go about doing this?

It's important to distinguish SIRT1, which is a gene, from the protein it encodes, sirtuin 1 (some papers use SIRT1 for the protein too, but to avoid confusion, I'll use Sirt1). When looking at compounds that affect gene expression, one method of measuring the effects is measuring the change in mRNA levels for that gene when the compound is administered, usually with northern blotting or qPCR. In this case, because the gene encodes a protein, quantifying the protein levels also tells us about the expression of the gene.

The SIRT1 gene doesn't have any regulatory role itself, so in this case, people are more interested in the Sirt1 protein. While we can measure protein concentrations, Sirt1 is an enzyme that deacetylates proteins and resveratrol has been proposed to directly alter the activity of the enzyme (it's also known to increase SIRT1 expression, but this isn't the controversial aspect), so it's important to examine the effects of resveratrol on Sirt1's ability to deacetylate certain proteins. The way this is done is to expose various proteins/peptides to Sirt1 and look at whether they deacetylate. One method of measuring this is to use a special peptide as the substrate to which a fluorophore can be attached if the acetyl group is cleaved.

The reason resveratrol's effects have been debated is that the early studies that showed large activation effects were later found to only occur because of the fluorophore on the substrate and not resveratrol alone. Some work since then using other detection methods have shown activation effects: This paper tried many different substrates on a microarray using fluorescently labelled antibodies as well as kinetic measurements by mass spectrometry and found that resveratrol increased Sirt1's activity for some substrates, decreased it for others, and left many the same, though they didn't offer much in the way of structural insights that might be responsible. I haven't done a thorough literature search, but it seems like the debate is less contentious now that people are using other assays to examine activity.

Just to summarize your questions, it's not hard to quantify whether a molecule increases the expression of SIRT1, but because Sirt1 is an enzyme, any kind of label you add to measure activity could have its own effects on that activity. Any detection method that can measure the deacetylation of substrates could potentially be used for studying Sirt1 activity, and there are many, but label-free techniques like mass spectrometry are especially useful because they reduce complications in simulating biological conditions.


What is the use of silver ion? I was going through the mechanism of Arndt-Eistert synthesis and got stuck in the step where silver ion is used in the rearrangement step.
If anyone can clarify the use of silver ion?
It would be a great help!

The step in question here is the so-called Wolff Rearrangement.

In this rearrangement, silver in the form of for example $\ce{Ag2O}$ is used as a catalyst. A catalyst facilitates a reaction (lowering the activation energy), but it comes out of the reaction unchanged (which doesn't mean it's not changed during the reaction, it's just changed back at the end ;))

So why silver? Well, transition metals are good catalysts most of the time for different reasons:

They have vacant d-orbitals which leads to
An easy tendency to exhibit variable oxidation states.
They can form reaction intermediates with reactants.
The presence of defects in their crystal lattices.

(I conveniently copied this list from a quora post, because im lazy)

What this essentially means is that due to their electronic structure, transition metals are often able to bind reactants to their surface and activate them for example. There are quite some catalyst reactions that we don't even understand today, but sometimes it's as easy as this

And why silver? Well probably Wolff just tried a couple of transition metal catalysts and chose the one with the best effect/cost ratio! Theoretical expectations don't always meet reality so often you just have to try and find the right reactants/catalysts.


Why does steel wool catch fire when rubbed with a battery? I was watching an episode of Man vs Wild where steel wool was rubbed with phone battery. To my surprise, the wool caught fire. What is the reason for this?

Steel wool is a mat of very fine steel fibres. Because of their small size, it's easy to heat them up quickly to the point of burning and their relatively large surface area to volume ratio lets them oxidize quickly. If you try it with a flame, you'll find that it's quite easy to burn steel wool. When you short the terminals a battery across steel wool, a lot of current passes producing heat (think about incandescent light bulbs), and especially if you touch the battery such that only a few fibres are touching, enough heat is produced to cause them to burn.


why disproportionation of free radicals does not happen in kolbe's electrolysis if we use secondary or tertiary carboxylic salt as substrate? In notes given by chemistry teacher, for secondary carboxylic salts the final products are alkane and ester.

but i guess, as the reaction intermediates involve secondary/tertiary free radicals, some radicals should also go under disproportionation reaction and produce alkene.

why this doesn't happen?

Because of the free-radical-scavenging activity of the secondary/tertiary  carboxylic salts?

Hence, radicals that could possibly undergo disproportionation reactions resulting in alkene are removed.

As a possible example, see 'Evaluation of the Free-Radical-Scavenging Activity of Diclofenac Acid on the Free-Radical-Induced Haemolysis of Human Erythrocytes'.


What is the mechanism for the reaction of an organic hydroperoxide with sulfuric acid? The question asked to react the organic hydroperoxide substrate, $\ce{R-OOH}$, with dilute sulfuric acid, $\ce{H2SO4}$.

In the last step, I removed a proton, making it act like a catalyst, along with the removal of $\ce{H2O2}$. But I am very unsure about its correctness.

Please confirm or correct if this is a wrong answer.

As Marko has commented it is like the Hock rearrangement, which is the migratory step in the cumene process.

It is somewhat similar to the Beckmann rearrangement except that there is an oxygen instead of a nitrogen.

The acid protonates the peroxide, generating a great leaving group on electron-deficient oxygen.
Migration towards electron-deficient oxygen occurs with loss of water to give an oxonium ion.
The migratory aptitude here is determined in the same way as for the Baeyer-Villiger reaction. There is a buildup of positive charge on the migrating carbon during the migratory step, and therefore the group that can best stabilise the charge preferentially migrates. The methoxy-substituted ring is electron-rich and the positive charge is more stabilised on it than on the unsubstituted phenyl group. The methyl group is even worse at stabilising positive charge.
Under the dilute acid conditions this oxonium ion is hydrolysed.


Why heat of reaction under neither constant pressure or constant volume is considered (in my textbook, or probably somewhere else)? So my textbook gives me this random equation:
 $\Delta U = q + w$. Then consider two conditions, under constant pressure and under constant volume. Because $w = -P\Delta V$, constant volume will make $\Delta U = q$. And constant pressure equation is basically the same.

But why is isn't heat under neither condition considered? Because it is of significance?

The constant pressure and constant volume scenarios are presented when you first study thermodynamics because they present an easy way to simplify the calculation of energy changes ($\Delta U,\ \Delta H,$ etc.) based on the change of one easy to measure phenomenon.

  Internal Energy $U$

So, the change in internal energy must equal the sum of the work done $w$ and the heat transferred $q$. This is not random, it is a consequence of the Law of Conservation of Energy. If the internal energy of a system has changed then energy must be exchanged with the surroundings. There are only a few ways this can happen (heat, work, electromagnetic radiation, ...), and heat and work are the most important for most thermodynamic scenarios.

Thus, $\Delta U = q + w$. We need to be able to relate $q$ and $w$ to things we can measure. We calculate $q$ based on the temperature change of the system $\Delta T$ and the heat capacity of the system $c$. Thus, $q=c\Delta T$. Work is defined (in one way) as the change in volume of the system times the pressure. The negative sign is present to make sure that an expanding system is doing work on the surroundings (and thus losing energy): $w=-P\Delta V$.

Now we have $\Delta U = c\Delta T -P\Delta V$

  Constant Volume

If volume is constant $\Delta V = 0$, which means that $w=0$, regardless of pressure. Thus, $\Delta U = q = c\Delta T$.

  Constant Pressure

This one is trickier, but we can do it for a gaseous system.

We do not need to measure volume, since $PV=nRT$, but it is easier to measure volume than $n$, which may also be changing. If $n$ is constant, then, the equation for change in internal energy becomes:

$$\Delta U = c\Delta T - nR\Delta T$$

If $n$ is not constant, then we need to measure that change in volume and temperature.

If the process is in another state, like in solution, $\Delta V$ might be very challenging to measure. Let's hope we can measure $\Delta n$.

  Enthalpy $H$

Because of the challenge of measuring volume changes in solution, we invented a new thermodynamic potential $H = U + PV$. Total differentiation and whatnot gives us:
$$\Delta H = \Delta U +P\Delta V + V\Delta P = c\Delta T - p\Delta V + P\Delta V + V\Delta P$$
$$\Delta H = c\Delta T + V\Delta P$$

Thus, at constant pressure, $\Delta H = q = c\Delta T$, which is easy to measure. Enthalpy is similarly challenging at constant volume as internal energy is at constant pressure.

  Constant pressure and volume.

Under these conditions, $\Delta U = \Delta H = q = c\Delta T$.

  Pressure and volume are not constant

Okay, so if $w=-P\Delta V$, how do we calculate work? If pressure is changing, which value of $P$ do we use? It turns out we use all of them and we need some calculus to do it. This case is often left out because many introductory textbooks consider the likelihood that many students beginning their study of chemistry may not have taken calculus.

Here we go. We need to replace the total change in $V$ with the infinitessimal change in $V$ at each moment $\Delta V \rightarrow  dV$. $\Delta V$ is the sum of all of the infitinessimal $dV$ values. Now, we can multiply each $dV$ with the pressure at that moment and add them up to get work (which involves integration):

$$w = \sum_{j=1}^n P_j dV_j=\int_{V_i}^{V_f}PdV$$

However, we cannot leave well enough alone, because $P$ is changing, so we need to substitute for $P=\frac{nRT}{V}$.

$$w=\int_{V_i}^{V_f} \dfrac{nRT}{V}dV=nRT\dfrac{\ln V_f}{\ln V_i}$$

Now the above maths only work if temperature is constant. What if it's not?


Comparing the Lattice energy of different compounds I am confused about some concepts related to Lattice energy.

First of all,how to compare lattice energy between $2$ compounds belonging to different groups and periods?

1.Between $\ce{AlF3}$ and $\ce{MgO}$ (In $\ce{AlF3}$ product of ionic charge is $3\times1$ and in MgO it is $2\times 2$. So,according to me $\ce{MgO}$ should have higher lattice energy but the opposite is true.

2.Between $\ce{KF}$ and $\ce{LiF}$ (According to Fajan's rule $\ce{KF}$ is more ionic than $\ce{LiF}$ and lattice energy depends directly on ionic character).But, value of lattice energy of $\ce{KF}$ is $\pu{808 kJ mol-1}$ and $\ce{LiF}$ is $\pu{1030 kJ mol-1}$. How is this possible?

So,can you please explain me how to compare lattice energies of different compounds and which factor is more dominating size of cation and anion or charge on ions.

You might want to look at

https://scilearn.sydney.edu.au/fychemistry/prelab/e12.shtml

and the related page

https://scilearn.sydney.edu.au/fychemistry/calculators/lattice_energy.shtml

which has a nice lattice energy calculator that allows you to play with the parameters and see how the lattice energy varies, but I'll try to summarise those the argument below.

Here we are dealing with the ionic model - everything is totally ionic, there is total charge separation, all binding is electrostatic. Thus our system consists of a set of point charges. We can write the energy of such a system as

$$
E=N_AM \frac{z_1 z_2 e^2}{4 \pi \epsilon_0 r}
$$

where $N_A$ is the Avogardro constant, $z_1$ the charge on the first ion, $z_2$ the charge on the second, r is the closest separation of the ions and M the Madelung constant. Now note the Madelung constant only depends upon the structure of the crystal - all Sodium Chloride type crystals, whether it be NaCl, KF, or FeO, all use the same value off M. Looking at the formula we can see that FOR A GIVEN STRUCTURE (i.e. with the same M)

High charges on the ions mean high lattice energy
Small separation means high lattice energy

Thus point 2 addresses point 2 in your question, K is bigger than Li, hence the separation is bigger in KF than LiF, hence KF has a lower lattice energy than LiF. The first point explains why MgO has a higher lattice energy than NaF. However it doesn't cover your first example, as $\ce{AlF3}$ and MgO have different structures (and actually I don't think it is a very good question).

Comparing different structures is very difficult. About the best we can do is the electrostatic term from the Kapustinskii equation (https://en.wikipedia.org/wiki/Kapustinskii_equation)
$$
U_{L}={K}\cdot {\frac {\nu \cdot |z^{+}|\cdot |z^{-}|}{r^{+}+r^{-}}}
$$
This is an approximation to the above. Here K is a constant which is INDEPENDENT of structure and $\nu$ is the number of ions in the empirical formula, 2 for MgO, 4 for $\ce{AlF_3}$. Thus assuming ${r^{+}+r^{-}}$ is the same for both of these $\ce{AlF3}$ has a bigger lattice energy because $\nu\cdot |z^{+}|\cdot |z^{-}|$ is bigger - For $\ce{AlF3}$ it is 4*3*1=12, while for MgO it is 2*2*2=8. What this is really saying is that the $\ce{AlF3}$ structure has a much larger Madelung constant than the NaCl one, enough to overcome the charge differences, and thus it has a higher lattice energy.


How to determine the order of elution in silica gel column chromatography? Benzofuranoids ustusoranes A (1), B (2), and E (3) were isolated from Aspergillus ustus 094102. If these compounds were separated by column chromatography on silica gel, what would be the elution order? (top — comes off the column first, bottom — comes off last).

As I get it, column chromatography is like TLC where in a silica gel the nonpolar compound comes out first.

Your conclusion is absolutely correct. Column chromatography can be thought of as three-dimensional version of TLC (and vice-versa). So the most polar compound which interacts with silica gel most elutes slowest and the least polar compound leaves the column first.

I also support your suggested order as an initial starting point. However, As Klaus mentioned in a comment, unless one has performed the same separation multiple times one should never rely on paper-chemical polarity analysis alone and always confirm structures independently — typically that means ‘measure an NMR of all fractions’.


Conductivity of water and chlorine? I am a novice in chemistry and i have a question: I know that water is not a good conductive material (since there is a perfect bond in water I think) but I am thinking that if we add chlorine which has 7 valence electrons this means that there is a hole and this increases conductivity... am I right?

In a nutshell, water is very much not like semiconductors. Its conductivity is due to altogether different mechanism, and follows different trends.

Typical semiconductor is a rigid network of atoms held together with bonds. Nearly any "wrong" atom would provide either extra electrons or holes (that is, missing electrons). Holes and electrons are the charge carriers here. Being very small, they move fast, so you only need a tiny amount of those impurity atoms (say, a few ppm) to create significant conductivity.

Water is quite different. You can't have free electrons in water, to begin with; they will instantly react with something and become ions. Ditto for holes. Ions are the only allowed charge carriers in water. Ions are as big as atoms, and each carries a thick coat of water molecules stuck to it, so effectively they are even bigger than atoms. An ion would move in an electric field, but with great difficulty, much like a man pushing his way through a busy crowd. (Compared to it, the electrons would be like little sparrows darting overhead.) Therefore you need a lot of ions to create some real conductivity.

Things that produce ions in water (dissociate, as we say) are called electrolytes, and there is plenty of them around. For example, table salt dissociates into $\ce{Na+}$ and $\ce{Cl-}$; sulphuric acid dissociates into $\ce{H+}$ and $\ce{SO4^2-}$. Elemental chlorine ($\ce{Cl2}$), being a neutral molecule, does not dissociate; also, you can't add much of it, for it is a gas, and its solubility is not that great. Over time, it would react with water to produce certain compounds which do dissociate, so you'll have some conductivity after all, but that's a weird and inefficient way of achieving it. Using any soluble salt is much more straightforward.

The choice of particular electrolyte depends on your ultimate purpose.


Meaning of an ideal chemical system What is meant by a an ideal chemical system?
Moreover, is it true that only in an ideal chemical system the equation

$$\mathrm{d}U = \mathrm{d}H - \mathrm{d}(pV),$$

i.e.

$$\Delta U = \Delta H - \Delta (pV)$$

can be written as

$$\Delta U = ΔH - pΔV?$$

If so, why?

I also presume that non-ideal chemical systems do not maintain a mechanical equilibrium. If I am wrong, please correct me.

For myself, ideal means it is composed of ideal gas particles. This means that the ideal gas law applies. This means that there is no accounting for particles interacting which is a grave mistake for liquids and solids, and often for gases too.
$dU = dH - d(pV)$ is always true. In an real system however, the variables involved may take on different quantitative values for a given state point, thus the final change in eternal energy $\Delta U$ may be different.

In general $\Delta U = \Delta  H -V \Delta p - p \Delta V - \Delta p \Delta V$

If pressure is constant during the change then $\Delta p \Delta V = V \Delta p = 0$ which then leads to the term

$$\Delta U = \Delta H - p \Delta V $$

If volume is constant during the change then $\Delta p \Delta V = p \Delta V = 0$ which would then lead to

$$\Delta U = \Delta H - V \Delta p $$

If neither volume or pressure is constant the simplest form is

$$\Delta U = \Delta  H -V \Delta p - p \Delta V - \Delta p \Delta V$$

a.k.a.

$$\Delta U = \Delta H - (p_f V_f - p_i V_i) $$

see Andrew's answer for more details on the last step.

And I do correct you, non-ideal systems maintain equilibrium just as ideal systems do. If they didn't, we wouldn't be able to use thermodynamics on them since thermo only applies to equilibrium.


What's the difference between atomic orbitals and energy levels? If I'm drawing a Bohr model of Neon, the middle is the nucleus, the first energy level contains 2  electrons. The second level contains 8 electrons. 8+2= 10, Neon's atomic number. But then what are the atomic orbitals, like s, p, d, and f. Where do they come in? Are they completely different from the energy level?

Neon's atomic number. But then what are the atomic orbitals, like s,
  p, d, and f. Where do they come in? Are they completely different from
  the energy level?

Atomic theory was developed in stages almost a century ago. Bohr's model is the simplest version of hydrogen atom. General chemistry textbooks write rarely mention the Sommerfeld's model from which the concept of orbital angular momentum originated on elliptical paths (recall that a circle of Bohr's model is a special ellipse). Sommerfeld model model explained more features of the hydrogen spectrum. After Sommerfeld the next person was Schrodinger who proposed the H atom in terms of probabilities of finding electrons rather than fixed orbits.

In short, s, p, d, f come into play in more advanced models of Sommerfeld. Bohr was the first one to talk about electronic configurations for multielectron atoms.

Sommerfeld' model


Electrolysis of Anhydrous Sodium Hypochlorite Considering liquid anhydrous sodium hypochlorite:

$$\ce{NaClO (l)}$$

Assuming the ions are free, What would be produced if this were electrolysed?

As I've found lots of information for it's hydrated pair (a.k.a. bleach). And while I know the half equation for sodium:

$$\ce{Na+  + e- -> Na}$$

I cannot find any half equation for the hypochlorite ion: $\ce{ClO-}$, apart from in acid, or even what would be the product (maybe $\ce{Cl2O}$ or $\ce{Cl2O2}$?)

Obviously, sodium does not have a choice other than getting reduced. Oxidising it to sodium(II) is not an easy feat considering that the electron would have to be extracted from a core orbital. That means, the hypochlorite must be oxidised.

Usually, an oxidation of hypochlorite will yield chlorate, $\ce{ClO3-}$. If we attempt to form a redox half-reaction out of that, we can only use $\ce{ClO-}$ as a charge balancing agent:

$$\begin{align}\ce{ClO- \phantom{\ce{+ 4 ClO-}} &-> ClO3- + 4e-}\tag{Ox1}\\
\ce{ClO- + 4 ClO- &-> ClO3- + 4 e-}\tag{Ox2}\end{align}$$

If we do this, we have four extraneous chlorine atoms and two extraneous oxygen atoms. Thus, it makes most sense to do the mass balance by adding $\ce{Cl2O}$:

$$\ce{ClO- + 4 ClO- -> ClO3- + 4 e- + 2 Cl2O}\tag{Ox3}$$

However, it will be difficult confirming this. Sodium hypochlorite is not stable, especially not at elevated temperatures, especially not in pure form, especially not anhydrous sodium hypochlorite (sensing a pattern?). You would need a certain amount of recklessness to attempt this and sufficient luck to be able to report the result. Which is possibly why I cannot find any scientific publication on the topic; so take the equation with two grains (or more) of salt.


Why are alpha carbonyl halides most reactive towards SN2 reactions? Relative rates of some compounds towards SN2 is given below

The reason for reactivity of alpha carbonyl  halides is given as

It’s not clear what exactly is happening.

I have asked a question previously which concerns the same electrophilic system with the exception that nucleophilic attack happens on the carbonyl carbon and not on the α-carbon.

From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891:

The $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But, if $\ce{X}$ is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the $\ce{C-X}$ and $\ce{C=O}$ bonds are perpendicular so that the orbitals align correctly.

This interaction when considered in the context of an attack on the carbonyl is the principle of the polar Felkin-Anh model used to describe a certain diastereoselectivity. However it can likewise be used to explain the lowering of the $\unicode{x3c3}^*(\ce{C-Br})$ orbital which is attacked in a nucleophilic substitution reaction. It may seem as though these statements are contradictory, but the orbital interaction will create a three-centre π-type double-antibonding orbital with only the contribution on oxygen being neglegible due to oxygen’s electronegativity being highest and thus its contribution to high-energy orbitals being lowest.
It will depend on your nucleophile what will actually happen. It can attack both on the carbonyl carbon and on the α-carbon. If the attack on the carbonyl carbon is reversible, only the attack on the α carbon leading to an observed $\mathrm{S_N2}$ process will be observed after the reaction — even if the carbonyl attack is faster. Only if the attack on the carbonyl is not reversible as is the case e.g. with Grignard reagents, a mixture of two products will be observed.
Naturally, if $\ce{X}$ is a very poor leaving group — e.g. $\ce{OTBS}$ — the attack on the carbonyl will dominate this way or that.

To sum up:
The reaction rate is increased because the $\unicode{x3c3}^*(\ce{C-X})$ orbital and the $\unicode{x3c0}^*(\ce{C=O})$ orbital linear combine to give a lower-energy orbital. Since this is the LUMO, the LUMO energy is lowered and nucleophilic attacks are facilitated.


Oxidation of Sodium Hydroxide in Ethanol? So I read recently that:

  Alcoholic solutions of sodium hydroxide will oxidize in air, turning brown.

First is this true?

And if so what is oxidizing?

I can't think of what sodium hydroxide would oxidize to, and ethanol well wouldn't that oxidize to water and CO2!

[Answer based on comments from @KemonoChen, @IvanNeretin and @NilayGhosh. Thanks for your input :D]

  In the presence of alkali (or acid, for that matter) aldehyde will
  turn against itself, creating all matters of ugly polycondensate
  products. See Aldol
  condensation and
  note that it ends up in another aldehyde, so the process can and will
  be repeated

So yes it can happen.

And is is most likely ethanol oxidation to an aldehyde, which has (aldol) condensed into an 'ugly polycondensate'

It may also be Sodium Alkoxide, which is known to turn brown.


What is dielectric constant and how does it explain the reason why water is one of the best solvents When i was reading solublity of ionic compounds,they mentioned that polar compunds are fairly soluble in polar solvents and water is better solvent due its high dielectric constant.Can anybody explain what is it and how does it affect solublity.

The dielectric constant $\epsilon$ (known also as permittivity) is a measure of the extent to which a substance is polarized under an applied (external) electric field. Polarization amounts to net separation of charge across the substance. A "dielectric" is the common name used to refer to a substance placed between the plates of a capacitor and used to store electrical energy in the form of polarization, that is, as charges separated across the material. A capacitor is the non-chemical analogue of a voltaic cell, that is, a form of storing electrical energy.

A substance with a high dielectric constant is easily polarized. Polarization allows countercharges to be placed around an ion, resulting in Coulombic interactions between solvent and ion, promoting solubilization of the ion by competing with interionic interactions. Similarly, a polar solvent - one with a high dielectric constant (relative permittivity $\epsilon_r>\approx15$) - will solubilize polar molecules by forming stabilizing interactions with the solute that compete with solute-solute interactions.

Water is a good solvent for polar molecules for the above-mentioned reasons. Water is sometimes referred to as the "universal" solvent, but it is generally a good solvent for polar solutes, not apolar ones.


In Chemistry, is it possible to use the old representation of the atom as a nucleus surrounded by orbiting particles? In Physics, Newton's laws are enough for many applications. Sometimes, relativity must be used instead or complementary, and sometimes, Quantum Mechanics.

Is this the same case with Chemistry? Are there areas of Chemistry where the simplification of electrons as little particles orbiting the nucleus at distances proportional to their energy level is possible? Or does all of modern Chemistry must use Quantum Mechanics? At what point Chemistry "breaks" if the particles and orbits model is used?

When the fixed-orbit particle model is taught in general chemistry classes, it is usually presented as Bohr's fixed circular orbit model.

However, this was not the culmination of the deterministic-orbit particle model.  In 1916 Sommerfeld expanded the model to include precessing elliptical orbits, and made the theory relativistic. The varying of the eccenticity of the ellipses permits multiple angular momentum values for a given energy.

The Sommerfeld relativistic model correctly predicts the fine structure of the hydrogen atom.

In this one aspect, the deterministic-orbit particle model is superior to the non-relativistic Schrodinger model.  The Sommerfeld model correctly predicts the fine structure, whereas the Schrodinger model does not.


Why is phosphorus pentachloride not used in the synthesis of chlorophenols? The best method (as I have read) for preparation of chlorobenzene is by reacting it with $\ce{Cl2}$/$\ce{FeCl3}$. Why its not the best practice to react phenol with $\ce{PCl5}$ ?

Well you see, phenol do not react with PCl5 to give chlorobenzene as it undergoes EAS mechanism.In order to make the reaction possible you will have to activate phenol with nitro group.


Thermodynamics of DNA double helix formation? The entropy of a system decreases when 2 single stranded DNA molecules come together and form a double stranded molecule. For this to occur the total entropy of the universe must increase. Thus, heat is released when the double stranded molecule forms. My question is: Where does this heat come from?

The entropy of a system decreases when 2 single stranded DNA molecules come together and form a double stranded molecule.

Are you sure?

I would argue that association of two single-stranded DNA strands to form a double-strand is energetically favourable — for entropic reasons. This should be a combination of

the hydrophobic effect
the liberation of a few water molecules per base that are no longer needed in hydrogen bonding.

Point 2 is explained quickly. Upon association of two single strands, the characteristic Watson-Crick base pairs form and hydrogen bonds are established between the DNA bases. These hydrogen-bond donors and acceptors were previously associated with water molecules keeping them close to the DNA single strands and thus rather ordered. These water molecules are now liberated; and since there are at least two per base, this partial process is entropically favourable.

Point 1 may not be as clear as point 2, but it again boils down to the fact that water molecules are liberated from a rather well-defined structure. Large parts of DNA — the area around the bases — are not as hydrophilic as other parts of the molecule or water itself. While they still are somewhat polar, their interaction with water is not ideal. Thus, the ‘surface’ water molecules will form a kind of defined boundary around the molecule. This boundary becomes lesser overall if the surface to volume ratio decreases. And upon association, you have almost double the volume in practically the same surface (i.e. you have halfed your effective surface). Therefore again, a large number of water molecules are liberated.

If DNA association is entropically favoured, there is no need to invoke any heat.


Calculating heat capacity of a calorimeter I am struggling on this question :

The heat of neutralization of $\ce{NaOH}$ and $\ce{HCl}$ is $-57.1\:\mathrm{kJ}$, calculate the numbers of moles of $\ce{HCl}$ and if $x$ moles of $\ce{HCl}$ were used how many $\:\mathrm{kJ}$ would be given out if one mole results in $57.1 \:\mathrm{kJ}$

I already got the moles as $0.05$ by doing $n=\frac{v}{1000}  c$ where I had $50\:\mathrm{mL}$ of $2\:\mathrm{M}$ $\ce{HCl}$

First, is this a question from your book, and do you have the correct answer to check your work.

Do you know how to use dimensional analysis?

Start with 50 mL. You have 2 Molarity of some subtance, how do you write that as a quotient and what are the units of the denominator and numerator? You will also need to know how many mL are in a liter and you should get the correct moles of that substance. It is very important that you write this down on paper correctly and make sure your units will cancel leaving only moles. It is hard to be wrong when your units cancel correctly.

Do you mean the heat of neutralization is $-51.7\ \frac{\mathrm{kJ}}{\mathrm{mol}}$

The more information you give us, the more help you will get.


How can I- be both a nucleophile and a weak base? I am just confused as to how $\ce{I-}$ cannot attract $\ce{H+}$, but it can attack a carbocation. It is sharing electrons in both cases, so what is the difference?

As Jori pointed out, nucleophilicity is a kinetic property while basicity is a thermodynamic property. Nucleophilicity in part is about the rates of chemical reactions; thermodynamics is in part about the favorability of chemical reactions.

A reaction may be very favorable, i.e. it has a very negative free energy change, but it might take a long time to actually complete. For example, the rusting of iron is favorable from a thermodynamic standpoint. The products, iron oxides, are very stable. However, the process is relatively slow from a kinetic standpoint; putting water on iron isn't going to instantly result in a familiar reddish-brown hue. It might take a day or so.

On the other hand, if you squirt something such as tert-butyllithium into the air, you'll probably catch yourself or something on fire. This is a carboanion salt. Carboanions are strong (Bronsted, and therefore, Lewis) bases. Acid-base reactions also tend to be relatively fast (kinetics) since they involve the transfer of something so small - a hydrogen proton. A graduate student unfortunately exposed some of this carbocation salt to the air and the resulting fireball instantly caught her sweater on fire. The carboanion salt likely grabbed protons from atmospheric moisture - a very favorable process from a thermodynamic standpoint - and kinetically, a very fast process. And so the resulting heat likely ignited the resulting protonated carboanion (alkane in this case) on fire.

In your case, $\ce{I-}$ is a good nucleophile in both protic and non-protic solvents. It's a rather big ion - the biggest of the halide anions. That means its negative charge is relatively dilute. This in turn means it's not as strongly solvated by the solvent molecules; it doesn't have as many tenacious solvent molecules hindering its trek toward a carbocation. So $\ce{I-}$ can reach carbocations relatively quickly. Note that terms dealing with time usually indicate that the writer is speaking about a chemical's kinetic properties.

For the same reason that $\ce{I-}$ is a good nucleophile it's a poor base. It's dilute negative charge (due to its size) means that $\ce{H+}$ has better things to do than be attracted to $\ce{I-}$. $\ce{I-}$ has a dilute amount of negative charge so it cannot effectively stabilize the hydrogen proton as well as many solvents might - especially water, which extensively associates with $\ce{H+}$; $\ce{H3O+}$ only begins to describe how well the proton is associated with water molecules; the proton has been found to be associated with 2, 3, 4, or more water molecules!


Are petroleum-based products prone to produce hard residues with black powder? A common controversy about black powder is lubricant composition. Some people tell this is a bad idea to use any petroleum based products in such lube because it will produce hard, difficult to clean residues.

As far as I know, in the old days, gunners used mostly mutton tallow in order to lube their bullets, and most animal-based grease have been found an easy (available) and efficiant way for that use.

Now we can avoid the problems of stinking, liquifying or decaying animal greases, and some people tell they used petroleum jelly or lithium grease for decades with no particular problems. Unfortunetaly, other people disagree, and the exact recipes are so numerous nobody can give me rationnal and definitive answer.
Looking at classic lube recipes teach us there is no consensus on such questions :

  Historical Black Powder Bullet Lubricants

  Composition of Extensively Used Bullet Lubricants

  (E.H. Harrison, American Rifleman, Jul. 1965)

  U.S. Army 1855 - 1 beeswax, 3 tallow.
  U.S. Army 1861 - 8 beeswax, 1 tallow.
  U.S. Army 1873 - 8 bayberry wax, 1 graphite.
  U.S. Army 1880 and thereafter - Japan wax.
  Sharps Rifle Co., 1878 - 1 beeswax, 2 sperm oil.
  Massachusetts Arms Co. (Maynard rifle), 1890 - 1 beeswax, 3 tallow.
  Marlin Firearms Co., 1891 - 1 beeswax, 4 tallow.
  Smith & Wesson, 1891 - tallow.
  H.M. Pope, about 1900 - 3 mutton tallow, 2 bay wax, 1 beeswax, 1 steam cylinder oil, .2 of 1 acheson graphite. The bay wax could be
  omitted.
  Automobile door latch stick lubricant, U.S. Patent 1,920,161
  (1931) - 5 paraffin wax, 3 petroleum jelly, 2 oil.
  A large police department, 1962 - 1 beeswax, 1 paraffin wax, 1 cosmoline. Notes: "Cosmolene" in this context refers to dark
  petrolatum with no anti-corrosion additives. Refined yellow petrolatum
  (petroleum jelly, Vaseline) may be substituted.
  Any mixture containing paraffin wax must include a plasticizer, such as petrolatum. Microcrystalline petroleum waxes may be used
  as-is.
  The 1:3 beeswax/tallow mixture (or any composition composed mainly of tallow) is probably the most traditional choice for "primitive"
  shooters. The 8:1 mixture is rather stiff, and better suited to
  conicals, paper cartridges, and the like. For paper-patched bullets,
  I'd be inclined to try the Sharps formula, substituting Dexron II/III
  automatic transmission fluid for the sperm oil.

Could someone explain or quote some good source explaining if and why petroleum based lubricant will produce such disagreement with black powder?

The answer may be more simple than some may expect, I start with Black powder, which contains, to quote:

Black powder is a granular mixture of

a nitrate, typically potassium nitrate (KNO3), which supplies oxygen for the reaction;

charcoal, which provides carbon and other fuel for the reaction, simplified as carbon (C);

sulfur (S), which, while also serving as a fuel, lowers the temperature required to ignite the mixture, thereby increasing the rate of combustion.

Next, the warming of an organic together with sulfur can produce a volatile sulfur-containing compound (see, for example, 'Reactions of sulfur with organic compounds XXI. The action of sulfur on 2-thienylchloromethane and di-2-thenyl sulfide').
Further, mixing of the new compound with carbon powder could account for the observed product especially after evaporation of all volatiles.
Note, per the comment: "problems of stinking, liquifying or decaying animal greases, and some people tell they used petroleum jelly or lithium grease for decades with no particular problems", the precise sulfur-based product, if any, created could account for the differences reported.
Also, this proposed explanation requires an element of heating, which likely accounts for the development of the reported problem over time.
So, to precisely answer the question: "Are petroleum-based products prone to produce hard residues with black powder? Yes, per my assessment.


Are methyl n-propyl ether and methyl iso-propyl ether metamers? Are methyl n-propyl ether and methyl iso-propyl ether metamers?
It says so here.

So the difference must be in number of carbon atoms or the distribution?

Definition(http://dictionary.reference.com/browse/metamerism)

The term metamer is equivalent to isomer.  Isomers differ in connectivites of atoms between molecules with the identical number and type of atoms.

Methyl n-propoyl Ether and methyl iso-propyl ether have the same numbers of carbon, oxygen and hydrogen atoms.

As an aside, I have never heard the term metamer before, I expect it is archaic.


Mixing of molecular orbitals in polyatomic molecules I am trying to solve the following Hamiltonian,$$H=-\frac{1}{2}\nabla^{2}-\sum_{i=1}^{10}\frac{1}{|\vec{r}-\vec{R_{i}}|}$$ Here $Z=1$ . To get all ther molecular orbitals of this system. I took the following linear combination to construct the hamiltonian matrix,$$\psi(\vec{r})=\sum_{i=1}^{10}c_{i}1s(\vec{r}-\vec{R_{i}})+\sum_{i=1}^{10}a_{i}2p_{x}(\vec{r}-\vec{R_{i}})+\sum_{i=1}^{10}b_{i}2p_{y}(\vec{r}-\vec{R_{i}})+\sum_{i=1}^{10}g_{i}2p_{z}(\vec{r}-\vec{R_{i}})$$

I took this wave functions and solved for $E$ using the following equation. Usually in polyatomic molecules, one would look for symmetry to construct various molecular atomic orbitals. However, here there is no symmetry. How do I get the molecular orbitals? What combination should I have taken to solve this problem. Please suggest some references where I can read the theory of molecular orbitals for polyatomic molecules with no symmetry

The standard way is to solve for the eigenvectors of your Fock matrix. Since you have a single electron, that's your Hamiltonian. Any number of programming languages have algorithms for this implemented. Each of your coefficients would correspond to an entry in the eigenvector.


What is the reasoning for higher pH values in this scene? In this article regarding low-boiling point amine-catalyzed biodiesel transesterification reactions, it is stated that,

  "From Table 1, we can also see that, for the isomers, the
  branched amines always showed higher yields of methyl acetate
  than the normal amines. For example, the yields of methyl
  acetate for n-dipropylamine, n-butylamine, and n-propylamine
  were 25.3%, 46.1%, and 48.2%, respectively, while those for
  diisopropylamine, tert-butylamine, and isopropylamine were
  42.1%, 49.6%, and 50.7%, respectively. This was probably due
  to the high pH values of the branched amines than the normal
  amines."

Why is this the case? Why do high pH values correlate to higher yields? Does the geometry (branched or linear) of the amine play a role in this?

Branched amines are more basic than their linear counterparts because inductive effect of alkanes places more electron density in the adjacent carbon atom. The higher the electron density near/at the amine group, the higher its pH will be.


If hydrogen bonds are chemical bonds, and these bonds are broken when water changes phases, why doesn't this classify as a "chemical" change? This is more of a terminology question. How are phase changes considered a physical property, not a chemical property when hydrogen bonds break. A chemical property can only be observed during a chemical change. Aren't hydrogen bonds a type of chemical bond, and those bonds are being broken during a phase change from ice to liquid water, for example. I understand that no intramolecular bonds are being broken. Water molecules remain intact. But we teach that a chemical bond has to break during a chemical change, but phase changes are considered to be physical.

You must realize that the line between physical and chemical properties is very thin. Most bonds work via the same principle, which has mostly to do with electronegativity. When looked at in reactions it has probably also to do with the according entropy and enthalpy change. Hydrogenbonds are basically bonds based on electronegative attraction between two atoms (which is a simplified few of the reality). A phase change basically when a certain energy point has been reached: since heating op a substance is equal to giving it more kinetic energy (since heat is the same as the movement of individual atoms or molecules). When you reach a certain heat there is too much kinetic energy to form hydrogenbonds the same way it as it did before it changed phase. You probably can agree that molecules or atoms moving quickly is an obvious physical property. When you look at great detail to how hydrogenbonds work, it has something in common with many other way of two objects attracting each other (for example: the formula for the gravitional force between to large heavenly bodies, is very similar to the formula for the electrical force between an electron and a nucleus).

In simple: Many of these chemical properties seem very physical when you look at them in detail. So saying that a certain property should actually be chemical because the changing of this property seems to correspond with the changing of chemical bonds, is not a realistic claim looking at how "physical"/physic based these "chemical" bonds are.


Why are spherical compounds with a half filled HOMO concidered aromatic? In planar annulenes, huckel's rule effectively states that compounds with a full HOMO for their conjugated $\pi$ systems are considered aromatic while those with a half filled outer shell are antiaromatic. Aromatic compounds are stabilized both by delocalization across their $\pi$ complex and possessing a full HOMO spread across the ring. On the other hand, antiaromatic compounds break degeneracy among their orbitals, preventing resonance, as it would only result in a half filled outer shell. Despite this, according to this source, a half filled HOMO in a spherical compound will result in aromaticity on a sphere. Is this because there is so much potential for delocalization in spherical compounds, since they contain so many conjugated pi bonds, or can this be explained with something else?

OK, let's get this straight. There is no such thing as spherical aromaticity. The so-called Hirsch's rule is sheer nonsense and never works.

Now don't get me wrong, the premises were sound. When we have dozens of atoms arranged kind of uniformly on the surface of a sphere, the whole thing starts to look mighty like continuous - reminiscent of a huge "atom" consisting of surface only, without any depth. Naturally, the eigenfunctions (AKA molecular orbitals) of such a system would be the pure angular functions without the radial part, and their energy will be determined by the angular quantum number $\ell$. The bottommost orbital would be like $s$ (a sphere of one sign), then would go three $p$-like orbitals (each with one change of sign), then five $d$-like orbitals, and so on. Look at the lowest energy levels of $C_{60}$ (pictured), or any other fullerene, for that matter; they would go just like that - one, three, five... (With any other fullerene, the correspondence would be approximate; with $C_{60}$, because of its high symmetry, it is exact). And of course, when we count the electrons up to the $n$'th level, we get $2n^2$.

Sounds good, does it?

Well, it just does not work. There are too many crude approximations on top of each other. Most importantly, when we get all the way up to the HOMO, our eigenfunctions start changing sign too often, so we can't any longer pretend that the surface of a fullerene is continuous. Consequently, the pattern breaks and all conclusions fall apart.

One might point out that pretty much any rule in chemistry is like that, including the Hückel's rule itself. It is also built of crude approximations, isn't it?

That's right, it is, but there is a huge difference, too. Hückel's rule works. Hirsch's rule doesn't. The former predicts the outstanding stability of benzene, and not-so-great stability of cyclooctatetraene, and that cyclopentadienyl would prefer to be an anion, and so on - and they all obey. The latter predicts... well, see for yourself.

What are the magic numbers of electrons for the presumed spherical aromaticity? $2,\,8,\,18,\,32,\,50,\,72,\,98$ and so on. The first five are too small for a fullerene, so ignore them. 60 is not magic, so one would expect it to be unstable. WRONG. It is pretty stable. The paper you mentioned says it is strained. Now what was that about? Of course it is strained, as are all fullerenes and nanotubes; if you don't strain that $sp^2$ carbon, it would just form a flat sheet known as graphene. But what in the world does this have to do with [anti]aromaticity? Think again of the old good Hückel's aromaticity. Think of the cyclopropenyl cation; is it aromatic? Positively yes (pun intended). And yet it is like ten times more strained than fullerene. Aromaticity and strain are totally unrelated things; if people bring up one of them to judge the other, they are either clueless or deliberately misleading, even if they managed to get a paper published in a peer-reviewed journal.

Antiaromaticity, if anything, is about having a half-filled HOMO and thus being prone to Jahn-Teller distortion, radical reactions, and other unpleasant things. Is this the case with $C_{60}$? Look again at those energy levels. Look at the completely filled HOMO, and the completely empty LUMO, and that huge gap between them. Look and think twice before calling this guy antiaromatic.

Now comes the "magic" number 72. That would be $C_{72}$. Does it exist? Well, yeah, it kinda does, like any $C_n$ where $n$ is even and $\ge70$. Is it particularly stable? No, it does not seem so; it is way less abundant than the smaller $C_{70}$ and the bigger $C_{76}$, which themselves are way less abundant than $C_{60}$. Also, starting at $C_{76}$, the fullerenes would have multiple isomers, and as soon as $C_{80}$, we find an example where one isomer is unstable for electroniс reasons (antiaromatic indeed, if you'd like) and the other is stable. So the stability is not determined by the number of electrons alone, after all.

Now what would you call a seemingly reasonable rule which works $\bf{0\%}$ of the time?


Confusion on light absorption by electrons Say an atom's electron can jump between energy levels n = 2 and n = 3 with x joules of energy. If a light beam of frequency f provides x+y joules (where y is a positive number), that is, more than enough to excite the electron to the next energy state but not enough to reach n = 4. The electron should accept this quantized quantity of energy and then
later emit it. On emission spectra, however, the extra "y" J of energy are not accounted for, only the minimum value x to transition between n = 2 and n = 3 (the energy difference between orbital levels). Where, then, does the extra y energy go? It cannot simply "not be absorbed" because it exists along with the rest of the energy, x, in quantized packets. If it contributes to raiseing the KE of the electron, it must at some point be emitted, but that would give way to an incorrect value on the emission spectra. What am I missing?

Depending on how large the value $y$ is, the premise of your question can be wrong.

On the other hand, if you jump between two electronic energy levels where $y$ is on the order of vibrational states, it is entirely possible this extra energy will be stored in vibrations of the molecule on the excited state. This energy is then typically radiated away quite quickly until you reach the ground state of the higher electronic energy level, and then the emission takes place where you could measure the energy $x$ again but in an emission spectrum.

It is entirely possible, however, that the emission process will happen from above the $v=0$ state on the excited surface. It is also possible that you will emit down to different vibrational energy levels on the ground state electronic surface. That is, $0(v=0)\leftarrow1(v=0)$, $0(v=1)\leftarrow1(v=0)$, $0(v=2)\leftarrow1(v=0)$, etc. are all possible and will be observed with varying intensities.

So, in general if you pump in some amount of energy to excite an electron, you don't need to see that same amount of energy coming back out from electronic transitions. You would need to look quite closely at infrared radiation and other energy to make it add up.

As another point, it is actually possible for this energy $x+y$ to be absorbed even if $x+y$ is very far away from the energy of a state. This is how scattering Raman spectroscopy works. You have to think about it differently, however, because the state you are excited to is not a stationary state, but rather a virtual state. See the question I asked here.

So, the short answer is that energy very frequently gets added to other degrees of freedom besides the electronic ones. The same thing goes for vibrational transitions where the excess energy gets put into rotational states. There are rules for how these transitions (and how they couple) will take place.

The other answer, which is less interesting, is that it's totally possible for the molecule to simply not absorb this $x+y$ precisely because there is nowhere to put this extra $y$.


What are coexistent metals? In ACS Sustainable Chem. Eng. 2018, 6 (10), 12542–12561, I saw the term "coexistent metals", but I could not find a definition for that on the Internet. So what does it mean?

Karapinar et al. utilized ammonium-loaded zeolite as a seeding material for precipitating calcium phosphate and suggested the possibility of using the precipitate-ammonium loaded zeolites as fertilizer. The efficiency of chemical precipitation is affected by alkalinity, organic matter content and the coexistent metals.

I see no special meaning there, aside of the common English. It means eventual metals present in the solution as side components or impurities, aside of calcium. Like alkali metals, magnesium, heavy metals as iron, manganese, copper atc.


selective hydrolysis problem I am aware that esters can be selectively hydrolysed depending on the type of alkyl group present e.g. Me ester vs t-butyl ester, the latter is reactive to acidic conditions and not basic because of its inability to form tetrahedral intermediate. But what reagent could I use to carry out the following transformation?

I see it is the O-C bond that breaks, so it is acyl cleavage (Ac). Acidic conditions would destroy my Me ester as would basic, so how can I get around this? Could a big bulky base, say LDA, take it down a B(Ac)2 route (generate CO2?)

If you use methanol as a solvent and add a suitable base, you should end up with the correct product, since any MeOH attacking the protecting group you want to keep will simply re-form the methyl ester, whereas the acetate will cleave to give the unprotected alcohol.

You could potentially also achieve this using sodium methoxide/methanol to speed the reaction up, but standard potassium carbonate/methanol is usually just fine.


Nano coatings reaction with Al bottles I have a question regarding nano coatings based on the formula "modified $\ce{SiO2}$".

I have been told that $\ce{Al}$ bottles react with substances with a $\mathrm{pH}$ value below 3 or above 9. When the substances are not within these limits, it will form $\ce{Al^3+}$ which will cause the $\ce{Al}$ bottle to corrode, if I understand it correctly.

My question is therefore, when does this corrosion begin, how much time does it take, and can it ruin my nano coating effects?

My supplier of the $\ce{Al}$ bottles also has a bottle that they call "Chemical pesticide insecticide agrochemical pharmaceutical Aluminum bottle", does this bottle create any difference regarding the possible reaction with $\ce{Al}$? Perhaps it corrodes at a lower or higher $\mathrm{pH}$ value?

Also when my supplier of the coatings indicates the products has a chemical stability between 1-13, what does that mean?

A cited a pH value below 3 or above 9 is in line with the usually citation for the stability of the Al2O3 protective layer. Per a source on the effect of pH, to quote:

  It is well known that aluminum resistance is related to the thin and compact layer of naturally formed oxide on aluminium surface, but this oxide layer is stable only in pH range 4-8. Lower or higher pH values caused prominent destroying of protective layer and thus the significant metal dissolution [15].

So, outside the pH range, your nano SiO2 coating formulation may be subject to attack. The presence of active Aluminum may serve to introduce some new chemistry.

Most likely, in my opinion, the unprotective Al can react even with water to form hydrogen and aluminum hydroxide. The latter reacts with so-called reactive silica per Science Direct:

  Silica forms complex precipitates with iron, aluminum, and magnesium hydroxides.

which disrupts the SiO2.

Another but unlikely path starts with some atomic hydrogen ingressed into the Aluminum metal (see, for example, this Science Direct discussion and also a 2008 thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari discussing surface atomic hydrogen formation on Aluminum). Suggested reaction mechanics could then be if such a reaction is favorable (unlikely):

$\ce{.H ⇌ H+ + e-}$

$\ce{4 .H + SiO2 (nano) -(?)-> Si + 2 H2O}$

potentially directly destroying the nano SiO2. Here relatedly to the reaction above is a hydrometallurgy source discussing a commercial leaching application with $\ce{.H}$ on Al metal acting on $\ce{PbS}$ at the bottom of Page 218. However, the heat of formation of PbS is 10 times less than SiO2, hence this reaction path is not likely.

Proposed Solution: Test Copper vessels for suitability (see this reference).


Stability variation in heavy nuclei The heavier the atom, the more unstable it gets, right?

That is not true about Uranium and we know it.
I wondered why. A brief explanation stated that since more neutrons are there in the nucleus there is more nuclear force to 238U than 235U. Seems legitimate, but then that should have been true about the other radioactive species. Again, we know that's not true.

So, either the explanation about the Uranium exception is not totally true or the principle of the heavier-atom instability (of course, exceptions occur at elements like U) is not a principle. Or maybe there is a relation between these two (nuclear force and atomic weight) which is beyond me.

My hunch is that the latter is true; searched some sites which had an answer to the question, but no acceptable results gained.

The principle that "heavier nuclei are more stable" is a very crude approximation. Theoretical explanations of nuclear stability are complex. While the transuranium elements become progressively more unstable with increasing atomic mass, this trend is not predicted to continue indefinitely; there is an island of stability around copernicium and flerovium, and the as-of-yet unsynthesized copernicium-291 has been predicted to have a half-life of 1200 years or so. You may be interested this paper for details; figure one is a nice illustration of how stability is affected by neutron/proton number towards the end of the periodic table.


Is there any explosion hazard with an enclosed graphite crucible melting silver? the crucible volume is about 2 ounces.   The final design of the crucible will be enclosed tightly except for a small .5mm diameter hole at the bottom and a 1mm diameter hole at the top through which silver solder will feed.

I was wondering.  In my experiments should I worry about the thing exploding?

Silver will absorb oxygen readily from air while molten and spatter as it cools. This is hardly an explosion, but some care should be taken to avoid being spattered with hot particles ejected from the crucible or plugging the holes.


How to make an aluminumoxide coating on an electrode? The Questions is above.
I am not a chemist but that is what I know:

Aluminum has an Oxide Layer normally
Aluminum-Oxide Layer can be broken down with NaOH
Zink coatings are done by galvanic coating using a certain voltage

Another way to ask: Can I do this with Aluminum?

Yes, it is definitely possible, but also relatively difficult as compared to electroplating zinc onto another metal.

It is a significantly important industrial process. Although aluminium itself is reactive, it develops an impervious oxide layer that quite much renders it unreactive.

  However, comparatively larger negative standard electrode potential of Al/Al(III) couple (− 1.67 V vs. SHE) precludes its deposition from the aqueous baths.

So it is usually considered impossible in aqueous (water-based) solution so is conducted in either organic liquids or fused (molten) state.

Fused: Usually a bath of an eutectic mixture of $\ce{LiBr, KBr}$ & $\ce{CsBr}$ is taken along with molten $\ce{AlBr3}$ (melts at about 94-98°C). A concentration of $80\%$ (by wt.) of $\ce{AlBr3}$ is reported to give a shiny finish.

Source $1$ reports that a satisfactory result was obtained with constant current of $1-3$ $\pu{A}$ passed for $1$ hour.

Non-aqueous solution:

From Source $2$ (emphasis mine)

  ..the commercial processes for electroplating of Al are based on organic solvents such as SIGAL (Siemens-Galvano-Aluminium) and REAL (Room-temperature Electroplated Aluminium) processes. In the SIGAL process which was developed in Siemens laboratories Al is electrodeposited at 100 °C from alkylaluminium compounds in toluene. In the REAL process which was developed in the laboratories of Philips $\ce{Al}$ is deposited at room temperature from $\ce{AlCl3}$ and $\ce{LiAlH4}$ in tetrahydrofurane.

The process must be conducted avoiding any traces of water, or the water would be electolysed first, leading to observable evolution of gas.

TL;DR: Electroplating aluminium is not something I would recommend trying at home.

References:

Aluminum electroplating on steel from a fused bromide electrolyte, P. K. Tripathy et al, Surf. Coat. Tech., Vol. 258, 15 Nov. 2014, 652-663. DOI:10.1016/j.surfcoat.2014.08.021
Electroplating of mild steel by aluminium in a first generation ionic liquid: A green alternative to commercial Al-plating in organic solvents, S. Zein El Abedin et al, Surf. Coat. Tech., Vol. 201, Issues 3–4, 5 Oct. 2006, 1352-1356. DOI:10.1016/j.surfcoat.2006.01.065


How are the Van der Waals constants a and b related to each other? I know that the constant $a$ represents the attractive forces between the molecules while the constant $b$ represents the size/repulsion between them. Does that mean that $a$ and $b$ are "inversely proportional" or are they independent of each other? If we have two gases and we can compare their sizes and attractive natures, can we compare $a$ and $b$ for the two gases? If $a$ is greater for gas 1 will $b$ always be lesser for gas 1 as compared to gas 2?

Yes, they are certainly correlated as shown in the following plot based on data from the Wikipedia Van der Waals constants data page

The correlation should not be entirely surprising: the covolume b is a measure of the size of the gas molecules, whereas a is a measure of the strength of intermolecular interactions which is expected to depend through polarizability and dispersion interactions on the size of the molecules (molecular volume).

For a given gas parameters a and b tend to increase together. The shown correlation is empirical but since both a and b depend on molecular size and apparently in a very similar functional way, the result is a linear correlation.

Following the hint from Karl, here is a version of the plot with log(a) along the abscissa, and with superimposed labels.