Why are DCM and chloroform so resistant towards nucleophilic substitution?

Introduction (and Abstract, TLDR)

In very short words you can say, that the anomeric effect is responsible for the lack of reactiveness. The electronic effect may very well be compensating for the the steric effect, that could come from the methyl moiety. In any way, most of the steric effects can often been seen as electronic effects in disguise.

Analysis of Molecular Orbitals

I will analyse the bonding picture based on calculations at the density fitted density functional level of theory, with a fairly large basis set: DF-BP86/def2-TZVPP. As model compounds I have chosen chloromethane, dichloromethane, chloroform and chloroethane.

First of all let me state, that the bond lengths are a little larger at this level, however, the general trend for shortening can also be observed. In this sense, chloroethane behaves like chloromethane. An attempt to explain this will be given at the end of this article.

\begin{array}{lr}\hline
\text{Compound} & \mathbf{d}(\ce{C-Cl})\\\hline
\ce{ClCH3}      &  1.797\\
\ce{Cl2CH2}     &  1.786\\
\ce{Cl3CH}      &  1.783\\\hline
\ce{ClCH2CH3}   &  1.797\\\hline
\end{array}

In the canonical bonding picture, it is fairly obvious, that the electronic effects dominate and are responsible for the lack of reactivity. In other words, the lowest unoccupied molecular orbital is very well delocalised in the dichloromethane and chloroform case. This is effectively leaving no angle to attack the antibonding orbitals.

In the  mono substituted cases there is a large coefficient at the carbon, where a nucleophile can readily attack.

One can also analyse the bonding situation in terms of localised orbitals. Here I make use of a Natural Bond Orbital (NBO) analysis, that transforms the canonical orbitals into hybrid orbitals, which all have an occupation of about two electrons. Due to the nature of the approach, it is no longer possible to speak of HOMO or LUMO, when analysing the orbitals. Due to the nature of the calculations, i.e. there are polarisation functions, the values do not necessarily add up to 100%. The deviation is so small, that it can be omitted.
The following table shows the composition (in $\%$)of the carbon chloro bond and anti bond.
\begin{array}{lrr}\hline
\text{Compound} &\sigma-\ce{C-Cl} & \sigma^*-\ce{C-Cl}\\\hline
\ce{ClCH3}      &  45\ce{C}(21s79p) 55\ce{Cl}(14s85p)
                &  55\ce{C}(21s79p) 45\ce{Cl}(14s85p)\\
\ce{Cl2CH2}     &  46\ce{C}(22s77p) 54\ce{Cl}(14s85p)
                &  54\ce{C}(22s77p) 46\ce{Cl}(14s85p)\\
\ce{Cl3CH}      &  48\ce{C}(24s76p) 52\ce{Cl}(14s86p)
                &  52\ce{C}(24s76p) 48\ce{Cl}(14s86p)\\\hline
\ce{ClCH2CH3}   &  44\ce{C}(19s81p) 56\ce{Cl}(14s85p)
                &  56\ce{C}(19s81p) 44\ce{Cl}(14s85p)\\\hline
\end{array}
As we go from mono- to di- to trisubstituted methane, the carbon contribution increases slightly, along with the percentage of $s$ character. More $s$ character usually means also a stronger bond, which often results in a shorter bond distance. Of course, delocalization will have a similar effect on its own.

The reason, why dichloromethane and chloroform are fairly unreactive versus nucleophiles, has already been pointed out in terms of localised bonding. But we can have a look at these orbitals as well.
In the case of chloromethane, the LUMO has more or less the same scope of the canonical orbital, with the highest contribution from the carbon. If we compare this antibonding orbital to an analogous orbital in dichloromethane or chloroform, we can expect the same form. We soon run into trouble, because of the localised $p$ lone pairs of chlorine. Not necessarily overlapping, but certainly in the way of the "backside" of the bonding orbital. In the case of chloroethane we can observe hyperconjugation. However, this effect is probably less strong, and from the canonical bonding picture we could also assume, that this increases the polarisation of the antibonding orbital in favour of carbon.
In the following pictures, occupied orbitals are coloured red and yellow, while virtual orbitals are coloured purple and orange.

(Note that in chloroform two lone pair orbitals are shown.)

Conclusion

Even though this article does not use the Valence Bond Approach, one can clearly see the qualitative manifestation of Bent's Rule (compare also: Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?). A higher $s$ character means a shorter bond. The lack of reactivity towards nucleophiles can be explained electronically with a delocalised LUMO. In terms of localised bonding, the lone pairs of any additional chlorine atom would provide sufficient electron density, to shield the backside attack on the carbon.


Why can't oxalate ion donate two pairs of electrons from the two double-bonded oxygen atoms

Oxalate is a planar molecule. The four oxygen atoms share the pi electron density through resonance, as noted in comments, and thus are all identical. Imagine those four oxygen as four corners of a table. Now you are a Lewis acid, perhaps a transition metal cation, and you walk up to the table. How many of the four corners ligate you? Perhaps you imagine the planar ligand dropping on top of you, so that the four corners are around you. For effective overlap with all four corners of the table, you'd need to be a pretty big cation. In almost all cases you'd achieve better orbital overlap, and a more energetically favorable coordination complex, if you just cozy up to two of the donor oxygens.


Why are trans fats worse than saturated fats?

Someone may very well give a more thorough answer but from what I understand, most naturally occurring unsaturated fats have their double bonds in the cis conformation which is generally higher in energy than the trans conformation.

Thus, the idea is that when you ingest trans fats (i.e. fats with a trans double bond) your body has to somehow hydrogenate this double bond before it can break it down further. This is more difficult for your body to do because it must either isomerize to a cis bond and then hydrogenate, or just hydrogenate the trans bond which is not easy for your body to do effectively. In some cases hydrogenation of trans fats can't actually happen because nature basically makes cis double bonds in fats (which sounds weird but is true) so your body has ways of breaking these fats down but not trans fats. So, the fat just gets stored or excreted without doing anything to it.


How to calculate the probability of finding an electron in a box between 0.25L and 0.75L?

You have the reasoning correct, so I suspect you made a mistake in your integral calculation. The probably of finding the particle between $a$ and $b$, for the $n$-th state, is indeed given by:

$$P_n(a,b) = \int_a^b \left|\psi_n(x)\right|^2\, \mathrm{d}x $$

but if I do the calculation for the ground state ($n = 1$), I find the same results as your textbook:

$$ P_1\left(\frac{L}{4},\frac{3L}{4}\right) = \frac{1}{2} + \frac{1}{\pi} \approx 0.818 $$


Why is picric acid more explosive than TNT?

You're trying to define the stronger explosive by comparing the reagents, but in this case comparing the products is far more important.

Trinitrotoluene is a rather oxygen-poor explosive, to the point that when detonated as a pure substance, it produces carbon soot and hydrogen gas:

$$\ce{2C7H5N3O6 -> 3N2 + 5H2 + 12CO + 2C}$$

The production of $\ce{CO}$, $\ce{CO2}$ and $\ce{H2O}$ is highly exothermic, so a lot of extra energy could be obtained if there were more oxygen to burn the soot and hydrogen. However, in detonations there's not nearly enough time for atmospheric oxygen to partake extensively in the reaction. This means that a fair amount of the explosive potential of TNT is wasted, decreasing its detonation yield per gram of substance. One way to compensate this yield loss in oxygen-poor explosives is to mix them with an oxidiser, usually an oxygen-rich substance such as $\ce{KClO3}$.

Now, on to picric acid. The structure and molar mass are very similar to TNT, but replacing the methyl group with a hydroxyl means the oxygen content of picric acid is higher. While still overall oxygen-poor, the detonation proceeds with much less soot and hydrogen generation, meaning a more complete burn, a higher energy release, and consequently a greater explosive yield.

$$\ce{2C6H3N3O7 -> 3N2 + H2 + 2H2O + 12 CO}$$

In reality, the decomposition reactions are more complex. The balanced equations above follow a rule of thumb presented here. However, a more thorough analysis should agree with this simplified result.


Melting post-1982 minted U.S. pennies and aluminum: Why does the slag turn blue?

Zinc burns in air forming zinc oxide with a bright blue flame. The bit of copper in US cents will add a greenish color to the flame.

N.B. Fumes of burning zinc and some other metals are toxic. Avoid contact or inhaling the smoke.


Is it possible to use Vmax as Kcat in a Michaelis-Menten reaction?

$V_{max}=k_{cat}[E_0]$ where $[E_0]$ is the total enzyme concentration and not just the free enzyme. See my answer here.

You cannot use the value of $V_{max}$ as $k_{cat}$ because while the latter is the property of the enzyme (actually the enzyme-substrate pair), the former is the property of the chemical sample i.e. it is a system parameter.

Obviously, the $V_{max}$ will be same as $k_{cat}$ for a unit-concentration of total enzyme (unit concentration does not mean units of enzyme. It means, for whatever you are using as the units of concentration in your equation, lets say unit-x, the concentration of the enzyme should be 1 unit-x)


What does "parts by weight" or "parts by mass" specifically mean?

It seems that the unusual wording comes from the fact that we are used to having all of the atomic masses available to us simply because we have measured all of them. In the early days of chemistry, however, things were not so simple. Thus, the equivalent weight (technically mass since it has units of $\mathrm{kg}$ rather than $\mathrm{N}$) was used as a method for determining a weight relative to some predefined standard. In this case, hydrogen.

Some of the simplest reactions are acid/base reactions, and this explains why the scale is measured relative to equivalents (where one equivalent is one mole) of hydrogen. That is, in your quote from the wikipedia page, one mole of hydrogen, which has a mass of 1.008 grams, can react with 8 grams of oxygen or 35.5 grams of chlorine to form $\ce{H2O}$ and $\ce{HCl}$ respectively. Another way of saying this is that every hydrogen atom reacted with two oxygen atoms and every hydrogen atom reacted with one chlorine atom.$^1$ Thus, we see what is meant by the term "parts by mass". Notice this is not the same as the phrase "parts per mass". "Parts by mass" could be stated more clearly as the "parts of X measured by mass". That is, we are indirectly measuring how many parts of oxygen, chlorine, or whatever other X goes into a reaction by measuring the final mass of a substance which reacts with a known mass of 1.008 grams of hydrogen.

The wikipedia article notes that not everything reacts with hydrogen, so in order to find equivalent weights for some materials, you must measure its equivalent weight to a known mass of oxygen and then use the 8:1 conversion to hydrogen equivalent weights.

I hope this clarifies the idea. Don't get worried though, this does not contain any more information than simple atomic weights.

$^1$This is kind of cheating though because we know the weights of oxygen and chlorine, while chemists in the 1800s may not have known this.


Plot Binary Eutectic from Temperature of Melting and Enthalpy

The reason that you can't figure out how to solve for the eutectic based on the melting temperature and enthalpy of fusion of the two constituents is because, well, you can't. Those bits of information, while useful in constructing the phase diagram (well, at least the melting temperatures), are not what determines the eutectic. You will have a eutectic if and only if there are interactions between the constituents in the liquid phase that reduce the Gibbs free energy below the ideal value.

Lets make this more concrete. Take elements A and B, each with a solid and liquid phase, no intermediate phases. Taking just one phase (solid or liquid) for the moment, what happens to the Gibbs free energy when we mix them together? The first bit you get is the weighted combination of the pure phase free energies, that is:
$G_{mixture} = x_{A}G_{A} + x_{B}G_{B}$.  But is that enough? Well, no, we forgot the entropy of mixing term. So, for an ideal mixture,

$G_{ideal} = x_{A}G_{A} + x_{B}G_{B} + RT(x_{A}\ln x_{A} + x_{B} \ln x_{B})$

What does this look like? Well, taking 'fake' elements, A that melts at 950K and B that melts at 1000K, just using ideal mixtures for solid and liquid, one gets a pretty boring diagram that looks like:

with the slight separation in solidus and liquidus from the slight asymmetry in the melting points. Hmmmm... - no eutectic (since the ideal mixing impacts both liquid and solid in the same way, that is to be expected).

OK, now lets move to a regular solution model. The difference is the addition of an enthalpy term, that describes whether an A atom would rather see A or B around it (and vice versa). In the simplest way, this can be represented by a heat of mixing term using one parameter, $\Omega$. $\Omega$ is positive if the atoms don't want to mix (so, it can cause a miscibility gap), or negative if they actually prefer to mix. So, for a regular solution one gets:

$G_{regular} = x_{A}G_{A} + x_{B}G_{B} + \Omega x_A x_B + RT(x_{A}\ln x_{A} + x_{B} \ln x_{B})$

Setting $\Omega$ to -1kJ/mol, the previous diagram turns in to:

and you see that a eutectic has now formed at about $x_B = 0.25$.

Now, back to the parameters you thought were important:
The melting temperature is kind of important, in so far as an asymmetry between them shifts the eutectic of the regular solution towards the lower melting element.

The enthalpy of fusion is not a factor at all in determining where phase boundaries are.

The interaction term(s) are highly relevant, since they determine where and how deep the eutectic is.


How can I determine if there are π-π interactions between an amide and an aromatic ring in a protein?

It seems that NickT was looking for an experimental solution. My post deals, however, with a computational solution.

  How would I quantify how significant the interaction is?

Determining the interaction energy between two defined monomers such as your aromatic triazole and amide is a rather straightforward process.  This process is referred to as the supramolecular approach.  I'll point you to a paper that analyzes the benzene dimer.  This method is strictly a computational one so a bit of knowledge in computational chemistry is necessary.

The Supramolecular Approach

The What

The supramolecular approach boils down to this.

$$E_{\mathrm{int}} =  E_{\mathrm{dimer}} - (E_{\mathrm{mon1}} + E_{\mathrm{mon2}})$$

Here we have some interaction energy ($E_{\mathrm{int}}$) determined from the difference of a dimer energy ($E_{\mathrm{dimer}}$) and the sum of the two monomers ($E_{\mathrm{monomer}}$).  If both monomers were equivalent (say, you were interested in the interaction energy of the benzene dimer where each monomer was a benzene ring), you could simplify the summation to two times the energy of one monomer ($2\times E_{\mathrm{mon}}$). In your particular case, you have two different monomers.

The whole idea is if I have two interacting molecules, I can determine the energy of each molecule individually (as if they were separated at infinite distance) as well as their complex.  So as you bring these molecules closer and closer together, the energy starts to go down (the interaction energy).

The How

NOTE: Your geometry is from a crystal structure.  Do NOT modify this geometry.  You will want to keep everything exactly as is.  This means you don't want to optimize your system.  You do not want to eyeball monomer placement.  Take everything from your known structure and be careful not to change it otherwise it can ruin this process.  We are modifying the structure by truncating and capping but intermolecular parameters must stay the same for whatever it is you are trying to model.  You will want to optimize your cap meaning, run an optimization on your capped-monomer but freeze everything but the thing you are using to cap.

Define your monomers.  You will need to determine what part of your 'dimer' system is important for describing this weak interaction.  I recommend keeping the aromatic ring and truncating the ring with something similar to what is being truncated.  You could cap your monomer with a hydrogen or a methyl group for example.  If the piece you've cut out is highly polarizable, cap with something with a similar property.  If your truncated piece is neutral in charge, cap with something that is neutral.  You get the idea.
Define your dimer.  Your dimer is simply a combination of your two defined monomers.
Determine the method you want to implement.  Wavefunction based or density functional theory (DFT) will work for this.  The former will be prohibitively expensive.
Define your basis set.  For aromatic systems your best bet would be to use Dunning-Hunzaga's correlation consistent family of basis sets.  I recommend using aug-cc-pVTZ for good results.  Whatever you decide, be sure your basis set includes polarization and diffuse functions.  The suggested basis set does this (augmented means diffuse on all atoms whereas pVTZ means 'polarized-valence triple zeta').
Determine the energies of your monomers and dimer.  You will want to run a single-point energy calculation on your monomers and your dimer. Optimize your monomers first but freeze all atoms except those you added to cap your monomer.
Determine the interaction energy.  Plug your energies into the equation given above and determine the interaction energy.  Convert to whatever units you wish to use (I prefer kJ/mol but most people use kcal/mol so you may want to use that).

I hope this helps.  Looks like an interesting project to say the least.

A quick note about the 'valence' basis set:

Typically core electrons do not participate in any sort of interesting chemistry (they are so much lower in energy than the valence electrons that they do not play a role in things like chemical bonding).  When you perform these energy computations, you will be invoking the frozen-core approximation (default on many quantum software packages but be sure to check the documentation first).  The pVTZ basis set expands the basis functions out to triple-zeta quality for valence electrons only.  The core electrons are given s-type functions and that is it.  If you ever find yourself in a situation where the core electrons are important to your system of interest, you would want to use the pCVXZ basis sets (where X = D,T,Q, etc.) which stands for polarized core-valence X zeta.  Here the core electrons are given the full set of X-zeta quality basis functions.  However this means you've just made your computation much more expensive.  The frozen-core approximation is typically used.


The exact choice of electrode really depends on the goal of the experiment and what will be in your cell:

The most basic requirement is that the solvent or electrode material not decompose at the potential you want to work at.
For each solvent/electrode combination, there is a potential window in which you can safely work.
In the most common situation—working in aqueous solutions, the choice of working electrode mainly depends on whether you want to observe oxidation or reduction.
Electrodes like platinum or gold are commonly used for oxidation reductions because they have high overpotential for oxygen evolution in aqueous solutions (platinum in particular, performs poorly for reduction as it is rather good at catalysing the hydrogen evolution process).
Conversely, for reduction reactions, mercury and bismuth have potential windows that extend to quite negative potentials, but will themselves oxidize as you move towards positive potentials.
There are also some materials that work pretty well for both positive and negative potentials, such as glassy carbon and boron-doped diamond.

For analytical purposes, you can get away with (and often prefer) small electrodes, so cost is less of a concern, but if the application is mass electrolysis, large electrodes are required to have a decent reaction throughput and more expensive materials may be impractical.
Cheap materials like stainless steel and graphite can be used as electrodes in some cases, though they are more limited in the systems in which they may be used.

As far as the reason to choose platinum over lead oxide, gold, or silver is concerned, the standard electrode potential isn't a primary consideration. (Silver's is not higher than platinum's, incidentally)
The standard electrode potential reflects the thermodynamic situation of a particular half-reaction, but is not terribly informative of how the material will behave in a given cell as it does not account for the kinetics of the reaction or any other factors.
Silver is usually not a good choice for a working electrode as it oxidizes quite easily in aqueous solutions, gold is actually a fairly common choice as it has a wider potential window than platinum in aqueous solutions, but it is easily fouled by organic/biological contaminants.
Platinum doesn't have the widest potential window, but is generally resistant to fouling and being reacted itself (giving it a long lifetime).

In a nutshell, electrode choice is dictated by many competing factors: cost, lifetime, potential range, mechanical considerations, etc.
In an analytical lab, screen-printed gold electrodes are quite common because they are relatively inexpensive, suitable for many different reactions, and can be discarded and replaced when fouled.
In an aluminum refinery, you'll find carbon and coke electrodes used in the Hall-Héroult process, as they need to be very large and withstand the high temperatures of molten cryolite.
In the cloralkali process, the anode is usually made from titanium as it must resist the highly oxidizing conditions that produce chlorine, but the cathode can be made from much cheaper nickel as that half of the cell is only used to reduce water to hydroxide and hydrogen gas.


How do ionic, polar, hydrogen-bonding and non-polar compounds compare in TLC?

No, your analysis is wrong.

The nice thing about TLC is that you can pick the "paper" and the solvent to suit your analysis. So if you use benzene as the solvent then it won't "drag" ionized compounds along behind the solvent front very well.  If you use water ionized compounds do get dragged along nicely.


Using a significantly large metallic sphere, could we draw out the electrons from the negative terminal of a battery?

Yes, this is possible, you have effectively described using a battery to charge a very weak capacitor, then discharging the capacitor.  The charge stored will be minimal (possibly immeasurable), pico-coulombs or less and the number of cycles to fully discharge then battery may be on the order of billions or trillions, but in theory the process described is possible (more properly: plausible).


Designing a theoretical synthesis of methamphetamine from phenylpropyne

I think the synthetic route you've proposed is quite good for the most part. However, I think your use of $\ce{NaH}$ to remove the tosylate is incorrect. $\ce{NaH}$ is a very strong base, but not a good reductive hydride donor in most instances. Here, it will almost certainly just react by eliminating the tosylate to give a double bond. However, LAH (and various similar reagents) should achieve the desired effect. Alternatively, one of the more popular dehydroxylation reactions is the Barton-McCombie deoxygenation.

An alternative route might begin with hydroboration-oxidation of the alkyne directly. Generally, when the two ends of the alkyne are equally substituted, one obtains a mixture of isomers, and electronic factors are usually more decisive than sterics. That being said, in this case, I suspect that it might be possible to control the regiochemistry by selecting a sterically hindered borane, perhaps 9-BBN or catecholborane. I'm not certain, however. Assuming one can indeed install the ketone at the end of the alkyne further from the phenyl group, then reductive amination should give the desired product. Direct reductive amination should be possible in this case (e.g., with $\ce{NH2Me}$, catalytic acid, and $\ce{NaBH3CN}$), allowing for a one-pot reduction.

Hence, the proposed reaction scheme would be:
\begin{align}\ce{
Ph-C#C-CH3 &->[\mathrm{1)~9-BBN}][\mathrm{2)}~\ce{H2O2,~NaOH}] Ph-CH2-C(O)CH3\\\\
           &->[\ce{NH2Me,~[H+]}][\ce{NaBH3CN}] Ph-CH2-CH(CH3)NHMe
}\end{align}

An additional point in favor of this approach is that it can be made highly enantioselective by use of asymmetric catalysts (q.v., Noyori asymmetric hydrogenation).

Addendum

In response to your second reaction scheme, jerepierre's comment above is definitely correct about a competing elimination reaction in the last step being a problem. It's especially likely because the proton transfer is fast, $\ce{NaNHMe}$ is extremely basic, the elimination is highly entropically favored, and the resulting $\pi$-bond is conjugated to the ring, giving an especially stable product (the major elimination product by a large margin should be the Zaitsev one). In sum, I think both kinetic and thermodynamic factors are heavily against you here. If you really want to go the route of nucleophilic substitution to attach your amine, better approaches are probably the Delépine reaction, any of the azide reduction methods, or the Gabriel synthesis (approximately in that order, I think). All of those suffer from the fact that you would then need to methylate the resulting primary amine, and avoiding extensive polyalkylation is more or less impossible, as far as I know.

There's also an issue before that stage, in that hydroboration-oxidation will not predominantly install the alcohol at the correct position, giving mainly the benzylic alcohol instead, unless you use a suitably bulky borane to achieve the desired regioselectivity. Since you only want to replace the alcohol with a bromine, I think $\ce{HBr}$ in the presence of peroxide (or some other radical initiator) should work, with no intermediate step and a reasonably low proportion of unwanted side-products.


Formula for rate constant for the first order reaction

I am attaching a written answer as typing the log and other exponents is hard :(


As for your first question, I have personally had better luck only aging 24-48 hours (actually unsealed, just covered with Kimwipe to prevent contamination, but allowing for some of the solution to evaporate).  We're getting good $\ce{ZnO}$ XRD peaks on this process, and are currently developing an SOP/Optimization for $\ce{ZnO}$ by sol-gel technique. I'll let you know when we publish.

When you say you're getting $\ce{ZnO}$ and Zinc peaks, do you mean you're seeing XRD peaks matching Zinc as crystalline metal?  To your question precisely: "What possible reactions would produce $\ce{Zn}$ from $\ce{ZnO}$, and how likely are they?" Remember that your sol-gel is not necessarily $\ce{ZnO}$ yet.  The reaction that produces $\ce{ZnO}$ begins with the dissociation of $\ce{Zn}^{2+}$ and $\ce2{CH_3COO}^{-}$ ions in ethanol, followed by a hydrolysis. As to how you're getting the extra electron to precipitate metal, I don't know. What we have seen has been the precipitation of $\ce{Zn(CH_3COO)_2}$ after about 96-120 hours after originally mixing.  (We're pretty sure it's acetate because it smells like vinegar when we wash it out of the flasks). There is a very good article going over the chemistry and technique of $\ce{ZnO}$ sol-gel: article


After reading the inputs from @Curt F., @PH13, @Martin and @Mithoron, I infer that N won't be involved in nitrogen inversion due to hindrance caused by connected rings and attached sterically demanding groups. Hence, 'N' pointed to by the red arrow will be considered as a stereocenter in ampicillin.


What's the Gibbs free energy of a reaction at zero temperature and pressure?

$G^\circ = U(0K) = E^{ele}+ZPE$, i.e. the sum of the electronic energy (electrons and nuclei interacting) and the zero point vibrational energy.

You can't use $K=e^{-\Delta G^\circ/RT}$ at 0K because the derivation assumes that a certain amount of thermal energy is available. But you don't need the equation to figure out the relative amount of two molecules. At equilibrium at 0K all molecules are in the lowest possible state.


As Loong said, the up-to-date IUPAC recommendation stipulates that the chain with the longest continuous chain of carbon atoms should be the parent chain. For this reason, I would agree with your textbook's answer.


API formulations - are excipients required

The ICH guidelines give some details about the presence of excipients, and the forms they may take (along with acceptable types/classes of excipients);
“The excipients chosen, their concentration, and the characteristics that can influence the drug product performance (e.g., stability, bioavailability) or manufacturability should be discussed relative to the respective function of each excipient. This should include all substances used in the manufacture of the drug product, whether they appear in the finished product or not (e.g., processing aids). Compatibility of excipients with other excipients, where relevant (for example, combination of preservatives in a dual preservative system), should be established. The ability of excipients (e.g., antioxidants, penetration enhancers, disintegrants, release controlling agents) to provide their intended functionality, and to perform throughout the intended drug product shelf life, should also be demonstrated. The information on excipient performance can be used, as appropriate, to justify the choice and quality attributes of the excipient, and to support the justification of the drug product specification (3.2.P.5.6).”
The major use of excipients is often just as bulk material in tablets. Many modern drugs are active at doses of 10-100 mg — these quantities are often so small that fillers are needed to make a tablet of acceptable size to be acceptable to the patient (compare to some older drugs where doses were often in the gram range).


Chemical reactions as exponential functions

Intuitively, we can understand why only a first order reaction (at least for noncomplex reaction orders) displays exponential decay because a first order reaction implies that the rate of decay ($d[A]\over dt$) is proportional to the concentration ($[A]$). This relationship precisely describes an exponential function.


Will octahedral splitting occur in tribromidotricyanidocobaltate(III)?

In reality it would perhaps boil down to experimental reasoning, but if it's an exam asking you this, I'd like you to remember a rule of thumb as you'd call it:

If the complex has a coordination number of $6$, with $3$ strong field ligands and $3$ weak field ligands, then consider all the ligands to be strong field ligands.
If the complex has a coordination number of $4$, with $2$ strong field ligands and $2$ weak field ligands, then consider all the ligands to be weak field ligands.


First, how much compound do you want to make?
Second, how much cis-3-hexanol is there in grass? It is probably around 0.01%, so be ready to process a lot of grass.
Third, it might be reactive (that is it can decompose).

Your best bet is to:
(1) Figure out theoretical yield
(2) Do steam distillation rather than extraction (bp is 156.5 °C - good for distillation).
(3) Extract collected fraction (mostly water+ some organics including hexenol) with some light organic solvent (hexane, ether). You can try ethyl acetate (aka MEK substitute, bp = 77C), but that is the highest bp you want to go for.
(4) Split organic solution in several fractions and try washing it with water solutions of 5% baking soda and then brine (saturated solution of NaCl). Hexenol should stay in organic solvent unless it decomposes.
(5) Concentrate it under reduced pressure. If you don't have access to rotavap - let it sit in open air. You can direct fan at it if you will, but you might catch fire if you are not accurate.

You will get a drop of something that is a mix of 20 compounds. How you need to do column chromatography, which is pain even with equipment.

It doesn't sound like you have access to chemistry lab. In this case for practice purposes try extracting eugenol from cloves first. Grind cloves, place in distillation apparatus, add water, distill. Oil on top is eugenol.

Steam distillation: Shred grass, add to a gallon of water and distill it off using a lab distillary apparatus.


This is perhaps my biased view as a theorist, but I've collaborated quite a bit with a condensed-phase 2D-IR group and a little with a UV resonance Raman group, so I have some understanding of the experimental perspective as well.

For anything outside of the simplest models, like a 2- or 3-state system you can scribble on an envelope, or the simplest molecules that would display configurational dynamics, like short peptides or single amino acids, experiment strongly benefits from computation. Have fun searching the literature for tables of IR peak frequencies, looking at the assigned normal modes, (which are sometimes compound motions), and then seeing if any of those match your hopefully very similar structure. If the molecular structures are known, then for simple cases a few of the conformers can be made by hand using your favorite computer program or a partially-automated conformer search, have their geometries optimized, and harmonic vibrational frequencies calculated to at least qualitative accuracy in an afternoon. Then it's a matter of clicking around in a visualization tool to see which interesting peaks correspond to which normal modes.

If populated conformers are harder to find, such as in systems that are larger with many rotatable bonds or have substantial librational motion, an exhaustive conformer search may be too expensive in terms of computer time, and running molecular dynamics trajectories with some sort of post-trajectory clustering analysis is an alternative, since IR and Raman spectra can be generated from trajectories via velocity, dipole, and polarizability autocorrelation functions.

Example

Here is an example where a lithium cation is interacting with L-proline in two ways: 1. with the backbone-bound carboxylic acid, and 2. with the ring. Perhaps this is cheating, because it isn't a single molecule, it's two different coordination modes, but you would see something similar with an intramolecular conformational change or rearrangement such as keto-enol tautomerization.

Here are their respective vibrational spectra within the harmonic approximation.

Already without any assistance, we can see huge changes outside of the fingerprint region. This is similar to how hydrogen bonding in water changes the IR spectra when moving from molecules to the bulk. The three most notable changes going from backbone- to ring- bound are

a blueshift (increase in energy) of the carbonyl peak, moving it from 1760 to 1852 wavenumbers,
the intensity increase of the ring $\ce{C-H}$ stretches below 3000 wavenumbers, and
a few changes above 3500 wavenumbers.

1 is probably due to charge transfer into an antibonding orbital localized on the carbonyl group. However, it's also strongly coupled to the $\ce{-OH}$, so interactions there will have an indirect effect. I'm less sure about the $\ce{C-H}$ stretches, but they belong to the three hydrogens closest to the $\ce{Li+}$ (7, 11, and 14), and Mulliken charges don't support a CT idea, which isn't surprising and doesn't rule it out. For number 3, the $\ce{-OH}$ stretch moves from 3686 to 3439. This redshift is probably a result of the blueshift to the coupled mode; assuming that the total vibrational energy between conformers is similar ($E_{\text{ZPE}} = 90.100$ vs. $\pu{90.182 kcal/mol}$ here), vibrational energy tends to be redistributed evenly and in ways that make physical sense; see here for a discussion about how this simplified two-mode coupling works which might also be applicable here.

Frequencies are in wavenumbers (cm$^{-1}$) and intensities are in units of km/mol. Modes below 1200 wavenumbers are commented out.

$$
\begin{array}{rrrr}%
\hline%
\text{backbone frequencies} & \text{backbone intensities} & \text{ring frequencies} & \text{ring intensities} \\
\hline%
%% 30.91 & 2.37 & 62.75 & 4.92 \\
%% 88.36 & 6.77 & 91.13 & 2.54 \\
%% 168.98 & 24.56 & 146.67 & 15.42 \\
%% 200.22 & 35.17 & 239.15 & 53.54 \\
%% 301.33 & 5.59 & 299.49 & 10.23 \\
%% 323.29 & 67.62 & 399.53 & 47.70 \\
%% 393.10 & 18.41 & 436.61 & 77.33 \\
%% 423.78 & 49.21 & 454.79 & 10.29 \\
%% 485.16 & 69.90 & 517.44 & 4.46 \\
%% 534.48 & 56.37 & 541.24 & 57.13 \\
%% 565.71 & 45.28 & 582.68 & 14.18 \\
%% 625.82 & 27.87 & 622.71 & 9.03 \\
%% 663.98 & 12.16 & 703.37 & 26.78 \\
%% 754.48 & 22.99 & 758.60 & 12.72 \\
%% 802.57 & 16.27 & 808.16 & 50.74 \\
%% 842.01 & 5.53 & 823.92 & 48.36 \\
%% 858.14 & 14.85 & 863.84 & 7.69 \\
%% 875.87 & 27.62 & 875.21 & 13.13 \\
%% 897.13 & 13.53 & 912.31 & 2.55 \\
%% 919.58 & 0.12 & 928.34 & 12.11 \\
%% 966.10 & 12.84 & 952.79 & 7.90 \\
%% 1007.42 & 0.12 & 1017.17 & 0.53 \\
%% 1034.74 & 23.77 & 1056.15 & 1.11 \\
%% 1070.65 & 26.45 & 1116.38 & 40.56 \\
%% 1154.24 & 13.04 & 1119.11 & 9.28 \\
%% 1161.61 & 3.40 & 1148.10 & 7.38 \\
%% 1181.00 & 74.31 & 1173.28 & 7.56 \\
1213.61 & 4.92 & 1203.94 & 6.81 \\
1222.70 & 2.73 & 1228.83 & 4.07 \\
1263.58 & 152.20 & 1261.21 & 39.44 \\
1279.48 & 0.73 & 1272.66 & 12.99 \\
1314.79 & 1.01 & 1321.29 & 57.22 \\
1328.39 & 58.17 & 1342.94 & 314.05 \\
1366.92 & 47.01 & 1365.77 & 74.02 \\
1390.24 & 127.09 & 1402.33 & 19.65 \\
1424.08 & 13.57 & 1522.07 & 16.68 \\
1432.59 & 5.88 & 1561.70 & 0.40 \\
1460.15 & 3.30 & 1572.71 & 24.88 \\
1760.04 & 284.35 & 1851.86 & 270.80 \\
2976.69 & 25.42 & 2607.42 & 324.10 \\
2983.82 & 7.16 & 2806.11 & 147.60 \\
2993.71 & 4.29 & 2900.28 & 54.24 \\
3000.08 & 8.94 & 3024.05 & 5.13 \\
3039.46 & 13.36 & 3027.04 & 2.76 \\
3063.00 & 5.22 & 3061.92 & 2.04 \\
3075.69 & 4.87 & 3074.89 & 3.80 \\
3477.77 & 29.07 & 3438.98 & 235.23 \\
3685.91 & 93.01 & 3527.08 & 36.70 \\%
\hline%
\end{array}%
$$

All calculations were performed using B97-D/def2-SVP. See source for XYZ structures.


Algorithm used by PubChem to determine the 2D layout assignment for Compound entries?

According to Xemistry, PubChem uses Cactvs for 2D layout coordinate computation.
A friend pointed this out to me. If you scroll to the bottom of the CACTVS Toolkit FAQ (http://www.xemistry.com/docs.htm), it says the following:

Interesting. I did not know that PubChem is built with Cactvs. I heard it was implemented with OpenEye's OEChem library.
Well, that is not wrong, but only part of the story. Both Cactvs and OpenEye software are used, in combination with the in-house NCBI toolkit. Software built with Cactvs is in charge of structure submission pre-parsing, structure input, structure ASN.1 encoding, 2D layout coordinate computation, structure rendering, most property computations, structure unification, the complete structure search system and Web-based structure sketching for queries. The OpenEye toolkit is used for structure standardization, some property computation (names, canonic SMILES), and SD file output. There are applications at NCBI which link to both OpenEye and Xemistry libraries - they can peacefully co-exist.


Why is the entropy of solidifying solution higher than that of a pure solvent?

1) How do we know the particles in a solution are more separated?

When a solution forms, intermolecular bonds in the solid solute are broken and replaced by interactions with the solvent. The solute is then free to diffuse in the solvent. If the solute molecules still interacted in a highly favorable way they would tend to bond and precipitate out. This is exactly what happens when a solute has a low solubility.

  2) How do the solvent molecules actually separate the solute particles?

What drives the dissolution process is an overall favorable combination of the change in the energy of intermolecular interactions and motional freedom. This is described by the Gibbs free energy. The solvent provides an environment for the solute to move (diffuse), which increases the solute's entropy, and the solvent forms weak interactions with the solute that make up (perhaps only partially) for the lost interactions when in the solid. In general, there is an enthalpic component (bond-breaking and making) and an entropic component (changes in solute configurational freedom) to the free energy. If the energy of the interactions of a solute with itself in the solid is identical to the interaction energy between solute and solvent in the solution, then the change in free energy is described entirely by the entropic contribution.

  3) Also, we know that the entropy of solid is less than that of a liquid, so shouldn't the addition of solute particle (which are obviously solid) actually lead to lesser entropy than pure liquid?

The solid particle is no longer "solid" when dissolved, it is in a new state. The entropy in solution reflects the new state of the solute upon dissolution. The solute does not retain a memory of what it was before. At low concentrations, the solute is diffusing with few other solute molecules nearby. The new entropy is not an average over the entropy of the starting states (solid and solvent).

(That is unless you are referring to a dispersion of solid particles, but that is called a colloidal solution and is another matter.)


Why is there syn-addition in this bromination of an alkene?

In this example, it is impossible to tell from the product whether the mechanism was syn- or anti-addition. Let's look at the two possible products. After anti- or syn-addition of bromine, two compounds are produced that look different. However, facile bond rotation around the C1-C2 bond shows that they are in fact the same compound.

How would one go about proving that anti-addition is actually occurring in 4-methyl-1-pentene. One solution would be to use an isotopically labeled variant as shown below. Starting from the deuterated version of 4-methyl-1-pentene, we can draw the anti- and syn-products again. Similar rotation around the C1-C2 bond makes it clear that the products are not the same but instead are diastereomers. It would be possible to distinguish these products by some analytical technique (1H NMR) and show that the anti-product is formed selectively.


The OH group of phenol is indeed more polarized - and a stronger acid - than the OH group of an aliphatic alcohol - maybe that is what you are asking. There are two contributions to this:

(1) inductive: the sp2 hybridized carbon participating in the sigma bond to the OH group is more electron-withdrawing than the sp3-hybridized carbon of an alcohol bonded to OH; and

(2) mesomeric: the phenyl group allows non-bonding electrons on oxygen to partially delocalize into the ring via pi bonding.


Which procedure(s) will allow a student to differentiate between solid sodium sulfate and solid sodium sulfite?

Sodium sulfite and hydrochloric acid partially react to $\ce{SO2}$ which is gaseous and thus will form bubbles if not dissolved immediately.

When sodium sulfate is added to hydrochloric acid it will dissolve. The solution then contains sulphate and hydrogensulphate ions. No gaseous product will be formed.


How to assign oxidation number of sulfur in thiocyanate anion?

In the end the oxidation state of sulfur does not really matter for balancing equations.  Rather, identifying the real oxidizing and reducing agents is.

Clearly the iodine in the iodate ion is the oxidizing agent, dropping from $+5$ to $+1$.  But copper, sulfur, and even carbon are all being oxidized or reduced from the cuprous thiocyanate.   We have multiple parts of the $\ce{CuSCN}$ participating in the electron exchange.

Why not then identify the whole $\ce{CuSCN}$ molecule as the reducing agent, thus covering all the reducing agent atoms in their proper proportions?  With this approach you see that the sum of oxidation states for this reductant goes from $0$ to $+7$ ($+2$ for $\ce{Cu}$ in the copper sulfate, $+6$ for the sulfur in the same compound, and $-1$ for $\ce{CN}$ in the hydrogen cyanide).  Then $\ce{CuSCN}$ gives $7$ electrons, iodate takes $4$ electrons, and thus you get $\ce{4 CuSCN + 7 KIO3}$, just like with the matrix method.


Is there any specific formula to find periodic table periods and valence electrons

The number of electrons in each orbital is double an odd number (the Pauli exclusion princilel states electrons can be in the same energy only if of opposite spin, so there can be two otherwise identical electron energy states [which can be separated in an EM field]).

s (sharp spectral lines): 2 (twice 1)
p (principal spectral lines): 6 (twice 3)
d (diffuse spectral line): 10 (twice 5)
f (fine, or "fundamental", spectral lines): 14 (twice 7)

Using that information, and the fact that the first row has only s electrons, second row s & p, usw., you should be able to calculate the first few rows easily. However, rules become much more complex for transition metals and heavier elements, where an outer shell may be added before all inner orbitals are filled.

Primary valence for the first few rows is simply the number of electrons needed to either complete the current shell (e.g. oxygen is two e- short), or revert to the previous complete, noble gas, configuration (e.g. Na minus one e- is neon). Again, there are complications, such as transition metals with multiple valences, fluorine reacting with some noble gases, usw.


What is the use of "pyro" in the naming of organic compound?

So, as @long and @permeakra stated, the name pyrogallol came from the fact that it is obtained by heating gallic acid. The IUPAC goldbook clearly states that:-

  "pyro" is used as a prefix designating compounds formed by heating a compound,
  usually with the elimination of water, carbon dioxide, or other simple
  molecule, e.g. pyroglutamic acid from glutamic acid.

So as per rule, it is pyrogallol and not gallol.


The most important factor is the nucleophilicity versus basicity of the attacking species.

Taking the trimethylcarbenium cation as a classic example:

Sterically hindered bases such as lithium diisopropyl amide (LDA) will favour elimination because it is physically much easier for them to remove a small proton from the outside of the cation than to reach the carbon atom at the centre surrounded by three methyl groups and form a new bond there. On the other hand, small species such as cyanide and water will favour substitution because because it is physically easier for them to reach the central carbon.

Additionally the strength of the base influences which reaction takes place. LDA is a strong base and the methyl hydrogens are quite acidic means that the elimination reaction is much more likely. However, weak bases such as iodide and water will favour substitution because they are less basic and so they are less likely to abstract an acidic proton from one of the methyl groups.

It should also be noted that the nature of the solvent has a strong effect both on the stability of the cation but also on the nucleophilicity of the attacking species. This page has some discussion on the effects of solvents on nucleophilicity and the website also has good discussions of elimination and substitution mechanisms.

http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/alhalrx1.htm#hal1


Is the reaction between thiosulfate and hydronium an acid-base reaction?

Two of the three acid-base theories (Brønsted and Lewis) also define an acid-base reaction while the third theory (Arrhenius) does not — it merely defines what acids and bases are.

In the Lewis theory, an acid-base reaction occurs when a lone pair donor interacts with a lone pair acceptor to form a new bond without a simultaneous redox reaction happening. The less complicated Brønsted-Lowry theory describes an acid-base reaction as a reaction in which a proton $\ce{H+}$ is transferred from an acid to a base, giving us a conjugate base and a conjugate acid, respectively.

As has been pointed out, your reaction is not a one-step reaction. Rather, we have to break it down into its constituent steps. The first step, a reaction of hydronium $\ce{H3O+}$ with thiosulfate $\ce{S2O3^2-}$ has to be an acid-base reaction because that is practically the only reaction hydronium undergoes.

$$\ce{2H3O+ + S2O3^2- -> H2S2O3 + 2 H2O}\tag{1}$$

The acid is the species that donates the transferred proton, the base is the species that accepts it. Therefore, we can assign the acid and base roles as shown in equation $(1')$.

$$\ce{2 \underset{acid}{H3O+} + \underset{base}{S2O3^2-} -> 2 \underset{conjugate base}{H2O} + \underset{conjugate acid}{H2S2O3}}\tag{$1'$}$$

The second reaction is a redox reaction. Thiosulfate (and therefore also its conjugate acid) has two sulfur atoms in the oxidation states of $\mathrm{+V}$ and $\mathrm{-I}$, respectively. These undergo a redox reaction leading to compounds of the oxidation state $\mathrm{+IV}$ (sulfurous acid $\ce{H2SO3}$) and $\pm0$ (elemental sulfur). We can use the full redox reaction determination algorithm, but I’m skipping that since it is a simple one-electron transfer.

$$\ce{H2S2O3 -> H2SO3 + 1/8 S8}\tag{2}$$

And finally, we have another decomposition namely that of sulfurous acid to sulfur dioxide $\ce{SO2}$ and water. This can be seen as the back-reaction of a Lewis acid-base reaction wherein water is the liberated Lewis base and sulfur dioxide the liberated Lewis acid. Indeed, the forward Lewis acid-base reaction (the reverse direction of this reaction) occurs, too. This reaction is in equilibrium, driven to the product side because of the volatility of $\ce{SO2}$.

$$\ce{H2SO3 (aq) <=>> H2O (l)  + SO2 (g) ^}\tag{3}$$

Therefore, the overall reaction is the sum of one Brønsted acid-base reaction, one redox reaction, and one Lewis acid-base reverse reaction.

$$\ce{2 H3O+ (aq) + S2O3^2- (aq) -> 3 H2O (l) + 1/8 S8 (s) + SO2 (g) ^}\tag{4}$$


Does Wurtz-Fittig-reaction involves Sn1 or Sn2?

Aryl carbons cannot go through SN1 nor SN2 substitution reactions. SNAr involves the creation of benzyne intermediate. The mechanism is this:

The benzyne has sp hybridized carbons, which should be linear, but forced to be bent. Therefore it can be attacked by a nucleophile, and become sp2 which isn't linear.

Notice that the carbanion can also attack the adjacent carbon. It's a special phenomena for this type of reaction. The mechanism was confirmed by carbon-14 labeling (Organic Chemistry/Paula Bruice, sixth edition, page 692).


Can you concentrate alcohol by evaporating the aqueous solution at room temp and collecting on a cold surface?

Yes you can do that, this technique is called 'distillation' (it requires a minimum difference of 10 degrees celsius to seperate two liquids with different boiling points, for efficiency), I don't know what temeperature the fish tank heater runs at, but you should know that ethanol i,e. what you are trying to separate evaporates at 37 degrees celsius same as normal body temperature, and the temperature of the cooling plate should be < 27 degrees for efficiency

Edit -
yes, exactly the boiling point of ethanol is 78.3 degrees but ethanol is highly volatile and evaporates very fast at 37 degrees (without taking much water vapour with it) as far as my observation is concerned(with room temperature around, 1 atm pressure). the values I mentioned are the optimal conditions with experimental background. your goal as I can percieve is to avoid as much water as possible, if you take the condenser temperature to below 10 degrees the water vapour present in air will also start to condense at a good rate


What is the rotational symmetry number used to calculate the rotational entropy from an ORCA frequency calculation?

Your #2 is exactly right, as far as I know, which answers your #1.

A key thing to remember is that the rotational symmetry number doesn't include reflections, so molecules in groups like $C_s$ still have $s_n = 1$ since there are no superimposable orientations achievable only with rotations.

For reference: Table II of the NIST CCCBDB introduction to thermochemistry is a nice, concise summary of the symmetry numbers for the common point groups observed in chemistry.


contradiction between Gibbs energy variation and redox potential

Looking up the redox potentials in tables vs. Hydrogen as standard at zero volts the redox for Oxygen is

$$\ce{O2 +  4H^+ + 4e <=> 2H2O} \qquad \qquad E^\circ = +1.229~\mathrm{V}$$

so the tables you use may be vs. calomel electrode as standard.

The redox potential for the nitrate reaction has to be constructed from others as

$$\begin{align}
\ce{NO3^- + 3H^+ + 4e}\, &\ce{<=> HNO2 + H2O} & E_1^\circ &= +0.934~\mathrm{V} \\
\ce{2HNO2 + 4H^+ + 4e}\, &\ce{<=> N2O + 3H2O} & E_2^\circ &= +1.297~\mathrm{V} \\
\ce{N2O + 2H^+ + 2e}\, &\ce{<=> N2 + H2O} & E_3^\circ &= +1.766~\mathrm{V}
\end{align}$$

The overall half-reaction, after multiplying the nitrate reaction by 2 (to balance) and adding is

$$\ce{2NO_3^- + 12H^+ + 10 e <=> N2 + 6H2O} \tag{1}$$

The redox potential for the reaction $(1)$ is calculated from $n_\mathrm{r}E_\mathrm{r}^\circ = n_1E_1^\circ + n_2E_2^\circ +....$ where the n's are the number of electrons in each half reaction and $n_\mathrm{r}$ for the new reaction.

The $\ce{NO_3^- to N2}$ redox potential is therefore $1.246~\mathrm{V}$, calculated as

$$E_\mathrm{r}^\circ = \frac{(4 \times 0.934) + (4 \times 1.297) + (2 \times 1.766)}{10}$$

The two values, $1.229~\mathrm{V}$ and $1.246~\mathrm{V}$, would now seem to agree with your $\Delta G$ values. They may be the same within error as may your $\Delta G$ values.

(Redox Values are from the CRC Handbook of Chemistry and Physics, 75 ed.)


How a titration curve is affected when a poorly soluble salt is formed?

If by "I think that titration curve don't be affected" you mean that the graph of pH vs volume of strong base will not be affected by the precipitations, then you are right. This is because the species that form the precipitate are not part of any other equilibrium reaction.

  how to write a mass balance expression to take account into the precipitation of BA?

In general terms, the mass balance (really the amount balance) can be expressed in concentrations if the system is homogeneous (all in solution). For example, it could look like this:

$$[E] + [F] + [G] = \mathrm{constant}$$

If there is a solid species $H$ that you have to consider, use an expression based on amount of substance:

$$n_E + n_F + n_G + n_H= \mathrm{another\ constant}$$

If you want to, you can express parts of this as concentration again:

$$([E] + [F] + [G]) \cdot V_\mathrm{solution} + n_H= \mathrm{another\ constant}$$


Why does mesitylene absorb at a longer wavelength than benzene (UV-Vis)?

The general observation is that alkyl substitution intensifies and shifts the benzene spectrum to longer wavelengths. The effect is similar to that of groups with lone pairs such as $\ce{NH2}$ and O-alkyl. So I don't think that opposite effects are observed as you suggest. Some data are given below.

The 0-0 band transition, from  v=0 vibrational level to v' =0 in the first excited state of benzene is at 262 nm (due to symmetry this is a forbidden transition so its position has to be calculated. The maximum in the S0-S1 transition in benzene spectrum is at 245nm, and S0-S2 at 203 nm).  Toluene does not have a forbidden 0-0 transition, and has a 0-0 band at approx 266 nm (with maxima at 261 nm and S0-S2 at 207 nm), 0-zylene 0-0 is at approx 273 nm, so the substituents do stabilise the excited state very slightly. This has been attributed to 'hyperconjugation', i.e. some charge being donated by the methyl group to the ring, or the alkyl groups behaves as it it had a lone pair. (The idea of hyperconjugation is still hotly contested). Toluene shows a slight increase in intensity as the symmetry is broken $\ce{D_{6h} to C_{2v}}$ but otherwise the effect on the electronic spectrum is slight. Aniline has a relatively broad and intense  absorption maximum (not 0-0) at 280 nm, nitrobenzene max very broad spectrum with a maximum at approx 268nm, but with a long shoulder in absorption stretching out to almost 400 nm, which is thought to cover the n-$\pi^*$ transition.

With very broad and structureless spectra (almost all the molecules you mention except benzene and toluene) the maximum intensity does usually always correspond to the 0-0 energy of the transition, due to Franck-Condon factors (overlap of wavefunctions which control intensity of transitions) and the fact that extra groups add considerable vibrational intensity to transitions and so broaden them. Thus comparing absorption peaks is not a reliable way of comparing state energies. (Experimentally the only ways are to measure vapour  phase absorption spectra or measure in a solid matrix  at 4k if possible.) Secondly spectra are stabilised differently in different solvents, even relatively non-polar ones and this can lead to spectral shifts.

The general sort of handwaving explanation, that is nevertheless useful,  is to suppose that the molecule has energy levels that can be described like those by a  particle on a ring. The bigger the ring the lower the energy levels become and so the lower the energy of the electronic transition. Thus incorporating a substituent that involves conjugating the electron over more atoms, increases the ring size very slightly and so  produces a longer wavelength transition.


In school I was taught that an atomic orbital is the 3-dimensional region in which the electron is located with a probability of 90%.

This is not correct.

An atomic orbital is a solution to Schrödinger's equation given the electrostatic potential. In the simplest case, we solve the one electron atom and perturb that with more electrons. An exact solution to the multielectron problem is intractable due to the chaotic nature of 3-body systems.

The concept you are referring to is a boundary surface. (See this question). Basically, it's a surface that has equal probability density for every point on the surface, and the surface encloses a region of space where the sum of probabilities is the desired value.

Because the surface is defined to have equal probability everywhere, and the wave function is smooth, it should be unique.

The boundary surface is usually how we represent the atomic orbitals that are familiar to us ($s, p, d, f$) but also molecular orbitals. Usually, we take something at 90% probability, for example, and plot the boundary, giving the orbital "shape." Otherwise, visualization can be tricky because the orbital only goes to zero probability at infinite distance from the nucleus, so there is a nonzero chance that an electron will be quite far from the nucleus.


Why does increasing orbital energy mean an electron can be more easily lost?

Why is the higher the orbital energy, the easier the electron lost?

Orbital energies are measured relative to an electron at infinity (i.e. one which has been ionized), which corresponds to an energy of 0 eV. An electron which is lower in energy than this (i.e. bound to an atom) has a negative energy.

The first six energy levels for hydrogen are shown above. As you can see, the energies converge to 0 as n, the principal quantum number, goes to infinity. In order to ionize an electron it must be given enough energy to promote it to the n = ∞ energy level. Therefore the higher the energy of the electron, the less energy is required to ionize it.

Also, if the distance between the electron and nucleus is getting father，the electronegativity is getting weaker, the electron is getting easier to lost?

The most likely distance of the electron from the nucleus (the maximum in the orbital's radial distribution function) increases with n and so the electron is in a higher energy orbital and is more easily lost.


Bonding, multiplicity and quantum chemistry for FeP2

Admittedly it is vacuum phase, but are you certain about your oxidation state assignments?  "nearly temperature independent susceptibility of -(8.8±0.7)centerdot10-6SI (at room temperature)"

doi:10.1016/j.jpcs.2013.04.015
https://sciborg.uwaterloo.ca/~lfnazar/publications/Electrochem_Solid-State_Lett_6_8_A162-A165_2003.pdf
doi:10.1021/ja301173q
http://www.staff.uni-bayreuth.de/~bt220235/refs/HPR_29_235.pdf
doi:10.1088/0031-8949/4/3/010 (given)


How to calculate Debye length of a 2-1 electrolyte

If we follow the derivation of the Debye length model, we would arrive at the following adapted model (doesn't assume uniform charge distribution):
$$\lambda_{D}^{2}=\frac{\kappa k_{B}T}{4\pi Z_{1}Z_{2}e^{2}c_{0}k_{c}}$$

In our equation:

$\lambda_{D}$ is the Debye length;
$Z_{1}$ represents the charge on the cation;
$Z_{2}$ represents the charge on the anion;
$\kappa$ is the dielectric constant;
$k_{B}$ is the Boltzmann constant;
$T$ is the absolute temperature; and
$c_{0}$ represents the number of ions per a cubic metre of solution.

Note that the general equation assumes that $Z_{1}=Z_{2}$, hence why this is usually written as $Z^{2}$.

Keep in mind this is only applicable for calculating the Debye length for a single dissociated species in solution! Further alterations would need to be made in order to consider a mixture.


It does exist. At least Google shows one supplier situated in New Jersey.

  ID 1130   Deuterium Peroxide (30% in D2O) D2O2    98  5   5.0g / 10.0g

Though for some reason I could not find any reputable supplier like Cambridge Isotopes or Sigma-Aldrich (strange!). As far as physical properties are concerned, as pointed out by @snurden there exists a significant amount of literature. Here is one article from the Canadian Journal of Chemistry that includes study of many properties (behind a pay wall though).

Abstract:

And to answer final part of your question:


Why do unripe papayas taste less bitter after being boiled?

Unripe papaya contains a high concentration of Papain enzyme, which is an enzyme that breaks down peptide bonds and therefore "digests" proteins, including potentially human tissues.  Enzymes themselves are proteins, and proteins tend to have a bitter taste.

Papain can self digest into amino acids over time, and this process could be accelerated by heating, as chemical digestion reactions will approximately double in rate for every 10K increase in temperature.  So one mechanism is that Papain may break itself down quickly when heated.  The other is that at a high enough temperature, Papain may denature.


Can an iodine clock reaction work without using a strong acid?

You wrote the answer yourself:  the hydrogen peroxide consumes hydrogen ion according to the reaction

$\ce{H2O2 + 2 I^- + 2\color{blue}{H^+} -> I2 + 2 H2O}$

You need the acid to provide the hydrogen ions, especially since if you allow the solution to become basic then the iodine disproportionates (see "Chemistry and Compounds" section, "Iodine Oxides and Oxoacids" subsection).

In terms of pure chemistry, maybe you could have used a moderately strong acid such as phosphoric acid or a bisulfate salt.  But then the limited dissociation of such species might make the iodine formation reaction slower and thus slow down the "clock".

Looking at the Wikipedia article on this reaction, one alternative is to use a peroxydisulfate salt instead of hydrogen peroxide as your oxidant.  This reacts with iodide ion according to

$\ce{S2O8^{2-} + 2I^- -> 2 SO4^{2-} +I2}$

with thiosulfate again as the reducing agent to regenerate iodide ion.  Unlike hydrogen peroxide or the other oxidants, the above reaction does not consume hydrogen ions and is your best bet to run the reaction without a strong acid addition.  However, peroxydisulfates have their own hazards (one of which is requiring a strong sulfuric acid solution to make it in the first place), and if you make the common choice of potassium peroxydisulfate rather than the ammonium salt you could have solubility issues.


Self tan mousse cannot contain UV, right?

That's right, it can't contain UV, because UV light is not a thing you can put in a bottle. The rest I'd rather dismiss as some marketing gibberish akin to this:

(sold back in the days when radium was all the hype, this thing is probably less awful than it seems, because it could not realistically have contained more than a few atoms of radium.)

Then again, chances are that the gibberish bears some resemblance to actual facts; so what? Fluorescence is just the ability to absorb and re-emit light. Whatever its composition may be, this thing would not shine UV on you all by itself.

As to the possible dangers, you either trust the regulatory body acting in your jurisdiction, or you don't. At least these guys supposedly did test the product, which I didn't.


Can a six member ring expand to achieve octet completion to stabilize a carbocation?

The reaction which we are talking about here is called Tiffeneau–Demjanov rearrangement.

According to Wikipedia,

  the Tiffeneau–Demjanov rearrangement (TDR) is the chemical reaction of a 1-aminomethyl-cycloalkanol with nitrous acid to form an enlarged cycloketone. The Tiffeneau–Demjanov ring expansion, Tiffeneau–Demjanov rearrangement, or TDR, provides an easy way to increase amino-substituted cycloalkanes and cycloalkanols in size by one carbon. Ring sizes from cyclopropane through cyclooctane are able to undergo Tiffeneau–Demjanov ring expansion with some degree of success. Yields decrease as initial ring size increases, and the ideal use of TDR is for synthesis of five, six, and seven membered rings. A principal synthetic application of Tiffeneau–Demjanov ring expansion is to bicyclic or polycyclic systems.

According to The Encyclopedia of Reagents for Organic Synthesis (DOI: 10.1002/047084289X), the yield of your product will be around 40-42%:

You can check out the Wikipedia page for more details.


Artificial foods — Could you make them? And would they compare to their "organic" counterpart?

No, it is not currently possible to create whole foods (like an apple) (as opposed to processed foods, like sugar or vinegar*) from their component chemicals.  The reason is simple:  All whole foods are composed of tissues, which are in turn made of cells, and we aren't close to being able to synthesize cells, especially eukaryotic cells, from their component chemicals. Their structures are far too complex. [https://en.wikipedia.org/wiki/Artificial_cell]
*In this sense I would also consider honey a processed food, though it is processed by bees instead of humans.
The closest we can come to creating artificial tissues is to clone existing cells; but that wasn't your question.
So then the question becomes whether we could fully characterize all the components of a whole food (again, let's use the apple as an example), and create a mixture containing all those components—and whether, in so doing, we would have something nutritionally equivalent.
The answer to that is not known, and I would say the burden of proof lies with those who wish to claim it would be nutritionally equivalent.  Even if we could characterize every possible micronutrient in an apple, we still wouldn't necessarily know, much less be able to recreate, the states in which they exist (in particular, their states of association with other compounds) within the apple.
Certainly there is a difference between a micronutrient that is complexed with other micronutrients, and then contained within a structure and then contained within another structure, and one that is mixed within a paste and thus complexed with compunds with which it might not be naturally associated in the intact apple. Of course, we do "paste" the apple ourselves when we chew it, but that process likely does not lead directly to the type of paste you would have if you tried simply mixing all the apple's constituent compounds together.
And it's not just how we directly would react to the difference.  There's also how the organisms of our microbiome would react.  It's certainly possible that consuming intact apples would favor one species of microorganism, while consuming synthetic apple paste would favor another, thus leading to changes in the balance of our microbiome.
Then we have to combine all the above uncertainty with the enormous added uncertainty that exists surrounding human physiology and nutrition.


You have the right name unless you need to specify the geometric isomer.

If I may stretch you a little, consider whether to specify the relative positions of the atoms at the double-bonds, e.g. (1Z,6Z)-cyclodeca-1,6-dien-3-yne. It's not possible to tell which geometric isomer it is from your drawing, because you would need a clear depiction of the bond angles around the double-bonds. If this drawing is all you have to go on, you can't specify the isomer and your answer is spot on.

The below image is a model of your molecule showing the (1Z,6E)-cyclodeca-1,6-dien-3-yne form. You can see the difference between E ("opposite" or in this case "trans"), on the left, and Z ("together" or in this case "cis"), on the right, by looking at the shape of the chain of the four carbons nearest each double-bond. Of course, there is more to geometric isomerism than this but I hope this is a useful introduction if you've not encountered it before.


Thermodynamics behind organic chemical reactions?

Whether or not the product is "more stable" would really depend on the sense in which you consider its reactivity. The $\mathrm{p}K_\mathrm{a}$ of a $\beta$-keto ester is about 11, so in that sense, it is much more reactive than either of the reactants. Once the $\beta$-keto ester is formed it is quickly deprotonated. It remains predominantly in this form because the anion is so stable, and the reaction is incapable of reversing. Thermodynamics does not always determine the result of an organic reaction. Many times, kinetics is the deciding factor.

The Claisen condensation is catalyzed by the base, as it is recovered in the end, and though I am sure it occurs spontaneously without it, I doubt it happens very often. I don't think ethyl acetate would be as popular a solvent as it is if it quickly and spontaneously converted to ethyl acetoacetate.


Acidic acid (vinegar) does not react with water, and it is not soluble in water. The gel and white bloom may come from a reaction the acidic acid with minerals such as calcium that may be present in the dehumidifier. the gel could also come from the breakdown of the chitin cell wall of the fungi. If the white bloom appeared within a minute or two of the addition of acidic acid it was most likely a reaction with minerals or other compounds.


Hyperconjugation, stability of molecules, and heat of formation

Hyperconjugation is an electronic effect which is similar to resonance (one of the most fundamental concepts of organic chemistry). Here's how it works:
$$\ce{H3C-CH2+ <-> H+H2C=CH2}$$

You can write the above structure three times for each hydrogen replaced. Note that only α-hydrogens (on the adjacent sp3 carbon atom) can be replaced.

Now, answering your first question, hyperconjugation leads to stability because of the delocalisation of charge. The positive charge gets diffused away from the carbon atom, and starts to occupy a larger volume, decreasing the overall energy of the system caused by electronic repulsions. As a matter of fact, the greater the number of structures you can draw due to hyperconjugation, the greater the stabilisation. That explains why tertiary carbocations are much more stable than primary ones.
$$\ce{(H3C)3C^+ <-> H+H2C=C(CH3)2}$$

If you look closely there are nine replaceable hydrogens, so you can draw nine structures (explaining its stability).

The Wikipedia page says:

  the heat of formation of molecules with hyperconjugation are greater than the sum of their bond energies

Here, heat of formation refers to the heat released on the formation of these molecules. So it is negative.

In the absence of hyperconjugation, you would expect that the heat released on formation is equal to the sum of all bond energies. But because of stabilisation due to hyperconjugation, the product molecule has lower energy than expected, so overall, more heat is released (higher magnitude enthalpy of formation).
It's important to understand that enthalpy of formation (including its sign) decreases on increasing the stability of the molecule no matter what. But in the case that the enthalpy of formation is positive (energy absorbed), a decrease in the magnitude of heat of formation is observed, while if the enthalpy of formation is negative (energy released), an increase in magnitude is observed. In your other post, you include a list of carbocations, and their heats of formation are seen to be positive, while those of the non-ionic species are negative. So all the ionic species need to absorb heat in order to form (indicating a loss in stability), but that loss in stability (and hence the heat of formation) would be less if they were stabilised due to hyperconjugation.


At what voltage does the electrodeposition of the metal start?

We can't write a relation for V, V1 and V2, because we can apply virtually any V, and still have V1 and V2 constant :) Though, if V signifies the minimum applied potential for starting reactions, then it's ok.
Instead, V may be compared to total potential of cell, which is V2-V1 or V1-V2 (see [3] below about the sign), but not V1+V2, as you assumed.
From wikipedia:
E°cell = E°cathode − E°anode.
The exact sign of (V1-V2) is hard to remember and a subject to mess, so I would better remember the processes involved:
positive voltage means lack if electrons, i.e. oxidation. But (citation)
since the electrode potentials are conventionally defined as reduction potentials (literally: ability to be reduced), the sign of the potential for the metal electrode being oxidized must be reversed when calculating the overall cell potential.
In order to have a spontaneous reaction (ΔG° < 0), E°cell must be positive (see link above). The gold would deposit on the cathode spontaneously if you short external circuit and provide conditions for anode reaction with lower E°. This may be oxidation of OH- (alkaline solution) to O2 on inert electrode. Why do you wonder? Au3+ is powerful oxidizer.
Other note: for high-ionic mix (1M + 1M + 1M + 1M + anions, probably such a mixture is not possible at all) the real redox kinetic will easily be different from simple potential calculation because of polarization effects and other factors. For example, 1-molar Au3+ is probably stable only at low pH, which shifts redox potentials.


Rearrangement with sodium iodide (ring opening)

I think the final product will also contain some Bromine atoms attached to it. and the no. of double bonds will be one less than that of the compound given in the picture. I may be wrong but with all of my known possibility of Organic reactions, I can actually reach to the final product somewhat different from that given in the picture  and I don't think the final product given in the picture can ever be got from the given starting material. I will be happy if someone else can find flaws in the logic  given below.

$\ce{NaI}$ in acetone is the reagent used for Finkelstein reaction which is nothing but a standard substitution reaction with $\ce{I-}$ ions. But in presence of vicinal dibromides, or any special structure where there is a chance of elimination of hallide ions even further, the iodide ions can actually make the molecule undergo preferably an $E_2$ elimination (in most of the cases).

In this starting compound also, there are substitutable $\ce{Br}$ atoms present. So, the iodide ions first performs a substitution reaction, and then undergo an elimination via bond migration. This same procedure occurs for two times. and finally we are left with a product which has $\ce{Br}$ atoms attached to $sp^2$ carbon atoms, and therefore they are highly reluctant to undergo any substitution or elimination reaction. So, that's why, logically the end product should be, $\text{ (1E,3Z,5E,7E,9Z,11E)-2,5,8,11-tetrabromo-6a,12a-dihydrooctalene}$. Here is the structure of the possible end product.

$\textbf{Mechanism:-}$

As far as I see, there can not be any more further reaction to the final compound. According to me, the given compound in the picture should not be the final product if we take the starting material to be the correct one. Instead, the end product in the mechanism should be the one produced with higher yield at the end.


How do I synthesize 3-hydroxydihydrofuran-2-one from formaldehyde?

Starting from pyruvic acid which is readily available and not very expensive: react pyruvic acid (only one enolizable side) with a Mannich base, then react the resulting amide with $\ce{MeI}$ and base to obtain the exo-methylene product. Brominate the exo-methylene. Now cyclization will prefer the five-membered ring over the strained four-membered ring. Reduce the ketone with a reletively mild reducing agent like $\ce{NaBH4}$ and then eliminate $\ce{HBr}$ in a E1cB type reaction, yielding 3-hydroxydihydrofuran-2-one.

This synthesis involves a Mannich reaction as you asked, but it can be done much simpler without it I think.

Note: I'm not an expert, just an enthusiastic amateur. Before trying anything in the lab  make sure your synthesis is checked by a supervisor.


As I mentioned above, during an isotope labelled mechanistic study, the formation of epoxide intermediate is suggested to explain the scramble isotopic carbon (between carbonyl carbon and neighboring carbon). The detailed arrow pushing mechanism is given bellow:

These steps are subjected to change according to many factors including the reaction conditions and nature of the diol used as correctly pointed out by Zhe and aforementioned MSU website in your comments, which states:

Many factors must be considered when analyzing the course of a pinacol rearrangement. These include:
• Which hydroxyl group is lost as water? or Which intermediate carbocation is more stable?
• What is the inherent shifting tendency (migratory aptitude) of different substituent groups?
• What is the influence of steric hindrance and other strain factors on the rearrangement?
• Are epoxides formed as intermediates in the pinacol rearrangement?
• Does product stability govern the outcome of competing rearrangements?
• Do the reaction conditions (i.e. type of acid, concentration, solvent and temperature) influence the course of rearrangement?

A reference for the rearrangement through an epoxy intermediate:
C. A. Bunton,  T. Hadwick,  D. R. Llewellyn,  Y. Pocker, Tracer studies in alcohols. Part III. Intermediates in the pinacol–pinacone rearrangement, J. Chem. Soc., 1958, 403-408 [DOI: 10.1039/JR9580000403] (http://pubs.rsc.org/en/Content/ArticleLanding/1958/JR/JR9580000403#!divAbstract).


Power is the energy transferred per unit time ($\frac{J}{s}$) and is determined by the potential drop (V) and the current (I) at a given point in time as $P = IV$ so to get the current, solve for $I=\frac{P}{V}$. This gives you the current at a given point in time.

Your question has a few points that might be worth addressing. As Dr Moishe Pippik commented, "current transferred" is a little confusing, either instantaneous current ($I$), or total charge transferred ($Q$), which is the current integrated over time: $\sum_{t} I(t)\Delta t$ in your case. So if what you really want is the total charge transferred, you would have $Q = \sum_{t}\frac{P\Delta t}{V(t)} $. Strictly speaking you should exclude t=0 point to avoid double counting.

You also asked about both the anode and the cathode. The current for both will be the same number but with opposite sign.


How to calculate the amount of citric acid to neutralise the pH within wastewater?

Gotta start with my favorite formula:
$$Q = CV$$

$$C_\textrm{eff}V_\textrm{eff}+C_\textrm{acid}V_\textrm{acid} = C_\textrm{out}V_\textrm{out}$$

$$(10^{-10.5}~\mathrm{mol/L})(10^5~\mathrm{L})+(10^{-2.2}~\mathrm{mol/L})V_\textrm{acid} = (10^{-7}~\mathrm{mol/L})(10^5+V_\textrm{acid})$$

$$V_\textrm{acid} = 1.51~\textrm{L}$$


Why does my coffee creamer have titanium dioxide in it and is it safe?

$\ce{TiO2}$ is is the most widely used white pigment out there, and it is considered to be non toxic. You'll find it used as food coloring commonly in candies, but in this instance  it's in your coconut sweetener. Aside from food, it's used as a primary pigment in paint (the more $\ce{TiO2}$ generally the higher the quality of paint, because it results in higher opacity), solar cells, toothpaste, and as you've mentioned, sunscreen.

The only health concerns I found were from inhalation

  Titanium dioxide dust, when inhaled, has been classified by the International Agency for Research on Cancer (IARC) as an IARC Group 2B carcinogen, meaning it is possibly carcinogenic to humans. The findings of the IARC are based on the discovery that high concentrations of pigment-grade (powdered) and ultrafine titanium dioxide dust caused respiratory tract cancer in rats exposed by inhalation and intratracheal instillation. The series of biological events or steps that produce the rat lung cancers have also been seen in people working in dusty environments. Therefore, the observations of cancer in animals were considered, by IARC, as relevant to people doing jobs with exposures to titanium dioxide dust.


Hybridisation of similar energy orbitals

I would suggest that this entire factory of bogus hybridization stories should be eliminated from general and organic chemistry curricula. The word mixing is worse, as it is a jug where one can add different orbitals and run a blender and here is the mixture. This stuff is good for rote memorization only.
Only after a thorough QM course and mathematical foundation, these ideas should be introduced with an example along with historical development. Neither the textbook author, nor the teacher know what it is. The only real experimental observable is only electron densities via X-ray diffraction or other techniques.


Why are we not supposed to heat Erlenmyer flasks?

This really depends on what you mean by heating.

One of the main uses for Erlenmeyer flasks is for recrystallization, which of course involves heating the container and contents. Typically this is only moderate heating, such as over a steam bath or with a hair dryer. Some manufacturers specify that their Erlenmeyer flasks are ideal for heating applications.

Traditionally, boiling flasks have more uniform and thicker walls, which allow for much greater heating limits, and a more uniform heating distribution. They are designed to handle more intense localised heating, and manage thermal and physical shock distribution. Erlenmeyer flasks usually have thinner walls and greater wall thickness variation, and are less able to handle sudden thermal changes, such as heating with a high energy source like a Bunsen, or heat gun. Using them in this way may well lead to cracking, fracturing or worse, and is unsafe.

So, although Erlenmeyer flasks can be heated, they are not specifically designed for high temperature, constant heating applications. For this, you should use a boiling flask.


Does 2,2-dimethylpropan-1-ol give a positive Lucas test?

No. It does not.

In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions

In chemical kinetics, there are two ways to deal with multiple-step mechanisms

Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate-determining step can be compared to the neck of a funnel. (1)
Steady-state approximation method- This is used when no step is the clear slowest. The steady-state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction is constant. (2)

So, in the Lucas test, the formation of the carbocation is the clearly slowest first step. This governs the reaction rate and the rest of the steps don't really matter. So, although it might seem deceptive at times if the R.D.S. is slow then the whole net reaction in slow.

Hence, the rate of formation of the one-degree carbocation is so less, that the entire rection does not proceed. So what you can theoretically say will be the only difference in say a Neopentyl alcohol and a 1-alkanol would simply be that after the super slow first step, the rest of the reaction will be much faster in the case of Neopentyl alcohol. But, due to the slow first step your given example will give an absolutely accurate negative Lucas test


Single electron reduction of ketone by Na or Zn

Let the carbonyl compound under consideration be acetone.

Case I: Hydrogen Abstraction

The species formed are NaH and an allyl-type radical. The resonating system includes an electronegative oxygen atom, which also has an incomplete octet in one of the canonical forms.

In essence, products are not really stable.

Case II: Attack on Carbonyl

The species formed is a radical anion, which are relatively stable as the filled orbital of the oxygen relieves some electron deficiency of the radical. This effect is stronger than the one offered the pi-bond in Case I.


Determination of molecular formula of hydrocarbons by combustion analysis

The combustion can be described by the following reaction:

$$\ce{C_xH_y (g) + ($x + y/4$) O2 (g) ->[\Delta] x CO2(g) + $y/2$ H2O (g)}$$

All products are gaseous at first, but as the system cools down to the room temperature, water condenses attributing to the decrease in $\pu{30 mL}$. Further decrease in $\pu{40 mL}$ is due to reaction of alkali with the remaining gaseous product, $\ce{CO2}$:

$$\ce{CO2 (g) + 2 M^IOH (aq) -> M^I2CO3 (aq) + H2O}$$

From the ideal gas law it's obvious that amounts of gaseous substances are proportional to their volumes:

$$n = \frac{p}{RT}V \quad\to\quad n\propto V$$

Let's rewrite the first equation with assigned volumes:

$$\ce{\underset{\pu{20 mL}}{C_xH_y (g)} + ($x + y/4$) O2 (g) ->[\Delta] \underset{\pu{40 mL}}{x CO2(g)} + \underset{\pu{30 mL}}{$y/2$ H2O (g)}}$$

Since $n\propto V$, and for the balanced chemical equation amounts of reactants and products are equal, so are their volumes. Hence:

  Q: Do they mean that the combined volumes of the reactants and the resultants are the same? That's an incorrect inference, isn't it?

A: Yes; no, this is correct for the given reaction.

That "strange line"

  $$\pu{20 mL} + 20(x + y/4)~\pu{mL} - \pu{20 mL} = \pu{30mL}$$

is a direct consequence of that as it can also be rewritten as

$$\underset{\ce{C_xH_y}}{\pu{20 mL}} + \underset{\ce{O2}}{20(x + y/4)~\pu{mL}} = \underset{\ce{CO2}}{\pu{20 mL}} + \underset{\ce{H2O}}{\pu{30mL}}.$$

However, as this is a single equation with two unknowns, it only illustrates the preservation of the total amounts of products and reactants. In order to find the formula of unknown hydrocarbon, you just have to equate corresponding coefficients with the ratio between $\ce{C_xH_y}$, $\ce{CO2}$ and $\ce{H2O}$ ($20:40:30 = 1:2:1.5$):

$$x = 2$$
$$y/2 = 1.5 \to y = 3$$

which results in empirical formula $\ce{C2H3}$. In order to find out the molecular formula you would also need auxiliary data such as molecular weight of the hydrocarbon.


Why is (NH4)2[CuCl4] square planar complex but Cs2[CuCl4] is tetrahedral?

In solution, the $\ce{[CuCl4]}$2- is expected to exhibit terahedral, or nearly tetrahedral geometry.

The Jahn-Teller theorem states that degenerate orbitals cannot be unequally occupied. Molecules with unequally occupied degenerate orbitals will distort so as to render the orbitals non-degenerate, with more electrons occupying the lower energy states. Generally, this distortion leads to a reduction in symmetry. In the case of Cu (II), it is a d9 cation with a tetrahedral electronic configuration of (e)4(t2)5, so the three degenerate t2 orbitals are unequally occupied, therefore a distortion towards the square planar geometry of (eg)4(b2g)2(a1g)2(b1g)1, with no unequally occupied degenerate orbitals, is electronically favoured. Nonetheless, experimental evidence (1) indicates that this d-orbital stabilization energy is very small. Since copper is a first-row transition metal (not very big), steric effects dominate electronic effects allowing $\ce{[CuCl4]}$2- to assume a tetrahedral geometry. Most spectroscopic methods I've read involve looking at the molecule in crystal form (because that is where this effect gets interesting), but DFT studies of the $\ce{[CuCl4]}$2- hydrate in concentrated aqueous solution indicate that the anion's tetrahedral structure is more stable (2).

The issue of $\ce{(Cs)2[CuCl4]}$ vs $\ce{(NH4)2[CuCl4]}$ comes up when the complex precipitates and crystals are formed. Due to the borderline electronic effects, the $\ce{[CuCl4]}$2- anion exhibits a great degree of structural flexibility. With large cations ($\ce{Cs+}$ radius is 0.174 nm while $\ce{NH4+}$ radius is 0.143 nm) , it crystallizes as a distorted ("squashed") tetrahedron.

$\hskip1.4in$ (3)

Square-planar co-coordination occurs in such a way that you effectively get octahedra with tetragonal distortion (4). I would imagine that this phenomenon is simply a result of the smaller cation allowing the unit cell to crystallize like that. The general idea with unit cells is that the cation is never allowed to "rattle" around, so the structure will find a way to fill in the space. Given the preexisting electronic favourability of square planar geometry, $\ce{(NH4)2[CuCl4]}$ this distortion favoured by close-packing is made even more favourable.

$\hskip2.1in$ (4)

The canonical reference for $\ce{(NH4)2[CuCl4]}$ 's crystal structure is (5) if you want to take a look at the unit cell.

(1) Emeléus & Sharpe. Advances in Inorganic Chemistry and Radiochemistry, vol. 21, p. 128

(2) https://www.ncbi.nlm.nih.gov/pubmed/21462945

(3) L. Helmholz & R. F. Kruh, J. Am. Chem. Soc., 1952, 74, 1176

(4) Molecular Structure by Diffraction Methods, vol. 1, The Chemical Society, London, 1973, p. 642

(5) R. D. Willet, J. Chem. Phys., 1964, 1964, 41, 2243


How is the Felkin-Anh model of stereoinduction correctly explained with MO theory?

From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891:

  The $\pi^{*}$(C=O) and $\sigma^{*}$(C–X) orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But,
  if X is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the C–X and C=O bonds are perpendicular so that the orbitals align correctly.

  What does this mean for stereoselectivity? Conformations of the chiral carbonyl compound that place an electronegative atom perpendicular to the C=O bond will be more reactive-size doesn't matter.

Update

In response to the requests specified in the bounty description:

  1) What about the mixing of sigma(C-X) and pi*(C=O)? 2) What about pi(C=O) and sigma*(C-X)? Shouldn't that raise the sigma*'s energy?

As you probably know, the above treatment of the effect of one fragment orbital interacting with another one is basically just a pictorial way of applying perturbation theory to the problem. From this theory you get the result that the interaction strength of two fragment orbitals gets stronger the better the overlap and the closer the orbitals are in energy - a very(!) detailed derivation of the fundamental equations involved has been done by @Martin here (an answer he truly deserves much more upvotes for :)). Overlap will not be an issue, but the energy differences will be. So, when thinking about which orbital interactions are most important for the problem at hand we have to look at their supposed energy levels. Thus, I've done a rough sketch, taking into account that:

the $\ce{C}$ atomic orbitals are lower in energy than the $\ce{O}$ or $\ce{X}$ atomic orbitals but that the $\ce{O}$ and $\ce{X}$ orbitals will be rather close in energy (similar electronegativities etc.) and
$\pi$-interactions are weaker than $\sigma$-interactions due to less overlap thus resulting in a lower level-splitting between the bonding and antibonding orbitals.

On the right side of the picture I added arrows indicating the energy differences between the most important orbitals.

As you can see the energy differences between $\sigma(\ce{C-X})$ and $\pi^{*}(\ce{C=O})$ and between $\pi(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ are very large, much larger than the one between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$. Thus the two interactions you ask about will only have a very minor influence on the LUMO and the single most important interaction is the one between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$. Certainly, some mixing will occur (see below for some more on this) but when employing this method you are mostly interested in the big effects at work and can ignore the small ones.

  3) Shouldn't we better consider the entire thing as an allyl cation system?

The system at hand has much lower symmetry than the allyl cation because it consists of 3 different atoms/fragments. That makes drawing analogies between this and the simple pure carbon case difficult. It's certainly possible but the treatment shown in the book by Clayden is much easier. But even if you do a more rigorous perturbational treatment not only considering the first order interactions between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ but also the second order mixing of $\pi(\ce{C=O})$ (which in effect would be like doing something similar to the allyl cation) you get qualitatively the same picture as drawn above in the book by Clayden. Basically the only effect the mixing of $\pi(\ce{C=O})$ has is a slight down-shift in energy of the low-lying $\pi(\ce{C=O})$ orbital and a slight up-shift in energy of the new LUMO you get from Clayden's method, i.e. the bonding combination of $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$.

  4) If you can perform calculations on a simple system or know where to find one that would be greatly appreciated

You could try to do calculations for the following molecule

whereby you do one calculation with the $\ce{F}$ of the $\ce{CH2F}$ group perpendicular to the $\ce{C=O}$ plane and one with one of the $\ce{H}$ atoms of said group in that position. Then you can compare the LUMOs of both conformers and see if the position of $\ce{F}$ atom actually has an energy-lowering effect.


Although the acid and its salts are unknown, orthonitrite esters have been found, and from a natural source at that [1].  The orthonitrite function, which is trivalent rather than monovalent as nitrate would be, is incorporated into a macrocycle that hinders its decomposition.

The picture below is from Ref. 1.

Reference:

1.
Leanne Murray,* Tang K. Lim,* Graeme CurrieB and Robert J., "Aplidites (A-G): Macrocyclic Orthonitrites from
an Australian Tunicate, Aplidium sp.", Aust. J. Chem., 1995, 48, 1253-1266.Link


Is solubility in Qsp affected by coefficient?

You don't need to multiply s, since $Q_\text{sp}$ is the result of multiplication of ions concentration (and you don't need to multiply with the coefficient in any $Q$). So the right way to answer $Q_\text{sp}$ is

$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[\ce{Ag^{+}}]^{2}[\ce{CrO_{4}^{2-}}]$$
$$Q_{\text{sp}}\:(\ce{Ag_{2}CrO_{4}})=[0.01]^{2}[0.01] = 10^{-6}$$


How does the Mentos-Coke explosion work?

It's actually not a chemical reaction, but a physical reaction.  The surface of mentos acts as a nucleation site for the dissolved carbon dioxide in soda, allowing the gas to come out of solution quite rapidly.  This rapid release of gas is what causes the coke geyser.  Despite mentos appearing to have a smooth surface to the naked eye, there are actually a multitude of microscopic crevices, providing a large amount of nucleation sites.  As far as diet coke working better than regular coke, I am unsure (perhaps diet coke has more carbonation?  Or perhaps there are chemicals in diet coke which make it form foam much better than the other sodas).  I always thought that it was the mint vs fruity mentos that made the largest difference (mint seems to have a lot more nucleation sites which explains the larger geyser in comparison to fruity mentos).


Simple experiments involving enantiomer synthesis

If you have some way of measuring the enantiomeric excess of the product mixture, I suggest trying proline catalysis of an aldol reaction. The materials are cheap / readily available, and the reaction setup is straightforward. Proline is probably the cheapest chiral catalyst available.

Here is a procedure for an organic chemistry lab run at Rutgers that might get you started:
http://chem.rutgers.edu/sites/default/files/coursefiles/Courses_f09/310/Procedure_Proline_Catalysis.pdf


How long does it take to change states of matter?

Part of the answer is that the rate of freezing depends on the heat of fusion of the material being frozen (relatively large for water, the chief ingredient of uncooked chicken), but smaller for other substances).

Another piece of the answer is that an object can lose heat by convection, conduction, and/or radiation. You would increase the rate of heat loss by:

(1) increasing the surface area of the chicken as much as possible (flattening it),

(2) placing it against a surface held as cold as possible and made of a material such as metal with high heat capacity;

(3) hitting it with a continuous blast of extremely cold air or gas at the highest possible wind speed (remember wind chill factor?) and maximizing radiant heat loss by previously having painted the chicken with flat, black paint and conducting the experiment in a mirrored, insulated chamber. Some unusual materials can be induced to cool by irradiation with a laser (this is how we have obtained the lowest recorded temperatures ever) but this will probably not work with chicken.

I do not think there is an absolute limit on cooling rate, just diminishing returns for increasing energy expenditure. But I may be wrong. I hope you pursue this until you get an answer that satisfies you. That is how new discoveries are made.