What is the major product in the reaction of 2,3-dichlorobutane with sodium amide in liquid ammonia?

These are both reasonable mechanisms, and the question outlines well the factors favoring each. In favor of mechanism I:

Low temperature suggests kinetic deprotonation
Statistically more terminal hydrogens than internal hydrogens

In favor of mechanism II:

Small base suggests thermodynamic deprotonation

In these cases where there are conflicting factors, I agree with @Lighthart that the best answer is to actually look at the experimental results. That will give an indication of how to weigh the various factors.

In this case, I do not find an exact hit in SciFinder, but the reaction of  2,3-dibromobutane does come up in two older reviews by Bergstrom and Franke. I do not have access to Franke's review, although SciFinder has a note that "Bromine represents any halogen."

The SciFinder result for the Bergstrom review is deceptive. There is a comment "generalized reaction, halogen and R-groups can vary," and I don't find the exact result (or reasonable analogs) in the review.

More broadly, there are seven results for an internal, vicinal (1,2) dibromide reacting with sodium amide to give an alkyne. One reliable example is the synthesis of stearolic acid from Organic Syntheses.

Similar SciFinder searches for the 1,3-diene product come up empty.

Based on this research, I believe the alkyne would be the predominant product.


The acylium ion is stabilized by the positive mesomeric effect of the lone pair of electrons on the oxygen atom onto the empty p-orbital of the $\ce{C+}$. This helps achieve octet completion of the carbon atom, and thus imparts stability to the carbocation.

(Image by Jü CC BY-SA 4.0, from Wikimedia Commons)

Resonance effects mostly dominate the weak inductive effect, so here the destabilizing negative inductive effect of oxygen atom has only a negligible role.

An alkyl carbocation has no such positive stabilizing mesomeric effect, rather only hyperconjugative effects, which are not able to provide as much stability as is possible in the acylium ion.


How was Mendeleev able to develop his table?

This quote from the wikipedia article on History of Chemistry might provide some insight into the determination of atomic mass:

  Chemists in different countries developed several different incompatible atomistic systems. A paper that suggested a way out of this difficult situation was published as early as 1811 by the Italian physicist Amedeo Avogadro (1776-1856), who hypothesized that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, from which it followed that relative molecular weights of any two gases are the same as the ratio of the densities of the two gases under the same conditions of temperature and pressure.

So this implies that you could take equal volumes of different gases at the same temperature and pressure, then weigh them and take the ratios. This would allow you to get relative masses for $\ce{H2}$, $\ce{N2}$, $\ce{O2}$, $\ce{Cl2}$, $\ce{Br2}$, $\ce{I2}$. Other gases could be used, but they wouldn't be as useful here. You could then determine that $\ce{H2}$ is the lightest gas, and that $\ce{O2}$ is 16 times heavier, $\ce{N2}$ is 14 times heavier, and so on. You don't even need to know that these are diatomic gases in order to make these measurements. Once you had these determined you could move on to gases like $\ce{CO2}$, or $\ce{H2S}$. Once you knew the mass of the gas molecule and what elements were in it, you could calculate how much of each element must be there, and by subtracting the mass of the known element, the remaining mass must be the atomic mass for the other element. As you accumulated more data you could approach substances that don't easily form gases, such as $\ce{NaCl}$. If you know the mass for $\ce{Cl}$ you could determine the mass for $\ce{Na}$.

What I suggest might not be the way atomic masses were determined, but it would probably work. Looking at this image of Mendeleev's original table shows they already had pretty good estimates of atomic mass by 1869.


When we lower the temperature of a substance, we are decreasing the average molecular kinetic energy (they're slowing down). As mentioned in the comments on the question, this doesn't really affect the electron cloud.

So, the substance has been cooled down and molecules are moving more slowly. At this point, it's not that the dispersion forces because stronger, but rather that they become comparatively stronger. While in the gaseous phase, the molecules have velocities of several hundred meters per second. At those speeds, London forces don't really have much of an effect. But, as the molecules are cooled and slowed down (this seems to suggest this is around 10m/s), London forces can cause more interactions between molecules. If the substance is cooled far enough, those forces are able to overcome the kinetic energy of the molecules and they condense into a liquid. As the cooling continues, the molecules move around less and less until they take on a definite shape, at which point it is considered a solid.

The primary issue with your intuition is merely understanding what cooling the substance down really does.


Actually, it is, but a very weak one.

The lone electron pairs in dinitrogen are in $\mathrm{sp}$ hybridization state, and nitrogen is quite electronegative, so they are not easily donated. Still, if a Lewis acid is strong enough, dinitrogen may form a complex.

It is known that dinitrogen adsorbs on acidic $\ce{OH}$-groups in zeolites, forming complex with IR-active frequency. The nature of the complex may be debated, as no full proton transfer occurs, but the electron pairs are definitely involved.

Diazonium salts may be considered as a lewis-acid complex of a carbocation and dinitrogen molecule. While no direct synthetic route exists from dinitrogen to diazonium salt (because typically a stronger Lewis base is available), they typically decompose with formation of dinitrogen molecule.

Significant amount of dinitrogen complexes with transition elements are known. Typically they are obtained indirectly, by forming dinitrogen ligand in-place, though.

Finally, adsorption of dinitrogen on the surface of ammonia synthesis catalysts starts as a Lewis-interaction. Similarly, nitrogen adsorption on cations in zeolites is known.


Strength of Dipoles, Covalent bonds and H-Bonds

It means that of the three forces, dipole-dipole forces are the weakest, hydrogen bonding is the second weakest, and covalent bonds are the strongest. Now, dipole-dipole attraction happens between an ion and a polar molecule. Due to a net dipole movement, this is weak. In comparison to a covalent bond, it has only $\frac{1}{100}$ the efficacy. A hydrogen bond is a force between H and either N, O, or F, the three most electronegative elements. These are classified as d-d forces, but due to the differing electronegativities, they are a tad stronger. Covalent bonds are much stronger than either, $20$ times more than the H-bond and $2000$ times stronger than the d-d force. A covalent bond involves the sharing of electron pairs between atoms, generally two non-metals. There are also ionic bonds and metallic bonds, an ionic bond involving the transfer of electrons, such as $\ce{NaCl}$ and a metallic bond being between metals.


How to get the reducible representations of the B-B and the B-Cl bond of B2Cl4

There seem to be two problems: the first is that your reducible representation for the $\ce{B-B}$ bond is wrong but your reducible representation for the $\ce{B-Cl}$ bonds seems to be correct and the second is that your book is apparently completely wrong. If I didn't make a mistake and my second assumption is correct then maybe you should think about switching to another book as it seems to raise more questions than it answers. But I'm not working regularly with group theory, so my analysis of the problem might be faulty, but maybe someone else can support or refute my answer.

Ok, here is my analysis:
First, make yourself familiar with the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group. Have a look at this video for example. Now, I looked at the effect each symmetry operation has on the bonds I'm looking at. For that I will use the following depiction of $\ce{B2Cl4}$:

For the $\ce{B-B}$ bond I used an arrow to indicate the direction of the bond for better emphasis as I will need that soon. In the following I will assume that you are familiar with the way characters are related to your reducible representation and how you get those characters from analyzing the changes of your basis vectors (in this case the bonds) under the symmetry operations of the group since you seem to have already used it to determine the reducible representations in your question.
Furthermore, I use the following notation: The reducible representation of the $\ce{B-Cl}$ bonds will be denoted as $\Gamma_{\ce{B-Cl}}$ and the character of the symmetry operation $C_{2}$ for this reducible representation will be denoted as $\Gamma_{\ce{B-Cl}}(C_{2})$.

Ok, let's see what the symmetry operations of the $\mathrm{D}_{2\mathrm{d}}$ group do to this structure, so that we can get the reducible representations:
The $E$ operation does nothing to the structure, i.e. all four $\ce{B-Cl}$ bonds and the one $\ce{B-B}$ bond stay where they are. Thus you get a character of $4$ for $\Gamma_{\ce{B-Cl}}(E)$ and a character of $1$ for $\Gamma_{\ce{B-B}}(E)$.

Now, we come to the improper axis $S_{4}$. It rotates along the long the $\ce{B-B}$ bond and mirrors perpendicular to it:

As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(S_{4}) = 0$. Since the mirror plane is perpedicular to the $\ce{B-B}$ bond the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(S_{4}) = -1$.

Next, we look at the principle $C_{2}$ rotation along the $\ce{B-B}$ bond. It converts your molecule to the following:

As you can see the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}) = 0$. But as the rotation is conducted along the $\ce{B-B}$ bond axis the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}) = 1$.

Now, for the two $C_{2}^{\prime}$ axes which are perpendicular to $C_{2}$. One disects the dihedral angle between the dark red $\ce{Cl}$ and the cyan $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other disects the dihedral angle between the red $\ce{Cl}$ and the blue $\ce{Cl}$ - its effect on the structure is shown in the right picture:

$\qquad \qquad \qquad$

Again the red and the dark red $\ce{Cl}$ atoms switched places and the same is true for the blue and the cyan $\ce{Cl}$ atoms. So the $\ce{B-Cl}$ bonds are completely rearranged, none keeps its location or direction, and thus you get $\Gamma_{\ce{B-Cl}}(C_{2}^{\prime}) = 0$. But now the rotation is conducted along an axis perpedicular to the $\ce{B-B}$ bond and so the two $\ce{B}$ atoms' positions are exchanged and thus the $\ce{B-B}$ bond - while keeping its location - reverses its direction, and thus you get $\Gamma_{\ce{B-B}}(C_{2}^{\prime}) = -1$.

Finally there are the two $\sigma_{\mathrm{d}}$ planes which lie along the $C_{2}$. The one plane contains the dark red and the red $\ce{Cl}$ while disecting the angle betweeen the  cyan and the blue $\ce{Cl}$ - its effect on the structure is shown in the left picture - and the other contains the cyan and the blue $\ce{Cl}$ while disecting the angle betweeen the dark red and the red $\ce{Cl}$ - its effect on the structure is shown in the right picture:

$\qquad \qquad \qquad$

In this case either the red and dark red $\ce{Cl}$ atoms or the blue and cyan ones remain unchanged while the other $\ce{Cl}$ atoms change their places. So, two $\ce{B-Cl}$ bonds keep their location and direction which gives $\Gamma_{\ce{B-Cl}}(\sigma_{\mathrm{d}}) = 2$.
The $\ce{B-B}$ bond axis lies within the mirror plane and so the two $\ce{B}$ atoms' positions are unchanged and thus the $\ce{B-B}$ bond keeps its location and direction, and thus you get $\Gamma_{\ce{B-B}}(\sigma_{\mathrm{d}}) = 1$.

So, in conclusion you end up with the reducible representations (I kept the order of the symmetry elements you chose, i.e. $E$, $2 \, S_4$, $C_2$, $2 \, C_{2}^{'}$, $2 \, \sigma_{d}$):

\begin{align}
  \Gamma_{\ce{B-Cl}} &= 4, 0, 0, 0, 2 \\
  \Gamma_{\ce{B-B}}  &= 1, -1, 1, -1, 1 \\
\end{align}

and they reduce to

\begin{align}
  \Gamma_{\ce{B-Cl}} &= A_1 + B_2 + E \\
  \Gamma_{\ce{B-B}}  &= A_2 \\
\end{align}

As you can see both results differ from the ones your book provides.


Among hydrogen, helium and carbon dioxide, which gas would behave most like ideal gas and why?

In the simplest model, a gas is called ideal when its particles are point-like (no volume) and have no interactions. Real gases behave like ideal gases at low pressure (where the particle volume is neglible compared to the total volume) and high temperature (where condensed phases, i.e. interatomic or intermolecular interactions are disfavored).

The size-comparison between helium and dihydrogen is straightforward: Dihydrogen is larger. As for the strength of interparticle interactions, we can compare normal boiling points: Helium's is 4 Kelvin and dihydrogen's is 24 Kelvin.

So this would suggest that helium is "more ideal" as it has the lower boiling point and the smaller size.


Why is ATP, but not ADP or AMP, a sole source of energy?

Let me add to iad22agp's excellent answer.

Often times in introductory biochemistry classes, the difference between phosphoanhydride and phosphate ester bonds is not adequately explained.

ATP (and ADP) contain phosphoanhydride bonds.  In these bonds, two phosphoric acid groups are condensed into one:

$\ce{R~-H2PO3 + R'~-H2PO3 -> R-HO2P-O-PO2H-R' + H2O}$

In contrast, AMP and many other phosphate-containing biomolecules such as glucose-6-phosphate glyceraldehyde-3-phosphate, contain phosphate ester bonds, where a phosphate group is condensed with an alcohol:

$\ce{R~-H2PO3 + R'OH -> R-HPO2-O-R' + H2O}$

Just as "regular" (i.e. carboxylate) esters like ethyl acetate are more stable to hydrolysis than carboxylic anhydrides (e.g. acetic anhydride), phosphate esters are much more stable and "lower energy" than phosphoanhydride bonds.  They don't have enough energy to power many of the key reactions of biosynthesis.  So that explains why AMP isn't a good energy source.

(By the way, mixed phosphate-carboxylate anhydrides are as high or higher in energy as phosphoanhydride bonds.  Examples from biochemistry include acetyl phosphate or phosphoenolpyruvate.)

So ATP has two phosphoanhydride bonds.  And as iad22agp mentioned, the electrophilic nature of the central P atom makes the outermost phosphoanhydride bond especially prone to hydrolysis.  However, ADP still contains a high-energy phosphoanhydride bond.  Consider this reaction:

$\ce{ATP + AMP <=> 2 ADP}$

The equilibrium constant for this reaction is about $2.82$, meaning the rightward "forward" direction is favored, but not by a whole lot.  So the "extra" energy of ATP's particularly high-energy phosphoanhydride bond is not really that much higher than that of ADP's regular run-of-the-mill phosphoanhydride bond.

There are somewhat rare examples where ADP powers key metabolic reactions but ATP does not.  That shows it is nearly as good an energy source as ATP.

The reason that ATP is so prevalent is probably because of evolutionary reasons.  Also, another trick that enzymes used with ATP is sometimes to hydrolyze the interior phosphoanhydride bond, leading to AMP and pyrophosphate.  Because pyrophosphatase enzymes quickly degrade pyrophosphate to two free ortho-phosphate  groups, this trick effectively makes what would have been a reversible reaction effectively irreversible.  That trick doesn't work with ADP because it has only one phosphoanhydride bond.


If you're sure that the initial formula is $\ce{H3COOCH3}$ (and that's a big if), then I disagree with Richard's answer in comments. $\ce{H3COOCH3}$ is dimethyl peroxide. While peroxides are not the most stable of molecules, and decompose rapidly (often by oxidizing other compounds), they do not spontaneously transform in acid conditions.

If the formula were $\ce{H3C–COOCH3}$ (note the added C), then yes, it's be acid-catalysed ester hydrolysis.


Does space in between have any significance in IUPAC nomenclature?

$\boxed{\textbf{d}}$ is correct. It is the standard IUPAC name.

However, "2,4,6-trimethyl cyclohexanone" will often be understood as 2,4,6-trimethylcyclohexanone, similar to how, in theory, neopentane, 2,2-methylpropane and 1,1,1,1-methylmethane are all equivalent.


How to balance the redox reaction between bismuth(III) sulfide and nitrate anion?

While in general a reaction involving three independent redox pairs may not have unique coefficients, this particular one contains one coupled oxidation half-reaction and one regular reduction half-reaction.

The key is to note that both elements in $\ce{Bi2S3}$ are oxidized, and hence we have a stoichiometric relationship between the two oxidation half-reactions. This suggests we write the oxidation half-reaction as follows: $$\ce{18H2O + Bi2S3 -> 2BiO3- + 3SO4^{2-} + 36H+ + 28e-,}$$ and we can proceed with the usual algorithm for balancing half-reactions.


Why do electrodes need to share an electrolyte in a lemon battery?

Like @Pritt said, the circle has to be closed, 2 electrons are traveling from one half-cell to the other, means 2 negative charges are missing, the half-cell needs to replace it by other 2 negative ions or send 2 positive charges, since charges cannot exist separately.


What is the difference between cofactors and coenzymes?

Enzyme cofactors are a collective name of all the chemical compunds or elements associated with the enzyme to increase its efficiency. There are mainly two types- inorganic ions and organic compounds. Inorganic ions are also known as enzyme activators (Cl ion in salivary amylase). Organic compunds are again divided into two- co-enzymes and prosthetic groups. Co-enzymes are loosely associated with the enzyme. They can be removed from the enzyme and undergo cyclic reactions to again combine with enzyme (EX: NAD, NADP). But prosthetic groups are tightly associated with enzyme molecules(EX : FAD). Hope this helps.


Are Hartree-Fock solutions eigenfunctions of the electronic hamiltonian?

$\newcommand{\ket}[1]{\left|#1\right>}$
$\newcommand{\bra}[1]{\left<#1\right|}$

  (1) Is this all correct so far?

Yep.

  (2) If so, is it justified to use it in numerical derivations as an
  eigenfunction of the Hamiltonian, i.e. can we write
  $\hat{H}\ket\Phi\overset{?}{=}E_{0}\ket\Phi$ or are we restricted to
  $\bra\Phi\hat{H}\ket\Phi=E_{0}$?

First, you can write $\hat{H}\ket\Phi = E\ket\Phi$ where $\Phi$ is the HF wave function but then $\hat{H}$ has to interpreted as the so-called mean-field Hamiltonian. It is the approximate Hamiltonian in which the exact potential which represents the energy of the Coulomb repulsions between the
electrons ($\hat{V}_\mathrm{ee}$) is replaced with some approximate mean-field potential ($\hat{V}_\mathrm{MF}$) which describes the model system in which electrons do not instantaneously interact with each other, but rather each and every electron interacts with the average, or mean, electric field created by all other electrons. That is basically the definition of the HF wave function, i.e. it is an eigenfunction of the mean-field Hamiltonian:
$$
\hat{H}_\mathrm{MF} \ket\Phi = E_\mathrm{MF} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{MF} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{MF} \, .
$$

Secondly, despite the fact that the HF wave function is not an eigenfunction of the exact electronic Hamiltonian
$$
\hat{H}_\mathrm{e} \ket\Phi \neq E_\mathrm{e} \ket\Phi \, ,
\quad \text{where} \quad
\hat{H}_\mathrm{e} = \hat{T}_\mathrm{e} + \hat{V}_\mathrm{en} + \hat{V}_\mathrm{ee} \, ,
$$
we could evaluate it energy $\bra\Phi\hat{H}_\mathrm{e}\ket\Phi$ using Slater rules and minimize it to find an upper bound to the ground state energy
$$
\bra\Phi\hat{H}_\mathrm{e}\ket\Phi \geq E_\mathrm{e0}
$$
But the resulting Slater determinant obtained by minimizing the energy is only an approximation to the ground state wave function.

Finally, yes, can we use $\ket\Phi$ as our trial wave function despite the fact that it is not an eigenfunction of $\hat{H}_\mathrm{e}$ since it is not required by the variation method. $\ket\Phi$ is normalizable, it (presumably) satisfies same boundary conditions as the exact wave function, and it has adjustable parameters. That is basically enough to qualify as the trial wave function. And at the end of the day, starting from a known solution for a simpler problem is a very usual practice for variation method.


As long as your experimental setup did not contain any leakages or impurities through which air or moisture could have gotten into the mixture, the reaction should have worked fine under these conditions. The color of the hexaamminsodium electride solution is actually deep blue and intense, so at higher concentrations it can appear black at first sight (see picture below).


Why does graphite have a high melting point even though its intermolecular forces are weak?

It's not usual to consider graphite as a material composed of "molecules" in the typical sense, though it could be viewed as a kind of polymer with two-dimensional macromolecules. Regardless, it is fruitful to analyse the individual sheets in graphite as the limit of increasingly large polycyclic aromatic hydrocarbons (PAHs). The sequence goes: benzene ($\ce{C6H6}$) → coronene ("superbenzene", $\ce{C24H12}$) → $\ce{C54H18}$ → ...

The main kind of intermolecular interaction relevant to this sequence of compounds is pi-stacking. For the smallest example in the sequence, benzene, it appears that the strength of this intermolecular interaction is only about $\mathrm{10\ kJ\ mol^{-1}}$. This represents only a modest attraction; a hydrogen bond can easily be twice as strong even though it involves less atoms. In this sense, when "normalised" by the number of atoms taking part, pi-stacking is indeed a comparatively weak intermolecular interaction.

Furthermore, $\mathrm{10\ kJ\ mol^{-1}}$ is comparable to the average thermal energy of particles in ambient conditions ($\mathrm{k_BT_{amb}=2.5\ kJ\ mol^{-1}}$), so it doesn't take too much effort to pull apart benzene molecules. Indeed, benzene melts at $\mathrm{5.5\ ^oC}$ and boils at $\mathrm{80\ ^oC}$ under one atmosphere.

However, the next compound in the sequence, coronene, already melts at $\mathrm{437\ ^oC}$ and boils at $\mathrm{525\ ^oC}$. Larger PAHs would almost certainly have even greater values, eventually reaching the limit of graphite, which melts around $\mathrm{4000\ ^oC}$ under pressure. The type of intermolecular interaction hasn't changed, so why are these molecules suddenly so difficult to pull apart? The answer comes from realising that, though individual sections of each molecule interact weakly, the sum of many weak intermolecular interactions over an entire molecule leads to a very strong intermolecular interaction overall.

Crudely speaking, imagine that each individual aromatic ring contributes $\mathrm{10\ kJ\ mol^{-1}}$ worth of intermolecular attraction. Coronene contains 7 fused aromatic rings, which would lead to a total interaction of $\mathrm{70\ kJ\ mol^{-1}}$ between two molecules. As the molecules get larger, this value increases further and further. Eventually the total intermolecular interaction between two very large PAH molecules becomes enormous. For the substance to liquefy, it is only necessary to "break" a fraction of these intermolecular interactions (breaking all of them is turning the material into a gas), but even a small fraction eventually represents a huge amount of energy, so melting only happens at very high temperatures.

It is interesting to note how often chemists make the mistake of neglecting weak long-distance interactions (e.g. van der Waals), especially in the presence of stronger ones. For example, van der Waals interactions are fundamental to the stability of alkyl-substituted hexaphenylethane derivatives. In proteins, weak interactions are often overlooked in favour of hydrogen bonding ($\alpha$-helices and $\beta$-sheets), though they can be decisive in determining the correct conformation of an enzyme or how a protein interact with medicinal compounds.

As a last slight tangent, I just want to point out that in undergraduate-level chemistry, often one finds statements such as "boiling points for covalent compounds increase with their molecular weight. Now it should be evident this is not strictly true. It merely happens that compounds with higher molecular weights tend to be larger and allow a greater amount of intermolecular interactions per molecule, thus leading to higher boiling points.


Rate of solvolysis of allyl and alkyl halides

Resonance effect is more stronger than hyperconjugation irrespective of the number of αH. In first case allylic carbocation is formed which is stabilised by resonance. So solvolysis rate is more for the first compound


I am afraid, direct acid-base titration is not the right way to analyze aspirin exactly because of the you stated - hydrolysis. The rule number no of any titration is that there should be no side reaction and it should go to completion almost instantly. The direct titration of aspirin is problematic because hydrolyzes pretty fast to salicylic acid- an unwanted side reaction which may or may not go to completion.

In order to circumvent this, analytical chemists do a back-titration. You add an excess of base, heat the mixture so that neutralization as well as hydrolysis are complete. The remaining base is titrated. The rest of the calculations are normally done.


Twisting stereoisomers with rings to determine R/S

Rotating structures is a recipe for disaster because errors in transcription will invariable occur. Use the best chirality tool you have---your hands. Leave the structure where it is! Use either hand to make a "thumbs up" gesture. Point your thumb in the direction of the lowest priority group. The remaining fingers should point in the direction of the highest to next to lowest priority group (OH>C2H5 >CH3). If not, you are using the wrong hand. Left hand = S. Right hand = R. Try to rotate structures 1-3 and see how it compares with the "hand method". Structures 1-3 are of the S-configuration. Cyclopentene 4 is Jon's structure with substituents. It is of the R-configuration (=CH>CH2>CH3), a conclusion reached without redrawing the structure.


Why is fluorine the most electronegative atom?

Fluorine is the most electronegative element because the definition of electronegativity makes it so. The electronengativity scales are defined based on experimentally determined properties of the elements. Fluorine has appropriate values for all of the common scales to ensure it has the highest electronegativity.

The Pauling scale, which is the first electronegativity scale proposed, relates the difference in electronegativity $\Delta\chi_{A,B}$ between two atoms $(A\ \& \ B)$ is related to the bond dissociation energies $(E_d)$ of the diatomic species $AA,\ BB,\ \& \ AB$:

$$\Delta \chi_{A,B}=|\chi_A - \chi_B| = (\text{eV})^{-1/2}\sqrt{E_d(AB)-[E_d(AA)+E_d(BB)]/2}$$

Since only differences can be determined, one element (hydrogen) was chosen to have an arbitrarily set value. On the Pauling Scale Hydrogen has a value of $2.20$. Since the $\ce{H-F}$ bond is much stronger $(E_d=586\ \text{kJ/mol})$ than the average of the  $\ce{H-H}$ bond $(E_d=436\ \text{kJ/mol})$ and the $\ce{F-F}$ bond $(E_d=157\ \text{kJ/mol})$. Data from here. In general, since fluorine forms stronger bonds to all other elements than it does to itself, fluorine will have the highest electronegativity.

The Mulliken scale relates electronegativity to two other properties: electron affinity $E_{ea}$ and ionization energy $E_i$:

$$\chi_A = \frac{E_{ea}(A)+E_i(A)}{2}$$

Fluorine has a high electron affinity (easy to gain an electron) and a high ionization energy (hard to lose an electron).

There are other scales, including the Allred-Rochow scale, based on effective nuclear charge and covalent radius, the Sanderson scale based on atomic size, and the Allen scale based on the average energy of the valence electrons.


What do you mean by "stable"? In a vacuum, $\ce{Li+}$ is quite stable. In water, Li ionizes readily to $\ce{Li+}$, with $\ce{OH-}$ accepting that electron. Of course, in vacuo, if an $\ce{Li+}$ ion were to be placed near an $\ce{Li-}$ ion, they'd quickly redistribute the charge.

There is also a difference between chemical (thermodynamic) stability and kinetic stability. Lifting a ball up a hill, or lifting an electron to a higher level, gives it potential energy, but if the ball rests in a cup on the hill, or the electron is in a metastable state, it will not go back on its own without a push.

Thermoluminescence is a case in point. Over time, some materials such as ceramics absorb radiation from cosmic sources and nearby radioactive minerals, putting atoms into excited states. Upon reheating, electrons hop back to lower states, emitting light. This is used to date ancient pottery, since the metastable states are preserved for millenia.

Another example of metastability is diamond. It takes considerable pressure and a specific temperature range to convert carbon, e.g. graphite, to diamond -- but once that pressure is removed, graphite is the preferred state. However, once formed, diamond is stable at room temperature for eons.

See Difference between thermodynamic and kinetic stability, answered by thomij, for a much more complete answer.


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$^\circ$.


At what pKa value do we consider an acid strong?

Your question is not as easy answered as you might want it to be. The reasons for that are mainly given in the answer by Zen to What is a "Strong" Acid?
The main problem is, that there hardly can be an absolute acid strength and you will always have to take the reference system into account. Another problem arises, that there is no consistent (or official by IUPAC) definition of the acid constants. Usually, when you are talking about $\mathrm{p}K_\mathrm{a}$ values, then you are most likely already implying aqueous solution.

It is a quite accepted classification of strong acids (in water), to state that they completely dissociate. Hence in the equilibrium is entirely shifted to the right side:
$$\ce{HA (aq) + H2O (aq) <=> A- (aq) + H3O+ (aq)}.$$
The equilibrium constants for this reaction is
\begin{align}
K  &=       \frac{a(\ce{A-})\,a(\ce{H3O+})}{a(\ce{HA})\,a(\ce{H2O})}
    \approx \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})\,c(\ce{H2O})}
    \approx \frac{K_\mathrm{a}}{c(\ce{H2O})}\\
K_\mathrm{a} &=
    \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})}\\
\mathrm{p}K_\mathrm{a} &= -\lg K_\mathrm{a}.
\end{align}

For a strong acid we can see that the concentration of undissociated acid is (nearly) zero. Therefore we write
\begin{align}
\lim_{c(\ce{HA})\to0} \left(K_\mathrm{a}\right) &= \infty&
&\therefore&
\lim_{c(\ce{HA})\to0}\left(\mathrm{p}K_\mathrm{a}\right) &= -\infty.
\end{align}

Strictly speaking, in aqueous solutions the acidity constants cannot be measured, since the hydronium ion is the strongest acid in this medium.

So in priniciple, the value that separates strong from weak acids would be the $\mathrm{p}K_\mathrm{a}$ of the hydronium ion. With that we arrive at a completely new problem, which is quite efficiently discussed here: What is the pKa of the hydronium, or oxonium, ion (H3O+)?

From this question we can take home the suggestion the value of $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = -0.7$ for the purpose of comparing it to other acids. In other words $\mathrm{p}K_\mathrm{a}(\text{strong acid}) < -0.7$ (in water). Taking this literal, however, might lead to many misconceptions, and using a relative approach of “this acid is stronger than this acid” should  be preferred in any case.


Why is the enthalpy of combustion higher for n-butane as compared to 2-methylpropane?

Take 4 moles of carbon and 5 moles of H2. Synthesize a mole of n-butane and measure the heat evolved (125.6 kJ). Then repeat the process, but make 2-methyl propane (you'll measure 134.2 kJ). Then, with sufficient oxygen,  prepare 4 moles of CO2 and 5 moles of H2O from each of the previous two procedures and measure the heat of combustion (more for n-butane).

The total heat evolved to form the final oxidation products from carbon and H2 will be the same, but since the paths are different, the heats evolved in the two steps of each procedure will be different. n-Butane evolves less heat upon formation, therefore can (must) evolve more heat in the next step when it is oxidized to final products.


There are a great many of each type of these filters.

Interference filters are optimized for different wavelengths, different bandwidths, transmission vs. reflection etc. by controlling the materials and the thicknesses of the layers. A commercial supplier, Edmund Optics, shows how to specify and order filters.

In a similar manner, absorptive filters depend on the chemicals used to make them. For example, a solution of $\ce{CuSO4}$, copper sulfate, in water removes wavelengths between ~600-750 nm, but letting blue light through..


Excess hydrazine hydrate to neutralise the AcOH and ensure complete reaction of the diketone.

Excess AcOH to protonate all anions arising from the LiHMDS, mostly to completely protonate the anion of the diketone.

Excess enolate (and LiHMDS) to ensure all the enolate reacts. The diketone product is more acidic than the enolate and will hence protonate the enolate, so you need 2 eqs or more of base/enolate to get complete reaction.

How much NaOH for the wash at the end? More than eqs of AcOH. Don't see why they specify a base wash as everything you want to remove into the aq phase would be aq soluble anyway.


One of the most important issues in understanding bond energy in chemistry is the question: energy relative to what?

The formation of bonds from (mostly hypothetical) atomic elements releases energy (or they wouldn't be bonded.) Solid carbon (graphite or diamond) has less energy than a cloud of carbon atoms so it could be said graphite has less energy relative to a cloud of atomic carbon atoms. We would describe this energy as "bond energy". But this is the energy released by the hypothetical process of forming lists of C-C bonds from isolated carbon atoms.

But there is another subtle complication if we want to understand reactions. sometimes, when chemists write a reaction scheme they don't specify all the detail. So C +O2->CO2 doesn't actually mean atomic carbon atoms plus oxygen molecules give carbon dioxide. Most chemists would assume that the carbon starts in the state it is normally found as in laboratories: graphite (so not isolated atoms but a molecular solid containing lost of C-C bonds). This terminology seems to have confused some people. But chemists don't like to waste space in formulae redundantly specifying things most of their colleagues will already know. The default state of an ingredient in a reaction is assumed unless specified otherwise and this usually means the form the compound or element takes in a bottle on the lab shelf.

So how can we say that carbon stores energy? Well we have to ask the question, relative to what? In the case of the question the answer is relative to carbon dioxide gas. The point here is that the double bonds in CO2 are stronger than the combined bonds in oxygen gas and graphite (so they are lower energy) This means that the reaction between carbon and oxygen releases energy (it takes less energy to break all the O-O and C-C bonds than is released when new C=O bonds are formed. So elemental carbon "stores" energy relative to carbon dioxide.

The claim that carbon "stores" solar energy absorbed by plants is a little simplistic (mainly because plants don't store carbon but usually store more complicated molecules like sugars or poly-sugars like lignin or cellulose). Dead plants are converted to more carbon like compounds (oil is mostly hydrocarbons, but coal is mostly carbon) under some geological conditions. What plants actually do at the start of this process is to convert carbon dioxide into sugars using energy input from sunlight. They do this to store energy and to grow by converting the sugars into the structures of their leaves, stems and trunks.

We can reverse that process by burning the plants or their degradation produce like oil and coal in air to reproduce the carbon dioxide and release energy. In this highly simplified sense, carbon is a store of "solar" energy.

But the key to understanding what is meant is to understand the chemical reaction involved in the creation or burning of the carbon. Then you can understand what energy is stored and how it is stored.


Le Chatelier’s principle when pressure of gases increased with the same stoichiometric coefficient

Your thinking is on the right track. However, what you really need to look out for is number of gas moles.

In order to understand this, think in terms of the equilibrium constant. The equilibrium constant for this reaction is:
$$K_\text{p}=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$
Remember, since carbon is a solid, changing the volume/pressure is not going to change anything about it, so it is not in the equilibrium constant.

Since we are introducing a supposed change into the system, we have to write its reaction quotient. The reaction quotient, $Q$, is just a way to determine which way the reaction will have to shift in order to return to equilibrium. We write it in pretty much the same way:
$$Q=\frac{P_{\text{CO}_2}P_{\text{H}_2}}{P_{\text{H}_2\text{O}}}$$
Even though it looks the same, remember that we are calculating this after introducing a change. So what we really plug into this equation is the following, where $x$ is the factor by which we increase pressure (suppose that $P_{\text{CO}_2}$, $P_{\text{H}_2}$, and $P_{\text{H}_2\text{O}}$ are the equilibrium pressures):
$$Q=\frac{(P_{\text{CO}_2}\cdot x)(P_{\text{H}_2}\cdot x)}{(P_{\text{H}_2\text{O}}\cdot x)}$$
$$Q=\frac{x^2P_{\text{CO}_2}P_{\text{H}_2}}{xP_{\text{H}_2\text{O}}}$$
Since we are increasing pressure, $x$ is obviously positive. Thus, $Q>K_\text{p}$.

Due to Le Chatelier's principle, since we introduced a stress into the system, the system will shift away from that stress. $Q$ and $K_\text{p}$ are a good way of mathematically representing this stress. $Q$ is essentially a representation of the system's current state, and $K_\text{p}$ is the representation of the system's ideal state.

So, how do we reach this ideal state? It's simple; we want $Q$ to equal $K_\text{p}$. Obviously this will happen through a change in pressure. Since $Q$ is too big, we need to decrease the numerator in order to make it equal to $K_\text{p}$. The numerator represents the pressure of the products, so the reaction will shift away from the products, using up more products and creating more reactants, until it reaches equilibrium.

Hope that helps.


Water purification procedure that does not involve a flocculant of any kind?

This of course depends on the scale and intent of the purification/treatment process, but flocculation is but one step in the water treatment process. So, I'm not sure whether you are asking if you can purify water without flocculation, or if you can get flocculation/aggregation to occur without the addition of floccing agents.

The simplest answer to water purification, on a small scale, is filtration and distillation. Flocculation serves to remove solid particles from the water which contribute to turbidity, but which may also include pathogens that are of real concern to human safety. On larger scales (household to community), incredibly effective methods for removing these types of contaminants include various forms of filtration. Some of the first successful water treatment plants on a large scale used sand filtration, and fast and slow sand filters are still used widely today. There are many commercially available drinking water filters for the removal of sediment, microbes and chemicals which are highly effective. Of course, on an industrial scale, filters of this nature would require frequent changing, and this is where sediment removal via flocculation is most effective in reducing costs. You could Google 'water purification in developing worlds' for a range of water purification methods that do not involve flocculation.

On the other hand, inducing flocculation without the addition of flocculant again depends on nature of the contaminant you wish to remove. The biofuel industries are pursuing this goal, developing methods of harvesting valuable microalgae stocks. Autoflocculation (or bioflocculation) of microalgae is known to occur as a function of pH, which can be controlled by restricting CO2 levels. Again a Google search on 'microalgae autoflocculation' will give you many hits. Another method is the light induced autoflocculation of microalgae. There is an interesting post on ResearchGate that may be of some interest to you if you are focused on flocculation of microalgae (if not, then ignore!)


Your diagram is correct.  Here is a link to an on-line energy diagram for propane and here is the diagram itself


How to cheaply extract iodine from an alcoholic solution?

Alcoholic solution of iodine

What firstly comes to mind is that you meant "iodine tincture".
This question must have been frequently asked, since answers are available out there with different degrees of reliability. You must first follow the safety instructions:

Do your experiment somewhere that the noxious fumes won't be able to do harm to anyone. It's best to work under a fume hood.
You're gonna be dealing with strong acids. They're very corrosive when in contact with skin, metals or anything you can imagine (except glass).
Iodine tends to sublime. If not contained wisely, you'll get irritated skin, along with damages to your eyes (See about iodine hazards here).

In short, what you should do is:

Pour the tincture in a beaker, or any container. The important thing to note is to have "control" over what's happening in there.
Add water to the solution. The volume of the water you add should preferably be equal to that of tincture.
Add $\ce{HCl}$ to the solution, $\approx \pu{5-7 ml}$.
Add bleach to the solution, preferably equal or more than that amount of $\ce{HCl}$ you added.
Wait...wait...wait. For a relative amount of time, you should leave the solution for the reactions to happen.
You've past the hard part! Congratulations! Now, filter the dilute solution with a coffee filter.

Additional tips:

Iodine crystals sublime, so you shouldn't leave them in open air for long.
You should at least be wearing appropriate eye and skin protection. If you don't want them, do not do the experiment!

Wikihow has additional tips in case of any failures.
Sources:
http://www.wikihow.com/Extract-Iodine
http://www.sci-spot.com/Chemistry/iodine.htm


The effect of HCL concentration on the Solubility of Hydroxyapatite

The $Ksp$ expression for hydroxyapatite is: $\ce{ Ca5(PO4)3(OH)_{(s)} -> 5Ca^{2+}_{(aq)} + 3PO4^{3-}_{(aq)} + OH-_{(aq)}}$.

When $\ce{HCl_{(aq)}}$ is added, it reacts with the dissolved $\ce{OH-}$ to form water, thereby decreasing the $\ce{OH-}$ concentration. Le Chatlier's principle predicts that more hydroxyapatite will dissolve (shift to the right).


Conditions for the reduction of carboxylic acids by LiAlH4

When the reaction is done, you need to quench it to (1) destroy the remaining LiAlH4 (2) protonate the lithium alkoxide to give you the alcohol. The aluminium salts formed can be separated from the alcohol. If you don't quench the reaction it makes separation essentially impossible.

Of course, if you add the water before the reaction is done, then that defeats the purpose of using an anhydrous environment.


Assumptions to be taken when rearranging terms in first law of thermodynamics

This equation can not be applied in general because it can't describe the amount of heat transferred to an incompressible solid or liquid at constant pressure.
For these cases, it would predict that $\mathrm dQ=0$ (which is not necessarily correct).  If it is not generic, it also cannot describe real gases beyond the ideal gas region.  Therefore, the equation must apply exclusively to an ideal gas. The presence of the ideal gas constant $R$ in the equation is a dead giveaway.
For a reversible path,
$$\mathrm dQ=T\mathrm dS=C_v\mathrm dT+P\mathrm dV$$
But, for an ideal gas,
\begin{align}
&\mathrm dT =\frac{P\mathrm dV+V\mathrm dP}{R}\\
&C_v+R = C_p
\end{align}
Therefore,
\begin{align}\mathrm dQ&=C_v\frac{P\mathrm dV+V\mathrm dP}{R}+P\mathrm dV \\ &=\frac{C_v}{R}V\mathrm dP+\frac{(C_v+R)}{R}P\mathrm dV \\
\implies\mathrm dQ&=\frac{C_v}{R}V\mathrm dP+\frac{C_p}{R}P\mathrm dV
\end{align}
This equation applies only to an ideal gas and a reversible path.


How do amines react with flue gases to absorb CO₂?

In addition to the reaction in permeakra's answer, some amines may react with $\ce{CO2}$ in absence of water also. This type of reaction is a Lewis acid-base reaction forming a zwitterionic complex: $\ce{R3N + CO2 -> R3NCO2}$.

MDEA adds some stability to this complex. The alcohol groups can stabilize the carboxylate group through hydrogen bonding.


Why can there not be more than one sigma bond in a set of bonds?

There can be, even in simple carbon compounds. Bent bonds, tau bonds or banana bonds; whatever you might like to call them were proposed by Linus Pauling; Erich Hückel proposed the alternative $\sigma - \pi$ bonding formalism. Hückel's description is the one commonly seen in introductory texts, but both methods produce equivalent descriptions of the electron distribution in a molecule.

In order to better understand the bent bond model let's first consider its application to cyclopropane and then move to ethylene.

In cyclopropane it has been found that significant electron density lies off the internuclear axis, rather than along the axis.

(image source)

Further, the $\ce{H-C-H}$ angle in cyclopropane has been measured and found to be 114 degrees.  From this, and using Coulson's Theorem $$\ce{1+\lambda^2cos(114)=0}$$ where $\ce{\lambda^2}$ represents the hybridization index of the bond, the $\ce{C-H}$ bonds in cyclopropane can be deduced to be $\ce{sp^{2.46}}$ hybridized.  Now, using the equation $$\ce{\frac{2}{1+\lambda^_{C-H}^2}+\frac{2}{1+\lambda_{C-C}^2}=1}$$ (which says that summing the "s" character in all bonds at a given carbon must total to 1) we find that $\ce{\lambda_{c-c}^2~=~}$$\mathrm{3.74}$, or the $\ce{C-C}$ bond is $\ce{sp^{3.74}}$ hybridized.

Pictorially, the bonds look as follows.  They are bent (hence the strain in cyclopropane) and concentrate their electron density off of the internuclear axis as experimentally observed.

We can apply these same concepts to the description of ethylene. Using the known $\ce{H-C-H}$ bond angle of 117 degrees, Coulson's theorem and assuming that we have one p-orbital on each carbon (Hückel's $\sigma - \pi$ formalism), we would conclude that the carbon orbitals involved in the $\ce{C-H}$ bond are $\ce{sp^{2.2}}$ hybridized and the one carbon orbital involved in the $\ce{C-C}$ sigma bond is $\ce{sp^{1.7}}$ hybridized (plus there is the unhybridized p-orbital). This is the "$\ce{sp^2}$" description we see in most textbooks.

Alternately, if we only change our assumption of one pi bond and one sigma bond between the two carbon atoms to two equivalent sigma bonds (Pauling's bent bond formalism), we would find that the carbon orbitals involved in the $\ce{C-H}$ bond are again $\ce{sp^{2.2}}$ hybridized, but the two equivalent $\ce{C-C}$ sigma bonds are $\ce{sp^{4.3}}$ hybridized. We have constructed a two-membered ring cycloalkane analogue!

Hybridization is just a mathematical construct, another model to help us understand and describe molecular bonding.  As shown in the ethylene example above, we can mix carbon's atomic s- and p-orbitals in different ways to describe ethylene. However, the two different ways we mixed these orbitals

one s-orbital plus two p-orbitals plus one unmixed p-orbital (Hückel's  $\sigma - \pi$), or
two s-orbitals plus 2 p-orbitals (Pauling's bent bond)

must lead to equivalent electronic descriptions of ethylene. As noted in the Wikipedia article (first link at the beginning of this answer),

  There is still some debate as to which of the two representations is
  better,[10] although both models are mathematically equivalent. In a
  1996 review, Kenneth B. Wiberg concluded that "although a conclusive
  statement cannot be made on the basis of the currently available
  information, it seems likely that we can continue to consider the σ/π
  and bent-bond descriptions of ethylene to be equivalent.2 Ian
  Fleming goes further in a 2010 textbook, noting that "the overall
  distribution of electrons [...] is exactly the same" in the two
  models.[11]

The bent bond model has the advantage of explaining both cyclopropane (and other strained molecules) as well as olefins.  The strain in cyclopropane and ethylene (e.g. heats of hydrogenation) also make intuitive sense with the bent bond model where the term "bent bond" conjures up an image of strain.

So, bent bonds are a mix of both sigma and pi properties and multiple bent bonds  can be used to describe the bonding between adjacent atoms.


Is pigment grade titanium dioxide likely to contain any impurities that are toxic?

According to the MSDS for titanium dioxide, the substance itself is suspected of causing cancer. See, for example, this link.

My opinion is that, in general, it's not a good idea to use reagents from unknown sources for any purpose, especially something like sunblock.


E2 step in dehydration of primary alcohol using concentrated sulfuric acid

What actually happens, as always, depends on the exact conditions and concentrations which you did not add to your question. However, you do state whether the catalyst is regenerated; as such, I assume you are talking about the following reaction:

$$\ce{C2H5OH ->[\ce{[H2SO4]}] C2H4 + H2O}\tag{1}$$

I.e. a dehydration of ethanol using catalytic sulfuric acid.

The first thing we have to consider is the $\mathrm{p}K_\mathrm{a}$ values of sulfuric acid and ethanol (and water; which will be present anyway unless you accidentally found dry sulfuric acid). Using a very quick ballpark, we can assume: $$\mathrm{p}K_\mathrm{a}(\ce{H2O}) \approx \mathrm{p}K_\mathrm{a} (\ce{C2H5OH}) > \mathrm{p}K_\mathrm{a}(\ce{H2SO4})\tag{2}$$

Since sulfuric acid is catalytic, we can assume that the first (complete; non-equilibrious) step will be:

$$\ce{C2H5OH + H2SO4 -> C2H5OH2+ + HSO4-}\tag{3}$$

$\mathrm{p}K_\mathrm{a,2} (\ce{H2SO4}) \approx 2$, which is only two units less than water. Therefore, — under the assumption that we added $10~\mathrm{mol}\text{-}\%\ \ce{H2SO4}$ — we might assume that some additional molecules of ethanol (and/or water) are protonated under equilibrium conditions:

$$\ce{C2H5OH + HSO4- <<=> C2H5OH2+ + SO4^2-}\tag{4}$$

Thus, part of the sulfuric acid will be present as sulfate, the majority as hydrogen sulfate. Sulfate is a stronger base (albeit still weak) and thus the final mechanistic step is probably better written using sulfate rather than hydrogen sulfate:

$$\ce{H-C2H4-OH2+ + SO4^2- -> HSO4- + H2O + C2H4 ^}\tag{5}$$

Of all the bases present at equilibrium, sulfate is the strongest as its conjugate acid’s $\mathrm{p}K_\mathrm{a} > 0$. The general order of basicity of all compounds in solution is:

$$\ce{SO4^2-} > \ce{H2O} \approx \ce{C2H5OH} > \ce{HSO4-}\tag{6}$$

Finally, note that the reaction does not have to proceed via $\mathrm{E2}$ — an $\mathrm{E1}$ mechanism is almost equally viable.


At any time, the number of decays per reasonable$^1$ time unit is proportional to the amount of the isotope present. (This is expressed by the first-order rate equation for a one-decay process.) Therefore, it does not matter whether one looks at the amount or the number of decays. It may be easier to measure the number of decays by e.g. a Geiger counter and prefer it for that reason.

It now matters what you understand to be the rate: the number of decays per second or the percentage of nuclei decaying at any given time. The first becomes less over time, the second one is a constant for each isotope.

$^1$ You must have a reasonable number of decays within your time unit. If the half-life is long, it makes little sense to look at millisecond slices.


What is the relation between chemical thermodynamics and chemical kinetics?

No. Both I think are equally theoritical or experimental. You must study both seperately. Thermodynamics doesn't speak about rates. While kinetics doesn't tell you stability.
If a reaction produces more than one product, thermodynamics will tell you which is more stable, while that doesn't necessarily mean it is the major product. It may happen that rate of formation of other product is more under certain conditionds. Read about thermodynamic vs kinetic product and you will get it.


1-bromo-2-fluorobenzene reaction with magnesium and furan

The overall pathway described is shown below. Treatment of the aryl bromide with magnesium gives a Grignard, which is equivalent to a carbanion. The carbanion can eliminate fluoride to give benzyne. Being highly strained, benzyne is a fantastic dienophile and reacts with furan through a Diels-Alder reaction.

Most of the techniques to generate benzyne use the same strategy: generate a carbanion next to a leaving group. The classic method taught in introductory organic chemistry is by deprotonation of an aryl halide with a very strong base (sodium amide), which eliminates the halide. The strategy shown here generates the carbanion using the milder method of forming a Grignard.

The elimination is possible because carbanions are very good at kicking out leaving groups. It's commonly taught that fluoride is a poor leaving group, but the end of that sentence is that it's a poor leaving group for SN2 reaction or compared to chloride, bromide, and iodide. Later, it's taught that enolates (a reasonably stable class of carbanions) are reactive enough to eliminate alkoxides/hydroxide. So it makes sense that an unstabilized carbanion as here should be able to kick out a relatively poor (but not that terrible) leaving group.

The Stoltz group has a nice summary of benzyne's history, generation, and reactivity.


Why does the light catalyzed radical addition of tetrachloromethane to an alkene produce only a single regioisomer?

While this reaction is highly regioselective, it is not 100% regioselective. Consider the nearly identical reactions:

$$\begin{align}
\ce{CCl4 + H2C=CH-(CH2)5CH3 ->[(PhCO2)2][90\text{-}105ºC] Cl3C-CH2-CHCl-(CH2)5CH3} \hspace{.63cm} \text{75%}^{[1]}\\
 \ce{CCl4 + H2C=CH-COH-(CH3)2 ->[(PhCO2)2][80ºC] Cl3C-CH2-CHCl-CH(OH)-(CH2)5CH3} \hspace{.63cm} \text{70%}^{[2]}\\
\end{align}
$$

Granted these reactions use radical initiators instead of light, but that wouldn't make the difference between 100% and 70-75% regioselectivity. The reality is that at the start of the reaction, $\ce{.CCl3}$ and $\ce{.Cl}$ are both competing to attack the alkene, and $\ce{.Cl}$ does in fact attack faster. Looking at the activation energies for the abstraction of hydrogen atoms by different radicals, we gain insight into their relative stability:

$$ \small
\begin{array}{lcc}
\hline
\text{Radical} & \ce{CH3-H} & \ce{CH3CH2-H} & \ce{(CH3)2CH-H} & \ce{(CH3)C-H}^{[3]} \\
\hline
\ce{.F} & \text{1-1.5} & \text{<1} & & \text{<1} \\
\ce{.Cl} & \text{3.4} & \text{1.1} \\
\ce{.Br} & \text{17.5} & \text{13.0} & \text{9.5} & \text{6.9} \\
\ce{.CH3} & \text{14.0} & \text{11.6} & \text{9.6} & \text{8.1} \\
\ce{.CF3} & \text{10.9} & \text{8.0} & \text{6.5} & \text{4.9} \\
\ce{.CCl3} & \text{17.9} & \text{14.2} & \text{10.6} & \text{7.7*} \\
\hline
_{^* \text{units are } \mathrm{kcal\ mol^{-1}}}
\end{array}
$$

As this reaction continues, however, chlorine atoms are abstracted from $\ce{CCl4}$, creating the product along with $\ce{.CCl3}$ radicals. Abstraction of $\ce{.CCl3}$ is highly disfavored due to steric effects. Thus, the radical chain reaction is as follows:

$$\ce{CCl4 ->[$h\nu$] .CCl3 + Cl.}$$

$$\ce{.CCl3 + CH2=CH-R ->  Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R}$$

$$\ce{Cl3C-CH2-\stackrel{\displaystyle .}{\ce{C}}H-R + CCl4 ->  Cl3C-CH2-CHCl-R + .CCl3}$$

The less than ideal yields of the radical addition of $\ce{CCl4}$ can be attributed to the strength of the $\ce{C-Cl}$ bond (higher activation energies correspond to slower rates of reactions). Because $\ce{CCl4}$ is in competition with the polymerization of the $\ce{.CCl3}$-alkene adduct, $\ce{BrCCl3}$ and $\ce{CBr4}$ are generally preferred for their higher selectivity, owed to the lower strength of the $\ce{C-Br}$ bond. One such example is shown below.

$$\ce{BrCCl3 + H2C=C(C2H5)2 ->[$h \nu$] Cl3C-CH2-CBr(C2H5)2} \hspace{1cm} {\text{91%}^{[4]}}$$

$^{[1]}$M. S. Kharasch, E. W. Jensen, and W. H. Urry, J. Am. Chem. Soc., 69, 1100 (1947).
$^{[2]}$P. D. Klemmensen, H. Kolind-Andersen, H. B. Madsen, and A. Svendsen, J. Org. Chem., 44, 416 (1979).
$^{[3]}$F. A. Carey, R. J. Sundberg, Advanced Organic Chemistry Part A: Structure and Mechanisms, Springer Science, New York, 2007, pg. 1034.
$^{[4]}$M. S. Kharasch and M. Sage, J. Org. Chem., 14, 537 (1949).


Exocyclic vs. endocyclic double bonds in E1 elimination

What will be the major product of this elimination reaction?

In the acid-catalyzed dehydration of 1-methylcyclohexanol, both 1-methylcyclohexene and methylenecyclohexane are formed in ca. an 85:15 ratio.  In your example we have two more substituents on the exocyclic double bond which will serve to further stabilize the exocyclic double bond and increase the amount of the exocyclic isomer. I don't know for sure how much it will push the equilibrium towards the exocyclic isomer, but I might guess that the product mixture would be at least 50:50, if not more of the exocyclic isomer - but I would expect a mixture.

  what influence will kinetics have on this reaction?

Well you won't see the kinetics unless you go out of your way and look for it.  Factors that influence which isomer is kinetically favored include:

statistical factors - there are 4 ring hydrogens that can be eliminated vs. only 1 on the isopropyl group; this favors kinetic formation of the endocyclic isomer
alignment of the hydrogen with the p-orbital on the carbocation carbon - the hydrogen on the isopropyl group can better align with the p-orbital than those hydrogens in the ring, however the cyclohexane ring can easily distort; so while this isn't a major factor it favors the exocyclic isomer
steric factors - I agree with your analysis, but would guess that the difference is small; still sterics would slightly favor the endocyclic isomer

It seems pretty close to call, but if I had to guess I'd say that the statistical factor wins out (and the steric argument further helps) and the endocyclic isomer would likely be the the kinetic product.

But again, in acid this is a fast equilibrium; the products will quickly equilibrate and the thermodynamically controlled product distribution will be observed.  Maybe if you run the reaction at low temperature with a strong dehydrating agent ($\ce{POCl3}$) you could capture some meaningful kinetic information.


Water is already very 'happy' with his own guys due to strong hydrogen bonds . While van der waals force is so weak that its presence is so little for it to be seen while there is hydrogen bond. So even though there is some dipole induced it isn't enough.

  the total energy is lower than when they are mixed. The system is more stable if unmixed. -Chin Yeh


Derivative of Gibbs Energy wrt Temperature at constant Internal Energy

The starting point is the two equations $$dG=\left(\frac{\partial G}{\partial T}\right)_VdT+\left(\frac{\partial G}{\partial V}\right)_TdV$$ and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$From the second equation, you determine the partial of V with respect to T and constant U by setting dU=0.  I leave the rest of the math up to you.

Chet


Consider the energy required to elevate electrons from one "shell" to another, and the frequency of visible light that has this amount of energy.  The substance is more likely to absorb wavelengths of light matching the energy difference of that band gap, and reflect or transmit others.  The colors NOT absorbed can reach your eye as an observer, so (assuming a white light source) you see white (which is all the colors together) minus the colors absorbed by the material (matching the band gap).  For example, gold (Au) absorbs the blues and greens and leaves just the colors adding up to what you now think of as "gold."  If your light source is not white, what your eyes see will also be missing any colors your light source was missing.
The crystal structure can also make a difference, e.g. compare the colors of carbon (C) as graphite vs. diamond.

Useful links: See also this same question on ResearchGate, WebExhibits, and Quora.


Is the transformation of pi bonds to sigma bonds always energetically favorable?

When thinking about questions along these lines, I always try to remember the bent-bond description of ethylene.  In all aspects related to physical reality (structure, electron distribution, etc.) the bent bond model pictured below on the right is equivalent to the pi model shown on the left (at least within the scope of all measurements made to date).

The "nice" thing about the bent bond model is that it casts ethylene as a highly-strained, two-membered ring.  When wouldn't releasing this strain and forming sigma bonds be an energy lowering process.  Aside from conjugated examples, I think that transformation of a strained, ethylenic system to a saturated, sigma-bonded system would always be energy lowering.

EDIT: Further Thoughts

Instead of analyzing hydrogenations or other reactions where the starting material and product have a different number of atoms, let's consider the following isomerization reaction.

Thermodynamically, cyclohexane is the most stable $\ce{C6H12}$ isomer in this series by a significant margin.  All we've done is convert a C-C pi bond into a C-C sigma bond and convert 3 $\ce{C(sp^2)-H}$ bonds into 3 $\ce{C(sp^3)-H}$ bonds.  Which (at least as I read it) was the OP's premise.  So I would say that, in carbon-hydrogen systems, pi bonds generally prefer to transform into sigma bonds, as pi bonds are inherently destablilized due to poor overlap (pi model) or strain (bent bond model).

That said, I retract what I wrote above that carbon-carbon pi to sigma transformations would "always" be energy lowering.  Consider the photolysis of benzene to produce dewar benzene, prismane and benzvalene.  These conversions are examples where one or more pi bonds have been converted to one or more sigma bonds and the reaction is not energetically (thermodynamically) favorable.

The products are much less stable (due to the strain in the small rings that have been created) than the starting benzene.  Because of these relative stabilities, the forward reaction does not occur thermally (in fact, thermally, each of those 3 products will revert to benzene), but by pumping in excess energy photochemically, the transformation can be brought about.

To summarize, I would maintain that generally, in model, carbon-hydrogen systems, ethylenic linkages are destabilized (due to poor overlap in the pi model, or strain in the bent bond model) and will thermodynamically prefer to exist as thermodynamically more-stable saturated linkages.


Angle of Light Source for Absorbance Spectroscopy

Yes you could do what you propose provided that the light is properly collimated so that it all reaches the detector. It is quite common to place mirrors inside the sample housing of a spectrometer to  take the light out into the lab, through a sample (usually a weakly absorbing gas ) and then use a mirrors to get it back into the spectrometer. Multi-pass cells can also be used to do this.

In atmospherical studies very long path lengths (100's metres) are sometimes used to identify trace amounts of various species. Of course if you do something like this you have to make sure extraneous light does not reach your detector.


Are these three molecules optical isomers?

There is no exocyclic methylene in structure 1.  Is it not an optical isomer of 2 and 3.

Likely there is a pericyclic rearrangement between 1 and 2/3 which would formally render them tautomers.

2 and 3 are diastereomers.


Franck-Condon factors in photoelectron spectroscopy

$\psi^\prime_{vib, emitted}$ is an energy normalized continuum wavefunction. The Franck–Condon factors involving bound ($\psi_v$) and continuum states ($\psi_E$) are actually Franck–Condon densities that are functions of the energy.

The Franck–Condon factors and densities are related by a closure relation
$$ \sum_{v^{\prime\prime}}|\langle \psi_{v^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 + \int |\langle\psi_{E^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 dE^{\prime\prime} = 1$$

References:

[1] J. Tellinghuisen, The Franck-Condon Principle in Bound-Free Transitions, in Photodissociation and Photoionization, edited by K.P. Lawley (John Wiley & Sons, Ltd, 1985).

[2] R.J. Le Roy, Comput. Phys. Commun. 52, 383 (1989).


Where does the extra electron go in ionic compounds?

In $\ce{OH-}$, that extra electron is on the Oxygen atom.  Oxygen has a very large electronegativity, so you can be sure that extra electron is with the $\ce{O}$ not the $\ce{H}$.  A lone oxygen has 6 electrons in its valence shell (2s2, 2p4), and wants to get to 8 (2s2, 2p6).  In the $\ce{OH-}$ ion, both atoms are relatively "happy", the $\ce{H}$ has 2 electrons in its valence shell (1s2) shared with Oxygen in a bond, and the Oxygen has 8 electrons in its valence shell (2s2, 2p6), 6 unpaired electrons and 2 shared with hydrogen.

In general, in an anion the extra electron will be going onto the most electronegative atom.  In a cation the positive charge will go to the lest electronegative atom.

Here's a helpful picture, your extra electron is in red:

picture reference: http://chemistrypartner.blogspot.com/2009/10/drawing-lewis-electron-dot-structure-or.html

As for your question about $\ce{OH+}$, it would never be stable because that oxygen is going to tear some electrons off of the first thing it sees, because it REALLY wants that full octet.  If you somehow made some $\ce{OH+}$, my guess would be that the $\ce{H+}$ would just dissociate, and your $\ce{O}$ would find another $\ce{OH+}$'s $\ce{O}$ and make $\ce{O2}$ immediately.


Transition state - multidimensional surfaces

The picture below shows a simple potential energy surface with a transition state at the maximum point on the reaction path, which is at the 'saddle point'. The reaction path is the solid black line. (Oscillations in this show that the reactant and product are vibrating). The contours are at different energy with large negative being lowest. This picture is based on a calculation using known experimental parameters.

Along the reaction path the energy is as small as it can be, even though it increases as we go along, because by moving to the right or left (perpendicular to the general direction of the path)  the energy is even higher. So the saddle point is just like a pass between high mountain peaks. At the highest point on the reaction path the only way to lower the energy is to go down, either to return on the original path or to continue to products. The time it takes to cross the transition state is definitely sub picosecond, < $10^{-12}$ s and possibly up to a 100 times faster.

It is not possible to visualise a multidimensional reaction path, so we have to make do with 2D plots, as below, of one reaction coordinate vs. the other with contours at different energy. In principle many similar plots could be needed for a complex reaction but very often most of the atoms in the molecule are bystanders and it possible then to only look at the separation and energy of the atoms in bonds being broken and made.

Trajectory of $\ce {Cl + H2}$ in a gas phase reaction. The H atom approaches the chlorine molecule at the right. The yellow circle shows the transition state.


Good software to perform simulations on polymer (absorption spectra)?

It depends very significantly on what you want to accomplish.

I'm not going to suggest software, since there are many programs that can (and are) used to predict optical absorption spectra of conjugated polymers.

First, you can either approximate the properties of the polymer by running several calculations on oligomers (e.g., dimer, trimer, tetramer, etc.) and extrapolating the polymer band gap. Because conjugated polymers are similar to the particle-in-a-box model, you can roughly plot the HOMO-LUMO gap or excitation energy vs. 1/(# of repeat units) and get an approximately linear fit.

If you want more information, you could instead run a calculation on a polymer in a periodic unit cell, which will give you the electronic structure of the infinite system (but no information on disorder).

As suggested in the comment above, you can use multiple programs and different computational approaches depending on:

What kind of polymer
What kind of information you want
What computational resources you have
How accurately you want the information

The two most common methods for predicting optical absorption energies are the ZINDO/S semiempirical method, and TDDFT. The latter is more accurate particularly for predicting absorption intensity from the oscillator strength of the excitation energies.


A periodic table coupled with simple electrostatic arguments should suffice to provide at least a rough ordering of ions and atoms. For the ions:

We expect POSITIVE ions to be SMALLER than the corresponding atom because there are fewer electrons than protons => an excess of POSITIVE charge => the electron cloud is more attracted to the nucleus on average and becomes more 'concentrated' => $\ce{Na+}$ is 'small' and $\ce{Mg^{2+}}$ is 'very small'.
We expect NEGATIVE ions to be LARGER than the corresponding atom because there are more electrons than protons => an excess of NEGATIVE charge => the electron cloud becomes more diffuse to compensate, lowering the Coulombic interactions => $\ce{F-}$ is 'large', $\ce{O^{2-}}$ is 'very large'.

Even though $\ce{Na+}$ has a lower principal quantum number for its outer shell than $\ce{Mg^{2+}}$, Mg is twice as charged and I believe this can be more important. $\ce{F-}$ and $\ce{O^{2-}}$ have the same principal quantum number for their outer shells, but $\ce{O^{2-}}$ is more charged and accordingly, we expect that $\ce{O^{2-}}$ has a larger radius.

$\ce{Ne}$ carries no unbalanced charge and has comparable valence principal quantum number to the ions, so we expect its radius to fall somewhere between the positive ions (which are smaller) and the negative ions (which are larger). Go over these arguments in your head and try to put the pieces together yourself from here.

(Note: I wouldn't be willing to stake my life on this, so some community validation or corrections would be much appreciated)


Why does bubbling carbon dioxide through calcium hydroxide result in a precipitate?

The important thing to consider is the chemistry of carbon dioxide in water, forming carbonic acid. Note also that calcium hydroxide is at edge between being a strong or a weak base.

$$\begin{align}\
\ce{CO2 ^ + H2O &<=> CO2 . H2O} \tag{1} \\
\ce{CO2 . H2O &<=> H2CO3} \tag{2} \\
\ce{H2CO3 &<=> H+ + HCO3^-} \tag{3} \\
\ce{HCO3^- &<=>  H+  + CO3^2- } \tag{4} \\
\ce{ Ca(OH)2 &<=> CaOH+ + OH- } \tag{5} \\
\ce{CaOH+ &<=> Ca^2+ + OH-} \tag{6} \\
\ce{Ca^2+ + CO3^2-  &<=> CaCO3 v} \tag{7}
\end{align}$$

Yes, bonds in $\ce{CO3^2-}$ are equivalent, with strength and length somewhere between a single and a double bond. It is similar as  for $\ce{SO4^2-}$, $\ce{SO3^2-}$, $\ce{NO3^-}$, $\ce{ClO4^-}$, with sharing the "doublebondness" within the former two, and sharing the charge in all of them.


It looks like you are not the first to get gummed up by the mono-acetoxy iodinane byproduct of this reaction. A seemingly simple solution is given in this article about converting an alcohol to ketone via DMP:

  The mono-acetoxy iodinane by-pdt can be converted to easier to remove by-pdts during workup by treatment with aq $\ce{NaOH}$ or $\ce{Na2S2O3}$.

This is not a procedure I have experience with specifically, so I hope this a do-able solution for your problem as I understand it.  Good luck regardless.


Handling vectors for double-bond graphics

I'm not sure I exactly know what you want, but such code is implemented in Avogadro and in other places.

The first thing to realize is that the unmarked atom in your plane is also your friend. You can calculate the plane of best fit using all 4 atoms.

You're making your work too hard, though. If you had a single bond, you simply draw a line or cylinder between atom i and j.

So the question at hand is how to displace two cylinders slightly in the plane of the bond. There are a few ways to do this. (Avogadro computes the plane of best fit for the molecule. Some people project towards the plane of the screen.)

My recommendation is to realize that atom k in your diagram is in one direction, and the unmarked atom is in the other. Let's call that other atom l. You can compute the vector between them: k - l. In practice, atom i is midway along this vector. So you want to displace the b vector in your diagram by a small amount, say 0.1 in one direction along that k - l vector, and 0.1 in the opposite direction.


This is a method to generate alternate Lewis structures of the same molecule. Here are two examples from the cumulative dissertation by Sascha Urbaczek[1]:

If you were to search for the left molecule in panel (1) using an image search or a SMILES string, you might miss the right molecule in that panel.

According to the RDkit document cited in the question, the software routinely generates the alternate position of double bonds, and then (in a second step they call "aromatization") labels the ring as aromatic. In panel (2), there are three possible Lewis structures contributing to the actual structure (i.e. there is resonance), so the software would have to generate all three to be able to search for identical structures.

Urbaczek, Sascha. A consistent cheminformatics framework for automated virtual screening. Ph.D. Thesis, Universität Hamburg, August 2014. URL: http://ediss.sub.uni-hamburg.de/volltexte/2015/7349/; URN: urn:nbn:de:gbv:18-73491; PDF via Semantic Scholar


Derivation of time probability distributions for single molecule kinetics

To see how this calculation is done, let us consider the simpler case of just the second step shown above, ie $\ce{AB ->[k_{f2}] C}$. Since we are interested in the situation where no $C$ has yet formed, the reverse reaction can be ignored. I will assume that this reaction is first order in $AB$, so $k_{f2}$ has units of $s^{-1}$.

In this case, the rate constant $k_{f2}$ represents the average frequency of the reaction.

Now let's consider that we establish an infinite number of evenly spaced time intervals of length $\Delta t$. When we consider the probability distribution, we need to consider two factors:

1) The probability of reaction occurring in an interval $t_n + \Delta t$ and

2) The probability that the reaction has not yet occurred at $t_n$

The first value is simply the product of the rate constant $k_{f2}$ and the length of the time interval: $k_{f2}\Delta t$.

The second probability is slightly more complicated, since it is the chance of the reaction $not$ occurring in each of the $n$ intervals preceding $t_n$. That probability is given by $1-(k_{f2}\Delta t)^n$. Conveniently, as our time interval $\Delta t$ goes to $0$, this probability becomes a continuous function $e^{-k_{f2}t}$.

To get our final probability, we simply multiply these two probabilities:

$[\text{total probability of reaction occurring in interval }t+dt] = [\text{probability of occurring in interval }t+dt] \times [\text{probability that reaction has not occurred by time }t]=k_{f2}dt\times e^{-k_{f2}t}$

The probability as a continuous function of $t$ therefore can be written as

$p_{\ce{AB->C}}(t)=k_{f2}e^{-k_{f2}t}$.

As expected, integration from $t=0$ to $t=\infty$ gives an overall probability of $1$.

Reference: Adapted from the lecture notes published by Dmitrii Makarov here: http://makarov.cm.utexas.edu/resources/Lecture-notes,-tutorials-etc./single-vs-bulk.pdf

For the two-step process, if $\ce{A+B <=> AB}$ establishes equilibrium very fast relative to the $\ce{AB->C}$ reaction, then the above is a reasonable approximation (after adjustment of $k_{f2}$ to reflect the equilibrium concentration of $AB$). If the equilibrium is not established quickly, then you would use the same reasoning to determine the probability first of $AB$ having formed by time $t$ and then the probability of it reacting based on how long it has been in existence.

I have not read Makarov's textbook on single molecule kinetics, but I assume it covers these kinetics in much greater detail and might be a good resource:

Single Molecule Science: Physical Principles and Models
1st Edition by Dmitrii E. Makarov; CRC Press 2015; ISBN 9781466559516


Aurora borealis - less favored energy transitions higher in atmosphere?

Why don't you see a mix of the two colours? After all, from the collision rate argument you would expect fast transitions (like the green of atomic oxygen) to become even more prominent.

There can be a mix of colors.  However, according to Auroral electron energy derived from ratio of spectroscopic emissions Journal of Geophysical Research Volume 79, Pages 5181–5186, the ratio of the atomic oxygen versus molecular nitrogen bands are a function of "a characteristic energy of the precipitating electron flux".

  Why is it that some transitions are less likely as you increase in height?

Well, consider Atmospheric Composition and Vertical Structure.  Figure 3 shows how dramatically pressure drops as altitude increases.  This means a change in altitude of 15 kilometers could reduce the concentration of the species of interest by a factor of 10.

In addition, dissociation of the $\ce{O2}$ and $\ce{N2}$ to atomic O and N increases with altitude.  See especially Fig. 6 which plots the points where monoatomic oxygen starts to exceed diatomic oxygen, and where monoatomic oxygen starts to exceed diatomic nitrogen in concentration.


Simply speaking: The rising bubbles in your pot shrink wherever it is too cold (< 100 °C), and grow whereever it is too hot (> 100 °C), equillibrating the temperature, by exchanging some heat of evaporation. If you keep your thermometer at a distance above the heat source, it will show (mostly, see below) exactly 100 °C, or whatever is the boiling point at the given outside pressure.

Very small bubbles have to have a measurably increased inside pressure, but while the water is boiling, the bubbles grow, and so this difference vanishes. You can calculate it with the surface tension of water (0.06 N/m at 100 °C) using the https://en.wikipedia.org/wiki/Kelvin_equation, and it is negligible for bubbles of millimeter size.

Of course with increasing distance to the water surface, the temperature rises  because of the pressure increase of ~ 0.1 bar/m. Slightly less, because water at 100 °C is less dense.


Which electrons will be taken when oxidations take place and how does the resulting orbital diagram look like?

For transition metals (specially for those on the second and third rows), like chromium, or iron, which electrons are taken away in oxidation is not straightforwardly deduced. The ones that will be lost are those which are higher in energy and lead to a more stable situation.

For iron, which is $\mathrm{[Ar]3d^6 4s^2}$, the higher energy electrons are $\mathrm{4s^2}$, and those that will be removed on oxidation, leading to a $\mathrm{d^6}$ configuration for iron (II).


Hydrolysis of 4-chloro-3,3-dimethylbut-1-ene forms which products much faster than primary alkyl halides?

Apparently, (B) is wrong for obvious reasons.

The product formed fastest will be kinetically determined. Shifting of bonds takes time so the fastest formed product will be the the one where minimum migrations occur hence (A).
As @orthocresol stated below, a stable compound similar to cyclopropylmethyl carbocation is forming here. "The double bond helps to kick out the chloride ion to give a stabilised carbocation, which makes it react much faster than usual". Hence product (A) will be favored thermodynamically too (refer).


They should both work. I would use oxalyl chloride with a solvent like DCM. DMF should act as a catalyst here for the acid chloride formation. (I don't have access to the paper but this is most likely what they have done) The oxalyl chloride reaction is maybe easier to work-up although I don't think there is a big difference when working in lab scale. In this case you can skip work-up (assuming you have used DCM as solvent and 1-2 equivalents of oxalyl chloride) and just add $\ce{AlCl3}$ for the cyclisation step.

EDIT: You can read the experimental in the supporting info -page S7


Most dense hydrogen containing composition

A few binary metal hydride candidates:

Magnesium hydride gives about 0.11 grams of hydrogen per cubic centimeter, and unlike water it can release its hydrogen leaving the element behibd on simple heating.  It does, however, require 287°C to decompose, limiting reversibility.

Lithium hydride offers nearly the same hydrogen density as the magnesium compound but decomposes at a much higher temperature, limiting its utility.

Beryllium hydride approaches approaching 0.12 grams hydrogen per cubic centimeter.  But it has several drawbacks; beryllium is expensive, it's a serious health hazard and the element does not react directly with hydrogen, making an element/hydride cycle impossible.

Aluminum hydride, with a favorable stoichiometry, comes inat nearly 0.15 grams hydrogen per cubic centimeter and also releases its hydrogen at 150°C.  Like magnesium hydride it has been considered for hydrogen storage.  However, reversible cycling between the element and the hydride is again inhibited because aluminum and hydrogen do not react directly except at very high pressure (10GPa for aluminum hydride versus 20 MPa for a direct synthesis of magnesium hydride).  As with beryllium hydride, getting the aluminum back to hydride form requires indirect methods with other hydride sources.

A ternary compound with 0.15 grams of hydrogen per cubic centimeter is $\ce{Mg2FeH6}$, which can be made from the elements at 2-12 MPa pressure after compressing the metals into pellets.  This compound can also release its hydrogen thermally and is one of many ternary hydride candidates for hydrigen storage.


Well I guess that is not a perfect solution but I think you will improve it easely if your are studying organic chemistry. I"ll try to make a beautiful picture with chemdraw when i'll have time.

Remember that you can always draw an enol when you have a ketone. If you have no idea at first with the ketone, try it with the enol.


What makes banana bonds possible in diborane?

Look carefully, it's (distorted) tetrahedral--four groups at nearly symmetrically positions in 3D space{*}. So the hybridization is $sp^3$.

As you can see, the shape is distorted, but it's tetrahedral. Technically, the banana bonds can be said to be made up of orbitals similar to $sp^3$ but not exactly (like two $sp^{3.1}$ and two $sp^{2.9}$ orbitals--since hybridization is just addition of wavefunctions, we can always change the coefficients to give proper geometry). I'm not too sure of this though.

$\ce{B}$ has an $2s^22p^1$ valence shell, so three covalent bonds gives it an incomplete octet. $\ce{BH3}$ has an empty $2p$ orbital. This orbital overlaps the existing  $\ce{B-H}$ $\sigma$ bond cloud (in a nearby $\ce{BH3}$), and forms a 3c2e bond.

It seems that there are a lot more compounds with 3c2e geometry. I'd completely forgotten that there were entire homologous series' under 'boranes' which all have 3c2e bonds (though not the same structure)

And there are Indium and Gallium compounds as well. Still group IIIA, though these are metals. I guess they, like $\ce{Al}$, still form covalent bonds.

So the basic reason for this happening is due to an incomplete octet wanting to fill itself.

Note that "banana" is not necessarily only for 3c2e bonds. Any bent bond is called a "banana" bond.

Regarding similar structures, $\ce{BeCl2}$ and $\ce{AlCl3}$ come to mind, but both of them have the structure via dative(coordinate) bonds. Additionally, $\ce{BeCl2}$ is planar.

Sneaks off and checks Wikipedia. Wikipedia says $\ce{Al2(CH3)6}$ is similar in structure and bond type.

I guess we have less such compounds because there are comparatively few elements ($\ce{B}$ group pretty much) with $\leq3$ valence electrons which form covalent bonds(criteria for the empty orbital). Additionally, $\ce{Al}$ is an iffy case--it like both covalent and ionic bonds. Also, for this geometry (either by banana bonds or by dative bonds), I suppose the relative sizes matter as well--since $\ce{BCl3}$ is a monomer even though $\ce{Cl}$ has a lone pair and can form a dative bond.

*Maybe you're used to the view of tetrahedral structure with an atom at the top? Mentally tilt the boron atom till a hydrogen is up top. You should realize that this is tetrahedral as well.


What is the correct nomenclature for diphenyl methane?

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

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

  Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name. In general nomenclature, a ring or a chain can be the parent hydride (see P-44.1.2.2).

Method (1): “1,1'-methylenedibenzene” (preferred IUPAC name; ring preferred to chain)

Method (2): “diphenylmethane” (treats phenyl groups alike as substituents)

(This example is is explicitly mentioned in the Blue Book.)


The source for the N-N bond dissociation energy of 240 kJ/mol in the table

Your question about the source of Table 7.1 is correctly answered by the andselisk's comment. The value of $\pu{240 kJ\:mol-1}$ for $\ce{N-N}$ bond is probably initiated from Encyclopedia of Inorganic Chemistry, which has listed bond energies of wide variety of bonds and their respective bond lengths. It has given two values for $D_\circ$ of $\ce{N-N}$ bond: ~$\pu{167 kJ\:mol-1}$ for $\ce{N2H4}$ in general and $\pu{247\pm 13 kJ\:mol-1}$ for $\ce{H2N-NH2}$ in particular.

The reason is given in the very first paragraph of the article. In their words:

  There are two different ways to define the bond energy even for the simplest diatomic molecule (see the diagram below). $\Delta D_{\circ'}$ is the dissociation energy measured from the very bottom of the potential energy well. However, the molecule possesses zero-point energy and thus, the experimentally measured dissociation energy $\Delta D_{\circ}$ is somewhat less:

In poly atomic compound there are other considerations. For example, the mean bond energy of $\ce{N-H}$ is the average of three bond energies associated with sequential fission of the three $\ce{N-H}$ bonds in ammonia ($\ce{NH3}$). These three values are different from each other and also different from their average, the mean bond energy of $\ce{N-H}$. Using this mean bond energy of $\ce{N-H}$ in $\ce{NH3}$ and the heat of atomization of $\ce{H2N-NH2}$, a value of $\pu{159 kJ\:mol-1}$ has been found for the $\ce{N-N}$ single bond energy. Then again, using the mean bond energy of $\ce{N-F}$ in $\ce{NF3}$ and the heat of atomization of $\ce{F2N-NF2}$, a value of $\pu{172 kJ\:mol-1}$ has been found for the $\ce{N-N}$ single bond energy. These two values are in somewhat good agreement. However, the dissociation of the two molecules are quite different:

$$\ce{H2N-NH2 -> 2 H2N^.} \qquad \pu{247 kJ\:mol-1}$$
$$\ce{F2N-NF2 -> 2 F2N^.} \qquad \pu{88 kJ\:mol-1}$$

Therefore, it is warned that care should be exercised in using any values of the bond energies.

Source:

R. Bruce King (Editor in chief), "Bond Energies," Encyclopedia of Inorganic Chemistry; 2nd Edition, John Wiley & Sons, Ltd.: New York, NY, 2005 (ISBN: 978-0-470-86078-6). Bond Energies is available online: https://doi.org/10.1002/0470862106.id098


Why is formyl chloride unstable while higher acyl chlorides are stable?

The simplest stable acyl chloride is ethanoyl chloride or acetyl chloride; methanoyl chloride (formyl chloride) is not stable at room temperature, although it can be prepared at –60 °C or below. (Wikipedia)

The instability of $\ce{HCOCl}$ is caused by ease of elimination of HCl from its molecules. Cl is decent leaving group and after it's gone, remaining acylium cation has a very acidic hydrogen instead of alkyl present in other acyl halides. Thus easier breaking of C-H bond vs C-C bond causes drastic difference in thermal stability.

Note also the formyl chloride can be considered not only the derivate of formic acid, but also a derivate of formaldehyde.


Do non-English speaking countries use the same element symbols?

Both North Korea and South Korea use the periodic table. The names, however, are different. Here is a periodic table in North Korea

The title translates to "Mendeleev's atomic periodic table." The names of the elements are different for some.

In South Korea
Hydrogen is known as and pronounced as soo-saw.
Phosphorus is known as and pronounced as in.
Potassium is known as and pronounced as kallium.
etc.

But some are the same.

In South Korea
Magnesium is known as and pronounced as magnesium.
Argon is known as and pronounced as argon.
Fluorine is known and pronounced as fluorine.

I haven't taken classes in North Korea or visited any research facilities. But it is likely that they have further modified the names  to be "less western".

All of these are names, and do not replace the symbols used in the periodic table.


For the same reason $\ce{O2}$ is paramagnetic: degeneracy.
The filled orbitals are $\sigma(\mathrm{1s})^2$, $\sigma^*(\mathrm{1s})^2$, $\sigma(\mathrm{2s})^2$, $\sigma^*(\mathrm{2s})^2$.  The half-filled orbitals are $\pi(\mathrm{2p}_x)^1$ and $\pi(\mathrm{2p}_y)^1$.

But you're right to think that this is unexpected:  why are the $\pi(\mathrm{2p})$ orbitals  lower than the $\sigma(\mathrm{2p})$? That's the reverse of what we are told to expect! I've never encountered a rationalization that satisfied me.

See Matt Law's lecture notes for the expected and observed ordering:
MO Diagrams for Diatomic Molecules; Chapter 5; Friday, October 9, 2015; PDF via the Internet Archive.


Difference between shells, subshells and orbitals

Here's a graphic I use to explain the difference in my general chemistry courses:

All electrons that have the same value for $n$ (the principle quantum number) are in the same shell
Within a shell (same $n$), all electrons that share the same $l$ (the angular momentum quantum number, or orbital shape) are in the same sub-shell
When electrons share the same $n$, $l$, and $m_l$, we say they are in the same orbital (they have the same energy level, shape, and orientation)

So to summarize:

same $n$ - shell
same $n$ and $l$ - sub-shell
same $n$, $l$, and $m_l$ - orbital

Now, in the other answer, there is some discussion about spin-orbitals, meaning that each electron would exist in its own orbital. For practical purposes, you don't need to worry about that - by the time those sorts of distinctions matter to you, there won't be any confusion about what people mean by "shells" and "sub-shells." For you, for now, orbital means "place where up to two electrons can exist," and they will both share the same $n$, $l$, and $m_l$ values, but have opposite spins ($m_s$).


What is the oxidation state of aluminium in NaAlO₂?

For every compound the sum of the oxidation states of the constituent atoms must be equal to the charge of the compound. So, since your compound is neutral the individual oxidation states have to add up to $0$. And as you have already determined the oxidation states of 2 atoms in the system the last one must be $+3$, since $1 + 3 - 2 \cdot 2 = 0$.


What's the chemical nature of a dishwasher's rinsing agent?

According to the MSDS for Jet Dry (a common brand name) it contains:

Sodium 2-phenylpropane-2-sulfonate (Sodium cumene sulfonate)
Propane-1,2-diol
Citric acid

I think it may be safe to assume that the citric acid is in there for a "clean" smell and also to soften the water (thanks to the commenter).  The sulfonate is presumably a surfactant to help break the surface tension of the water, and promote evaporation, and the propylene glycol may help serve as a coolant (and is favored due to low toxicity).


In the case of the water solubility what you want to know is how powerful the secondary interaction  of the compound that you are going to dissolve with  water?If the compound that you wish to dissolve have hydrogen bonding then it's definitely dissolve in water .It does not mean others  are  insoluble  in water.

But  the compounds do not have hydrogen bonding are  less soluble in water when compared to    the ones which have hydrogen bonding.Because ones which have hydrogen bonding interact with water molecules easily than others which do not have hydrogen bonding

Now take a look at the image   below,

It shows two water molecules  which posses the hydrogen bonding interaction.

Now give the eye on compounds that you have got,

The  compounds that you going to dissolve  may form  interaction force  with water as above image. If that interaction force is hydrogen bonding ($\ce{C5H11OH}$) then it's solubility is greater than ones which do not have hydrogen bonding($\ce{C5H11Cl}$) (dipole dipole interactions)

So because of that solubility of $\ce{C5H11Cl}$ is less than the solubility of ($\ce{C5H11OH}$) with water.

Then the bottom line is ,the solubility of a compound is vary according the  secondary interaction between the solute and solvent.In the case of water acting as  the solvent solubility is highest in  compounds which having H bonding , than which haaving dipole-dipole forces and  weak dispersion forces.