Heat of formation of aqueous H⁺ Heat of formation of aqueous $\ce{H+}$ is taken to be zero, by convention. Heat of formation is energy released or absorbed when 1 mole of a compound is formed from its elements in their standard states.
$$\ce{1/2 H2(g) -> H(g) -> H+(g) -> H+(aq)}$$
Since heat of formation of $\ce{H+(aq)}$ is assigned to be zero, this would imply that this reaction has $ΔH = 0$. Wouldn't this imply that:
$$\text{Enthalpy of atomization} + \text{Ionization energy} + \text{Heat of hydration} = 0~?$$
However, this doesn't seem to be the case:
$$435 + 1312 - 1091 = \pu{656 kJ/mol}$$
Where am I going wrong here?

Where am I going wrong here?

Your chemical reaction is not balanced. You wrote

$$\ce{1/2 H2(g) -> H(g) -> H+(g) -> H+(aq)}$$

but it should be

$$\ce{1/2 H2(g) -> H(g) -> H+(g) + e- (?)-> H+(aq) + e- (?)}$$

I put a question mark for the physical state of the electron because it's not clear where that electron should go or what enthalpy of formation we would assign it. You could instead involve a second element in your reaction, e.g.:

$$\ce{H2 + Cl2 -> 2H+ + 2Cl-}$$

You could assign an enthalpy of formation for the mixture of hydrogen and chloride ions, but you would not know how to assign it to the individual components of the mixture.

  In my book it is given that by convention heat of formation of aqueous H+ is taken to be zero.

Yes, that's what we do. It would equally be possible to assign it a value of 100 kJ/mol. All the enthalpies of formation of ions would change (for example, that of chloride ions would by 100 kJ/mol lower to compensate). However, all the calculations of enthalpies of reactions would have the same result because it all cancels out in a balanced equation. Setting the enthalpies of formation of elements to zero is also a bit arbitrary (not once we specify that to measure them, we are making the compounds from the elements).

Updating the definition

Instead of saying the enthalpy of formation of a compound is the enthalpy of making that compound from the elements, we could widen the definition to say:

"The enthalpy of formation of a species is the enthalpy of the reaction making the species where all other reactants and products are either elements or the hydrogen ion""

This definition would include ions. For chloride ions, the following reaction would work:

$$\ce{1/2H2 + 1/2Cl2 -H+ -> Cl-}$$

For sodium ions, the following reaction would work:

$$\ce{Na + H+ - 1/2H2 -> Na+}$$

Why the hydrogen ion?

Again, it is arbitrary and any other ion would have been fine as well. The hydrogen ion is special for two reasons, though. It is the smallest ion, and it is fairly common in reactions, so it saves some arithmetic effort when doing calculations.


On the stability of deuterium The isotope of hydrogen, tritium, is indeed radioactive resulting in $\ce{^3He}$ through beta decay.

But what of the radioactively 'stable' isotope deuterium?

Are all deuterium atoms destined to be deuterium for eternity or is there any process other than radioactive decay that could lead to transition such as say $\ce{^1H}$?

No, deuterium is completely stable. I found the answer at Hyperphysics, and it has to do with the mass energies of the products and reactants of this hypothetical reaction.

The decay of deuterium would be
$$\ce{D -> P + N + e + \bar{\nu}}_e$$
where $\ce{D}$ is deuterium, $\ce{P}$ is a proton, $\ce{N}$ is a neutron, $\ce{e}$ is an electron and $\ce{\bar{\nu}}_e$ is an electron antineutrino. The combined mass energies total to $1877.05\,\mathrm{MeV}$ (mega electron volts). But the mass of $\ce{D}$ is $1875.6\,\mathrm{MeV}$, so the decay cannot happen because conservation of mass/energy would be broken. There's no way to get around this, not even imbuing the initial deuterium atom with extra energy of some other sort. Deuterium will never decay unless, for some reason, the proton turns out to be unstable.

If, for some reason, deuterium did decay, it could go to $\ce{^2He}$ … which would then decay back to deuterium.


Is 1 ppb equal to 1 μg/kg? In an article I recently submitted, a reviewer asked that I provide a concentration in μg/kg instead of ppb (parts per billion), and mentions that the later is not correct. I am not a chemist, and I thought that 1 μg/kg = 1 ppb.

Is 1 ppb equal to 1 μg/kg ? What is a reason to consider ppb as incorrect ?

You are correct suggesting that 1 μg/kg implies 1 ppb, however the reverse is not true. For instance, 1 ppb can also be 1 nmol/mol, and the reader will never have a chance to deduce which one is it unless you explicitly define the usage of the "parts per something" in the text.
This clutters the manuscript with redundant notes and causes overall confusion.

IUPAC also lists all similar symbols (ppm, ppt, ppb etc.) as deprecated; from IUPAC's “Green Book” [1, p. 98]:

  Although ppm, ppb, ppt and alike are widely used in various applications of
  analytical and environmental chemistry, it is suggested to abandon completely their use because of the ambiguities involved. These units are unnecessary and can be easily replaced by SI-compatible quantities such as pmol/mol (picomole per mole), which are unambiguous. The last column contains suggested replacements (similar replacements can be formulated as mg/g, μg/g, pg/g etc.).

  $$
\begin{array}{lllll}
\hline
\text{Name} & \text{Symbol} & \text{Value} & \text{Examples} & \text{Replacement} \\
\hline
\ldots & & & & \\
\text{part per billion} & \text{ppb} & 10^{-9} & \text{The air quality standard for ozone is a} & \pu{nmol/mol} \\
  & & & \text{volume fraction of}~\varphi = 120~\text{ppb} & \\
\ldots & & & & \\
\hline
\end{array}
$$

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. (PDF)


Cyanide from seeds breaking down in a pond I have a large loquat tree that overhangs a pond. The loquat seeds apparently contain cyanide of some sort, so I'm trying to find the effect the seeds have when they fall in the pond.
Over time I imagine the seeds would break down.
What happens to the cyanide? Does it escape into the air, or remain in the water?

Loquat seeds contain Amygdaline, which is also found in apricot kernels. Amygdaline belongs to the group of cyanogenic glycosides.

In these compounds, a potentially cyanide releasing aglycon, such as mandelic nitrile, is bound to a carbohydrate (= a "sugar"). These compounds are quite abundant in nature; other known examples are Prunasin from the japanese bush cherry or Sambunigrin from elderberry seeds.

In some cases, cyanide is released by enzymatic cleavage of the glycoside, that is when the plant material is digested by animals. Consequently, some of the plants (or at least some parts of them) are known to be toxic. In other cases, the glycosides do degrade in water.

If hydrogen cyanide is released to the pond, I'd assume it stays there - $\ce{HCN}$ is highly soluble in water.

However, I wouldn't be very concerned. Seeds aren't like paper bags full of toxic material.

It will likely take a while until the seeds break up to release Amygdaline.
To my knowledge, Amygdaline is rather stable in water, hydrogen cyanide isn't released until a sugar-splitting enzyme or acid acts on the molecule.


Mechanism of initiation in contact explosives? It's well known that undiluted nitroglycerin will explode when subjected to a physical shock.  What is happening as a result of the physical shock that sets off the explosive reaction?  In particular, why doesn't it explode spontaneously at room temperature without a physical shock?

My guess would be cavitation, presumably 'non-inertial cavitation':

Non-inertial cavitation is the process in which a bubble in a fluid is forced to oscillate in size or shape due to some form of energy input, such as an acoustic field. Such cavitation is often employed in ultrasonic cleaning baths and can also be observed in pumps, propellers, etc.

The energetics of cavitation are astonishingly out-sized with respect to the tiny size of bubbles, probably leading to more than enough energy to cross the initial activation barrier of the combustion reaction.  As one example, the abstract here summarizes other reports' results indicating that the temperature inside collapsing cavitation bubbles may reach between 10,000 K and 100,000,000 K.
I don't have access to the full text to determine whether or not they met with success in attempting to ascertain cavitation's role in nitroglycerin initiation, but these authors apparently investigated the question almost fifty years ago.  The abstract (references excised):

It is known that in some circumstances liquid explosives can be initiated with unexpected ease, and in other circumstances only with great diffculty.
Thus, Winning has shown that nitroglycerine (NG) free of gaseous inclusions and poured into a vessel so as to leave no wall surfaces free of liquid is not exploded even by the action of a fairly strong shock wave from a detonator immersed in the NG.
On the other hand, in handling liquid explosives there have been quite a few cases in which relatively weak vibrations or impacts have led to unexpected explosions, which have sometimes had serious consequences. For example, a British report describes an unfortunate accident that resulted from dropping a polyethylene bottle containing NG. Upon hitting the ground the NG exploded.
The initiation of explosion by “hot spots” resulting from the adiabatic compression of gaseous inclusions even before the arrival of the shock wave has been reliably demonstrated in numerous experiments. However, some cases of initiation of liquid explosives simply cannot be attributed to the heating of such gaseous inclusions, since in these cases the adiabatic compression temperatures of the gas are so small that it is not possible to talk of a “hot spot. “Such puzzling cases include, for example, the above-mentioned explosion of NG in a polyethylene bottle. In other experiments the role of gaseous inclusions has been completely eliminated by first subjecting the liquid explosive to a constant high pressure. This so reduced the degree of compression of the gaseous inclusions by a weak shock that strong heating of the gas in the bubbles, if any were present in the liquid explosives, was completely excluded. Nonetheless, there was no reduction in the sensitivity of the explosive to weak shocks.
In attempting to explain such puzzling cases it is usually pointed out that explosion can be initiated by cavitation, which may develop in a liquid even as a result of a weak impact or vibration. So far, however, no one has offered any direct experimental evidence of the possibility of cavitational initiation of explosion in liquid explosives. The object of our research was to fill that gap.

Clearly they published their results, so I figure the odds are good that they reached some sort of affirmative conclusion.


What does the equilibrium constant mean for reactions with very small concentrations? When you calculate the equilibrium constant for a reaction with a higher concentration of reactants, the answer is fairly intuitive. For example, for the reaction:
$$\ce{A + B <=> C}$$
Where we have $0.10 M$ of A, $0.30 M$ of B and $0.45$ M of C the equation for the equilibrium constant is:
$$ K = \frac{[0.45]}{[0.10][0.30]} = 15$$
This is relatively intuitive, we can see that the concentration of C is larger than the product of A and B and so saying the products are favoured makes sense.
However, an issue comes about when you consider really small amounts of all of them. For example, for the same reaction
$$\ce{A + B <=> C}$$
If we have $2.4*10^{-4} M$ of A, $2.4*10^{-4} M$ of B and $7.5*10^{-6} M$ of C we have the equation:
$$ K = \frac{[7.5*10^{-6}]}{[2.4*10^{-4}][2.4*10^{-4}]} = 130$$
In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products. The constant makes sense since multiplying two numbers less than 1 will get a smaller number but it's confusing when you consider the implications that has on the equilibrium. My immediate assumption is that I am misunderstanding the relationship between the equilibrium constant and the concentrations of species in the reaction and if this is the case is there a more intuitive way of looking at this?

"In this case, the equilibrium constant suggests the products are favoured but if you look at how much product we have compared to reactants there's so much more reactants than products."  This is what I take as the heart of your question.  To that point:
I believe you need to compare the concentration of all the reactantS, not just one of the products.  In this case, the concentration of all of the reactants is 5.76 X 10 -8 which is a lot less than the 2.4 X 10 -4 which you might have been comparing to the product in the equation.


Is +6 really the most stable form of sulfur? I think everyone might be aware that +6 is considered the most stable form of sulfur. But I think why this reaction happens very slowly and at such a high temperature and even the help of a catalyst is required, but in almost all redox reactions sulfur goes to +6:

(controversial where +6 is forms very forcefully):
$$\ce{SO2 +O2->[\ce{Pt/V2O5}][\text{high temp.}]SO3}$$
And also does it reduces easily (controversial where +6 is reduced easily):
$$\ce{P +SO3->P4O10 +SO2}$$
The reactions in my favour are(where stable +6 is formed):
$$\ce{SO2 +Cr2O7^2- ->SO4^2- +Cr^3+}$$
$$\ce{BaSO3- ->[\text{warming/$\ce{Br2-}$water}][\ce{/HNO3/H2O2/$\text{dil.}$HCl}]BaSO4 }$$
$$\ce{[AgSO3]- ->[\text{boil}]Ag + SO4^2- +SO2}$$
$$\ce{Ag2SO3+H2O ->Ag + SO4^2- +H+}$$
$$\ce{SO3^2- +MnO4- ->Mn^2+ +SO4^2- }$$
$$\ce{PbSO3->[\text{boil}]PbSO4}$$
$$\ce{S +HNO3->[\text{boil}]SO4^2-}$$
etc.

I think it's a matter of activation energy and reaction kinetics. Although the product is much more thermodynamically stable than the reagent(s), the reaction path could feature a high activation energy which requires those harsh conditions and even a catalyst to make it happen. As shown in the graph below, despite having a favourable difference of free energy ($\Delta G < 0$), the activation energy $E_a$ required to go from reactants to products could be not negligible and hard to overcome.

The catalysts allows the reaction to happen more easily by lowering the free energy of the transition state (by stabilising it). The activation energy is also reduced by the high temperature, according to the Arrhenius equation:

$$k = A e^{{-E_a}/{RT}}$$


Why is the major product of the reaction of 1-naphtol with benzene diazonium chloride not the ortho product? If 1-napthol reacts with benzene diazonium chloride, it gives the para product with respect to the hydroxyl group. However, shouldn't the major product be the ortho one, because there will be intramolecular $\ce{H}$-bonding between the $\ce{H}$ atom and $\ce{N}$?

When benzene diazonium chloride attacks the ortho position with respect to -OH it(the benzene ring) experiences a lot of steric hindrances,along with that,the triple bond also repels the lone pairs of oxygen,therefore the nitrogen atom will attack para with respect to the -OH group.If the temperature is increased i think that an ortho product will be formed as it is the thermodynamicly stable product,while the para product is the kinetically stable product.


Which d orbitals are involved in "d3s hybridisation"? I was reading about hybridisation and came across a point saying that $\mathrm{d^3s}$ hybridisation involves only $\mathrm d_{xy}$, $\mathrm d_{yz}$, and $\mathrm d_{zx}$ orbitals and not $\mathrm d_{x^2-y^2}$ nor $\mathrm d_{z^2}$. Why is that so?

First off, I'll reiterate what's mentioned in the comments - that hybrid orbitals are not a particularly good approach for transition metal complexes. But with that said, if you are going to apply this approach to a tetrahedral transition metal complex, here's how it works. For simplicity, I'm going to assume that only sigma bonding is relevent. You can do the same analysis including pi bonding, but it's more complicated.
Short answer: only the off-axis d orbitals have the right symmetry to match the ligands in a tetrahedral point group.
Long answer: The mathematical basis for constructing both delocalized and localized hybrid bond orbitals comes from group theory. If you haven't gotten to group theory in your studies, this explanation might not make a lot of sense, but at least you'll have to starting point to look deeper.
A tetrahedral molecule with four identical ligands is said to be in point group Td, which is just an abbreviation of "tetrahedral". Every point group has symmetries associated with it based on what operations can be performed that return the original structure. In point group Td, these are named A1, A2, E, T1 and T2.
When we build molecular orbitals, we combine the orbitals on the four ligands and look for possible symmetry groups. Then we look at the symmetry of the orbitals on the metal center and match the metal orbitals and "ligand group orbitals" based on symmetry. (An alternative approach is to look at the symmetry transformation of the bond vectors, which is described here: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Modules_and_Websites_(Inorganic_Chemistry)/Advanced_Inorganic_Chemistry_(Wikibook)/01%3A_Chapters/1.09%3A_Td_Molecular_Orbitals. The result is the same.)
Making a long story short, the tetrahedral ligand group orbitals for a sigma-bonded complex fall into two symmetry groups. One orbital, in which all four ligand orbitals have the same phase, is A1. A second group of three orbitals in which the ligand orbitals are pairwise in opposite phases matches the T2 symmetry of the Td group.
On the metal center, our valence atomic orbitals are 4s, 3d and 4p. The 4s orbital matches the A1 symmetry, while the 4p and 3d (xy, yz and xz) match the T2 symmetry. The 3d(x2-y2) and 3d(z2) orbitals, on the contrary, match the E symmetry from group Td. Since the E symmetry doesn't match either of our ligand group orbitals, we can't use these two orbitals for bonding.
Since the 4s orbital is the only match for the A1 symmetry, we must involve that orbital in bonding. For the T2 symmetry, however, we have two choices. We can either use the three 4p orbitals and ultimately make four sp3 hybrids, or we can use the three T2 symmetry 3d orbitals and end up with d3s hybrids. The sp3 hybridization is much more familiar to most people because they have learned it for organic compounds such as methane, but that's not the right choice here. If we consider the relative energy levels of the orbitals on a transition metal, we see that 3d is lower than 4p and so is closer to the ligand orbitals (which are lower in energy than both). Furthermore, the off-axis d orbitals are pointed more directly at the ligands in a tetrahedral complex (which are also off-axis), so this intuitively makes more sense as well.
Finally, to your question about $\ce{CrCl2O2}$, that molecule belongs to point group C2v, so you can apply the analysis above to that point group to get the allowable hybridizations in the same manner as for the Td case.


Clarification on Recrystallization concept Any help you could provide me would be greatly appreciated and I thank you for reading (even if you don't comment)

I am in an organic laboratory and as a 3rd chemistry major these concepts usually are pretty basic but recrystallization is kind of leaving me scratching my head. When you dissolve your compound with a small impurity in it, dissolve it in a solvent and heat, I am just confused on this part:

What I understand is that the solubility constant changes with temperature and in the case of organic compounds, solubility almost always increases with increasing temperature. So, if the compound is put in a solvent at room temperature and is heated until it is fully dissolved (solubility constant increased) then it is cooled and the solubility constant decreases and consequently solute starts precipating (crystal formation) My only issue is what is that makes the impurity go away? Thinking through it here, I know I'm so close to the Ah-hah moment but as of right now I'm mental blocked — I suppose.

You note that you are pretty close to the answer and you are. Let’s assume for a minute (I’m pretty sure the assumption is false, but it’ll get us a good way) that the solubility of the impurity is independent of the amount of desired product dissolved and vice-versa. Let’s also assume we have a mixture of $95~\%$ desired product and $5~\%$ side product which you want to purify by recrystallisation. The desired product has a solubility of $1\,\mathrm{\frac{mol}{l}}$ at $4\,\mathrm{^\circ C}$ and $10\,\mathrm{\frac{mol}{l}}$ at $80\,\mathrm{^\circ C}$; and the side product has the same solubility. And finally, let’s assume that you have $100\,\mathrm{mmol}$ total product (side and desired).

You then add $9.5\,\mathrm{ml}$ of your recrystallisation solvent and heat to $80\,\mathrm{^\circ C}$. That would be enough to dissolve both your $95\,\mathrm{mmol}$ desired product and your $5\,\mathrm{mmol}$ side product. You observe a clear (hopefully boiling because best solubility) solution.*

You then let your solution cool down and put it in the fridge overnight. Due to the decreased temperature, the solubility of both desired product and side product drop. The same volume of solution now dissolves only $9.5\,\mathrm{mmol}$ of both. Since the desired product is supersaturated ($95 > 9.5$), $85.5\,\mathrm{mmol}$ of the desired product will precipitate or crystallise. But the side product is still not supersaturated. We have $5\,\mathrm{mmol}$ of the side product dissolved, but the solvent could dissolve $9.5\,\mathrm{mmol}$. So all of the side product will remain in solution (ideally).

Next step is usually filtration and washing. The filtrate will now contain the saturated product solution which is undersaturated with respect to the side product. The side product didn’t disappear, it merely remained in solution. Wash your pure product, record an NMR spectra and rejoice upon its purity.

Of course, as I stated in the beginning, the assumption that solubilities be independent is wrong. It is also tendencially wrong to assume similar solubilities of desired product and impurity. One would attempt to select a solvent so that the product is not well soluble at low temperatures, but the impurity ideally is. However, the basic principle remains valid: The desired product will form a supersaturated solution and precipitate/crystallise while whichever impurity you have should still be soluble to not co-precipitate.

*: A note on how to do it practically: Usually, you would put your crude product into a flask into an oil bath with a refluxing condenser on top. You would then slowly, dropwise add solvent through the condenser until the solution is boiling and everything is dissolved. One wouldn’t go about saying ‘I should have $x\,\mathrm{mmol}$ so I should need $y\,\mathrm{ml}$ of solvent.’


By what reaction does (–)-sparteine decompose in ambient conditions? In its pure form (–)-sparteine is an only-slightly-yellow clear, viscous liquid. Yet, after only a week of being kept in dry, normal atmosphere (in a round bottom flask covered by a septa), some of it reacts with something to form a sticky, orange-brown substance. I've had to distill the sparteine I have twice due to whatever reaction is occurring, and I would like to know exactly what the reaction is so I don't have to distill every time I need it. The literature I found mentioned that it readily absorbs carbon dioxide in a 1-1 molar ratio, and I know that tertiary amines do sometimes react with CO2, but I haven't been able to piece it together as something that could happen in ambient conditions.

Amines left out in the atmosphere will air-oxidize to give N-oxides, which typically have a yellow color. These N-oxides can decompose to give further byproducts. I would store your sparteine under nitrogen or argon atmosphere.

[edit to give examples and mechanism]
The reaction proceeds through a radical chain mechanism with molecular oxygen. Here is one study of the mechanism [from first page of Google search for amines air-oxidation] .

From wikipedia article for triethylenetetramine. "This oily liquid is colourless but, like many amines, assumes a yellowish color due to impurities resulting from air-oxidation."

"It can be observed that arylamines develop a yellow or black color if left exposed to air for prolonged periods". Ref Note - I realize that sparteine is not an aromatic amine, but the result is the same.

If you have access to triethylamine or other amines in your lab, do a control experiment: leave one vial open to air, and leave the other under identical conditions except sealed under an inert atmosphere. A few days worth of observation will convince you of the role in oxygen in the yellowing of amines.


Dipole and size of a thioether relative to hydrocarbon I am an enzymologist and I would like to check whether the consensus of how enzymes tell thioethers from hydrocarbons on substrates is a fact or fiction.
So are any of these wrong?

S–π interactions are the sole discriminant (in enzymes at least)
They have the same dipole
They are the same size
A thioether and a hydrocarbon have the same angle

From what I can tell...
Point #1 is not really in question: even if it does not account for much, aromatic residues are often found with thiols and thioethers (1/3 of protein have a methione involved in a S–π interaction).
From what I can tell of point #2, 3.55 vs. 3.58 units in electronegativity (ref. wikipedia: electronegativity) won't make a difference. From point #3, while, yes, sulfide is 360 pm and methane is 380 pm (ref. wikipedia: kinetic diameter), but the former is R-S-R, R-CH2-R, so it might be different.

I am not saying they are very different: thialysine (thioether) is a toxic analog of lysine (alkane) and norleucine (alkane) of methionine (thioether; toxic for SAM usage only) due this reason. But nature has evolved to discern similar compounds.

An example, cystathionine (pictured below) is a thioether of cysteine + homoserine or homocysteine + serine (both routes exist) and one enzyme (cystathionine beta-lyase, CβL) can bind the amine on the shorter side, deprotonate the α-carbon and eliminate the thiol (homocysteine) and release the imine (hydrolysed to pyruvate), while a different enzyme (cystathione gamma-lyase, CγL) acts on the other (cysteine and ketobutyrate). The two enzymes do not catalyse both reacts as it would be wasteful (not much just a ATP>AMP). The evolution from one to the other has not be published, but from what I can tell S–π interaction is the driving force. AFAIK, the closest literature to this on cystathionine elimination is my own paper, so I know there is no answer there.

I just realised that I got the dipole business utterly wrong as sulfur has two lone pairs.
I checked dimethyl sulfide, propane and dimethyl ether and was surprised how differently they behaved. The former has a total dipole moment of 1.5 D ref, while the second near zero (0.1 D) as expected from a non-polar molecule, while the last has 1.3 D.
Whereas there is only a 0.03 difference in electronegativity between C and S (vs. .89), the geometry is different, which I did not know about. Wikipedia on says:

thioether article: "C–S–C angle approaching 90°. The C–S bonds are about 180 pm."
ether article: "C–O–C linkage defined by a bond angle of about 110° and C–O distances of about 140 pm".
alkane article: "154 [pm] for a C–C bond [and] with an angle of 109.47° between them."

So the side chain of methionine should actually be able to hydrogen bond with stuff (not electrostatically, but dispersively). So I was surprised to find papers I had not found prior (ex1, ex2).

S-π are probably the strongest force nevertheless (the bond has an enthalpy of 2.6, 1.1 or 0.7 kcal mol-1 depending on the angleref, which is comparable to N−H···:O hydrogen bond (1.9 kcal mol-1))


What are solvolytic displacement reactions? High school student here, I recently purchased a book titled: "Solvolytic Displacement Reactions" by Andrew Streitwieser published in the early 1960's. I found it at a book store and it looks like one of those incredibly valuable (in terms of knowledge) books. I tried reading the first few pages (have a basic background in chemistry) but I do not see "the big picture". What are solvolytic displacement reactions, how are they used and what are they used for?

A book could be written on this subject!

A solvolysis reaction is an $\ce{S_{N}1}$ reaction where the solvent acts as the nucleophile and replaces the leaving group.  The following reaction is a solvolysis reaction run in ethanol.

I suspect there are some industrial uses of solvolysis reactions and they have been used in labs and by drug companies as a step in a synthetic sequence to prepare some complex molecules.

But I suspect the biggest contribution from the study of solvolysis reactions is the impact they have had on our understanding of carbocation chemistry.  From our study of solvolysis reactions we have learned how and why some carbocations rearrange before being captured by solvent.

While studying such rearrangements the solvolysis of the norbornyl system was examined.

It turns out that

norbornyl compound 1 solvolyzes several hundred times faster than compound 3 - why?
if compound 1 is prepared optically active, the products are racemic - why?
If compound 1 is radiolabeled, the label is scrambled in the products - why

It took decades to answer these and other questions.

These investigations led to a whole new field of carbocation chemistry: the classical - non-classical carbocation debate (see the norbornyl link above for a discussion of this topic).  These investigations added a whole new section to physical organic chemistry textbooks.  What was learned from these solvolysis studies completely changed our thinking about carbocation chemistry.


Is there a way to dissolve glue? Background: I have a journal from my childhood. I applied glue (uhu brand. this one) to the papers almost 10 years ago.

Q: Is there a way to dissolve the glue and make the pages readable?

You might try different solvents on a spot that does not have writing, to see if it removes the glue, and on glue with non-critical writing, to see if it removes or smears the writing as well as the glue.

Since it described as a clear synthetic resin, you might try the following sequence, from milder solvent to stronger:

Comparatively mild non-polar organic solvents, such as paint thinner or kerosene. If that does not dissolve the glue, try
Some Polar solvents such as acetone, methyl ethyl ketone or nail-polish remover. If that does not remove the glue, try
PVC pipe solvent (tetrahydrofuran).

N.B. All solvents listed are flammable, and some are fairly toxic. Use them in small quantities, closing the containers after getting a bit out, to prevent flash-back of flames, and use in well-ventilated areas.


Why is H₂O V shaped? We know that the molecule of H₂O is V-shaped. This is what makes it a dipole.

But why is that? I mean, if the hydrogens have a partial positive charge, then they should try to get away from each other, until they are diametrically opposite to the oxygen molecule. But that doesn't happen. Why is this?

The oxygen atom fills its octet rule by forming two bonds. It shares 1 electron in a covalent bond with each hydrogen and has 4 remaining valence electrons. It is  sp$^3$ hybridized and has the 4 non-bonding electrons in two lone pairs. An sp$^3$ hybridized atom has four attachment points spaced approximately 109$^\circ$ apart and has the shape of a pyramid with a triangular base. See this image for an example. You don't typically see the water molecule drawn in 3 dimensional space with the lone pairs of electrons, all you see is a planar molecule with H-O-H bond angle of ~109$^\circ$.


Does combustion occur more easily at higher oxygen concentrations or just higher oxygen partial pressure? I saw a trailer for The Martian and started thinking about space flight.  I know that spacecraft and spacesuits often operate at low pressure with high oxygen content (near 100% $\ce{O2}$ at about 20 kPa), which is harmless to humans and keeps the mass of the space vehicle lower.  Humans do just fine with an oxygen partial pressure of 20-30 kPa as long as other gases present are in safe amounts.

But an issue I hadn't considered is the effect on combustion.  A lot of people mistakenly believe that the Apollo 1 fire was due to using pure oxygen in the test, but that's only part of the issue, and astronauts routinely use nearly pure oxygen at low pressure even today.

So my question is, is flammability affected by the total concentration of oxygen in the atmosphere, just its partial pressure, or some combination of the two?

As a specific example, what affect would a 30 kPa almost pure oxygen atmosphere have on flammability?

When you're dealing with questions like "how well does this burn", you are intrinsically dealing with the reaction rate of the combustion reaction.

When looking at chemical reaction rates, they are typically proportional to the "chemical activity" of the species reacting. (There's complications due to the stoichiometry of the reaction, but those corrections are confined solely due to the nature (formula) of the reaction, rather than any particular concentration/pressure conditions.) The definition of chemical activity can get complex in certain cases, but for most normal reaction conditions, the "chemical activity" of a reactant is proportional to its molar concentration - or in other words, how many molecules of the reactant are in a given volume.

For solution reactions, we typically use the concentration directly, but for gaseous reactants, we can take advantage of the ideal gas law and realize that at a fixed temperature, the concentration of a substance is proportional to it's partial pressure. Which means the chemical activity of a particular gaseous reactant will be proportional to its partial pressure.

So in determining the reaction rates of such things as combustion, it's the partial pressures of oxygen and other reacting gases that matter, rather than the total pressure. Absent any side reactions, or considerations such as thermal transfer, adding or removing an inert gas should not affect the rate of reactions.

(But note for combustion, which is highly exothermic, thermal transfer effects can't necessarily be neglected. The ability of the inert gas to either carry away excess heat - or act as a blanket to keep heat near the reaction point - could substantially change how quickly a fire burns in practice. But I'd expect it to be a complex relationship based on air circulation, convection considerations, etc.)


About dative covalent bonding and expanded octet rule I'm beginning my studies of chemistry at high school (I'm not a native English speaker, though) and I have some basic questions:

1)  Consider the compound xenon tetroxide. In this molecule, does the xenon have its valence layer expanded? In another words, the oxygen atoms only have double covalent bonds with the Xenon?

I've tried to look for this answer but I'm still in doubt whether xenon does four dative covalent bonds (with each oxygen) or four double covalent bonds. In the first case, does not occur the expanded octet rule for xenon, right?

2) Another question; is it possible for a dative covalent bond to occur with just one electron? I mean, the atom which has more electrons in its valence layer "donates" just one electron instead of two? If the answer is "yes", can you provide me an example, please?

I'm sorry for the English mistakes. I started to learn this "chemistry vocabulary" just now.

To answer your first question: the concept of octet expansion for main group metals is no longer generally considered correct and only upheld by the more traditional organic chemists that don’t deal with inorganic bonding situations on a day-to-day basis — or anybody else who didn’t need to recheck what they were taught decades ago. Nowadays, these seemingly octet-expanding structures such as $\ce{XeO4}$ or $\ce{XeF2}$ are explained by dative bonds and/or four-electron-three-centre bonds; both have the beauty of not requiring d orbitals (whose energy would be comparable to the next shell’s s orbital; remember the aufbau principle!).

To answer your second question: no. A dative bond always requires two electrons that are given to one atom upon supply of the bond dissociation energy. It doesn’t matter how a certain bond is created, it only matters how it dissociates; and if it dissociates and was a bond, it will have to have had an electron pair.


Which formula can be used to calculate the exact hydronium concentration present in sodium hydrogen carbonate solution? Which formula  can be used to calculate the exact  hydronium concentration present in  sodium hydrogen carbonate solution ?

$$\ce{[H3O+]}=\sqrt{K_\mathrm{a1}K_\mathrm{a2}}$$
$$[\ce{H3O+}]=\sqrt{\frac{K_\mathrm{a1}K_\mathrm{w}}{[\ce{NaHCO3}]}}$$
$$[\ce{H3O+}] = \sqrt{\frac{K_\mathrm{a1}K_\mathrm{a2}\times[\ce{NaHCO3}]+K_\mathrm{a1}K_\mathrm{w}}{[\ce{NaHCO3}]+K_\mathrm{a1}}}$$

I  tried to order the series  of reactions once sodium hydrogen carbonate  is dissociated according  the following sequence :

Taking into consideration  equation two and equation four, and  taking the assumption: Degree of ionization=degree of hydrolysis, an approximate calculation hydronium concentration can be done by the  formula  one.
Recognizing  equation four and equation two  , that for every one mole of carbonate- formed, one mole of hydronium was formed, and  for every mole of carbonic acid formed, one mole of hydronium  was used up and when comparing K value between both of these equation, I can ignore the contribution of equation four. And because of the low concentration of carbonic acid  in equation three and low carbonate ion  concentration in equation five , I can ignore the contribution of  both equation. So I take into my consideration equation two to derive formula two.
I tried to  derive formula three but  I don't know the assumptions beyond this formula and how to derive it.
I need to know how  to calculate concentrations  of  all species are present in  sodium hydrogen carbonate solution ?
I appreciate any help.

Wording of the Question...

The question asks: Which formula can be used to calculate the exact hydronium concentration present in sodium hydrogen carbonate solution?

What does "exact" mean?

The word exact gives me pause. In general there isn't usually an exact answer, but rather only an answer that is within the required experimental error. The values of $K_\mathrm{a1}$, $K_\mathrm{b1}$, and $K_\mathrm{b2}$ are given to three significant figures. The value for $K_\mathrm{a2}$ is only given to two significant figures, but

$$K_\mathrm{a2} = \dfrac{K_\mathrm{w}}{K_\mathrm{b2}} = \dfrac{1.00\times10^{-14}}{2.13\times10^{-4}} = 4.69483\times10^{-11}\tag{SF1}$$

So it seems reasonable to assume that $K_\mathrm{a2} = 4.70\times10^{-11}$ and that exact means to within 1 part per thousand (ppt).

Since the pH of pure water is taken to be 7.000, this means that for any pH between 4.000 and 10.000 the autoionization of water will have to be taken into account to get results within 1 ppt. (Remember that the characteristic of the pH doesn't count in significant figures, only the mantissa does.)

Realistically that is terribly precise. At 25 $^\circ \mathrm{C}$ $K_w \approx 13.995$ and the change is about 0.033 per $^\circ \mathrm{C}$. So a 1 ppt change is only a 0.03 $^\circ \mathrm{C}$ change in temperature. For such calculations I'd probably expect the typical error would be on the order of ± 5% of reality. But the problem poses 1 ppt precision and so that is how the calculations "should" be done. If the problem poser wanted less precision them the equilibrium constants should have been specified with fewer significant figures.

Setup

Assumptions

Ignoring relationship between $\ce{CO2}$ gas phase and dissolved $\ce{CO2}$ in the liquid phase. Mass balance would require relative volumes of gas and liquid phases.
Ignoring relationship between dissolved $\ce{CO2}$ and carbonic acid. Accept  that $K_{\mathrm{a1}}$ is the "combined" equilibrium constant which takes that relationship into account.
The equilibriums will be calculated using concentrations instead of the more appropriate activities. So above 1 millimolar concentration of sodium bicarbonate the calculated results are suspect.

Starting with the basic relationships.

$$K_{\mathrm{a1}} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\tag{S1}$$

$$K_{\mathrm{a2}} = \dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}}\tag{S2}$$

$$K_\mathrm{w} = \ce{[H^-][OH^-]}\tag{S3}$$

Letting $\mathrm{[C_T]}$ be the total concentration of all the various carbonate species (which we know from is nominal molarity of the sodium bicarbonate).

$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$

But this leaves us with five unknowns ($\ce{[H+], [OH^-], [H2CO3], [HCO3^-]}$ and $\ce{[CO3^{2-}]}$) but only four equations (S1), (S2), (S3) and (S4). So another equation is needed relating the five unknowns to completely numerically solve the system.

Fraction of Species

The fraction of each species can be calculated for a particular pH value using equations (F1), (F2), and (F3). Basically you substitute values into (S4) as a function of $\ce{[H+]}$ and whatever species ($\ce{[H2CO3], [HCO3^-], [CO3^{2-}]}$) that you're interest in. The equations for the fractions of the species are exact.

$$\dfrac{\ce{[H_2CO_3]}}{\ce{[C_T]}} = \dfrac{\ce{[H^+]^2}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F1}$$

$$\dfrac{\ce{[HCO3^-]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}[H^+]}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F2}$$

$$\dfrac{\ce{[CO3^{2-}]}}{\ce{[C_T]}} = \dfrac{\ce{K_\mathrm{a1}K_\mathrm{a2}}}{\ce{[H^+]^2 + K_\mathrm{a1}[H^+] + K_\mathrm{a1}K_\mathrm{a2}}}\tag{F3}$$

Think of dissolving $\ce{H2CO3}$ ($\ce{C_T}$) and then adding between 0 and 2 equivalents of $\ce{NaOH}$. The equations relate the three variables pH, $\ce{[C_T]}$ and the $x$ equivalents of $\ce{NaOH}$. Knowing any two fixes the third.

Another interesting fact is found by differentiating equation (F2) for $\ce{[H^+]}$ which shows that the maximum of $\ce{[HCO3^-]/[C_T]}$ (or any diprotic acid...) occurs when

$$\ce{[H^+] = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}}\tag{F4}$$

It is the pH as which the bicarbonate solution has its maximum buffering capacity. Also at that pH $\ce{[H2CO3] = [CO3^{2-}]}$.

For the bicarbonate system the maximum concentration of $\ce{HCO3^-}$ is 97.99% which occurs at a pH of 8.340. At that pH $\ce{[H2CO3] = [CO3^{2-}] = 1.01\%}$. So for an accuracy of 1 ppt the hydrolysis of water and the multiple species cannot be neglected.

If the pH is above 7.648, then the concentration of the bicarbonate species will be at least 95%.

\begin{array}{|c|c|c|c|c|}\hline
\ce{[H+]} & \mathrm{pH} & \ce{F(H2CO3)} & \ce{F(HCO3^-)}&\ce{F(CO3^{2-})}\\ \hline
1.00E+00 & 0.000E+00 & 1.000E+00 & 4.450E-07 & 2.091E-17 \\ \hline
8.00E-01 & 9.691E-02 & 1.000E+00 & 5.562E-07 & 3.268E-17 \\ \hline
6.00E-01 & 2.218E-01 & 1.000E+00 & 7.417E-07 & 5.810E-17 \\ \hline
4.00E-01 & 3.979E-01 & 1.000E+00 & 1.112E-06 & 1.307E-16 \\ \hline
2.00E-01 & 6.990E-01 & 1.000E+00 & 2.225E-06 & 5.229E-16 \\ \hline
1.00E-01 & 1.000E+00 & 1.000E+00 & 4.450E-06 & 2.091E-15 \\ \hline
8.00E-02 & 1.097E+00 & 1.000E+00 & 5.562E-06 & 3.268E-15 \\ \hline
6.00E-02 & 1.222E+00 & 1.000E+00 & 7.417E-06 & 5.810E-15 \\ \hline
4.00E-02 & 1.398E+00 & 1.000E+00 & 1.112E-05 & 1.307E-14 \\ \hline
2.00E-02 & 1.699E+00 & 1.000E+00 & 2.225E-05 & 5.229E-14 \\ \hline
1.00E-02 & 2.000E+00 & 1.000E+00 & 4.450E-05 & 2.091E-13 \\ \hline
8.00E-03 & 2.097E+00 & 9.999E-01 & 5.562E-05 & 3.268E-13 \\ \hline
6.00E-03 & 2.222E+00 & 9.999E-01 & 7.416E-05 & 5.809E-13 \\ \hline
4.00E-03 & 2.398E+00 & 9.999E-01 & 1.112E-04 & 1.307E-12 \\ \hline
2.00E-03 & 2.699E+00 & 9.998E-01 & 2.225E-04 & 5.228E-12 \\ \hline
1.00E-03 & 3.000E+00 & 9.996E-01 & 4.448E-04 & 2.091E-11 \\ \hline
8.00E-04 & 3.097E+00 & 9.994E-01 & 5.559E-04 & 3.266E-11 \\ \hline
6.00E-04 & 3.222E+00 & 9.993E-01 & 7.411E-04 & 5.805E-11 \\ \hline
4.00E-04 & 3.398E+00 & 9.989E-01 & 1.111E-03 & 1.306E-10 \\ \hline
2.00E-04 & 3.699E+00 & 9.978E-01 & 2.220E-03 & 5.217E-10 \\ \hline
1.00E-04 & 4.000E+00 & 9.956E-01 & 4.430E-03 & 2.082E-09 \\ \hline
8.00E-05 & 4.097E+00 & 9.945E-01 & 5.532E-03 & 3.250E-09 \\ \hline
6.00E-05 & 4.222E+00 & 9.926E-01 & 7.362E-03 & 5.767E-09 \\ \hline
4.00E-05 & 4.398E+00 & 9.890E-01 & 1.100E-02 & 1.293E-08 \\ \hline
2.00E-05 & 4.699E+00 & 9.782E-01 & 2.177E-02 & 5.115E-08 \\ \hline
1.00E-05 & 5.000E+00 & 9.574E-01 & 4.260E-02 & 2.002E-07 \\ \hline
8.00E-06 & 5.097E+00 & 9.473E-01 & 5.269E-02 & 3.096E-07 \\ \hline
6.00E-06 & 5.222E+00 & 9.310E-01 & 6.905E-02 & 5.409E-07 \\ \hline
4.00E-06 & 5.398E+00 & 8.999E-01 & 1.001E-01 & 1.176E-06 \\ \hline
2.00E-06 & 5.699E+00 & 8.180E-01 & 1.820E-01 & 4.277E-06 \\ \hline
1.00E-06 & 6.000E+00 & 6.920E-01 & 3.080E-01 & 1.447E-05 \\ \hline
8.00E-07 & 6.097E+00 & 6.426E-01 & 3.574E-01 & 2.100E-05 \\ \hline
6.00E-07 & 6.222E+00 & 5.741E-01 & 4.258E-01 & 3.336E-05 \\ \hline
4.00E-07 & 6.398E+00 & 4.733E-01 & 5.266E-01 & 6.187E-05 \\ \hline
2.00E-07 & 6.699E+00 & 3.100E-01 & 6.898E-01 & 1.621E-04 \\ \hline
1.00E-07 & 7.000E+00 & 1.834E-01 & 8.162E-01 & 3.836E-04 \\ \hline
8.00E-08 & 7.097E+00 & 1.523E-01 & 8.472E-01 & 4.977E-04 \\ \hline
6.00E-08 & 7.222E+00 & 1.187E-01 & 8.806E-01 & 6.898E-04 \\ \hline
4.00E-08 & 7.398E+00 & 8.239E-02 & 9.165E-01 & 1.077E-03 \\ \hline
2.25E-08 & 7.648E+00 & 4.803E-02 & 9.500E-01 & 1.984E-03 \\ \hline
2.00E-08 & 7.699E+00 & 4.291E-02 & 9.548E-01 & 2.244E-03 \\ \hline
1.00E-08 & 8.000E+00 & 2.188E-02 & 9.735E-01 & 4.576E-03 \\ \hline
8.00E-09 & 8.097E+00 & 1.756E-02 & 9.767E-01 & 5.738E-03 \\ \hline
6.00E-09 & 8.222E+00 & 1.320E-02 & 9.791E-01 & 7.670E-03 \\ \hline
4.57E-09 & 8.340E+00 & 1.006E-02 & 9.799E-01 & 1.008E-02 \\ \hline
4.00E-09 & 8.398E+00 & 8.806E-03 & 9.797E-01 & 1.151E-02 \\ \hline
2.00E-09 & 8.699E+00 & 4.372E-03 & 9.728E-01 & 2.286E-02 \\ \hline
1.00E-09 & 9.000E+00 & 2.142E-03 & 9.531E-01 & 4.479E-02 \\ \hline
8.00E-10 & 9.097E+00 & 1.695E-03 & 9.429E-01 & 5.540E-02 \\ \hline
6.00E-10 & 9.222E+00 & 1.249E-03 & 9.262E-01 & 7.255E-02 \\ \hline
4.00E-10 & 9.398E+00 & 8.037E-04 & 8.941E-01 & 1.051E-01 \\ \hline
2.00E-10 & 9.699E+00 & 3.638E-04 & 8.094E-01 & 1.902E-01 \\ \hline
1.00E-10 & 1.000E+01 & 1.528E-04 & 6.802E-01 & 3.197E-01 \\ \hline
8.00E-11 & 1.010E+01 & 1.132E-04 & 6.298E-01 & 3.700E-01 \\ \hline
6.00E-11 & 1.022E+01 & 7.560E-05 & 5.607E-01 & 4.392E-01 \\ \hline
4.00E-11 & 1.040E+01 & 4.133E-05 & 4.598E-01 & 5.402E-01 \\ \hline
2.00E-11 & 1.070E+01 & 1.342E-05 & 2.985E-01 & 7.015E-01 \\ \hline
1.00E-11 & 1.100E+01 & 3.942E-06 & 1.754E-01 & 8.246E-01 \\ \hline
\end{array}

The fourth equation

The fourth equation comes from the charge balance in the solution. The solution must be electrically neutral so the total charge on cations and anions must be equal.

$$\ce{[Na^+] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C1}$$

But in a solution of $\ce{[NaHCO3]}$ the $\ce{[Na^+] = [C_T]}$ so:

$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$

Which yields and exact solution as a quartic equation. But rather than solve only that particular quartic equation, let's consider the more general case. Assume we start with $\ce{H2CO3}$ in water and add amounts of $\ce{NaOH} according to the equation:

$$\ce{H2CO3 + xNaOH -> Na_xH_{2-x}CO3 + xH2O}\tag{C3}$$

where $x$ is a real number bounded by $ 0 \le x \le 2$. The special cases of $x = 0,1,2$ correspond to adding $\ce{H2CO3, NaHCO3}$ or $\ce{Na2CO3}$ to pure water. Then it is intuitively obvious to the casual observer that (a la analytical chemist Dr. Peter Carr, University of Ga.) the general equation is:

$$\begin{multline}
  [\ce{H+}]^4
  + (K_\mathrm{a1} + xC_T)[\ce{H+}]^3
  + (K_\mathrm{a1} K_\mathrm{a2} + (x - 1) C_T K_\mathrm{a1}
    - K_\mathrm{w})[\ce{H+}]^2\\
  + K_\mathrm{a1} ((x - 2)C_T K_\mathrm{a2} - K_\mathrm{w})[\ce{H+}]
  - K_\mathrm{a1} K_\mathrm{a2} K_\mathrm{w}
  \tag{C4}
\end{multline}$$

It is not hard for a computer to solve the quartic equation numerically, but to do so by hand is extremely tedious. Hence books never(?) required such a solution in problem sets.

DERIVATION OF EQUATION 1 $\ce{[H^+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}$

$$K_\mathrm{a1}K_\mathrm{a2} = \dfrac{\ce{[H+][HCO3^-]}}{\ce{[H2CO3]}}\dfrac{\ce{[H+][CO3^{2-}]}}{\ce{[HCO3^-]}} = \dfrac{\ce{[H+]^2[CO3^{2-}]}}{\ce{[H2CO3]}}\tag{EQ1-1}$$

rearrange (EQ1-1)

$$\ce{[H+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[H2CO3]}}{\ce{[CO3^{2-}]}}}\tag{EQ1-2}$$

assume $\ce{[H2CO3] = [CO3^{2-}]}$

Then

$$\ce{[H+]} = \sqrt{K_\mathrm{a1}K_\mathrm{a2}}\tag{EQ1-3}$$

DERIVATION OF EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}$

Sodium is a spectator ion. Thus it is be possible to model the system as a monoprotic Brønsted–Lowry acid which has $\ce{H^+}$ as the acid and $\ce{HCO3^-}$ as the conjugate base.

This derivation has two problems. First it ignores the autoionization of water. Second the equation models the system as a monoprotic base, but the carbonate system has three species - carbonic acid, bicarbonate, and carbonate, and all three are likely to be present in significant quantities in alkaline solutions.

Start with the reaction of the conjugate base

$$\ce{HCO3^- + H2O <=> H2CO3 + OH^-}\tag{EQ2-1}$$

and the equilibrium expression

$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[H2CO3][OH^-]}}{\ce{[HCO3^-]}}\tag{EQ2-2}$$

Assume that little of $\ce{HCO3^-}$ accepts a proton, thus $\ce{[HCO3^-] = [C_T]}$ and $\ce{[H2CO3] = [OH^-]}$. Now substituting:

$$\ce{K_\mathrm{b1}} = \dfrac{\ce{[OH^-]^2}}{\ce{[C_T]}}\tag{EQ2-3}$$

but $\ce{K_\mathrm{b1}} = K_w / K_{a1}$ and $\ce{[OH^-] = K_w / [H^+]}$. Now making more substitutions

$$\dfrac{K_\mathrm{w}}{K_\mathrm{a1}} = \dfrac{(\frac{K_\mathrm{w}}{\ce{[H^+]}})^2}{[C_T]}\tag{EQ2-4}$$

rearranging and solving for $\ce{[H^+]}$ gives the desired result.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T]}}}\tag{EQ2-5}$$

The desired result could have been obtained more expeditiously by starting with equation (2) and letting $\ce{[H2CO3] = [OH^-] = K_w/[H^+]}$, but that route doesn't seem as clear chemically and seems more like math magic.

DERIVATION OF EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$

Starting with equation (C2) for the charge balance of a solution of $\ce{NaHCO3}$ where $\ce{[C_T] = [Na^+]}$

$$\ce{[C_T] + [H^+] = [OH^-] + [HCO3^-] + 2[CO3^{2-}]}\tag{C2}$$

subtract equation (S4) which is for mass balance of all the carbonate species

$$\mathrm{[C_T]} = \ce{[H2CO3] + [HCO3^-] + [CO3^{2-}]}\tag{S4}$$

yielding

$$\ce{[H^+] = [CO3^{2-}] - [H2CO3] + [OH^-]}\tag{EQ3-1}$$

rearranging

$$\ce{[OH^-] = [H2CO3] - [CO3^{2-}] + [H^+]}\tag{EQ3-2}$$

substituting $K_\mathrm{w} / \ce{[H^+] = [OH^-]}$ and rearrange

$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} \ce{+ [CO3^{2-}] = [H2CO3] + [H^+]}\tag{EQ3-3}$$

Solve equation (S1) for $\ce{[H2CO3]}$ and solve equation (S2) for $\ce{[CO3^{2-}]}$, and then substitute those results into equation (EQ3-3)

$$\dfrac{K_\mathrm{w}}{\ce{[H^+]}} + \dfrac{K_\mathrm{a2}\ce{[HCO3^-]}}{\ce{[H^+]}} = \dfrac{\ce{[H^+][HCO3^-]}}{K_\mathrm{a1}} + \ce{[H^+]}
\tag{EQ3-4}$$

Multiplying both sides by $\ce{[H^+]}$, then $K_\mathrm{a1}$, and finally collecting terms

$$K_\mathrm{a1}K_\mathrm{w} + K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} = \ce{[H^+]^2([HCO3^-] + K_\mathrm{a1})}
\tag{EQ3-5}$$

rearranging and solving for $\ce{[H^+]}$ yields the exact equation for $\ce{[H^+]}$ in terms of the particular species $\ce{[HCO3^-]}$ when $\ce{NaHCO3}$ is dissolved in pure water.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[HCO3^-]} + \mathrm{K_{a1}K_w}}{\ce{[HCO3^-] + \mathrm{K_{a1}}}}}\tag{EQ3-6}$$

As indicated previously the above equation is exact in terms of the relationship between $\ce{[H^+]}$ and the carbonate species $\ce{HCO3^-}$. Over the whole pH range $\ce{[C_T] \ne [HCO3^-]}$, however for the range $ 7.648 \lt \mathrm{pH} \lt 9.00$ the species $\ce{HCO3^-}$ is at least 95% of $\ce{[C_T]}$. So we can "reasonably" substitute $\ce{[C_T] = [NaHCO3]}$ and get the desired equation even though the equation is not exact.

$$\ce{[H^+]} = \sqrt{\dfrac{K_\mathrm{a1}K_\mathrm{a2}\ce{[C_T]} + K_\mathrm{a1}K_\mathrm{w}}{\ce{[C_T] + K_\mathrm{a1}}}}\tag{EQ3-7}$$

!! CHECKS !!

Why aren't students taught to do checks any more!?!

The column $\ce{[C_T]}$ is the nominal molarity of the sodium bicarbonate solution. It is also the molarity of the spectator cation $\ce{[Na^+]}$.

The column $\Sigma \space \text{cations}$ should equal the column $\Sigma \space \text{anions}$.

The column % Diff is a measure of the inconsistency of the calculations of the concentration of anions and cations and is calculated by

$$\text{% Diff} = 200*\dfrac{\Sigma\text{ cations} - \Sigma\text{ anions}}{\Sigma\text{ cations} + \Sigma\text{ anions}}$$

Again, these calculations use concentrations instead of the more appropriate activities. Concentrations are probably fine for these solutions below 1 millimolar. Above that and activities should really be used.

EQUATION 1 $\ce{[H^+]} = \sqrt{K_{a1}K_{a2}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline
1.00E-01 & 4.5733E-09 &  8.340 &  1.0000E-01&  2.1866E-06 &9.7986E-02 &2.0140E-03 &1.0000E-01 & 0.0 \\ \hline
1.00E-02 &4.5733E-09 &  8.340 &1.0000E-02 &2.1866E-06 &9.7986E-03 &2.0140E-04 &1.0002E-02 & 0.0 \\ \hline
1.00E-03 &4.5733E-09 &  8.340 &1.0000E-03 &2.1866E-06 &9.7986E-04 &2.0140E-05 &1.0022E-03 & -0.2 \\ \hline
1.00E-04 &4.5733E-09 &  8.340 &1.0000E-04 &2.1866E-06 &9.7986E-05 &2.0140E-06 &1.0219E-04 & -2.2\\ \hline
1.00E-05 &4.5733E-09 &  8.340 &1.0005E-05 &2.1866E-06 &9.7986E-06 &2.0140E-07 &1.2187E-05 & -19.7\\ \hline
1.00E-06 &4.5733E-09 &  8.340 &1.0046E-06 &2.1866E-06 &9.7986E-07 &2.0140E-08 &3.1866E-06 & -104 \\ \hline
1.00E-07 &4.5733E-09 &  8.340 &1.0457E-07 &2.1866E-06 &9.7986E-08 &2.0140E-09 &2.2866E-06 & -183 \\ \hline
\end{array}

So assuming that the pH is constant only works for concentrated solutions (greater than  1 millimolar) for which activities should really be used...

EQUATION 2 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_w}{\ce{[C_T]}}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H^+] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff\\ \hline
1.00E-01 &2.1095E-10 & 9.676 &1.0000E-01 &4.7405E-05 &8.1748E-02 &3.6427E-02 &1.1822E-01 & -16.7 \\ \hline
1.00E-02 &6.6708E-10 & 9.176 &1.0000E-02 &1.4991E-05 &9.3287E-03 &1.3145E-03 &1.0658E-02 & -6.4\\ \hline
1.00E-03 &2.1095E-09 & 8.676 &1.0000E-03 &4.7405E-06 &9.7369E-04 &4.3388E-05 &1.0218E-03 & -2.2\\ \hline
1.00E-04 &6.6708E-09 & 8.176 &1.0001E-04 &1.4991E-06 &9.7844E-05 &1.3787E-06 &1.0072E-04 & -0.7 \\ \hline
1.00E-05 &2.1095E-08 & 7.676 &1.0021E-05 &4.7405E-07 &9.5271E-06 &4.2453E-08 &1.0044E-05 & -0.2 \\ \hline
1.00E-06 &6.6708E-08 & 7.176 &1.0667E-06 &1.4991E-07 &8.6910E-07 &1.2247E-09 &1.0202E-06 & +4.5\\ \hline
1.00E-07 &2.1095E-07 & 6.676 &3.1095E-07 &4.7405E-08 &6.7830E-08 &3.0225E-11 &1.1527E-07 & +91.8 \\ \hline
\end{array}

If 1 ppt is required, then Equation 2 doesn't work. For a +/- 5% error, the equation works in the range from $1\times10^{-2}$ to $1\times10^{-6}$ molar.

EQUATION 3 $\ce{[H^+]} = \sqrt{\dfrac{K_{a1}K_{a2}C_T + K_{a1}K_w}{C_T+K_{a1}}}$

\begin{array}{|c|c|c|c|c|c|c|}\hline
\ce{[C_T]} & calc \space [H] & pH &\Sigma \space cations &\ce{[OH^-]} &\ce{[HCO3^-]} &2*\ce{[CO3^{2-}]} &\Sigma \space anions & \% \space Diff \\ \hline
1.00E-01 &4.5781E-09 & 8.339 &1.0000E-01 &2.1843E-06 &9.7986E-02 &2.0119E-03 &1.0000E-01 & 0.0 \\ \hline
1.00E-02 &4.6216E-09 & 8.335 &1.0000E-02 &2.1638E-06 &9.7986E-03 &1.9930E-04 &1.0000E-02 & 0.0 \\ \hline
1.00E-03 &5.0352E-09 & 8.298 &1.0000E-03 &1.9860E-06 &9.7977E-04 &1.8291E-05 &1.0000E-03 & 0.0 \\ \hline
1.00E-04 &8.0700E-09 & 8.093 &1.0001E-04 &1.2392E-06 &9.7660E-05 &1.1376E-06 &1.0004E-04 & 0.0 \\ \hline
1.00E-05 &2.1120E-08 & 7.675 &1.0021E-05 &4.7348E-07 &9.5267E-06 &4.2400E-08 &1.0043E-05 & -0.2\\ \hline
1.00E-06 &5.5624E-08 & 7.255 &1.0556E-06 &1.7978E-07 &8.8822E-07 &1.5010E-09 &1.0695E-06 & -1.3\\ \hline
1.00E-07 &9.0382E-08 & 7.044 &1.9038E-07 &1.1064E-07 &8.3082E-08 &8.6408E-11 &1.9381E-07 & -1.8 \\ \hline
1.00E-08 &9.8897E-08 & 7.005 &1.0890E-07 &1.0111E-07 &8.1785E-09 &7.7735E-12 &1.0930E-07 & -0.4\\ \hline
\end{array}

If 1 ppt is required, then Equation 3 works for concentrations grater than  $1\times10^{-5}$ molar. For a +/- 5% error, the equation works in the range from $0.1$ to $1\times10^{-8}$ molar.

Summary of $\ce{[H^+]}$ values

the table below is a summary of the results for the three derived equations, and the exact solution using equation (C4) by Wolfram Alpha.

\begin{array}{|c|c|c|c|c|}\hline
\ce{[C_T]} & Eq(1)& Eq(2) & Eq(3) & exact\\ \hline
1.00E-01  & 4.5733E-09 & 2.1095E-10 & 4.5781E-09 & 4.5782E-9 \\ \hline
1.00E-02  & 4.5733E-09 & 6.6708E-10 & 4.6216E-09 & 4.6226E-9 \\ \hline
1.00E-03  & 4.5733E-09 & 2.1095E-09 & 5.0352E-09 & 5.0443E-9 \\ \hline
1.00E-04  & 4.5733E-09 & 6.6708E-09 & 8.0700E-09 & 8.1354E-9 \\ \hline
1.00E-05  & 4.5733E-09 & 2.1095E-08 & 2.1120E-08 & 2.1602E-8 \\ \hline
1.00E-06  & 4.5733E-09 & 6.6708E-08  & 5.5624E-08 & 5.7983E-8 \\ \hline
1.00E-07  & 4.5733E-09 & 2.1095E-07  & 9.0382E-08 & 9.1835E-8 \\ \hline
1.00E-08  & -- & -- & 9.8897E-08 & 9.9096E-8 \\ \hline
\end{array}


Caffeine boiling and melting points According to both caffeine's PubChem page and ChemSpider page its boiling point lies at $\pu{173 °C}$ and its melting point at $\pu{\sim 235 °C}.$ How can it melt at that temperature if it already sublimated at $\pu{173 °C}?$ Or is that the temperature of the triple point? If so, how come none of those pages gives any pressure specifications for that temperature?
I heated up some anhydrous caffeine at home, and at my home atmospheric pressure it appears to melt (not sublimate) and starts creating vapors immediately after (but not before) melting. This is based just on what I can see with the naked eye.

Anhydrous caffeine occurs in two different polymorphic crystal forms, with a transition between the two forms occurring at $\pu{136 °C}.$ The high temperature form has a melting point of $\pu{236 °C}.$
The vapor pressure of the high temperature solid form is described by the equation:
$$\log P[\pu{Pa}] = -\frac{5477}{T[\pu{K}]} + 14.395,\tag{1}$$
and for the liquid
$$\log P[\pu{Pa}] = -\frac{3918}{T[\pu{K}]} + 11.14.\tag{2}$$
So, $\pu{173 °C}$ is incorrect; the boiling point is more like $\pu{365 °C}.$
References

Griesser, U. J.; Szelagiewicz, M.; Hofmeier, U. Ch.; Pitt, C.; Cianferani, S. Vapor Pressure and Heat of Sublimation of Crystal Polymorphs. Journal of Thermal Analysis and Calorimetry 1999, 57 (1), 45–60. DOI: 10.1023/A:1010188923713.
Bothe, H.; Cammenga, H. K. Phase Transitions and Thermodynamic Properties of Anhydrous Caffeine. Journal of Thermal Analysis 1979, 16 (2), 267–275. DOI: 10.1007/BF01910688.


Can there be any substance whose number of electrons and protons are different at neutral state I want to ask that if there any substance that has the different number of electrons and protons at its neutral state

It's certainly possible, you just need other charged particles that aren't electrons or protons.

For example, a positron is positively charged, so something that's made out of an electron plus a positron would work. See e.g. https://en.wikipedia.org/wiki/Onium, or this answer on SE, for a few examples.


Explaining why s and p orbitals do not interact While going through molecular orbital theory, my chemistry teacher mentioned that the reason why there is no significant interaction between $\ce {s}$ orbitals on one atom and $\ce {p_x}$ and $\ce {p_y}$ orbitals on another atom is that the $\ce {s}$ orbitals interact with both lobes of the $\ce {p}$ orbitals, having both in phase and anti phase interactions with these $\ce {p}$ orbitals, taking the z axis to be the internuclear axis.

However, I read that orbitals which do not have the right symmetry simply cannot interact, i.e. their overlap integral is necessarily $\ce {0}$. Thus, is the reasoning given by my chemistry teacher actually valid?

According to my chem textbook:
"Molecular orbitals obtained from 2px and 2py orbitals are not symmetrical around the bond axis because of the positive lobes above and below the molecular plane."
Your instructor might be pointing out to the bond-symmetry thing in MO formation between pz and s orbitals extrapolated in most chemistry textbooks.
But orbital overlap is a result of constructive and destructive interference. As s orbitals are spherical, they have infinite possibilities of orbital overlap with other orbital shapes and thus even overlap with d orbitals (as explained by VSEPR: Valence Shell Electron Pair Repulsion Theory in a similar manner).
Thus s and px and s and py orbitals do overlap; if you take px to be of lower energy than py and pz, s-px overlapping occurs in boranes (formed by boron, which has one electron in 2px orbital, and hydrogen, which has one electron in 1s orbital), that often dimerize. This is also the case with methane, where px, py-s overlapping occurs, assuming that in sp3-hybridized state px, py orbitals in C bond with H 1s orbitals.


Is honey really a supersaturated solution? Does heating to un-crystalize redissolve it or melt it? In the SciShow video Honey: Bacteria's Worst Enemy after about 00:30 the narrator says:

Honey is only about 17% water. Most, but not all of what remains is sugar. The two main types of sugar in honey are glucose and fructose. Like all sugars, glucose and fructose are sticky — they attract water.
Honey is technically a supersaturated solution, meaning it contains more sugar than would normally dissolve at that temperature. That’s why it eventually gets all crystally in the pantry — over time, sugar comes out of the solution.

In my experience when old honey "gets all crystally in the pantry" I've placed the jar in a hot water bath, and eventually the honey's viscous liquidy consistency is restored.
When I do this, am I redissolving the sugar, or melting it?
I'm asking because even hot, with only 17% water it's hard to imagine it can become an unsaturated solution.

Honey is indeed a complex mixture containing more than hundred compounds.

As for Wikipedia and depending on the point of view it is

a supersatured liquid solution
a viscous supercooled liquid (in the sense that it can get so viscous as to appear solid, without affecting its status of being a supersatured solution, and undergoes glass transition).

https://en.wikipedia.org/wiki/Honey#Physical_and_chemical_properties

The facts that you describe in the question are less surprising if we consider that:

with respect to sugars crystallization, of which at least  there are two different ones, glucose and fructose, the rest of the non-sugar components must be considered to be impurities
impurities, even in traces, often hamper the crystallisation of a compound, even in simple mixture of a single compound and the above traces. This is common after organic synthesis, in which often a viscous "oil" is attained that might crystallize only upon prolonged storage or a careful removal of the disturbing trace compound(s)
finally and most important, the solubility of sugars in water is very high, and very sensitive to temperature. For instance, at room temperature glucose is already soluble in the reason of 90 g per 100 ml of water, that means a saturated solution already contains about fifty percent weight per weight of sugar

A table is here (I didn't cross check the values):

http://www.mpcfaculty.net/mark_bishop/supersaturated.htm

All this, viscosity included, makes the attainment of a supersatured solution particularly easy, as in the kitchen in the case of sucrose:

https://sciencing.com/make-supersaturated-solution-sugar-6199355.html

As such, heating crystallized honey does indeed dissolve the sugar, and a supersaturated solution is attained upon subsequent cooling.


Why do vinegar and other acids remove rust? I have started a project on restoring a very rusty knife. I had used vinegar to remove the rust. A question dawned on me: why does vinegar remove rust? So being the thoughtful person I am, I googled it. Unfortunately, Dr. Google could only provide me with what it is doing (which I already know).

But what I really want to know is: what chemical reaction causes the rust to disappear?

The acid can dissolve rust (e.g. $\ce{FeOOH}$). The chemical reaction is:

$\ce{3CH_3COOH + FeOOH -> Fe(CH_3COO)_3 + 2H_2O}$

EDIT: ($\ce{Fe(OAc)_3}$is NOT soluble in water) The acid forms a water-insoluble salt with the iron oxide, which then probably just crumbles from the rust layer and precipitates in the container.

Mind that rust also contains for example $\ce{FeO(OH)_3}$ and $\ce{Fe_2O_3·nH_2O}$. The mechanism is the same.

Also, the acid oxidizes the upper layer of the iron under the rust to form a rust-resisting oxide-layer.


Why is 2-fluorobenzoic acid more acidic than 4-fluorobenzoic acid despite its intramolecular hydrogen bonding? I was reading about $\mathrm{p}K_\mathrm{a}$ values of 2-fluorobenzoic acid and 4-fluorobenzoic acid and found that the latter had a higher value which was surprising to me. I think that intramolecular hydrogen bonding should dominate all the other effects in 2-fluorobenzoic acid as mentioned in the following figure and thus it should have a higher value:

What is wrong with my logic?

You seem to be asking the question "Why does this intramolecular hydrogen bonding not cause a considerable effect to the $\mathrm pK_\mathrm a$ values of these compounds?"

TL;DR That is because this type of bonding is only present in a specific conformer only and this type of conformer of the compound 2-fluorobenzoic acid exists in low population levels, 2%$^{[1]}$ to not affect the ${\mathrm pK_\mathrm a}$ of the compound considerably.

Regarding the conformers of the 2-FluoroBenzoic Acid

A throughout conformational search on the B3LYP/6-311++G(d,p) potential energy surfaces of the
studied molecules (the mono- and di- ortho fluoro- or/and chloro- substituted benzoic acids, plus the
parent benzoic acid, for completeness) was undertaken. The identified conformers are presented$^{[2]}$

Here both the cis conformers are planar and low energy while only one trans conformer is planar while both of them are relatively high energy of which Trans II is of relatively lower energy.

The two cis conformers have rather similar energies, with cis-II (where
the fluoro-substituent stays the same side of the molecule as the acid oxygen atom and a C-H . . . O=
stabilizing interaction exists), being the most stable form. The cis-I conformer has a relative energy of
2.82 kJ·mol−1
. In this conformer, the C-H . . . O= interaction present in cis-II is replaced by a weaker
stabilizing C-H . . . OH interaction. The C-H . . . O= interaction is more efficient in stabilizing cis-II than
the C-H . . . OH interaction in stabilizing cis-I due to the more favorable localization of the interacting
carbonyl oxygen lone electron pair, which stays in the molecular plane, whereas in cis-I both lone
electron pairs of the acid oxygen atom are out of the molecular plane. On the other hand, the interactions
between the fluoro-substituent and the oxygen atoms, which are of repulsive nature, also favor a lower
energy for cis-II compared to cis-I. This can be rationalized in similar terms as for the interactions
involving the ortho hydrogen atom discussed above. In cis-I, the F . . . O= repulsion is stronger than the
F
. . . OH repulsion in cis-II because the interacting lone electron pair of the carbonyl oxygen atom is in
the same plane as the fluorine atom in the former conformer, while the lone electron pairs of the acid
oxygen atom in cis-II are both out of the plane in relation to the fluoro-substituent.$^{[2]}$

It is important to note that though the population level of Trans II are low, it is still a stable compound and nevertheless has been detected.$^{[3]}$

So.. Why is 2-fluorobenzoic acid more acidic than 4-fluorobenzoic acid?

I could not find any research for this but I could make an educated guess that the acidity of 2-fluorobenzoic acid comes from the similar stabilization as that in salicylic acid.$^{[4]}$ in the conformer cis II as shown below:-

References:

Kuş, N.; Fausto, R. Effects of the matrix and intramolecular interactions on the stability of the higher-energy conformers of 2-fluorobenzoic acid. J. Chem. Phys. 2017, 146 (12), 124305. [DOI: 10.1063/1.4978794]
(https://doi.org/10.1063/1.4978794).
Ogruc Ildiz, G.; Fausto, R. Structural Aspects of the Ortho Chloro- and Fluoro- Substituted Benzoic Acids: Implications on Chemical Properties. Molecules 2020, 25 (21), 4908. DOI: 10.3390/molecules25214908.
Daly, A. M.; Carey, S. J.; Pejlovas, A. M.; Li, K.; Kang, L.; Kukolich, S. G. Gas phase measurements of mono-fluoro-benzoic acids and the dimer of 3-fluoro-benzoic acid. J. Chem. Phys. 2015, 142 (14), 144303. DOI: 10.1063/1.4917031.
Dunn, G. E.; Penner, T. L. Effect of intramolecular hydrogen bonding on the relative acidities of substituted salicylic acids in benzene solution. Can. J. Chem. 1967, 45 (14), 1699–1706. DOI: 10.1139/v67-274.


How do you name an alkene using IUPAC rules if the longest carbon chain in this alkene does not include the double bond? For example, which is the right way to name this compound:

Is it 2-bromo-4-methylenehexane or 4-bromo-2-ethylpent-1-ene?

Does the larger number of carbon in a chain take priority, as in the former, or does the carbon-carbon double bond take priority, as in the latter?

The IUPAC recommendations on this matter have been changed. In the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed:

  (…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

For example, a preferred IUPAC name is 3-methylidenehexane (not 2-ethylpent-1-ene) since the longer chain now supersedes a shorter unsaturated chain. (This example is given in the Blue Book.)

Besides, the prefix ‘methylene’ is no longer recommended in IUPAC nomenclature to designate the $\ce{=CH2}$ group. The preferred prefix is ‘methylidene’.

Therefore, the correct name for your example is 2-bromo-4-methylidenehexane.


Is sublimation caused by an environmental difference or an element's composition? I was having a conversation when the topic of sublimation came up. Prior to the discussion, I had always thought that sublimation was caused by a great pressure/temperature difference, but another person stated that whether or not an element sublimates depends on its composition.

Will any element sublimate under the right conditions? Or are there elements which will always pass through a liquid state?

To a first approximation, any pure substance can sublimate. However, complex substances may decompose or change their structure before any appropriate sublimation conditions are reached.

The only (and very remarkable) exception is helium (in fact, helium is so special that helium-3 and helium-4 have quite different phase diagrams even though they're isotopes!). Note that due to the presence of superfluid phases, there is no direct transition between the solid and gas phases in either isotope.

As an aside, there is nothing prohibiting a substance to exhibit a solid-gas transition curve crossing a region of mild temperature and pressure conditions. Though unusual, an interesting example is iodine heptafluoride. It has a very narrow liquid range at ambient pressures (meaning a small decrease in pressure is enough to go below the triple point pressure), and the liquid range happens to fall near very reasonable temperatures.


Why is substitution of halogens to alkanes a chain reaction? During the propagation step 2, the alkyl radical always react with the halogen molecule according to my textbook. However, isn't the alkyl radical react with the halogen radical formed in the initiation step much better since it lowers the potential energy?

If there is a chance that the 2 radicals combine, won't the chain reaction terminate rapidly as less and less halogen radical is formed?

You must realize that the number of atoms created in the first step is smaller than one percent of the whole number of Halogen molecules. So the first reaction produces a tiny amount of HCl and a tiny amount of an alkyl radical. The probability that such a radical touches the second halogen radical is extremely small. It will collide much more frequently another halogen molecule, and produce a new halogenated compound plus a new halogen radical, etc.


Why are enynes named as -en-yne and not -yn-ene? The preferred IUPAC name for the compound above is hept-1-en-6-yne. I understand that the numbers are chosen in order to give the alkene the lower locant, as explained in Should double or triple bonds have preference for low locants in IUPAC nomenclature?.

However, I'm wondering why the name has the form -en-yne instead of -yn-ene. In other words, why is it not "hept-6-yn-1-ene" instead?

Furthermore, if this is an IUPAC specification, is there any rationale for why en-yne should be preferred to yn-ene?

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.
  (…)

Since, in names, the ending ‘ene’ always precedes ‘yne’ (with elision of the final letter ‘e’ in ‘ene’), the given compound is a ‘hept-x-en-y-yne’.

Since the lowest sets of locants in ‘hept-1-en-6-yne’ and ‘hept-6-en-1-yne’ are equivalent (1, 6), preference for low locants is given to the double bond. Therefore, locant 1 is assigned to the double bond and locant 6 is assigned to the triple bond.

Thus, the correct name is hept-1-en-6-yne.


Nonsubstituted Cycloalkene HBr Reaction If I have a cyclohexene with no substituents and add HBr, how do I know what side of the former double bond to place the Br anion? I think I understand Markovnikov's rule, but in my chemistry book they always show one carbon specifically that the Br attaches to and I seem to always pick the other. Does it matter or are they equally likely? Is there some way to decide which carbon is more substituted if there are no other groups attached to it and it's just a carbon ring?

Hopefully I've worded this in a way that makes sense, thanks for any help!

Cyclohexene is very symmetric. At first approximation, you can draw two planes of symmetry: one containing the double bond and one orthogonal, bisecting the double bond.

These two planes of symmetry mean, that both sides of the double bond are equal as are both carbon atoms. Therefore, the hydrogen atom protonating the double bond in $\ce{HBr}$ addition cannot sense any difference between either of the four possible attack pathways and chooses one at random. The bromide then has to choose the other carbon, but can again choose the plane from which it attacks (‘above’ or ‘below’ — in inverted commas because they are equivalent!) at random.

Even after the reaction, the product bromocyclohexane is highly symmetric. You can again draw a plane of symmetry through it; this time, the plane must include the $\ce{C-Br}$ bond and also bisect the carbon atom on the other side of the ring. The presence of a plane of symmetry means that the molecule cannot exist in enantiomeric forms, and since you cannot say on which side the hydrogen was previously added (they, too, are all identical), you only get one possible product.

We might consider a more complicated case and decide to add $\ce{DBr}$ instead of $\ce{HBr}$. In this case, we would indeed observe more than one type of product since a deuterium atom cannot be mapped onto a hydrogen atom by symmetry. However, we would obtain all symmetry-equivalent products at equal ratios because of the high symmetry of the starting compound.

In general, if you can find planes of symmetry then both sides are equally likely to react and you cannot predict a preference.


Silver electrorefining Ok, so let's say I have raw silver which is polluted with a bit of gold, copper and lead.
So through electrorefining I want to make pure silver out of it.

Let's say I have it all in an acidic solution. What happens when I turn on the Voltage?
How would I prevent the other elements from getting electrolyzed?
Because when I have the electrorefining of copper it gets clean even when polluted with silver, so why doesn't the copper get electrolyzed as well?

It depeds on the acid. If it is HCl then you will get Wohlwill Process only the silver will be not disolved. So baisiclly what you want.


Determination of major product in crossed aldol condensation The title speaks for itself really.

I know that if we have a conventional ketone (with 3+ α-hydrogens) and an aldehyde with less than or equal to 3 α-hydrogens, the ketone acts as the nucleophile and attacks the aldehyde, which consequently forms the major product.

Let us take an example of a case where we have acetophenone and acetaldehyde. Both have 3 α-hydrogens. Also, the acetophenone is sterically hindered due to the phenyl group. Which is the major product here?

There are two main steps in any aldol condensation:-
1. Formation of enolate ion
2. Using the enolate ion as a nucleophile.

The 1st step is a reversible step. If you think that the kinetically favoured enolate ion will be formed fast then you are wrong as it also disappears faster. So, in any case the one which is more thermally stable will form in more amount except in the cases where the formation of stable enolate is too slow or the backward reaction of kinetic favoured enolate is blocked or not enough time has been given to let the reaction reach equilibrium.

Now, I will tell you about above strategies to block reverse reaction or make formation of stable enolate too slow:-

1. Use a strong electron withdrawing group at alpha position to the newly formed CH2(-)COR so that the electron can spend most of time away from the original carbon.
-This blocks reverse reaction

2. Use a bulky base or somehow hinder the alpha carbon to make the extraction of proton hard which will in turn make the reaction slow.

3. Using Non-Protic Solvent stops the protonation of the enolate ion which in turn blocks the reverse reaction.

Now, the answer for your question, the conjugate base of acetaldehyde is more stable when compared to the former but the attack of conjugate base of acetophenone is more easier as compared to latter because of steric reasons and the more electrophilic character of aldehyde and when you make cannonical structures, the number of structures which has negative charge on oxygen is greater than being away which makes it a good nucleophile when you take weighted average of all. So, the major product should be 1-phenyl-but2en1one.


Calculating Gibbs Free Energy This is a question from Atkin's Physical Chemistry Book.

Question

Suppose that 3mmol $\mathrm{N_2\,(g)}$ occupies $\mathrm{36~cm^3}$ at $\mathrm{300~K}$ and expands to $\mathrm{60~cm^3}$. Calculate $\Delta G$ for the process.

My Attempt

I decided to split this process up into two separate process (I am not sure if this allowed. Please correct me if this wrong). The first process is an reversible isothermal expansion of the gas to $\mathrm{60~cm^3}$. Then the second step is an adiabatic cooling of the gas. I was planning to calculate $\Delta G$ for each step then add them up to get the answer. For the first step, $\Delta U = 0$, therefore:
$$q = -w = n~R~T~\ln\frac{60}{36}$$
$$\Delta S = \frac{q}{T} = n~R~\ln\frac{60}{36}$$

Since pressure is constant, $$\Delta H = q = n~R~T~\ln\frac{60}{36}$$
Therefore: $$\Delta G = \Delta H - T\Delta S = 0$$
I stopped her because I am sure that this is incorrect as Gibbs Free Energy shouldn't be equal to $0$.

Can someone please clarify if the above procedure is correct and perhaps a better way to attempt this question?

I'm quite certain about the following, because it's from the related (german) woorkbook to the 4th (german) edition. The beneath numbering is from the linked online version, which should be the 9th english edition.

Your way to the solution is part of subsection 3.9 Properties of the Gibbs energy and subsubsection 3.9 c) The variation of the Gibbs energy with pressure.

I seriously entrust you to read this subsubsection and try to solve it with the given information. If this does not help, the following "hidden" section reveals what you need to solve it.

   You need to insert the equation for the molar volume of an ideal gas $V_m = R~T/p$ into (3-57b) $$G_m(p_f) = G_m(p_i)+\int_{p_i}^{p_f}V_m~\mathrm{d}p \tag{3-57.b}$$ which yields one of the two equations that you need: $$G_m(p_f)=G_m(p_i)+R~T\int_{p_i}^{p_f}\frac{\mathrm{d}p}{p}=G_m(p_i)+R~T\ln\frac{p_f}{p_i}\tag{3-59}$$ The other one is the Boyle's law.

Good luck.


Can we define a new unit instead of specific heat that is constant with temperature rise? Specific heat plays an important role in calculating energy and thus, heat, enthalpy and etc.
But as I sighted the definition for calorie, I was surprised.

  A calorie is the amount of heat needed to alter the temperature of 1 gram of 14.5 °C water to 15.5 °C water in the pressure of one atmosphere.

Though the unit was redefined later, the question remained still. Why were there those temperatures? It was just then that I realized specific heat capacity varies with temperature differences and it is due to vibration movements of matter and internal energy ....
Two questions: Can there be a unit which is constant with temperature alterations that can be used instead of S.heat capacity? If so, why isn't it replacing SHC?
The resource for the quote The resource for the reason of the increase that happens to SHC with temperature rise

There isn't a simple redefinition that would work in all cases, and this is probably why nobody does this.  Of course, you COULD define such a unit for a given system by defining it as a function of temperature, but it would be applicable only to the specific system.

As you noted, the physics behind the temperature dependence of a material's heat capacity lies in the quantum states which become available for excitation at different energies.  Consider an insulator which stores heat primarily in the lattice vibrations.  As you increase the temperature, more vibrational modes are accessible.  This is often described using the Debye model which gives the heat capacity as such:

$C_V = k(\frac{T}{\theta})^3$

where $C_V$ is the heat capacity, $k$ is a constant, $T$ is the temperature in Kelvin, and $\theta$ is a constant pertinent to a specific solid.[1]

However, a metal can also store a significant amount of heat in the motions of electrons since conduction electrons are not bound to specific sites.  Heat capacity for the electron 'gas' is often described like this:

$C_V' = k' \frac{T}{\theta'}$

where $C_V' \neq C_V$, $k' \neq k$ and $\theta' \neq \theta$.[2]  Depending on $\theta$ and $\theta'$ for a given material, at low temperatures (often cryogenic) the electron contribution can be the dominant mode of storing heat in the solid.  Thus, if you chose a unit which was sensible for an insulator, it wouldn't apply to a conductor (and vice versa).

Even within a specific material, the heat capacity on either side of a phase transition is often different since the phase transition has changed the quantum states available in the material.  Therefore, your new unit would become discontinuous at phase transitions and would need a different mathematical expression for each segment of the heat capacity function, not to mention a different parameterization for each material.  Hence, no real gain.

[1] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 106, eq. 47b

[2] See Kittel, Kroemer, Thermal Physics Second Edition; W. H. Freeman and Company, New York, 1980, p. 193, eq. 37.  Note that folks don't actually use the symbol $\theta'$ in this case, but choosing so makes the explanation above clearer.


What does the newest IUPAC nomenclature have against unsaturation? IUPAC nomenclature 2013 changed a couple of rules very commonly used in naming simple acyclic compounds. As a result, depending on which nomencalture we choose, the answer to this question varies greatly, as demonstrated in the comments. Educational textbooks, hence, would propose an outdated name that is not PIN (Preferred IUPAC Name) according to the newest standards.

This change has been illustrated in the draft, page 442 (page 58 of P-4, section P-44.3), which reads:

  P-44.3 The principal chain

  In an acyclic compound, or in a compound composed of chains and rings, the chain on
  which the nomenclature and numbering is based is called the ‘principal chain’. When there is a
  choice for the principal chain, the following criteria are applied, in the order listed, until a
  decision is reached.

  $$\mbox{A change to the traditional order of seniority criteria is recommended;}\\ \mbox{the length
of the chain is senior to unsaturation.}$$

What was the reason for this change?

Normally, there doesn't need to be a reason for changes. However, a change as drastic and influential as this must have had a reason behind it. I'm thinking a group of [newly discovered] compounds needed a new nomenclature, and I remember I read some news that said vaguely the same thing. Was there a need for revision of the criteria? Why would a nomenclature in which unsaturation is senior be inferior to the new system of IUPAC nomenclature?

I think it is important to note that these are recommendation and not rules (in the sense of which you must follow) and hence you can choose for yourself whether you like to implement them or not.

A generated unambiguous name still conforms to the general recommendations of the IUPAC, hence what you call outdated is still a valid name, it just isn't the preferred IUPAC name, which is a concept introduced in 2013. Hence names generated according to the 1993 recommendations will still be valid names according to the new recommendations; the new recommendations are backwards compatible.

The reasoning can be found in the Introduction of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) (It is also in the cited draft):

  P-10 INTRODUCTION
  [...]
  A major new principle is elaborated in these Recommendations; the concept of preferred IUPAC names' (PINs) is developed and systematically applied. Up to now, the nomenclature developed and recommended by IUPAC has emphasized the generation of unambiguous names in accord with the historical development of the subject. In 1993, due to the explosion in the circulation of information and the globalization of human activities, it was deemed necessary to have a common language that would prove important in legal situations, with manifestations in patents, export-import regulations, environmental health, and safety information, etc. However, rather than recommend only a single unique name for each structure, we have developed rules for assigning preferred IUPAC names, while continuing to allow alternative names in order to preserve the diversity and adaptability of the nomenclature to daily activities in chemistry and in science in general.

  Thus, the existence of preferred IUPAC names does not prevent the use of other names to take into account a specific context or to emphasize structural features common to a series of compounds. Preferred IUPAC names (PINs) belong to a 'preferred IUPAC nomenclature'. Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a general IUPAC name, in the context of a 'general IUPAC nomenclature'.
  [...]

If nothing more, the new guidelines provide a more robust nomenclature. In my very personal opinion that is a very good change. Let's look at a more practical approach.
Upon a chemical reaction which modifies a compound (in many if not most case) the principle chain will be retained according to the new change. Since unsaturation often provides the possibility of relatively easy modification, similar to a functional group (and some would even consider it one), a reaction at these bond will likely involve that the principle chain changes according to the old recommendations. Let's stick to the example of Loong's answer in the linked post; and let's just add $\ce{HBr}$ to the double bond:

According to the old recommendations the reader is likely to imply that there was a modification hexene to heptane, and without further inspection one could assume that there was an extension of the carbon skeleton, i.e. a carbon-carbon bond formation.
In the new guidelines it is much more obvious what the modification actually is, a simple addition. This obviously is of little concern for these simple compounds, but when it comes to natural product synthesis, it removes some of the ambiguity of what actually happens, highlighting the change of the functional group rather than implying a modification to the carbon skeleton.

There are probably instance where you deliberately want to deviate from PINs. A simple example would be the protection of an alcohol:

It might be a necessary step in your synthesis, it might even be an important one, for the chemistry involved and expressing what you are actually doing, PINs will be a nuisance. I personally would prefer the names in parentheses. However, when you are eventually writing your patent, you might want to stick to a PIN in order to avoid any ambiguity, after all that's why they were introduces in the first place.


Do bidendates occupy cis position always? https://chem.libretexts.org/Core/Inorganic_Chemistry/Coordination_Chemistry/Properties_of_Coordination_Compounds/Isomers/Stereoisomers%3A_Geometric_Isomers_in_Transition_Metal_Complexes

In example 1.1, it is written that en which occupy only adjacent positions.....is it true for any bidendate for octahedral isomer?

Its not impossible to have a trans-spanning ligand, but they seem to be rarer than cis-ligands due to their increased length and greater potential for steric strain. While en is too small of a ligand to reach trans-positions, it is possible with significantly longer ligands.

It seems octahedral ones are even rarer than their square planar counterparts, but there has been at least one reported synthesis of a trans-spanned octahedral$^1$ that I could find. In the linked paper, they report the synthesis of trans-$\ce{[Ru(Cl)(trpy)(C3SPAN)]}$ where trpy = 2,2:6,2-terpyridine and C3SPAN=$\ce{Ph2PC6H4CH2O(CO)(CH2)3(CO)OC6H4PPh2}$.

$\hspace{25ex}$

C. Bessel, et al. Chem. Soc., Dalton Trans., 1999, 2281–2292


Bromination of but-3-en-1-ol in basic medium and Baldwin's rules After formation of the bromocarbocation the alkoxide anion attacks the bromocarbocation which leads to ring formation and can proceed through two pathways

5 endo tet cyclization is disfavored but 4 exo tet cyclization is favored according to Baldwin's rules; however, the answer given in my text book is the 5 membered ring.

Your reasoning is totally correct according to Baldwin's Rule - the 4-exo-product would be more stable than the 5-endo-product ( see : This for more details).
But your method may be wrong in the given question the product A was first formed and then a ring closure step was performed using $\ce{KOH}$ as shown below :

here the 5-exo-trig ring closure is favoured according to Baldwin's rule .
Let me also address something that may be a doubt ( It was mine ) i.e. why doesn't the oxygen atom attack the bromonium Ion in the first step ? this is because even if it will attack you will also form a acid $\ce{HBr}$ which will cause the cleavage in the cyclic ester we just formed returning us to the original reagent hence it doesn't attack.


What is the most compact way to store Oxygen? What is the most compact way to store Oxygen? Perhaps I should qualify that as being at room temperature since the most compact form of oxygen is probably liquid oxygen. At room temperature would it be better to store it as rust or compressed air or something else, like hydrogen peroxide?

Solid ozone has a density of 1.728 g/cm3 according to The Density of Solid Ozone J. Am. Chem. Soc., 1959, 81, pp 805–806.

Best I can think of at atmospheric pressure, assuming your goal is to store and recover as much pure oxygen as possible in a given volume.

If your goal is just to pack as many oxygen atoms in a given volume, BeO, which has a density of 3.01 g/cm3 is better.

At high pressure there is metallic oxygen.


Do the spins of outer orbitals block measurement of spin of inner orbitals? Do the outermost s1 and s2 electrons block an outside observer from measuring the larger inner group orbitals for elements heavier than Nitrogen (for instance, would Chromium's 4s1 electron or Manganese's 4s2 electrons prevent you from detecting the spin of the 3d5 electrons in either case?)

I edited this into an individual question from the original list of them per moderator instruction.

With regards to your first question, as you get to higher energy levels, the distance between each one increases and as a consequence the shells get closer together and begin to overlap. This is also seen with 5s and 4d iirc.


Correct nomenclature for reaction types What is the correct name for a reaction like this?

$$ \ce{ 4MnO2 ->[500\ ^\circ \text{C}] 2Mn2O3 + O2 ^}$$

Is it a synthesis, or a Decomposition?
I tried to find it out with http://en.wikipedia.org/wiki/Chemical_reaction but i did not get the link between the formula and the nomenclature as it seems to be both.

This is a decomposition reaction.  Synthesis requires a minimum of two reactants.  Decomposition is where a reactant breaks down into two or more products.  Here, one reactant (MnO2) goes to two products.  It seems like it may not be decomposition because MnO2 is becoming MnO3, but it is still one reactant to two products.


Why is para-aminobenzoic acid more acidic than ortho-aminobenzoic acid? Compare the acidic strength of o-, m-, p-aminobenzoic acids.

I got that meta will be the most acidic as it won't be able to show R+ effect of $\ce{NH2}.$
But among the other two, i.e. ortho and para, shouldn't ortho be more acidic as it will cause ortho effect?
But according to my book and the data I found online:
$$
\begin{array}{lc}
\hline
\text{Compound} & \mathrm{p}K_\mathrm{a}\text{(amino)} \\
\hline
\textit{o}\text{-Aminobenzoic acid} & 4.89 \\
\textit{m}\text{-Aminobenzoic acid} & 4.79 \\
\textit{p}\text{-Aminobenzoic acid} & 4.77 \\
\hline
\end{array}
$$

The problem in ortho-aminobenzoic acid is that the acidic hydrogen of carboxylic group is H-bonded with the lone pair of nitrogen in amino group. As a result it is more difficult to extract it compare to that in para-aminobenzoic acid since the H-bond must also be broken during acid-base reaction. Para-aminobenzoic acid does not have a H-bond due to the distance between the two groups.

I would like to add that since the hydrogen of the $\ce{-COOH}$ group is more acidic and $\ce{N}$ has a lone pair. This is why H-bond occurs with $\ce{-NH2}$ group as electron donor and not the other way round.
Also, I presume that by ortho effect you intended to tell that due to repulsion of $\ce{-NH2}$ group $\ce{-COOH}$ group would go out of the plane. However, as we have seen, it is difficult as it is already H-bonded with the nitrogen atom and moving out of the plane would put strain into the H-bond. Moreover, the repulsion by $\ce{-NH2}$ group is not too intense so it would (in most part) avoid the strain.


How does electron localization/delocalization work? I can't find a simplified explanation on this topic. My textbook just adds to the confusion and all I'm really left with are more questions rather than answers.

And adding to that question, does polarity have an effect on electron localization/delocalization?

Draw the LCAO orbitals.  Delocalization is the summation of discrete electron occupations of orbitals, such as a 1,3 shift (a pi-bond flip).  Remember that LCAO is a terrifically wrong black box. Its only virtue is that it simply, beautifully works (until the Woodward-Hoffmann rules).

Given propene, abstract a methyl proton.  The LCAO anion is half at both ends by a 1,3-shift of the pi bond.  MO shows degenerate orbital occupation still with (nearly) no additional density at the central carbon.  Abstract an allylic proton from 1-butene.  The allylic ends are no longer 50:50 occupied by the anion for the different electronic properties of two attached protons versus a methyl group and a proton.