The correct answer is

  (d) none of these

In the usual substitutive IUPAC nomenclature, simple halogen compounds are always expressed by prefixes (‘bromo’, ‘chloro’, etc.). Hence, the senior parent structure of such compounds is not determined by a principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name).

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ring or the chain can be the senior parent structure; however, for the preferred IUPAC name, the ring is always selected as the senior parent structure.

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

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

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

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

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

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

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

Therefore, the benzene ring is selected as the senior parent structure for the preferred IUPAC name of the compound given in the question.

According to P-29.1.2, The chloromethyl ($\ce{-CH2Cl}$) substituent is a compound substituent group, i.e. a substituent group consisting of two parts, a simple substituent group (the parent substituent group, here: methyl) to which is attached one or more simple substituent groups (here: chloro).

According to P-16.5.1.1, parentheses are used around such compound prefixes. Therefore, the preferred IUPAC name for the compound that is given in the question is (chloromethyl)benzene.


alpha-hydrogen atom and keto-enol tautomerism

In general, the "alpha" hydrogen is a hydrogen attached to a carbon that is "alpha" (adjacent to) the substituent.

image source

So in this case, an "alpha" hydrogen is a hydrogen attached to the carbon alpha to the carbonyl.

image source

The following drawing represents a typical keto-enol equilibrium.  If we removed the last alpha hydrogen and replaced it with a substituent (alkyl group, phenyl group, bromine, etc.) then we could not form a double bond with that alpha carbon, it would have 4 non-removable substituents.  If we can't form a double bond that means we can't form an enol.

image source


Are there ways to predict which chemicals will kill certain bacteria?

The basic idea is sound, and is used to some degree; however, there are significant difficulties inherent in "tailoring" a chemical to selectively target certain organisms and not others.

There are three things that a chemical can do that make it toxic to cellular life:

Rip some vital chemical to shreds. Chlorine and other light halogens (fluorine, iodine) are huge here, because they have such high electron affinity that they can tear apart the otherwise relatively stable bonds within most organics, destroying that molecule. The most effective ones don't form any stable compounds; fluorine, for instance, is very effective at shredding organic compounds because it isn't happy just barging in on any one organic molecule and making itself at home; it'll deprotonate this one, split that one in half, before finally binding tightly and insolubly to calcium (robbing your body of that essential mineral).
Bind to a vital chemical and not let go. Carbon monoxide is a deadly danger to most aerobic organisms because the hemoglobin (or related compound) that loosely binds oxygen for transport to cells will bind more stably with CO, so the chemical can't transport oxygen and can't get rid of the CO. A lot of more selective poisons attach themselves to enzymes, not necessarily destroying them, but preventing the enzyme doing the work it's supposed to. Cyanide is poisonous specifically because it works like (well, it is) a nitrile group, of which there are very many in a lot of inorganic biochemicals, and specifically it targets a key enzyme involved in cell respiration and programmed cell death.
Mimic a chemical's interaction with a key receptor, causing it to go haywire. Many other "selective" poisons and antibiotics work this way; they fit into the receptor sites that a neurotransmitter or enzyme would, triggering some behavior of the cell, and by flooding the cellular environment with this chemical you cause the underlying programmed reaction to go haywire (or stop completely). A lot of insecticides and pesticides do this; they increase or shut down cell metabolism, or in creatures with a nervous system they mimic or trigger the release of massive amounts of neurotransmitters, basically causing cardiac or respiratory arrest because any normal message to control these actions is completely overridden.

The problem with all this comes in identifying which bacteria have a particular enzyme, receptor or other targeted chemical process, and thus would be susceptible to poisoning by the chemical. In addition, the chemical must not be toxic to human biochemistry (otherwise, no matter how beneficial it might be as a disinfecting agent, it will never see approval as an antibiotic medicine).

Antibiotics must have selective toxicity; while they do not affect eukaryotic cells (such as the ones that make up most plants and animals, thus allowing the chemical to be taken internally or directly applied), they have a much more pronounced effect on bacterial and archaeal cells. Usually, the chemical targets something common to a large class of bacteria (such as the thick peptidoglycan mesh forming the cell wall of Gram-positive bacteria), or most/all bacterial cells (such as the S30 robosomal subunit, key to bacterial protein synthesis and thus found in nearly all bacteria, but completely absent from eukaryote cells which use a different mechanism). However, even within an antibiotic's "target class", there are bacteria that are naturally resistant (perhaps they don't accept the chemical through the cell membrane due to some variation unique to the genus) or develop genetic resistance through overuse or misuse.

Also, bacteria are fickle; most are at least harmless if not beneficial in certain areas (E. coli, for instance, produces the essential vitamin K2 in your intestine), but lethal in others (ingestion of e-coli or introduction through an open cut can cause a fatal infection). Staph is relatively harmless on the skin, but fatal in the spine. Antibiotics don't stop and question bacteria; they kill any bacteria that comes across them, and so overuse of antibiotics to kill one harmful bacteria will kill the bacteria in beneficial places as well.

Lastly, not every infection is bacterial. Fungi and protozoa produce a significant portion of food-borne illnesses and poisonings, as well as other types of infection. These life forms are eukaryotes and unaffected by most antibacterial agents (in fact, penicillin was discovered when a fungus contaminated a bacterial culture; the fungus produces the antibiotic chemical as a natural defense).


Refrigerators work by exploiting some properties of compressible gases.

To simplify greatly, gases tend to heat up when they are compressed and liquefied. And, when the liquid revapourises this process absorbs energy. |In a refrigerator those two processes are physically separated and this enables the transfer of heat from one place (the inside, which gets colder) to the outside (which gets warmer).

The way this works in a typical refrigerator is to have a compressor on the outside of the cold area that takes gas and compresses it while letting the extra heat dissipate into the outside environment. This combined process of compression and cooling (or, more accurately shedding the extra heat generated by compression) liquefy the gas. This liquid is then allowed to flow into a heat exchanger on the inside of the refrigerator where it is allowed to evaporate. This evaporation absorbs energy from inside the fridge making that environment colder. This gas is then allowed to flow into the compressor, completing the cycle.

This allows the fridge to convert mechanical energy in the compressor into a clever way of transferring energy from the inside of the fridge to the outside. This amounts to exploiting a phase change in the refrigerant into a pump that moves energy from cold to hot. Though this only works because we can separate the inside from the outside thermally enabling the two side of the process to behave normally (heat flows from hot to cold on both sides) but the net flow ends up taking energy from cold to hot by exploiting the energy used to cause a phase change in the refrigerant.


Why is 2-methylpropene less in energy than its alkene counterparts?

There are two things we need to understand before we can answer the question.

1) More highly substituted double bonds are generally more stable than less substituted double bonds.

This is because the $\ce{sp^3}$ hybridized carbon in the alkyl group is electron donating towards the $\ce{sp^2}$  hybridized carbon in the double bond. Electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower the energy of the electrons in that orbital.  As the electrons flow from the $\ce{sp^3}$ carbon to the $\ce{sp^2}$ carbon the energy of the electrons (and therefore the energy of the molecule) is lowered.

A "shorthand" way to illustrate this electron flow is by drawing hyperconjugated resonance structures like the following.

Consequently 2-butene (where we can draw 6 hyperconjugated resonance structures) is more stable than 1-butene (where we can only draw 2 hyperconjugated resonance structures).

2) Carbanion stability decreases in the order primary > secondary > tertiary

As we've just shown above, alkyl groups are electron releasing.  If we have a negative charge on a carbon (a carbanion), the last thing we want is to have more alkyl groups attached to that carbon releasing more electron density onto an already negative carbon atom.

Now we can return to your question and try to understand why 2-methylpropene is more stable than trans-2-butene.   Carbon-hydrogen hyperconjugation is at the center of the answer.  Let's compare some of the hyperconjugated resonance structures that can be drawn for these two molecules.

Remembering that a) more highly substituted double bonds are more stable and b) that primary carbanions are more stable than secondary carbanions, we can compare hyperconjugated resonance structures A and B. Resonance structure A contains a disubstituted double (more stable) and a primary carbanion (more stable), while resonance structure B contains a monosubstituted double bond (less stable) and a secondary carbanion (less stable). Therefore, resonance structure A should be more stable than resonance structure B.  Since resonance structure A is more stable than resonance structure B, 2-methylpropene will be more stable than trans-2-butene.

Finally, we note that cis-2-butene is less stable that trans-2-butene for steric reasons.  This allows us to place these $\ce{C_4}$ isomers in the following order of stability:

2-methylpropene is more stable than trans-2-butene which is more stable than cis-2-butene, which is more stable than 1-butene. If we examine the heats of combustion for these isomers, we see that our order of stability is confirmed.

image source


Degeneracy of second excited state of H-?

I think it is important to understand that for hydrogen atom (or any other one-electron system) all orbitals from the same shell have same energy. For instance, $E_\mathrm{2s} = E_\mathrm{2p}$, $E_\mathrm{3s} = E_\mathrm{3p} = E_\mathrm{3d}$, etc. Thus,

The first excited state of hydrogen atom would be one in which either $\mathrm{2s}$ or one of the three $\mathrm{2p}$ orbitals is occupied and it will be 4-fold degenerate: $\mathrm{1s^0 2s^1}$, $\mathrm{1s^0 2p_x^1}$, $\mathrm{1s^0 2p_y^1}$, $\mathrm{1s^0 2p_z^1}$.
Analogously, the second excited state of hydrogen atom would be one in which either $\mathrm{3s}$ or one of the three $\mathrm{3p}$ or one of the five $\mathrm{3d}$ orbitals is occupied and it will be 9-fold degenerate: $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_x^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_y^1}$, $\mathrm{1s^0 2s^0 2p_x^0 2p_y^0 2p_z^0 3s^0 3p_z^1}$, plus 5 configurations with only $\mathrm{3d}$-orbitals occupied.

For hydride the situation is different: it is not a one-electron system, so different orbitals from the same shell do not have same energy anymore. For instance, $E_\mathrm{2s} < E_\mathrm{2p}$, $E_\mathrm{3s} < E_\mathrm{3p} < E_\mathrm{3d}$, etc. Thus,

The first excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another $\mathrm{2s}$ one, i.e. a non-degenerate $\mathrm{1s^1 2s^1}$ state.
The second excited state of the hydride would be one in which one electron populates $\mathrm{1s}$ orbital and another one of the three $\mathrm{2p}$ ones, i.e. a 3-fold degenerate state: $\mathrm{1s^1 2s^0 2p_x^1}$, $\mathrm{1s^1 2s^0 2p_y^1}$, $\mathrm{1s^1 2s^0 2p_z^1}$.


Detecting the concentration of chlorpromazine

A cursory online search furnished

Graphene paste electrode for detection of chlorpromazine by Mohammad Hadi Parvin in Electrochemistry Communications, 2011, 13, 366-369 (DOI).
A multipurpose Voltammetric Sensor for the Determination of
Chlorpromazine in Presence of Acetaminophen, Uric Acid,
Dopamine and Ascorbic Acid by Binesh Unnikrishnan, Pei-Chi Hsu, and Shen-Ming Chen in Int. J. Electrochem. Sci., 2012, 7, 11414-11425 (PDF).


In principle, alkanes may be converted to alkyl bromides by photoinitiated bromination.

However, taking statistics (2 $\ce{CH3}$ vs 4 $\ce{CH2}$ groups) and stabilities of the radicals resulting from H abstraction by a bromine radical from your starting material into account, this approach is pretty worthless and will most likely yield 1,6-dibromohexane as a minor product in a complex mixture.

Consequently, your reasonable options are a) a different starting material (and a different synthesis) or b) simply buy 1,6-dibromohexane (around 20 Euro for 100 g).


True, Indigo can form hydrogen bonds, but with what? Is it water? Nope.

It makes hydrogen bonds with itself.

This makes the molecule incapable of bonding with water. Not only that, but the molecule itself is quite symmetric, with oppositely oriented polar bonds, that cancel out each others' dipole moments. Notice the molecule's similarity to an ace of diamonds:

This type of symmetry is called center of symmetry as it is about it's center we can observe the symmetry.

So we got it, it can't associate itself with water, and it's nonpolar. What do we expect? It doesn't dissolve in water.


Why does butter get soft with heat, but not ice?

The main difference between warming 'solid' butter and warming solid ice is that one is a mixture and one is a pure substance.

Ice has a crystal structure set in place by the shape and polarity of the water molecule. It requires quite a bit of energy to be removed from 0 degree water to cause the formation of the solid phase. The structure has two options - remain water, or, once cooled enough, become solid.

Butter is a mixture of fat and water and other components. As this mixture warms, the fat component changes viscosity and becomes more liquid and less sludgy. The water does not change much, but is trapped in the layers of the fats. At 40 F, just above refrigerator temperature, butter is 'hard' because the fats are not flowing. As the temperature warms, the fats start to move, thus softening the butter.

The butter softening behavior can be seen in some metals as they get heated close to the melting point. Iron becomes 'soft' and malleable, and can be forged. Ice has a particular, rigid structure, so never has a 'soft solid' stage.


Mechanism of lithium-halogen exchange of primary alkyl iodide

With primary alkyl iodides the (simplified) mechanism is likely similar to a $\mathrm{S_N2}$ process, but with attack of $\ce{t-Bu}$ at iodine instead of carbon.

Note that this is not necessarily the case for other substrates, e.g. aryl iodides, or aryl/alkyl bromides. In general lithium-halogen exchange can occur either via a polar mechanism (such as the one shown above) or via a radical mechanism (by two successive electron transfers from RLi to RX). There is also a solvent and temperature dependence. Why does the solvent matter? We know that ethereal solvents such as Et2O or THF are capable of acting as Lewis bases - or ligands for lithium - and possibly assist in deaggregation of the alkyllithium. This sort of thing can't be depicted easily with curly arrows, which is why I add the disclaimer that the mechanism above is (heavily) simplified.

In any case, this particular substrate (Tetrahedron Lett. 1986, 27 (17), 1861–1864) is fairly interesting because if the Li/I exchange were to occur via a radical pathway, the radical formed would cyclise onto the alkene via a 5-exo-trig pathway (even at −78 °C):

However, if you quench the reaction at −78 °C with methanol, no methylcyclopentane is observed, which rules out the intermediacy of a radical. [If you warm up the reaction before quenching then the organolithium can cyclise by nucleophilic attack onto the alkene.]

This particular case (and general mechanisms for Li/Hal exchange) are described quite thoroughly in Clayden's Organolithiums: Selectivity for Synthesis, Chapter 3.

There is also an excellent set of slides on Li/Hal exchange reactions, courtesy of the MacMillan group at Princeton; and also another set, courtesy of the Denmark group at the University of Illinois.


Gases produced by pyrolysis of cellulose

During pyrolysis, organic compounds are thermally decomposed in the absence of oxygen. The pyrolysis products are classified into categories based on their physical state of existence: char (solid), bio-oil (liquid) and non-condensable gases (gas). The relative proportions of these three product fractions significantly vary depending upon the process conditions, as is shown in the table below.

$$ \small
\begin{array}{lcc}
\hline
\text{Pyrolosis Technology} & \text{Residence Time} & \text{Heating Rate} & \text{Temperature} & \text{Char} & \text{Bio-Oil} & \text{Gases} \\
\hline
\text{Conventional} & \text{5-30}\ \mathrm{min} & \text{<50} ^\circ \mathrm{C\ min^{-1}} & \text{400-600} ^\circ \mathrm{C} & \text{<35%} & \text{<30%} & \text{<40%}\\
\text{Fast Pyrolysis} & \text{<5}\ \mathrm{s} & \text{~1000} ^\circ \mathrm{C\ s^{-1}} & \text{400-600} ^\circ \mathrm{C} & \text{<25%} & \text{<75%} & \text{<20%}\\
\text{Flash Pyrolosis} & \text{<0.1}\ \mathrm{s} & \text{~1000} ^\circ \mathrm{C\ s^{-1}} & \text{650-900} ^\circ \mathrm{C} & \text{<20%} & \text{<20%} & \text{<70%}^{[1]}\\
\hline
\end{array}
$$

The exact compositions of the products of cellulose pyrolysis at different temperatures can be seen below.

$$ \small
\begin{array}{lcc}
\hline
\text{Products} & \text{Peak Temp,}\ 500 ^\circ \mathrm{C} & \text{Holding Temp,}\ 400 ^\circ \mathrm{C} & \text{Peak Temp,}\ 750 ^\circ \mathrm{C} & \text{Peak Temp,}\ 1000 ^\circ \mathrm{C}\\
\hline
\ce{CO} & \text{0.99%} & \text{0.25%} & \text{15.82%} & \text{22.57%}\\
\ce{CO2} & \text{0.3%} & \text{1.45%} & \text{2.38%} & \text{3.36%}\\
\ce{H2O} & \text{3.55%} & \text{6.49%} & \text{8.72%} & \text{9.22%}\\
\ce{CH4} & \text{0%} & \text{0%} & \text{1.11%} & \text{2.62%}\\
\ce{C2H4} & \text{0%} & \text{0%} & \text{1.05%} & \text{2.18%}\\
\ce{C2H6} & \text{0%} & \text{0%} & \text{0.17%} & \text{0.28%}\\
\ce{C3H6} & \text{0%} & \text{0%} & \text{0.70%} & \text{0.80%}\\
\ce{H2} & \text{0%} & \text{0%} & \text{0.36%} & \text{1.18%}\\
\ce{CH3OH} & \text{0.25%} & \text{0.21%} & \text{1.03%} & \text{0.98%}\\
\ce{CH3CHO} & \text{0.01%} & \text{0.05%} & \text{1.58%} & \text{1.7%}\\
\text{tar} & \text{16.37%} & \text{83.35%} & \text{59.92%} & \text{49.12%}\\
\text{char} & \text{83.63%} & \text{6.17%} & \text{3.32%} & \text{3.91%}\\
\text{other} & \text{0.19%} & \text{0.16%} & \text{2.14%} & \text{1.78%}\\
\text{total} & \text{105.25%} & \text{98.36%} & \text{98.8%} & \text{99.86%}\\
\hline
\end{array}
$$

The holding time for each of these reactions was $30\ \mathrm{s}^{[2]}$. As shown in the table, $\ce{CO}$, $\ce{H2O}$, and $\ce{CO2}$ are the major gaseous products, with $\ce{H2}$ and hydrocarbons being produced in considerably smaller proportion.

$^{[1]}$ Patwardhan, Pushkaraj Ramchandra, "Understanding the product distribution from biomass fast pyrolysis" (2010). Graduate Theses
and Dissertations. Paper 11767.
$^{[2]}$ Hajaligol, M. R.; Howard, J. B.; Longwell, J. P.; Peters, W. A. Product Compositions and Kinetics for Rapid Pyrolysis of Cellulose. Industrial & Engineering Chemistry Process Design and Development Ind. Eng. Chem. Proc. Des. Dev. 1982, 21, 457–465.


Milk is mostly water, and diluting it (roughly speaking) with sulfuric acid to a concentration of ~50% milk and 50% sulfuric acid will result in a very acidic, but still aqueous, solution.  We can examine the likely behavior of other milk components using basic chemical principles.

Sugars.  Milk is rich a variety of di- and oligo-saccharides such as the eponymous lactose, as well as many other more complex molecules:

  [Free oligosaccharides are natural constituents of all placental mammals' milk. Human milk contains 7–12 g/L oligosaccharides, making the oligosaccharide fraction a major component of human milk. Compared with human milk, the concentration of oligosaccharides in the milk of the most relevant domestic mammals is smaller by a factor of 10 to 100.]

Fortunately, the effect of sulfuric acid on the sugars is easy to predict: they will simply stay dissolved in aqueous solution.  Hydrolysis of the sugars is possible, but will not happen to any great degree without heating or prolonged incubation.
Proteins.  Milk contains thousands of unique proteins.  However, by far the most predominant proteins are called caseins.  Caseins are relatively hydrophobic, proline-rich molecules and in milk are present as tiny aggregates -- a colloid -- of molecules.  Adding sulfuric acid will denature the casein.  As a result, the casein molecules are likely to stick to each other and form a centrifugable pellet that goes to the bottom of a centrifuge tube after spinning.  Proteins contain amino acids linked via amide bonds that are unlikely to hydrolyze in the acid, unless very high temperatures or very long incubation times are used.
Phospholipids.  Phospholipids in milk include glycerophospholipids and sphingolipids.  Polar lipids play an important role in stabilizing the triglyceride (fat) emulsions that keep fat in solution.  They do this by forming bilayers and other structures on the surface of triglyceride microdroplets.  To do this job, phospholipids have a phosphate head group attached to a fatty acid tail.  The fatty part of the phospholipid sticks to the triglyceride, and the polar head group interacts with aqueous solution.  The polarity of the head group is tunable by pH.  At neutral to mildly acidic pH, the phosphate head groups are (negatively) charged, greatly strengthening their propensity to interact with water and maintain emulsion stability.

Phospholipids may be partially hydrolyzed by the acid, but without very high temperatures or prolonged incubation, most molecules will remain intact.

The strong acid conditions that arise from adding $\ce{H2SO4}$ will protonate these head groups, neutralizing the negative charge and vastly weakening their emulsifying properties.
Triglycerides.  Triglycerides are the molecules that provide most of the fat content of milk.  They are triacylglycerols.  Unlike phospholipids, they have no polar head groups, and thus they do not dissolve in water.  They would "want" to form a separate, oily phase from the water (just like mixing vegetable oil and water), but the fact that they are usually bound in tiny droplets by phospholipids prevents this separation.  Because triglyceride oils are less dense than water, if they did ever get the chance to coalesce into large droplets, they would rise to the top of the milk.  This happens when cream naturally separates from fresh (raw) milk, but in homogenized milk, the droplets are artificially forced to much smaller sizes, effectively preventing the droplets from coalescing or rising.

Upond acidification, the loss of phospolipid surfactant activity breaks down the stability of the fatty droplets, allowing them to rapidly coalesce, so they rise to the top of the tube, a process that is only accelerated by centrifugation.

Hydrolysis of triglycerides in the acid is possible, but unlikely.  In addition to required prolonged incubation or high temperatures, keep in mind that the separated triglyceride phase is actually not exposed to the acidic solution very much at all.


The SN2 reaction involves an attacking group and a leaving group moving in concert. A pentacoordinate transition state is involved.  Typically, there is no rearrangement in an SN2 reaction.  Similarly for an E2, as a base removes a proton. a leaving group is ejected in concert to create an olefin, no rearrangement products are observed.

It is the SN1 and E1 reactions that proceed stepwise through distinct carbocation intermediates.  As the carbocation is being formed in these reactions, a hydride or alkyl group can migrate to produce a more stable cation.

The leaving group can be things other than a hydroxyl group, for example chloro, bromo, or iodo groups make good leaving groups.  Back in the early days of physical-organic chemistry many alcohols were used in reactions simply because they were readily available as natural products.  Alcohols could also be readily converted into acetates,  tosylates, brosylates, or other sulfonate esters, which were even better leaving groups and allowed for reactions to proceed under even milder conditions.


I find the top mechanism unlikely for two reasons.

1) In the top mechanism, the proton removed in the first step should be way less acidic than a typical hydroxyl since the intermediate is already negatively charged. Going to a di-anion should be very far uphill, and therefore, slow. If that di-anion was ever formed, the next step would be exceedingly fast

2) In the bottom mechanism, the first step should be fast. It's intramolecular and entropically favored. The second step is instantaneous.

I can't imagine any text book includes the top mechanism. Check your book for saponification of esters.


Surely there are both $\ce{H2O}$ and $\ce{OH-}$ which attacks the carbocation, as well as, heaven forbid, $\ce{H3O+}$ (and countless other species formed by the auto-ioniziation of water).

Usually textbooks record $\ce{H2O}$ as the attacker because it is much more abundant in the mixture.

After that, usually another $\ce{H2O}$ comes and take away the extra $\ce{H}$ with itself converting to $\ce{H3O+}$, but as I mentioned above, this step could also be done by $\ce{OH-}$ among many more species, but the concentration of those other species are too small that they are not worth mentioning.

PS: the last step can also be done by the product, i.e. t-butyl alcohol.


The idea of a 'rotatable bond' is somewhat arbitrary, given that under the right conditions basically any bond can be rotated. When thinking about concepts in medicinal chemistry like rotatable bonds, its therefore important to make sure you're working with the definition used when defining the rule.

The original Veber papers (J. Med. Chem. 2002, 45, 2615) define what they consider to be a rotatable bond:

  Rotatable bonds were defined as any single bond, not in a ring, bound to a nonterminal heavy (i.e., non-hydrogen) atom. Excluded from the count were amide C-N bonds because of their high rotational energy barrier.

Based on this rule, the Ar-Ar single bond would count as a rotatable bond, as would the Ar-N bond. The sulfonamide is generally excluded for the same reason as amide C-N bonds (although I can't find a quote by Veber that actually discusses this).

In reality, these properties are usually calculated automatically, for instance using ChEMBL.


You assumed that coupling to the three chlorides would yield some type of quartet. This is correct in principle. However, chlorine is one of the many quadrupole nuclei that are basically unobservable by NMR due to their rapid relaxation. I hope another answer is around explaining that as I am not good at it.

I can, however, help you interpret the $1:1:1$ triplet of your spectrum. It is not, as you have assumed, the signal for $\ce{CHCl3}$ but that for $\ce{CDCl3}$. The difference between the two in NMR terms is that chloroform contains the spin ½ nucleus protium while deuterochloroform contains the spin 1 nucleus deuterium — and the latter does not relax rapidly as the chlorine isotopes do. Hence, deuterium is well-observable by NMR.

Now your first thought should be something like the following:

  But I am only coupling to one nucleus. That should give me a $1:1$ doublet and not a $1:1:1$ triplet, shouldn’t it?

That simplification is only correct for spin ½ nuclei. The actual formula for calculating the multiplicity of a peak is $2\,n\,I + 1$, where $n$ is the number of chemically equivalent nuclei your observed nucleus is coupling to and $I$ is the spin of the coupling nuclei. Since deuterium is spin 1 ($I = 1$), the formula gives us:

$$2\,n\,I + 1 = 2 \times 1 \times 1 + 1 = 3$$

Or an expected triplet. Since this triplet does not derive from two equivalent couplings to spin ½ nuclei, it is not a $1:2:1$ triplet but a $1:1:1$ one.

Your second spectrum shows a $\ce{^13C}$ NMR of deuterochloroform which still contains residues of chloroform — at least, that is what it is supposed to show. For some reason unknown to me, the protochloroform signal has not been decoupled, i.e. the spectrum is in fact $\ce{^13C}$ and not $\ce{^13C\{^1H\}}$. Thus, you are observing a strong coupling to the single hydrogen nucleus giving a $1:1$ doublet as expected if chlorine is ignored.


How does the wax layer of a leaf get damaged by acid rain?

Being insoluble in water does not mean something cannot react with water or water-based solvents.

Plant waxes are made up of a number of constituents; amoung them esters, alcohols, alkenes, fatty acids and carbonyls. Especially the esters can be hydrolysed by acidic conditions, creating a more hydrophilic area which water (and more acid) can use to penetrate the layer.

Another probable reason is the fact that acidic proton transfers often involve the release of heat which will make the wax layer less viscous. All these things taken together mean that slowly but surely, the wax layer will be destroyed given enough acid rain.


Is it actually possible to dispose of a body with hydrofluoric acid?

Hydrofluoric acid is toxic and corrosive, but actually isn't that strong of an acid compared to other hydrohalic acids; the fluorine has a very good orbital overlap with hydrogen and is also not very polarizable, therefore it resists donating its proton, unlike other hydrohalic acids which are good proton donators. It will break down some tissues, but it will take a relatively long time and won't turn the entire body into stuff that can be rinsed down the drain. Hydrochloric acid is a much stronger acid, and as it has several uses from pH-balancing pool water to preparing concrete surfaces, it's available by the gallon from any hardware store. However, it isn't very good at dissolving bodies either; while it will eventually work by breaking down the connective tissues, it will make a huge stink and take several days to dissolve certain types of tissues and bones.

The standard body-dissolving chemical is lye aka sodium hydroxide. The main source is drain clog remover because most drain clogs are formed by hair and other bio-gunk that accumulates naturally when humans shower, exfoliate etc. It works, even though the body's overall chemistry is slightly to the basic side of neutral (about 7.35-7.4) because the hydroxide anion is a strong proton acceptor. That means that it strips hydrogen atoms off of organic molecules to form water (alkaline hydrolysis, aka saponification), and as a result, those organic molecules are turned into simpler molecules with lower melting points (triglycerides are turned into fatty acids, saturated fats are dehydrogenated to form unsaturated fats, alkanes become alcohols, etc). Sodium hydroxide is also a ready source of the sodium ion; sodium salts are always water-soluble (at least I can't think of a single one that isn't). The resulting compounds are thus either liquids or water-soluble alcohols and salts, which flush down the drain. What's left is the brittle, insoluble calcium "shell" of the skeleton; if hydrolyzed by sodium hydroxide, the resulting calcium hydroxide ("slaked lime") won't dissolve completely but is relatively easy to clean up.


Why do raisins bob up and down in carbonated water?

The raisin has nucleation sites on it that allow bubbles of $\ce{CO2}$ to form. The raisin is light enough to be lifted by the bubbles as they push their way to the surface. As the bubbles are released into the atmosphere, the raisin once again sinks until more bubbles form on it. This will continue until the soda water has lost the majority of its dissolved $\ce{CO2}$.


Generally speaking (without need to refer to your particular problem), the sign of the free energy change and thus the magnitude of maximum possible work does not depend on $\Delta H$ or $\Delta S$ alone, it depends on the combination:
$$w_{\mathrm{max},\text{non-}pV} = \Delta G = \Delta H - T \Delta S.\tag{1}$$
This equation describes the maximum amount of non-$pV$ work that can be obtained from the system or minimum amount that needs to be done if the process is performed reversibly at constant $T$ and $p.$ In addition, when work other than mechanical $(pV)$ is performed, the value of $\Delta H$ is no longer equal to the heat evolved, in fact then $\Delta H = q_p + w_{\mathrm{max},\text{non-}pV}.$ Finally, it remains that at constant $T$ the entropy change of the surroundings is related to the amount of heat evolved as
$$\Delta S_\mathrm{surr} = \frac{-q_p}{T}.\tag{2}$$
This allows the above to be rewritten in terms of entropies of the system and the surroundings (at constant $T$ and $P):$
$$w_{\mathrm{max},\text{non-}pV} =  w_{\mathrm{max},\text{non-}pV} + q_p - T \Delta S = w_{\mathrm{max},\text{non-}pV} -T(\Delta S_\mathrm{surr} + \Delta S_\mathrm{sys}) $$
Removing the cancelling terms we conclude that
$$\Delta S_\mathrm{surr} + \Delta S_\mathrm{sys} = 0\tag{3}$$
that is, the entropy of the universe remains constant, which is consistent with reversibility.
That clears up the question of how it is possible for a process to be formally endothermic (in the sense that $\Delta H > 0$) and yet heat is released $(q < 0).$ This does not however explain why the net process of glucose generation by photosynthesis, assumed ideal (amongst other things reversible) generates heat.
The answer is that the released heat is responsible for an increase in the entropy of the surroundings which offsets a reduction in the entropy of the system when glucose is generated, which can be shown by writing
$$\Delta S_\mathrm{sys} = -\Delta S_\mathrm{surr} = \frac{q_p}{T} < 0\tag{4}$$
Without that increase in the entropy of the surroundings life would not be possible.


Given the context of your question, I assume you must mean "the associated volume", not "the associated area".

Other than that, you are exactly correct.  For example, given the following parameters:

  aerosol emissions rate = 0.010 mg $\mathrm{hr^{-1}}$
  duration of aerosol emissions = 1 $\mathrm{hr}$
  container volume = 1 $\mathrm{m{^3}}$

Then the aerosol mass loading would be given as:

  total aerosol mass = 0.010 mg $\mathrm{hr^{-1}}$ $\mathrm{x}$ 1
  $\mathrm{hr}$ = 0.010 mg

  aerosol mass loading = 0.010 mg / 1 $\mathrm{m{^3}}$ = 0.010 mg
  $\mathrm{m{^{-3}}}$

If you really meant area and not volume, as in surface area of the aerosols, then your statement would be over-simplified and you would need to include other assumptions made concerning particle size and density etc. in order to determine whether the way your statement is correct as written.


Stuck with derivation of the integrated rate equation for a pseudo first order equilibrium reaction

You have made no mistake. Your solution and the solution from the paper are practically identical. The solution from the paper is

\begin{align}
  F(t)=c_1 \frac{1−e^{−c_2 t}}{c_2}=\frac{c_1}{c_2} - \frac{c_1}{c_2} e^{−c_2 t}
\end{align}

which is exactly the form your equation has, i.e.

\begin{align}
  F(t)=\frac{c_1}{c_2} + K e^{−c_2 t} \qquad \text{with} \qquad K = - \frac{c_1}{c_2} \ .
\end{align}

The integration constant $K$ is not defined the way it is but fixed via the boundary conditions of the reaction. In this case you get $K = - \frac{c_1}{c_2}$ by requiring that at the start of the reaction, i.e. $t=0$, the reactants have not yet reacted with each other and there is no product $\ce{AB}$ present initially, i.e. $F(t\!=\!0) = 0$. This leads to the desired result:

\begin{align}
  F(t\!=\!0) \overset{!}{=} 0 &=\frac{c_1}{c_2} + K e^{0} \qquad \Rightarrow \quad K=- \frac{c_1}{c_2} \ .
\end{align}


What is the process of connecting two molecules under removal of one molecule of water?

Since it is a removal of water, it cannot be hydration which is defined as the addition of water.

Similarly, hydrolysis includes the addition of water to something so that it degrades in some way.

Neutralisation does indeed generate water, but not always one molecule per reaction equivalent (can be two or three) and you do not connect molecules but ions.

Condensation not only has its physical definition you stated but it also means (in a chemical sense) connecting two molecules under formal exclusion of one molecule of water.

Note that in many to most processes, the water exclusion is entirely formal and what actually is excluded is something like $\ce{H2PO4-}$.


What does "atoms make no compromises" mean?

The subject we're looking at here is the quantization of energy absorbed and released by atoms, which is a fancy word describing the fact that atoms can only absorb or emit very specific amounts of energy (in the form of light) and nothing in between.  This curious fact has everything to do with how electrons orbit the nucleus.

The electrons whirling around the nucleus do so only at specific distances and in specific patterns.  Some orbit in spheres, others in dumbell-shapes, and still others in even odder patterns (a search for "orbital shapes" turns up a few relevant images).  On top of that, smaller spheres (and other shapes) are nested within larger ones, in a manner Niels Bohr compared to the orbits of planets in the solar system (https://en.wikipedia.org/wiki/Bohr_model).

Now here's where the energy comes in.  Electrons can "jump" between orbitals (unlike the planets, thank goodness!) when they absorb energy and move farther from the nucleus.  When electrons jump orbitals, they gain potential energy, just like when we move farther from the Earth, we gain potential energy.  The farther up we go, the more energy we have to expend on our way down: think jumping out of a 1st story window or a 2nd story window, or even a 3rd story window.  Note that the electrons can only jump between the first, second, third, etc. "floors" (orbitals) of an atom.  No mezzanines or mid-levels.

Obviously, the more electrons an atom has, the harder it is to map all of the specific energies it absorbs and emits.  Hydrogen atoms are easy, however, as this equation depicts.  Just plug in the starting shell #, the final shell #, and you've got your specific wavelength (energy) of light.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/graphics/6_2.gif

The logical question to ask now is "why do electrons orbit only in those specific levels?"
And that's a tough one, involving a crazy (interesting) amount of mathematics.  Equations we've worked out from quantum mechanical theory describe "wavefunctions" for electrons that predict the probability of them being in a certain area (orbital) around the nucleus.  They're convoluted, but brilliant.  Take a gander here if you're curious: https://en.wikipedia.org/wiki/Wave_function.


Symmetry and NMR linewidths of quadrupolar nuclei

I know that you asked for an intuitive explanation, but I'm afraid that if one wants to go beyond ron's comment, the technical aspects are somewhat necessary. It isn't incredibly important to understand the maths, but I have included it as I want to enable interested readers to go further, and it provides a basis for the last section, so I make no apology for it. I have also assumed some basic knowledge of NMR theory.1

Linewidths and relaxation

Following a pulse sequence, the spectrometer detects transverse magnetisation (i.e. magnetisation in the xy-plane). The signal that is detected is generally of the form $s(t) \propto \exp(\mathrm i \Omega t)\exp(t/T_2)$, where $T_2$ is the transverse or spin-spin relaxation time.

If one takes the Fourier transform of $s(t)$, one gets the spectrum $S(\Omega)$, and it can be shown that the widths of the Lorentzian peaks thus obtained are proportional to $1/T_2$. To be precise, the linewidth at half-height $\Delta \nu_{1/2}$ is given by

$$\Delta \nu_{1/2} = \frac{1}{\pi T_2}$$

Therefore, fast transverse relaxation (which corresponds to small $T_2$) leads to broad lines, and slow relaxation leads to narrow lines.

Quadrupoles and the electric field gradient

For quadrupolar nuclei ($I > 1/2$), a major source of relaxation arises due to interaction of the nuclear electric quadrupole moment with the electric field gradient at the nucleus.

An introduction to the above two concepts is as follows: a quadrupole moment reflects an asymmetry in the charge distribution of the nucleus, and can be thought of as two opposing dipoles placed next to each other. The electric field gradient refers to the derivative of the electric field $E$, which is itself the derivative of the electric potential $V$ (so, the field gradient is the second derivative of the potential). These derivatives are evaluated at the spatial position of the nucleus.

The electric field gradient therefore most generally has nine components, since two derivatives, each with respect to 3 axes, are taken. These are represented by $q_{ij}$ (where $i,j$ are one of $x,y,z$):

$$q_{ij} \equiv \left.\frac{\partial^2V}{\partial i\,\partial j}\right|_\text{at nucleus}$$

However, by a suitable transformation of coordinates, the cross derivatives $q_{ij}$ ($i \neq j$) can be set to zero (this essentially involves diagonalising the matrix $\mathbf{q}$). The remaining quantities $q_{xx}$, $q_{yy}$, $q_{zz}$ are referred to as the principal values, and without loss of generality we can adopt the convention $|q_{zz}| \geq |q_{yy}| \geq |q_{xx}|$. Since the electric potential obeys Laplace's equation $\nabla^2 V = 0$, this also means that $q_{xx} + q_{yy} + q_{zz}$ is necessarily equal to 0 (this is often referred to as "traceless", since the trace of the matrix $\mathbf{q}$ is 0). Therefore, there are only two independent parameters. These parameters are chosen to be $q_{zz}$, and $\eta$, a quantity called the biaxiality and defined as

$$\eta = \frac{q_{yy} - q_{xx}}{q_{zz}}.$$

Quadrupolar relaxation2

The interaction of the quadrupole moment with the electric field gradient contributes to the energy of the nucleus (exactly analogous to how an electric dipole moment interacts with an electric field).3 This interaction is modulated by molecular tumbling. That is to say, as the molecule rotates in solution, the magnitude of this interaction changes as well.4

If this interaction varies at the frequency corresponding to the transition between two spin states, then it is possible to induce this transition, thereby leading to spin relaxation. (More strictly speaking, it has to contain a component that oscillates at the transition frequency.) In this aspect, it is analogous to the more well-known case of how magnetic dipolar interactions between two nuclei can lead to relaxation, which is described quite thoroughly in Keeler's book.1 For spin-1/2 nuclei in solution-state NMR, dipolar interactions are the primary method of relaxation since these nuclei do not have quadrupole moments. For quadrupolar nuclei, though, quadrupolar relaxation is the primary method of relaxation, simply because this interaction term tends to have a large magnitude.

For our purposes, it is sufficient to use a derived result, which you seem to have already come across. I know that I am glossing over a lot, but as far as I can tell, the derivation of this is very involved:5

$$\frac{1}{T_1} = \frac{1}{T_2} = \frac{3\pi^2}{10}\frac{(2I+3)}{I^2(2I-1)} \left(\frac{e^2q_{zz}Q}{h}\right)^2 \left(1 + \frac{1}{3}\eta^2 \right) \tau_\mathrm{c}$$

Definitions:

$T_1$ and $T_2$ are the spin-lattice (longitudinal) and spin-spin (transverse) relaxation times respectively; these only account for relaxation via the above mechanism and not for other possible mechanisms
$I$ is the spin of the nucleus
$e$ is the elementary charge, $\pu{1.602 \times 10^-19 C}$
$q_{zz}$ is the main component of the electric field gradient, as explained above
$Q$ is the quadrupole moment of the nucleus
$h$ is Planck's constant, $\pu{6.626 \times 10^-34 J s}$
$\eta$ is the biaxiality, defined above
$\tau_\mathrm{c}$ is the correlation time, which is a constant that essentially measures the rate of molecular tumbling (see books in ref 1 for more information).

Please, no more maths

OK - the only remaining maths is to analyse the equation above and identify the factors which contribute to relaxation, and hence, linewidths. For the actual question, the relevant portion is $e^2q_{zz}Q/h$, which is often written as $\chi$ and termed the nuclear quadrupole coupling constant. Some representative values of $\chi$ are given in ref 2.

In particular, the value of $q_{zz}$ is mainly determined by the distribution of electrons close to the nucleus, and this provides an obvious link to the molecular geometry of the quadrupolar nucleus. For nuclei in tetrahedral, octahedral, cubic, or spherical environments, $q_{zz}$ is, by symmetry, equal to zero. Therefore, quadrupolar relaxation by the above mechanism does not take place, $T_2$ is long, and linewidths are small. One sees very narrow linewidths for the central atoms in, for example, $\ce{Cl- (aq)}$ (neglecting solvation effects, spherical symmetry); $\ce{NMe4+}$ and $\ce{SO4^2-}$ (tetrahedral), and $\ce{[Co(CN)6]^3-}$ (octahedral). On the other hand, if such symmetry is lacking, then $q_{zz} \neq 0$, there is usually fast quadrupolar relaxation by the above mechanism, and linewidths are large.

$q_{zz}$ is directly tied to the distribution of electrons and not the distribution of bonds. Although there is nearly always a correlation, there are some interesting cases where this is not true; in $\ce{Et2N-NO2}$, for example, the $\ce{NO2}$ nitrogen has an extremely sharp peak, despite the local molecular symmetry not intuitively suggesting that it would be the case.2 (The $\ce{Et2N}$ nitrogen, as expected, has a very broad peak.)

Since we went the full distance to obtain the above expression, we can also extract a bit more information from it. The rate of relaxation is also proportional to the nuclear quadrupole moment $Q$; some quadrupole moments are tabulated in ref 2. Some nuclei which have small quadrupole moments (namely deuterium, lithium-6, and cesium-133) have smaller linewidths and are much more amenable to NMR studies. Also, the factor of $I^2$ in the denominator also means that higher spins tend to give sharper lines.

A bonus: coupling to quadrupolar nuclei

$\ce{^13C}$ NMR experiments are often run with proton decoupling - this means that when the carbon signal is collected, the protons are irradiated with a radiofrequency field. This has the effect of inducing rapid transitions between the two spin states of the protons, and each carbon-13 nucleus effectively "sees" an average of the two spin states, causing the C–H couplings to be lost.

Since quadrupolar relaxation also induces rapid transitions between spin states of quadrupolar nuclei, it leads to what is effectively a "self-decoupling" of all quadrupolar nuclei. However, the couplings can still be seen when the quadrupolar nuclei are in highly symmetrical environments: for example, the $\ce{^19F}$ spectrum of $\ce{[BrF6]+}$ is shown here.6 Both $\ce{^79Br}$ and $\ce{^81Br}$ have roughly equal abundance (~50%) and a nuclear spin of $3/2$, so we expect to see two 1:1:1:1 quartets of equal intensity, one arising from $\ce{[^79BrF6]+}$ and one from $\ce{[^81BrF6]+}$.

A few other examples are presented on Hans Reich's page. Of course, these can be understood from the discussion above about how molecular symmetry affects quadrupolar relaxation.

In fact, because deuterium has a small quadrupole moment and undergoes slow quadrupolar relaxation, coupling to deuterium can often be seen. All organic chemists should recall that $\ce{CDCl3}$ appears as a 1:1:1 triplet - this is why!

Notes and references

For an exposition of basic NMR theory see either Keeler, Understanding NMR Spectroscopy, 2nd ed. (Wiley) or Hore, Nuclear Magnetic Resonance, 2nd ed. (OUP).
A necessarily mathematical, but still fairly accessible, introduction to the topic is provided in: Gerothanassis, I. P.; Tsanaktsidis, C. G. Nuclear electric quadrupole relaxation. Concepts Magn. Reson. 1996, 8 (1), 63–74. DOI: 10.1002/(SICI)1099-0534(1996)8:1<63::AID-CMR5>3.0.CO;2-N.
The full Hamiltonian is provided in Appendix A.7 of Levitt, Spin Dynamics, 2nd ed. (Wiley).
Abragam (ref 5) writes that this is best understood in terms of "a fluctuating electric field gradient acting on the quadrupole moment of the nucleus", and refers to: Bloembergen, N.; Purcell, E. M.; Pound, R. V. Relaxation Effects in Nuclear Magnetic Resonance Absorption. Phys. Rev. 1948, 73 (7), 679–712. DOI: 10.1103/PhysRev.73.679. (Non-paywall version from Harvard available here.)

However, as far as I can tell, Bloembergen et al. simply describe the case for dipolar relaxation of spin-1/2 nuclei, and then write that "the interaction of the electric quadrupole moment of the deuteron with a fluctuating inhomogeneous electric field can bring about thermal relaxation. We omit the analysis of this process, which parallels closely the treatment of dipole-dipole interaction".
This equation is given without explanation in Gunther, NMR Spectroscopy, 3rd ed. (Wiley). An explanation can be found in Abragam, The Principles of Nuclear Magnetism (OUP), including the conditions under which this equation holds true. Quadrupolar relaxation is dealt with on p 313, but going through the earlier sections is necessary to understand it. (I didn't do that.)
Gillespie, R. J.; Schrobilgen, G. J. Hexafluorobromine(VII) cation, $\ce{BrF6+}$. Preparation of hexafluorobromine(1+) hexafluoroarsenate(1-) and hexafluorobromine(1+)-fluorodecafluorodinabinonate(1-) and characterization by fluorine-19 nuclear magnetic resonance and Raman spectroscopy. Inorg. Chem. 1974, 13 (5), 1230–1235. DOI: 10.1021/ic50135a043.


Does Finkelstein's reaction proceed through an SN2 mechanism?

The Finkelstein reaction is a nucleophilic substitution reaction. As such, it generally has the choice of proceeding via the $\mathrm{S_N1}$ or the $\mathrm{S_N2}$ mechanism.

The reaction is typically conducted in acetone; a polar but not very protic solvent. The solvent is used because all the sodium halides except for sodium iodide are not very soluble in it and precipitate out. It does not show any real preference for either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.

Therefore, we should study the nucleophile and leaving group. Iodide is both a good nucleophile and a good leaving group; the former is important as it allows us to assume a viable $\mathrm{S_N2}$ mechanism. The other halides that you are substituting with iodide are not as good leaving groups but still rather okay. This again does nothing really to tilt the scales towards either $\mathrm{S_N1}$ or $\mathrm{S_N2}$.

The final choice we have is that it depends on the substrate. Probably, if you are replacing a primary halide chances are that the reaction will proceed according to the $\mathrm{S_N2}$ mechanism while an $\mathrm{S_N1}$ mechanism should predominate for tertiary halides.


How do you change atomic valence in Avogadro?

While there are some automatic valence features in Avogadro, you can always turn them off. Bonding is mostly a matter of appearance.

For example, let's import $\ce{IF7}$.

Notice that the central iodine doesn't show enough bonds. So we'll draw some in. Switch to the draw tool and turn "Adjust Hydrogens" off.

Now just draw connections between the atoms. Voila!


You're right that the first step involves protonation of the terminal carbon in the  styrene double to yield a resonance stabilized benzyl carbocation.  This carbocation then adds to another styrene molecule in the exact same fashion, reacting at the terminus of the styrene double bond to again yield a resonance stabilized benzyl carbocation.  This latter carbocation then alkylates itself (like a Friedel-Crafts alkylation).  Loss of a proton from this final carbocation yields the indicated product.


Increase in Yield of products in an exothermic reaction

Look, you say: "which was initially in equilibrium". Then the temperature are stable. In the reaction you have four mols in the reactants and 2 mols in the products. When you increase the temperature the system pressure increase too.

So to increase the stability the system "walk" for the reactants, because this decrease the pressure (2 mols in products and 4 mols in reactants).

If you remember the gas ideal law, the pressure is directly proportional to the mol numbers. So this is the reason because the temperature favor the products.


What atom is always paramagnetic and why?

Neon and argon are correctly eliminated, no questions asked. Henceforth, we do indeed need to pay attention to question semantics and the phrase ‘regardless of its electronic configuration’.

The key here is that following Hund’s rule will only give you one possible electronic configuration. For example for carbon, Hund’s rule will give $\mathrm{1s^2\, 2s^2\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^0$ or any other combination of two of the three equivalent p orbitals half-populated. However, in early organic chemistry courses you typically rapidly learn about the hybridisation approach which changes that electronic configuration to $\mathrm{1s^2\, 2s^1\,2p}_x{}^1\, \mathrm{2p}_y{}^1 2\mathrm{p}_z{}^1$ — four instead of two unpaired electrons. Likewise, it is possible to imagine an electronic configuration as follows: $\mathrm{1s^2\, 2s^2\,2p}_x{}^2\, \mathrm{2p}_y{}^0 2\mathrm{p}_z{}^0$. In this case, all electrons are paired and the atom is diamagnetic.

Performing this type of analysis with all the atoms at your disposal — carbon, nitrogen and oxygen — you will be able to succeed in pairing all electrons for carbon and oxygen but not for nitrogen since it has five valence electrons. Five, being an odd number, cannot be mapped pairwise. Thus, a nitrogen atom is always paramagnetic.


What mass of water is present as a liquid when equilibrium is established?

The given temperature of $T=50\ \mathrm{^\circ C}$ and the given pressure of $p=92.5\ \mathrm{mmHg}$ approximately correspond to equilibrium conditions for liquid water and steam. (Strictly speaking, the corresponding saturation pressure for a given temperature of $T=50.000\ \mathrm{^\circ C}$ is $p=92.647\ \mathrm{mmHg}$, which could be rounded to $p=92.6\ \mathrm{mmHg}$ and not $p=92.5\ \mathrm{mmHg}$; in the opposite direction, however, the corresponding saturation temperature for a given pressure of $p=92.500\ \mathrm{mmHg}$ is $T=49.968\ \mathrm{^\circ C}$, which could be rounded to $T=50\ \mathrm{^\circ C}$.) The question does not explain whether these values describe the initial or final state; however, since the values correspond to equilibrium conditions, we may assume that these values apply to the final state when equilibrium is established (e.g. this can be achieved by keeping the closed system at a constant temperature of $T=50\ \mathrm{^\circ C}$).

Strictly speaking, the question doesn’t even explain whether the added water initially is liquid or steam; however, this is not relevant for the final state. If the water is added as liquid, a part of it evaporates until the equilibrium state is reached in the closed container; if the water is introduced as steam, a part of it condenses until the equilibrium state is reached. Therefore, when the defined equilibrium is established, the container contains certain amounts of liquid water and steam irrespective of the initial conditions.

The question does not mention any air in the container. For simplicity’s sake, we may assume that the container has been evacuated before the experiment and contains only liquid water and vapour.

The available volume of the container is reduced by the volume of the liquid water. However, since the density of liquid water at the given temperature and pressure is about $\rho=988\ \mathrm{kg\ m^{-3}}$, the given mass of $m=1.00\ \mathrm g$ corresponds to a maximum volume of $V_\text{liquid}=1.01\ \mathrm{ml}=0.00101\ \mathrm l$, assuming that no water has evaporated. Considering that the total volume is given as $V=5.00\ \mathrm l$ (note the number of significant digits), the difference caused by the liquid water is not significant and may be neglected.

Note that the use of the non-SI unit “conventional millimetre of mercury” (unit symbol: mmHg) is deprecated; the use of SI units is to be preferred.
$$1\ \mathrm{mmHg}\approx 133.3224\ \mathrm{Pa}$$
You might want to convert the given values to coherent SI units, or find a value for the molar gas constant $R$ that is expressed in a suitable unit.

The following data are given:

Temperature $T=50\ \mathrm{^\circ C}=323.15\ \mathrm K$
Pressure $p=92.5\ \mathrm{mmHg}=12332.322\ \mathrm{Pa}$
Volume $V=5.00\ \mathrm l=0.00500\ \mathrm{m^3}$

The value of the molar gas constant is $R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}$.[source]

The amount of vapour $n$ in the container may be estimated using the ideal gas law
$$\begin{align}
p\cdot V&=n\cdot R\cdot T\\[6pt]
n&=\frac{p\cdot V}{R\cdot T}\\[6pt]
&=\frac{12332.322\ \mathrm{Pa}\times0.00500\ \mathrm{m^3}}{8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\times323.15\ \mathrm K}\\[6pt]
&=0.022949674\ \mathrm{mol}
\end{align}$$
(Note that this intermediate result is given with an excessive number of digits. It is deliberately not rounded in order to avoid round-off errors in subsequent calculations.)

Since the molar mass of water is $M=18.01528\ \mathrm{g\ mol^{-1}}$, the corresponding mass of vapour is
$$\begin{align}
m&=M\cdot n\\[6pt]
&=18.01528\ \mathrm{g\ mol^{-1}}\times0.022949674\ \mathrm{mol}\\[6pt]
&=0.413444803\ \mathrm g\\[6pt]
&\approx0.41\ \mathrm g
\end{align}$$
(Note that this final result is rounded to two significant digits. The number of significant digits is estimated in view of the fact that the mass of water $m=1.00\ \mathrm g$, the volume $V=5.00\ \mathrm l$, and the pressure $p=92.5\ \mathrm{mmHg}$ are given with three significant digits, but the temperature $T=50\ \mathrm{^\circ C}$ is only given with two significant digits.)

This result is the mass of vapour $m_\text{vapour}$ in the container, whereas the question is about the mass of liquid water $m_\text{liquid}$, which can be calculated from the given total mass of water $m_\text{total}=1.00\ \mathrm{g}$.
$$\begin{align}
m_\text{total}&=m_\text{vapour}+m_\text{liquid}\\[6pt]
m_\text{liquid}&=m_\text{total}-m_\text{vapour}\\[6pt]
&=1.00\ \mathrm g-0.41\ \mathrm g\\[6pt]
&=0.59\ \mathrm g
\end{align}$$

Therefore, the correct answer is option 3) $0.59\ \mathrm g$.

By way of comparison, precise engineering calculations for water and steam usually do not rely on the ideal gas law but use so-called steam tables. We may use such steam tables to check the accuracy of our estimate that has been based on the ideal gas law.

For example, for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium, we find the following values in REFPROP – NIST Standard Reference Database 23, Version 9.0:

Pressure $p=12352\ \mathrm{Pa}$
Liquid density $\rho_\text{liquid}=988.00\ \mathrm{kg\ m^{-3}}=988.00\ \mathrm{g\ l^{-1}}$
Vapour density $\rho_\text{vapour}=0.083147\ \mathrm{kg\ m^{-3}}=0.083147\ \mathrm{g\ l^{-1}}$

Furthermore, the database includes various other thermodynamic and transport properties, which are not required for the following calculations.

If you do not have access to professional steam tables, you may want to consider using the steam tables that are included in WolframAlpha. The corresponding results for a temperature of $T=50\ \mathrm{^\circ C}$ at equilibrium can be obtained using the input “water boiling at 50 °C”:

Pressure $p=12352\ \mathrm{Pa}$
Liquid density $\rho_\text{liquid}=988\ \mathrm{kg\ m^{-3}}=988\ \mathrm{g\ l^{-1}}$
Vapour density $\rho_\text{vapour}=0.08315\ \mathrm{kg\ m^{-3}}=0.08315\ \mathrm{g\ l^{-1}}$

Note: The indicated values for STP are not in accordance with IUPAC recommendations.

We know that the mass balance in the container is
$$m_\text{total}=m_\text{vapour}+m_\text{liquid}\tag{1}$$
where $m_\text{total}=1.0000\ \mathrm g$ is the total mass of water (liquid and vapour) in the container (in this example, the precision of this value is arbitrarily increased to five significant digits).

Since density is defined as
$$\rho=\frac mV$$
the corresponding volume balance can also be expressed in terms of mass:
$$\begin{align}
V_\text{total}&=V_\text{vapour}+V_\text{liquid}\\[6pt]
&=\frac{m_\text{vapour}}{\rho_\text{vapour}}+\frac{m_\text{liquid}}{\rho_\text{liquid}}\tag{2}
\end{align}$$
where $V_\text{total}=5.0000\ \mathrm l$ is the total volume of the container (in this example, the precision of this value is arbitrarily increased to five significant digits).

Solving the system of the equations $\text(1)$ and $\text(2)$ yields the solutions

$$\begin{align}
m_\text{vapour}&=\frac{\rho_\text{vapour}\cdot\left(V_\text{total}\cdot\rho_\text{liquid}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]
&=\frac{0.083147\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times988.00\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]
&=0.41569\ \mathrm g
\end{align}$$

and

$$\begin{align}
m_\text{liquid}&=-\frac{\rho_\text{liquid}\cdot\left(V_\text{total}\cdot\rho_\text{vapour}-m_\text{total}\right)}{\rho_\text{liquid}-\rho_\text{vapour}}\\[6pt]
&=-\frac{988.00\ \mathrm{g\ l^{-1}}\times\left(5.0000\ \mathrm l\times0.083147\ \mathrm{g\ l^{-1}}-1.0000\ \mathrm g\right)}{988.00\ \mathrm{g\ l^{-1}}-0.083147\ \mathrm{g\ l^{-1}}}\\[6pt]
&=0.58431\ \mathrm g
\end{align}$$

These results confirm that the above-mentioned estimates $m_\text{vapour}\approx0.41\ \mathrm g$ and $m_\text{liquid}\approx0.59\ \mathrm g$, which have been obtained using the ideal gas law and neglecting the volume of the liquid water, are reasonable and sufficient in order to answer the question.


Do you say "reduce something to polystyrene"?

Just write what you actually did: You used the calibration curve of your SEC column for polystyrene in solvent xx. Those phrases you cite are meant to sound impressive, but are actually just scientifically wrong, bad English, and therefore stupid.

Nothing gets reduced here like a reduced temperature or a chemical reduction. Also there is no "PS equivalent Mw", because it depends strongly on the solvent. "Relative" is as wrong, because that implies that there is simply a factor between the two scales, while in fact there are two calibration curves, one of which is unknown.


Electrolysis of molten Calcium Chloride: Identifying the product

I would hypothesize that the "black glassy stuff" is nothing more than glassified $\ce{CaCl_2}$. Metal is an excellent heat conductor and the liquid $\ce{CaCl_2}$ likely cooled rapidly to form glass when the cold wire was inserted.

Two things could be done to avoid this:

1.) heat the wire before inserting it into the mixture of molten $\ce{CaCl_2}$.

2.) heat the $\ce{CaCl_2}$ well above its melting point.

If you are determined to make Calcium metal it might be worth electrolysing an aqueous solution of $\ce{CaCl_2}$ to evolve the $\ce{Cl2(g)}$ (do this responsibly outside/in fume hood of course as $\ce{Cl2(g)}$ is poisonous.) Calcium Hydroxide should be the result. Evaporating the water will yield solid $\ce{CaOH}$, which should be much easier to electrolize using your method since it has a melting point of 580°C vs. 772°C for $\ce{CaCl_2}$.


Why is (1E,3Z,5E,7E,9Z,11E)-cyclotetradeca-1,3,5,7,9,11-hexaen-13-yne not antiaromatic?

The textbook answer to this is that the triple bond consists of two orthogonal π systems whose interaction with each other is symmetry-forbidden. So as the attached figure shows, the two p-orbitals marked in red cannot participate in the aromatic system if the molecule’s structure is as drawn; thus the two electrons do not count towards the Hückel rule, thus $4n+2$ is satisfied with $n=3$ and $4n+2 = 14$.

However, that is not the end of the story; merely its beginning. Professor Dimm should take the time to determine the actual crystal structure of the compound he synthesised. You might want to compare this with cyclodecapenta-1,3,5,7,9-ene. While it may look like a perfect Hückel-aromatic molecule, it does not adopt a planar structure due to ‘inverse Baeyer strain’ (the flat all-cis conformation would have $144^\circ$ angles, far away from the theoretical $120^\circ$ $\mathrm{sp^2}$ angles) or steric hindrance (the (1E,3Z,5E,7Z,9Z)-confomation is not flat due to steric strain of the internal hydrogens). A similar case can be made for (1E,3Z,5E,7E,9Z,11E)-cyclotetradeca-1,3,5,7,9,11-hexaen-13-yne as can be seen in the figure below.

The hydrogens coloured in red would overlap far too much to create a stable molecule. It will adopt a non-planar configuration to relieve this steric stress which is a greater source of destabilisation than giving up an aromatic system. We can only speculate on the molecules structure until we actually saw it. Unfortunately, it is also unknown as far as SciFinder is concerned. However, two geometric isomers of dehydro[14]annulene (as the molecule is also called) have been synthesised and isolated:

Interstingly, these compounds that were discovered in the 1960’s[1] show aromaticity by their chemical shifts. The internal hydrogens are shielded and display a chemical shift of approximately $-1~\mathrm{ppm}$ (symmetric isomer) and $0$ and $-1~\mathrm{ppm}$ (unsymmetric isomer). Similarly, the outer hydrogens appear as a group in the aromatic region.[2] Cresp and Sondheimer used low-temperature NMR studies to assign the correct structures.[2] And in 2013, Santander et al. performed calculations to determine the mechanism of interconversion between the two (‘by allowing a dilute ether solution to stand without protection from daylight, or, more rapidly by irradiation of a refluxing benzene solution with a $100~\mathrm{W}$ lamp.’[2]), which turns out to be via an interesting (in my opinion) Hückel-bond shift.[3]

Let’s hope that Professor Dimm actually synthesises the compound and properly reports its structure.

References:

[1]: F. Sondheimer, Proc. Roy. Soc. A, 1967, 297, 173. Taken directly from reference [2].

[2]: T. M. Cresp, F. Sondheimer, Tetrahedron Lett. 1982, 23, 1731. DOI: 10.1016/S0040-4039(00)87202-0.

[3]: M. V. Santander, M. B. Pastor, J. N. Nelson, C. Castro, W. L. Karney, J. Org. Chem. 2013, 78, 2033. DOI: 10.1021/jo302072p.


How can different no. of unpaired electrons lead to different colours of co-ordination compounds?

The crystal field splitting does definitely contribute.But it is not the only thing to contribute to the colour of the coordinate complex solution. The colour is equivalent to the spectra (emission or the absorption) of the solution, and hence by enumerating the factors contributing to the emission spectra of a solution, we can compare the colours of two different solutions.

It usually is an unpaired electron which shifts between the various available orbitals and paired electrons rarely do so, since the energy required to unpair acts as a barrier for its' transitions. Moreover, there might also be spin inversions (see Hyperfine structure) which might give additional lines in the spectra. Presence of unpaired electrons helps in producing additional energy levels in the electric and magnetic fields of the ligands (zeeman and stark effect), providing additional lines in the spectra. All these establish the uniqueness of the spectral distribution which the solution absorbs, and hence the colour.

The number of unpaired electrons also determines the resultant magnetic field of other species in the solution, which alters the energy levels of the solvating molecules.


What type of intermolecular forces will dominate Diethyl ether?

As you have already figured out, diethyl ether is a small dipole due to the electronegativity difference between carbon and oxygen, which can also be seen in this electron density plot. As both dipole moment vectors point roughly in the same direction, they do not cancel each other out like in the case of carbon dioxide, so the molecule has a net dipole moment and is therefore (weakly) polar.

London dispersion forces, which result from short-lived dipoles induced by fluctuations in the electron shell of molecules, are also present. They are responsible for the weak interactions between the alkyl chains, like in other alkanes and non-polar molecules. The permanent dipoles can also interact with each other, in the form of intermolecular attractive interactions between centers with $\delta+$ and $\delta-$ partial charge (and likewise repulsion between centers of equal partial charge).

Hydrogen bonding between C-H groups and oxygens, although possible in principle (see comments to your question) and, for example, found in protein structures, can be neglected in this case due to the low number of donors and the weakness of the interaction (< 17 kJ/mol). Nevertheless, the ether oxygen can act as a hydrogen bond acceptor in combination with better hydrogen bond donors like OH groups, which explains why diethyl ether is sparingly soluble in water despite its low polarity (69 g/l at 20 °C (source)). On the other hand, the absence of hydrogen bonding between ether molecules becomes apparent in the much lower boiling point of diethyl ether (34.6 °C, source) compared to water.


What's the difference between HOMO-LUMO and valence-conduction bands?

Because convention. Yes, that’s not the answer you wished for but it’s the way it is.

Coming from an organic perspective, one would think of concepts such as linear combination of atomic orbitals or others to arrive at molecular orbitals for entire molecules. Since organic chemistry always deals with rather small molecules and since these have discreet energy levels with discreet differences, it makes sense to speak of a single HOMO, i.e. one single highest occupied orbital (and likewise for a LUMO). The larger a molecule gets, the more orbitals it has and the less the difference between HOMO and LUMO might turn out to be, but from an organic perspective it is still a molecule and must begin and end somewhere.

(Note that this applies to a lesser extent to polymers and carbon nanotubes etc. This is where the shady grey area in which unicorns live begins.)

From an inorganic perspective, if you are dealing with solid materials, they will usually be crystalline in some way; hence they have infinite extents by definition. It no longer makes sense to think of discreet orbitals but rather one realises that certain levels contain a large number of orbitals — hence bands.

If organic molecules end up being sufficiently large, there is no longer a relevant physical difference between HOMO/LUMO designations or bands. But because nobody likes throwing their traditions and customs overboard, the two different names exist.


The way I would perform the dimensional analysis you give in your question would be slightly different. That is mainly due to the fact that I don't consider chemical names or formulae as units, but as designators.

As an aside, I think you were starting from a question like the following:

  How much $\ce{C7H4O}$ would you have to weigh in for it to fully react
  with $150~\mathrm{g}$ of $\ce{KNO3}$ if the reaction is described by
  the following equation? $$ \ce{C7H4O + 6KNO3 -> Fumes^  + Tar} $$

The result is $m_{\ce{C7H4O}}$, given is $m_{\ce{KNO3}} = 150~\mathrm{g}$.

The amount of $\ce{C7H4O}$ used per $\ce{KNO3}$ is
$$|\nu_{\ce{C7H4O}} / \nu_{\ce{KNO3}}| = \frac{1}{6}$$

As such, the result is composed as follows:
$$\begin{align}
n_{\ce{C7H4O}} &= \frac{1}{6} n_{\ce{KNO3}} \\
\tag{$n=m/M$}\frac{m_{\ce{C7H4O}}}{M_{\ce{C7H4O}}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} \\ \hline
m_{\ce{C7H4O}} &= \frac{1}{6} \frac{m_{\ce{KNO3}}}{M_{\ce{KNO3}}} M_{\ce{C7H4O}}
\end{align}$$

If we now apply dimension analysis to what we think is the result, we can very quickly see how the two molar masses cancel each others units out and only the mass of $\ce{KNO3}$ remains to give the right hand side of the equation a dimension of $\mathrm{g}$. As the only symbol left on the left hand side is a $m$ we know that it needs a dimension $\mathrm{g}$ and as such we can happily plug in the numbers into our calculator for the result.

Frankly, I am so used to "my" method that I had to look at your example quite closely to see what was actually going on. To me that method just seems unwieldy, as you clutter up equation space with chemical formulae.

That being said; find the method that works the best for you (meaning it is robust enough to also solve your chemical engineering, biology and other problems using dimensional analysis) and stick to it.


An acid or base can react with a solid. For instance HCl will dissolve $\ce{CaCO3}$ with much fizzing of $\ce{CO2}$ gas.

You also miss the point that $\ce{Cu(OH)2}$ has a solubility product. So there is a small amount of $\ce{Cu^{2+}}$ and $\ce{OH^-}$ in solution.

$\ce{K_{sp} = [Cu^{2+}][OH^-]^2}$

The perchlorate anion is a powerful oxidizing agent, but since $\ce{Cu^{2+}}$ is fully oxidized no redox reaction occurs.


Questions like these always appear simple, when in fact they are not. First of all it has to be clear that stability is no absolute concept. In this case one could ask which of the cations is most persistent, in other words is least reactive towards nucleophiles. With the exception of the tert-butyl cation, one could also ask the question, which of the remaining substituted cyclopropenium cations has the largest aromatic stabilisation. One would evaluate this traditionally with the hydrogenation energy. This obviously proves difficult, because it is a cation, it has to be neutral first.

Of course you can go ahead and apply approximations like +/-I or +/-M effect, and hyper- or  ordinary conjugation, or negative conjugation, and the there are steric effects to be considered, and dispersion effects, and don't neglect the solvent, and what about the anion. There are many things to be considered, and just naming them makes this problem a lot more complicated. It is our way of looking at these questions that make them appear simple.

While I wanted to use a more rigorous approach and I am generally interested in such weird molecules, I still have not come up with something proper, that suffices my standards (in a reasonable time frame).

First I thought that could be solved with calculating the ions and comparing them via the following isodesmic reaction scheme:

This is a conformational nightmare, a reaction that produces diastereomers, highly strained rings, etc. I therefore dropped this approach.

The other way to evaluate this is to simply add a hydride ion. While this reduces the complexity a little, it is still a conformational nightmare. It is in principle a little bit too much time to invest in such an exercise, and in the end the answer might not be helpful at all.

Coming back to your original question:

  What is the reason for the exceptional stability of tricyclopropyl-cyclopropenyl carbocation?

In extremely simple terms, it's exceptional stable because of the cyclopropenium core structure, because it is aromatic. Larger substituent moieties usually further stabilise such substructures, because they hinder an attack from nucleophiles. With a large enough counter ion (weakly coordinating ions, not necessarily as large as the ones mentioned on Wikipedia), it will be very persistent.

When you look at all the choices you have been given, the first conclusion should be, that they are all comparatively stable. We are not talking about primary, or secondary carbocations, or other clear cases. There really is only very little use in further analysing it.

For the sake of the argument:

Methyl groups can provide some more stabilisation (than hydrogen) because of hyperconjugation.
Phenyl groups are excellent at delocalising a positive charge when it overlaps with its π system. In that case the cyclopropenium ring is too small and the rings are too large, so they have to twist out of plane and therefore conjugation is not ideal. Here is the optimised structure of a DF-BP86/def2-SVP calculation:

Cyclopropanyl groups are rather small. They will usually attach in a way that the hydrogen is in plane with a π system. The group itself behaves like a π donor. (That's a factoid you might want to remember. See my answer on Why does cyclopropane react with bromine? and there are surely others to find. You can search in ron's answers first.)

To demonstrate this a little further, I have calculated the molecule at DF-BP86/def2-SVP level of theory and the π orbitals are below (click to enlarge).

In summary, the cyclopropanyl moiety has a good size and nice π-donating properties.

I still have to reiterate, that I find these questions not very helpful, especially if the limitations of such model views are not thoroughly discussed.


Walter Reppe developed a process for the "safe" trimerization of acetylene to benzene and cyclooctatetraene. Although completely superseded by more modern alternatives, this process was used before/during WW2 because starting from only coal, calcium carbonate and electricity one could synthetise calcium acetylide and therefore acetylene, having no need for oil derivatives, scarce during war.

The mechanism (in simplified terms - hold onto the stones!) is that on the surface of the catalyst (iron, copper or nickel depending on the variation of the process) there would be formation of transient "clusters" of three or four molecules of Fe/Cu/Ni-acetilyde. That close arrangement facilitates one of the electrons of the triple bond of the acetylide "attacking" one of the carbon atoms of one of the nearest acetylide neighbours (again, simplifications). That molecule with an excess of one electron would in turn atack its other in vicinity and finally this one would attack the first one (that became deficient in the first step).

Since the aromatic structure of benzene is rather stable compared to ethyne, and the product is 1/3 (or 1/4) the number of moles of reagents, the formation of product is favored under pressure.

Some sources:
Walter Reppe and the chemistry he developed: https://en.wikipedia.org/wiki/Walter_Reppe
Trimerization: https://en.wikipedia.org/wiki/Trimer_(chemistry)


Reaction of alkyne with sodium in ethanol

the reduction of alkynes using Na, proceeds through the anion radical, which can obtain trans conformation for the stability. Now considering the usage of NH3 and ethanol are proton donors. So, there is no effect on trans conformation altering the proton donors.


In atomic diffusion in Au/Cu metal contact, where do the atoms come from ? (need melting temperature to break bonding in lattice ?)

Yes, it takes energy to get an atom off a lattice site. However, as an activated process with an energy barrier, Arrhenius' equation shows you that even with that requirement, some number of atoms will be off their lattice sites in thermal equilibrium, making vacancies (empty sites) and interstitial atoms.

This was realized some time ago by Simmons and Balluffi. They investigated many metals, using precision measurements of the change in length of a bar of material in a furnace, coupled with x-ray diffraction to determine the lattice constant at the furnace temperature. The result is you can back out how many lattice sites are vacant from the difference in the change of length actually measured vs what you would expect from the change in lattice spacing.

The linked article focuses on gold, but has some data on other fcc metals as well. Just at the melting point of gold (not yet melted), the concentration of vacancies is about $7 \times 10^{-4}$, or 0.07% of lattice sites are vacant. The activation energy is given as 0.94 eV.

Bottom line - there are native populations of point defects even at room temperature, and those drive diffusivities. Certain point defects or defect complexes may be more mobile than others, and the concentration of a given defect will vary with temperature.


Chemistry has an organization called the International Union of Pure and Applied Chemistry (IUPAC), which sets standards for nomenclature (among other things). Chemists usually follow its recommendations.

IUPAC says

  Ionic charge is indicated by means of a right upper index, as in $\ce{A^n+}$ or $\ce{A^n-}$ (not $\ce{A^$+n$}$ or $\ce{A^$-n$}$).

and

  The iconic charge number is denoted by a right superscript, by the sign alone when the charge number is equal to plus one or minus one.
  ...
  $\ce{Al^3+}$ is commonly used in chemistry and recommended by [74 = see quote above]. The forms $\ce{Al^$+3$}$ and $\ce{S^$-2$}$, although widely used, are obsolete [74], as well as the old notations $\ce{Al^{+++}}$, $\ce{S^=}$, and $\ce{S^{--}}$.

No reason is given, but you'll find a historical account of sign conventions for charge and a defense of IUPAC's choice in this article. He says,

  [IUPAC's convention] avoids confusion with the conventional symbolism for inherently positive and negative numbers and maintains consistency in how we count physical entities. Thus, in counting apples, we say two apples, three apples, etc., not apples two, apples three – that is, the number always precedes the name of the entity being counted. Likewise, when counting charges, we should say two positive charges or three negative charges, not positive charges two or negative charges three. The IUPAC ruling was intended to make the charge number symbolism consistent with this verbal convention.

Personally, I think the charge should be written like any other signed number. Formulas typed from the keyboard without superscripting can be ambiguous; is X2+ to be interpreted as $\rm X^{2+}$ or $\rm X_2^+$? (You should type X(2+) or X2(+), but few people actually do that.) But it's pretty clear what X+2 would mean. Putting the sign first also makes it trivial to parse the charge from a formula typed this way without the need for parentheses.


What is a laboratory set-up for the synthesis of ethanol from ethene?

You should try bubbling the ethene gas through dilute sulfuric acid. It should form ethyl sulfate and get hydrated with water forming ethanol. As the gas will bubble through the solution, you should have an ample amount of liquid so that most of the gas can maximally react with the sulfuric acid before bubbling out as gas. You should do the reaction in a fume, hood too. Then I think that you could separate the ethanol from the water by distillation of the ethanol from the mixture, or you could separate it by exploiting the great difference in the freezing points of ethanol and water.

If you could connect tubes in such a way with special chemistry glassware, then maybe you could make some kind of reflux apparatus like an acid reflux apparatus where the gas funnels back into the bottom of the liquid and cycles through several times then you could increase your yield. Just if you use some cyclic mechanism, don't release the gas too quickly or you could break your apparatus.

I'm not exactly sure what the experimental design is. But due to the concentration of gas in the reaction chamber, it's important to have an escape valve for the gas for safety purposes of equalizing pressure. If you know what you want, an experienced glass-blower can make your apparatus.


Enthalpy of mixing of a non-ideal solution

There are two ways (at least) to solve this:
(1) Evaluate $\Delta S = -\left( \frac{\partial \Delta G}{\partial T} \right)_{p,n} $
and add $T\Delta S$ to $\Delta G$.
or
(1) Evaluate directly $\Delta H = -T^2 \left( \frac{\partial (\Delta G/T)}{\partial T} \right)_{p,n} $
Either way I obtained the result
$\Delta H = \chi n_1 \phi_2 - n_1\phi_2 T \left( \frac{\partial \chi }{\partial T} \right)_{p,n} $


The energy difference between $n=2$ and $n=3$ in Bohr's model is
\begin{equation}
\Delta E=13.6\ \mathrm{eV} \times \left(\frac{1}{4}-\frac{1}{9}\right) = 1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ} = 181.4\ \mathrm{kJ/mol}
\end{equation}
Thus, for this excitation to occur, a single energy quantum (photon) with an energy of $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ is required.

The excitation does not happen if:

the photon has more or less energy than $1.88\ \mathrm{eV}$ (The energy difference needs to be matched exactly!)
the same amount energy is distributed over multiple photons (The energy needs to be provided by a single photon!)

As already has been pointed out, the $181.4\ \mathrm{kJ/mol}$ itself is not a quantum, but one mole of quanta, where each quantum has $1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ}$ of energy. As each atom can only absorb this energy once (assuming it does not relax back to $n=2$ by emitting the energy), to absorb so much energy will also require one mole of atoms.


Why does O2NCl only have two resonance structures?

$\ce{O2NCl}$ does not have two or three resonance structures but rather almost infinitely many that have different probabilities. Resonance structures will never bring you close to the actual picture; for the true picture to evolve you will need to do molecular orbital calculations. That said, resonance structures are often a somewhat good approximation if weighted accordingly.
This last expression in italics is important. Not all resonance structures are equal. People teaching resonance structures usually use expressions like most probable resonance structures. To determine whether a resonance structure is more probable than another one, check the following list:

does it give all main group atoms a valence electron octet?
are formal charges minimised?
are formal charges distributed in accordance with electronegativity (i.e. do electronegative atoms have more formal negative charge than electropositive ones)?

The list is ordered from most important to not so important although traditional teaching sometimes weighs maximised amount of bonds higher and thus arrives at hyperoctet structures.
Now let’s check the structures of $\ce{O2NCl}$ that you are talking about:

We see that all structures fulfill rule 1. However, rule 2 clearly shows a difference: The two structures on the left contain $+1$ and $-1$ formal charges each while the rightmost structure has $+2$ and $-2$. We don’t need to check rule three, we can see that the rightmost structure is unfavourable and thus less close to the truth. Finally, the formal charges are distributed correctly on all three structures, because oxygen is more electronegative.
Therefore, the left two structures are equally probable while the rightmost one is not so; oftentimes in exams, only the left two will be considered ‘correct’. But as I said, the true picture (or something really close to the truth) can only be determined by quantum chemical calculations.


Does oxygen catalyze the decomposition of sucrose?

If any sucrose hydrolysis were occurring due to oxygen as a factor, it's more likely due to acids formed as it oxidizes species in the solution (for example, alcohol groups to carboxyls), which conceivably could include sucrose itself. But, it seems more likely that the $\ce{CO2}$ in air would be a factor because $\ce{CO2}$ dissolution increases the acidity of the solution, in addition to any other acids present (citric, for example).

However, sucrose hydrolysis is really slow unless the acidity and/or heat is significant ($\mathrm{pH} < 0$, or heat near boiling), so I doubt either of these explanations, unless these are really old, and/or had been heated.  More likely the glucose was there in the first place.

The way to properly answer this question would be to do control studies in which known amounts of sucrose and acid are initially present and tested with or without exposure to air having varying amounts of oxygen and carbon dioxide.


Potentiometric titration using a pH meter

In principle, you should know the pH of the buffer solution. Measuring it before an unknown sample is a straightforward way of calibrating the detector (pH detectors can get a bit funny after a while).

If you put the detector into a pH 5.5 solution and it reads 7, you're going to have to tweak the detector (if possible) or find another.

Measuring unknown samples with a detector that incorrectly measures a known buffer solution is a bad idea but an even worse idea would be to measure unknown samples without knowing your detector was incorrect in the first place.


Does a C=C double bond donate electron or withdraw it?

I guess it depends on the way you quantify "electron donation/withdrawal" (because it is a useful concept, but not really a measurable observable). I'm sure someone will find a counterexample, but if you consider methane and ethylene as Brønsted acids, you find that methane is a much weaker acid. Thus, ethylene is better at stabilizing the negative charge. (This is in line with your sp2 argument.)


Why Does Activation Energy not change with Temperature change

You have conceptual misunderstanding. In the activation energy context, energy of reactants, products and transition state are all potential energies, independent on the kinetic energy of molecules.
The kinetic energy of molecules is used to reach the peak of the potential energy, similarly as e.g. a acrobatic aircraft uses its kinetic energy to climb to the apex of an acrobatic loop.
Higher temperature means higher kinetic energy, but it does not mean the potential energy barrier is lower. It means there is higher probability to overcome it, leading to a faster reaction.
There is a famour Arrhenius' equation for the reaction rate:
$$k = A \exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$$
where $A$ is the frequency factor, describing the rate of molecular collisions, potentially able to take part of the reaction and
$\exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$ is the term from the Boltzmann distribution,
telling us the probability a molecule at temperature $T$ has sufficient energy, if the activation energy is $E_\mathrm{A}$, and $k$ is the Boltzmann constant.


The idea of boundary conditions is nice and intuitive, but this is one case in which intuition cannot really help us.

In general when you are solving the Schrödinger equation in spherical coordinates, you can separate the radial and angular parts (unless you have some pathological potential energy term, I think). See Griffiths' Introduction to Quantum Mechanics (2nd ed.), Chpt 4 for details. It doesn't matter whether you're solving the H atom, a particle on a sphere, or indeed a free particle: in all cases the potential energy has no angular dependence in the potential energy, so the angular component is the same problem.

When you solve the angular bit, you can (again) separate out the two angles, so you end up with two differential equations: one in $\theta$ and one in $\phi$. The $\phi$ bit is easy enough:

$$\frac{1}{g}\frac{\mathrm d^2 g}{\mathrm d\phi^2} = -m^2$$

where $m$ is unconstrained for now, and $g$ is a function of only $\phi$. This is easy to solve and applying the "common-sense" boundary condition that $g(\phi) = g(\phi + 2\pi)$, you can find that $g = \exp(\mathrm i m\phi)$ with $m$ an integer. So far so good.

The $\theta$ bit is substantially more difficult, but this differential equation is so important that it has been studied for centuries and has its own name, the Legendre equation:

$$(1 - x^2)\frac{\mathrm d^2\! f}{\mathrm dx^2} - 2x\frac{\mathrm df}{\mathrm dx} + \left[l(l + 1) - \frac{m}{1 - x^2}\right]f  = 0$$

where $f$ is again a function of only $x$, and we have made the substitution $x = \cos\theta$. In our case, it turns out that unless $(l, m)$ obey certain constraints (i.e. the now-familiar $l = 0, 1, 2\ldots$ and $m = -l, \ldots, +l$), then the solutions $f(x) = f(\cos\theta)$ go to $\pm\infty$ at $x = \pm 1$, i.e. $\theta = 0$ or $\theta = \pi$. That's not acceptable for a well-behaved wavefunction, so we have to reject all the other cases.

So, you could say that the boundary condition on $\theta$ is that the wavefunction  must be welll-behaved (i.e. not diverge to $\pm\infty$) at $\theta = 0$ and $\theta = \pi$. But that is not something we could say without delving into the maths. A priori, we could certainly say that it is important for the wavefunction to be well-behaved at all points. But we cannot pick out the two particularly crucial points $\theta = 0$ and $\theta = \pi$ just using common sense, or "by observation".

More details can be found in Griffiths (there is a fair bit of mathematical detail in there), or any other book that covers differential equations, really.

Bonus: if you multiply the solutions $f(\theta)$ with $g(\phi)$, then normalise the entire thing, you get what are known as the spherical harmonics, which are also extremely important functions in their own right; that's where the shapes of the atomic orbitals come from.


pH of a buffer made with boric acid and potassium dihydrogen borate

Update: Per the advice of @DavePhD , I decided to trust my work and I approached my teacher with the problem. He just emailed me to say there must have been some error in the online system; 8.92 is the correct answer.

Thanks so much to everyone who took the time to help me with this!


Why do aqueous solutions of sodium chloride and silver nitrate swap ions?

While solutions of silver nitrate and sodium chloride are in stable equilibrium on their own, once mixed the equilibrium is changed. When the salts are dissolved, you no longer have silver and nitrate or sodium and chloride ions associated with each other, you have a mixture of individual ions in solution.

But silver chloride has very low solubility, about 1 mg can dissolve in 1 L of water. So when a silver ion and chloride ion meet, they quickly fall out of solution. So if we had an equilibrium between the dissolved ions and the small amount of "soluble" silver chloride, that equilibrium would be driven toward the silver chloride by the precipitation. When compounds escape the solution by precipitating or evaporating, they are removed from the equilibrium, so trying to replace them to restore equilibrium requires converting more free ions into silver chloride and pushes the reaction to completion.


The answer is correct. After the first nucleophilic attack from $\ce{MeMgBr}$, the ethoxy group left and the ester is transformed into a ketone (Step 2 in the link), then another $\ce{MeMgBr}$ attacks and introduce an extra methyl group, resulting in a $3^\circ$ alcohol after hydrolysis. The process is well described in this page.


Comparison of chemical shift of methyl group attached to alcohol and carboxylic acid

It's a good idea to start with some data. The table below shows relevant $\ce{^1H}$ shifts in $\pu{ppm}$ for three compounds in various deuterated solvents (shift values are from Cambridge Isotope Labs): methanol (MetOH), formic acid (FormAc) and ethanoic or acetic acid (AcAc).

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Compound} & \ce{CDCl3} & \ce{(CD3)2SO} & \ce{C5D5N}\;\text{or}\;\ce{C5H5N}  & \ce{C6D6}\;\text{or}\;\ce{C6H6} & \ce{D2O} \\ \hline
 \text{Acetic acid} & 2.13 & 1.95 & 2.13  & 1.63 & 2.16\\ \hline
 \text{Formic acid} & 8.02 & 8.18 & 8.54 & 7.24 & 8.22\\ \hline
 \text{Methanol} & 3.48 & 3.20 & 3.57 & 3.09 & 3.35\\ \hline
\end{array}
$$

The protons in methanol are more deshielded than those in ethanoic acid because in methanol an electron withdrawing oxygen atom is directly bonded to the carbon, while in ethanoic acid there is an intervening carbon.

If you remove that intervening carbon things get interesting, as the case of formic acid illustrates. The carbon atom in a carboxylic group is $\mathrm{sp^2}$ hybridized, which makes it more electrophilic than the $\mathrm{sp^3}$ hybridized carbon in methanol. Of course the carbon in a carboxylic group is also bound to two electron withdrawing oxygen atoms, rather than one, which results in greater deshielding. The formic acid proton is also subject to deshielding magnetic anisotropy associated with the delocalized electrons of the carboxylic group. The anisotropy effect is less pronounced in ethanoic acid because the methyl group is further away from the carboxylic group. The ultimate result is that the formic acid proton is very strongly deshielded compared to protons in the other compounds.


How to remove fluorine from 2-[2-(3-bromophenyl)ethenyl]-5-fluoro-1,3-benzoxazole?

How to remove fluorine from this compound?

There's no need to try that!

Just use  the commercially available 2-methyl-1,3-benzoxazole and try to react it with 3-bromobenzaldehyde in N,N-dimethylformamide under reflux for 2-4 h using self-prepared $\ce{KF-Al2O3}$ as a catalyst.

The reaction has been described (DOI) for several benzaldehydes, such as $\ce{4-H, 4-Cl, 4-Me, 4-OCH3, 4-N(CH3)2}$, with yields > 80%.