Where does the name ‘aliphatic’ come from?

Olefins are hydrocarbons that contain at least a double bond. The term is used —sometimes misleadingly— interchangeably with alkenes.

  Although the nomenclature is not followed widely, according to IUPAC, alkenes are acyclic hydrocarbons with one double bond between carbon centers. Olefins comprise a larger collection of cyclic and acyclic alkenes as well as di- and polyenes. — Alkene nomenclature – Wikipedia

The Goldbook page Wikipedia cites says the final words:

  Olefins

  Acyclic and cyclic hydrocarbons having one or more carbon–carbon double bonds, apart from the formal ones in aromatic compounds. The class olefins subsumes alkenes and cycloalkenes and the corresponding polyenes.

While Oxford dictionary seems to agree with the origin thingy:

  Origin

  Mid 19th century: from French oléfiant 'oil-forming' (with reference to oily ethylene dichloride).

Funny thing is, "aliphatic" also means "producing fats". Goldbook says:

  aliphatic compounds

  Acyclic or cyclic, saturated or unsaturated carbon compounds, excluding aromatic compounds.

Some definitions exclude alicyclic compounds from the list of aliphatic compounds, but IUPAC doesn't agree, so. Note that aliphatic compounds can be saturated or unsaturated.

So, back the main answer part,

  Aliphatic; Etymology

  From Ancient Greek ἄλειφαρ (áleiphar, “oil, ointment”). — Wiktionary

There are literally no articles on the Net speaking of the first discovered or documented aliphatic compounds, but I read in some textbook somewhere which mentioned that since these compounds were first referred to as what animal fats used in primitive or medieval lamps were consisted of, they're referred to as "aliphatic".

The only thing about a "first aliphatic" I could find is mentioning "Muconic acid is the first aliphatic compound formed as a result of aromatic ring splitting, which is further oxidized to fumaric acid and then participates in the cell metabolism as a standard endogenous substrate." by this paper.


What is the stability order of the carbocation and carboanion?

Alkanes or $\mathrm{sp^3}$ hybridized carbons show +I effect.

Alkenes contain $\mathrm{sp^2}$ hybridized carbons. Due to greater $\mathrm{s}$  character they have higher electron withdrawing tendency. So they show -I effect.

Coming to the first question:

In the given figures we have $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized carbons in the given rings. Notice that in the second figure there are greater number of $\mathrm{sp^2}$ hybridized carbons. Hence, the net withdrawal of electron density is more, and accordingly, stability goes down. In the third figure there is no resonance, which is the more dominating effect than the inductive effect. So it is the least stable.

In the second question:

In figures 1 and 2, we have equal number of $\mathrm{sp^2}$ hybridized carbons. But notice in the first figure, the negative charge is on the ring and is closer to the $\mathrm{sp^3}$ carbons which are electron donating (Remember, inductive effects are distance dependent). In the third figure greater +I is being imposed on the anion, hence it has the least stability.


Why are naturally occuring liquids more rare than solids and gasses on Earth?

Your argument misses two major points.

First, a great number of solids are ionic and metallic substances. Those show a really strong bonding mechanism between elementar particles.

And second, to get your theory to work, you have to take into account the size of molecules. To evaporate or melt anything, you have to break or at least disturb all the bonds each molecule has to its surroundings.

For something as small as H2O, not even hydrogen bonding is strong enough to make it solid. A large molecule like Polyethylene, on the other hand, is only held by weak London forces, but has melting points as high as 145°C, and cannot be evaporated.


Which is better: Increase in collision frequency or increase collision energy

Collision energy is what matters

Both the frequency of collisions and the average energy of those collisions are increased at higher temperatures. But the frequency doesn't matter much for the rate of reaction because collisions are already very very common and are not the limiting factor in causing a reaction to happen.

The distribution of energy in moving particles (and therefore the energy involved in collisions) follows a statistical distribution. Simplifying a bit (by ignoring the orientation of colliding molecules and a bunch of other probabilistic factors) the reason a reaction follows a collision is that the collision has enough energy to overcome some threshold energy barrier for a reaction to take place (simplifying again, collisions without enough energy just cause molecules to bounce off each other perhaps exchanging some kinetic or vibrational energy in the process). Only when the net energy involved is sufficient to, for example, break a bond in one of the molecules, does a reaction result. A lot of collisions are happening all the time and most don't have enough energy to cause a reaction to happen. In fact, if the population of collisions with enough energy is too low there will be no reaction at all no matter how many collisions there are and increasing the number will make no difference at all.

As the temperature increases, more of the molecules are pushed into the "have sufficient energy" part of the distribution (which might be a very small proportion of all the molecules in the mixture). Then reactions start to happen.

But the number of collisions is very large compared to the number leading to a reaction so it is the number of molecules with enough energy that dominates the result.


What happens if the electronegativity difference is exactly 2.1?

As mentioned above there is no hard line between an ionic and a covalent bond, they are just two ends of a spectrum where a bond can have a certain degree of ionic or covalent character. At the lower end of this spectrum, bonds with small electronegativity differences between the atoms have very little ionic character because the charge distribution in the bond is almost symmetrical between the two atoms. As the EN difference increases the charge distribution becomes asymmetrical and this results in an increasingly polarised bond, which therefore has an increasing degree of ionic character. Eventually at the other end of the scale you have a bond where almost all of the charge is concentrated on one atom and this is generally termed to be an ionic bond.

So in answer to your question; what happens if the EN difference is exacly 2.1; nothing happens, you have a relatively strongly polarised bond which under some definitions might be classed as ionic and under others as strongly polar covalent but in reality it has a large degree of both ionic and covalent character.


What is the difference between statistical mechanics and quantum mechanics in terms of accuracy?

Quantum mechanics is about the physics of very small things, molecules and smaller.

Classical mechanics is about macroscopic things. Quantum mechanics covers the whole of classical mechanics as well, but in the macroscopic limit both become equivalent. For example discretized energy states become so close, that you can thing of them as a continuum of states.

Statistical mechanics can be employed to consider many, quantum or classical, like systems and how they evolve. One application is to get thermodynamic quantities from quantum mechanical energy states, by considering how these states are statistically populated.


Designing a device to separate water from water-ethanol mixture using dry ice baths

One suggestion: try cooling the alcohol-water mixture more slowly. Slower crystal growth may produce larger ice crystals that occlude less of the ethanol and are easier to separate by filtration. How to do it? Place the steel beaker in the outer bath before adding dry ice to the outer bath, then add the dry ice incrementally to gradually bring the temperature down to where you first observe ice crystals, then keep the temperature constant for a while as crystals form. Then lower the temperature a degree or two and repeat the cycle. Probably better not to stir the inner mixture. It will probably help to use a thermometer to carefully monitor the temperature of the alcohol mixture.


What is the oxidation state of copper and iron in chalcopyrite?

A number of papers report that the formal valency states of chalcopyrite are best considered as $\ce{Cu+Fe^3+S2}$, based on various computational and spectroscopic evidence.

References:

Klekovkina, V. V.; Gainov, R. R.; Vagizov, F. G.; Dooglav, A. V.; Golovanevskiy, V. A.; Pen’kov, I. N. Oxidation and magnetic states of chalcopyrite CuFeS2: A first principles calculation. Opt. Spectrosc. 2014, 116 (6), 885–888. DOI: 10.1134/S0030400X14060149.
Pearce, C.; Pattrick, R.; Vaughan, D.; Henderson, C.; van der laan, G. Copper oxidation state in chalcopyrite: Mixed Cu d9 and d10 characteristics. Geochim. Cosmochim. Acta 2006, 70 (18), 4635–4642. DOI: 10.1016/j.gca.2006.05.017.
Goh, S. W.; Buckley, A. N.; Lamb, R. N.; Rosenberg, R. A.; Moran, D. The oxidation states of copper and iron in mineral sulfides, and the oxides formed on initial exposure of chalcopyrite and bornite to air. Geochim. Cosmochim. Acta 2006, 70 (9), 2210–2228. DOI: 10.1016/j.gca.2006.02.007.


The authoritative source Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (Red Book) lists $\ce{H[AuCl4]}$ as an example of a salt in the subsection IR-4.4.3.4 Generalized salt formulae [1, pp. 61–62].
Further, introduction to the section IR-8 Inorganic Acids and Derivatives underlines that IUPAC nomenclature is established from composition and structure, not properties, hence the word "acid" should not be used in systematic names [1, p. 124]:

Certain inorganic and simple carbon-containing compounds are commonly given non-
systematic or semi-systematic names containing the word ‘acid’. Examples are boric acid or
orthoboric acid, metaboric acid, phosphoric acid, diphosphoric acid, cyclo-triphosphoric
acid, catena-triphosphoric acid, dithionous acid, peroxodisulfuric acid or peroxydisulfuric acid, etc. These names are unique in modern nomenclature in that, interpreted literally, they describe a particular chemical property of the compounds in question. Systematic names are otherwise based solely on composition and structure.
All such acids may also be given structure-based systematic names using principles
already described in preceding chapters on substitutive and additive nomenclature, so in that respect the ‘acid’-containing names are superfluous. Furthermore, many species which
would be classified as acids based on their chemical properties are never named as such,
e.g. aqua ions such as hexaaquaaluminium(3+), and hydrides and derivatives such as
ammonium, hydrogen sulfide (sulfane), etc. The term ‘acid’ is thus not used consistently.
Based on these considerations, the use of the word ‘acid’ in any new name in inorganic
nomenclature is discouraged. However, a number of the existing ‘acid’ names are so
commonly used (sulfuric acid, perchloric acid, etc.) that it would be unrealistic to suggest replacing them altogether by systematic alternatives.

According to Wikipedia, $\ce{H[AuCl4]}$ exists in aqueous solutions as square planar $\ce{AuCl4-}$ anion, which is also retained upon crystallization as tetrahydrate, but there is a rapid
exchange of all protons in the unit cell [2]. For compounds of this class where the structural aspects of protons are either not essential or cannot be precisely established, a hydrogen name can be used. From the section IR-8.4 HYDROGEN NAMES [1, pp. 134–136]:

An alternative nomenclature for hydrogen-containing compounds and ions is described here. The word ‘hydrogen’, with a multiplicative preﬁx if relevant, is joined (with no space) to an anion name formed by additive nomenclature and placed within appropriate enclosing marks (see Section IR-2.2). This construction is followed (again with no space) by a charge number indicating the total charge of the species or structural unit being named (except for neutral species/units).
Hydrogen names are useful when the connectivity (the positions of attachment of the
hydrons) in a hydron-containing compound or ion is unknown or not speciﬁed (i.e. when
which of two or more tautomers is not specified, or when one does not wish to specify
a complex connectivity, such as in network compounds).
[…]
Examples:
[…]

$\ce{H4[Fe(CN)6]}$
tetrahydrogen(hexacyanidoferrate)
$\ce{H2[PtCl6]·2H2O}$
dihydrogen(hexachloridoplatinate)—water (1/2)

[…]
The strict deﬁnition of hydrogen names proposed here is meant to eliminate such
confusion by imposing the requirements:
(i) that ‘hydrogen’ be attached to the rest of the name,
(ii) that the number of hydrogens must be speciﬁed by a multiplicative preﬁx,
(iii) that the anionic part be placed in enclosing marks, and
(iv) that the charge of the total structure being named is speciﬁed.
Hydrogen names constructed in this way cannot be mistaken for other types of name.

To sum it all up, I think the IUPAC name for $\ce{H[AuCl4]}$ is hydrogen(tetrachloridoaurate).
Note that new recommendations also suggest using chlorido instead of chloro for anionic ligands. Also, in names formal oxidation states are enclosed in parentheses, not square brackets.
Wikipedia might not always mention IUPAC names, and even those which are listed as such should not be automatically trusted. Also, acceptance of systematic names in chemical community is inertial; not every systematic name is immediately replacing the traditional one which everyone is familiar with. I suspect this is the case as to why chloroauric acid still gets a pass.
References

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.
O’Reilly, D. E.; Peterson, E. M.; Scheie, C. E.; Williams, J. M. Nuclear Magnetic Resonance of the Aquated Proton. II. Chloroauric Acid Tetrahydrate. Phase Transitions and Molecular Motion. J. Chem. Phys. 1971, 55 (12), 5629–5635. DOI: 10.1063/1.1675731.


Is Henry's Law sufficient for calculating CO2 solubility?

There are multiple forms of Henry's law (see this Wikipedia entry). The net or effective Henry's Law takes into account the other pH-dependent equilibria of $\ce{CO2}$ and carbonate species in water (chemical reactions with water in your question).

In the case of blood, as you said the picture is even more complex. You would have to be careful using, for example, published values of the effective Henry's Law solubility of $\ce{CO2}$ in that case.


Proof of Brillouin's theorem using the diagonality of the Fock matrix

Your question actually contains two linked inquiries. The proof itself depends on the nature of the Fock-operator to which I'll have a few words afterwards. For the proof itself it is only important, that the Fock matrix is diagonal. Let's restate Brillouin's theorem:

In case of the Hartree-Fock (HF) reference determinant $\Phi_0$, the Hamiltonian matrix element between it and any singly excited determinant $\Phi_a^r$ reduces to zero.
$%
\newcommand{\ll}{\left\langle}\newcommand{\rr}{\right\rangle}
\newcommand{\lb}{\left|}\newcommand{\rb}{\right|}
\newcommand{\op}[1]{\mathbf{#1}}$
$$\langle\Phi_0|\op{H}_\mathrm{el}|\Phi_a^r\rangle = 0$$

The proof of this then rather simple:
\begin{align}
&& E &= \ll \Phi_0 \rb \op{H}_\mathrm{el} \lb \Phi_a^r\rr\\
&& E &= \ll \phi_a \rb \op{H}^\mathrm{c} \lb \phi_r \rr +
     \sum_j^{N/2} \ll \phi_a\rb \left(2\op{J}_j -\op{K}_j \right) \lb \phi_r \rr\\
&& E &= \ll \phi_a \rb \op{F}_j \lb \phi_r\rr\\
\text{with}&& \op{F}_i\phi_i &= \varepsilon_i\phi_i\\
\text{and}&& \ll \phi_j \rb \op{F}_i \lb \phi_i \rr
          &= \varepsilon_i \ll \phi_j | \phi_i \rr = \varepsilon_i\delta_{ij} \\
&& E &=\varepsilon_j\delta_{ar} = 0
\end{align}

The direct consequence and therefore significance of this is that for any post-HF method the singly excited determinants only mix indirectly with the ground state. This simplifies a couple of the integral evaluations.

The remaining question to solve is why the Fock matrix needs to be diagonal. Even though the equation
$$\op{F}_i\phi_i = \varepsilon_i\phi_i
     \tag{1}\label{fock-pseudo}$$
suggests and eigenvalue problem, it is not. Remember the definition of the Fock operator and the operators contained
\begin{align}
  && \op{F}_i &= \op{H}^\mathrm{c} + \sum_j (\op{J}_j - \op{K}_j),\\
\text{with}&&
     \op{J}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_j(\op{x}_1) \rr \lb \phi_i(\op{x}_2) \rr,\\
\text{and}&&
     \op{K}_j\lb \phi_i\rr &=
       \ll \phi_j(\op{x}_1) \rb r_{12}^{-1}
       \lb \phi_i(\op{x}_1) \rr \lb \phi_j(\op{x}_2) \rr.
\end{align}
As you can see, the "one-electron" Fock operator depends on the solution of all "one-electron" Fock operators (ref. Szabó-Ostlund p. 115). The only way to achieve self-consistency and therefore $\eqref{fock-pseudo}$ is to use unitary transformation and diagonalise the Lagrange multipliers from
$$\sum_j \lambda_{ij} \lb \varphi_j \rr =
  \op{F}_i \lb \varphi_i \rr.\tag{2}\label{fock-lagrange}$$

In other words, the Fock operator is not actually associated with the total energy, but only with the variation of it (due to the Lagrange minimisation). To know one orbital you have to know all orbitals, and to know one orbital energy, you have to know all orbital energies. The Hamilton operator is not the sum of all Fock operators, and the total HF energy is not the sum of all orbital energies.

Therefore the Fock operator is also dependent only on the total wave function and any unitary transformations of the canonical orbitals will retain the total energy. However, the canonical orbitals are the only ones for which $\eqref{fock-pseudo}$ holds.


test tube cracked when added glycerin to potassium permanganate

I am not sure I should continue on sulfuric acid and potassium parmagnet test in a TEST TUBE after seeing this affect...

In case of doubt you should not. Anyway, I recommend to avoid doing it with any significant amount. It may and will spread around and even if not it produces manganese (IV) oxide smoke that settles down nearby and is quite hard to wash out.

  I want to know why, because it lab we some time directly heat test tube, it does not cracked so why this time..

Potassium permanganate is a strong oxidizer, glycerol is a reducer. Mixed together they quickly start to react. The exact nature of the products is debatable, but the mixture self-ignites if warm. The thermal shock is enough to crack glass. This is a common weakness of glass and ceramics.

Furthermore, in case of bad luck it may produce enough gases to spread the mix around. Consequently, safety precautions must be always employed (incombustible table, safety goggles etc). A movie how it may go http://www.youtube.com/watch?v=6F4cuQkKt2U . The reaction has some quirks, specifically if the glycerol is cold, it may not start immediately, so a rather well mixing is achievable. Now, once it warms up to room temperature, it suddenly bursts in flame, potentially delivering itself to nearby surfaces, including your face.


Regioselectivity in electrophilic substitution of pyrrole

The following figure shows the resonance structures we can draw to describe the intermediates produced by electrophilic at the 2- ($\alpha$) or 3- ($\beta$) positions in pyrrole.

image source

You can see that the intermediate produced by attack at the $\alpha$-position can be described by 3 resonance structures.  Whereas attack at the $\beta$-position produces an intermediate that can be described by only 2 resonance structures.

Therefore, the intermediate formed by $\alpha$ attack is more stable and the activation energy leading to it will be reduced.  Consequently, electrophilic attack at the $\alpha$ position in pyrrole is preferred.


Why does physisorption (physical adsorption) increase with ease of liquefaction?

It can be a probable answer but I am not sure about it.

You are correct to say that easily liquifiable gases have stronger intermolecular forces of attraction (van der Waals forces). As a result of which they are strongly bound to themselves. Consider physisorption, here, the forces between the adsorbate molecules and the adsorbent  are weak van der Waals forces. Once the surface area of the adsorbent is completely occupied, the next layer of adsorbate molecules will be bound weakly. So if the gas molecules have stronger van der Waals force of attraction, they would be adsorbed readily in preference to gas molecules having weaker van der Waals forces of attraction.


Why does reduced mass help when talking about two body problems?

You're transforming a problem in terms of two particles with individual masses and accelerations into a problem with a single fictitious particle. That simplifies the problem and lets you separate the relative motions and interactions of the particles from overall translation of their center of mass.

In step 3, you need the relative acceleration $a_{rel}$ because that's the acceleration the single fictitious particle has.

In step 4, you're calculating the mass $\mu$ of the fictitious particle.

Before, you had $F_{12} = m_1 a_1$ and $F_{21} = m_2 a_2$ (two forces, two masses, two accelerations). Now you've got $F_{12} = \mu \ a_{rel}$ (one force, one mass, one acceleration). The price is that we've thrown away information about the overall translation of the system through space; we're only looking at the relative motions of the two particles.

The coordinates of the fictitious particle are relative or internal coordinates $$x = x_2-x_1,\quad y = y_2-y_1, \quad z = z_2-z_1$$
This is useful because the potential energy $V$ of the interacting particles depends only on these relative coordinates. Making use of reduced masses and relative velocities and coordinates cleanly separates internal motion from translation of the system as a whole. The classical energy of the two body system will be
$$E = \left[\frac{1}{2\mu}(p_x^2+p_y^2+p_z^2)+V(x,y,z)\right] + \left[\frac{1}{2M}(p_X^2+p_Y^2+p_Z^2)\right]$$
where $p_i$ means momentum for motion along the $i$-th coordinate. The first term in brackets is the energy of our fictitious particle with mass $\mu$. The second term is the energy of a second fictitious particle with total mass $M=m_1+m_2$ and coordinates  $X$, $Y$, and $Z$ located at the center of mass of the system (e. g. $X = (m_1x_1 + m_2x_2)/M$).

In chemistry, we don't care about the second term. For example, in an atom, we'll assume that the nucleus is clamped in place so we can just look at interactions within the atom.


The short answer to the title question is the ubiquitous presence of geologic and biogenic calcium carbonate $(\mathrm{p}K_\mathrm{a} = 9)$.  The oceans as a whole can largely be thought of as residing on beds of calcium carbonate. The following are some excerpts from this Wikipedia page:

Eggshells, snail shells and  most seashells are predominantly calcium carbonate.
Carbonate is found frequently in geologic settings and constitute an enormous carbon reservoir. Calcium carbonate occurs as aragonite, calcite and dolomite. The carbonate minerals form the rock types: limestone, chalk, marble, travertine, tufa, and others.
Calcium carbonate contributors, including plankton (such as coccoliths and planktic foraminifera), coralline algae, sponges, brachiopods, echinoderms, bryozoa and mollusks, are typically found in shallow water environments where sunlight and filterable food are more abundant. Cold-water carbonates do exist at higher latitudes but have a very slow growth rate.

Regarding your question

Doesn't a compound need an hydroxide ion to be a base?

Here you go:
$$\ce{CO3^2- (aq) + H2O (l) <=> HCO3- (aq) + OH-(aq)}$$
Although calcium carbonate is only sparingly soluble $(K_\mathrm{sp} = 3.3×10^{−9})$, as stated above the oceans are simply rife with it.


Why does boiled water have a different taste as compared to cool water?

The changed taste that stays after cooling the heated water

Hot (boiling) water has a much lower capacity for dissolving gases. When boiling water, $\ce{CO2}$ is removed and through the equilibria behind ($\ce{CO2 + H2O <=> H2CO3}$ and $\ce{H2CO3 <=> H+ + HCO3-}$) so is $\ce{HCO3-}$.

Basically, this removes  $\ce{CO2}$, overall carbonic acid species and also acid equivalents: the resulting water will be more basic (less acidic).

$\ce{CO2}$, $\ce{H2CO3}$ and $\ce{HCO3-}$ are not only present in carbonated (sparkling, fizzy) water but also in tap water. For example, this analysis sheet of the Munich municipal water supply cites 300 mg/l $\ce{HCO3-}$, 15 mg/l $\ce{CO2}$ and 0.5 mg/l $\ce{CO3^2-}$ and a slightly basic pH.

It is known that mammals have $\ce{CO2}$ receptors in their acid taste cells, and this is thought to contribute to the water taste. (I don't have the full text, but Zocchi et al: The cellular mechanism for water detection in the mammalian taste system, Nature Neuroscience volume 20, pages 927–933 (2017) looks promising.)

BTW, two further candidates for that bland/flat/stale taste are
deionized water and reverse osmosis drinking water.

Warm vs. cold water taste

There will be further changes with temperature, including acid base equilibria and pH, partial pressure of volatile molecules and so on between cold and warm water. And this also applies to our sensing in the taste cells. Put this together with taste/smell sensory input being subject to further processing in our brain to produce the taste/smell impression, and I think it is very hard to compare taste/smell of cool and hot things.


A simple explanation might be that the ratio of carbon to hydrogen is higher in the cyclic hydrocarbons, and carbon is a heavier element than hydrogen.


Both are linear with a $C_{\infty}$ axis, but $D_{\infty h}$ has a center of inversion and $C_{\infty v}$ does not.


Why is naphthalene less stable than benzene according to per benzene ring?

A short answer is: you're forgetting that a double bond is shared between the two rings in naphthalene.
A long answer is given below.

In terms of heat of hydrogenation

Naphthalene can be hydrogenated to give tetralin.
Further hydrogenation gives decalin.

Scheme 1: hydrogenation of naphthalene.

On the other hand, the hydrogenation of benzene gives cyclohexane.

Scheme 2: hydrogenation of benzene.

In an old report it reads (Sherman, J. Oil Soap (1939) 16: 28):

  If benzene could be simply regarded as cyclohexatriene with no interaction between the double bonds, the heat of hydrogenation to form cyclohexane should be just three times the heat of hydrogenation of cyclohexene to form cyclohexane.
  From Table I, this is $3 \times -28.6 = -85.8$ kcal/mol.
  The experimental value is $-49.8$ kcal/mol.
  Hence benzene is more stable by an amount $-49.8 -(-85.8) = 36.0$ kcal/mol than it would be if it were completely unsaturated in character (with no interaction between the double bonds).
  From the theoretical viewpoint, this extra stability of benzene (and other aromatic compounds) is shown to be a consequence of the fact that the normal state of the molecule is not that corresponding to either Kekulé structure but is a sort of combination of the two.

This $36.0$ kcal/mol may be regarded as a direct measure of aromaticity and is cited in many places as resonance energy.

In particular, the resonance energy for naphthalene is $61$ kcal/mol.
This value is shifted by around $5 \times -28.6 = -143.0$ kcal/mol (five double bonds) from the actual heat of hydrogenation of naphthalene.
This means the heat of hydrogenation of naphthalene would be somewhere around $61 - (-143.0) = -82$ kcal/mol.
The actual value from NIST (to cis-decalin) is $-318$ kJ/mol, or $-76$ kcal/mol.

So it costs $-49.8$ kcal/mol to hydrogenate benzene to cyclohexane but only $-76$ kcal/mol to hydrogenate naphthalene to cis-decalin, less than twice a benzene.
Furthermore, part of this energy is due to the resonance energy, which is $36.0$ kcal/mol for benzene, but only $61$ kcal/mol for naphthalene, again less than twice a benzene.

What did we forget?

Naphthalene rings are fused, that is, a double bond is shared between two rings.
This makes the above comparisons unfair.

A better comparison would be the amounts of resonance energy per $\pi$ electron.
Benzene has 6 $\pi$ electrons, which makes its resonance energy equal to $36.0 / 6 = 6$ kcal/mol per $\pi$ electron.
Naphthalene (10 $\pi$ electrons) shows a remarkably similar value: $61 / 10 = 6.1$ kcal/mol.

Even comparison of heats of hydrogenation per double bond makes good numbers.
For benzene (hypothetically three double bonds) it costs $49.8 / 3 = 16.6$ kcal/mol to hydrogenate each double bond.
For naphthalene it would be $-76 / 5 = 15.2$ kcal/mol, again very similar.

What about tetralin?

Again NIST comes to our rescue.
The most recent value for the heat of hydrogenation from naphthalene to tetralin is $-125$ kJ/mol, or $-29.9$ kcal/mol (Scheme 1, first arrow).

As a per double bond value it gives us $-29.9 / 2 = -15.0$ kcal/mol, close to both benzene and naphthalene (all of them differ by less than $1.6$ kcal/mol).

In terms of electronic structure

  Both are aromatic in nature both have delocalised electrons but naphthalene has more number of $\pi$ bonds and hence more resonance structures and more delocalisation so overall it must be more stable.

I believe the highlighted sentence tells it all.
As you said, delocalisation is more significative in naphthalene.
A simple model for delocalisation is the particle in a box (in this case, more of a rectangle).

The $n$-th energy level for an electron confined in a single dimensional box (line segment) is

$$E_n = \frac{n^2 h^2} {8 m_e L^2}\text{,}$$

where $m_e$ is the mass of the electron, $h$ is the Planck's constant and $L$ is the length of the line segment.
So energy decreases with the square of the length of the confinement.
Results are analogous for other dimensions.

From this simple model, the more confined an electron, the higher will be its energy.
It is thus perfectly reasonable to rationalise that more delocalisation results in more stability.
Given its larger delocalisation, it seems rational that the energy levels of naphthalene have less energy in comparison to benzene.


Mirko: The mechanism you present for the early stages of the Wolff-Kishner Reduction includes hydroxide necessary for the decomposition of the yet to be formed hydrazone 6 of the ketone 1 (or aldehyde). One may preform the hydrazone and then do the actual reduction in the presence of hydroxide in a high boiling hydroxylic solvent (Huang-Minlon variation1).

The mechanism for hydrazone formation is a reversible reaction and is often conducted by removal of water to shift the equilibrium to the right.2 Realizing that most commercial hydrazine contains 15% water, it can act as the agent for proton exchange in adduct 3. One need only use the electron pair in structure 4 to eject hydroxide contrary to the elimination proposed in the last step of your mechanism. There is no need to be concerned about the anion of hydrazine as a leaving group. Hydroxide only has to deprotonate the acid 5.

1) Huang-Minlon, J. Am. Chem. Soc., 1946, 68, 2487.
2) Hydrazones and oximes are not as susceptible to hydrolysis as are imines owing to the resonance stabilization provided by the second heteroatom.


These species usually do not exist in nature, but they can be synthesized.

Silver has been reduced in liquid ammonia to give $\ce{Ag-}$.
A lot of anionic metal carbonyl complexes $\ce{M(CO)_{n}^{m-}}$ have been synthesized:

-1

$\ce{[V(CO)6]-}$, $\ce{[Nb(CO)6]-}$, $\ce{[Ta(CO)6]-}$, $\ce{[Mn(CO)5]-}$, $\ce{[Ir(CO)4]-}$, $\ce{[Co(CO)4]-}$, $\ce{[Rh(CO)4]-}$

-2

$\ce{[Ti(CO)6]^{2-}}$, $\ce{[Zr(CO)6]^{2-}}$, $\ce{[Hf(CO)6]^{2-}}$, $\ce{[Fe(CO)4]^{2-}}$, $\ce{[Ru(CO)4]^{2-}}$, $\ce{[Os(CO)4]^{2-}}$

-3

$\ce{[V(CO)5]^{3-}}$, $\ce{[Nb(CO)5]^{3-}}$, $\ce{[Ta(CO)5]^{3-}}$, $\ce{[Mn(CO)4]^{3-}}$, $\ce{[Re(CO)4]^{3-}}$

-4

$\ce{[Cr(CO)4]^{4-}}$, $\ce{[Mo(CO)4]^{4-}}$

There's a review by John E. Ellis: Highly Reduced Metal Carbonyl Anions: Synthesis, Characteri's zation, and Chemical Properties on these compounds.

Edit

While Christian Aichinger's comment concerning $\pi$-backbonding in metal carbonyls is correct, it is important to realize that $\ce{CO}$ is a neutral ligand!

With other words: The Mond process: $\ce{Ni(s) + 4CO -> Ni(CO)4}$, using nickel powder and carbon monoxide, is not a redox reaction and the oxidation state of the metal still is $0$ in the metal carbonyl.


Does aqueous component contribute to the equilibrium constant?

If you are talking about a species dissolved in a aqueous solution then yes it would feature in the constant .
However if you are referring to the solvent itself then no it wouldn't feature in the constant this is because the concentration of the solvent is constant , as well as the fact that it is present on both sides of the equation since both reactants and products are dissolved in the same solvent
If you are referring to a partial pressure equillibrium constant , then a hybrid one will have to be created that has the partial pressure of the gases and the concentration of species that are not gases
Using a Kc value would mean simply taking the concentration of both the gases and the solutes , to end up with a constant , however solvents are excluded because the concentration of solvents is constant and is thereby thought as being absorbed in the eq constant itself
Derivation of the equilibrium constant requires probability and is quite lengthy so maybe you can check the video on khan academy that discusses the derivation


How to solve 1-deuterio-3-methylcyclohexene dehydrohalogenation?

As you noted, the chlorine and the other group being eliminated must be antiperiplanar.  In the case of the chair form of cyclohexane, this means that the chlorine and other group being eliminated must be trans-diaxial.  In the figure below I've drawn the 2 possible conformations (interconverted by a chair flip) for your deuterated cyclohexane.  Only in the top conformation is the chlorine axial, and the hydrogen on the adjacent (non-methylated) carbon is also positioned trans and axial.  Only this hydrogen can be eliminated as it is the only one that is axial and antiperiplanar to the chlorine.  Therefore, the only product produced when this material is treated with base and an E2 occurs, is the pictured 1-deuterio, 3-methylcyclohexene.  The hydrogen on the methylated carbon cannot be eliminated since it is never axial when the chlorine is axial.


Pople Basis Set Abnormality: 3rd row 6-311G

From the Gaussian[1] webpage:

  6-311G: Specifies the 6-311G basis for first-row atoms and the McLean-Chandler (12s,9p) → (621111,52111) basis sets for second-row atoms [McLean80, Raghavachari80b] (note that the basis sets for P, S, and Cl are those called negative ion basis sets by McLean and Chandler; these were deemed to give better results for neutral molecules as well).

I have no idea exactly why they still call it 6-311G: I suspect it is because this configuration provides a more accurate guess for a triple-zeta Pople basis set, and to create an entirely new basis set would be impractical.

What I do know is that when you use the '6-311G' basis set, you're not necessarily getting this configuration. What you're getting is a basis set which is roughly of this quality (computational effort, accuracy).

Whilst this explains the abnormality, it still doesn't explain the exact configuration: the original paper[2] suggests a (621111,52111) set whilst the set used is (621111,42111). I hope somebody can contribute to this and complete my answer as, whilst I recognise the abnormality, further investigation has led me to this confusion.

[1] http://gaussian.com,

[2] J. Chem. Phys. 72, 5639 (1980).


Avogadro's law says that $V/n = \text{constant}$ at constant temperature and pressure.   That means your volumes can be used "as if" they were moles in the chemical reaction stoichiometry.   So, if the mole ratio is 1:1 for H$_2$ and Cl$_2$ reacting, then the limiting reagent is obviously the smaller volume of 10.0 L for Cl$_2$.

For (b) you can easily calculate the volume of HCl since 1 mole of Cl$_2$ produces 2 moles of HCl.  So, you get 20.0 L of HCl.

For (c) the excess reagent is the H$_2$.  Since it reacts with 10.0 L of Cl$_2$ in a 1:1 mole ratio, then you have 15.0L - 10.0 L = 5.0 L of H$_2$ left over.   Since this all ends up at STP, you can use the ideal gas law to calculate the moles of H$_2$ left over:

$n_{H_2} = \displaystyle \frac{p V}{R T} = \frac{(1 \text{atm})(5.0 \text{L})}{R \times 273.15 \text{K}} = 0.2230749373505515$ mol

Don't forget sig. figs., which in this case is determined by the 2 sig. figs. in the volume of 5.0 L, giving you

$n_{H_2} = 0.22$ mol

For Review check out:
gas laws: http://www.grandinetti.org/Teaching/Chem122/Lectures/Gases

If you have a mac, check out PhySyCalc. It will automatically do all the unit conversions in your calculations.

https://itunes.apple.com/us/app/physycalc/id644105995


To illustrate Oscar's answer further, it should be noted that boron, despite existing in more than ten allotropic modifications, is in general quite chemically inert (especially in crystalline form).

For example, boron doesn't react with hydrogen directly and all boranes are synthesized by other means.
Although boron may be oxidized by fluorine at room temperature, other halogens (even chlorine) require elevated temperatures above $\pu{400 °C}$:

$$\ce{2 B(s) + 3 E2(g) -> 2 BE3}\qquad(\pu{30 °C}: \ce{E} = \ce{F}; \quad\text{above}~\pu{400 °C}: \ce{E} = \ce{Cl}, \ce{Br}, \ce{I})$$

There is no preference in reactivity towards acids or bases either.
Boron indeed doesn't appreciably react either with non-oxidizing acids or with alkali solutions under ambient conditions.

However, concentrated oxidizing acids such as nitric and sulfuric acids as well as aqua regia react all right — especially at elevated temperatures — yielding boric acid:

$$\ce{B(s) + 3 HNO3(aq) ->[Δ] B(OH)3(aq) + 3 NO2(g)}$$

Amorphous boron reacts with concentrated alkali solutions:

$$\ce{2 B (am) + 2 NaOH (conc) + 6 H2O(l) -> 2 Na[B(OH)4](aq) + 3 H2(g)}$$

When fused in air with alkali, $\ce{Na2O2}$ or $\ce{KNO3}$ + $\ce{Na2CO3}$ mix, borates are formed:

$$\ce{4 B(s) + 4 NaOH(s) + 3 O2(g) ->[\pu{350-400 °C}] 4 NaBO2(s) + 2 H2O(g)}$$

All chemical reactions have been adapted from [1, p. 74].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)


Column order, signs of MO coefficients, and density matrix construction in Hartree-Fock

In the future, this should really be two separate questions, but I will try and tackle both.

MO coefficient column ordering and signs

The electronic energy is invariant to orbital swapping within subspaces. If you interchange one occupied orbital with another, the total energy of a molecular won't change. The same goes for virtual-virtual. Now, you know that the MO coefficients are by convention shaped as $C_{\mu i}$, where the rows are the AOs and the columns are the MOs. What this means is that for $\ce{H2O}$ in a STO-3G basis, where there are 7 AOs forming 7 MOs where 5 are doubly-occupied and 2 unoccupied, switching the order of the first 5 columns should not change the electronic energy. The same goes for the final 2 columns.

As you say, the $\epsilon_i$ follow the same ordering as the columns in $C_{\mu i}$; this is because they are the matching eigenvalues and eigenvectors from diagonalizing the Fock operator. The eigenvectors are not going to be affected by a sign change, which doesn't change how the space is spanned.

Regarding why the ordering is not what you expect, there is no guarantee made by NumPy or the underlying LAPACK routine DGEEV that the results will be sorted, just that the vector/value pairs will be in the same order. If you want to sort the results, you can use argsort:

idx = energies.argsort()
energies = energies[idx]
eigvecs = eigvecs[:, idx]

Make sure the ordering of the MO energies is maintained along with the ordering of the MO coefficient matrix columns.

Constructing the density matrix

This is a very irritating difference in convention among many sources. See later, for calculating the electronic part of the dipole moment:

  $$
\left<\vec{\mu}\right> = 2 \sum_{\mu\nu} D_{\mu\nu} \left< \phi_{\mu} | \hat{\mu} | \phi_{\nu} \right>,
$$
  ...

  The factor 2 appearing above arises because the definition of the density used in this project differs from that used in Szabo & Ostlund.

This can also have an effect on whether or not there is a 1/2 in front of the $2J - K$ portion of the Fock operator.


Can the electron configuration of Te be written that way?

The second configuration allows Te to take advantage of the relative stability of half-full shells. Having the 5p and 6s orbitals both half-populated is more stable than having 4 electrons in the 5p orbital.

Similar behavior is what allows carbon to form 4 bonds. The $\ce{s^1p^3}$ configuration is more stable than the $\ce{s^2p^2}$.

Tellurium shows the behavior while oxygen does not as a result of the overlap between the 5p and 6s orbitals in the larger atoms. The same degree of overlap does not occur with the 2p and 3s orbitals.


Why does (3E)-3,4-dimethylhex-3-ene yield a meso dihalide when reacted with Br2/CCl4 at room temperature?

A meso compound is a molecule with multiple stereocenters that is superimposable on its mirror image. Meso compounds are easily identified by their internal plane of symmetry.

Using this knowledge, and given that $\ce{Br2}$ undergoes anti addition to alkenes, there are only really two answers that could possibly be meso compounds, B and D. Let us consider the products:

Notice how only B has an internal plane of symmetry, and is therefore a meso compound.


Why does Le Chatelier's principle not apply to pure solids and liquids?

Le Châtelier does apply to pure solids and liquids - what happens is that pure solids and liquids experience almost no change in concentration[*] in chemical equilibria, so for all intents and purposes they are not affected by Le Châtelier considerations.

Heterogeneous systems

Imagine a system with two phases; for instance, a gas phase and a solid phase, and we have some sort of equilibrium between them; for instance if we put water in a closed container (remember that Le Châtelier only properly applies to closed systems), it reaches an equilibrium with its own vapour:

$\ce{H2O(l) <=> H2O(g)}$

Here we have two phases: a liquid phase and a gas phase. The molecules in both are the same but in different aggregation states, which crucially differ in a fundamental behaviour regarding volume: in the liquid, molecules occupy a fixed volume each in certain conditions. If you have twice the number of molecules, you get twice the volume, and the density remains constant. So, if the equilibrium above is displaced to the left, the concentration of water molecules in the liquid phase doesn't change: the size of the liquid phase does. You simply get more volume of liquid. The same is true about solids.

For the gas it's different: the gas phase will always occupy the whole free volume in the container, and what changes is its pressure (which is to say, its concentration). If you double the number of gas molecules, the volume doesn't change - it still fits the whole container - but their concentration doubles.

This is a simplified picture; formally, the concentration of real solids and liquids is not exactly constant. However, it is much, much less sensitive to conditions than gases or solutes, so for all practical purposes it's a constant. See the next section for a numerical example that also applies here.

Dilute homogeneous systems

When we deal with dilute homogeneous systems (only one phase), the concentration of the solvent can be safely treated as constant because it's so much larger than anything else that changes simply don't impact it.

Imagine that you are applying Le Châtelier to the self-ionisation equilibrium of water; if you have a sample of pure water, a very small fraction of the molecules dissociate per the following reaction:

$\ce{2H2O<=>H3O+ + OH-}$

If you have a pure water in equilibrium and you add $\ce{OH-}$ ions, Le Châtelier predicts that this will displace the equilibrium to the left, and some of the previously dissociated water molecules will associate again to form $\ce{H2O}$.

However, water itself is the medium of the reaction, so its concentration is pretty much constant. Let's calculate some actual figures to get an idea. The self-ionisation constant is $\mathrm{K_w=10^{-14}}$, so for a chemically pure sample of water we'd get these concentrations:

$[\ce{H3O+}] \approx 10^{-7} \mathrm{M}$

$[\ce{OH-}] \approx 10^{-7} \mathrm{M}$

$[\ce{H2O}] \approx 55.55 \mathrm{M}$

Now, imagine we add a lot of $\ce{OH-}$ ions - and we make their concentration a thousand times bigger than before - that will have a big impact on the concentrations of both $\ce{H3O+}$ and $\ce{OH-}$. When we reach equilibrium we'd have

$[\ce{H3O+}] \approx 10^{-10} \mathrm{M}$

$[\ce{OH-}] \approx 10^{-4} \mathrm{M}$

Note that this means that 99.9% of the dissociated molecules have been displaced to the left and formed water. However, this number of new water molecules (around 0.0000000999 M) is simply irrelevant compared with the previous concentration of water (55.55 + 0.0000000999 is still 55.55 for all practical purposes). Even further, as with heterogeneous systems, even if the addition of more water were relevant, its impact would be increasing the volume of the phase, not changing its concentration.

That's why the concentration of water isn't even included in the self-ionisation constant - yes, formally the equilibrium depends on $[\ce{H2O}]$, but since that's pretty much always the same, it's easier to factor it into the constant and write the equilibrium as $\mathrm{K_w=[\ce{H3O+}][\ce{OH-}]}$.

[*] More precisely, activity


What kind of equilibrium constant we use for Gibbs free energy and Van't Hoff equation?

The first equation actually contains the definition of the standard equilibrium constant:
$$K^\circ = \exp\left\{\frac{−\Delta_r G^\circ}{R T}\right\}$$
With this definition the equilibrium constant is dimensionless.

Under standard conditions the van't Hoff equation is
$$\frac{\mathrm{d} \ln K^\circ}{\mathrm{d}T} = \frac{\Delta H^\circ}{R T^2},$$
and therefore uses the same constant. The integrated variant is therefore already an approximation and may be correct using a different definition of the equilibrium constant.
$$\ln \left( {\frac{{K_{T_2} }}{{K_{T_1} }}} \right) =
  \frac{{\ - \Delta H^\circ }}{R}
  \left( {\frac{1}{{T_2 }} - \frac{1}{{T_1 }}} \right)$$

Now the ordinary equilibrium constant may be defined in various forms:
$$K_x = \prod_B x_B^{\nu_B}.$$

Probably one of the best representations for the standard equilibrium constant involves relative activities, for an arbitrary reaction,
$$\ce{\nu_{A}A + \nu_{B}B -> \nu_{C}C + \nu_{D}D},$$
this resolves in
$$K^\circ = \frac{a^{\nu_{\ce{C}}}(\ce{C})\cdot{}a^{\nu_{\ce{D}}}(\ce{D})}{a^{\nu_{\ce{A}}}(\ce{A})\cdot{}a^{\nu_{\ce{B}}}(\ce{B})}.$$

The concentration is connected to the activity via
$$a(\ce{A})= \gamma_{c,\ce{A}}\cdot{}\frac{c(\ce{A})}{c^\circ},$$
where the standard concentration is $c^\circ = \pu{1 mol/L}$.
At reasonable concentrations it is therefore fair to assume that activities can be substituted by concentrations, as
$$\lim_{c(\ce{A})\to\pu{0 mol/L}}\left(\gamma_{c,\ce{A}}\right)=1.$$
See also a very detailed answer of Philipp.

The partial pressure is connected to the activity via
$$a(\ce{A}) =
\frac{f(\ce{A})}{p^{\circ}} = \phi_{\ce{A}} y_{\ce{A}} \frac{p}{p^{\circ}},$$
with the fugacity $f$ and the fugacity coefficient $\phi$ and the fraction occupied by the gas $y$, the total pressure $p$, as well as the standard pressure $p^\circ=\pu{1 bar}$ or traditional use of $p^\circ=\pu{1 atm}$.
For low pressures it is also fair to assume that you can rewrite the activity with the partial pressure $p(\ce{A})$, since
\begin{align}
\lim_{p\to\pu{0 bar}}\left(\phi_{\ce{A}}\right) &=1, &
p(\ce{A}) &= y_{\ce{A}}\cdot{}p.
\end{align}

Of course concentrations and partial pressures are connected via the ideal gas
\begin{aligned}
pV\ &=nRT\\
p\ &\propto \text{const.} \cdot c,
\end{aligned}
and therefore it is valid to write:
$$K_c\propto \text{const.} \cdot K_p.$$
It is important to note, that the two expressions are not necessarily being equal, and can scale by powers of $(\mathcal{R}T)^{\sum{}\nu}$.

While using these equations it is always necessary to keep in mind, that there are a lot of approximations involved, so it depends very much on what you are looking for. Either use might be fine, as all these functions are related - some might lead to simple, some may lead to complicated solutions.


In the study of amino acids (and peptides, proteins etc), it IS standard practice to label the alpha carbon as such. More carbons leading into the R-group may also be labeled with Greek letters, but this is not as necessary as having a label for the stereogenic center, the alpha carbon. Personally, I've found Greek letters helpful in remembering the structures of the different amino acids, for example I always think of Lysine as "the one with the epsilon-amino side chain". Note that in chemistry in general, whose nomenclature follows IUPAC guidelines, it is NOT "standard" to use Greek lettering, however it is still generally recognized. The practice is probably most prominent in the study of fatty acids (another biochemistry subfield), which results in such Greek -rich categorizations as "omega-3 unsaturated fats".


Justifying the tremendous −I effect of trimethylammonium group by molecular orbital theory

Powerful electron withdrawing substituents, like $\ce{-NO2}$ can be interpreted as having a large effect on the $\pu{\pi}$-system of an aromatic ring. Resonance structures can be drawn that put a positive charge at the ortho and para positions and put the electron out in the nitro group, on an oxygen.

This must be adjusted for $\ce{-N(CH3)3+}$, because it has no $\pu{\pi}$ electrons, and we can't draw structures with a double bond between the nitrogen and the ring carbon. It works thru a very strong $\ce{-I}$ effect. However, if we use the concept of hyperconjugation, and donate a $\pu{\pi}$ electron from the ring to the nitrogen and then a methyl group (making it $\ce{-CH3}$), unbounded to any other atom, we can imagine the oxygens in the picture replaced with methyls and will get the same distribution of charge in the ring: a positive charge at the ortho and para positions. This will result in meta substitution if any reaction does occur.

For $\ce{-N(CH3)3}$, it's like hyperconjugation in reverse. Interestingly, while $\ce{-N(CH3)3}$ makes it very difficult to substitute, and always meta, $\ce{-NH3+}$ is only a weak meta director.

Reference

Advanced Organic Chemistry, L.F. Fieser, 1961, p628


Effect of evaporation on the concentration of a saturated solution

When a solution is saturated, it contains the maximum amount of solute which can be dissolved at that particular temperature. So effectively it is at its "maximum concentration" (for a given temperature).

This means that if your solvent is evaporating and the temperature remains constant, then solute will precipitate out of solution. The concentration of the solution will still be at its maximum as you are concerned with the amount of solute dissolved.

If you are heating up the solution in order to evaporate the solvent then the increase in temperature might mean that your solute solubility increases. This will depend on what solvent/solute you have.

However, a saturated solution which cools down without being disturbed may form a super-saturated solution whereby its concentration has gone beyond the solubility of the solute at that temperature.


Preference of grignard between aldehyde and double bond?

The Grignard will first attack the aldehyde carbon, giving you a secondary alcohol. Since the alcohol is strongly polarized, it would be difficult for the double bond to be attacked.


Planar transition-state of proline racemization facilitated by proline racemase

As others have written in the comments, there are large parts of your question that can't be answered by experiment. However, it is possible to come up with a mechanistic hypothesis (and even better, also an alternative hypothesis) and gather evidence that rejects one hypothesis and is consistent with another.

Structural data supports the mechanism you sketched for the PLP-independent racemases. The structure of proline racemase in complex with the transition-state analog (PDB ID 1W61) shows fairly symmetric binding interactions between the enzyme active site and the ligand:

In the figure (a screen grab from https://www.rcsb.org/3d-view/1W61?preset=ligandInteraction&sele=PYC), the inhibitor is labeled [PYC]700, and it has similar interactions on the left and the right (for example two main chain hydrogen bond interactions with the carboxylic acid group each, and one cysteine residue each in the vicinity of the inhibitor.

The binding pocket is fairly symmetric (view from the "backside" with respect to previous image):

In the paper reporting the structure, the authors point out "two Cys residues optimally located to perform acid/base catalysis through a carbanion stabilization mechanism". They also suggest that the sulfur atoms sandwich proline when it binds, "squeezing" it into a conformation closer to planar than it would be in solution.

However, as the proline-bound structure of a different, putative racemase shows, the active site does accommate proline with a non-planar alpha carbon:

http://www.rcsb.org/3d-view/6J7C?preset=ligandInteraction&sele=PRO

  When I was first reading about this, I did not believe it because of the negative charge on the alpha-carbon. I decided to draw the resonance structure of the transition state, to find a double bond between the alpha and carboxyl carbons.

As the enzyme is stepping through catalysis, the enzyme active site binds to the molecule in different charge states, and itself changes charges as the cysteine residues accept and donate protons. In the cited paper, it is left open whether the reaction proceed in a concerted manner (one proton abstracted on one side while one is donated on the other) or via a carbanion intermediate. They do point out the the pKa of the alpha carbon is  "in the range 21–32", arguing against a long-lived intermediate.

  But my question is, how "planar" is the transition state, and how might one determine the exact bond angles at the alpha-carbon?

As you can read from the excellent comments, there is no good experiment to get at the exact structure of the transition state, and it is not a given that the transition state is planar. However, if the carboxylate group is on one side of the ring before the reaction and on the other side of the ring after the reaction (and stays attached to the alpha carbon throughout), at one point on the reaction coordinate the structure around the alpha carbon has to be planar (or some version of trigonal bipyramidal for a concerted mechanism). Therefore, the active site has to be able to accommodate that structure.

It would be possible to study this system using computational methods, to at least get a sense of the energy landscape in the absence of enzyme. The calculation with enzyme is complicated by the fact that the enzyme undergoes fairly large conformational change upon ligand binding.


Which is more stable Ethyl carbocation or Neopentyl carbocation?

In that case ethyl carbocation has two alpha hydrogen atoms as CH2 is bonded to C+.And in case of neopentynyl carbocation CH3 is attached to C+ .more alpha hydrogen atoms ,more hyperconjugative structures,and more stability.and there are three donating groups in neopentynyl carbocation.these may be the reasons


The standard reduction potentials for the following half reactions can be found here.

$$
\begin{align}
\ce{Fe^3+(aq) + e- &-> Fe^2+(aq)} &\quad E^\circ &= \pu{+0.77 V} \\
\ce{I2(s) + 2 e- &-> 2 I-(aq)} &\quad E^\circ &= \pu{+0.54V} \\
\ce{Br2(l) + 2 e- &-> 2 Br-(aq)} &\quad E^\circ &= \pu{+1.07V} \\
\ce{Cl2(g) + 2 e- &-> 2 Cl-(aq)} &\quad E^\circ &= \pu{+1.36V}
\end{align}
$$

You can see from this that only iodide is a strong enough reducing agent to reduce $\ce{Fe^3+}$ to $\ce{Fe^2+}$ at standard conditions. Even with non-standard concentrations it will be very difficult to get bromide to do the reduction because the difference in electrode potential is large.

The trend in electrode potentials for the halogens can be explained in terms of the increasing electronegativity going from iodine to chlorine which increases the first electron affinity. It just so happens that the crossover point with the iron reduction is between iodine and bromine.


Is the neutral di-µ-oxo-bridged manganese dimer with eight ammonia ligands a reasonable structure?

The structure is extremely reasonable to the point where I would ask why any other di-oxido-bridged manganese(II) ammine complex would not show this structure if it were proposed. To describe why, I am going to be lazy and just copy my molecular orbital scheme of a general octahedral 3d-metal complex that I have used many times before.

Scheme 1: Octahedral $\ce{[ML6]}$ complex with no π interactions. Image copied from this answer and originally taken from Professor Klüfers’ internet scriptum to his coordination chemistry course.

As you can see, the four empty orbitals ($\mathrm{a_{1g}}$ and $\mathrm{t_{1u}}$) are indeed present and part of the bond-forming orbitals, but the fact that the $\mathrm{e_g}$ orbitals are also half-filled does not prevent them from participating, too. In fact, these orbitals turn into $\mathrm{e_g^*}$ orbitals in the resulting complex, formally antibonding. This, however, doesn’t matter, because there are still many more electrons in bonding orbitals ($\mathrm{t_{1u}, a_{1g}}$ and $\mathrm{e_g}$) that this small amount of lost energy is more than offset. Typically, for a high-spin $\mathrm{d^5}$ complex, the bond order would be somewhere around 5/6. Still well within the overall bonding range.

If one looks at the kinetic stability of various complexes, one will find that $\mathrm{d^3}$ and low-spin $\mathrm{d^6}$ often give very inert octahedral complexes as no antibonding orbitals (of the simplified scheme) are populated—but even a $\mathrm{d^{10}}$ or $\mathrm{d^0}$ metal will form these complexes, their structures just won’t be as long-lasting. (Zinc has been shown to equilibrate between tetra, penta and hexacoordinated complexes in aquaeous solution as I was told in my coordination chemistry lecture.)

So yes, this octahedral complex is perfectly fine.


Which is more stable? CF2 (singlet) or CF2 (triplet)

So generally singlet is less stable as triplet has half-filled orbitals which leads to stability.

However whenever a group that can give  electrons by back donation (examples being lone pair carriers like $\ce{-OH},$ $\ce{-NH2},$ etc. are attached, they stabilise the singlet by back donation of their lone pair into the vacant orbital of the singlet carbene.

This is similar to back bonding in $\ce{BF3}.$


Calcium carbonate and hydrogen peroxide reaction

The effect of aqueous hydrogen peroxide on natural limestone has been examined in a different context (removal of moss and algae from the walls of frequently visited and therefore illuminated show-caves).

The results were published in Environ Pollut., 2003, 122, 417-422 (PubMed).

The authors exposed etatons of limestone to various aqueous solutions for about four months and examined possible damages using a scanning electron microscope. They found:

  Based on SEM study, the polished surfaces of the carbonate etalons did not show any changes after 135-day exposure to rainwater, 15% peroxide, and 30% peroxide, respectively. No etched pits or any other visible
  effects of dissolution [...] were detected.


Weight per volume percent. It's basically the density.

$$ w/v = \frac{mass_{solute}}{volume_{solution}} * 100 $$

Usually it includes units too, but you didn't specify. I'm guessing it means grams per liter?

For more, see http://www.ausetute.com.au/wtvol.html


Configurations of Reactants in Second Order Nucleophilic Substitutions

I tried to pinpoint you in the right direction through comments, but apparantly I failed. The bottom line is that the molecule you drew and the one in the textbook answer are identical. Here’s why:

In the top row, you can see the (S)-2-chloropentane, which I have drawn as you have on the left, slowly rotating around an axis equivalent to the reaction arrows to give you the same molecule drawn in a different orientation.

Now, if you do an $\mathrm{S}_{\mathrm{N}}2$ reaction, all you need to do is replace the chlorine atom and invert the stereochemistry which I have done for both cases. You can now see that both sides are identical.

It does not (and must not) matter whether you draw the 2-chloropentane with the substituted carbon to the right or to the left. Both ways are identical. Usually, you will chose an orientation that is consistent with further or earlier steps of your synthesis, but if you only have one one reaction that point is moot. (It matters a lot more in total synthesis, where it helps to draw reactands in a way that show their conformation in the final molecule.) Furthermore, there is a IUPAC style guide out there somewhere, which probably says one of the two ways is to be preferred if no other reasons speak for the other … but that’s more relevant for journal publications, not for practising or for exam answers.

So just for total consistency’s sake: Everything you did was correct; just remember that molecules can be drawn with different orientations, and learn to turn them around before your inner eye.


Numerical simulation and calculations of pH

As an answer to your overall question: it is indeed possible to do numerical simulations of pH in different solutions or mixtures, and there is a number of programs or spreadsheets available if ou search the net. My personal favorite is Curtipot made by Prof. Gutz: http://www.iq.usp.br/gutz/Curtipot_.html


Will gaseous ionic compounds be free moving ions?

Usually not. Boiling point rarely exceeds 4-5 thousand kelvin. A typical ionic bound energy is about 5 eV. 1 eV is roughly 11 thousand kelvin, so ions in low-temperature vapors exist as molecules. When the temperature becomes enough to break ionic molecules, it is enough to strip one or two electrons from atoms, so high-temperature vapors will be plasma with electrons and positive ions.


The most important factor in these reactions are nucleophiles. Here, the use of strong base (Sodium methoxide) in both of the reactions favors the bi-molecular mechanisms ($\ce{S_N2}$ and $\ce{E2}$).

Now, the second factor would be degree of halogen-linked carbon. Lower degree carbons ($\ce{\text{deg.} < 2}$) favor $\ce{S_N2}$, while higher degree favors $\ce{E2, E1, S_N1}$. So, $\ce{S_N2}$ is favored in first and $\ce{E2}$ in second.

Temperature conditions and nature of solvent are least important factors, and should be looked up when above two factors are similar.


What is the constant of integration in a series reaction?

There might or there might not be a problem with your final equation. Either you have made a mistake while using the integrating factor to solve your differential equation or you made some further substitutions so that your constant $C$ contains something more than just the "standard stuff" I would have expected from the method. Thus it's best I go through the derivation from the beginning so that it'll become clear what I mean.

The method of solving differential equation using an integrating factor is explained well here. You have used the correct integrating factor $u(t)$. For the differential equation at hand, i.e.

\begin{align}
  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= k_1[\ce{A}] - k_2[\ce{B}] \\
  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} + k_2[\ce{B}] &= k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \ ,
\end{align}

it is indeed

\begin{align}
  u(t) = \exp\biggl( \int_{0}^{t} k_2 \, \mathrm{d} t \biggr) = \mathrm{e}^{k_2 t} \ .
\end{align}

With the integrating factor you can get $[\ce{B}]$ according to

\begin{align}
  u(t) [\ce{B}] &= \int_{0}^{t} u(t) k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \, \mathrm{d} t + \tilde{C} \\
  [\ce{B}] \mathrm{e}^{k_2 t} &= \int_{0}^{t} k_1 [\ce{A}]_{0} \mathrm{e}^{(k_2 - k_1) t} \, \mathrm{d} t + \tilde{C} \\
  &= k_1 [\ce{A}]_{0} \biggl[\frac{1}{k_2 - k_1} \mathrm{e}^{(k_2 - k_1) t} \biggr]_{0}^{t} + \tilde{C} \\
  &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{(k_2 - k_1) t} - 1 \bigr) + \tilde{C}
\end{align}

Rearranging for $[\ce{B}]$ gives

\begin{align}
  [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \underbrace{\frac{ \mathrm{e}^{(k_2 - k_1) t} - 1 }{\mathrm{e}^{k_2 t}}}_{= \, \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t}}  + \frac{\tilde{C}}{\mathrm{e}^{k_2 t}} \\
 &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) + \tilde{C} \mathrm{e}^{-k_2 t} \ .
\end{align}

This equation differs from yours a little if your integration constant $C$ would be the same as my $\tilde{C}$. But if you used some substitutions so that your integration constant is $C = \frac{k_2 - k_1}{k_1 [\ce{A}]_{0}}  \tilde{C} - 1$ then your solution and mine would be equivalent. If you indeed made a mistake then I hope my derivation cleared things up a bit and if you had intended the equation to look that way by introducing the substitution then at least you know what integration constant I mean when I am going to explain the method for determining its value in the following.

The integration constant can now be determined by enforcing boundary conditions. In this particular case this means that at the start of your reaction ($t=0$) you require that $[\ce{B}]$ is equal to the initial concentration of your reactant $\ce{B}$, i.e. $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$. So, plugging $t=0$ and $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$ into the equation for $[\ce{B}]$ you get

\begin{align}
  [\ce{B}]_0 &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{0} - \mathrm{e}^{0} \bigr) + \tilde{C} \mathrm{e}^{0} \\
  [\ce{B}]_0 &= \tilde{C} \ .
\end{align}

In the case of your reaction $\ce{A -> B -> C}$ the initial concentration of $[\ce{B}]$ will usually be $0$ as $\ce{B}$ is only an intermediate product, so $\tilde{C} = [\ce{B}]_0 = 0$ and the final equation is

\begin{align}
  [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) \ .
\end{align}


Why does the AraC arabinose sensing system only respond to the presence of L-arabinose?

Your question is really about the origin of specificity in molecular recognition.

The simple answer to this is that two molecules (like the AraC protein and arabinose) bind to each other because they make favorable interactions with each other. ("Favorable" here means that free energy is released in the "reaction" of going from separate molecules to the complex.) There's a number of different types of interactions, but the ones you typically talk about with non-covalent protein binding are van der Waals,  hydrogen bonding, and electrostatics.

The reason AraC binds to arabinose is that it has functional groups in the binding pockets which exactly match three dimensional organization of corresponding atoms in the arabinose molecule (see the experimental structure of the AraC/arabinose complex). You can see a schematic representation of these interactions here.

The thing about these interactions is that they only happen if the three dimensional geometry matches up -- we might write molecules in a 2D schematic form, but molecules are intrinsically 3D. So when you have the hydrogen bond between, for example, Tyr82 and the C4 hydroxyl of arabinose, it can only form that favorable interaction if the C4 hydroxyl ends up in the correct 3D position.

If instead of L-arabinose you have D-arabinose (or L-ribose), then you have a different spatial orientation of atoms. Your C4 hydroxyl no longer goes "down", but instead goes "up". Since Tyr82 is still in the same place, it can no longer form a favorable hydrogen bond to the C4 hydroxyl - the distance is too far. There's no way to put the 3D structure of D-arabinose into the binding pocket of AraC and keep all the interactions. If you lose enough of these interactions (and sometimes one is enough), the total energy of binding is no longer favorable, and there's no driving force for binding.

Keep in mind that it's not always an all-or-nothing thing. It all depends on which interactions are destroyed by changing the chemical structure of the ligand. For example, AraC has no difficulty binding D-fucose. The difference between L-arabinose and D-fucose is just an addition of a C6 methyl group. For AraC this addition doesn't disturb the existing interactions enough to significantly perturb them. Therefore, you still get a favorable energy of interaction when the AraC binds to fucose.


Reaction of 1 equivalent of Mg in ether with 1,4-dibromobut-2-yne

Cyclobutyne cannot exist.  There is no way to enforce 90° bends at both ends of a triple bond, but the carbanion formed by the Grignard reagent synthesis finds an easy way around this problem.

Let's say the magnesium reacts at position 1.  Then the negative charge would couple with the conjugated triple bond to spread to position 3:

$\ce{Br\overset{+}{Mg}\overset{-}{C}H2-C#C-CH2Br <-> Br\overset{+}{Mg}CH2=C=\overset{-}{C}-CH2Br}$

Then the carbon atom at position 3 can act as a nucleophile, displacing the remaining bromine as a bromide ion to give the product named in the book.


Acetophenone reacted with LDA and diethyl carbonate. Why do I need HCl?

If you draw what you have carefully, without omitting any charges or adding magically appearing protons, notice that the actual product you initially get is either the ethanolate addition product of ethyl benzoylacetate, a mixture of ethyl benzoylacetate and lithium ethanolate or the lithium enolate anion of ethyl benzoylacetate and ethanol. In any case, you still have an anionic product.

For isolation and purification, you want to get rid of that negative charge (initially introduced by the strongest base you added: LDA). This is called quenching and typically performed with weak acids. I personally would use ammonium chloride for this type of reaction, but carefully adding dilute $\ce{HCl}$ will do the trick just as well. You should use $\ce{HCl}$’s proton to remove any remaining negative charges and eliminate $\ce{LiCl}$.


Does the amount of solvent interaction energy depend on the absolute gas phase energy?

One way this can happen is if the different conformers have different dipole moments. Those with largest dipoles might have higher energy due to greater charge separation. However those conformations with larger dipoles will also interact most strongly with the surrounding solvent medium (dielectric), and thereby the solvent will reduce their energy by a greater amount.


Calculating change in state function following different paths

You wrote, in one of your comments, "My biggest question is how to demonstrate the proper use of $\Delta H = n C_p \Delta T$irrespective of the path...".
So:
In a closed system (no flow of matter across the boundaries) of constant composition (no chemical reactions),
$$dH = \left(\frac{\partial H}{\partial T}\right)_ p dT +  \left(\frac{\partial H}{\partial p}\right)_ T dp$$
$$= C_ p dT +  \left(\frac{\partial H}{\partial p}\right)_ T dp$$
$$= C_ p dT +  \left(V - \alpha T V \right) dp$$
$$= C_ p dT +  \left(V - T \left(\frac{\partial V}{\partial T}\right)_ p  \right) dp$$
Thus, at constant $p$,  $dp =0$, and
$$dH_p = C_ p dT \textit{, for all substances.*}$$
It is worth noting, as Chet Miller mentioned in the comments, that $C_p \equiv C_p(T,p)$ (i.e., $C_p \text{ is a function of both } T \text{and } p$ for real substances).
If there are no inter-particle interactions (e.g., an ideal gas), then $V = T \left(\frac{\partial V}{\partial T}\right)_ p $, and thus
$$\left(\frac{\partial H}{\partial p}\right)_ T = 0$$
You can confirm this yourself for an ideal gas by substituting $V = \frac{n R T}{p}$ for the second occurrence of $V$ in: $\left(V - T \left(\frac{\partial V}{\partial T}\right)_ p  \right)$.
Hence, for an ideal gas,
$$dH = C_ p dT \text{, always,} \textit{ even if the pressure isn't constant.}$$
In summary, if the substance is an ideal gas, $dH = C_ p dT$ regardless of whether there is a change in pressure.  If the substance is real, $dH = C_p dT$ only if the pressure is constant.

*Note that this relationship holds for all types of work.  If we add the constraint that the only work is $pV\text{-work}$, then we have the additional nice relationship that:
$$dH_p = C_ p dT = q_p \textit{, for all substances.}$$
If we keep the constant-$p$, closed system, and $pV\text{-work}$-only constraints, but remove the constant-composition constraint (allowing chemical reactions), then it will no longer be the case that $dH_p = C_ p dT$, but it will be the case that:
$$dH_p = q_p \textit{, for all substances.}$$


Ca energy release when gaining an electron

Two sign conventions: (1) the more common one states that a positive
  electron affinity value represents energy release when an electron is
  added to an atom; (2) the other states that a negative electron
  affinity represents a release of energy.

http://www.rose-hulman.edu/~brandt/Chem251/GenChem_Review_v3.pdf

  Recommended values for these electron affinities, in the units
  commonly employed in introductory texts and with the sign convention
  used here, are 2.37, 5.03, and 13.95 kJ/mol for Ca, Sr, and Ba,
  respectively. The endothermic electron affinities often quoted for Be
  and Mg are also too large and should be reported simply as ">0". An
  argument for a return to the original sign convention for the electron
  affinity is presented in this paper.

http://pubs.acs.org/doi/abs/10.1021/ed074p123

  OP: I know that the electron will be settled in the 3d orbital

Actually, as explained in the above article, the electron occupies 4p.

  OP: but still, why would energy be released?

Very little energy is released and the lifetime of Ca- is on the microsecond scale in practical experiments.  The reference Contributions to the electron affinity of calcium and scandium may be the best source for theoretical models that do and don't predict a slight energy release.  According to Atomic negative ions: structure, dynamics and collisions, core electrons need to be considered in addition to the valence electrons in calculating the stability of Ca-.  Early calculations that considered only valence electrons predicted that Ca- was unstable.  Configuration-interaction study of differential correlation energies in Ca-, Ca, and Ca+ presents detailed calculations of energy levels.


Which double bond will react with m-CPBA to form an epoxide?

When two or more double bond are present then the more electron rich double bond will epoxidise. In above example the double bond containing the methyl group  has  electron donating  inductive  effect while the double bond containing the methoxy group  will have an electron donating resonance effect. The resonance effect is dominant over the inductive effect. Therefore the double bond containing the methoxy group is more electron rich. Therefore epoxidation occur on it.


Maximum Concentration of Benzene possible

Sorry, I figured it out- just substitute mols for mass/MW and density  for volume and rearrange until you get density/MW


metal salt solutions g/l means hydrate or anhydrous

Unless the experimental part of the paper provides additional details, such as equivalents or moles or the colour of the nickel chloride used ($\ce{NiCl2*6H2O}$ is green, while the anhydrous chloride is yellowish) you're up to guessing.

However, if the experiment you're trying to repeat is conducted in aqueous solution, it is safe to assume that you can use the hexahydrate.

Btw, gloves and proper waste management are a good idea here.


Atomic coordinates for DNA bases as they appear in a B-form or A-form duplex

I played around with a program called GROMACS about 4 years ago and don't remember much, but I do remember doing a tutorial simulating a DNA double helix and part of it included making the DNA by giving it a sequence and running a script. It can probably make A or B form.

EDIT: This was a long time ago and I forgot. It wasn't GROMACS, it was Amber. Look here http://ambermd.org/tutorials/ Tutorial B1 looks like the same tutorial I did with the DNA.


Many related questions have been asked on this topic before. I suggest you read these previous posts:

Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?

How can one explain Niobium’s weird electronic configuration?

The Aufbau Principle is a general rule of thumb to find the electronic configuration of an element but usually many exceptions arise in the d and f block elements due to electron-electron interactions.

The concept of valence electrons does get a little fuzzy when you consider $d$ block elements since the energies of the $\ce{3d}$ and $\ce{4s}$ subshells are quite similar so I think its better to avoid it. $\ce{d}$ block elements can show a staggeringly large number of oxidation states (for example manganese which shows almost all oxidation states from 0 to +7). This can't be explained using the concept of valence electrons and the octet rule.

$\ce{Fe}$ itself can lose 2 electrons to form $\ce{Fe^{2+}}$ or 3 electrons to form $\ce{Fe^{3+}}$ so now again you can see that you can't explain this by saying that $\ce{Fe}$ has 8 valence electrons.

Other exceptional configurations: (there are quite a few):
niobium, platinum, gold, cerium, gadolinium, etc.


Is there a consensus how to report coupling patterns greater than quartets?

There is no absolute consensus for reporting abbreviations for splitting patterns. Even Magnetic Resonance in Chemistry, a journal that I generally take as the best guide for reporting NMR data, does not have recommended abbreviations for multiplicities above quartets. Further, there are general inconsistencies in reporting formats within this journal, with abbreviations such as s, sept, spt being used for septet. Additionally, the actual term used to describe the splitting varies. For instance, both pentet and quintet are commonly used to describe a 1:4:6:4:1 splitting.

What is consistent, however, is that the abbreviations are defined within the paper.

Commonly used terms and abbreviations that I have come across include:

singlet (s)
doublet (d)
triplet (t)
quartet (q)
pentet (p, pent, pnt, pentet) or quintet (q, quint, qui, qnt, quintet)
sextet (sext, sxt, sextet)
setpet (s, sept, spt, septet) or heptet (hept, hpt, heptet)
octet (oct, octet)
nonet (non, nonet)

While splittings of higher multiplicity are less common within mainstream CHN organic molecules (septet for isopropyl substituents are a common exception, and 'apparent' splittings are excluded), they are more common with heteronuclear substitutions and organometallic systems, where spin numbers greater than 1/2 are frequently encountered. Reporting 'apparent' splittings of, say, (CH3)2CHCH2R as a nonet (where Jab ~ Jac) is not strictly correct according to the 2nI+1 rule, although in non-NMR journals still occurs, and is certainly taught at undergraduate NMR level.

As for reporting multiple couplings (doublet of septets, for example), again there is no strict rule to adhere to, although the commonly adopted practice is to report the larger coupling first. So, a doublet coupling of 10Hz and septet coupling of 2Hz would be a doublet of septets. If it were doublet (2Hz) and septet (10Hz), one would report a septet of doublets.


Can thionyl bromide produce alkyl bromides?

Yes, thionyl bromide can be used with alcohols for synthesis of alkyl bromides.
No, the process is not as efficient.  The reasons for the inefficiency include are primarily (a) thionyl bromide is much less stable than thionyl chloride, and (b) that a bromide-tribromide redox equilibrium is far more relevant than a chloride/trichloride equilibrium, which leads to undesired side pathways instead of the desired reaction.

For more information, see this 1955 paper from Frazer & Gerrard.

  On the other hand, pyridinium tribromide is readily formed during the addition of thionyl bromide to base hydrobromide, and its intervention weakens the analogy between the thionyl bromide and the thionyl chloride system.

In the same paper, a yield of 75% is reported for the synthesis of 2-bromobutane from butan-2-ol using pyridinium tribromide (which itself would be synthesized from thionyl bromide and pyridinium bromide).


What is the mechanism of functionalisation of nanoparticle using long chain alcohols

The longer the alykylchains, the more likely you can form self-assembled monolayers (SAMs), which will hold themselves together by covalent interactions between chains.  The nanoparticle analog would be miceles.

However, alcohols are a fairly weak ligand for silver, so I'm not sure this is feasible.  SAMs are known to form on silver with carboxylate ligands.


What is a rigorous definition of gas volume, and how is the Van der Waals equation derived?

I edited the first van der Waals equations in your question, because it was incorrect.

First, the volume available to the gas is pretty much was you think: it's the space left for it to occupy, i.e. the volume delimited by the container. If you think of a gas tank, it's the interior volume of the tank. For systems of macroscopic dimensions, there is no real ambiguity as to its definition, which can be purely geometrical. For nanoscale systems, such as accessible volume in porous media, the definition is much harder: it is typically done by constructing a Gibbs dividing surface between the two phases considered (here, a solid and a gas).

Second, importance of both terms in a real-life example. Let's write the van der Waals equation in terms of density, rather than volume and quantity:

$$ \left(P+a\rho^2\right)\left(\frac{1}{\rho}-b\right)=RT $$

Let's take a practical example of a gas with a large quadrupolar moment, CO2. We have $a$ and $b$ parameters for it from this source, and we can plot $\rho(P)$ at $T=300$ K:

(black: ideal gas; red: van der Waals gas; green: only $b$; blue: only $a$). We see that excluded-volume correction by itself is not much of a gain, but it's actually crucial once you include the attraction, because otherwise the phase transition (the cusp in the blue curve) occurs at much too low pressure.

Finally, regarding the derivation of the van der Waals equation, this is typically a topic of statistical thermodynamics and there are many such demonstrations. My favorite is the classical one by Hill, but it may not use the notations in modern textbooks. I'm also partial to that in McQuarrie’s book. All statistical physics derivation introduce the excluded volume in a very natural fashion, but they show more variation on the way they arrive at the $a$ term (so of them find the exact analytical form of the equation, while in some derivations you find a series expansion whose first term is the van der Waals equation).


Which of the following is the more stable carbocation?

Vinyl cations can also be generated by the protonation of acetylenes in strong acid.  The vinyl cations are captured by a nucleophile to produce a stable alkene with the nucleophile attached to the double bond.

  Which of the following is a more stable carbocation?

This route to vinyl cations has been well studied.  As shown in the following diagram, protonation of phenyl acetylene can proceed in two different ways.  The acetylenic carbon attached to the phenyl ring may be protonated or the terminal acetylenic carbon may be protonated.  As you can see, these two routes produce the two vinyl cations you want to compare.

Upon examining the product(s) found in this reaction, we find that only alkene A2 is produced.  Therefore, the reaction proceeds through the intermediacy of vinyl cation A1, not B1.  Consequently, the pathway involving A1 must be lower in energy, or said differently, vinyl cation A1 is lower in energy than vinyl cation B1.

Reasons Why Vinyl Cation A1 Is More Stable Than Vinyl Cation B1

The barrier to rotate the phenyl group is very low. Hence, in A1 the phenyl group can readily adopt a conformation where the aromatic p-orbitals are aligned with the empty p-orbital on the adjacent carbocation center. Additional resonance structures can now be drawn with the positive charge delocalized into the aromatic ring.  There is no such resonance stabilization with the aromatic ring possible for vinyl cation B1. This suggests that vinyl cation A1 is more stable than vinyl cation B1.
In addition to the resonance structures involving the aromatic ring, as shown at the bottom of the diagram, one can draw 2 more hyperconjugated resonance resonance structures for vinyl cation A1, but only 1 hyperconjugated resonance structure can be drawn for B1; again suggesting that vinyl cation A1 is more stable than vinyl cation B1.
Resonance effects are generally stronger and more controlling than inductive effects.  Therefore, the inductive arguments presented in @orthocresol 's answer play only a minor role, being outweighed by the stronger resonance effects described above.


For $\ce{HA <=> A- + H+}$  with a particular $K_a$ value at a particular temperature (Also, remember it usually applies to weak acid as $K_a$ of strong acid is rather meaningless),
Consider its equilibrium expression: $$K_a=\frac{\ce{[A^{-}][H+]}}{\ce{[HA]}}$$ when water (of the same temperature) is added, all three component concentration will decrease with the same extent.

Think about whether the numerator and denominator has a bigger decrease and you will find the answer (temperature is relatively constant). And for your free energy concern, being out of equilibrium necessarily means there is a favorable driving force pushing the reaction in one direction to reach back eqm.

Acquire a physical picture about the initial and equilibrium state and the dynamic state of a reaction in terms of collision theory. Think equilibrium in the form of competing forward and backward reaction and equilibrium constant is actually derived from rate constant.

Another reminder: Temperature and concentration/pressure all affect free energy. At the initial state (when you just add water and no reaction has taken place yet), there is only change in concentration but not temperature. So even if the temperature is changed upon reaction, the percentage of ionization will also be higher in all circumstances.


how many electrons are in the second subshell of this element?

You're answer is correct. The s-shell has 2 electrons in the 1s subshell, 2 electrons in the 2s subshell, 2 electrons in the 3s subshell and 1 electron in the 4s subshell. The d-shell has 5 electrons in the 3d subshell.

The element with atomic number 24 is chromium, who's electron configuration can be shown diagramatically as follows:

Note that the 3d shell is 'half-filled' before the 4s subshell is completely filled.

Chromium is one of the 'exceptions to the predicted electron configurations' when applying Madelung's rule for quantum numbers and electron configuration. The 'Aufbau' principle (represented in diagram below) can be used to determine the electron structure of relatively simple atoms, by hand, as long as one remembers the anomalies (such as chromium and copper). The 'anomaly' is typically 'explained' by stating that the a 'half-filled' or 'completely filled' d-subshell is more stable and will therefore happen before 'completely filling' the 4s subshell.

In modern quantum computational chemistry, various numerical methods (such as Dirac Hyper Hartree–Fock method and density functional theory) have been devised to iteratively solve the Schrodinger equation for more accurate calculation of electronic structures of larger atoms and molecules.


What is the heat of reaction of weak acid reacting with weak base compared to strong acid reacting with strong base?

Strong acid and strong base

First, lets see where the enthalpy of $\pu{-57.3 kJ/mol}$ comes from. In aqueous solution, strong acids and bases react completely to form $\ce{H+}$ and $\ce{OH-}$, respectively. The neutralization of a strong acid with a strong base then is always the same reaction:

$$\ce{H+(aq) + OH-(aq) -> H2O(l)}$$

We can calculate the enthalpy of reaction from the enthalpies of formation at a temperature of $\pu{25 ^\circ C}$.

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{H+(aq)}) &= 0 \\
\Delta H_\mathrm{f}(\ce{OH-(aq)}) &= \pu{−230.0 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{H2O(l)}) &= \pu{−285.8 kJ/mol}
\end{align}
$$

Comparing enthalpies of formation for the products with those of the reactants, we get  reaction enthalpy of $\pu{−55.8 kJ/mol}$. The value given in the question is slightly different (maybe for a different temperature, which is not given).

Weak acid and strong base

For this case, the reaction is the following (with $\ce{AH}$ as the weak acid):

$$\ce{AH(aq) + OH-(aq) -> A-(aq) + H2O(l)}$$

The reaction enthalpy for this reaction depends on the nature of the weak acid. Specifically, the answer depends on the difference in the enthalpy of formation of $\ce{A-(aq)}$ vs $\ce{AH(aq)}$. Conceptually, it takes energy to break the $\ce{A-H}$ bond, and the interaction with the solvent water changes when you go from $\ce{AH(aq)}$ to $\ce{A-(aq)}$. Here are two examples:

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) &= \pu{−484.3 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) &= \pu{−486.0 kJ/mol}
\end{align}
$$

Difference is $\pu{-1.7 kJ/mol}$.

$$
\begin{align}
\Delta H_\mathrm{f}(\ce{NH4+(aq)}) &= \pu{−132.5 kJ/mol} \\
\Delta H_\mathrm{f}(\ce{NH3(aq)}) &= \pu{-80.893 kJ/mol}
\end{align}
$$

Difference is $\pu{51.6 kJ/mol}$.

So sometimes the enthalpy is higher, sometimes lower than strong acid/strong base neutralization.

Weak acid and weak base

The weak acid/strong base consideration showed us that once you have a weak acid, the enthalpy depends on the nature of the acid. So can we make a general statement about the enthalpy of a reaction between a weak acid and a weak base? We start by writing down the chemical equation (weak base shown as $\ce{B}$):

$$\ce{AH(aq) + B(aq) <=> A-(aq) + BH+(aq)}$$

The enthalpy of this reaction could be positive or negative (if you read it backwards like this:

$$\ce{BH+(aq) + A-(aq) <=> AH(aq) + B(aq)},$$

you can call $\ce{BH+(aq)}$ the weak acid and $\ce{A-(aq)}$ the weak base, and the sign of the reaction enthalpy switches.

Notice that this is written as equilibrium. Depending on the $\mathrm{p}K_\mathrm{a}$ values of $\ce{AH}$ and $\ce{BH+}$, this reaction will have its equilibrium more on the left or on the right. If the reaction does not go to completion, the enthalpy (based on amount of starting materials) will be less.

Let's take the neutralization of acetic acid and ammonia as a specific example:

$$\ce{CH3COOH(aq) + NH3(aq) <=> CH3COO-(aq) + NH4+(aq)}$$

The $\mathrm{p}K_\mathrm{a}$ values are separated by more than 4 units, so this reaction goes almost to completion. The reaction enthalpy is:

$$
\begin{align}
\Delta H_\mathrm{r} &= \Delta H_\mathrm{f}(\ce{CH3COO-(aq)}) + \Delta H_\mathrm{f}(\ce{NH4+(aq)}) - \Delta H_\mathrm{f}(\ce{CH3COOH(aq)}) - \Delta H_\mathrm{f}(\ce{NH4+(aq)}) \\
  &= \pu{−486.0 kJ/mol} - \pu{132.5 kJ/mol} + \pu{484.3 kJ/mol} + \pu{80.893 kJ/mol} \\
  &= \pu{-53.3 kJ/mol}
\end{align}
$$

That is pretty close to the strong acid/strong base case.

Back to the question

To prove that

  "weak acid and weak base reaction will produce lesser heat than constant enthalpy" of strong acid and strong base neutralization

you would have to do this calculation for every weak acid and weak base combination out there. I would be careful making a statement like this. I think the idea was that because weak acids and bases dissociate much less, less heat is given off (per amount of starting material). However, when you combine a weak acid and a weak base, sometimes the reaction goes almost to completion. As you can see from the example I gave, there are cases for which the enthalpy is very close to that of the strong acid/strong base combination.

Gibbs energy of reaction

For the Gibbs energy of reaction, the situation is much simpler. Weak and strong are defined by the magnitude of acid dissociation equilibrium constants ($K_\mathrm{a}$, and related $\mathrm{p}K_\mathrm{a}$). The difference in $\mathrm{p}K_\mathrm{a}$ of the acid and the conjugate acid of the base is a direct measure of the standard Gibbs energy of reaction: the larger the difference, the larger the Gibbs energy (with appropriate signs, so the strong acid/strong base combination would have the most negative Gibbs energy of reaction, the strong acid/weak base or weak acid/strong base combination would also have negative values, and for the weak acid/weak base combination if would depend on the specifics).


Procedure/flow chart is a short, often diagrammatic, summary of a reaction sequence with reagents, solvents, temperature & yields. It is expected that a competent synthetic chemist would be familiar with the reactions shown and would be able to understand how the product was made.

Synthetic method contains all the details of the reaction. Sources and quantities of reagents and solvents, times and temperatures of reaction, workup and purification procedures, characterisation of product.


The reduction of a ketone all the way down to an alkane using catalytic hydrogenation over palladium on charcoal is not a general/widely used process.

Quoting from Comprehensive Organic Functional Group Transformations III:

  ...while the reduction of benzylic ketones by hydrogenation over palladium is often an efficient procedure , it ... sees limited use. An isolated report of the reduction of aliphatic ketones by hydrogenation over platinum on Montmorillonite K-10 merits ...attention here and further study

The majority of examples of this reaction occur using benzylic ketones, carrying out the reaction in the presence of aqueous acid in order to facilitate the loss of the hydroxyl caused upon initial reduction of the ketone. The adjacent aromatic ring helps stabilise the developing (+) charge as the C-O bond cleaves, which is what makes benzylic ketones good substrates for this reaction.

  Ref: J. Med. Chem. 2001, 44, 3424

In the context of lab scale chemistry (i.e. not industrial processes), several good methods exist including the Wolff–Kishner (or variants as Jan pointed out in the comments), the Clemmensen, and the Mozingo reduction. Some success is also noted via initial reduction, mesyaltion, and treatment with a hydride source.


What happens to energy changes in non-reversible processes?

$\Delta U = q + w$ and $dU = dq + dw$

This is the first law of thermodynamics.  It is not an inequality.

$dU\leq TdS-PdV$ is correct when there is only PV work. It is incorrect when there is other work such as electrical or gravitational.

  the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.

The PdV term alone represents the reversible work, the TdS term represents reversible heat

  Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?

the system must lose internal energy in a spontaneous change


Predict the number of peaks in H-1 NMR of sesamolin

For sesamolin, I think the pairs of protons at A and N would have the same chemical shift. In the spectrum this would appear as a singlet with an area corresponding to 4 protons due to the accidental equivalence of the A and N site. It looks like you did this in assignment of sesamin, which looks good to me.

Assigning shifts to geminal protons can be tricky. For example, the differing chemical shifts of the axial and equatorial protons in cyclohexane can only be resolved at low temperature:

Some more examples below:

Both of the images and more information on the topic are available here:
https://www.chem.wisc.edu/areas/reich/nmr/05-hmr-02-delta.htm


Grignard vs organolithium reagent opening cyclohexene oxide

$\ce{RLi}$ is much more reactive than $\ce{RMgBr}$ and attacks through the SN2 process, thus inversion of configuration occurs. But in case of $\ce{RMgBr}$, $\ce{MgBr-}$ coordinates with epoxide O atom and forms carbocation on the adjacent carbon. Then a $\ce{C-C}$ bond migrates and the above product is formed.


Electrochemical manufacturing of small metal components

Nice idea, but difficult to do.

The copper cations must continuously enter the "mold" electrode to grow there, but they must not be able to grow outside.
If there are holes in the "mold" the crystals will grow towards them because there is where the cations will come from.
A permeable "mold" would let cations in over all the surface, but nothing would stop them to continue growing outside the membrane.

If the crystals tend to grow to the other electrode in the shortest possible path, you could place a very small electrode just above the center of the triangle of the other electrode.
The shape will, however, not be very defined.
Maybe in combination with a permeable "mold"...

Trapping of liquids inside the piece will be almost certainly a problem.
You are not growing a single crystal.
Many crystals will start growing all at the same time in different places of the electrode.
When they join somewhere, they will not fit nicely together, so some liquid will be trapped.
This occurs also when big crystals are made in a normal crystallization, especially when there is no stirring.


Amount of Energy Needed for Boiling a Litre of Water

It is simple. Density of water is $1\;\frac{\text{g}}{\text{ml}}$ as we have 1 kg water.

Now according to the laws of calorimetry the amount of heat required

  $Q = c \cdot m \cdot \left( {100{\text{ }^\circ\text{C}} - T} \right) + \Delta {h_{\text{vap}}} \cdot m$

Here

$c$ = specific heat capacity of water, which is 1 cal/(g °C)
$m$ = mass of water; i.e. 1 kg
$T$ = the room temperature in °C
$\Delta {h_{\text{vap}}}$ = specific enthalpy of vaporization of water; 540 cal/g

So you will get the amount of energy needed (in cal) to boil a litre of water

To get answer in Joule multiply the answer in cal by 4.184 J/cal. And be careful with units!


Which of sodium or lithium will be more reactive with oxygen?

Reactivity of Group 1 and 2 elements increases as you go down the periodic table.
So sodium is more reactive than lithium. Sodium will react with oxygen forming $\ce{Na2O}$ (sodium oxide). Lithium forms lithium oxide $\ce{(Li2O)}$.

Basically, there are general trends (reactivity with acids, air, etc.) down a group or along a period.