Carbonyl carbon electropositivity? Is the carbon in a carbonyl functional group electropositive? I had a professor tell me so, but I immediately refuted him, because he defined electronegativity earlier as the ability to stabilize negative charge ... and taking electronegativity to be a converse property, the strong partial positive character on the carbon probably isn't able to stabilize any further positive charge.

Similarly, I know that lone pairs are defined as having an electronegativity of 0 because electrons cannot stabilize other electrons.

The guy agreed with me. So, to confirm, am I correct, and, as an extension, can we describe the electron deficient carbon as electronegative instead? I could very well see a electron-deficient carbon stabilizing negative charge.

However, if we stick with the IUPAC definition of electronegativity - which is the ability to attract electrons, the carbonyl carbon still seems to lose to oxygen. A charge-separated resonance structure can be drawn for any carbonyl group with a + on carbon and a - on oxygen.

So that leads me to another question - is electronegativity the ability to stabilize negative charge or the ability to attract electrons within a covalent framework? I guess the two go hand in hand - i.e. you wouldn't attract electrons if you weren't able to stabilize them ...

As a side question: can we calculate EN for ions?

Is the carbon in a carbonyl functional group electropositive?

Electronegativity and electropositivity are usually used in a relative sense, X is more electronegative than Y.  The carbonyl carbon in acetone is more electropositive than the central carbon atom in isobutane, and less electropositive than the methyl carbocation carbon.

  because he defined electronegativity earlier as the ability to
  stabilize negative charge

Weak definition, wouldn't you say; that's why you won that argument.  Atoms can be electronegative in the abstract, charge doesn't need to be involved.  The statement, "fluorine is more electronegative than carbon according to the Pauling Electronegativty scale" is accurate and understandable, yet no one is describing a situation where charge stabilization is involved.

  because electrons cannot stabilize other electrons.

I understand what you mean, but you should know by now that you can't make such sweeping statements.  Don't the electrons in one of the double bonds in 1,3-butadiene interact with and stabilize the electrons in the other double bond?

  Can we describe the electron deficient carbon as electronegative

Again, I understand what you mean, but... Yes, it is more electronegative than the carbon in an alkane, but not as electronegative as $\ce{F+}$.

  Is electronegativity the ability to stabilize negative charge or the
  ability to attract electrons within a covalent framework? I guess the
  two go hand in hand - i.e. you wouldn't attract electrons if you
  weren't able to stabilize them

Yes, the two go hand in hand.  More electronegative atoms are better able to stabilize negative charge than less electronegative atoms.  It's all about lowering the overall system energy.

  As a side question: can we calculate EN for ions?

I'm sure there is a way.  For example, using studying the $\ce{^{13}C}$-nmr spectrum of various carbocations leads to a relationship of (somewhere around) 180 ppm shift/ electron.  So you can estimate the fractional positive charge on any carbon atom, the carbonyl carbon in acetone for example.  Once you've estimated the amount of charge on an atom, I suspect there is a way to relate it to EN.


The Lewis structure of HNO3 So I was checking the structure of nitric acid in Wikipedia, however I couldn't quite fathom why it looked like that because it seemed to contradict the following statement:

  A Lewis structure with small or no formal charges is preferred over a Lewis structure with large formal charges.

If we draw it like the one on the right, we get rid of formal charges and the structure is said to be more "stable". Why does this concept not work in this case?

Nitrogen is in the second row with no $d$ orbital in the valence shell.  It obeys the octet rule and cannot have more than 8 electron.

There are exceptions to the octet rule.  Having less than 8 electron is less preferable but still possible, and is commonly seen in free radicals and cations. On the other hand, having more than 8 electron is extremely unfavorable for second period atoms. Such electronic structures can be found in extremely unstable species or excited states, such as the CH5 radical.

As a comparison, first shell atoms obey the duet rule while atoms in the 3rd shell and beyond may obey 18-electron rule, 12-electron rule or 8-electron rule.   However, 18-electron rule and 12-electron rules are much less strict than 8-electron rule and violations are common place.

According to valence bond theory, the electronic structure of a molecule is a combination of all possible resonance structure that you can write down, including structures with all possible formal charges and strange electron counts.  However, their contributions are not even.  some of them are more favorable than others.

For HNO3, in order to satisfy the octet rule, the nitrogen atom would form 1 double bond and 2 single bonds.  Based on octet rule alone, there are 3 possible resonance structures that are favorable.

However, the first two resonance structures are significantly more favorable than the third, because they have smaller amount of formal charges.  As a result, we usually only write the two dominant structures. The bond between OH and N is close to a normal single bond. The other two N-O bonds have bond order near 1.5.

It is also common to write a mixture of resonance structures as the hybrid form

Note that this representation is not a single Lewis structure, but a convenient way to represent many resonance structures in the same figure. It gives no information about the exact bond order or formal charges on individual atoms.  The dotted lines indicates that in some structures where that bond is a single bond and in others a double bond, and the order is somewhere between 1 and 2.


Why is hydroxide ion the strongest base only in water? My professor mentioned that hydroxide ion is the strongest base when we only consider aqueous solutions. Then why don't the strongest bases outside of aqueous solution, like superbase, remain the strongest in water?

If you add a superbase to water in small enough quantity that it still makes sense to think of water as the solvent, all (or essentially all) of your superbase will be converted to its conjugate acid by the reaction

$$\ce{B-}+\ce{H2O} \to \ce{HB} + \ce{OH-}$$

This will have a very large equilibrium constant (by the assumption that $\ce{B-}$ is a superbase) and will also be driven heavily to the right by the high concentration of water. If you add enough of your superbase, eventually you can have appreciable amounts of $\ce{B-}$ sticking around in the solution, but at this point it no longer really makes sense to call the solution "aqueous".


Stabilization of multiple bonds through inductive effects In the conversion of m-chloroanisole to m-anisidine using $\ce{NaNH2}$ and liquid $\ce{NH3}$, apparently the reason the $\ce{NH2-}$ goes for the meta- position instead of the para position is that, in the reaction mechanism for the conversion to m-anisidine, the benzyne intermediate involved has the triple bond closer to the $\ce{-OCH3}$ group than in the reaction to the para product, thus stabilizing the intermediate due to the stronger inductive effects of $\ce{-OCH3}$.
I know the triple bond in benzyne isn't exactly your typical triple bond, but is there always some(if negligible) stabilization of all multiple bonds through electron-withdrawing inductive effects?

Since electron-withdrawing groups (EWGs) lower orbital energy (we can rationalise this in terms of reduced electron density leading to reduced shielding), they will stabilise multiple bonds - you see this in the Diels-Alder reaction, where EWGs on the dienophile reduce the energy of its LUMO and so accelerate the reaction by reducing the gap between the dienophile LUMO and diene HOMO (conversely, electron-donating groups on the diene destabilise its HOMO by the same logic and so also speed up the reaction).


Principle of detailed balance $$\ce{A -> B -> C -> A}$$

In Alberty's Physical Chemistry (7th ed.), the author says that this reaction mechanism is not possible. The reason given is:

  Although this mechanism at first sight appears reasonable, no actual process follows this mechanism because it violates the principle of detailed balance. According to this principle, at equilibrium the forward rate of each step is equal to the reverse rate of that step. Thus equilibrium cannot be maintained by this mechanism.

I understand the statement of the principle ("at equilibrium the forward rate of each step is equal to the reverse rate of that step"), but I don't get how it implies that equilibrium cannot be maintained, or that this reaction is not possible.

In this triangular scheme, taking it literally, it is clear that all the population will end up in species C. If $A \rightarrow B$ is mathematically irreversible (rate constant back is zero)  it implies that B is at an energy infinitely below that of A, and the energy of C infinitely below that of B, so clearly C cannot revert to A without the input of an infinite amount of energy. Such an argument hardly makes sense in chemistry, however.

If we now assume that the reaction is reversible but with a very small back rate constant a similar conclusion is obtained, i.e. that the overwhelming (but not total) population ends up in C.

Suppose, without loss of generality and for simplicity, that the activation energy A/B and B/C are the same at $E_A$ and that B is at energy E below A, and C at E below B. Now let the forward rate constant A to B be $k_f=k_0\exp(-E_A/RT) $ and the back rate constant is $k_{BA}=k_0\exp(-E_A/RT)\exp(-E/RT)$ and which is going to be smaller, say it is $10^{-3}$ smaller. The rate constant from C to A is therefore $k_{CA}=k_0\exp(-E_A/RT)\exp(-2E/RT)$ which is now $10^{-6}$ times smaller that the rate constant A to B.


What is the theoretical quantum ground state energy of helium hydride positive ion? What is the theoretical quantum ground state energy of helium hydride positive ion?

I am speaking of the usual assumptions such as the approximate modeling of the two nuclei here - one being a proton, the other being 2 protons plus 2 neutrons - as two infinitely massive particles alone in the universe at some equilibrium distance, in the non-relativistic simple case alone considering no spin dependent terms etc.

So actually only the charge of the nuclei is relevant in this approximation unless the vibrational energy - perhaps modeled in first approximation as a simple harmonic oscillator with some force constant - is also taken into account - which would be the positive 'zero' point value of harmonic oscillator.

I would assume that it would have to be lower (that is greater bonding magnitude or absolute value of the negative energy) than that of the neutral helium atom, but I am not sure. But one place on internet gives $\pu{3600 cm^-1}$ as a fundamental frequency (but of what?) which - if I am not mistaken comes - to $\pu{1.64E-2}$ Hartrees or atomic units - so that cannot be correct or the value i am looking for as the helium atom is about -2.904 Hartrees or atomic units. I suppose it would be almost impossible to practically devise any experiment to arrive at any experimental value? Though, if so, what is that value?

You could get a quick estimate of the ground state energy with paper and pencil by using Heitler-London method, since all you need to do is change the charge on the second nucleus.

I would assume the vibrational energy absorption energy quoted referred to the He-H bond vibration.


Why are most medicines in salt form? Is there a reason for why most medicines are in salt form?
Why not have them in conjugate base form?

In general, salts are more soluble in water than organic acids or organic bases, because they are made of ions. It is a general trend. There are plenty of exceptions. So usually salts are quickly dissolved in the stomach and absorbed in the intestine later on.


Newer ionic radii data than Shannon's 1976 compilation? Shannon, R. D., 1976, Revised effective ionic radii and systematic studies of interatomic distances in halides and chalcogenides: Acta Crystallographica Section A, v. 32, p. 751–767. DOI: 10.1107/S0567739476001551 provides a very detailed list of ionic radii for just about any element, for various oxidation states and coordination numbers.

This paper was published in 1976. Is there a newer compilation with new data?

tldr; it's very difficult to determine consistent additive ionic radii
For various reasons, I've gone back to look at covalent radii recently (e.g., for estimating bond lengths).
In that case, there seem to be fairly useful recent work, e.g. by Pekka Pyykkö:
"Additive Covalent Radii for Single-, Double-, and Triple-Bonded Molecules and Tetrahedrally Bonded Crystals: A Summary"
J. Phys. Chem. A 2015, 119, 11, 2326–2337
For covalent radii, it's clear that one can fit experimental and theoretical interatomic distance r(A-B) to a sum of two covalent radii with systematic accuracy. Notably, Pyykkö defines not only single-bond but double-bond and triple-bond radii for almost all elements.
(N.B.: I'm not sure it makes sense to consider a triple bond to $\ce{Br}$ or $\ce{I}$ but if you want them, it's in the article.)
There are also recent systematic van der Waals radii, e.g., Cramer and Truhlar:
"Consistent van der Waals Radii for the Whole Main Group"
J. Phys. Chem. A 2009, 113, 19, 5806–5812
So what about ionic radii?
There's a really nice article by Liu, Schwartz, and Li, cited by Pyykkö about the problems with ionic radii.
"On Two Different Objectives of the Concepts of Ionic Radii"
Chem (Eur. J.) 2013, 19, 44, 14758-14767
The Liu, Schwartz, and Li point out the competing problems with ionic radii:

Since “bonded atoms” are evidently neither rigid nor spherical, quantum
computational chemists apply “valence multiple zeta and
polarized” basis sets in their molecular calculations. Conversely, X-ray ED researchers use the procrystal model of independent rigid spherical atoms for starting and screening purposes in their software for the processing and analysis of the observed diffraction data.

In other words, we like to consider atoms as having consistent, rigid spherical shapes. For ionic compounds, that's hard to determine, because of charge transfer. By definition, $\ce{A+B-}$ indicates cation $\ce{A+}$ has given electron density to $\ce{B-}$ .. but how much? Is there any covalent character and shared electron density? How do you divide?
Pyykkö also mentions the problem in his covalent radii work. His atomic radii work:

… provided that the A–B bond is not too ionic, or the coordination numbers of the two atoms too different from the original input data, used in the fit.

In short, covalent radii are easy to define because most of the electron density is shared and covalent bonds are thus fairly predictable. In contrast, almost no compound is purely ionic so it's hard to perform the division for ionic species.
Here are two examples from Liu, Schwartz, and Li:

Consider interatomic distances from alkali halide crystals. (The portion is performed by Bader's AIM analysis.)

Note that there's no "vertical" or "horizontal" line for well-defined anion or cation radii. Maybe you can come up with a correction based on electronegativity difference .. but the cation radii particularly depend on the anion.

The same problem shows up for monomeric alkali halide molecules in the gas phase.

It makes some sense. It's difficult to rigorously know how much charge transfers from anion to cation (vs. shared), which is why calculation methods define multiple algorithms to determine it. (Atomic partial charges are something of an eternal debate.)
I wouldn't be surprised to find someone creating a correction using electronegativity differences or covalent character. And it's possible someone will sweep this away with a compelling article - that's the point of science. But it won't be easy, because we expect ionic radii to serve a lot of different purposes.


How to convert ethanal to 2-hydroxy-3 butenoic acid? How to convert ethanal to 2-hydroxy-3 butenoic acid?

I tried crossed aldol condensation with formaldehyde followed by acid addition.After that how should I proceed ?

Once you have got your acrolein from the aldol condensation of ethanal and formaldehyde you can react it with $\ce{NaCN, HCN}$ under kinetic control to get direct addition of the cyanide to the enone, forming a cyanohydrin[1]. The nitrile can then be hydrolysed in aqueous acid to yield the carboxylic acid.


Dilution calculations $\pu{5 mL}$ of $\pu{0.012 M}\,\ce{Pb(NO3)2}$ is mixed with $\pu{5 mL}$ of $\pu{0.030 M}\,\ce{KI}$. $\ce{PbI2}$ forms. Calculate the diluted concentrations of $\ce{Pb^{2+}}$ and $\ce{I-}$ ions.

I couldn't figure out whether it is an equilibrium question, or something else entirely.

I have currently tried using the moles of the limiting reactant (Pb(NO3)2) to obtain the moles of PbI2 formed and using the molar ratio to get concentration of each ion. However I have no idea whether I have done the correct calculation for ion concentration. The values I am getting are 0.003M Pb2+ and 0.012M I-

update: I have looked at the questions that follow and am able to say for certain this is not an equilibrium question.

I haven't checked your numbers but you're assuming the $\ce{PbI2}$ to be perfectly insoluble. That's a reasonable assumption but in reality its solubility product is $K_{sp}=4.41\times 10^{-9}$ (Wikipedia), so it does have some solubility.

To take this into account it becomes an equilibrium problem. Assume the initial concentrations (just after mixing)  to be $\ce{C(Pb^2+)}$ and $\ce{C(I^-)}$.

After precipitation the lead concentration has been reduced by$X$ and the iodide concentration by $2X$.

With the solubility product after precipation:

$$K_{sp}=[\ce{Pb^2+}]\times [\ce{I^-}]^2$$

So we have:

$$K_{sp}=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

Note that this is a third degree polynomial equation in $X$. Solve for $X$ to find the real equilibrium concentrations $[\ce{Pb^2+}]$ and $[\ce{I^-}]$.

We can also use the equation for the case where $K_{sp}=0$. Then:

$$0=\Big(\ce{C(Pb^2+)}-X\Big)\times \Big(\ce{C(I^-)}-2X\Big)^2$$

First case: $\ce{C(Pb^2+)}-X=0 \implies \ce{C(Pb^2+)}=X \implies [\ce{Pb^2+}]=0$

Also: $[\ce{I^-}]=\ce{C(I^-)}-2\times\ce{C(Pb^2+)}$

In the second case we assume $\Big(\ce{C(I^-)}-2X\Big)^2=0$ and proceed as above.

Which case applies depends on which reagent is the limiting one.


Covalently bonding amide to glass I am unfortunately not very good at chemistry, so I am having trouble figuring out where to start on this project.  I need to have some way of get some amide functional groups stuck on a substrate (most likely some glass beads or slides) so I can try to show some interactions of the amide with a molecule I am interested in.

Does anyone know of any literature that shows a good way to functionalize silica with amides?

Thanks for any help with this, I really appreciate it.

EDIT: I found some of this stuff ((3-Aminopropyl)triethoxysilane) in our lab.  Could I just put a glass slide in it and leave it over night?
Then maybe come back and wash acetyl chloride over it as in this reaction?

Ethanol solution, soaked over night of APTEOS will work.  You won't get great coverage, but it will be likely good enough for what you want to do.

The subsequent chemistry is a bit more touchy.  The problem is acid byproduct which forms ammonium salt with the amine functions that is difficult to remove.  This leaves latent acid catalyst at your surface.

You are better served forming the amide from your APTEOS and coating with that product.


Calculating the molar mass of a gas given state function values Question:

  A gas has a volume $255\ \mathrm{mL}$, a pressure $733\ \mathrm{mmHg}$, a temperature $548\ \mathrm{^\circ C}$, and a mass $1.22~\mathrm{g}$. What is the molar mass?

My attempt:

$$pV = nRT$$

Convert $733\ \mathrm{mmHg}$ to $\mathrm{atm} = 733~\mathrm{mmHg} \cdot 1\ \mathrm{atm}/760~\mathrm{mmHg} =   0.9645\ \mathrm{atm}$

Convert $\mathrm{^\circ C}$ to $\mathrm{K}$: $548\ \mathrm{^\circ C} + 273.15\ \mathrm{K} = 821.15~\mathrm{K}$

Convert $\mathrm{mL}$ to $\mathrm{L}$: $255~\mathrm{mL} \times \frac{1~\mathrm{L}}{1000~\mathrm{mL}} = 0.255~\mathrm{L}$

$$0.9645\ \mathrm{atm} \times 0.255~\mathrm{L}= n \times \frac{0.8206~\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}}  821.15~\mathrm{K}$$

$$\frac{0.2459}{673.84} = n \times \frac{673.84}{673.84}$$

$$0.000364\ \mathrm{mol} = n$$

$$M = \frac{m}{n} = \frac{1.22~\mathrm{g}}{0.000364~\mathrm{mol}} =   3351.648$$

The mistake you made was the value of $R$. If pressure and volume are taken in $\mathrm{atm}$ and $\mathrm{L}$, then the value of $R$ is $0.0821$. (Not $0.82..$).

Now, $$n=\frac{0.9645\times 0.255}{0.0821\times 821.15}=0.0036$$

$$M=\frac{1.22}{0.0036}=334.412\mathrm{g}$$


Why does crystalline solids have well defined 3d geometry Crystalline solids have well defined geometry because the intermolecular forces are uniform throughout the solid unlike amorphous solid where it varies from place to place.

But still the question hold why does some solids have this property like why $\ce{NaCl}$ forms a crystalline solid but not glass or something else.

I thought it is because crystalline solids are ionic but diamond is an counterexample among many.

Question :

Why does some solids have crystalline structure while others don't ?

Keep two concepts separate: the solid crystalline state is not bound to occur on ionic compounds only.

In a simple view, crystalline describes a regular arrangement of matter, with predictable order both in short, as well in long range.  For the crystalline state, it is unimportant if the constituents of the crystal are atoms (like metal atoms), charged particles (like your example of NaCl), or neutral molecules (like benzoic acid) or even a mixture of neutral and charged molecules (many protein crystals contain water).

Some materials are easier packed in such a regular pattern, than others.  In the instance of metals, their (on first approximation) sphere like atomic geometry and similar radius eases such a settlement.  Inorganic salts tend to be a bit more complicated, as often smaller ions are localized in gaps built by larger ions (octahedral / tetrahedral gaps).

Organic molecules may equally build crystals, yet their shape and conformational flexibility around single bonds (for example in alkyl chains) and multiple intermolecular interactions may keep them over a long time in a glassy state (like supercooled benzoic acid, or some brands of honey).  More than for the aforementioned examples: formation of a nice crystal may be thermodynamically favourable, but it is not enough "just to cool down" below the melting temperature to reach this energetic minimum.  There must be seeds of crystallization, and typical conditions of their formation (temperature, concentration, etc.) are not necessarily the same where nicely shaped crystals growth further.

It is not surprising crystals attract attention, for example in material science and crystallography, and that there are journals with focus on the relationship between molecular properties, and macroscopic / crystallographic ones; like CrystEngComm or Crystal Growth & Design.


How many electrons does each shell hold? why some times 2nd shell onward it can hold 8 electrons and other times it is like they can hold K2 L8 M18 N32? when do i have to use first method(K2 L8 M8 N8) and when to use second method (k2 L8 M18)? I'm really confused.

This is a great question that allows us to learn from the periodic table. The previous answer asked if you have learned about orbitals, and if you haven't this is probably why you are having difficulty.
If you look at the periodic table, you will see that the elements are arranged in two ways, vertically (group) and horizontally (period), there is good reason for this. Electrons can only occupy space based on mathematical probability; each of these probability functions is what we refer to as orbitals. Each orbital can only hold a maximum number of electrons ( the equation for total electrons is 4l+2 electrons where l is a quantum number starting at l=0 for s, l=1 for p etc.) in the valence shell (for 1, it can hold 2 Starting on the left side of the periodic table with Hydrogen we know that it is in group 1, which means that it has 1 electron in its valence orbital. Element number 2 (He) is a bit of a special case because although it only has 2 electrons in its valence orbital, it is full because the orbital number is 1. What I mean by this is that if you look at period 2 elements, starting with Li, you see that it falls under Hydrogen, and has 1 valence electron. This is because the 1s orbital is full with 2 electrons, and the additional electron goes into the next lowest energy orbital which is called the 2s orbital. Be has 4 electrons, 2 of them in the valence orbital, and 2 of them in the 1s orbital. When we move to B(oron) the 1s and 2s orbitals are full, and we start to fill the 2p orbital. The p orbital can hold a total of 6 (4l+2 where l = 1) electrons and so the boron atom has the following configuration (1s)2(2s)2(2p)1. What you see is that the highest energy orbital (in this case 2 (s + p) is the valence orbital and so B(oron) has 3 valence electrons, but also has 2 electrons in the much lower energy 1s orbital. Think of these orbitals like layers of an onion, where the further from the center, the greater the energy of the orbital. There are rules to how the orbitals are filled and the rules are dictated by quantum mechanics but generally speaking the orbitals are filled as follows 1s 2s 2p 3s 3p 4s 3d 4p 4d using what is know as the Madelung rule. s orbitals can hold 2 electrons, p orbitals can hold 6 electrons, d orbitals can hold 10 electrons. As you work your way from low atomic number on the period table to high you will see why it is organized the way it is. The break between H and He is because of electrons, when you move to period 2 you start filling valence shell 2, when you move to period 3 you start filling valence shell 3 and so on.  I hope this helps, and let me know if you need clarification on anything!


Comparing purity grades for ethanol I am trying to determine if the following two grades of ethanol are equivalent:

Ethanol absolute, 99.8% Normapur AR (Analytical Reagent)
Ethanol alcohol, pure ACS reagent, ≥99.5%

Information on the technical specifications of these grades would also be helpful for comparison.

The important aspects of reagent purity are the level of purity with respect to compound(s) X, and the purity with respect to compound(s) X required for your particular use case.

For example, if you are conducting a reaction that requires anhydrous reagents but is unaffected by trace metals, then you would want to see a certificate of analysis or equivalent showing an acceptable limit for water in the reagent. Whereas if you require sub-ppm levels of trace metals, then that's the parameter of interest to you and you probably don't care if there is 0.5% water present.

In other words, a 99.5% pure ethanol product and a 99.8% ethanol product could be drastically different in their composition with respect to your needs; i.e. the apparently slightly more pure compound could have orders of magnitude more trace metals present than the slightly lower overall purity product. Furthermore, terminology regarding reagent purity is often similar between different vendors (i.e. pure, ultrapure, milipure, etc.), but frequently has very different implications on the aspect of purity that is important to you.

Summary, TL;DR:
Although you can frequently make generalizations about the meaning of "ACS Reagent" grade or "[pick-your-prefix]pure" grade, you need to inspect either the specifications on the bottle or certificate of analysis for each reagent to know for certainty that it meets your needs. The simplest reliable source for such information is the manufacturer, most of which have online resources providing this information for their products.


Triphenylphosphine vs. trimethylphosphine in reduction reactions Triphenylphosphine is used typically as a reducing agent for organic peroxides as it is a good nucleophile. So, if I wanted to computationally model a reaction like: $$\ce{PR3 + ROOH -> OPR3 + ROH}$$

Can something like a trimethylphosphine be used instead of triphenylphosphine, since it is an even better nucleophile, and the strength of the P=O is what's driving the reaction/reduction? Or am I missing some information?

Yes it can. It also has the advantage that the trimethyl phosphine oxide produced is water soluble and thus easier to remove from a reaction mixture than triphenylphosphine oxide which is notoriously difficult to get rid of. I have used it for a Staudinger reduction of an alkyl azide when triphenyl phosphine was unreactive.

However, trimethyl phosphine is more expensive, pyrophoric, has an unpleasant odour and is toxic requiring full personal protective equipment. Safety data here: https://www.alfa.com/en/content/msds/english/30143.pdf


Why are 1,3-diols protected with aldehydes but 1,2-diols are protected with ketones? 1,3 diols are usually reacted with benzaldehyde when protection of the OH groups is desirable, but acetone is used when the relation between the hydroxyl groups is 1,2. Why is it so?

Let's look at the protection of a typical 1,3-diol with benzaldehyde and acetone:

The theory is that the formation of the six-membered cyclic acetal is less favourable with acetone than it is for benzaldehyde, because with acetone you cannot avoid having one methyl group being axial.

The next-most stable thing after a six-membered ring is a five-membered ring, which can be formed by the protection of a 1,2-diol instead of a 1,3-diol. So, acetone happily reacts with those 1,2-diols to form five-membered cyclic acetals, whereas benzaldehyde is more selective for 1,3-diols since it can enjoy the stability of a six-membered ring.

When there is a choice, like in glucose, you get rather different results:

However, I am not sure how much this principle extends to protection of compounds with only two hydroxyl groups. I am guessing it should be possible to protect 1,3-diols with acetone too - after all, a methyl group is hardly the worst axial substituent to have.


All resonance structures of the hydrogen molecule This answer written by Jan says that "a professor of mine claimed to have drawn some 18 (or was it 80?) different resonance Lewis structures."

Please reproduce those resonance structures of the hydrogen molecule.

If nobody is able to come up with that many, the answer with the most resonance structures would be accepted.

This is a reference kick-starter contestant solution.

There are 6 resonance structures:

$\ce{H:H}$ (the usual one)
$\ce{:H- H+}$
$\ce{H+ H:^-}$
$\ce{*H*H}$
$\ce{H*H*}$
$\ce{*H H*}$


Galvanic cell chemistry I made a galvanic cell with the anode electrode being aluminum and the electrolyte being aluminum sulfate. The salt bride was magnesium chloride. I am a little confused on what my oxidation equation would be for this set up.

So I did some research and I found out that aluminum sulfate in water will go to aluminum hydroxide and sulfuric acid, so I proposed this half cell equation:

$$\ce{2Al_{(s)} + Al2(SO4)3_{(aq)} + 6Cl- + 6H2O_{(l)} -> 3H2SO4_{(aq)} + 2Al(OH)3_{(s)} + 2AlCl3 + 6e-}$$

So this would then become:

$$\ce{2Al_{(s)} + 2Al^3+ + 3(SO4)^2- + 6Cl- + 6H+ + 6OH- ->
6H+ + 3(SO4)^2- + Al(OH)3_{(s)} + 2Al^3+ + 6Cl- + 6e-}$$

The spectator ions cancel off to give:

$$\ce{2Al_{(s)} + 6OH- -> 2Al(OH)3_{(s)} + 6e-}$$

This equation makes sense but aluminum hydroxide should form a precipitate. However, when I ran my cell, there was no visible precipitate, so it makes me question whether my anode half cell reaction is correct.

What am I doing wrong? Is the process I used to describe the half reaction right?

The aluminum may not hydrolyze all the way to a neutral hydroxide.  It could form hydroxo complexes that remain in solution, especially if the solution is made acidic by the same hydrolysis.

Consider a half-reaction in which the Al is combined with only one or two hydroxide ions.

Some other tips:

1)  Since you have an acidic solution don't put hydroxide ions on the left.  Put water and balance with hydrogen ions on the right.

2)  Sulfate ion is a weak base.  It will combine with hydrogen ions to make bisulfate.  So add sulfate on the left and incorporate the hydrogen ions noted above into bisulfate ions.


Will an alloy leaching in an acid act as both anode and cathode? If an alloy is placed in an acid where selective leaching is occurring, will the alloy act as both anode and cathode?
If the alloy does act as both anode and cathode then will the potential difference be zero as the cathode and anode are shorted together?

You can setup a situation where one metal is acting as a cathode and the other metal acting as an anode. There will be a small potential difference due to their differences in Fermi Levels much like a thermocouple but the alloy will be at an electrochemical potential that is constant.

This is not that different than a single metal(like a piece of iron) in solution. In the corrosion reaction, the counter reaction may happen somewhere else on the iron. So the stick of iron may be the anode and the cathode at the same time.


Confusion between electronegativity and electron affinity Electronegativity is a chemical property that says how well an atom can attract electrons towards itself.

The electron affinity of an atom or molecule is defined as the amount of energy released when an electron is added to a neutral atom or molecule in the gaseous state to form a negative ion.

However,fluorine is taken as the most electronegative element but chlorine is accepted as the element with most electron affinity.

But shouldn't the atom which attracts the electron the strongest also release the most energy?

I wouldn't shoot this down so quickly - consider, e.g. Mulliken electronegativity, defined as the average of the electron affinity and ionisation energy. If you think about it, electronegativity depends not only on how strongly an atom pulls other electrons to itself, but also how willing it is to give up its own electrons. The former can be somewhat represented by the EA, the latter by the IE.

One point that hasn't been mentioned is that electronegativity is not a physical parameter which can be measured. It is a human construct, which is often helpful in explaining chemical reactivity, but at the end of the day it is not something which can be derived from physics. Therefore, there is no real reason to expect that there should be any one-to-one correlation between the EN and the EA (unless the EN scale is explicitly defined to follow the EA). For example, if you use the more common Pauling scale (defined by bond dissociation energies), there's no real reason why the trend in EN should match the trend in EA exactly.

Broadly speaking, the trends match (because of what you've said) - e.g. both EN and EA generally increase going across a period. However, the small details like chlorine vs. fluorine won't necessarily match.


Identifying food, quick and cheap? I’m asking this question in the most technical perspective of identifying substances, so I’m not considering the use of chemical tests like Benedict’s solution, biuret solution, Sudan III stain, or indicator dichlorophenolindophenol, as valid answers since these methods rely on identifying the color by observation. The answers I’m really looking for are based less on physical and more on chemical properties without the need of human observation or intervention.

(At this point, if you have an answer please reply… the rest is clarification to my question)
With that being said spectroscopy, chromatography, electrophoresis, and biochemical analysis are more along the lines of the answers I’m expecting. However, I’m really asking which of these methods or others in connection with other techniques, are the most convenient, smallest in size (in terms of equipment), and cost-effective in identifying food sources.

So in attempt to sum up my question, here's a scenario: I have a yellow onion. I extract juice from the onion and I give it to a skilled chemist/engineer labeled unknown. Using an analytical technique, in about a few seconds, the chemist supplies a prediction based on the components found within the unknown food (amino acids, peptides, proteins, enzymes, lipids, phospholipids, carbohydrates, alcohols, fat-soluble vitamins, water-soluble vitamins, organic acids, organic bases, phenolic compounds, bittering substances, pigments, aroma compounds, and dietary fiber)

What is/are the analytical techniques used? Is there a device that does such a thing?

Unfortunately, your confidence in what a chemist can, is quite exaggerated. Though flattering.

Analyzing the entire range of parameters that you list, requires several analytical techniques, including sample preparations like homogenization, extraction etc.

A technique like Near-infrared spectroscopy can give an easy and fairly good estimate of some major components (protein, fat etc) in selected types of food. FOSS Analytical (and probably several others) purchases equipment for this.

Routine methods exist for analyzing components like vitamins, pigments or metals, but they require a laboratory setup and often very expensive equipment.


Why does ice water get colder when salt is added? It is well known that when you add salt to ice, the ice not only melts but will actually get colder. From chemistry books, I've learned that salt will lower the freezing point of water. But I’m a little confused as to why it results in a drop in temperature instead of just ending up with water at 0 °C.

What is occurring when salt melts the ice to make the temperature lower?

When you add salt to an ice cube, you end up with an ice cube whose temperature is above its melting point.

This ice cube will do what any ice cube above its melting point will do: it will melt. As it melts, it cools down, since energy is being used to break bonds in the solid state.

(Note that the above point can be confusing if you're new to thinking about phase transitions. An ice cube melting will take up energy, while an ice cube freezing will give off energy. I like to think of it in terms of Le Chatelier's principle: if you need to lower the temperature to freeze an ice cube, this means that the water gives off heat as it freezes.)

The cooling you get, therefore, comes from the fact that some of the bonds in the ice are broken to form water, taking energy with them. The loss of energy from the ice cube is what causes it to cool.


How does chromium(III) react to form chromium(VI) with hydrogen peroxide? $$\ce{2Cr(OH)3 + 4NaOH + 3H2O2 -> 2Na2CrO4 + 8H2O}$$

How does this reaction occur? We suddenly get sodium chromate in the place of chromium hydroxide. I can't understand it. We mix two hydroxides and get a salt?

What role does the peroxide play?

The hydrogen peroxide is a strong oxidiser. In basic conditions, this is partly due to $$\ce{H2O2->H2O + O\ \ }[*].$$ The monooxygen is very eager to gain electrons. Therefore, $$\ce{Cr^{III}->[\ce{[O]}]Cr^{VI} +3e-}.$$Chromium(III) hydroxide$^{[**]}$ is amphoteric, it reacts with bases, too.

$$\ce{Cr(OH)3 + 3NaOH <=>Na3[Cr(OH)6]}$$

This complex is then oxidised. For common chromium(VI) compounds, one would expect to see the equilibrium $$\ce{2CrO4^{2}^- <=>[\ \ \ce{H3O+}][\ \ \ce{OH-}]Cr2O7^{2}^{-}} \ \ \text{[***].}$$

Thus, in (strongly) basic conditions, the chromate ion $\ce{CrO4^{2}^-}$ would be favoured.

Extra

One thing that might be confusing are the equations themselves. In other words, they simply show

the most likely process (not always the case),
the end result (there are exceptions),
the formulae are not true to the actual structure (cannot be sure).

To make inorganic chemistry more tolerable, it is best to accept that there are multiple reactions occurring concurrently. Then it is easier to arrive at the "correct" end result by taking into account reaction conditions and stability.

[*] This is a simplified explanation taken only as a model for basic understanding. The actual decomposition mechanism is more complicated, depending on pH, other ions in solution, etc. For more information on specific situations, look at Kinetic Studies and Mechanism of Hydrogen Peroxide Catalytic Decomposition [---]. There you can find many other great references as well.

[**] Chromium(III) hydroxide might actually be $\ce{CrOOH*H2O}$, or $\ce{Cr2O3*}$ $n$$\ce{H2O}$. Even if we were to be dealing with the hydroxide, it actually comprises of $\ce{Cr(OH)6}$ octahedrons, and is claimed to be inactive (active form is the oxide hydrate). [N. S. Ahmetov's Inorganic chemistry]

I would make the claim that the hydroxide does not actually exist at all, and is, like for iron, mistaken for the oxidehydroxide hydrate. Furthermore, the reaction might not be direct, it could be a case of Le Châtelier – Braun principle.

[***] The overall transition is $$\ce{2CrO4^{2}^{-} +2H3O+ <=>Cr2O7^{2}^{-} +3H2O}$$

from where it is easy to see that adding hydroxide ions tilts the equilibrium to the left. Also, you might wish to look up different aqueous complexes of chromium and the structures of the ions above. That also helps to visualise the process.

Suggestion: Chromium complexes tend to polymerise ;).

Even if $\ce{CrO3}$ would form, as you suggested, it reacts rather well with water via

$$\ce{CrO3 + H2O \rightarrow H2CrO4, \\ 2CrO3 + H2O \rightarrow H2Cr2O7,}$$

and hydroxides:

$$\ce{CrO3 + 2NaOH \rightarrow Na2CrO4 + H2O}.$$


Williamson Ether and Carboxylic Acids What would happen if you tried a Williamson Ether Synthesis on something that has both an alcohol and a carboxylic acid? Would both -OH groups be functionalized, where the alcohol would become an ether and the carboxylic acid become an ester?

There is a general principle known as “last in-first out” that I think applies here. If a molecule contains multiple acidic positions with significantly different pKa, you can selectively react on the least acidic position (in your case, the alcohol).

Specifically, I would try treating your molecule with 2 eq. Of strong base such as NaH or LDA. The first equivalent will deprotonate all of your acid to form the quite stable metal acetate. The second equivalent removes the alcohol hydrogen to generate a less stable alkoxide. If you add 1.0 eq. Of the electrophile it ought to react selectively on the more reactive alkoxide nucleophile, leaving the metal acetate unaffected, then you can just neutralise in work up.


Second Peak in X-Ray Diffraction Pattern of bcc Metal By Bragg’s law,
$$2d\sin\theta=n\lambda$$
and
$$d=\frac{a}{\sqrt{h^2+k^2+l^2}}$$

Combining the two gives:
$$\sin\theta=\frac{n\lambda\sqrt{h^2+k^2+l^2}}{2a}$$

By the reflection rule for bcc, there will be destructive interference when $\sqrt{h^2+k^2+l^2}=\sqrt{1}$, on the (100) plane. The first peak will thus occur when $n=1$ and $\sqrt{h^2+k^2+l^2}=\sqrt{2}$, on the (110) plane. However, I’m confused about the second peak. It will be impossible to get $\sqrt3$ while satisfying the reflection rule for bcc, so I’ve moved on to $\sqrt4$, which is equal to 2. However, mathematically there is no difference between having $n=2,\sqrt{h^2+k^2+l^2}=1$ ((100) plane) and $n=1,\sqrt{h^2+k^2+l^2}=2$ ((200) plane). Which one does the second peak represent, then? Does it represent both? Why then is this peak so much smaller than the other peaks in most diffraction patterns that I see? For example, for $\alpha$-iron, the diffraction pattern is

Image Link: http://questions.transtutors.com/Transtutors001/Images/Transtutors001_6804131b-7ca5-47cd-984a-b824a4926c3a.PNG

The reflection with $n=2,\;(h,k,l)=(1,0,0)$ is the same as $n=1,(h,k,l)=(2,0,0)$. Yes, I said "same", not "similar"; they are one thing, there are no "both". For this reason, you may just as well abandon using $n$ altogether.

As for why the peak is so small compared to others you've seen, let me respectfully suggest that maybe you haven't seen all that many. Indeed, there is no law that all peaks on the diffraction pattern must be about the same size. On the contrary, they are essentially proportional to the squares of Fourier coefficients of electron density, and as such, can be anything from 0 to the maximum. The early textbook exercises usually employ some mercifully simple cubic or tetragonal structures with few well-resolved and well-developed peaks. Not all compounds are like that. Look at the graphite diffraction pattern, for instance: you'll basically see a lonely (002) peak, large as life, and only after the most careful inspection you'll start noticing the rest of them amidst the background noise.

As for the (100) peak, it couldn't be there, since it fell victim to the systematic absences of the bcc lattice. Ditto for (111) and (210). Only those with even $h+k+l$ survived, and that's precisely what you see on the picture.


Does the boiling boint of miscible liquids follow a linear relationship? I want to create a steam bath for a reaction the proceeds well at 105℃.  I cannot use pure (demineralized, anyway) water for this, since it would boil at 100℃ at STP.

As such I figured that maybe I could get the required temperature by mixing water with another liquid (with which water is miscible) which has a higher boiling point (e.g. vinegar at BP around 118.5℃ or vegetable glycerin with BP at 190℃).

Does the boiling point of the mixture of liquids proceed in a linear manner between the two temperatures?  E.G. would a volume of 50% water & 50% VG boil at 145℃?

As it is being done right on the boiling point of the combined liquids, I expect to lose some of the liquid to steam and need to add more. Will the lower temperature liquid boil out first?

Except if molecular composition of your two liquids are similar, the boiling point of the mixture will not proceeds linearly but like this for instance :

With the lower blue curve the boiling curve. In between the two curve you'll then have a mixture of vapor and liquid and only vapor on the upper part the binary diagram.

So you should find out an already drawn binary diagram for your mixture or draw it out by yourself first.


Why is the acoustic absorption coefficient of hexane in air less than air? Why is the acoustic absorption coefficient of hexane in air less than air? A 120 kHz pulse train exhibits a larger amplitude after traveling in a hexane/air mix than in pure air.

My hypothesis is that the hexane molecule has fewer degree of freedom than that of $\ce{N2/O2}$. Is this plausible?

Best hypothesis so far is that as gas density increases the transducer couples more efficiently to the medium, giving the impression of a decrease in absorption coefficient due to the increased amplitude at the receiver.


Why does mercury in mercuric acetate have a lone pair? The configuration of mercury ends with $6s^{2}$. In mercuric acetate, mercury is in the +2 oxidation state, so does it still have a lone pair?

TLDR:

Mercury in its (+2) salts does not have lone pairs readily accessible for covalent bonding. However, its 5d electrons do affect its properties,  resulting in a relatively high polarisability.

The long story:

Mercury is the last 5d element and has electron configuration $[\ce{Xe}] 4f^{14} 5d^{10} 6s^2$. The $6s$ electrons are the outermost electrons and readily available for covalent bonding. $4f$ electrons are virtually inert since they are very deep. $5d$ electrons belong to the previous shell, but are the outermost of them and once mercury looses its $4s$ electrons, $5d$ electrons become the outermost and define its properties. They are still too strongly bound to the nucleus to participate in covalent bonds under ordinary conditions ($\ce{HgF4}$ was only observed spectroscopically in inert gas matrix). , but are far enough to wiggle around and participate in non-covalent interactions, such as van-der-waals interactions. $6p$ orbitals are also available for covalent bonding, though extent of their use is often debatable and/or unobvious.


Werner's theory and complex salts According to Werner's theory, metal ions have two types of valency-primary and secondary, where the primary valency is said to be satisfied by negative ions only, and  the secondary valency can be satisfied by positive ligand, negative or neutral molecule.

My question here is:
While formation of complex compounds, which of these valencies is satisfied?

What I'm not able to interpret from this is whether there is a 'possibility' for both the valencies being satisfied or is it necessary that both are satisfied?

Also if the later is true how is it possible? Can it be explained taking some example as reference.

I think I'm not able to interpret it and understand its details.

A certain number of ions, atoms or molecules closely associate around a central atom leading to the formation of distinct entity called coordination complex.
The groups atoms or molecules linked to the central atom are said to be coordinated with the latter and their number gives the coordination number.

The non-ionisable units lie in the primary zone or sphere, whereas the  ionisable  units lie in the secondary sphere and appear as ions in the solution.

In hexammine platinic chloride $\ce{[Pt(NH3)6]Cl4}$  and in potassium ferrocyanide $\ce{K4[Fe(CN)6]}$ four chlorine ions of the former and four potassium ions of the latter ionize in the aqueous solution, whereas six ammonia molecules of the former and six cyanogen groups of the latter belong to the primary sphere or non-ionizable sphere and thus fail to respond to their characteristic tests or fail to ionize. Or you could also say that the chlorine ions in the former and the potassium ions in the latter occupy the primary valencies and hence are ionizable and the ammonia and cyanogen group occupy the secondary valencies  and are non-ionizable, while the $\ce{Fe}$ atom in the latter is present in the primary Sphere and thus fails to ionize.

As per your question during the formation of a compound say $\ce{[Pt(NH3)5Cl]Cl3}$, five molecules of ammonia and the chlorine ion (linked by secondary valencies) are present in the primary Sphere and the rest of the three chlorine atoms lie in the secondary sphere. Thus the coordination number of the compound would be  six as five ammonia molecules and one chlorine atom are attached to the central metal atom.

Now, every element tends to satisfy both its primary and secondary valencies. A negative ion tends to satisfy both these valencies i.e. primary as well as the secondary valencies.
The presence of a negative ion in the coordination sphere reduces the amount of charge on the complex ion by the amount of charge present on it while the negative ions in the primary sphere or the coordination sphere are not ionized.


Cyclopentanone is treated with piperidine cyclopentanone is treated with piperidine in acid catalyzed conditions to form the organic product (A). (A) on reaction with benzyl bromide followed by hydrolysis gives another organic product (B) containing aromatic ring. If the sum of degrees of unsaturation of (A) and (B) is x and the difference of the number of carbon atoms in (A) and (B) is y, then find the value of x/y.

According to me compound (A) is an enamine, but I'm not sure about (B)!

So according to this, my answer is x/y = 10/7
But the actual answer is x/y = 4.5

I don't know where am I going wrong!

As per the reaction sequence what you said about A is absolutely correct and in B I think you didn't consider hydrolysis of the product. The product what you have written is correct before hydrolysis, and after hydrolysis the carbonyl reforms and the other component becomes piperidine again. So,the final product B is 2-(phenylmethyl)cyclopentan-1-one. Hope that helps!


About possibly healthy salt substitutes I noticed Heinz Ketchup has a salt substitute version that uses potassium chloride instead of the usual salt. If this is safe and tastes like salt could it be a better additive than sodium chloride? Since it has potassium wouldn't this be a benefit for Heart health or against stroke if not used too much?

The taste of potassium chloride is not close enough to sodium chloride so the salt substitutes usually contain a mix of potassium and sodium chloride.  This is healthier than pure sodium chloride.


Sulphate ion- confirmatory test On adding $\ce{BaCl2}$ solution to $\ce{ SO4^{2-}}$ salt aqueous solution $\ce{ BaSO4}$ is formed as a white precipitate, which confirms the presence of the anion.

While doing the test in my school laboratory, we were given a list of procedures to follow; for $\ce{ SO4^{2-}}$ it read:

'Experiment: Aqueous solution of sample + dilute $\ce{ HCl}$ + $\ce{ BaCl2}$ solution

Observation: white ppt'

Why do we need to add $\ce{ HCl}$?

The method consists
of slowly adding a dilute solution of barium chloride to a hot solution of the
sulphate slightly acidified with hydrochloric acid

It is  customary to carry out the precipitation in weakly acid solution in order
to prevent the possible formation of the barium salts of such anions as chromate,
carbonate, and phosphate, which are insoluble in neutral solutions; moreover,
the precipitate so obtained consists of large crystals, and is therefore more easily filtered

Reference: Vogel's Textbook of 'Quantitative Chemical
Analysis 11.72 Sulphate: Determination of sulphate as barium sulphate


Would ionic compounds dissolve in methanol? Question:

  Determine whether liquid hexane ($\ce{C6H14}$) or liquid methanol ($\ce{CH3OH}$) is the more appropriate solvent for the substances grease ($\ce{C20H42}$) and potassium iodide ($\ce{KI}$).

It seems logical that $\Delta H$ to dissociate $\ce{C6H14}$ would be small, since it's non-polar, that $\Delta H$ to dissociate $\ce{C20H42}$ would also be small, since it's non-polar, which would be conducive to a solution forming between those two.

The question that I'm stuck on is if the $\Delta H$ of formation of a solution between $\ce{K+}$, $\ce{I-}$, and the mildly polar $\ce{CH3OH}$ would be greater than the amount of energy it would take to dissociate potassium iodide. How can I estimate how much energy will be released by the attraction of the $\ce{I-}$ to the slightly positive hydrogen on the hydroxide group is greater than energy required to dissociate $\ce{KI}$?

Am I missing something?

You don't need to do compare the said enthalpies. Maybe the process of dissolution of $\ce{KI}$ in $\ce{CH3OH}$ is energetically favorable, maybe it isn't (the latter seems more likely to me); that's irrelevant. Thanks to entropy, everything dissolves in everything, if only in minuscule amounts.

You just need to compare solubilities of $\ce{KI}$ in methanol and hexane, which basically boils down to the comparison of solvation enthalpy in methanol vs. that in hexane, and here the difference in their polarity provides a clear and obvious answer.


Enolate ion opening an epoxide? Does anyone know if it is possible to open an epoxide and do a 2-C extension using an enolate ion, say, of acetone? Here's an example of what I mean:

Based on what I know of enolate chemistry, this shouldn't be problematic, but I've never seen it.

I have done some preliminary literature searching and checked out masterorganicchemistry.com, etc. but have not been able to find anything about this.

Never say never! :-) Here is the answer for your question (Ref.1):

  Abstract: We report here that (1) enolate anions of five- to seven-membered cycloalkanones nucleophilically open cyclopentene and cyclohexene oxides in 57−76% yields and with 4−8:1 diastereoselectivity; (2) enolate anions formed regiospecifically via kinetic deprotonation of 2-cyclohexenone and 2-cycloheptenone open cyclohexene oxide in 60−62% yields and with 32−95:1 diastereoselectivity; and (3) an aryl methyl ketone enolate anion opens a monosubstituted epoxide as the key step in a short synthesis of the $\gamma$-hydroxyketone (GHK) aglycon of the natural product curculigine.

Graphical Abstract:

The product(s) and selectivity are depicted in following scheme:

Reference:

G. H. Posner, J. P. Maxwell†, M. Kahraman, "Mild, Fast, and Stereoselective Epoxide Opening by Ketone Enolate Anions. Application to Synthesis of the Norlignan Curculigine," J. Org. Chem. 2003, 68(8), 3049–3054 (DOI: 10.1021/jo020744q).


Why do we integrate square of wave function over N-1 electron coordinates? Quote form F.Jensen's Introduction to computational chemistry:

"The electron density is the square of the wavefunction, integrated over N-1 electron coordinates..."

Why do we integrate over N-1 instead of N electorn coordinates?
What would happen if we integrated over N coordinates?
I suppose we would get a norm if we integrated over N coordinates, or 1 if wavefunction is normed.
But why N-1?

$\newcommand{\el}{_\mathrm{e}}
\newcommand{\dif}{\mathrm{d}}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}$Consider an $n$-electron system in an arbitrary electronic state represented by the electronic wave function $\psi\el(\vec{q}_{1}, \vec{q}_{1}, \dotsc, \vec{q}_{n})$, where $\vec{q}_{i}$ stands for joint spin-spatial coordinate of $i$-th electron, $\vec{q}_{i} = \{ m_{si}, \vec{r}_{i} \}$.

Integrating the square of the absolute value of the electronic wave function over the spatial coordinates of all the electrons except for electron-one and summing up over the spin coordinates of all of them without any exception gives the probability density of finding electron-one with arbitrary spin at $\vec{r}_{1}$ while having the remaining $n - 1$ electrons with arbitrary spins at arbitrary positions,
\begin{equation*}
    \rho_{\mathrm{e-1}}(\vec{r}_{1})
    =
    \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsc \dif \vec{r}_{n} \, .
\end{equation*}
And yes, if we "integrate" over all coordinates in this expression we'll just get an inner product of the wave function with itself,
$$
    \braket{\psi\el}{\psi\el}
    =
    \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{1} \dif \vec{r}_{2} \dif \vec{r}_{3} \dotsc \dif \vec{r}_{n} \, ,
$$
which is usually equals to 1 since the wave function is usually normalized to unity.

Electrons are, of course, indistinguishable, so that their labeling is arbitrary and can be changed any time.
Thus, the expression above should work equally well for calculating the probability density of finding electron-two, electron-three, etc., or, in other words, it gives the probability density of finding one particular electron, no matter which, with arbitrary spin at $\vec{r}_{1}$ while having the remaining $n - 1$ electrons with arbitrary spins at arbitrary positions.

The electron density, defined as the probability density of finding any of the $n$ electrons with arbitrary spin at $\vec{r}_{1}$, is evidently $n$ times as large as the probability of finding one particular electron,
\begin{equation} \label{eq:electron_density}
    \rho(\vec{r}_{1})
    =
    n \sum_{m_{s1}} \sum_{m_{s2}} \dotsi \sum_{m_{sn}} \iint \dotsi \int
    \left| ψ\el(\vec{q}_{1}, \vec{q}_{2}, \dotsc, \vec{q}_{n}) \right|^{2}
    \dif \vec{r}_{2} \vec{r}_{3} \dotsb \dif \vec{r}_{n} \, .
\end{equation}
Note here that, for a wave function that is normalized to unity, if we "integrate" over all coordinates in this expression we'll just get the number of electrons $n$, which is one of the well-known properties of the electron density,
\begin{equation}
    \int \rho(\vec{r}_{1}) \dif \vec{r}_{1} = n \, .
\end{equation}


How are pi-orbitals delocalized in a porphyrin chain? According to this paper, the orbitals appear delocalized mostly from left to right, and only along 1-2 units of the chain.

Is that really the case? If so,

Why is the delocalization only along 1-2 units of the chain and not throughout the chain?
Why is there no delocalization in the up-down direction?

What you are seeing are molecular orbitals which, by definition, extend across the entire molecule, restricted only by inherent symmetry. It is important to note and is not clearly visible in your image that you are looking at a $\ce{H2-Fe-Zn-H2}$ system, i.e. from one side to the other the centres of the porphyrin system contain two hydrogens, an iron ion, a zinc ion and another two hydrogens. Replace the zinc with iron or vice-versa and you would get an axis of symmetry with molecular orbitals that are either perfectly symmetric or perfectly antisymmetric to this axis of symmetry.

As it is drawn, the molecule is $C_\sigma$, i.e. there is a single plain of symmetry along the plain of the molecule. All orbitals must be symmetric or antisymmetric to this plain, and since we are looking at a $\unicode[Times]{x3C0}$-system, it will be antisymmetric.

All the rest is the result of a calculation! So we cannot really answer why, all we can say is ‘because that is what maths gives us’. But we can slightly rationalise it. First, there is a difference between the top and the bottom, it is just not as easily visible. This is because the only difference between the top and the bottom is ‘where exactly are the two hydrogens and the double bonds?’ Second, the two ions are clearly different ones with different electronegativities that create entirely different environments. Therefore, it is understandable that one orbital is centred more around one of the two ions and the other around the other. And third, the LUMO$+1$ is clearly delocalised across the entire molecule.


Finding the repeating unit of polymerisation given two constituent molecules The compounds $\ce{H2NCH2CH2NH2}$ and $\ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $\require{enclose}\ce{\enclose{horizontalstrike}{(}HNCH2CONHCH2CH2NHCO\enclose{horizontalstrike}{)}}$
B. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2CO\enclose{horizontalstrike}{)}}$
C. $\ce{\enclose{horizontalstrike}{(}OCCH2CONHCH2NHCO\enclose{horizontalstrike}{)}}$
D. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NHCOCH2NH\enclose{horizontalstrike}{)}}$

I came across this question which requires me to find the structure of the repeating unit of the polymer made by the reaction of two different compounds. However, I am not sure as to which repeating unit would suffice for this particular polymer.
As far as I know, there should be double bonds present at the ends of the repeating unit, so that the pi bond can break and allow the formation of additional bonds, and hence, a chain of molecules. However, using that line of thought I cannot seem to figure out the answer to the question.
I just need a direction, a hint or a clue as to how I should approach problems like such.

You are starting out with amino groups on one monomer and carboxylate groups on the other monomer. As William R. Ebenezer states in the comment, these combine via condensation (to form an amide, which is written as -NHCO- or -CONH- in condensed formulas).
The linkages in the answer choices between repeating units are not visible until you write another repeating unit next to the ones shown. A. and B. show amide linkages between repeating units, while C. and D. show something else.
Here is the exercise rewritten to emphasize the amide bonds in the repeating units:

The compounds $\ce{H2NCH2CH2NH2}$ and $\ce{HOOCCH2COOH}$ react to form a polymer. What is the structure of the repeating unit of the polymer?
A. $\require{enclose}\ce{\enclose{horizontalstrike}{(}HNCH2CO-NHCH2CH2NH-CO\enclose{horizontalstrike}{)}}$
B. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2CO\enclose{horizontalstrike}{)}}$
C. $\ce{\enclose{horizontalstrike}{(}OCCH2CO-NHCH2NH-CO\enclose{horizontalstrike}{)}}$
D. $\ce{\enclose{horizontalstrike}{(}HNCH2CH2NH-COCH2NH\enclose{horizontalstrike}{)}}$

Now that we emphasized the amide linkages, you can see that the first monomer in A, the second monomer in C and the second monomer in D don't match your starting materials. We can also exclude C. and D. because repeating units are not linked via amides. The answer has to be B.


Why does dichloridobisethylenediaminecobalt(III) have only three optical isomers? Sorry for the bad drawing and having $\ce{NH}$ instead of $\ce{NH2}$

In the first structure, the chlorine atoms are in trans position on the plane.

In the second structure, the chlorine atoms are in cis position on the plane.

In the third structure, one chlorine is in the plane and the other chlorine is occupying one axial position.

Now all three of these will have one enantiomer making the number of isomers in total $6$.

So $\ce{[Co(en)2Cl2]+}$ should have $6$ optical isomers, but the answer in the book says that $\ce{[Co(en)2Cl2]+}$ has $3$ optical isomers. How?

In simplistic terms, we can see $\ce{[Co(en)2Cl2]^+}$ as an $\ce{[MA4B2]}$ pseudooctahedral complex. We can initially consider the $\ce{A\bond{->}M}$ and $\ce{B\bond{->}M}$ bonds identical. Trying this out on a model octahedron or molecular modelling kit will show you that only two possible different configurations exist: cis and trans.

Here, however, we have one of the ligand types actually being a briding ligand. Because the bridge is too short to allow for a trans configuration, the two coordination sites of these bridging ligands will always be cis to each other.

For the complex in which the two chlorido ligands are trans to each other, only one possible arrangement of the $\ce{en}$ ligands is possible. At first approximation, the entire complex should have $C_\mathrm{2v}$ symmetry if not more: a $C_2$ axis with two perpendicular $\sigma_\mathrm{v}$ planes; one bisecting the $\ce{en}$ ligands through the $\ce{C-C}$ bond, the other being orthogonal and mapping each $\ce{en}$ onto the other. Due to the presence of planes of symmetry, this complex cannot be chiral.

In the complex in which the two chlorido ligands are cis to each other, one could think of up to three potential arrangements for the bridges; however, as I pointed out, it is not possible to have the two coordinating atoms of one $\ce{en}$ ligand be trans to each other, reducing the total down to two. These two possibilities do not differ in whether one of the chlorides is in an ‘axial’ position (remember, we are assuming equal bond lengths initially so there is no defined ‘axial’ position). Rather, in one case you have a leftwards-turning screw and in the other case a rightwards-turning screw if you assume the $\ce{en}$ ligand to be a plane and wanted to screw the entire thing into your paper.

Neither of these two structures have a plane of symmetry (or higher symmetry) and both are therefore optically active. Careful analysis will reveal, that they are in fact mirror images and thus enantiomers of one another. They are called the Δ and Λ isomers, respectively. All three isomers are attached in the image below.

Figure 1: isomers of $\ce{[Co(en)2Cl2]^+}$. From left to right: trans, cis-Λ, cis-Δ.

Thus, there are three isomers of the compound, two of which are optical isomers of each other (the third is a geometrical isomer).

Finally on the topic of whether axial bonds are actually longer, i.e. whether there is a Jahn-Teller type distortion or not. The answer is no. $\ce{Co^{III}}$ is a $\mathrm{d^6}$ complex. It could be assumed to be high or low spin. In case it accidentally ends up to be low spin, the $\mathrm{t_{2g}}$ orbitals are fully populated while the $\mathrm{e_g}$ orbitals are entirely unpopulated; there is no need for Jahn-Teller distortion.

In the potentially more likely high-spin case, one of the $\mathrm{t_{2g}}$ orbitals is fully populated while all other orbitals are half-populated. We might expect a Jahn-Teller distortion to remove the degeneracy of the $\mathrm{t_{2g}}$ orbitals. However, this distortion would be minor at best: the $\mathrm{t_{2g}}$ orbitals do not interact well with the ligands since they are oriented to point in-between the ligand orbitals. (On a group theory basis, the $\mathrm{t_{2g}}$ orbitals are nonbonding.) Therefore, any symmetry reducing effect will be very minor at best and not lead to a noticeable distortion. Thus, it is reasonable to assume near-identical bond lengths and no principal (‘axial’) direction.


Ambigious nature of aldehydic group in glucose In aqueous solution, α-anomer and β-anomer of glucose
  remain in equilibrium with each other along with a small amount of
  glucose in open chain form ($0.02~\%$). The open chain form in the
  solution contains a free aldehydic group and hence gives several
  reactions involving free aldehyde.

Though little, the open chain form of glucose reacts with many reagents which are used to treat free aldehydic groups. Therefore, glucose in aqueous solution also reacts with those reagents.

$$\ce{glucose_{(aq)} + HCN -> gluco cyanohydrin} $$

$$\ce{glucose_{(aq)} + H2NOH -> gluco oxime} $$

$$\ce{glucose_{(aq)} + H2NNHPh -> gluco phenylhydrozone} $$

$$\ce{glucose_{(aq)} + 3H2NNHPh-> gluco osazone} $$

$$\ce{glucose_{(aq)} + Br2_{(aq)}-> gluconic acid} $$

Glucose also answers Tollen's reagent test, Fehling's solution test and Benedict's solution test.

However, as interesting it may seem, glucose is a two-face molecule. Here are some reactions which glucose does not like:

$$\ce{glucose_{(aq)} + NaHSO3 -> no reaction} $$

$$\ce{glucose_{(aq)} + {2,4-DNP} -> no reaction} $$

$$\ce{glucose_{(aq)} + {Schiff's reagent} -> no reaction} $$

Why is that glucose does not react with $\ce{NaHSO3}$ or $\ce{{2,4-DNP}}$ or Schiff's reagent while it answers many other reactions where the free aldehydic group is involved?

I am looking for a convincing answer. Most answers in web tell that some are strong and some aren't strong enough. If you give enough time, all the carbonyl reactions should work with glucose because there will always be $0.02~\%$ in open chain form. If some of it gets used up, then the equilibrium will shift and hence more open chain form of glucose will be formed.

The distal hydroxy groups are in a perfect distance to form pyranose (six-membered oxane) or furanose (five-membered oxolane) rings. Any equilibrium reaction that leaves the carbonyl group liberated for extended periods is prone to lose free aldehydes to the (unreactive) hemiacetal forms. Any reaction whose equilibrium is strongly product-sided or which proceeds somewhat irreversibly due to precipitation etc. will not be affected by that effect.

This explains why Fehling’s and Tollens’ solutions react with glucose: in both cases one of the products precipitates which drives the reaction to completion. None of the aldehyde that is used up in the test reaction will be regenerated. The same is true for the osazone formation which involves a redox step or the reaction with bromine which is also a redox reaction.

The reaction with sulphite to give sulphonic acids is an equilibrium: sulphite can attach but is equally quickly liberated again. However, sulphite is a rather soft nucleophile and an acceptable leaving group. The reaction is reversible and due to the formation of hemiacetals not much of the sulphonic acid will be observed. In general sulphite testing, the precipitation of the strongly hydrophilic sulphonate drives the reaction to completion but sugars themselves are already strongly hydrophilic so this driving force cannot exist.

A cyanohydrin, however, is a much more stable intermediate because cyanide is a much worse leaving group. Thus, here again we are driving a reaction to completion due to a practically irreversible step — and contrary to sulphite which is based on the same mechanism but is much more reversible.

The same difficulty occurs when attempting to generate imines: these are also prone to attack by the distal hydroxy functions giving O,N-acetals. While these may be stable products, they do not have the same features as the imines usually generated by the reaction of carbonyl compounds with amines and thus are not captured by the tests. This also prevents productive reaction of glucose with Schiff’s reagent — and additionally, Schiff’s test requires the addition of a sulphite which we learnt already is unfavourable.

The big question remains why some hydrazines react with glucose according to your list but some don’t. I have no satisfactory explanation why that would be the case. The only difference between 2,4-DNPH and phenyl hydrazine is the former’s electron-withdrawing groups that could lead to an overall slower reaction or less tendency to undergo the required redox step to form an oxazone. It is not a great explanation but maybe a starting point.


Calculating KC Given Equilibrium Concentrations Ex. N2O4(g) 2NO2(g)
• If you place 0.0350 mol N2O4 in 1 L flask at
equilibrium, what is KC?
• [N2O4]eq = 0.0292 M
• [NO2]eq = 0.0116 M

i want to know from where this coming

• [N2O4]eq = 0.0292 M

• [NO2]eq = 0.0116 M

Those are the concentrations of $\mathbf{N_2O_4}$ and $\mathbf{NO_2}$ at equilibrium. They are just given to you as part of the question for you to use in calculating $K_c$.


Apparent paradox in the formation of ice at room temperature The formation of ice out of liquid water can be written down like this:
$$\ce{H2O (l) <=> H2O(s)}$$
We can calculate the change in standard Gibbs free energy (per mol substance) in the following way:
$$\Delta G^\circ = \Delta H^\circ - T \Delta S^\circ$$
If we do this at room temperature $(\pu{298 K}),$ we get: $\Delta G^\circ = \pu{546 J mol^-1}.$
So we can see that this process is very non-spontaneous, which is what you'd expect, you never see ice forming at room temperature. However if we calculate the equilibrium constant $K_\mathrm{eq}$ for this reaction we get:
$$K_\mathrm{eq} = \exp\left(-\frac{\Delta G^\circ}{RT}\right) \approx \mathrm e^{-0.22} \approx 0.80$$
It's obvious that this is wrong, because at room temperature we never see liquid water in equilibrium with a decent amount of ice. I just don't know what I did wrong, I have checked my units and calculations but can't seem to find what I did wrong.

Now that you've got $K_{eq}$, you need to take a look at the expression for the equilibrium constant, which is $$K_{eq} = \frac{a_{\text{ice}}}{a_{\text{water}}},$$ where $a$ is the activity of each species.
For a pure liquid and a pure solid, the activity is defined as 1, so the expression on the right hand side of that equation (usually referred to as the reaction quotient $Q$) can only take on three values:
1 if ice and water are both present
0 if water is present but not ice
undefined if water is absent, whether or not there is ice.
So if a mixture of ice and water is put in a room temperature environment, the ice melts because the reaction quotient of 1 is greater than the equilibrium constant of 0.8. Only when all of the ice melts and it makes the step change to $Q=0$ does it get below the equilibrium constant. Then we have a situation where anytime any infinitesimally small amount of ice crystallizes, then $Q$ exceeds $K_{eq}$ and the crystal returns to liquid form.


Can we prepare alkyl Fluoride by Finkelstein reaction? My book states that:

  Finkelstein reaction is particularly useful for preparing iodoalkanes. The iodoalkanes are obtained by heating chloro or bromoalkanes with a concentrated solution of sodium iodide in dry acetone.
  $$\ce{R-X~+~NaI->[\ce{acetone,~reflux}]R-I~+NaX}$$
  $(\ce{X=Cl,~Br;~R=alkyl ~group})$
  Sodium chloride and sodium bromide being less soluble in acetone get precipitated from the solution and can be removed by filtration. This also prevents the backward reaction to occur according to Le Chartelier's principle. The reaction gives the best results with primary halides.

My question:

Can we prepare alkyl fluoride (not aryl fluoride) by Finkelstein reaction?

My book mentions about another reaction called Swarts reaction, through which alkyl fluoride is prepared. In Swarts reaction, alkyl fluorides are prepared by heating alkyl chlorides or bromides with salts such as mercurous fluoride ($\ce{Hg2F2}$), silver fluoride ($\ce{AgF}$), cobalt fluoride ($\ce{CoF3}$) or antimony trifluoride ($\ce{SbF3}$). Why can't we prepare alkyl fluoride by treating sodium fluoride with alkyl halide (Finkelstein reaction)?

In the future I recommend consulting March's Advanced Organic Chemistry before asking such questions. It is a priceless reference work for organic synthesis methods. The following text is result of brief consultation with the book.

The equilibrium shift in Finkelstein reaction is governed by relative nucleofility of halogenide-ions. In general, in polar solvents iodine is the strongest nucleophile, because it is poorly solvated, while in nonpolar solvents chlorine is the strongest solvent as it is smaller.

Fluoride in non-polar solvents should be the strongest nucleophile of all halogenide-ions, but it so strongly solvates in protonic solvents that it almost do not migrate in organic phase even if phase-transfer catalists are present. Strong basicity of fluoride does not help either as beta-eliminations becomes significant side-reaction.

This article claims to develop practically meaningful procedure of clean fluoro-de-halogenations.


Solubility of EDTA in water Why is EDTA ($\ce{C10H16N2O8}$) poorly soluble in water?
I am interested in analyzing the hardness of water in a sample due to the presence of $\ce{Ca^2+}$ cations. I attempted to do this by titrating with ethylene diamine tetraacetic acid ($\ce{C10H16N2O8}$) however these attempts have proven to be futile.

I ran into the same problem myself when I bought (foolishly, it turned out) $\ce{H4EDTA}$ instead of $\ce{Na2H2EDTA}$ or $\ce{Na4EDTA}$ for some of the experiments for my doctorate.  It really confused me, because carboxylic acids are generally quite soluble and, with four of the things, you'd think it'd dissolve right away.  But, no.  There are even amines in there, too, but they apparently don't help much either.

Based on the abstract of a 1973 article by Ladd and Povey1, my guess is the apparent poor solubility of $\ce{H4EDTA}$ is actually a kinetic effect, due to a network of intra- and inter-molecular hydrogen bonds present in the solid (emphasis added):

  The [EDTA] molecule exists in a cis conformation and possesses two-fold symmetry, the diad axis passing through the central $\ce{C-C}$ bond. The nitrogen atom is protonated, and takes part in a bifurcated intramolecular hydrogen bond. A short $\mathbf{\left(2.46~Å\right)}$ hydrogen bond links the oxygen atoms of adjacent molecules.

I don't have the necessary thermodynamic data at hand to argue with certainty, but my guess is that the activation energy for solvent penetration into this hydrogen-bond network is rather high, and thus the rate of solvation is slow even though the overall process is (presumably) exergonic $\left(\Delta G < 0\right)$.

Reference: Ladd, M. F. C.; Povey, D. C. Crystallographic and spectroscopic studies on ethylenediaminetetraacetic acid (edta). J. Crystal Molec. Struct. 1973, 3 (1), 15–23. DOI: 10.1007/BF01270899.


Does lowering temperature of a gas make it easier to compress? My teacher said that for ideal gases,

  the lower the temperature, the lower the kinetic energy of a gas will be, and it will be easier to compress the gas.

Using the ideal gas equation $PV=nRT$, if the volume is constant, a decrease in temperature will cause a decrease in the pressure of the gas. That will make it easier to compress. Will the volume remain constant, or will it change?

Compressing a gas by applying an external force means to work against the force the gas exerts on the walls of its container, i.e. its pressure.

The definition of compressibility

$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p} \tag{1}$$

for an ideal gas with the well-known equation of state

$$p V = n R T \tag{2}$$

gives

$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p} = \frac{1}{V} \frac{n R T}{p^2} = \frac{1}{p} \tag{3}$$

which is independent of the temperature $T$.

For non-ideal gases there will be higher order terms that also depend on temperature.

It is also worth noting that at the critical point the compressibility of a real gas is infinite. So if you are below the critical temperature you will be able to compress the gas more easily at a state (the critical point) with higher temperature.


Synthetic options for selective hydroxylation at terminal position Are there any synthetic options to add a hydroxyl group at the terminal position to the bicyclo compound below? What might be reactions worth trying, any idea? Or is this a lost cause?

Or is this a lost cause?

I don't think so. This looks like a good candidate for an allylic oxidation with selenium dioxide, look here for some examples.


What are the predominant acids in sphagnum peat moss? I'm trying to figure out what the predominant acidic substances in peat moss are. Peat moss is said to have a pH of approximately 4.0. So, I'm curious what is making it acidic.

I know not all acids bind to various elements in the same ways. So, I'm trying to find out which acids (and/or acidic salts) are present in peat moss in order to aid me in my research to discover approximately how long it might take to acidify soils with it compared with the traditional methods of adding various forms of sulfur to soil. You don't need to tell me how long it takes, by any means. I can do more research after I find out which acids or acidic salts are in peat moss, in what levels.

I'm not familiar with this particular SE site's customs. So, if this is the wrong place to ask, or if I used the wrong tag(s), feel free to let me know. I figured people might be more likely to know here than on Gardening and Landscaping, though.

This 1992 article by Gagnon and Glime suggests that the ability of Sphagnum to lower pH depends on the presence of cations in the surrounding environment and is not (contrary to prevalent belief) due to organic acids in the plant itself. Here is a more recent article from New Phytologist amplifying that cation exchange is responsible for lowering of pH.

If you want to acidify soil using sphagnum/peat--assuming the authors are right--it's more important what cations you put into the soil. The sphagnum will then exchange them for $H^+,$ lowering the pH of the soil.

This does not mean there aren't organic acids in moss. You would have to look in specialized journals to determine what acids are found in the plant, but they are not necessarily very acidic and are, according to both articles, not the cause of acidity of the soil associated with the plants.


Why are these two methods of converting from ppm to g/m3 equivalent? I've found two methods to convert from $ppm$ to $g/m^3$. They both appear to give equivalent results, but I don't understand why. Can anyone join the dots for me and explain why they are equivalent methods, or show a derivation for Method 1? I've found a derivation for Method 2.
Method 1 - Definition
$$C[g/m^3] = \frac{C[ppm] \cdot \rho}{1000}$$
Where $C[ppm]$ and $C[g/m^3]$ represent gas concentration in units of $ppm$ and $g/m^3$ respectively and $\rho$ represents gas density.
Method 2 - Definition (As derived here)
$$C[g/m^3] = \frac{C[ppm] \cdot M \cdot P}{R \cdot T \cdot 10^6}$$
Where $M$, $P$, $R$ and $T$ represent molar mass, pressure, ideal gas constant and temperature respectively.
Method 1 - Example - Carbon Monoxide (CO)
Assuming $C[ppm]$ as $100ppm$ and CO density as $1.15 kg/m^3$ at $20^{\circ}C$ and $100000Pa$
$$C[g/m^3] = \frac{C[ppm] \cdot \rho}{1000}$$
$$C[g/m^3] = \frac{100 \cdot 1.15}{1000}$$
$$C[g/m^3] = 0.115$$
Method 2 - Example - Carbon Monoxide (CO)
Assuming $C[ppm]$ as $100ppm$ at $20^{\circ}C$ and $100000Pa$.
$$C[g/m^3] = \frac{C[ppm] \cdot M \cdot P}{R \cdot T \cdot 10^6}$$
$$C[g/m^3] = \frac{100 \cdot 28 \cdot 100000}{8.3145 \cdot (20+273.15) \cdot 10^6}$$
$$C[g/m^3] = 0.115$$

Using the ideal gas law
$$
pV = nRT = \frac{m}{M}RT
$$
you can express the density $\rho = \frac m V$ in terms of molar mass, pressure, ideal gas constant, and temperature:
$$
\rho = \frac{pM}{RT}
$$
With the units you use in your second example, this yields a density in $\mathrm{g\,m^{-3}}$, whereas you insert a density in $\mathrm{kg\,m^{-3}}$ into your first equation. Thus, I will write your first equation as
$$
C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6}
$$
where the density is in $\mathrm{g\,m^{-3}}$.
Inserting the expression for $\rho$ derived above into this equation, we get
$$
C[\mathrm{g\,m^{-3}}] = \frac{C[\mathrm{ppm}] \cdot \rho}{10^6} = \frac{C[\mathrm{ppm}]}{10^6}\frac{pM}{RT}
$$
which is identical to your second equation.


What nuclei need to be NMR active in an HMBC experiment? I recently had a discussion where I was told that in an $\ce{^1H}/\ce{^13C}$-HMBC experiment coupling through an $\ce{C-O-C}$ ether bond would not be observed with the reasoning that O is not NMR active. That somehow surprised me as I thought only the "end nodes" in the coupling needed to be NMR active. So I have the following two questions:

Do the nuclei in-between need to be NMR-active?
Does the end node X need to be NMR-active?

It is incorrect to say that oxygen block the HMBC coupling because it be magnetically inactive.

The theoretical explanation is that the coupling of spins $^nJ_{\ce{XY}}$ is mediated by the electrons i.e. transferred completely independently of the nuclei on the way. And I can provide you with multiple examples that show how coupling crosses over magnetically inactive nuclei:

I myself in my bachelor’s thesis have distinguished the two different ortho-nitrobenzyl ethers of 1-(4-bromophenyl)-1,2-ethandiol by HMBC coupling. It was always possible to locate a coupling from one side’s hydrogen to the other side’s carbon in HMBC in spite of the magnetically inactive oxygen in-between.
During my Ph.D. thesis, I performed a number of $\ce{^29Si}$-HMBC experiments to determine which hydroxy groups were protected by the silyl group used. Again, this transferred across a magnetically inactive oxygen.
Remember that $\ce{^12C}$ itself is magnetically inactive. In your standard $^3J_{\ce{CH}}$ 3-bond-HMBC-experiment, the coupling and thus the cross-peak is transferred along $\ce{H-C-C-^13C}$. I didn’t spell it out explicitly but considering each carbon has only a $1~\%$ chance of being $\ce{^13C}$, it’s much more likely for the chain to be $\ce{H-^12C-^12C-^13C}$, i.e. involving only two magnetically active nuclei.
Finally, your standard hydrogen spectrum exhibits a lot of $^3J_{\ce{HH}}$ coupling. Again, this crosses over two carbon nuclei; the probability of both carbons being $\ce{^13C}$ is $0.01~\%$. If only magnetically active nuclei were able to transfer coupling, then the uncoupled hydrogen signals should amount to $99.99~\%$ of a signal with the coupling signal having only $0.01~\%$ of the intensity. Clearly not what is being observed. In fact, $98~\%$ of the coupled hydrogen signal is due to molecules without magnetically active carbons between the vicinal hydrogens.

The final node has to be NMR-active, otherwise you wouldn’t be able to observe magnetisation.


What's the biggest organic molecule that could have a smell? For a molecule to have a smell it's necessary that the molecule be volatile enough to be in the air. So I think that excludes molecules which are solid at room temperature and atmospheric pressure. Maybe the question then is equivalent to: what is the highest molecular weight organic compound which is liquid at room temperature and atmospheric pressure?

I'll quote from$\ce{^{[1]}}$:

  The general requirements for an odorant are that it should be
  volatile, hydrophobic and have a molecular weight less than
  approximately 300 daltons. Ohloff (1994) has stated that the largest
  known odorant is a labdane with a molecular weight of 296. The first
  two requirements make physical sense, for the molecule has to reach
  the nose and may need to cross membranes. The size requirement appears
  to be a biological constraint. To be sure, vapor pressure (volatility)
  falls rapidly with molecular size, but that cannot be the reason why
  larger molecules have no smell, since some of the strongest odorants
  (e.g. some steroids) are large molecules. In addition, the cut-off is
  very sharp indeed: for example, substitution of the slightly larger
  silicon atom for a carbon in a benzenoid musk causes it to become
  odorless (Wrobel and Wannagat, 1982d).

  A further indication that the size limit has something to do with the
  chemoreception mechanism comes from the fact that specific anosmias
  become more frequent as molecular size increases. At the “ragged edge”
  of the size limit, subjects become anosmic to large numbers of
  molecules. An informal poll among perfumers, for example has elicited
  the fact that most of them are completely anosmic to one or more musks
  (e.g. Galaxolide® mw 244.38) or, less commonly, ambergris odorants
  such as Ambrox®, or the larger esters of salicylic acid.

  One can probably infer from this that the receptors cannot accommodate
  molecules larger than a certain size, and that this size is
  genetically determined (Whissel-Buechy and Amoore, 1973) and varies
  from individual to individual.

References

$\ce{[1]}$: "Structure-odor relationships: a modern perspective", by Luca Turin (Dept of Physiology, University College London, UK) and Fumiko Yoshii (Graduate School of Science and Technology, Niigata University, Japan), which appears as chapter 13 of: Handbook of Olfaction and Gustation. Richard L. Doty (ed.). 2nd ed., Marcel Dekker, 2003.

$\ce{[2]}$: Ohloff, G. (1994) Scent and fragrances: the fascination of odors and their chemical perspectives [Berlin, Springer].

$\ce{[3]}$: Whissell-Buechy D. and Amoore J.E.(1973) Odour-blindness to musk: simple recessive inheritance Nature Sep 21;245 (5421):157-8.


Is chirality absolute or relative? While going through some questions on chirality and optical isomerism, I came across this compound:

Note that I have marked 3 carbons as $C_1,C_2,C_3$
Its obvious that $C_1,C_3$ are chiral carbons.
$C_2$ is not a chiral carbon as it has the same group above and below it( With respect to the picture)
However a thought struck me that chiral carbon meant that 4 different groups are attached to it.
But what exactly does same group mean?
Does it mean that even the spatial arrangement of the groups should be different?
Or does it not matter?
So if spatial arrangement mattered, then the compound I have mentioned would have a chiral $C_2$ carbon for specific configurations around $C_1$ and $C_3$ carbons.

Yes, spatial configuration of attached groups matters. Meaning different stereo configuration of $C_1$ and $C_3$ may render $C_2$ either chiral or not:

Also because you have more than 1 stereo center you're not limited to just enantiomers, you can get diastereomers - the molecules will be different even though they're not mirror images of each other:


At what temperature CO2 becomes solid at normal pressure? I have met contradicting evidence. The phase diagrams show that $\ce{CO2}$ solidifies at $-78.5\ ^{\circ} \mathrm{C}$:

Wikipedia confirms this

  At 1 atmosphere (near mean sea level pressure), the gas deposits directly to a solid at temperatures below $−78.5\ \mathrm{^\circ C}$ ($−109.3\ \mathrm{^\circ F}; 194.7 ~\mathrm{K}$) and the solid sublimes directly to a gas above $−78.5\ \mathrm{^\circ C}$.

Yet this article claims it happens at $133\ \mathrm{K}$ (otherwise $\ce{CO2}$ would showfall at Earth's poles).

Which claim is correct?

The phase diagram you have shown is correct and shows the sublimation point at $1~\mathrm{atm}$ as $-78.5~\mathrm{^\circ C}$ . So if you have a pure $\ce{CO2}$ atmosphere, it will solidify at $-78.5~\mathrm{^\circ C}$.

Temperatues in Antarctica are known to get below $-78.5~\mathrm{^\circ C}$ so you might wonder why whe don't find dry ice in Antartica (like we do at the Martian poles [Note: they are a lot colder]). The reason is that the partial pressure of $\ce{CO2}$ in the atmosphere is so low that at $-78.5~\mathrm{^\circ C}$, dry ice is forming but it is subliming faster than it forms because the vapor pressure of $\ce{CO2}$ in the atmosphere above it is well below the saturation vapor pressure required for ice to accumulate.

  The vapor pressure must reach the saturation vapor pressure for dew or frost to form. This happens at the dew point or frost point temperature, which is dependent on atmospheric pressure and the absolute amount of vapor in the air. As atmospheric temperature increases, the dew/frost point temperature increases. As atmospheric pressure increases, the vapor pressure increases. At very low temperatures, the dew/frost point temperature is very low.

http://wattsupwiththat.com/2009/06/13/results-lab-experiment-regarding-co2-snow-in-antarctica-at-113%C2%B0f-80-5%C2%B0c-not-possible/

https://stevengoddard.wordpress.com/2014/06/03/antarctica-gets-cold-enough-to-freeze-co2/


Peroxide in acidic conditions being reacted with iron sulphate mechanism For the proposed mechanism, I would like to think the $\ce{O-O}$ bond would be cleaved first, thus creating two radicals (an $\ce{OH}$ radical and the radical situated on the other Oxygen), then radical on the oxygen will go into the ring to form a double bond in which the alkene double bond will donate 1 electron towards that bond and 1 electron to the carbon on the end of the opened ring.
My mechanism would be incomplete as I cannot seem to figure out how this reaction occurs.
However, I have another suggestion in which there is an epoxide formation. Can anyone elaborate as to which one is correct or can give me some insight?

At least for the second part of the product formation, you have pretty much given the answer yourself.

The methylcyclopentane moiety of the products results from two different 5-exo-trig radical ring closures, that is from the intramolecular addition of two different radicals to $\ce{R-\color{\red}{C}H=CH2}$.

How are these two different radicals generated?

Have a look at the starting materials! Isn't the combination of the hydroperoxide $\ce{R-OOH}$ and $\ce{FeSO4}$ very similar to the conditions of a Fenton reaction?

Let's therefore postulate the following: $\ce{R-OOH + Fe^2+ -> R-O* +\quad{}OH-  + Fe^3+}$

The resulting 1-methylcylohexan-1-ol derived radical can form methyl ketones by cleaving two different $\ce{C-C}$ bonds; the resulting carbon-centered radicals then give rise to the different cyclization products.


How can we get the toxicity level of a plant without using animals? We need to know the answer for our research paper. We will appreciate it so much if you will answer our question.

Short answer
You can't!
Longer answer

Toxicity is not defined for a plant (or animal), but for specific compounds which are either produced or accumulated in these "toxic organisms". The concentration of toxic compounds may vary during the vegetation cycle and can depend, particularly in the case of accumulation, on the growth conditions.

Toxicity of particular compounds is typically given against specific endpoints, such as daphnids, rats, etc.

You asked for "the toxicity level". Which one do you mean?
$\mathrm{LD_{50}}$, $\mathrm{LC_{50}}$, NOAEL/NOAEC, NOEL/NOEC, DNEL,...

If you cannot measure in vivo for whatever reasons, use established QSAR models for specific endpoints.


Elimination reaction on 2-fluorobutane using alc.KOH Question: Find the major product formed when 2-fluorobutane reacts with alcoholic $\ce{KOH}$

My answer: Elimination of $\ce{HF}$ takes place and a more substituted alkene should have formed. (Zaitsev product). I think the mechanism is E2 since strong base is used.

but answer given is a less substituted 1-butene. Is the answer wrong or am I making any mistake?

E1CB elimination takes place because of a poor leaving group. This is because of the high $\ce{C - F}$ bond strength. Stronger bonds are more difficult to break, making them less reactive.

If the reaction was E2, then zaitsev product would have been formed. But since the mechanism is E1CB, the reaction proceeds via carbanion.

  From wikipedia: The first step of an E1cB mechanism is the deprotonation of the of β-carbon, resulting in the formation of an anionic transition state, such as a carbanion. The greater the stability of this transition state, the more the mechanism will favor an E1cB mechanism.

I hope you know that for carbanions, the stability order is primary > secondary > Tertiary.

Now that you know the stability order of carbanions, the preferred position for carbanion is a primary carbon. So base abstracts a proton from a primary carbon.