Why does hydrogen burn with a pop sound? Our teacher made us to do an experiment with zinc granules and hydrochloric acid.

$$\ce{2HCl + Zn -> ZnCl2 + H2}$$

We collected the hydrogen gas that was evolved in a test tube and on ignition, it burned with a "pop" sound. Why does hydrogen show this behavior? Why doesn't it burn more quietly like other fuels (gasoline, coal, wood, etc.)?

We may agree that during the combustion, regardless if the one of wood, gasoline, or in your example, hydrogen gas, is an exothermic reaction.  In a simplified view, products of these reaction are gaseous water; in the processes burning organic matter varying relative amounts of carbon monoxide and carbon dioxide are to be considered, too.

The gaseous products formed take up a volume, much more than their starting materials in the condensed state (liquid, or solid). Because of the local heat, these gases aim to expand rapidly.  In contrast to burning wood for a bonfire, the volume offered by a test tube is confined except for the mouth of the test tube, where a decompression to reach ambient pressure may take place.  This small scale explosion initially occurs at the interface of air's oxygen at the mouth of your test tube filled with hydrogen, yet this frontier actually propagates up to the closed end of the test tube, too.  Sometimes, especially with lengthier test tubes, this may be seen from the outside, too.

The combustion of hydrogen is worth a special note, because -- as correctly spot by airhuff -- formally there is volume contraction along

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

which is more than compensated by the thermally caused expansion of the gaseous water.

Gasoline driven combustion engines work by the same principle; spark plugs locally ignite a mixture of finely dispersed fuel and air, triggering a rapid combustion (explosion) where the reaction products again aim to expand. As they operate at a larger scale as the test tube experiment, the generated sound is much more intense, and these vehicles need a silencer, too.

While your bonfire equally yields $\ce{CO2}$ and water, these products may expand easily; there is no pressure generation, hence there is no sudden expansion (at least, not at this scale), and hence we may perceive the bonfire as (in comparison to the other examples) as peacefully calm.

I suggest to take a view on videos about explosions of hydrogen gas, and an appealing demonstration how it looks in the inner of such an Otto engine here.  In the later example (the display of the combustion starts around 2:51 min and again at 6:00 min), gasoline was replaced and a high speed camera looks through a special lid, yet the example still shows nicely the propagation of this frontier of explosion.


Order of Nucleophilicity I know this question has been asked many times but I need to confirm this doubt which has been pestering me for the past few days.

Suppose in an $\mathrm{S_N2}$ reaction (in polar aprotic solvent) we have two nucleophiles:

ethoxide $(\ce{EtO-})$ ion;
ethyl thiolate $(\ce{EtS-})$ ion.

Now, since we know fluoride $(\ce{F-})$ is stronger nucleophile than iodide $(\ce{I-})$ in polar aprotic solvent, i.e. nucleophilicity parallels basicity.
We should expect the alkoxide ion to be more nucleophilic than thiolate since alcohols are weaker acids than thiols.

But my book says the opposite. Is this a exceptional case? If so, could it be predicted in advance that thiolates are better than alkoxide even though fluoride is stronger than iodide?

Any help would be much appreciated.

Please remember that nucleophilic substitution at a saturated carbon atom does not directly correlate with pKa.

The pKa of a substance (in the Bronsted-Lowry definition) relates to the substance's ability to have strong interactions with $\ce{H^+}$. Thus, this interaction is mainly electrostatic in nature; since $\ce{H^+}$ is a very small, weakly polarizable ion, according to hard-soft acid base theory, strong interactions with this ion are largely electrostatic in nature, i.e. proportional to charge density.

However, nucleophilic substitution at a saturated carbon atom occurs mainly through orbital interactions, not through electrostatic interactions. Namely, the highest occupied molecular orbital (HOMO) of the nucleophile interacts with the LUMO (sigma antibonding orbital) of the substrate (usually alkyl halide).

Since higher energy HOMOs interact better with LUMOs, molecules or ions with higher energy HOMOs are more nucleophilic.

With the specific case that you gave ($\ce{EtO-}$ and $\ce{EtS-}$), consider that sulfur is a larger atom, and the electrons in HOMO in $\ce{EtS-}$ are going to be less tighly held by nuclear charge than electrons in the HOMO in $\ce{EtO-}$. Alternatively, you can consider that the highest energy level orbital in S is 3p while that of oxygen is 2p --> electrons are more tighly held and lower in energy in $\ce{EtO-}$ --> $\ce{EtO-}$ is less nucleophilic.

Thus, if you consider the relative energy levels HOMOs in related molecules and ions, you get the general trend below:

References

Clayden, J., Greeves, N., Warren, S. Organic chemistry, 2nd ed.; Oxford University Press: New York, 2012.


Counting basis sets in quantum chemistry I have to use STO-nG (a "minimal basis set" - meaning that only one basis function is used for each atomic orbital in the atoms of which the molecule is made from). Let's take the example of a water molecule.

The water molecule has two H atoms and one O atom. Thus, we have a total of 7 orbitals (two 1s of H, one 1s of O, one 2s of O and three 2p of O). So when using STO-nG, would that then mean that three 1s type basis functions, one 2s type basis function, and 3 p-type basis functions are used, totalling at 7 basis functions, each being made up from a linear combination (LC) of n simple gaussians?

Yes, you have counted the number of basis functions correctly:

    #1s    #2s    #2px    #2py    #2pz    #Total
H1   1                                      1
H2   1                                      1
O    1      1       1       1       1       5

Which as you also counted is 7 basis functions.
As you also correctly state, each of the basis functions will be a linear combination of $n$ primitive Gaussians:

$$\phi(r)=\sum_{i=1}^nc_i\exp\left(-\alpha_i\left(r-r_0\right)^2\right)$$

As you have observed that is what the $n$ in STO-$n$G denotes.


Which mechanics or law is predominately seen on the nanoscale? Nanotechnology and sciences are the study or the manipulation of matter on an atomic, molecular, and supramolecular scale especially at nano meter range. We know that at macro molecular level a system can be described using Newtonian/classical mechanics (eg: $F=ma$) whereas for an electronic level quantum mechanics are used. In nanotechnology and science which mechanics is seen predominantly; is it Newtonian, quantum or both and why?

Its depence on the size of the system your are looking at. To understand spectroscopy of molecules, you usually use quantum mechanics. In molecular dynamics simulation, you'll need both: On the one hand, there is a potential needed based on QM-interactions, on the other hand, the classical equations of motion are good enought to move your atoms in a QM-force field.

Based on your problem, you have to decide how much classical mechanics you can justify. But in the nanoscale everything is made up with quantum mechanics and classical mechanics are used to approach a problem, which is not solvable in quantum mechanic way in an accetable periode of time.


How many stereoisomers can exist for 4,5-bis(3-hydroxyoct-1-en-1-yl)cyclopentane-1,3-diol? Q.10 For the given compound $\ce{X}$, the total number of optically active stereoisomers is ____.

This is from JEE Advanced 2018 second question paper.
I thought the answer to be at least 32 because of exactly 5 stereocentres present.  One of the double bond cannot show geometrical isomerism possibly.
But the answer given in the answer key is 7:

I did not find any plane of symmetry because of the different kind of double bonds.

The question is tricky, because

the two ends of the wavy bond are both stereogenic units (chiral cyclopentane ring carbon atom, and the double bond), so there is some additional unintentional ambuguity, but it's not a big deal, because the legend at right explains the situation somewhat;
it in fact asks “How many … stereoisomers does this structure represent?” (because of some stereogenic units with unspecified configuration) instead of “how many … stereoisomers does it have?”.

The unspecified-configuration stereogenic units are two chiral carbons and one double bond, which give total 8 stereoisomers.

The first one (top left) has a plane of symmetry, the other 7 stereoisomers are optically active.


What is the mechanism for the oxidation of benzene to maleic anhydride using vanadium pentoxide? What is the mechanism for the oxidation of benzene to maleic anhydride using vanadium pentoxide, $\ce{V2O5}$?

A talk given by Ken Waugh named The Detailed Kinetics of the Partial Oxidation Reactions: Benzene/Butane to Maleic Anhydride, Ethylene to Ethylene Oxide,[1] proposes the following mechanism on page 4:

[1]: Acessible as pdf via rsc.org


Why can't we use orbital correlation diagrams for sigmatropic rearrangements? In Ian Fleming's Molecular Orbitals and Organic Chemical Reactions book, he describe the use of correlation diagram as a method for predicting whether pericyclic reactions will be allowed/forbidden on symmetry argument (he calls this a basis for the Woodward–Hoffmann rules).

  Correlation diagrams have given us a convincing sense of where the barriers come from for those reactions that we have been calling forbidden.

Whilst there are given examples of cycloadditions and electrocyclic reactions there are no examples of sigmatropic rearrangements, and on the internet a few sources say the diagram is not appropriate for these sigmatropic rearrangements.

I am curious why a correlation diagram does not work for sigmatropic rearrangements, since the the Woodward–Hoffmann rules are able predict which are allowed and forbidden.

The reason that comes up again and again is the one below, taken from Orbital Interaction Theory of Organic Chemistry.

  Orbital correlation diagrams are useful for cycloadditions and electrocyclic reactions but not for sigmatropic rearrangements since no element of symmetry is preserved.

  Source:   Orbital Interaction Theory of Organic Chemistry. Arvi Rauk.

In order to (easily) draw a correlation diagram you need to find some symmetry element that is preserved throughout the reaction. A correlation diagram is shown below for an electrocyclic ring closing reaction, note how the C2 axis is preserved during the process, allowing the various correlations to be made.

  Source: Dr Wallace's web notes on pericyclic reactions

Since you mention Fleming's book, I'll quote him on this matter:

  Step 3. Identify any symmetry elements maintained throughout the course of the reaction.

  Source: Molecular Orbitals and Organic Chemical Reactions: Reference ed. Ian Fleming.

Essentially, you have to be able to look at the orbitals directly involved in the reaction (i.e. we dont care about random groups hanging off the side that play no role).

In a sigmatropic rearrangement there is rarely (possibly never but I'm sure there are exceptions to the rule) a preserved symmetry element, which makes it impossible to draw a correlation diagram.

There are, of course other approaches, each of which have advantages and limitations depending on the reaction in question:

Just apply the Woodward–Hoffmann rules
Use frontier orbital theory to look at the HOMO–LUMO interactions
Use a Zimmerman approach and look if the transition state is aromatic, or anti-aromatic.


Inductive effect of -NHMe Why does $\ce{-NHMe}$ show -I effect. I think it should be opposite. I know that $\ce{-Me}$ groups are electron donating in nature. So the nitrogen atom will have a small $\delta^-$ charge. So then why would it still be electron withdrawing?

The inductive effect is reflective of the comparison of electronegativity between an -R group and carbon. There is an -I effect because the nitrogen is more electronegative than carbon, so it will inductively withdraw electrons. I will also note that the inductive effect is measured relative to the -R group being hydrogen.


Sephadex and purification of products Sephadex is useful for the separation and purification of many natural products.
 Is it also useful for the separation and purification of artificial chemicals (in organic synthesis ) or not and why ?

First, a conceptual but important point: there is nothing that sets artificial (a better word is synthetic) chemicals from natural products, chemically speaking. What we call "natural products" tend to be larger, complex molecules largely because we reserve the term for compounds for which extracting from natural sources is easier than synthesis, and these tend to be very large organic or bioorganic macromolecules, but there's no fundamental difference in behaviour between natural and synthetic molecules of those compounds, and in principle we could synthesise any natural product and reproduce exactly the same chemical properties that the same compounds have when extracted from natural sources. I state this because your use of the categories of natural and artificial can suggest some essentialist distinction while there is none.

Now, as to whether Sephadex can be used to separate and purify synthetic compounds, the short answer is yes, if these compounds are suitable. As @Waylander mentioned, and as with any separation medium, the main question is whether the compounds you want to separate are in the fractionation range of the medium. Sephadex can be manufactured with different bead sizes, which result in different separation ranges (these are marketed as G-10, G-25, ..., G-200). Different Sephadex gels can are used to separate proteins in the range from a few kDa to hundreds of kDa - and in principle they will be useful to separate synthetic organic compounds in the same molecular weight range through size exclusion. Commercial products are generally considered capable of discriminating molecules above ca. 700 Da.

Another factor to consider is functionalisation: Sephadex is highly functionalisable, which can produce specific affinities and retain desired compounds. This allows for a large range of possibilities for purification or separation, depending the composition of your mixture, what you want to retain and what you want to remove.

In practical terms, it is probably wise to begin by looking at separation procedures involving compounds of similar molecular weight and functional groups to those of your compounds of interest that are already described in the literature and work from there.


What happens to a quaternary ammonium cation at a pH above its pKa? I am working with a lipid and I have firm reasons to believe that its pKa is around 8. This lipid is a cationic lipid where the nitrogen is a quaternary amine with two lipid chains and two methyl groups.

Now, my question is that at a pH above the pKa of this lipid, the quaternary ammonium ion should get deprotonated, or basically act as an acid. But the quaternary ammonium cation doesn't have any protons to lose, so what would happen to it? My guess would be it accepts a lone pair and becomes negatively charged.

Amine compounds in which the nitrogen is bound to four carbon atoms through covalent bonds are known as quaternary ammonium salts. They are stable compounds that will not converted back to their relevant amines by treatment with base. Generally, they are very soluble in water because of their ionic nature, and also soluble in commonly used organic solvents due to (long chain) organic substituents (e.g.: phase-transfer catalysts). Because they do not contain a free electron pair on nitrogen, in general, the quaternary ammonium ions are inert towards chemical reactions. Thus, pH changes course only them to change their counter ions.

In fact, not only quaternary ammonium salts are relatively unreactive, but also, they do not diffuse readily across biological membranes when used in medicine, because they exist as permanent cations.


Software for chemical reaction simulation in ICE I want to study what happens in a internal combustion engine in which fossil fuels are mixed along with other chemical additives.
Example:
Acetone drastically reduces surface tension.
Most fuel molecules are sluggish when it comes to their natural frequency. But acetone has inherent molecular vibration that "stirs up" the fuel molecules, to break the surface tension.
This results in a more complete vaporization with the various other factors remaining the same. More complete vaporization means less wasted fuel, hence the increased gas mileage from the increased thermal efficiency.
Along with this I also want to study the emission of $\ce{CO2,\  O2\,\ H2O}$ and more.
Is it possible to carry out the above phenomena/process in a virtual simulator ?

I am not sure about the phenomena you are describing but simulators are just tools with built-in models and thermodynamic data together.

What you need is, for example, to find some kinetic (combustion kinetic) that it has in its equation a parameter regarding the acetone concentration.

Try to find at the literature some of these kinetic equations, if you are student you can access with your university credentials into sciencedirect.com papers.

If not, check arxiv.org or google ;)

Once you have these, then you can use simulators to reproduce your "special" kinetic.

Another possibility is to describe better your system behaviour (including graphs) and try to fit them suing a correlation or a polynomial fit. Or ask the mathematicians, they could recognise models just seeing the patterns and give you an equation type that you can use.

Check this paper as an example;

http://www.combustion.org.uk/ECM_2007/ecm2007_papers/1-20.pdf


Can the 5+ and 4+ oxidation state of Vanadium ion exist independently? From my class text book:

  The oxidation states of Vanadium ions 5+ and 4+ are shown in the compounds $\ce{VO2^{+}}$ and $\ce{VO^{2+}}$  respectively, whereas $\ce{V^{3+}}$ and $\ce{V^{2+}}$ are shown like this in the redox reaction (reduction half equation).

Why is this? cannot $\ce{V^{5+}}$ and $\ce{V^{4+}}$ exist indepepently as ions like $\ce{V^{3+}}$ and $\ce{V^{2+}}$ (when measuring the reduction potential)?

When an atom $M$ looses one or more electrons, it becomes positively charged and becomes the ion ${M^{z+}}$. In aqueous solution this positive charge attracts the unused doublets of the Oxygen atom from $H_2O$, and it repells the proton of $H^+$.

If $Z$ = $1$, the ion $M^+$ attracts several molecules water around it, and it makes a Debye layer of oriented $H_2O$ molecules around the ion, and no chemical changes.

If $Z = 2$, the attraction is strong enough to repell one proton $H^+$ from one of the molecules adsorbed around the $Z^{2+}$ ion. The solution becomes a little bit acidic, and the $M^{2+}$ion is partially hydrolyzed into the ion [${M(OH)]^+}$. It is an equilibrium :$$M^{2+} + H_2O \ce{<=>}[M(OH)]^+  +  H^+$$

If $Z = 3$, the attraction between $Z^{3+}$ and water is strong enough to at least transform all $M^{3+}$ions into $[M(OH)]^{2+}$, and part of them to [$M(OH)_2]^+$ with the two reactions :
$$M^{3+} + H_2O\ce{->}   [M(OH)]^{2+}  +  H^+$$
$$[M(OH)]^{2+} + H_2O \ce{<=>} [M(OH)_2]^+  +  H^+$$
This is the case if $M^{3+}$ is $Al^{3+}$ or $Fe^{3+}$

If $Z = 4$, all ions $M^{4+}$ are at least hydrolyzed into $[M(OH)_3]^+$ and even into $M(OH)_4$, according to :  $$M^{4+} + 4 H_2O \ce{->}M(OH)_4  +  4 H^+$$It should be mentioned that $M(OH)_4$ has a tendency to loose water and produce a structure that has acidic properties. So it is often presented under the formula $H_4MO_4$ or $H_2MO_3$. That is the case with $M = Si$. Silicic acid is often presented as $H_4SiO_4$ or $H_2SiO_3$

With Vanadium $V^{4+}$, it goes another way. The molecule $V(OH)_4$ does not exist, as the ion $V^{4+}$ is too small to fix $4$ groups $OH^-$. As a consequence, $V^{4+}$ gives another positive ion $VO^{2+}$ according to :
$$V^{4+} +  H_2O \ce{->}  [VO]^{2+}+2H^+$$ Strangely enough, it looks as if the non-existent $V(OH)_4$ behaves like a basic hydroxide and gets dissociated according to : $$V(OH)_4\ce{->}[VO]^{2+}+ 2 OH^–$$
If $Z=5$, the ion $M^{5+}$ can be hydrolyzed according to $$M^{5+} + 3 H_2O \ce{->}  [MO_3]^{-}+6H^+$$$$M^{5+} + 4 H_2O \ce{->}  [MO_4]^{3-}+8H^+$$
But with Vanadium $V$, the ion $V^{5+}$ may be hydrolyzed  according to :$$V^{5+} + 2 H_2O \ce{->}  [VO_2]^{+}+4H^+$$$$V^{5+} + 3 H_2O \ce{->}  [VO_3]^{-}+6H^+$$The second reaction is favored at high pH.

This theoretical development is taken from General Chemistry, by Linus Pauling, W. F. Freemann and Co., San Francisco, 1953, Chapter XXI.


Are all ionic compounds salts? According to Wikipedia:

  A salt is an ionic compound that can be formed by the neutralization
  reaction of an acid and a base.

Are all ionic compounds salts? Are all salts ionic compounds?

Interestingly, IUPAC states that a "salt" is "a chemical compound consisting of an assembly of cations and anions". Under this definition, all ionic compounds are salts, and all salts are ionic compounds.

Therefore, something like sodium hydroxide ($\ce{Na+OH-}$, definitely an ionic compound) could actually be correctly called a salt. This clashes with the commonly taught high-school level definition of a salt ("the product of an acid-base reaction"), unless you consider very general definitions of acids and bases such as the Usanovich definition, whereby sodium metal $\ce{Na^0}$ is an electron donor (and therefore a base) and water is an electron acceptor (and therefore an acid).

That said, the high-school definition is too simplistic. It is common for compounds to be an acid, a base and a salt all at the same time; consider for example sodium bicarbonate ($\ce{Na+HCO3-}$). It is made of cations and anions, and therefore is definitely a salt. Furthermore, it can act as both a Brønsted–Lowry acid ($\ce{NaHCO3 + OH- -> H2O + Na+ + CO3^2-}$) and as a Brønsted–Lowry base ($\ce{NaHCO3 + H+ -> Na+ + H2CO3}$). Another amusing example is hydrazinium sulfate, a salt, acid and base, where both the cation and anion are also both acids and bases!


Do acyl fluorides exhibit resonance donation or inductive withdrawal? I know that in acyl chlorides, there is partial triple bond character in the $\ce{C=O}$ bond not because the chlorine is electronically donating (due to size mismatch with carbon) but because the chlorine is highly electron withdrawing and this stimulates the oxygen into partially donating one of its lone pairs to the partially positive carbon.

This is supported by IR stretch data for the $\ce{C=O}$ bond in acid chlorides; we're looking at frequencies from 1810-1775 reciprocal centimeters - somewhat higher than a vanilla $\ce{C=O}$ bond stretch frequency.

However, in acid fluorides, in which the fluorine is of comparable size to the central carbon, it seems that resonance donation is at least a possibility. However, resonance must compete with the inductive withdrawal effect. So, which one wins out in acid fluorides, and does the $\ce{C=O}$ bond exhibit more single bond character or triple bond character?

I tried looking up acid fluorides, and I can't find any mention of them. So I guess a good starting point would be: are they stable enough to even exist?

Yes, acid fluorides exist.  You can purchase acetyl fluoride from Sigma-Aldrich.

Look back at these two questions for background on how the increased contribution from the triple-bonded carbonyl resonance structure (see below) will lead to an increase in the carbonyl ir stretching frequency (e.g. the triple-bonded carbonyl resonance structure will lead to a stiffening of the carbonyl bond and consequently shift it absorption to higher energy).

Ring Strain and C=O Stretching Frequency
Triple Bond Character in Acyl Chlorides

Here are the resonance structures we considered in the case of an acid chloride.

Resonance structure IV is what we're calling the "triple-bonded" carbonyl structure.

The carbonyl stretch for a acetyl chloride (I, R=Methyl) occurs at 1818 ${cm^{-1}}$.  In the case of acetyl fluoride (replace Cl with F in the above structures) this stretch occurs at 1869 ${cm^{-1}}$, significantly higher.  This suggests that resonance structure IV plays a larger role in the case of acid fluorides, despite resonance structure III presumably also playing a larger role compared to the acid chloride case.  The stability of $\ce{F-}$ vs. $\ce{Cl-}$ seems to be a very significant factor.


Why are nanoliposome encapsulated antibiotics still so unavailable despite being so effective? I wasnt sure how the phrase the headline question really nor if this is the right place to ask it, but i'm primarily interested in the chemical synthesis aspect and the experience of chemists regarding this subject.
In the antimicrobial literature specific liposome or nano particle encapsulations of antibiotics have been shown over and over again to show immense potential for quite a long time by now, enhancing efficacy, pharmacokinetics while at the same time reducing toxicities. My question basically is, why is this not being used really? Advanced liposomes can be found in some vaccines for example and the likes of it. But we continue to pump e.g. Aminogylcosides like Gentamicin into people which while very effective against certain hard to treat organisms can fail due to 1. pharmacokinetic limitations (low intracellular penetration etc.) and 2. toxicity and damage.
Im not interested in the discussion about efficacy for treatment in people, as that wouldnt be on the subject of chemistry and that has already been shown, but rather what is it that makes it so hard about this being available? I found literature ranging from 20 years ago showing specific PH sensitive liposome encapsulations of Gentamicin vastly increased efficacy against gram negative intracellular infections in vitro and in vivo and immensely reduced toxicity towards human cells - yet its just nowhere to be seen.
Is it that hard to synthesize/make these compounds? Does it cost a castle made of gold to produce a single vial of it? I would like to hear chemists views on the complexity/cost ratio of the production if possible. I do no have an advanced background in chemistry but it is just hard for me to imagine it would take half the world to produce some specifically charged or ph sensitive liposomes - especially if the science on what and how has already been done.

Liposomes are the first nano-drug delivery systems that have been successfully translated into real-time clinical applications. So far, all liposomal drugs are designed for intravenous applications with control release. Most of them are designed to certain targets such as cancer cells. That's why most of liposomal drugs are used in chemotherapy (Ref.1):

The main fact is those liposomal drugs are used mostly in hospitals with control of appropriate medical personals. For example, all seven liposomal drugs so far in chemotherapy are Doxil®, Depocyt®, DaunoXome®, Mepact®, Marqibo®, Myocet®, and Onivyde® as listed in Ref.1 used in cancer patients (in- or out-patients, but under the supervision of their medical doctors). Morphine sulfate extended-release liposome injection, DepoDur™, and non-opioid, bupivacaine liposome (trade name Exparel®) are analgesics used in pain management, specially after postsurgical procedures (e.g., bupivacaine liposome is a postsurgical analgesic used in the management of postsurgical pain and It can also numb an area of the body to relieve pain during surgery or medical procedures). In addition, VYXEOS® is recently FDA approved  (First approved August 3, 2017) liposomal drug for the treatment of adults with newly-diagnosed therapy-related acute myeloid leukemia (t-AML) or AML with myelodysplasia-related changes (AML-MRC). VYXEOS®, a liposomal daunorubicin/Cytarabine, is a fixed-dose combination of liposomal bound cytarabine and daunorubicin that delivers the two medications in a 5:1 molar ratio. Further, it is the ONLY treatment recommended by National Comprehensive Cancer Network® (NCCN®) for induction in sAML patients $\ge \pu{60 years}$. Also, in addition to its use in chemotherapy for leukemia, DaunoXome®, a liposomal daunorubicin, is also FDA approved to treat AIDS related Kaposi's sarcoma.
Based on these facts, therefore, it is safe to say that liposomal drugs are in use without aware of general public. Regardless to say that those treatments are very expensive as you have guessed.
I think main drawback to not available liposomal antibiotics is their high vulnerability to acidic conditions, thus no oral applications available (almost all liposome broke open in the stomach). So far, only literature available for acid stable liposome is Ref.2 and 3. However, we have to stop the development in that direction due to the lack of funding. Therefore, do not believe what you are hearing. Liposomal applications and their oral applications are coming pretty soon (someone else may follow our direction). Nonetheless, now we have a vaccine for COVID-19 within a year, don't we?
References:

Upendra Bulbake, Sindhu Doppalapudi, Nagavendra Kommineni, Wahid Khan, "Liposomal Formulations in Clinical Use: An Updated Review," Pharmaceutics 2017, 9(2), 12 (33 pages) (doi: 10.3390/pharmaceutics9020012)(PDF).
Adelphe M. Mfuh, Mathew P. D. Mahindaratne, Maritza V. Quintero, Frederick J. Lakner, Ande Bao, Beth A. Goins, William T. Phillips, George R. Negrete, "Novel asparagine-derived lipid enhances distearoylphosphatidylcholine bilayer resistance to acidic conditions," Langmuir 2011, 27(8), 4447-4455 (DOI: https://doi.org/10.1021/la105085k).
George R. Negrete, Mathew P. D. Mahindaratne, Adelph M. Mfuh, Maritza V. Quintero, "Compositions and Methods Related to Acid Stable Lipid Nanospheres," US20110268653A1 (https://patents.google.com/patent/US20110268653A1/en) (PDF).


Why should 2,6-Dimethyl-4-nitrophenol be more acidic than 3,5-Dimethyl-4-nitrophenol I was trying to compare acidity of some organic compounds but can't understand some:

Why should 2,6-dimethyl-4-nitrophenol be more acidic than 3,5-dimethyl-4-nitrophenol?

I have been trying to figure it out using various factors like hyperconjugation and induction effect of Me group but its not working. Suppose i take H.C effect 2,6-dimethyl-4-nitrophenol be "less" acidic than 3,5-dimethyl-4-nitrophenol since it destabilizes the negative charge on conjugate base. But this logic is not working.

In the 3,5-dimethyl isomer, steric effects come into play and influence the acidity.

Resonance structures like B in the figure below, that stabilize the phenoxide anion explain why 4-nitrophenol ($\ce{R=H}$) is more acidic than phenol.  Resonance structure B requires that the nitro group be planar with the aromatic ring.  If methyl groups are placed adjacent to the nitro group ($\ce{R=CH3}$), the steric interactions between the two methyl groups and the nitro group becomes large and forces the nitro group to rotate out of the aromatic plane,  Hence resonance structure B can no longer contribute and the compound is less acidic.


Which tautomer of hexane-2,4-dione is more stable? Which tautomer will be more stable?

Both 1 and 2 show intramolecular hydrogen bonding and form 6 membered cyclic ring, thus increasing the stability.

I think I need to compare the effect of the two alkyl groups ,ie, methyl and ethyl  but how?

TL;DR
The question you are asking is not worth asking. Systems like these are hard to judge and the equilibrium between these structures is extremely fast. Considering any structure as more stable than another is certainly a mistake, because it completely ignores that the molecule is not static.

The double minimum of the tautomers of acetylacetone
 Long time it was believed, that for the equilibrium of acetylacetone (AcAc) the enol tautomer (4-hydroxypent-3-en-2-one) has a symmetric hydrogen bond and shows therefore only one minimum, like the following scheme by DMacks from Wikipedia suggests:

A quite recent publication by Howard et. al. nicely explores the potential energy surface of this molecule and manages to explain the transitional barrier.[1] They find the barrier to be around 10.5 kJ/mol (2.5 kcal/mol) which translates to a rate of 1011 s-1, which is ultrafast. At room temperature you are not able to distinguish between those structures.[2,3]

The equilibrium of hexane-2,4-dione
When we move away from the symmetrically substituted case, we can expect things to be similar, but a bit more complicated. First of all, we would expect a transition barrier to be roughly around the same size, so the transition from one form to the other is still ultrafast.[4]

I did a few, very rudimentary calculations on the subject and I have to admit, that they were far more complicated than I have anticipated. For reasons I will explain in the notes section, I choose to only discuss the electronic energies on the DF-BP86/def2-SVP level of theory.[5]

First of all it should be noted, that the dione open form 3 is more than 35 kJ/mol higher in energy and will play an insignificant role in the whole mixture. However, I chose this to be the reference state.
The (4Z)-5-hydroxyhex-4-en-3-one 1 is therefore 39.3 kJ/mol lower in energy, while (3Z)-4-hydroxyhex-3-en-2-one 2 is 37.2 kJ/mol lower in energy. This separates the two states by just 2.1 kJ/mol. At room temperature it is impossible to really distinguish those two. It is an incredibly shallow potential energy surface. The transition state 4 is at -36.5 kJ/mol, which makes the activation barrier less than one kJ/mol. Now we should not trust these values absolute, but they give us a rough idea about the whole process. The virtual mode in the transition state is depicted below.

The same mode in 1 looks like this:

And in 2 it looks like this:

You can see how these modes are all essentially the same. Again you will probably have to cool your system to below 50 K to actually see a separation of those.

One remarkable feature is that against the predictions of judging the +I effect, this level of theory would actually predict 1 being more stable than 2. This is not even surprising since the ethylene chain is also rotating out of plane in this structure.[6]

Conclusion
The tautomers of hexane-2,4-one are in a very fast equilibrium with each other. Based on simple models, like +I effect, it is impossible to judge which tautomer would be more stable, i.e. which tautomer has the lowest absolute energy. Not even higher level approximations like density functional theory are reliable enough to predict this.

Notes and References

Daryl L. Howard, Henrik G. Kjaergaard, Jing Huang, and Markus Meuwly, J. Phys. Chem. A 2015, 119 (29), 7980–7990.
Assuming that every transition leads to the formation of the associated structure, i.e. $\kappa =1$, which is reasonable for this small a barrier, we can approximate via transition state theory the rate for a reaction with a barrier of around 10 kJ/mol:
$$ k = \kappa\frac{k_\mathrm{B}T}{h}e^{\frac{- \Delta G^{\ddagger }}{RT}}\approx 1\times10^{11}~\mathrm{\frac{1}{s}}.$$
You might need to cool the system to about 50 K to actually see the formation of two minima. Of course for the symmetric case these two would be identical, so you would still need to mark one of the sides with an heavier isotope. This of course will influence the measurement, but I think you can get where this is going to.
If you are as confused as me about the correct IUPAC nomenclature of the compound see: How to name the tautomers of hexane-2,4-dione correctly according to IUPAC rules?
This system is incredibly hard to treat with DFT. It might be also very hard with more elaborate quantum chemical methods. The transition state is easily obtained, but a caluculation following the intrinsic reaction coordinate seems to be impossible. The reason for this is that the transition state itself is in very close proximity to the structure of 2. The position of the proton does differ by less than 0.2 Å. See a figure of the superimposed structures.
Accordingly all calculations involving frequencies will fail, since the harmonic treatment is simply not enough for this system. This is actually an interesting case, where the structures are in very close proximity, so that adding thermal contributions leads to completely nonsensical results. Here it is, that the activation Gibbs energy between the tautomers is negative.
Fast equilibria like these cannot any longer be described accurately in the Born-Oppenheimer approximation, hence molecular dynamics need to be involved, or at least high level geometry optimisations on the coupled-cluster level with a triple zeta basis set. (The last time I tried that, it already took forever.)
Reasons for this to happen are plenty. It is partially discussed in Why is the eclipsed conformer more stable than the bisected conformer in 1-butene? Also related to this is Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered?


Halogen and inductive effect background :

  As proved by rate of EAS, mesomeric or simply conjugated resonance is weaker for halogens than inductive and make them electron taking, with fluorine almost doing nothing due to strongest mesomeric and strongest inductive among halogens, So, halogens should take electrons away from any system if they do it in a specially conjugated system known for many resonating structures

If we have simple carbocation as shown in the chlorine and carbocation, being right next to chlorine Inductive effect must overpower resonance/mesomeric but why the opposite happens?

Simply put, the energy costs for a sextet carbocation are too high to allow chlorine's strong inductive effects to dominate the situation.

The driving for for a carbon octet is very high.


Can the 1st generation Grubbs Catalyst be used for ring closing metathesis on non-terminal alkenes? I'm wondering if the 1st generation Grubbs catalyst can be used to do ring closing metathesis on a non-terminal alkene.

Yes, it appears that that ring closing metathesis of non-terminal alkenes has been performed using the 1st generation Grubbs catalyst several times:

Asymmetric syntheses of enantiopure C(5)-substituted transpentacins
via diastereoselective Ireland–Claisen rearrangements
Stereoselective Synthesis of Deuterium-Labeled (2S)-Cyclohexenyl
Alanines, Biosynthetic Intermediates of Cinnabaramide
Ru-catalyzed metathesis of octadienylether xyloside Synthesis
of (−)-Agelastatin A by [3.3] Sigmatropic Rearrangement of Allyl
Cyanate
A chiral auxiliary cleavable by ring-closing alkene
metathesis — Efficient synthesis of chiral nonracemic
cycloalkenes
Conversion of chiral unsaturated cyanohydrins into
chiral carba- and heterocycles via ring-closing metathesis
Total Synthesis of Pancratistatin Relying on the [3,3]-Sigmatropic
Rearrangement
Stereoselective Chelate-Controlled Addition of Grignard Reagents to Unsaturated Medium-Ring Heterocycles


Rate of Fenton Reaction I want to determine rate of Fenton reaction with different concentration of $\ce{Fe^2+}$. These are some questions that I have encountered:

How long does it usually take for the reaction to complete?

What concentration of $\ce{Fe^2+}$ is usually used for the reaction?

I am planning to use titration of $\ce{H2O2}$ with $\ce{KMnO4}$ to determine the rate of reaction. Is it possible to quench to reaction mixture by adding an excess of EDTA solution and would it affect the titration?

NEW:
How can I determine the rate of Fenton reaction? I proposed to titrate $\ce{H2O2}$ with $\ce{KMnO4}$ to determine change in $\ce{H2O2}$ as a mean of determining rate of reaction. However, I realised that $\ce{KMnO4}$ would also react with $\ce{Fe2+}$, so I need to either find another way to measure the reaction rate or find a way to remove all $\ce{Fe2+}$. Also, I am not sure if there would be any reaction between $\ce{KMnO4}$ and the hydroxyl radicals.
I am also wondering about adding in some wastewater and determine the before and after COD to determine the rate of reaction. However, I am wondering if the water that don't really have high COD (as opposed to the industrial wastes used in other papers) would result in a very slow reaction rate. Also, it seems like the part involving oxidizing by hydroxyl radicals is very exothermic. If I don't want to add $\ce{H2O2}$ bit by bit to control rate of reaction, is it OK?
Do I have to add acid to $\ce{Fe2+}$ and $\ce{H2O2}$ for the reaction to occur?
Is it necessary to add $\ce{H2O2}$ bit by bit, if so how much should I add at each turn?

This is only a partial answer:

I am not sure about which step you're looking for or the substrate of your reaction, but the Fenton-based chemistry I work with is extremely fast. I use it to (supposedly) form a methyl radical from DMSO which then reacts with an α-iodoester to form another radical.

Literature suggests I add brine immediately after I've completed the hydrogen peroxide addition and work it up from there. Truth be told, I usually give it a few minutes, but the reaction is certainly quite fast. The whole process takes me less than 5 minutes, and if I wasn't using DMSO and skatole (so smelly), it would be my favorite reaction of the ones I do.

I use iron(II) sulfate heptahydrate as my catalyst, and I use 0.5 equiv. relative to the iodide. It ends up being about 0.1 g per milliliter of solvent.

As for the peroxide titration, I use a decent excess, so I've never needed to titrate it. That is why I said this was a partial answer. I will say this: don't quench large amounts of hydrogen peroxide with sodium thiosulfate at room temperature. You end up forming sulfuric acid in aqueous solution. Very exothermic (like the water in the flask can start boiling in your hand).


When writing a paper should I describe the procedure thoroughly? I have not written any article on a chemistry subject yet, and I want to use good practices from the start.
What I have noticed reading papers, is that a lot of times it is not specified the exact procedure that is followed in a reaction, for example by saying the exact tools that are used, or the parameters used in a technique, not even in supporting information documents.
At first I thought that some information was not really relevant, but as I read more articles I found more things like “X product is centrifugated” or “Y is sonicated”, without further detailing the exact procedure that was followed, the set parameters or even the devices that were used, which can obviously be very important for a given reaction.
My opinion would be that the more information the better and I think that by omitting that information you are wasting time and resources to any researcher that tries to replicate your experiment, as well as making it difficult to check the replicability of the procedure.
Then my question is, how far should I go in describing the procedure? Should I mention even which glassware I use for a given reaction? Maybe sometimes it is pretty obvious what should be used, but it is additional information that does not hurt to know and may be useful for obtaining the exact same results. However, it is as far as I have seen not a common practice and there must be a reason behind that.
Maybe omitting this information is preferable in articles as it uses less words and it avoids making articles excessively wrong, if that is the case, what about reports or thesis with a less restricting length limits?
Thank you for your time.

Use the following criterion in mind when you write a method in a paper: If you were reading this synthesis procedure from a book or a paper, could you replicate it or not? Do you feel anything is missing for proper duplication? When I review a methodology section of a papwe, I keep this criterion in mind: "Using this written information, can I reproduce his/her work?", otherwise I ask for more information.
Since you mentioned synthesis, a good sample procedure could be from Organic Synthesis http://www.orgsyn.org/
See how they write good procedures, and these methods are verified by independent person. Additionally, some journal put restrictions on word count, but still you want more details. For that, Supporting Information file exists on most good quality journals. I once saw someone even added pictures of their synthetic procedure/apparatus. Nobody will mind but it is not common.
Odd cases: Mentioning the source of glassware might be important if glassware surface were affecting the reaction. I "re-discovered" this only once. Sometimes the age of a solvents might affect the reaction. Tetrahydrofuran when used a solvent can have traces of peroxide which can show new unexpected products.
Now, there is a trend these days to intentionally hide some information so that competitor's cannot catch up with your work or write the method in a sloppy bare minimum way. This is really sinister but it happens and such practices must be curbed. Ayway scientists are not saints-they are human beings like everyone.


is there any difference between glycerin and glycerol? In laboratory setting, is there a difference between glycerin and glycerol?

There are some conflicting info on this topic.

Nope, there is no chemical difference between glycerol, glycerin or glycerine.  All 3 names refer to the same compound, propane-1,2,3-triol.


Why is zinc sulfate used in daniell cell Why is zinc sulfate needed in the construction of a daniell cell.  It seems that because the zinc is getting oxidized and not reduced there is no need to have a solution of zinc sulfate as it will produced when the zinc is reduced.  The sulfate ions would simply cross the salt bridge and balance out the charges.

Zinc sulfate is not needed. With any other neutral salt in its place, your cell will work just fine.

Without any salt at all, though, it will not work, since pure water is a poor conductor of electric current.


Could sodium in ethanol reduce amides to amines? Strong reducing agents like $\ce{LiAlH4}$ can reduce amides to amines. Sodium in ethanol (as a source of protons) is also a very strong reducing agent. Could it be therefore used to, for example, turn formamide into methylamine?

Yes, that is used as a pathway for synthesis of amines from amides, albeit rarely.

I quote from the abstract of a communication that reports the same: (emphasis mine)

  The reduction of amides to their corresponding amines has been used extensively in the synthesis of many drugs and their intermediates. This reaction has usually been carried out by means of metal hydrides (especially lithium aluminium hydride) or by catalytic hydrogenation over various catalysts. Metal and alcohol has been used for the reduction of ketones to their corresponding hydroxy compounds and oximes to their corresponding amines, but the reduction of amides under these conditions, although previously reported, is not well documented in major works of organic chemistry. This procedure is specially useful for industry since it is safer, less expensive, requires less time and provides higher yields (65-80 per cent) when compared with lithium aluminium hydride or catalytic reduction methods.

The emphasized text is written in a slightly confusing language but nonetheless confirms that amides can be reduced to amines using $\ce{Na/ROH}$, otherwise known as Bouveault-Blanc reduction.

References:

Bhandari, K. , Sharma, V. L. and Chatterjee, S. K. (1991), ChemInform Abstract: A Convenient Method for the Reduction of Amides to Their Corresponding Amines.. ChemInform, 22: no-no. DOI: 10.1002/chin.199118077


Reaction between phenol and nitrous acid My question is about the products formed when phenol reacts with nitrous acid. I googled it and reached this PubChem paper (DOI: 10.1016/S0045-6535(02)00857-3), which states:

  It was found that phenol reacts with nitrous acid to produce cyanide ions. Cyanide ion generation is attributed to the conversion of phenol to nitrosophenol through the well-known nitrosation reaction, and decomposition of benzoquinonoxim to form cyanide and aliphatic compound.

Is this correct? If not, then what is the actual reaction?

What the reference says

The reference gives only following figure to support their theory. A weakness of this mechanism is requiring a high energy intermediate like sp2-carbocation.

An alternative proposal

This alternative avoids high-energy intermediates and thus appears more likely:


How to find maximum and minimum oxidation number of an element I've researched around and some sources claim, that in order to find the minimum and maximum oxidation number you do this:

Maximum: the group the element

Minimum: the group of the element - 8

However I cannot get this to work for Fe? I'd appreciate a clarification. :)
It seems to work for Sulphur and what I've read about it's oxidations.

Maximum: 6
Minimum: 6 - 8 = -2

Don't metals always have a positive oxidation number?
Does this mean for Fe (iron) it's 0 to +3?

(I know this is a duplicate, sorry, however the others weren't answered and asked years ago).

The rule you quoted is generally true for main group elements — but only if you count the groups in the older main-group/transition-metals formality. The newer IUPAC terminology counts the s, p and d element groups one after the other, so that sulphur would be in group 16. For that nomenclature, you need to subtract 10 from the group number for the maximum oxidation state.

Example: Phosphorus (group 15 by current terminology; group V by older terminology). Maximum oxidation state: $\mathrm{+V}$. Minimum oxidation state $\mathrm{-III}$.[1]

This does have a general quantum chemical explanation. These oxidation states add up to eight, which is exactly the number of electrons that typically make up the outer (valence) shell — where chemistry happens.

If we start at phosphorus($\mathrm{V}$) and want to remove another electron, we would have to remove this electron from the core orbitals. These are already pretty stable (i.e. low in energy) in the ground state, but for every electron removed they get stablised further. Thus, they end up in a very stable state and removing gets all the much harder.

Likewise, every electron added to a neutral compound will destabilise all orbitals of that atom. If three are already added to phosphous, giving phosphorus($\mathrm{-III}$), the next electron would need to be added to a distant orbital, which again is not something easy. This is why there is typically a range of eight for the chemically accessable oxidation states of main group metals.

However, I am not willing to bet anything that the s-elements of periods 4 and higher (potassium and below and calcium and below) stick to that rule; for the reason see and interpret below. So far, only $\mathrm{+I}$ and $\mathrm{-I}$ are known for alkaline metals (group 1) and only $\mathrm{+II}$ and $\mathrm{+I}$ for alkaline earth metals are known (save the elemental oxidation state $\pm 0$).

Transition metals are a lot harder. In theory, you could assume a range of either ten (d-electrons only), twelve (d and s-electrons) or eighteen (d, s and p-electrons). In fact, experimentally a range of ten has been established for chromium, manganese and iron while a range of twelve has been established for osmium and iridium. (Source: Wikipedia)

So far, the highest oxidation state has been found for iridium ($\mathrm{+IX}$). Platinum($\mathrm{X}$) has been predicted. (Source: also Wikipedia) Note that these states are well within the ‘s and d’ idea I alluded to earlier. To the best of my knowledge, there are no pieces of evidence pointing towards the ‘s, p and d’ idea.

There are a large number of unknown oxidation states for transition metals, e.g. while chromium($\mathrm{-IV}$) and ($\mathrm{-II}$) are known, chromium($\mathrm{-III}$) is not.

The notion that metals could only have positive oxidation states is incorrect. As an example, $\ce{[Fe(CO)4]^2-}$ with an iron oxidation state of $\mathrm{-II}$ is known.

All of this complicates the analysis strongly. Until much more research has been performed, you should probably not attempt to predict maximum and minimum oxidation states of these elements.


Confusion about comparing the charge densities of benzene and fluorobenzene I understand that the functional group $\ce{-F}$ is an electron-withdrawing group. It is expected that the difficulties of an electrophile to attack the carbon atoms are $$\ce{PhH}<\ce{2,4of PhF}<\ce{3 of PhF}\tag{1}$$Therefore, I want to demonstrate it quantitatively by computing the charge densities of carbon atoms.
Using command # opt freq hf/3-21g pop=nbo geom=connectivity
and the molecule models in Gaussian 09,
I obtained the charge densities of $\ce{PhH}$ and $\ce{PhF}$. To my surprise, the result is:
$$\begin{array}c\text{molecule}&\text{ortho}&\text{meta}&\text{para}\\\ce{PhH}&-0.239&-0.239&-0.239\\\ce{PhF}&-0.306&-0.222&-0.261\end{array}$$
By observing this result, the fact that the electrophilic substitution of $\ce{PhF}$ is likely to occur in o- and p- is verified, or the right half of $(1)$ is verified.

  But why would the charge density of position 2,4 of $\ce{PhF}$ is more negative than that of benzene, given the fact that $\ce{-F}$ decreases rate of the electrophilic substitution reactions?

My speculation
i) $(1)$ may be false.
ii) The NBO charge densities between two different molecules cannot be compared.
iii) The charge density cannot indicate the difficulty of the attacking of electrophile.

Which one, or something else, can explain this strange observation? Can another method to order electron-withdrawing ability of some functional groups computationally be suggested?

This is the way of directing the electrophiles to the ortho and para places  of the benzene cycle. As you see the ortho and para places are more negatively charged.That's true. It is due to the ability to donate its electron pair to the benzene cycle.

But an electronegative element can attract the electron cloud to it. Fluorine is the most electonegative element, so it has a greater ability to attract the electron cloud of the benzene cycle. So that makes it relatively  resistant to an incoming electrophile.

This does not say that it is not a  ortho-para director. It's a ortho-para director but it deactivates the ortho and para places of the benzene cycle.


Why is methyl iodide more dense than ethyl iodide, even though the latter has more carbon atoms? It is understood that more carbon atoms present in a hydrocarbon, the greater the density of the hydrocarbon.

But then why does $\mathrm{CH_{3}I}$ have a greater density than $\mathrm{C_{2}H_{5}I}$?

For the same reason why a bucket of sand will weigh more than a bucket of rocks.
Smaller molecules will result in a more dense, thus more heavy substance.


Gibbs Free Energy Notation What's the difference between $G$, $\Delta G$, $\Delta G^\circ$, and $\Delta_\mathrm r G^\circ$?

I've seen the first two used interchangably, and seen $G$ specifically referred to as change in Gibbs Free energy. Is this common? Are there cases where $G$ refers to absolute Gibbs Free Energy?

For those with the circle ($^\circ$), I've seen that referred to as "standard", and this means that they are in reference to standard pressure or concentration. Does this make their use noticably different?

$G$ is the Gibbs free energy of a system. It is a conceptual quantity in the sense that there is no reference point that defines $G = 0$ for a substance (unlike entropy). Whenever you see a plot of $G$, notice that the scale is not labeled with absolute values (no indication of where the origin is).
$\Delta G$ is the change in Gibbs free energy, i.e. $G_\text{final} - G_\text{initial}$. Because it is a difference, there is no issue with the lack of reference point.
$\Delta_\mathrm r G$ is the change in $G$ as a reaction proceeds, i.e. $\Delta_\mathrm r G = \frac{\mathrm dG}{\mathrm d\xi}$ with $\xi$ the extent of reaction variable. Because of this definition, it is an intensive quantity with dimensions energy per amount of substance. It is called the Gibbs energy of reaction.
$\Delta_\mathrm r G^\circ$ is the standard Gibbs energy of reaction, for the special case when all reactants and products are at standard state.
$\Delta G_\mathrm f^\circ$ is the standard Gibbs energy of formation of a substance, the $\Delta_\mathrm r G^\circ$ for the reaction that makes the substance from the elements (where all products and reactants are at standard state).
$\Delta G^\circ$ is for any process at standard state (e.g. ice melting at a pressure of 1 bar). It is also sometimes used instead of $\Delta_\mathrm r G^\circ$, but that is confusing because the meaning and the dimensions are different.

  I've seen the first two used interchangably, and seen $G$ specifically referred to as change in Gibbs Free energy. Is this common? Are there cases where $G$ refers to absolute Gibbs Free Energy?

$G$ should always refer to absolute Gibbs energy, and never to a change in Gibbs energy.

  For those with the circle ($^\circ$), I've seen that referred to as "standard", and this means that they are in reference to standard pressure or concentration. Does this make their use noticably different?

Yes, because the Gibbs energy is concentration-dependent. With $Q$ as the reaction quotient,

$$\Delta_\mathrm r G = \Delta_\mathrm r G^\circ + R T \ln Q$$

As concentrations change in a reaction, $\Delta_\mathrm r G$ changes but $\Delta_\mathrm r G^\circ$ does not. When a reaction approaches equilibrium, $\Delta_r G$ approaches zero (but $\Delta_\mathrm r G^\circ$ doesn't change, as just mentioned).

The reason that $\Delta_\mathrm r G$ is concentration-dependent is that the Gibbs energy has an entropy component ($G = H - T S$), and there is an entropy of mixing that depends on concentrations (the $R \ln Q$ part).


How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenes It's easy to apply CIP rules to chiral carbons but what about in chiral molecules without these carbon cneters like biphenyls, sulphoxides and allenes? Are they applied at all?

Structures with axial chirality (e.g. an allene or a 2,2′,6,6′-tetrasubstituted biphenyl) with two pairs of substituents (ab and cd) are regarded as elongated tetrahedra and viewed along the chiral axis. The substituents (a, b, c, and d) are arranged in the elongated tetrahedral system.

In each pair (ab as well as cd), one higher-ranking substituent (a > b and c > d) is chosen using the sequence rule.

If the path going from a to b to c (while looking toward d) is clockwise, the chirality is described by the symbol Ra. The symbol is Sa if this path is anticlockwise.


Proper common notation for a ligand I've observed several notations for the ligand in the literature in general, say, for the typical complexes like $\ce{MX_n}$ ($\ce{M}$ - metal) where ligand $\ce{X}$ is denoted as (according to my subjective frequency of occurrence):

$\ce{L}$;
$\ce{Lig}$;
$\ce{Lg}$;
$\ce{Lgnd}$.

I can imagine $\ce{Lg}$ being a poor choice as it can be easily confused with leaving group, and $\ce{Lgnd}$ which looks more like a company's NASDAQ code, but what about the first two options? And what normative document governs the proper notation anyway?

Regarding the $\mathrm{L}$ use for leucine (see comments section here and a thread on Biology.SE What is the preferred way to abbreviate amino acids?), I encountered more sources using $\mathrm{Leu}$ for that matter rather than $\mathrm{L}$. I guess 1-letter code for 20 amino acids is quite suitable and eliminates confusion between leucine ($\mathrm{L}$) and lysine ($\mathrm{K}$), but for me handling $\mathrm{Leu}$ and $\mathrm{Lys}$ is mnemonically easier.

The ACS Style Guide suggests using $\ce{L}$ [1, p. 186] in APPENDIX 10-2 Abbreviations, Acronyms, and Symbols:

  \begin{array}{ll}
[\ldots] & \\
\small{L} & \text{absolute conﬁguration} \\
\mathrm{L} & \text{leucine} \\
 & \text{ligand} \\
 & \text{liter} \\
\mathrm{L_I} & \text {spectral line} \\
[\ldots] &
\end{array}

IUPAC "Green Book" Quantities, units, and symbols in physical chemistry also defines $\ce{L}$ [2, p. 160] in chapter 9 ABBREVIATIONS AND ACRONYMS:

  \begin{array}{ll}
[\ldots] & \\
\mathrm{KS} & \text{Kohn-Sham} \\ \\
\mathrm{L} & \text{ligand} \\
\mathrm{L2TOFMS} & \text {laser desorption laser photoionization time-of-flight mass spectroscopy} \\
[\ldots] &
\end{array}

Glossary of Class Names of Organic Compounds and Reactivity Intermediates Based on Structure (IUPAC Recommendations 1995) uses $\ce{L}$ [3, p. 1349]. For example, see this entry:

  metal-carbyne complexes:
  Metal complexes of the type $\ce{RCML_n}$, ($\ce{M}$ = metal, $\ce{L}$ = ligand) in which formally a carbyne is coordinated to a metal. E.g.

$\mathrm{Lig}$ and other notations are not mentioned in the above sources as suitable ligand notations.

Bibliography

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. DOI 10.1021/bk-2006-STYG.
IUPAC “Green Book” Quantities, units, and symbols in physical chemistry, 3rd ed.; Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Moss, G. P.; Smith, P. A. S.; Tavernier, D. Pure and Applied Chemistry 1995, 67 (8–9), 1307–1375 DOI: 10.1351/pac199567081307.


Chemisorption vs Physisorption? What is the fundamental difference between chemical adsorption and physical adsorption.

Why do some compounds chemisorb and others physisorb?

It appears to be one of degrees of bonding, without a clear distinction. As stated in this Wikipedia article:

  Physisorption, also called physical adsorption, is a process in which
  the electronic structure of the atom or molecule is barely perturbed
  upon adsorption.$\mathrm{^{[1][2][3]}}$

A more thorough description is given later in the article:

  "The fundamental interacting force of physisorption is caused by van
  der Waals force.   [...]   In comparison with chemisorption, in which
  the electronic structure of bonding atoms or molecules is changed and
  covalent or ionic bonds form, physisorption, generally speaking, can
  only be observed in the environment of low temperature (thermal energy
  at room temperature ~26 meV) and the absence of the relatively strong
  chemisorptions. In practice, the categorisation of a particular
  adsorption as physisorption or chemisorption depends principally on
  the binding energy of the adsorbate to the substrate."

1) K. Oura; et al. (2003), Surface Science, An Introduction, Berlin: Springer, ISBN 978-3-540-00545-2
2) M. C. Desjonqueres; et al. (1996. Corrected printing 1998), Concepts in surface physics (2nd ed.), New York: Springer-Verlag, ISBN 3-540-58622-9
3) Hans Luth; et al. (1993), Surfaces and interfaces of solids, Springer-Verlag, ISBN 978-3-540-56840-7


Why is ligand substitution only partial with copper(II) ions and ammonia? When studying ligand substitution (at UK year 13 level), the following example has been given:

\begin{align}
\ce{[Cu(H2O)_6]^2+ + 2NH3 &<=> [Cu(OH)_2(H2O)_6] + 2NH4^+}\\
\ce{[Cu(OH)_2(H2O)_4] + 4NH3 &<=> [Cu(NH3)_4(H2O)_2]^2+ +2OH- +2H2O}\\
\end{align}

There has been no explanation given of why the substitution is only partial, whereas the substitution of, say, $\ce{[Co(H2O)_6]^2+}$ is complete.

To me, it seems that this partial substitution wouldn't be favourable, as it produces a slightly distorted octahedral shape, which is less stable than a regular octahedral shape, surely? Moreover my textbook states ammonia is a better ligand than water, so why would $\ce{NH3}$ not replace those final two aqua ligands?

Metal ion complexes have stepwise stability constants:
\begin{align}
\ce{[Cu(H2O)6]^2+ + NH3 &<=>[$K_1$] [Cu(NH3)(H2O)5]^2+ + H2O}\\
\ce{[Cu(NH3)(H2O)5]^2+ + NH3 &<=>[$K_2$] [Cu(NH3)2(H2O)4]^2+ + H2O}\\
\ce{[Cu(NH3)2(H2O)4]^2+ + NH3 &<=>[$K_3$] [Cu(NH3)3(H2O)3]^2+ + H2O}\\
\ce{[Cu(NH3)3(H2O)3]^2+ + NH3 &<=>[$K_4$] [Cu(NH3)4(H2O)2]^2+ + H2O}\\
\ce{[Cu(NH3)4(H2O)2]^2+ + NH3 &<=>[$K_5$] [Cu(NH3)5(H2O)]^2+ + H2O}\\
\ce{[Cu(NH3)5(H2O)]^2+ + NH3 &<=>[$K_6$] [Cu(NH3)6]^2+ + H2O}\\
\end{align}
with

\begin{align}
K_1 &= \frac{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_2 &= \frac{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]}}{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_3 &= \frac{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]}}{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_4 &= \frac{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]}}{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_5 &= \frac{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]}}{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_6 &= \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]\left[ \ce{NH3} \right]}}\\
\end{align}

The overall stability constant is:

$$K_\text{B} = \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} = K_1 \cdot K_2 \cdot K_3 \cdot K_4 \cdot K_5 \cdot K_6$$

(In particular, note the exponent 6.)

Furthermore, usually $K_i/K_{i+1}>1$; thus, $K_1 > K_2 > K_3 > K_4 > K_5 > K_6$. Consequently, it gets more and more difficult to exchange further ligands.

Therefore, the predominating complex strongly depends on the concentration of $\ce{NH3}$.
And you probably work with dilute aqueous solutions.

Simply speaking, the missing $\ce{[Cu(NH3)6]^2+}$ in aqueous solution can be attributed to the high concentration of water which is competing with ammonia for the coordination sites.


Caclulating the Solubility of Fluorite in River Water Source: Chapter 4 of Brunlow's 1979 Geochemistry text.

  Calculate the solubility of fluorite in river water with the average composition given in table 4-1.  Assume that no fluoride complexes form.

Table 4-1 lists ppm, mEq/L and moles/L values for constituents, anions and cations found in the water sample.  There is no value for $\ce{F}^-$ but there is for $\ce{Ca}^{2+}$. The molarity of $\ce{Ca^{2+}}$ is given to be $\pu{0.375E-3M}$

My attempt::

$$\ce{CaF2 -> Ca^{2+} + 2Fl^{-}}$$

$K_\mathrm {sp}$ of $\ce{CaF2}$= $10^{-10.5}$ (from a table in the book)

$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{F_{\text{added}}}]^2$

$[\ce{F_{\text{added}}}] = [2\ce{Ca_{\text{added}}}]$

$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][2\ce{Ca_{\text{added}}}]^2 = 4[\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{Ca_{\text{added}}}]^2$

Am I on the right track?  I end up with a polynomial to solve for.  The answer in the back of the book is $10^{-4.64}$ which I can't arrive at.

The other method I've tried is to calculate the ppm of fluorite dissolved in pure water ($\pu{15.6ppm}$) and then subtract the ppm value for the river water $\ce{Ca^{2+}}$ given in the table ($\pu{15.0ppm}$) and then convert the result ($\pu{0.6ppm}$) to $\pu{7.69e-6 moles/L}$ which doesn't equal the answer in the back of the book ($10^{-4.64}$), so I believe that I'm missing something conceptually.

the ppm of fluorite in pure water (15.6ppm)

the ppm of fluorite in pure water is zero!

  $K_\mathrm{sp}=[\ce{Ca_{original}} + \ce{Ca_{added}}][2\ce{Ca_{added}}]^2 = 4[\ce{Ca_{original}} + \ce{Ca_{added}}][\ce{Ca_{added}}]^2$

Yes, this is correct and can be solved, you say you know $[\ce{Ca_{original}}]$ is $\pu{0.375E-3M}$ and  $K_\mathrm{sp}$ is $10^{-10.5}$ so there is one equation and one unknown, just a matter of solving a cubic equation, which can be solved exactly or by approximate methods.


What are the key differences between Raoult’s Law and Henry’s Law Raoult’s Law ($p_A = \chi_A \cdot p_A^\circ$) becomes a special case of Henry’s Law ($p_A = K_H^{px} \cdot \chi_A $) when $K_H^{px} = p^\circ_A$, but where do these laws differ from each other?

There’s a statement in my text book which I’m having trouble understanding.

The statement:

  As a real solution approaches the limit of infinite dilution its components behave more ideal. The solvent obeys Raoult's law whereas solute (minor component) obeys Henry's law for dilute solutions.

Raoult's law can be seen to be valid for ideal liquids. The assumption behind Raoult's law is that both the liquid phase and the vapour phase behave ideally. This means that the interaction between any two molecules in the liquid and the vapour have no interaction.
The Raolt's law just states the mole fraction of a component in the liquid phase is proportional to the mole fraction of the same component in the vapour phase. This is valid only for mixtures of ideal liquids.
When you add a very small amount of impurity in a solvent (equivalent to infinite dilution) the interaction between the impurity and the solvent molecule is minuscule or technically infinitesimal. Therefore, the liquid still behaves ideally. Hence Raoult's law is applicable for the solvent.

Whereas Henry's law is also defined for a ideal mixture only this time the assumption is that the mole fraction of the component is tending to zero. this means that when very few molecules a present in a vast space, the interaction between them is almost zero. This makes the system ideal. therefore, an infinitesimally small quantity of impurity does not feel any interaction with another molecule of the same impurity in the solvent. Therefore, Henry's law is used for the solute. That is why Henry's law constant have huge values. Huge values makes sure that even small amount of the solute is accounted for when calculating the mixture properties.

For further clarity, please refer to chapter 6 of 'Molecular Thermodynamics of Fluid phase equilibria' by John M.Prausnitz et. al.


Oxidization of Copper In Electrolysis So during school today, I am running an electrolysis experiment with water and sodium bicarbonate copper leads on both the anode and cathode.  I am noticing the usual $\ce{NaO}$ on the anode, and $\ce{H2C2}$ on the cathode.  However, on the anode, the copper is obviously oxidizing, and changing the hue of the water to a bright green-blue.  How can this be calculated into the chemical equation?

Also, is this equation correct?

$$\ce{H2O + NaC2 + energy  ->  NaO + H2C2}$$

(with the order of anode then cathode)

Yes, I am aware that this is the best possible situation, but in general, is this what is supposed to be happening?

Sodium compounds:

$\ce{Na2C2}$ is sodium carbide. It reacts violently with water. $\ce{Na2CO3}$ is sodium carbonate. $\ce{NaO}$ is not a valid formula for any of the oxides of sodium: $\ce{Na2O},\ \ce{Na2O2},$ and $\ce{NaO2}$ are all real compounds, and they all react with water to produce $\ce{NaOH}$, which is probably what is formed.

  "Hydrocarbons"

If indeed you had $\ce{Na2C2}$, then you would form $\ce{H2C2}$ and $\ce{NaOH}$ when it was placed in water. However, you would not need electrolysis to do so. This is an acid-base reaction that is spontaneous under normal circumstances. $\ce{H2C2}$ cannot be the brown sludge on your cathode. It is a gas.

$$\ce{Na2C2 + 2H2O -> 2NaOH + H2C2 ^}$$

  What actually happened?

In an electrolytic cell, the process that occurs (at least under standard conditions) is the one with the least negative standard potential $E^o_{cell}$.

The possible oxidations are either the oxidation of copper or the oxidation of water. Sodium carbonate is the supporting electrolyte.

$\ce{2Cu + 2OH- -> Cu2O + H2O +2e-}\ \ \ E^o=+0.36 \ \text{V}$
$\ce{Cu -> Cu^{2+} + 2e-} \ \ \ E^o=-0.34 \ \text{V}$
$\ce{4OH- ->O2 + 2H2O +4e-}\ \ \ E^o=-0.40 \ \text{V}$
$\ce{2H2O -> O2 + 4H+ +4e-} \ \ \ E^o=-1.23\ \text{V}$

Since the oxidation of copper to $\ce{Cu^{2+}}$ has the least negative $E^o$, it is the most likely oxidation. $\ce{Cu^{2+}}$ also explains the color, since aqueous solutions of $\ce{Cu^{2+}}$ are blue. The first equation is also likely, since it has a positive $E^o$, however it does not explain the color.

The possible reductions are the reduction of water/hydroxide and the reduction of $\ce{Cu^{2+}}$, since we know that it is being formed.

$\ce{2H2O + 2e- -> H2 + 2OH-} \ \ \ E^o=-0.83 \ \text{V}$
$\ce{Cu^{2+} + e- -> Cu+} \ \ \ E^o=+0.16 \ \text{V}$

The reduction of $\ce{Cu^{2+}}$ to $\ce{Cu^{+}}$ is most likely, but not initially. Initially, the is no $\ce{Cu^{2+}}$, so the only reduction that can happen at the cathode is the first one, which produces hydrogen gas. The redox then might be: $$\ce{Cu + 2H2O -> H2 + Cu^{2+} + 2OH-} \ \ \ E^o =-1.19 \ \text{V}$$

Once the $\ce{Cu^{2+}}$ diffuses over to the cathode, a different reduction starts to compete, with a different net redox reaction: $$\ce{2Cu + H2O -> H2 + Cu2O} \ \ \ E^o=-0.47 \ \text{V}$$

This redox reaction still produces hydrogen, but it has a much less negative $E^o_{cell}$. The other peculiarity is that $\ce{Cu2O}$ is produced at the cathode after $\ce{Cu^{2+}}$ has diffused through the cell. $\ce{Cu2O}$ is your brown sludge, since like rust, it is both brown and completely insoluble in water.


What exactly happens to mass during deuterium fusion? I know that the mass of Helium-4 is slightly less than the mass of 2x Deuterium it may be made of and that it is the reason fusion gives a lot of energy.

But where exactly is this missing mass taken from? Aren't those 2 protons, 2 neutrons and 2 electrons forming Helium-4 atom the same as they were in previous Deuterium atoms?

Fusion of H-2 does NOT give He-4.  It gives ~50:50 H-3 (1.01 MeV) + p (3.02 MeV) or He-3 (0.82 meV) + n (2.05 MeV).  Either way, the missing "mass" is the mass equivalent of binding energy, E/c^2.  Binding energy is negative.  A bound system masses less than its separated constituent parts.  You must input energy (that being mass-equivalent) to unbind it.

Fusion converts weakly bound nuclei into more strongly bound nuclei.  The net difference in binding energies comes out as the kinetic energies of the product nuclei and particles.  Count neutrons and protons going in versus neutrons and protons coming out.  If one convetrs to the other, there must be elecron neutrinos or electron antineutrinos emitted to balance the books for energy, momentum and lepton generation number.

http://en.wikipedia.org/wiki/Nuclear_fusion
   Middle


Understanding the rate of change in titration curves As I am looking at titration curves, a few things stand out. For now, I have a two part question:

When titrating an acid with a base, for instance, the pH rises more or less abruptly around the equivalence point. Does this have to do with the logarithmic nature of pH? In other words, is this caused by the fact that the number of particles involved in the jump from e.g. a pH of 2 to one of 3 is several orders of magnitude than the number of particles involved in the jump from 6 to 7?
When a weak acid is titrated, the rise to the equivalence point is a lot more shallow. Does this happen because of Le Chatelier's principle? That is to say, as the base is titrating the acid, some of the acid that has not dissociated will dissociate in an attempt to restore equilibrium. This should retard the titration to some degree. Is that the case?

This answer from a member of Chemical Forums who seems to have been quite useful to other posters in the past given his or her "karma" would appear to confirm that the shape of titration curves owes in part to their logarithmic nature:

http://www.chemicalforums.com/index.php?topic=6140.0


Oxidation of metals/halogens by oxygen gas in acidic aqueous solution I had some confusion about the correct answer to the following question:

  Which substance can be oxidized by $\ce{O2(g)}$ in acidic aqueous solution?

Two of the answer choices were metal cations (usually not oxidized further). The remaining three choices were:

  $\ce{Br2(l)}$, $\ce{Ag(s)}$ and $\ce{Br-(aq)}$.

The standard reduction potentials were given:

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$

The correct answer given was $\ce{Br2(l)}$, however, this did not make sense to me as the oxidation of bromine would result in a positively charged bromine ion. Am I reasoning correctly that $\ce{O2(g)}$ can oxidize both $\ce{Ag}$ and $\ce{Br-}$ in acidic solution?

Since the value of $E^\circ$ for $\ce{O2(g)}$ is greater than both of the other values, more energy is released when $\ce{O2}$ gains the electrons, and this will result in the oxidation of both $\ce{Ag}$ and $\ce{Br-}$, correct?

As noted in your question, the standard reduction potentials are given as such in tables.

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag+ + e- &→ Ag(s)} &\quad E^\circ &= \pu{+0.799 V} \\
\ce{Br2(l) + 2 e- &→ 2 Br-(aq)} &\quad E^\circ &= \pu{+1.065 V}
\end{align}
$$

To get the the standard oxidation potentials you reverse the products and reactants and you must also flip the sign of the reaction. So:

  $$
\begin{align}
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V}
\end{align}
$$

Now there are two possible overall reactions (ignoring the Nernst equation):

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{Ag(s) &→ Ag+ + e-  } &\quad E^\circ &= \pu{-0.799 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Ag(s)&→ 4Ag+ + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 0.799 = +0.43 V}
\end{align}
$$

and the other overall reaction is

  $$
\begin{align}
\ce{O2(g) + 4 H+(aq) + 4 e- &→ 2 H2O(l)} &\quad E^\circ &= \pu{+1.23 V} \\
\ce{2 Br-(aq) &→ Br2(l) + 2 e- } &\quad E^\circ &= \pu{-1.065 V} \\
\ce{O2(g) + 4 H+(aq) + 4 Br-(aq) &→ 2Br2(l) + 2 H2O(l)} &\quad E_\text{Total}^\circ &= \pu{+1.23 - 1.065 = +0.16 V}
\end{align}
$$

Since the EMF for both overall reactions is positive, the reactions should be spontaneous. If the overall EMF was negative then the reaction would proceed in the opposite direction (in other words, swap the products and reactants.)

Notes:

(1) I don't know why the silver reaction doesn't happen. The EMF for the cell indicates that silver should oxidize in these conditions.

(2) The EMF's calculated ignore the Nernst equation.

(3) The oxygen half cell reaction is a bit weird. Obviously the oxygen must be dissolved in the aqueous solution to react. I'm assuming that the notation is trying to point out that the dissolved oxygen is in equilibrium with the atmospheric oxygen which has a constant concentration which is inexhaustible.


What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)? The recent Chemical & Engineering News article Photocatalyst shreds drinking water contaminant PFOA report the following:

Cates and his colleagues made an accidental breakthrough while testing the photocatalyst bismuth phosphate. After finding that it could break down PFOA as effectively as a previously reported photocatalyst, gallium oxide, they tried to improve its performance by varying the pH of the reaction used to make the catalyst particles. The researchers had hoped this would change the particles’ size and shape, but the process inadvertently produced bismuth oxyhydroxyphosphate (BOHP). This new catalyst has the highest photocatalytic activity for PFOA degradation ever reported.
In lab tests under UV light, BOHP degraded all the PFOA in a 130 μM solution in about an hour, roughly 15 times faster than did bismuth phosphate or gallium oxide. BOHP performed almost as well over five repetitions of the experiment, suggesting it can be reused. To simulate PFOA-contaminated groundwater, the researchers then tested a more dilute solution of 1.2 μM PFOA accompanied by 500 μg dissolved organic carbon/L. In this case, the photocatalyst also degraded all the PFOA, though it took two hours. Strathmann calls it a “promising development” that advances the application of photocatalytic treatment of PFOA.

It also includes this SEM image of microcrystals of the "inadvertently produced bismuth oxyhydroxyphosphate (BOHP)."

The accidentally synthesized BOHP (shown) can break down the toxic industrial contaminant PFOA faster than any other photocatalyst. Credit: Environ. Sci. Technol. Lett.

A quick check with google doesn't send me to the materials Wikipedia page, and I'm wondering if that is because this is an unusual substance or the crystal structure hasn't been solved yet.
Question: What is the crystal structure of bismuth oxyhydroxyphosphate (BOHP)?


Computing i0 in the Butler-Volmer equation: How to interpret Wikipedia and Wolfram Alpha So far, when using the Butler-Volmer equation, I had j0 given and could then just use I = A * j0 * [exp(...) - exp(...)], with A the electrode surface area.

Now I have a material from which I don't know j0 but just the rate constant. According to Wikipedia on exchange current density I can compute j0 as

Wolfram-Alpha on Butler-Volmer equation however says:

In this equation, the expression F * k0 would correspond to the above j0. k0 is explicitly defined as the rate constant in m/s on that same page.

Is Wolfram-Alpha wrong by missing out the concentration? The F * k0 does not even result in the units of a current density, and the whole expression will not evaluate as a current.

k0 has units of frequency (1/time)
All concentrations are surface concentrations (mol/area)

Sometimes people use k0 with units length/time and volume concentrations (mol/volume)

On Wikipedia: Coxy stands for Ox concentration * at equilibrium *. Cred stands for Red concentration * at equilibrium *. It's for 1 electron. Multiply by n for n electrons.

On Wolfram: [Ox] stands for Ox concentration at electrode, [Red] stands for Red concentration at the electrode. These concentrations are not necessarily at equilibrium.


Statistical thermodynamics Entropy of indistinguishable particles $S$ is

$$S = k\ln W,$$

but in some places notation is different:

$$S = k\ln Q + \frac{U}{T}$$

Can you explain, please?

The only time I've ever seen that second notation is when referring to the substitution of the Helmholtz energy, as A = U-TS and therefore A = -kTlnQ. If you want an in depth explanation of how the thing is derived I suggest looking at Chapter 10 of Biomolecular Thermodynamics: From Theory to Application by Douglas E. Barrick.

Here are some photos of the key parts:


Synthesis Golf VI: Muscarine October's synthesis golf is rather late - apologies. I've chosen a slightly different-looking target this round, with the aim of inspiring a wider variety of routes.

Muscarine is a natural product found in some mushrooms. Most notably, it is a good agonist of one class of acetylcholine receptors - these receptors are therefore called muscarinic acetylcholine receptors.

The only conditions for this round are that:

the synthesis must be enantioselective; and
no more than one chiral centre may be purchased in the form of a building block. (Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.)

Otherwise, any commercially available starting material (as usual, in the Sigma–Aldrich catalogue) is fair game! There is also no restriction on the counterion.

1. Retrosynthesis
This proposed route relies on being able to make the tetrahydrofuran via some type of ring closing etherification.
Through disconnection of the C-O bond, a linear diol is revealed which may be accessed in several ways.
Initially, I'd thought about doing a Sharpless dihydroxylation of  (Z)-hex-1,4-diene, however the starting material is surprisingly expensive and there are few literature reports of doing this dihydroxylation (and in those instances where it has been carried out, the e.e. and regioselectivity has been moderate to poor).
A different disconnection, using the chiral pool (less elegant, perhaps, but the starting material is cheap and ensures good selectivity) allowsthe diol can be disconnected across the C-C bond, revealing an aldehyde which is derived from a lactate ester.

2. Forward synthesis

Reagents and conditions: (a) TIPSOTF, 2,6-lutidine (b) DIBAL; allyl-MgBr (c) BnTCA, PPTS (d) Shi epoxidation (e) HF-Py, Py (f) TsCl, Py (g) NaN3 (h) H2, Pd/C (i) MeI
The synthesis starts with protection of methyl lactate with a TIPS group - a silyl group is chosen here as upon deprotection under fluoride conditions later in the synthesis, the THF should spontaneously close.
The TIPS protected methyl lactate may then be treated with DIBAL to afford the lactaldehyde. This must be used immediately to prevent epimerisation. In all likelihood, this could be achieved by adding a solution of allyl magnesium bromide to the same pot as the DIBAL reduction but failing this, there is good precedence for a reagent controlled boron mediated allylation (Tet. Lett. 2003, 44, 1737).
The terminal olefin installed by allylation is now setup to do a Shi epoxidation (Sharpless is terrible at terminal olefins) providing the final stereocentre. Treatment of the substrate with HF-Py should then deprotect the TIPS ether with concomitant formation of the THF ring system.
Finally, functional group transformations provide the desired natural product as shown in the scheme. Some issues may be observed when attempting to methylate the amine in the presence of the alcohol however from experience of similar substrates, the amine is likely to quaternise faster than the alcohol methylates and as such careful control of stoichiometry and conditions should allow the desired product to be isolated as the major component.
Summary
A synthesis of muscarine is proposed in 9 steps (linear) starting from methyl lactate, with the remaining stereocentres being installed using a substrate controlled allylation and a reagent controlled asymmetric epoxidation.
Addition steps could likely be cut out by (1) not protecting the alcohol that is currently protected with a benzyl group and (2) making use of an asymmetric aziridination in place of the Shi epoxidation.


How does a catalytic poison reduce a catalyst's activity? How does a catalytic poison reduce a catalyst's activity, such as palladium used in the hydrogenation of alkenes or any other such catalyst?

Basically the catalytic poison reduces the activity by slowing or inhibiting the reaction by occupying the surface of the catalyst and therefore the surface area available for the reactants reduces. You can get a general idea from the following text.

Substances which themselves are not catalysts, but when mixed in small quantities with the catalysts increase their efficiency are called as promoters or activators.
Catalyst poisoning refers to the partial or total deactivation of a catalyst caused by exposure to a range of chemical compounds. Poisoning may be desirable when it results in improved selectivity (e.g.Lindlar's catalyst) but may be undesirable when the catalyst is rendered ineffective (e.g. in catalytic converters). Poisoning refers specifically to chemical deactivation, rather than other mechanism of catalyst degradation such as thermal decomposition or physical damage.
Poisoning involves compounds which bonds chemically to the active surface sites of a catalyst. This may have two effects: the total number of catalytic sites or the fraction of the total surface area that has the capability of promoting reaction always decreases, and the average distance that a reactant molecule must diffuse through the pore structure before undergoing reaction may increase. Poisoned sites can no longer accelerate the reaction with which the catalyst was supposed to catalyze. Large scale production of substances such as ammonia in the Haber–Bosch process include steps to remove potential poisons from the product stream.

For more info, read the Wikipedia article on Catalytic poisoning.


Why does a 57.15% ABV spirit (ethanol+water) have a density of 923 kg per cubic meter? According to the UK proof standard, a 100 proof spirit was defined as a spirit with gravity of $\frac{12}{13}$ that of water or $\pu{923 kg m-3}$, which is equivalent to $\mathrm{57.15\% \ ABV}$.

One liter of $\mathrm{57.15\% \ ABV}$ spirit contains $\pu{571.5 cm3}$ ethanol and $\pu{428.5 cm3}$ water. If the densities of ethanol and water at $\pu{20 ^\circ C}$ are $\pu{0.789 g cm-3}$ and $\pu{0.998 g cm-3}$, respectively, then the density  of spirit should be $\pu{0.8785 g cm-3}$ or $\pu{878.5 kg m-3}$.
Where did I go wrong with that? How can a $\mathrm{57.15\% \ ABV}$ spirit have a density of $\pu{923 kg m-3}$?

There is nothing wrong with your calculations: but some liquids don't play nice
Your calculations are, likely, correct.
But liquids, unlike ideal gases, don't necessarily obey simple rules when mixed. In particular the two components, ethanol and water, strongly interact with each other when mixed and those interactions mean that the volumes of the pure liquids are not additive when they are mixed.
The effect is well known and a chart of it looks like this:

This is a fairly common behaviour for liquid mixtures where the components of the mixture interact. But the effect for ethanol/water is particularly strong and of practical significance both to chemists, industry and the tax authorities.


Why is the carbon-carbon bond in hexachloroethane shorter than in ethane? How can we explain the shorter carbon-carbon bond length in $\ce{(CCl3)2}$ ($1.49 \,\mathrm{\mathring{A}}$) than that of $\ce{(CH3)2}$ ($1.534 \,\mathrm{\mathring{A}}$)?

One reasoning would be applying Bent's rule:

In ethane the $\ce{C-H}$ bonds are approximately composed from $\ce{61\% sp^{3.3} C}$ and $\ce{39\% s H}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{2.3} C}$ orbitals.
In hexachloroethane the $\ce{C-Cl}$ bonds are composed from $\ce{48\% sp^{3.7} C}$ and $\ce{52\% sp^{4.8} Cl}$ orbitals.
That results that the $\ce{C-C}$ bond is composed from $\ce{sp^{1.8} C}$ orbitals.

Higher $\ce{s}$-character in bonds usually means that the optimal overlap of these orbitals is at shorter inter-nuclear distances. Since the $\ce{s}$-character in the $\ce{C-C}$ bond in hexachloroethane is much higher than in ethane, the bond length is expected to be shorter.

(Values obtained from an Natural Bond Orbital Analysis on the DF-BP86/def2-SVP level of theory.)


Is a discharged battery in a state of chemical equilibrium? A reaction that has reached chemical equilibrium no longer shows changes in reactant and product concentrations, but forward and reverse reactions are still going on. I'm confused whether a completely discharged battery is in a state of equilibrium.

Argument that an equilibrium has been reached

If I measure the cell potential, it will be zero. That means that the Gibbs free energy of reaction is zero, which is a property of a chemical equilibrium. When I discharge a battery by shorting it, there are concentration changes in the battery that cease when the battery is discharged, which looks like a reaction going toward and then attaining equilibrium.

Argument that this is not an equilibrium state

One property of a chemical equilibrium is that it is dynamic, i.e. the forward and reverse reactions are ongoing but occur at the same rate. For the battery, the reaction does not happen unless I close the circuit by electrically connecting the electrodes (e.g. shorting the battery with a wire). Even if I connect the electrodes with a wire, no current will flow because the battery is dead. So I don't see forward and reverse reactions going on. This would be like placing two beakers with 1 M aqueous salt solutions next to each other and saying they are at equilibrium. However, if the solutions were both contacting a dialysis membrane, ions would flow in either direction (at the same rate) so it would be fair to call this a state of dynamic equilibrium.

Which argument wins?

Is the discharged battery at a state of equilibrium or not? Does it matter whether the electrodes are connected electrically to decide on the answer?  Does it make sense to discuss a state of equilibrium in the context of electrochemistry in the first place?

Post-script: After recently encountering a related post I re-read my answer here and realized I could have provided a much clearer and more useful definition of equilibrium (rather than my somewhat philosophical earlier answer) that encompasses the different cases you present: in thermodynamics, a system is in an equilibrium state when no lower free energy states are accessible.

Therefore, the two beakers, just like the exhausted battery, are at equilibrium, because there are no accessible lower energy states under the defined conditions.

You can ask yourself: if the universe is expanding (that is, if it is in a dynamic state), is thermodynamics truly ever applicable? The answer is that in science we need to take approximations, during measurement or when extracting information from data. So it is in thermodynamics, which works well enough even though on long enough scales the approximations fail (on timescales of the "in the long run we are all dead" sort).

This is of course a tricky point. We often need to distinguish between "real equilibrium states" and "kinetically trapped or metastable states". However it is useful to invoke a "duck-typing" rule applicable to thermodynamics problems: if the measured properties of the system remain constant and homogeneous during the interval of the experiment, then the system can be considered to be at equilibrium during that interval. This definition is particularly valuable since thermodynamics is an empirical science. Caveat: that doesn't mean you can be sure that the system is in the state you meant it to be in (!), since if you wait long enough you might end up in another state (which might be the one you actually wanted the system to be in!). Perhaps more important is reproducibility: can you put the system into the desired state, reproducibly?

To summarize: the concept of equilibrium, like that of reversibility, is subject to the practical condition that no variations be observed across the system or over time during the course of the experiment, when the system is in the proposed state.

As regards a battery, if you don't see a change in the potential, you can regard the system as being "at equilibrium", at least temporarily! Again, the definition of equilibrium is not as strict as we'd like to think. The materials in the battery, left exposed to the environment long enough, will rust and fall apart, leading eventually to  other "equilibrium" states.


Why has lithium bromide a slight solubility in organic solvents? Why has lithium bromide a slight solubility in organic solvents?

Theoretical, LiBr is an ionic compounds, so it should be only soluble in polar solvent like water. But why is it soluble in others?

Lithium bromide is a salt of small cation and large anion. They cannot interact effectively, so the crystal lattice is quite easy to break. On the other hand, lithium cation is small, so it can interact effectively with small negatively charged atoms, like oxygen. Indeed, lithium cation has highest hydration energy of all alkaly metal cations. Altogether, it means, that in solvents with oxygen atoms (alcohol, esters, acetone) lithium cation effectively bounds to solvent molecules, leaving crystall lattice, and bromide has to follow, resulting in some solubility of lithium bromide in solvents with negatively charged oxigen and to lesser extent, nitrogen (for example pyridine). Lithium bromide is still insoluble in organic solvents that cannot interact with lithium cation, like benzene.


Creating small amounts of cellulosic ethanol from bamboo I own a medium-sized farm and commercial aquaponics business in Pennsylvania. I try to make my business as green, sustainable, carbon-neutral and self-sufficient as possible.

As a way to cut out the need to buy expensive and/or fossil fuel derived fertilizers, I started composting my animal mortalities and  organic/green waste left over from our operation. But, in order to compost I needed a source of steady carbon for the composting process, so we planted around 60 acres of clumping moso-bamboo, 20 acres of "running" bamboo and bought a couple electric wood chippers to create our own carbon-negative sawdust replacement for creating our own "compost-tea" fertilizer on sight.

However, I found out that my yields of bamboo were so massive (we harvest the bamboo 2 to 3 times a year) that we're constantly trying to find new uses for it (turning it to feed for crickets, grasshoppers and grass carp which we use in turn to feed our tilapia and prawns, occasionally feeding it to our alpacas and gasification power during the winter months).

We added a "combined heat and power" 145 KW gasification + Waste Heat Recovery micro-steam turbine power generation system to make up for the loss of daylight and efficiency from our solar PV system during the winter months. And two years later we still have way too much excess bamboo.

So while looking up other ways to utilize it and I found out about Chinese researchers creating a "cost-effective" cellulose ethanol from it. So I decided to Google around if there are any pre-existing procedures or methods to create ethanol from bamboo but I'm pretty much getting squat.

So my questions are:

What's the basic process of creating fuel ethanol from bamboo? Do you know of any books or resources that describe this in more detail? If not, would there be any special procedures/material/equipment needed to create ethanol from it?
Would it be worth the capital investment of buying all the stuff required to produce only about around 3000-4500 gallons of a year? I was told that under form 5110.74, a small producer of ethanol can only produce 10,000 proof-gallons of it, which apparently translates to just over 5000 gallons of fuel or something.

I really don't want to have to sell my excess bamboo to flooring and paper companies that are likely still using environmentally detrimental ways to process and ship its derivative products because it would go against my desire to help protect the environment. But I also don't want to invest into a "money pit" that I will never see any ROI on (I'd rather just use it to expand my operation by hiring new employees, branching out into land-based Marine shrimp and seaweed farming, lithium-ion energy storage, etc).

A process for creating fuel ethanol from bamboo is described in this original paper and this follow-up. The basic steps these researchers took are:

Bamboo saccharification with sulfuric acid
Removal of color compounds with activated carbon
Separation of acid and sugar (using an ISMB system)
Hydrolysis of oligosaccharides
Continuous ethanol fermentation of saccharified liquid (with yeast strain)
Distillation

There are also other, slightly different processes to do this (e.g. enzymatic hydrolysis, but this is more expensive).

I'm no expert on this, but it's my understanding that creating ethanol from lignocellulosic biomass is still in the research phase. AFAIK there isn't a good and economically viable process yet to create ethanol on a large scale. Several steps still need to be optimized and costs need to be reduced, so I don't think it's very interesting for you to do small scale production.