Synthetic options for selective hydroxylation at terminal position

Or is this a lost cause?

I don't think so. This looks like a good candidate for an allylic oxidation with selenium dioxide, look here for some examples.


Which chemical reaction to use to remove carbon/peat from sieved earth/gravel?

It's easier to use flotation or froth flotation to separate out the less-dense peat (silt?) and carbon (I assume from the activated-carbon filter) than to use a chemical reaction..

Simple flotation would depend on making a dense solution (e.g. saturated Epsom salt, $\ce{MgSO4}$) and shaking the gravel mix in it... if the liquid's density is greater than the average density of silt with its entrained air, the gravel will sink and silt float, to be skimmed off. You can reuse the liquid for more separation.

Froth flotation relies on a surfactant wetting the silt and keeping it suspended while the gravel settles out. Try a bit of hand dish-washing
detergent.


How to convert Gaussian input file .gjf to SMILES?

It is not aiming for code golf, but the following offers the conversion to intermediate .xyz which may be processed further (cf. vide infra):

import fnmatch
import os
import sys

gjf_register = []

for file in os.listdir("."):
    # collect all .gjf:
    if fnmatch.fnmatch(file, "*.gjf"):
        gjf_register.append(file)
    gjf_register.sort()

for entry in gjf_register:
    # recover only the atom block:
    per_file_register = []
    with open(entry, mode="r") as source:
        for line in source:
            per_file_register.append(line)
    del per_file_register[:5]  # clip the first lines

    # the other mandatory entry in the .xyz file:
    number_of_atoms = len(per_file_register) - 1

    # write an intermediate .xyz file:
    intermediate_xyz = str(entry)[:-4] + str(".xyz")
    print("File {} written.".format(intermediate_xyz))

    with open(intermediate_xyz, mode="w") as newfile:
        newfile.write(str(number_of_atoms))
        newfile.write(str("\n"))

        newfile.write(str(intermediate_xyz))
        newfile.write(str("\n"))

        for entry in per_file_register[:-1]:
            newfile.write(entry)

sys.exit(0)

The conversion into .smi may then be delegated to obabel, e.g. by

obabel *.xyz -osmi -m

if there are multiple files to consider.

Proof:  The following .cif

data_EFAHAW
_cell_length_a 8.1819
_cell_length_b 10.4285
_cell_length_c 9.2965
_cell_angle_alpha 90
_cell_angle_beta 99.962
_cell_angle_gamma 90
_symmetry_space_group_name_H-M 'P 21/c'

loop_
_symmetry_equiv_pos_site_id
_symmetry_equiv_pos_as_xyz
1 x,y,z
2 -x,1/2+y,1/2-z
3 -x,-y,-z
4 x,1/2-y,1/2+z

loop_
_atom_site_label
_atom_site_type_symbol
_atom_site_fract_x
_atom_site_fract_y
_atom_site_fract_z
O3 O 0.19290 0.35524 0.46580
C9 C 0.16780 0.24532 0.42690
O2 O 0.05757 0.21719 0.30246
C7 C 0.05180 0.07991 0.27798
C1 C 0.16866 0.02605 0.40441
C6 C 0.23660 0.12584 0.49162
C3 C 0.35070 0.10279 0.61788
C5 C 0.39278 -0.02130 0.65283
C2 C 0.32245 -0.12294 0.56458
C4 C 0.20840 -0.10008 0.43826

was converted by

babel -icif EFAHAW.cif -ogjf test.gjf

into

#Put Keywords Here, check Charge and Multiplicity.

 EFAHAW

0  15
O           0.82917         3.70462         4.26502
C           0.68636         2.55832         3.90884
O          -0.01540         2.26497         2.76942
C          -0.02324         0.83334         2.54528
C           0.72957         0.27166         3.70291
C           1.14519         1.31232         4.50144
C           1.87569         1.07195         5.65752
C           2.16377        10.20637         5.97753
C           1.73027         9.14642         5.16948
C           1.00028         9.38482         4.01285

The above script converted it into .xyz

10
test.xyz
O           0.82917         3.70462         4.26502
C           0.68636         2.55832         3.90884
O          -0.01540         2.26497         2.76942
C          -0.02324         0.83334         2.54528
C           0.72957         0.27166         3.70291
C           1.14519         1.31232         4.50144
C           1.87569         1.07195         5.65752
C           2.16377        10.20637         5.97753
C           1.73027         9.14642         5.16948
C           1.00028         9.38482         4.01285

and already mentioned babel -ixyz test.xyz -osmi test.smiles offered the SMILES string

O=C1O[C][C]=C1[C].[C][C][C] test.xyz

Due to the definition of asymmetric unit and unit cell, other original input formats than .cif may work even better (i.e., may yield molecules which are not fragmented as here), e.g. «native» .mol2 or .sdf.


Where does the energy come from to lower the temperature of a brine solution?

The minimum temperature reached should be −21 °C at 29 % NaCl. Initially suppose that the ice,  water and water vapour are at 0 °C in a thermally insulated container. Say point A in the figure and some salt is now added.
However, the mixture is now not at equilibrium, and some ice melts and dissolves some salt. The resulting solution is now too concentrated to be in equilibrium with the ice so more ice melts, diluting the solution and so more ice melts. At the same time the temperature of the whole system spontaneously falls until after sufficient salt is added a temperature of −21 °C is reached.
This is the eutetic point (point E) and equilibrium has been reached. The temperature falls because for a phase change to occur the ice uses the heat of fusion it needs to melt by drawing that much heat from its surroundings.
Many other salts can be used instead of NaCl, such as ammonium chloride and calcium chloride.

Figure 1. Crude phase diagram of the ice/water/salt mixture.


What is the meaning of [2,1-c][1,4] in Pyrrolo[2,1-c][1,4]benzodiazepine?

Yes, the order of the numbers indicates the relative orientation of the fused ring system. The direction of the numbers of the attached component (e.g. ‘1,2’ or ‘2,1’) corresponds to the direction of the lettering of the parent component (i.e. a, b, c, etc.).

For example:

The correspondig Subsection P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), reads as follows:

  (…)
  Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (…) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)


The statement, "only compounds in aqueous state can be split or dissociated into their ions," is false in a number of ways.

As you state, molten salts are ionic, and there are room-temperature ionic liquids, such as ethylammonium nitrate , $\ce{(C2H5)NH3+·NO3-}$.
In addition to water, there are other ionic solvents, such as nitromethane, $\ce{CH3NO2}$. Salts in these solvents break up to form ions.
Further, not everything dissolved in water is ionized. Sugar (sucrose), for example, is not ionized.


My question is this: Does electrolysis work the same way when the cathode and anode are made of the same metal? It might be a stupid question, but I can't find anything that answers this question specifically.

The battery or the power source does not care about the electrode material. Its job is to pump the electrons in the circuit. Now the problems originate from the chemistry rather. By definition, if a direct current passes through an electrolyte, electrolysis (=chemical decomposition) has to occur. Something has to reduce at the cathode and something has to oxidize at the anode. Here, we enter the competition, as shown below in the case of your simple aqueous NaCl set up.
At the anode (the electrode connected to the positive terminal of the DC supply), you have to think...
(a) Will the electrode oxidize itself? Possible
(b) Will the solvent i.e., water oxidize to oxygen? Possible
(c) Will the chloride ion oxidize to chlorine? Possible
(d) Na+ ions, they should be repelled
(e) Can the electrode react with the electrolytic product? Very much possible
Similarly at the cathode (the electrode connected to the negative terminal of the DC supply), think about ...
(a) Can the electrode reduce itself? If it is a metal, it cannot reduce further.
(b) Will the solvent i.e., water reduce to hydrogen? Possible
(c) Can the Na+ ions be reduced to metallic sodium? Possible, but not in plain water.
(d) Cl- should be pushed away.
(e) Can the electrode react with the electrolytic products? Very much possible
So you can see here, that depending on the electrode material, yes it can be oxidized if used an anode. It can also be oxidized indirectly, if the evolving chlorine begins to attack the metal, which happened in your case.
In any case, electrolytic cell does not care about the electrodes, the electrodes can certainly be the same but they have to be inert. People always choose inert electrodes so that the metallic cathode or anode do not interfere chemically during the electrolysis. Platinum is a good choice but it is too expensive for home experiments. Carbon/graphite electrode is another choice. As you have seen, stainless steel is not a good choice as an anode. It is readily oxidized by chlorine.


There is likely a question that addresses this on the site already, but I couldn't find it either, so I'll add an answer here.

You are correct that these symbols relate to the symmetry of the molecular orbitals that they label. In chemistry, we often label molecules by their point group symmetry, which describes how the molecule behaves with respect to particular symmetry operations (e.g. rotations, reflections, inversions).

If a molecule has a particular point group symmetry, its orbitals (and some other properties like vibrations) can be divided up into irreducible representations, which are essentially just different ways in which the overall symmetry can occur. One can find these irreducible representations (often shortened to irreps) in a character table for that point group.

$\sigma_\mathrm{g},\sigma_\mathrm{u},\pi_\mathrm{g},\pi_\mathrm{u}$ are all irreps of the $D_{\infty\mathrm{h}}$ point group, which you will most commonly encounter as the symmetry of a homonuclear diatomic. Since you are just being introduced to these, you won't likely use a character table and will just be given these MO diagrams.

At the level you will need for the moment, $\sigma$ orbitals come from the overlap of a single lobe from each atomic orbital and $\pi$ orbitals come the overlap of two lobes from each atomic orbital. The $\mathrm{g}$ and $\mathrm{u}$ labels tell you whether the orbital is symmetric or antisymmetric under inversion, that is, if you move every point $(x,y,z)$ to $(-x,-y,-z)$, does the sign of the orbital remain the same or does it become the opposite sign. The * label is used to describe whether an orbital is bonding (overlap is between atomic orbitals/lobes of the same sign) or antibonding (overlap is between atomic orbitals/lobes of opposite sign).


One of the great properties of enzymes is their specificity. Even broadly specific enzymes such as lipases and proteases are picky compared to, say, a general base catalyst. As a result, if you are interested in degrading a single specific compound (cocaine in your example), it is very likely that you can find or engineer an enzyme to do the job. But if you want to degrade a wide variety of compounds with different structures, you will need a diverse set of enzymes, each specific for one or a few of the compounds.


If the rate law order is three, does that mean that three bodies collide at some elementary step?

Let's disprove that statement by counterexample. Here, the slow step is a collision of some intermediate and another thing, but the maths work out that the overall reaction order is third.

Let's have a two step reaction with the stoichiometry:

$$\ce{2A + B -> C}$$

The mechanism looks like this. The first step is fast and reversible and rapidly establishes equilibrium between $\ce{A + B}$ and an intermediate $\ce{I}$.

$$\ce{A + B <=>[k_1][k_{-1}] I}$$

The second step is slow, not so reversible and involves the intermediate $\ce{I}$ reacting with the other equivalent of $\ce{A}$.

$$\ce{I + A ->[k_2] C}$$

Now the rate of the reaction shall be written as the rate of appearance of $\ce{C}$, because it makes our lives a little easier.

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=k_2 [\ce{A}][\ce{I}]$$

However, it's messy to have the intermediate in the rate law. Since the first step was fast and in equilibrium, we can get an expression for $\ce{[I]}$ in terms of $[\ce{A}]$ and $[\ce{B}]$.

$$K_{eq}=\dfrac{k_1}{k_{-1}}=\dfrac{[\ce{I}]}{[\ce{A}][\ce{B}]}\\
[\ce{I}]=K_{eq}[\ce{A}][\ce{B}]$$

A little substituting and we get:

$$\mathrm{rate}=\dfrac{d[\ce{C}]}{dt}=\dfrac{k_2 k_1}{k_{-1}} [\ce{A}]^2 [\ce{B}]=k_{obs}[\ce{A}]^2 [\ce{B}]$$


A titration mixture having pottasium oxalate and pottasium hydrogen oxalate!

You are right, $1$ mole of $\ce{KHC2O4}$ reacts with $1$ mole of $\ce{KOH}$. From that first piece of information in your exercise, you can find out that, in $\pu{10mL}$ of solution, you have $\pu{0.05 * 0.01 = 5 * 10^{-4}}$ moles of $\ce{KHC2O4}$. The next piece of information is meant to let you how much substance there is in the solution. $10$ moles of both substances react with $4$ moles of acidified $\ce{KMnO4}$. Which means the sum of the quantities of both substances in that solution is $\pu{(0.06 * 0.01 * 10) / 4 = 15 * 10^-4}$. So you got $\pu{a = 5 * 10^{-4}}$ and $\pu{a + b = 15 * 10^{-4}}$. Do the math.

Please tell me if I was clear enough.


Is lithium L-threonate a potential brain medicine?

Lithium has a diagonal relathionship with magnesium, so in effect, it makes sense to compare it to magnesium compounds. It may also substitute for sodium ions in transporters. I doubt the counter ions is relevant, since most Li coumpounds are quantitatively dissociated. Stable concentrations of lithium are only achieved after several week of treatment. Lithium carbonate, citrate and sulfate are used to my knowledge. Plasma concentrations are monitored so as to avoid reaching toxic levels (more than 1 milli molar). Some physicians also test for the concentration inside red blood cells to get an idea of the concentration in the brain, but this has not been confirmed to be relevant. What is important is not the amount of it, but the clinical effect : is the patient responding to treatment ? does he/she support it ?

Lithium has not only neuroprotective effects, but also mood-stabilizing effects. It is mostly used in case of bipolar disorder (mood swings), but also as an antidepressant. The mechanism of action is still elusive. It may enhance serotonin pathway, and, by competing with magnesium ions, inhibit some magnesium-dependent enzymes (typically enzymes handling MgATP). One of these enzymes is called glycogen-synthase-kinase (GSK).


Relationship between Quantum Numbers and the Wave-function

The hydrogen atom
Indeed, three numbers cannot define a function. There are several more criteria, one of which you mentioned already: it must satisfy the time-independent Schrödinger equation. The Schrödinger equation is a nasty differential equation, which looks something like the following:
$$-\frac{\hbar^2}{2\mu}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) - \frac{e^2}{4\pi\varepsilon_0r}\psi = E\psi$$
On top of this, it must also satisfy the usual criteria for a wavefunction ("boundary conditions"): for example it must be square integrable (i.e. its integral over all space must be finite, meaning that it can be normalised):
$$\int |\psi|^2\,\mathrm{d}\tau < \infty$$
Once you put all of these conditions together, it turns out that not every plain old function $\psi(x,y,z)$ is a permissible atomic orbital. There is a family of functions $\psi$ which do, and they all have a certain form which is completely determined by the three quantum numbers $(n, l, m_l)$.

A simpler example
This is pretty abstract, so why don't we make it more concrete with a simpler example. (The hydrogen atom is not easy to solve explicitly.) Consider the one-dimensional function $f = f(\phi)$, where $\phi$ is an angle. You should recognise at this point that, given no other information, one quantum number is not sufficient to specify what $f$ is, as there is an infinite possible different $f$'s.*
However, assume now that $f$ obeys the differential equation
$$\frac{\mathrm{d}f}{\mathrm{d}\phi} = \mathrm{i}mf$$
The general solution of this differential equation is (after dropping the normalisation factor)
$$f = \exp(\mathrm{i}m\phi)$$
But note also that since $\phi$ is an angle, we have that for all $\phi$
$$f(\phi) = f(\phi + 2\pi)$$
meaning that
$$\begin{align}
\exp(\mathrm{i}m\phi) &= \exp[\mathrm{i}m(\phi + 2\pi)] \\
&= \exp(\mathrm{i}m\phi) \exp(\mathrm{i}m2\pi) \\
1 &= \exp(\mathrm{i}m2\pi)
\end{align}$$
which means that $m$ must be an integer. Consequently, the allowed $f$'s are
$$f = \exp(\mathrm{i}m\phi) \quad\quad m \in \mathbb{Z}$$
and so we can now label the $f$'s by the value of $m$. The quantum numbers $m$ do not represent actual values of $f$ at any point in space, nor do they represent actual values of $\phi$: they instead serve to parameterise the $f$'s. That is to say, the value of $f$ at any value of $\phi$ is completely determined by the value of $m$. We do not write this as $f(m)$, but rather $f_m$, and thus we have:
$$\begin{align}
&\,\,\,\vdots \\
f_{-2} &= \exp(-2\mathrm{i}\phi) \\
f_{-1} &= \exp(-\mathrm{i}\phi) \\
f_{0} &= 1 \\
f_{1} &= \exp(\mathrm{i}\phi) \\
f_{2} &= \exp(2\mathrm{i}\phi) \\
&\,\,\,\vdots
\end{align}$$
Indeed, $m$ can be said to be a quantum number here. Notice how we've used a simplified problem: it has one instead of three dimensions, and it has a much simpler differential equation. However, it illustrates that by enforcing additional conditions on the wavefunction — namely that it obey a differential equation and boundary conditions — we can get a family of solutions which can be labelled with quantum numbers.†

What is the energy?
When you find the allowed $\psi$'s that solve the Schrödinger equation, you can plug it back into the differential equation and evaluate the value of $E$ inside it. That will give you the energy of the atomic orbital.

Footnotes
* Technically, there is an uncountably infinite number of $f$'s. If it were countably infinite, then we could establish a one-to-one mapping between the different $f$'s and the natural numbers, which would allow us to label the $f$'s as $f_1, f_2, f_3, \cdots$ which is exactly what a quantum number does.
† With regards to the previous footnote, this set of permissible $f$'s is still infinite, but it is countably infinite this time. Also, this actually happens to be the exact way in which the quantum number $m_l$ is derived, so my choice of $m$ as the quantum number was not arbitrary.


Slight variation in NMR integrals of CH protons vs. CH2 or CH3

That's to be expected: Less protons as spatially close neighbours, less dipolar coupling, slower relaxation.

For a relaxation/recycle delay of $\pu{1 s}$ and $T_1$ of perhaps $\pu{3 s}$, the Ernst angle is already only $44^\circ$, so your measurement is far from optimal anyway. Use a shorter pulse or longer recycle delay.

At one second, your are anyway dangerously close to generating echoes which overlay the signal from further scans, distorting the accumulated spectrum.


How can the outermost electrons in silicon be 2s and 2p electrons?

The book means to say that there are 2 electrons in the  $\mathrm{3s}$ subshell and and 2 electrons in the  $\mathrm{3p}$ subshell. You can see this in the electron configuration as

  $\mathrm{...3s^2 3p^2}$

and that the maximum configuration has 2 electrons in the  $\mathrm{3s}$ subshell and 6 electrons in the  $\mathrm{3p}$ subshell which would be

  $\mathrm{...3s^2 3p^6}$


Why does screening effect decrease due to d-orbital?

The reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals.  This is because of something called penetration .

The mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals.  In gallium, you have $10$ electrons in the filled $3$d-subshell, and each of these electrons is doing a slightly worse job (relatively speaking) of screening the nuclear charge than the electrons in the s and p orbitals. Therefore, the effective nuclear charge in gallium is slightly higher than that in aluminum, so the increase in the radius is a quite a bit smaller than would be expected based on the difference between boron and aluminum, or gallium and indium.
The trend goes:

$\pu{82 pm}$ ($\ce{B}$)
$\pu{118 pm}$ $\ce{(Al)}$
$\pu{126 pm}$ $\ce{(Ga)}$
$\pu{144 pm \ce{(In)}}$

[covalent radii from www.webelements.com].

This effect is generally known as the d-block contraction.  (It can be more or less pronounced depending on how you define the atomic radii.)

A similar thing happens (in principle) when you go from indium to thallium; except in this case you are now dealing with adding a filled f-subshell to the valence shell.

Electrons in f-orbitals are even worse at screening nuclear charge than those in d-orbitals, therefore again, the effective nuclear charge in thallium is a bit larger than it is in indium, so again the jump in radius is fairly small (from $144$ to $148$ pm).  This effect (of the filled f-subshell) is generally known as the lanthanide contraction.


Why only one peak is observed in NMR spectrum of H2?

For starters, it is incorrect to think of the ground state being substantially more populated in NMR spectroscopy. Because the energy difference between aligned and misaligned nuclear spins in the external magnetic field is very low, we can say that both states are almost equally populated due to thermal excitation alone. For a $500~\mathrm{MHz}$ experiment, the energy difference can be calculated to be $0.21~\mathrm{J \cdot mol^{-1}}$ which corresponds to a population ratio as follows:

$$\frac{n_\mathrm{ex}}{n_\mathrm{g}} = \exp \left(\frac{\Delta E}{RT}\right) = 0.999916$$

Next, consider the extremely high symmetry of the $\ce{H2}$ molecule: It has the point group $D_{\infty\mathrm h}$, one of the most symmetric ones around. Due to this high symmetry, we cannot distinguish between either proton at all — both can be transformed into each other by a rotation as defined by the point group. Thus, the two nuclei are magnetically equivalent — not only must their chemical shift be identical but also the coupling to the other one.

Think of it this way: Assume, that the spin transition of one proton happened at a slightly different frequency than the other’s. That would mean that one proton had to be different from the other in some way (a different transition energy is equivalent to a different chemical shift is equivalent to a different environment). But how would you explain a different environment on one side of the $\ce{H2}$ molecule compared to the other? Correct: You cannot.

Now we could still assume that the energy required to excite a nuclear spin whose neighbour is parallel be different from one whose neighbour is antiparallel. However, no matter how we excite, the transition is always $\ce{parallel <=> antiparallel}$. And since we cannot distinguish between the hydrogens, we also cannot distinguish which one pointed in which direction, so those two energies must be identical. Therefore, we only see one peak and no $^1J_\ce{HH}$ coupling.


There are several different ways to rate carcinogens.  According to its MSDS, Pyridine belongs to Group 3, A3:  Known to cause cancer in animals, inadequate evidence in humans.

If it was just contact with the skin, your friend should wash it off with soap and water immediately for 15 minutes (this is again from the MSDS).  Short term exposure can cause "irritation (possibly severe), allergic reactions, sensitivity to light."

So, if any irritation occurs, it is probably not serious, but I would recommend seeing a doctor anyways.  I doubt there is anything to worry about, but cancer is not something to mess around with.


Why do lone pairs of electrons have a higher repulsive force?

Bonding pairs are stabilized between two atoms. Since there's no atom on the other end of a lone pair, it spreads out more than would be if it were in a bond. Lone pairs therefore repel more because the charge density is spread out more, that is, it's bigger.

  I have read that lone pairs have orbitals which are shorter and more round

This is technically true in the sense that you'd want to have higher $s$ character in a lone pair orbital. But you're comparing possible ways to hybridize orbitals for lone pairs, as opposed to comparing lone pair atomic orbitals with bonding orbitals.


Is 4-chlorotoluene also an IUPAC name for 1-chloro-4-methylbenzene?

The use of 'toluene' as a parent hydrocarbon in nomenclature is described in Rule P-22.1.3 in Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013.

  These names [toluene, xylene, and mesitylene] are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not freely substitutable; toluene is substitutable under certain conditions, but only for general nomenclature. (emphasis mine)

"General nomenclature" refers to any name generated by the use of IUPAC principles, rules, and conventions, whereas the "preferred IUPAC name" (PIN) refers to a name that is preferred among two or more names generated from two or more IUPAC recommendations. The introduction P-10 reads:

  Any name other than a preferred IUPAC name, as long as it is unambiguous and follows the principles of the IUPAC recommendations herein, is acceptable as a 'general IUPAC name", in the context of a 'general IUPAC nomenclature'.

As stated in P-22.1.3, in the construction of a PIN for the molecule below, the use of 'toluene' as the parent hydrocarbon is not permitted; the PIN is 1-chloro-4-methylbenzene.

Regarding the substitution of toluene in general nomenclature, P-15.1.8.2.2 lists the characteristic groups that can be used in general nomenclature:

  [...] halides $\ce{-Br, -Cl, -F, -I}$, pseudohalides $\ce{-N3, -NCO}$ (and chalcogen analogues), $\ce{-NC}$, substituent groups derived from the halogen oxo acids $\ce{-ClO, -ClO2, -ClO3}$ (similarly for groups in which $\ce{Cl}$ is replaced by $\ce{Br}$ or $\ce{I}$), $\ce{-NO2}$ and $\ce{-NO}$, and $\ce{-OR}$ ($\ce{R}$ = alkyl groups), and chalcogen analogues, and $\ce{-SO-R}$ and $\ce{-SO2-R}$, and $\ce{Se}$ and $\ce{Te}$ analogues.

Since chloride falls under this list, the name 4-chlorotoluene is permissible in general nomenclature, but it is not the preferred IUPAC name.

Note that there is no space between the prefix 'chloro' and the parent hydride name 'toluene'.

Also note that there is no such thing as '1-toluene'; in the name '4-chlorotoluene', for example, the methyl group of toluene is automatically assigned the locant '1' (see figure below from the Recommendations). Therefore, the name '4-chloro-1-toluene' is incorrect.


Oxidation of water — what is the half reaction for water?

Both your reactions are equivalent. Here’s why. We start from $(1)$.

$$\ce{4 OH- -> O2 + 2 H2O + 4 e-}\tag{1}$$

(I have omitted the phase descriptors for clarity.) In any chemical reaction, we can add or subtract spectators as we see fit; this is akin to mathematical equations where we can add or substract expressions as we see fit as long as we do the same to both sides. For example, I want to add four protons to $(1)$ to give $(2)$.

$$\ce{4 OH- + 4 H+ -> O2 + 2 H2O + 4 e- + 4 H+}\tag{2}$$

If this were now a mathematical equation, I could combine (or divide) parts as I see fit. In a chemical reaction, we need to be a bit more careful. But as long as our focus is not on proton transfers, we can combine anything basic enough with acidic protons. Here, you can combine the four hydroxides and four protons on the reactant side of $(2)$ to give $(3)$.

$$\ce{4 H2O -> O2 + 2 H2O + 4 e- + 4 H+}\tag{3}$$

Now, we notice that we again have spectators. Two water molecules turn up both on the reactant and on the product side (an additional two are unique to the reactant side). We can remove them like we did the protons before, giving us $(4)$. I will also perform some rearranging.

$$\ce{2 H2O -> 4 H+ + O2 + 4 e-}\tag{4}$$

It so turns out that $(4)$ is your second suggested reaction. It is the same equation as $(1)$ as I have just shown; the difference is merely in the type of water molecules present.

Now which one should you use? Notice that one reaction consumes hydroxide while the other produces protons (formally! The consumption of hydroxide is equivalent to the production of protons). Thus, if your reaction medium is basic, you should use $(1)$ for your redox reaction. If, however, your reaction mixture is acidic, $(4)$ is more appropriate. In short, choose the one with the same charged particle as is used in the other half-reaction to make your life simpler.


If some particles of liquid in a solution convert to vapour then why can't some liquid particles convert to solid?

If water molecules go into the gas phase from the liquid then they are in a different phase. But water molecules go into the gas phase one at a time. If there are the same number of water molecules leaving the liquid as coming into it then the atmosphere above the liquid would have 100% humidity and be in equilibrium with the liquid.

Now if some small number molecules of water happen to line up to form an ice crystal within the water itself what happens? Such an ice crystal is thermodynamically unstable and then "melts" within the liquid phase. Since the whole process takes place in the liquid phase it is virtually undetectable.

Now you also have to realize that form ice crystals big enough to detect visually requires more than a handful of molecules. You'd need billions of billions of molecules. That is just a statistical impossibility.


What is an algorithm for making a net neutral molecule from a set of guessed positive charges?

Although you did not mentioned specifically, it is likely that you want to perform some Molecular mechanics force field calculations. The charges are important quantity, so you need more than educated guess to obtain usable results. Nonetheless, the problem of charge equilibration is solved by "Polarizable force fields".

It should be noted, that they are often more expensive, less reliable and less accurate than standard FFs.


Xylan vs. Teflon for non-stick food pans?

Both Xylan and Teflon are chemically polytetrafluoroethylene. But a polymer just by name can mean a lot of different things if you consider the structure/arrangement, additives and impurities. These are highly process-dependent and, guess what, we do not know their chemistry for sure because they are industrially protected. For actual matters, both Xylan and Teflon are unsafe above 260 ⁰C (500 ⁰F), when they start to break down to toxic fluorine polymer fumes.


How to generate fully coloured electrostatic potential maps in Avogadro?

The answer is that the color scheme changed, due to user (and developer) feedback. More people preferred the red to white to blue scheme for electrostatics.

In Avogadro 2, we hope to have a full-featured gradient editor, allowing you to set the colors you wish.


Generating electron density from protein coordinate (PDB)

For the more recent structures, you can view the density (based on measured diffraction data and the model, so-called 2Fo-Fc density) directly in the protein data bank, e.g. http://www.rcsb.org/3d-view/6QU9?preset=electronDensityMaps:

For a theoretical model or a model without deposited diffraction data, you would first have to generate structure factors, and then calculate the electron density. In the simplest case, the electron density would just reflect the density of isolated atoms (i.e. no deformation density). Program suites such as CCP4 (originally in Fortran) or Phenix (in python) are available to do these steps.

In these calculations, the electron density of individual atoms is encoded in so-called atomic form factors (see e.g. http://lampx.tugraz.at/~hadley/ss1/crystaldiffraction/atomicformfactors/formfactors.php), which are fourier transform coefficients of the model electron density for each atom.


One of the first places to look at for advice is the Materials Safety Data Sheet (MSDS, or just SDS). Make sure these are kept in a handy place and are up to date. These should be provided by the supplier (but are easy enough to get online).

A pertinent and thorough guide is produced by the aptly named Methanol Institute, the document is called Methanol Safe Handling Manual, the chapter Health and Safety, especially the Safety Precautions subchapter outlines the following:

  Smoking must be prohibited.
  Vehicle access should be strictly controlled.
  Ventilation must be sufficient to cope with the maximum expected vapor levels in buildings.
  Positive pressure may be required for methanol-free areas, such as control, switch, and smoking rooms.
  Storage tank vents to atmosphere should be sized for fire-heated emergency vapor release.
  Electrical equipment must be explosion-proof to meet national electrical code requirements.
  Alcohol-resistant Aqueous Film-forming Foam (AR-AFFF) with 6% foam proportioning (with water) equipment is advised for use on methanol
  fires.
  Dry chemical extinguishers should be accessible for small fires. An adequate supply of handheld and wheeled types should be available.
  Hydrants should be strategically placed with adequate hoses.
  Small spills should be remediated with sand, earth, or other non-combustible absorbent material, and the area then flushed with
  water. Larger spills should be diluted with water and diked for later
  disposal.
  Lighting should be grounded. Tall vessels and structures should be fitted with lightning conductors that are securely grounded.

These kind of directives should be on signs that are easily read alongside hazard signage.

I would suggest having a readily accessed first aid kit and make a poster with first aid instructions, also included in the handbook - First Aid Measures

In terms of storage, the subchapter Methanol Storage, the handbook states that

  In general, fire protection for gasoline tanks is sufficient for
  methanol tanks, provided extra precaution is made for leak detection, toxic hazard, and
  availability of alcohol-compatible fire suppression foam.

In terms of proximity to the house, I would consider storing these tanks as far as feasible from the house.


Let me see if I can get at some of your questions. As mentioned above, it's much easier when you ask individual specific questions.

One problem with books on introductory quantum mechanics is that, put simply, the language of quantum mechanics is math. Specifically, most people use the Schrödinger equation which involves second derivatives and differential equations.

When I teach quantum chemistry to undergraduates, there are typically problems due to the mathematical nature.

There are multiple "interpretations" of quantum mechanics. That gets into philosophy. Wikipedia has some nicely-written descriptions.

As you mention, the most widely used interpretation (the Copenhagen interpretation) of the wave function centers on the square of the wave function, or rather $|\psi^*\psi|$ as a probability density. As you mention, the wave function could be imaginary or complex, so this notation indicates a mathematical way of getting a real number for a probability density.

You ask about the shape and why atomic orbitals have particular shapes.

One thing that's amazing about quantum mechanics is that we put in some relatively simple equations to describe the motion of an electron in a hydrogen atom, and the solutions are exactly these atomic orbitals (s, p, d, f, etc.).

So I disagree that "no one knows" why atomic orbitals have particular shapes. The answer is that they are the mathematical solution to the Schrödinger equation. That is, we want to know the energies of the system and there are only certain quantized solutions.

And yes, the whole notion of chemistry is that electrons (and hence orbitals) on different atoms interact. Otherwise, there is no bonding.

I'll leave you with this. Without the math, it can be exceptionally hard to understand quantum mechanics. When you do understand the math, I think the results are quite beautiful. The shapes of orbitals aren't arbitrary. This is actually just a reflection of the mathematical nature of quantum mechanics.


The detailed calculation for $\Delta G$ is here,  Derive expression for internal energy of mixing and entropy of mixing using statistical thermodynamics  which gives $$\Delta G = RT[n_1\ln(x_1)+n_2\ln(x_2)] +(n_1+n_2)x_1x_2w$$

and as $\Delta S= -d\Delta G/dT$ then

$$\Delta S = -R[n_1\ln(x_1)+n_2\ln(x_2)] - (n_1+n_2)x_1x_2\frac{dw}{dT}$$

where $w \approx (2E_{12}- E_{11} - E_{22})$ and $E_{12}$ is the interaction energy between molecules of type $1$ and $2$ when they are adjacent and similarly for $E_{11}$ and $E_{22}$ all of which are negative. From a consideration of the energies involved $w$ is expected to be positive.

The first term is the  Raoult's law expression for an ideal solution and the second term the deviation, which is sometimes also called the excess  entropy of mixing. For benzene cyclohexane when $n_1=n_2=x_1=x_2=1/2$  the excess entropy  $\displaystyle -\frac{1}{4}\frac{dw}{dT}\approx 1.7$ J/mol/K at $298$ K and $1.2$ at $343$ K.


Could sodium fluoride or aluminum accumulation in soils contribute to increased amounts of palmitic acid in food?

This article explains the effects of fluoride ion concentrations to Prosopis juliflora.

  Both catalase (3.2 folds) and peroxidase (2.7 folds) enzymes activity increased with increase in F concentration.

Checking out Wikipedia articles about these enzymes, neither catalase not peroxidase are involved in the fatty acid biosynthesis. So as far as I know, I don't think there's any effect of fluorides in the biosynthesis of palmitic acid.

Also, regarding aluminum ions, this article explains aluminum toxicity as a hindrance to cell division and elongation, but nothing is mentioned about fatty acid synthesis.

References:

Saini, P, et al. “Effects of Fluoride on Germination, Early Growth and Antioxidant Enzyme Activities of Legume Plant Species Prosopis Juliflora.” Journal of Environmental Biology., U.S. National Library of Medicine, 34 Mar. 2013, www.ncbi.nlm.nih.gov/pubmed/24620580.

Mossor-Pietraszewska, T. (2017). Effect of aluminium on plant growth and metabolism.. [online] Psjd.icm.edu.pl. Available at: http://psjd.icm.edu.pl/psjd/element/bwmeta1.element.bwnjournal-article-abpv48i3p673kz [Accessed 23 Aug. 2017].


I would be more worried about carbon dioxide (rather than water) being absorbed into the NaOH solution. A 1 molar NaOH solution is not going to be appreciably hygroscopic. But it will absorb CO2: two moles of hydroxide yield one mole of carbonate ion, which, depending on what you are titrating, may act as only deprotonate one mole of acid per mole base - reducing the effective normality of the NaOH solution. You could try protecting any air vents with a drying tube filled with sodium or calcium hydroxide, or use a regulated supply of chemically pure nitrogen as a blanket.


Why does vibrational cooling occur much faster than sponteneous emission in the condensed phase?

As you point out, there are many places for this relatively small amount of vibrational energy to go. There are both radiative and non-radiative transitions which can take place to vibrationally quench the system in the electronic excited state. The explanation for both parts of your question (the vibrational and electronic parts) can be understood by Fermi's Golden Rule. I won't go into a lot of detail of deriving this or anything else because you can find anything you wanna know about it by searching online.

Fermi's golden rule says that the transition rate, $\Gamma_{i\to f}$, from some initial state, $|i\rangle$, to a set of final states, $|f\rangle$, is given by,
$$
\Gamma_{i\to f}=\frac{2\pi}{\hbar}|\langle f|H'|i\rangle|^2\rho
$$
where $\rho$ is the density of final states of the system and $H'$ is a perturbing Hamiltonian of some kind. For instance, the electric field of a photon. This is what it would usually be for absorption. In the case of decay, it may also be the field of a photon, but I am not as familiar with using Fermi's golden rule in this case.

Notice that this rule has dimensions of inverse time. It can be shown that the inverse of this rate is equal to the mean lifetime of the excited state (see this page).

Basically all of the physics you are asking about is contained within the fact that this transition rate is proportional to the density of possible states to which the system will decay. To put it simply, there are many many more intramolecular and intermolecular transitions which can take place to redistribute this energy. The vibrational density of states for this problem could probably be approximated quite well as being continuous because there will be many vibrational levels which are slightly perturbed due to distortions from equilibrium geometries. Think of the width of vibrational bands in IR.

On the other hand, a very large amount of energy must be dissipated for electronic relaxation to take place, and the density of these high-energy states is very low. Especially if one excites to the lowest electronic state. If, however, excitation occurs to a higher electronic energy level, decay should take place faster because the density of electronic states becomes much larger.

These processes have almost nothing to do with the speed of nuclear motion compared to electronic motion, as the condition which must be satisfied is really conservation of energy, and this is not always easy to do when the energy levels are quantized. From this perspective, the result is not surprising because vibrational energy levels are certainly closer to being continuous than electronic energy levels. Also, the spacing of vibrational state is generally smaller on an excited state than the ground state, so this adds to the quickness of the decay to the ground vibrational state.

As an aside, this paper is of some interest as it shows that vibrational relaxation times in the gas phase (for this one molecule) are slower than vibrational relaxation in solution. This can also be understood by the density of states argument discussed above.


How to detect harmful gas occurence in carboy bottle

Gas sensors are probably too specific. Infrared imaging is going to have to take account of the carboy material itself. However, sampling the air and performing a gc/ms analysis could work - basically its headspace analysis.


You are looking at two different entropy changes. The one computed directly from the enthalpy change is the entropy change of the universe. The entropy change of the system is measured by computing the component due to heat transfer. But internal changes in the density of states can also result in a change in entropy, and this can take place without any transfer of heat.


How is red oxide of copper converted into black oxide & vice versa?

Red Copper oxide is Copper(I) oxide Cu2O. Black Copper oxide is Copper(II) oxide CuO. Copper(I) oxide is oxidised to Copper(II) oxide in moist air.

Copper(I) oxide is oxidised to Cu(II) by dissolving in dilute nitric acid which can be thermally decomposed to Copper (II) oxide by heating at 180C.

Copper(II) oxide can be reduced back to Copper metal by Hydrogen, Carbon Monoxide or Carbon. This may then be heated with oxygen to make Copper(I) oxide. Copper(I) oxide may also be made by reaction of solutions of Copper(II) salts with Sulfur Dioxide.

Source - Wikipedia entries for Copper(I) oxide Cu2O and Copper(II) oxide CuO


What happens if I make a hole in a container with superheated water?

If you have water at a temperature of $T=250\ \mathrm{^\circ C}$, you need a pressure of approximately $p=39.2\ \mathrm{atm}=4.0\ \mathrm{MPa}$ to keep it liquid. Liquid water at this temperature and pressure has a specific enthalpy of $h_0=1086\ \mathrm{kJ/kg}$.

If you open the container, the pressure drops to the ambient pressure of $p=1\ \mathrm{atm}=101\,325\ \mathrm{Pa}$. A part of the hot liquid water flashes to steam. The temperature drops to the boiling point ($T=99.974\ \mathrm{^\circ C}$) at the new pressure. A new equilibrium of liquid water and steam is established at the new temperature and pressure. The corresponding specific enthalpy of the steam is $h_\text{steam} = 2676\ \mathrm{kJ/kg}$; the specific enthalpy of the liquid water is $h_\text{liquid} = 419\ \mathrm{kJ/kg}$.

The flashing process is very quick, hence we may assume that there is no significant heat exchange with the environment; i.e. we may assume the following enthalpy balance in order to estimate the vaporized fraction $x$:

$$\begin{align}
h_0 &= x \cdot h_\text{steam} + (1-x) \cdot h_\text{liquid}\\
x &= \frac{h_0 - h_\text{liquid}}{h_\text{steam} - h_\text{liquid}}\\
x &= \frac{1086\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}{2676\ \mathrm{kJ/kg} - 419\ \mathrm{kJ/kg}}\\
x &= 0.296
\end{align}$$

This means that approximately $29.6\ \%$ of the superheated water flashes to steam when the container is opened.


What is the effect of oxygen on soap foam?

Hmmm. I don't know, but I'd say that if you significantly lowered the percentage of $\rm O_2$ in the air, the average molecular weight of the molecules in the bubble would drop (there'd be more low-molecular weight nitrogen and less higher-molecular weight oxygen), so the rate at which gas would diffuse out of the bubble would increase, and the foam would flatten faster.

Try blowing a foam with nitrogen, and compare it with the results of blowing a foam with air or oxygen---if I'm right, the foam blown with nitrogen will flatten a bit faster.

There'll be a bit more $\rm CO_2$ in the air from the candle, and that would diffuse more slowly out of the bubble (in fact, $\rm CO_2$ is used as a "blowing agent" to make foams, partially for that reason). But I'd think in this case that the small amount of $\rm CO_2$ produced by the candle would dissolve in the liquid, with gas dissolution rates enhanced by the lower surface tension and higher surface area of the foam.


How does VSEPR theory explain the formation of different bond angles in PCl₅?

There are only five Platonic solids:

tetrahedron,
octahedron,
cube,
dodecahedron, and
isocahedron.

Therefore five atoms cannot be arranged in five equivalent positions, unless all five lie in a plane.
Trigonal bipyramidal is the minimum energy arrangement of five charges on a sphere. See Kevin Brown's MathPages Min-Energy Configurations of Electrons On A Sphere (Stable link via The Internet Archive).

Also there are not five equal energy states as demonstrated in the lecture notes by Marcel Schlaf and Kathryn Preuss from the University of Guelph:
Chapter 5.5 (continued 2) MO Theory ICl4, PCl5. (pdf) (Stable pdf link via The Internet Archive)


There is a complete science behind your question called Analytical Chemistry. However the goal is not to identify what a given object is made of, people usually look for targeted compounds or elements of interest. Take chocolate for example, there are thousands of compounds there. Analyzing it completely from scratch may take more than a year! There you will learn plenty of techniques to identify elements. If you just wish to know what this glass is made of, one may use a technique called X-ray fluorescence, it is good for quickly detecting most of the elements except the lighter ones.


You are close. The triplet part is the coupling of the enantiotopic hydrogens at C4. The width of the C5 blowup is ~20 Hz in width which means J = ~7 Hz. The smaller coupling is due to allylic coupling with the methyls (C7 and C8). Assuming J5,7 and J5,8 are the same, then the six hydrogens produce a septet. You just can't see the outer peaks because they are too small. If coupling to the two methyls is different then the septet will appear broadened because of lack of resolution. A larger blowup of C5 would reveal the outer peaks (one and seven).


This is not a proper answer but can give you a clue. I arrived here looking for the same answer.

I didn't arrive to find the C6 rotation axis in bulk 2H-MoS2. On the other hand, if you look for the symmetry operations of the space group of 2H-MoS2 (P63/mmc, No. 194, D46h):

http://img.chem.ucl.ac.uk/sgp/large/194az1.htm

You can see that there are not any C6 rotation axis for this space group. Nevertheless, there are 6-fold screw axes (http://img.chem.ucl.ac.uk/sgp/misc/sixscrew.htm) and 6-fold inversion axes (http://img.chem.ucl.ac.uk/sgp/misc/barsix.htm).

You are right that the point group corresponding to this space group is D6h (https://en.wikipedia.org/wiki/Space_group#Table_of_space_groups_in_3_dimensions). And according to character tables a D6h point group should have C6 (http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=606&option=4).

Both things are contradictory. My background is Chemistry, and I am used to group theory applied to molecules rather than crystals. My guess is that the 6-fold screw axis, which is a symmetry element that is only possible in crystals (as it requires translational symmetry) might be related with labelling as D6h. But this is just my speculation.

In this article, they also talk about the symmetry operations of single-layer, multi-layer (odd and even layers) and bulk MoS2, and could be usefull for you too:

http://www1.spms.ntu.edu.sg/~qihuagroup/data/Xiong/Papers/nano%202013%20yanyuan.pdf


Predicted number of signals in proton NMR spectroscopy

Since there isn't free rotation around the C=C bond in the alkene, the two hydrogens, Ha and Hb are experiencing different electronic environments. Whenever two hydrogen atoms are experiencing non-identical electronic environments, they are going to possess differing chemical shifts in proton NMR. If you heat this compound, and it doesn't decompose, I would expect the two signals from Ha and Hb to coalesce into one as the hindered rotation about the alkene C=C bond becomes unhindered.

https://chem.libretexts.org/Textbook_Maps/Organic_Chemistry/Book%3A_Organic_Chemistry_with_a_Biological_Emphasis_(Soderberg)/Chapter_05%3A_Structure_Determination_II/5.2%3A_Chemical_equivalence


Will oxidation/reduction of copper still occur in an electrochemical cell if it is not in an aqueous environment?

It is certainly possible for metals, including copper, to corrode in non-aqueous environments.  The details depend on the precise nature of the solvent, electrolytes, and other potentially complexing species present.  For example, in polar protic solvents (like water) that don't have any complexing ligands for copper, the equilibrium potential for copper oxidation will likely be the same as in water.  But oxidation potentials will likely shift in solvents that are non-polar, in ways that depend strongly on the composition of the electrolyte.

However, if you are using potassium hydroxide as the electrolyte, your conditions might be more aqueous than you think.  You can think of "hydroxide" as a type of water that likes to exist at high pHs.  And hydroxide itself generates high pHs.

Elecrochemical water oxidation occurs via this reaction:

$$\ce{2 H2O -> O2 + 4 H+ + 4e-}$$

The pH-dependent equilibrium potential is $1.23~\rm{V} - 0.059\frac{\rm V}{pH}\times pH $, which at a pH of 14 comes out to 0.4 volts.  So you might also possibly be irreversibly oxidizing hydroxide to oxygen during your electrolysis.


Ligand exchange is in equilibrium, so it is driven by thermodynamics
There are many reasons one ligand would replace another, but it often has to do with Lewis basicity, HSAB (see trans effect), or the thermodynamics of a subsequent reaction.
Ligand Exchange is a broad term for a set of mechanisms that include various types of ligand association and dissociation, and these all occur for different kinetic reasons. I'll try to provide a brief overview, but the answers to all of these questions are varied depending on specific situations.

In the covalent model, both $\ce{1e-}$ "X" ligands and $\ce{2e-}$ "L" ligands are called "ligands", so it's important to point out that we're only talking about L ligands here.

Generally speaking, there are three types of ligand exchange mechanisms: association, interchange, and dissociation, and each mechanism is further broken into the associative and dissociative subclasses, depending on the exact character of the transition state. If the higher energy TS‡ of a ligand association has the geometric character of the dissociation, it is called a "dissociative association" (the Ad mechanism), for instance, but this is somewhat rare. The equilibrium expressions of the Dissociation, Interchange, and Association mechanisms are as follows, respectively:(1)

$$\begin{align}\ce{ML5X &<=>[$k_1$][$k_{-1}$] ML5 + X}\\
\ce{ML5 + Y &->[$k_2$] ML5Y}\\
\ce{ML5X + Y &->[$k_1$]ML5Y + X}\\
\ce{ML5X + Y &<=>[$k_1$][$k_{-1}$] ML5XY}\\
\ce{ML5XY &->[$k_2$] ML5Y + X}\end{align}$$

The exact nature of a ligand exchange mechanism is something that requires computations, and physical chemistry to determine, and you should look at (1) to read more about this. Really, the mechanisms are part of a continuum:

$$\mathrm{D_d \cdots I_d \cdots I \cdots I_a \cdots A_a}$$

Once the ligand is fully displaced, it is no longer part of the inner sphere of coordination, so the "'previous ligand" is no longer bound to the complex, but there exist many ligand exchange mechanisms that involve transient intermediate in which both the leaving and the incoming ligand are bound to the central atom. All associative mechanisms will have such an intermediate, while dissociative and interchange mechanisms will not. Another important example is the formation of a preassociation complex, in which "rapid equilibrium occurs between the incoming ligand and the 6-coordinate reactant to form an ion pair" (1) (or a neutral preassociation complex with strong dipole-dipole interactions, depending on oxidation states). In the case of a preassociation complex, the outgoing ligand is no longer in the inner sphere of coordination, but it is still "bound" to the complex, albeit through weaker noncovalent interactions. Such a reaction has equilibrium like this:

$$\begin{align}\ce{ML5X + Y &<=>[$k_1$][$k_{-1}$] ML5X . Y}\\
\ce{ML5X . Y &->[$k_2$] ML5Y + X}\end{align}$$

Ligand exchange kinetic and thermodynamic arguments also frequently invoke the chelate effect and the trans effect.

Another important concept is that the driving force for individual ligand dissociations and associations involved in catalytic cycles often implicates the driving force of the overall cycle. In catalytic reactions, the rate of ligand dissociation or association is not dependent not on the thermodynamics/kinetics of the individual reaction, but instead on the step in the cycle with the highest activation energy. This step is known as the  turnover limiting step (2). In a catalytic cycle, all equilibria have the same rate. "On a circular track, on average the same number of trains must pass each point per unit time." (2) For example; in this catalytic alkene hydrogenation, the turnover limiting step is the dissociation of the phosphine from Wilkinson's catalyst (3), so this is the step that determines the turnover frequency (TOF). The rate of association of the ethylene ligand later on is driven by the cycle's overall TOF, so such a ligand association would be indirectly driven by a different step in the cycle. If the final step of the cycle is irreversible, the substrate release might also be a thermodynamic driving force.

$\hskip1.4in$(3)

These are just some of the many possible answers to your questions.

(1) G. L. Miessler, P. J. Fischer, D. A. Tarr. Inorganic Chemistry (5th Ed.), Pearson, 2014, p. 442-444

(2) R. H. Crabtree, The Organometallic Chemistry of the Transition Metals (3rd Ed.), John Wiley & Sons, 2001 p. 225

(3) J. Luo, A. G. Oliver, J. S. McIndoe, Dalton Trans. 2013, 42, 11312-11318


Stereoselectivity of ring closure in intramolecular iodolactonisation

The results of the iodolactonization example presented in this question were published in 1982 by Bartlett, et al.1 While @Aniruddha Deb has captured the essence of the reaction, there is more than one product formed in the reaction. A more detailed mechanism is offered, an issue not addressed by the original authors.
Carbonate 1 is a racemate. No products derived from reaction at the remote 7,8-double bond were detected. Understandably, the rate of closure is faster for a 6-membered ring than a 7-membered ring. A transition state for ring closure for a reversibly formed iodonium species will be favored having the carbon chain R equatorially disposed in the transition state. Transition state 2 is favored over that of 5 because the latter encounters a gauche butane interaction. In the iodonium cation the C-C bond is shorter than the C-I bond thereby having the former of greater steric consequence than the latter. SN2 attack on the iodonium cation is favored by the more polarized carbonyl oxygen than the oxygen attached to the t-butyl group. In the latter instance, establishing a positive charge on this oxygen is counterproductive because it is adjacent to a partially positive carbonyl carbon. This objection is in keeping with why amides are preferentially protonated on oxygen and not nitrogen. The cations 3 and 6 are now poised to lose a t-butyl cation, which, given the solvent acetonitrile, may lead to Ritter intermediates besides t-butyl iodide. In the event, racemic cyclic carbonates  4 and 7 are produced in 69% yield with the cis-substituted isomer 4 predominating over the trans-isomer 7.

P.A. Bartlett,  J. D. Meadows, E. G. Brown, A. Morimoto and K. K. Jernstedt, J. Org. Chem., 1982, 47, 4013. https://doi.org/10.1021/jo00142a002


The acetal reaction is performed in $\ce{H+/ROH}$ condition (e.g., $\ce{H+/CH3OH}$). The general mechanism can be found in any Organic Chemistry textbook. The reaction is reversible. The first step is protonation of aldehyde (or ketone) carbonyl oxygen by an acid ($\ce{H+}$):

The protonation makes the carbonyl carbon more susceptible to nucleophilic attack by available alcohol (e.g., methanol in the scheme). This gives the tetrahedral protonated hemiactal intermediate, which undergoes deprotonation to give stable hemiacetal. All of these steps are reversible, but forward reaction is favored due to highly available alcohol molecules (Le Chatelier's Principle).

The hemiketal has two oxygen molecules, $\ce{OH}$ and $\ce{OR}$, both of which are susceptible to protonate. However, if $\ce{OR}$ is protonate, reaction would go backward. Of cause, it protonates because the reaction is reversible (see the scheme). Yet, more favored is $\ce{OH}$, which would subsequently eliminate as a $\ce{H2O}$ molecule. The resultant positively charged carbonyl compound undergoes second nucleophilic attack by another $\ce{CH3OH}$ molecule to give protonated acetal, which proceeds to give expected acetal by deprotonation. Since we are using the alcohol as the reaction solvent, forward reaction is favored by multiple times.

OP's suggestion of protonation of $\ce{OH}$ group is legitimate, but reaction goes backward for a moment, but forward reaction favorably proceed to the end (Recall Fischer esterification). If you want back your aldehyde (or ketone), you can make backward reaction favored by using water as your solvent ($\ce{H+/H2O}$).


Le Chatelier's principle: Are there any exceptions?

There are many exceptions to Le Chatelier's Principle (LCP) so it can certainly be violated. Recently, I have read many articles pertaining to this violation so I believe I should be able to provide an informed answer. Most of the violations of the principle can be seen in gaseous systems. I would not call them "violations", rather, just cases where LCP does not work so well

One of the most classic cases for such an exception, as presented by Cheung (2004), is the addition of nitrogen gas, at constant temperature and pressure, to an equilibrium mixture containing $\ce {N2}$, $\ce{H2}$ and $\ce {NH3}$. The mechanical approach invoking LCP would quickly conclude that there would be an increase in the amount of the gaseous product due to an increase in the amount of one of the reactant gases, favouring the forward reaction. However, this approach fails to consider the instantaneous increase in volume of the system, which would cause the partial pressure of $\ce {H2 (g)}$ to decrease. Now, if we consider the principle again, taking into account both changes, we would realise that the principle would tell us that the equilibrium would shift in both directions. Thus, we see that there is a possibility that LCP cannot give a definite prediction if the number of moles of gaseous products in the balanced equation is not equal to the number of moles of gaseous reactants.

Another example cited by Cheung (2004) is the dimerisation of $\ce {NO2}$. Consider an equilibrium mixture of $\ce {NO2}$ and $\ce {N2O4}$ in a sealed container. When we raise the temperature, we may expect that the position of equilibrium to shift to the left since the backward reaction to form $\ce {NO2}$ is endothermic. However, there is also an increase in the total pressure of the system when the mixture is heated. Again, by LCP, the pressure increase would likely shift the position of equilibrium to the right to the side with less moles of gaseous molecules. This is another case where there are indeterminate predictions given by LCP.

References

Cheung, D. (2004). The Scientific Inadequacies of Le Chatelier's Principle. Hong Kong Science Teachers' Journal, 22(1), 35-43. Retrieved July 27, 2018, from https://www3.fed.cuhk.edu.hk/chemistry/files/LCP.pdf.

Cheung, D. (2009). The Adverse Effects of Le Chatelier's Principle on Teacher Understanding of Chemical Equilibrium. Journal of Chemical Education, 86, 514-518. doi:10.1021/ed086p514

Quilez-Pardo, J., & Solaz-Portoles, J. J. (1995). Students' and teachers' misapplication of le chatelier's principle: Implications for the teaching of chemical equilibrium. Journal of Research in Science Teaching, 32(9), 939-957. doi:10.1002/tea.3660320906


What are the prerequisites in hydrogen bonding?

Actually, theoretically speaking hydrogen bond is the attractive force between the hydrogen attached to an electronegative atom of one molecule and an electronegative atom of a different molecule. Usually the electronegative atom is oxygen, nitrogen, or fluorine, which has a partial negative charge. A hydrogen bond may occur between $\ce{H-O}$, $\ce{H-N}$, $\ce{H-F}$ (first two are more common). Apart from that, these hydrogen-bond attractions can occur between molecules (intermolecular) or within different parts of a single molecule (intramolecular)

See also the Wikipedia article about hydrogen bonding
A hydrogen atom attached to a relatively electronegative atom will play the role of the hydrogen bond donor An electronegative atom such as fluorine, oxygen, or nitrogen will be the hydrogen bond acceptor, irrespective of whether it is bonded to a hydrogen atom or not. An example of a hydrogen bond acceptor that does not have a hydrogen atom bonded to it is the oxygen atom in diethyl ether.

http://chemwiki.ucdavis.edu/@api/deki/files/4680/image115.png?revision=1


Benzene diazonium chloride is stable upto 5 degree Celcius, while in Lassaigne's test we need to heat the organic compound with sodium till red hot, before breaking it by plunging into cold water. All the nitrogen will be liberated before it finds a chance to combine with Na and form any NaCN, which is responsible for the test ultimately.


Can I use an acid other than sulfuric for hydrogen peroxide determination?

Don't use the organics!

This will severely mess up your titration because both permanganates as well as H2O2 will oxidize the carboxylic acids into explosive or corrosive things like peroxyacids (-C(O)OOH). Carboxylic acids are also usually weak, and the peroxy versions can be up to 1000 times weaker.

On the bright side, HCl is a strong acid, and you shouldn't have to worry about the production of chlorine gas; the only chlorine product should be MnCl2. However, since so much O2 Is being produced, your MnCl2 might contain a small amount chlorate impurities, so be wary of that if you plan to use the manganese chloride in the future.


When we talk about energy bands, we are talking about mixing of orbitals on different atoms (AOs) with similar energies. First, assume simple models. In the solid, the structure is very regular and all atoms have AOs with identical matching energies. This makes the number of orbitals available for mixing in solids very large, and mixing results in bands. In a liquid, on the other hand, structure is highly irregular. Orbitals on different atoms often have energy or overlap mismatches, but occassionally match and mix, and this results in much local mixing (say due to hydrogen bonding) but not extended delocalized orbitals. Because of energy mismatch you can also end up with bands in liquids but their origin (in this model) is different than in solids.
The difference between the origin of bands in liquids and solids is similar to the different cause of homogeneous and inhomogeneous broadening. In both cases you get bands (a distribution of energies), but in one case broadening involves delocalization over resonantly matched orbitals on different atoms/molecules and in the latter it involves variations in the local environment due to randomly fluctuating perturbations caused by neighboring molecules  ("the solvent"). Extensive orbital mixing and Pauli's exclusion principle results in homogeneous broadening in the first case. Motional jostling in the case of a liquid results in inhomogeneous broadening.
Energy levels of individual molecules in liquid water are "perturbed" by neighboring molecules. These perturbations are small compared to the ionization energy of the valence electrons. The stability conferred by an H-bond in water is in the order of 20 kJ/mol. Compare that to the ionization energy, ~1200 kJ/mol. Therefore electrons remain bound to local waters even as their energies are slightly perturbed by interactions with other molecules. H-bonds do contain covalent character, that is, involve sharing of electrons between two molecules (specifically between H-bond donor and acceptor atoms). However this does not imply electrons in liquid water are extensively delocalized beyond local bonds, certainly not on the spatial scale possible in solids.
As a note of interest ice under pressure exhibits metallic phases.
Note perturbations due to the solvent are evident in the difference between vibrational spectra obtained in different phases.


What happens to the shielding of the proton when an alcohol is oxidized?

The alcohol proton becomes an acid proton. Potassium permanganate oxidizes primary alcohols to carboxylic acids. In the process, the hydroxyl proton goes from 2-5 ppm to >10 ppm (generally). Therefore, since it is shifted to a higher ppm, it is being more deshielded.


Are alkenes with bulky groups attached (like the one shown in the picture) chiral or achiral?

Is the compound in the picture chiral or achiral?

The answer in practice is: it depends. You say that there would be free rotation along the phenyl–vinyl bond. However, every adjacent carbon atom — the 2- and 6-positions of the phenyl ring and the 1-position of the double bond — are substituted. For free rotation to occur, these substituents must pass each other. Since they are all bulky, though, free rotation is very effectively hindered. A planar rotamer with full π conjugation between the phenyl ring and the double bond is not populated at any extend which can be shown by the NMR shifts of the corresponding carbon atoms (and could also be shown if the methyl group on the double bond were a proton, because it’s shift would be vastly different from one in styryl conjugation).

Under low enough temperatures, the compound you have been given will indeed form to separable enantiomers due to the bromine atom which destroys the molecule’s symmetry. These are examples of axial chirality; if the chlorine atom is above the phenyl ring’s plane, it is (aR), if it is below, the compound is (aS). The two enantiomers can be separated, e.g. by chiral HPLC, and they will show optical activity.

At higher temperatures, thermal energy will be enough to overcome the rotation barrier and an optically active single enantiomer will racemise. This temperature is typically above room temperature for a system such as your’s but well below the temperature required for alkene (E/Z) isomerisation.

This concept is called atropisomerism and is also the reason why BINOL and BINAP are chiral.


Is there a relationship between viscosity and enthalpy of vaporization?

Yes. The phase envelope (or equilibrium curve) of most liquids have a similar shape and can be approximated, to a reasonable degree, by using the Clausius-Clapeyron equation:

$$
\Delta h_v=T(v^V-v^L)\frac{dP^s}{dT}
$$

which requires finding ${dP^s}/{dT}$ from e.g. a vapor-pressure-temperature correlation like the Antoine equation, and also a separate estimate for $\Delta v$ before $\Delta h_v$ can be obtained. This is an exact thermodynamic relationship, but is often simplified by using the ideal gas law for $v^V$ and neglecting the liquid volume $v^L$ i.e. $v^V \gg v^L$, giving

$$
\Delta h_v=\frac{RT^2}{P^s}\frac{dP^s}{dT}
$$

However, it is very important to note that the term $(v^V-v^L)$ becomes increasingly decisive as you go above 1 bar, where the assumption $v^V \gg v^L$ is no longer valid. In such a case, separate correlations for $v^L$ and $v^V$ are required before the correct curvature is obtained. $v^L$, for instance, can be calculated from e.g. a density correlation like the Rackett equation, and $v^v$ from an equation of state like Peng-Robinson.


How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry

In your example, the stereodescriptors R and S are already given. The remaining question is about the double bond which gives rise to cis-trans isomerism.
In the hierarchical Sequence Rules, we finally find

P-92.1.3.5 Sequence Rule 5
An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seq Trans’.

Source: Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)
Therefore, the configuration would be ‘E’ since ‘R’ precedes ‘S’.
However, the ‘E’ and ‘Z’ stereodescriptors do not distinguish between double bonds whose descriptors are reflection invariant from the geometrically enantiomorphic double bonds whose stereodescriptors are reflection variant.
Following the recommendations for enantiomorphic double bonds, the configuration is ‘seqTrans’.


Na2HPO4 is amphoteric: write the two reactions

$\ce{Na2HPO4}$ is dissociated in solution. The amphoteric species is actually the hydrogen phosphate dianion:

Reaction as an acid:
$$\ce{HPO4^{2-} + H2O <=> PO4^{3-} + H3O+}$$

Reaction as a base:
$$\ce{HPO4^{2-} + H2O <=> H2PO4- + OH-}$$


Waving a vodka soaked cotton ball near the sensor is sufficient to generate the required fumes. It was the circuit that was problematic - I did a subsequent testing with analog reader (voltmeter to be specific).


What are the alternatives to the Periodic Table of the Elements?

One periodic table that apparently physicists find useful is ADOMAH periodic table, for its usefulness is finding electron configuration of a particular element. More information at this guide. A disadvantage to this table, along with any new table configuration you have not experienced, will be the lack of knowledge in the trends on your new table. There are many spiral looking tables, but the trends such as the change in atomic radii, or electronegative characteristics, might not be as obvious, and you might have to find a guide or relearn old tricks to methods you would learn in your first year of chemistry. At least the ADOMAH Table is built in such a way that these trends should still be recognizable, although clearly not the same as the normal table you find in any high-school science classroom.

Of particular interest to me are dynamic periodic tables. These are tables that are basically computer applications which will change according to what you are looking for. One outstanding example is the widely popular ptable.com which will maintain all the common trends you probably know already, and dynamically show you various states of the elements when changing the temperature. I learned from this table that rhenium has the highest known boiling temperature, while I thought it was tungsten (although tungsten has the highest melting temperature of the transition metals). And better yet, carbon has the highest melting temperature of all elements, very cool table! The disadvantage is obviously that you cannot carry it around in your notebook without some flavor of a mobile computer and/or internet connection.

I am sure there are many other great tables still to be found, but I hope you have learned something new from my post.


Using Le Chatelier's Principle to explain why percent dissociation of acids/bases increase with dilution

Le Chatelier's Prinicple also says that the system will adjust itself based on a change in concentration.

Concentration of an acid is (normally) expressed in moles/liter. If you were to increase the amount of solvent (water) which is on the reactant side of the equation, you would be diluting the acid solution that is in equilibrium and causing the system to adjust accordingly.

As far as the last portion of your request is concerned I am slightly confused. Le Chatelier's Prinicple is built into the equilibrium expression. If I wanted to show how the system were to react based on the addition of more reactants/products I would just use the equilibrium expression for the system.
Hope this helps, D


At what frequency does a non-polar molecule acquire a dipole able to participate in London Dispersion forces?

According to Prof. Loren Williams of Georgia Tech, the fluctuations occur on the femtosecond $\left(10^{-15}\ \mathrm{s}\right)$ time scale:

  Figure 13 shows how fluctuating dipoles of liquid Xenon (or Helium or Neon, etc) are coupled. Darker blue indicates higher electron density. The fluctuations are correlated and are very fast, on the femtosecond ($10^{-15}$ second) timescale. Adjacent Xenon atoms experience electrostatic attraction from the transient dipoles. Two different representations of fluctuating dipoles are shown.

This corresponds to a frequency of $1\ \mathrm{PHz}$ (petahertz), which is about five hundred thousand times faster than most modern computer processors.

Another estimate of the time scale (albeit one devised by me and thus of potentially suspect validity), is the SI value of the atomic unit for time, derived by dividing the reduced Planck constant by the atomic unit for energy, the Hartree (values are rounded a bit):

$$
\left\{atomic\ time\right\} = \frac{\hbar}{E_\mathrm{h}} = \frac{1.05457\times 10^{-34}\ \mathrm{J\cdot s}}{4.3597\times 10^{-18}\ \mathrm{J}} = 2.41888\times 10^{-17}\ \mathrm{s} \approx 0.024\ \mathrm{fs} = 24\ \mathrm{as}
$$

That last value is $24$ attoseconds, which is a well and truly short span of time. For comparison, this corresponds to a frequency of $41\ \mathrm{PHz}$.

Whatever the exact value, I have to believe the fluctuations are fast enough so as to be impossible to measure directly. (I would be fascinated to learn otherwise, though!)  Thus, even if different atoms/molecules do actually have different frequencies of London dipole oscillation, we have no way of finding out these values.


For which pair of species is the difference in radii the greatest?

Your argument that $r_\ce{Li} > r_\ce{F}$ is valid only for the neutral atoms.Consider the lithium ion. Its electronic configuration is exactly similar to helium but the nuclear charge is greater by one unit, and hence is even smaller than the helium atom. On gaining an electron, fluorine now has the electronic configuration of neon with lesser nuclear charge and hence is larger than a neon atom. Similar, oxide ion, with even lesser nuclear charge is the largest. The decrease in radius of lithium on losing one electron (reduction by one entire shell), more than compensates for the difference in the radii of neutral lithium and fluorine. Finally, the order appears to be:-

$$\ce{Li+ < He < F < Li < F- < O^{2-} }$$


Just like @QuantumAMERICCINO has pointed out, it's probably anti addition.

Furthermore, you seem to wonder why a syn compound doesn't form. It all lies in the mechanism of bromination:

The double bond in cis-2-butene polarises the $\ce{Br-Br}$ bond in a nearby $\ce{Br2}$ molecule. A $\ce{Br+}$ ion attacks the double bond and forms a bromonium ion.

The bromonium ion makes the adjacent carbon atoms electrophilic. The bromide may attack the most electrophilic out of them. Here, however, they are equivalent.

You should now be able to understand how the mechanism works.


Why is 4-hydroxypyridine more acidic than benzoic acid?

The following are two resonance forms of 4-pyridone,

So, when 4-pyridone gets de-protonated, the negative charge on nitrogen gets delocalised on more electronegative atom i.e., oxygen, resulting in more stable resonance structure!
Note that, this explanation is identical to that one which is given when we compare $\mathrm pK_{\mathrm a}$ of benzoic acid and phenol.

Ref: Stability of 4-pyridone vs 4-pyridinol


What is the difference between enthalpy of reaction and standard enthalpy?

Yes, the standard enthalpy of reaction ($\Delta_\mathrm{r}H^\circ$) is the enthalpy change that occurs in a system when matter is transformed by a given chemical reaction, when all reactants and products are in their standard states. The only condition is that the participants have to be in their standard states, ie. usually at 1 bar pressure.

On the other hand, enthalpy of reaction ($\Delta_\mathrm{r}H$) is amount of energy absorbed or released at any condition. Considering your example,
$$ \ce{N2 + O2 -> 2NO} \qquad \tag{$H = \pu{+181kJ}$}$$
Here, $\Delta$H represents the amount of heat absorbed in the formation of 2 moles of $\ce{NO}$. For the formation of one mole of a substance from constituent elements, it is called enthalpy of formation($\Delta_\mathrm{f}H$)and if in standard states, standard enthalpy of formation.
$$ \ce{ 1/2 N2 + 1/2 O2 -> NO} \tag{$\Delta_\mathrm{f}H$}$$
As we can see, $\Delta_\mathrm{f}H$ = $\frac{1}{2}\Delta H$. Hence it should be $\Delta_\mathrm{f}H$= +90.5 kJ/mol.

Note: Unit of enthalpy is that of energy but unit of enthalpy for a reaction is of energy per unit mass or amount.


Have "molecular clusters" for azeotropes been identified?

No, in the absence of extra data, there is no reason to suppose that there is any vapor-phase cluster formation.

Cluster formation in the gas phase would demand very, very strong departures from ideal-gas behavior.  To the contrary, the ideal gas law is an excellent descriptor of gas phase mixtures of ethanol and water.

Check out a Wolfram Demonstration for the ethanol-water system.  It says:

  You can vary the pressure $P$ to any value between 50 kPa and 200 kPa  (i.e., low to moderate pressure so that the ideal gas-phase assumption holds).

If the ideal-gas assumption holds, then there is no significant structure formation in the vapor phase.  The "ideal" gas law describes negligibly small particles that have no attraction or repulsion to each other.  Structure formation means that molecules must be strongly attracted to each other in order for arrangement into a persistent structure to occur.

An "extended" form of Raoult's law that is valid for non-ideal vapor as well as non-ideal liquids, and thus is applicable to azeotropes, is

$y_i \phi_iP = x_i \gamma_i p_{i, \mathrm{sat}^{\star}}$

Here, $\phi_i$ is the fugacity coefficient and takes into account vapor-phase non-idealities (i.e. deviations from the ideal gas law), and $\gamma_i$ is an activity coefficient and  takes into account liquid-phase non-idealities.

For many, many systems of interest, $\gamma_i$ is the driver of non-ideality, including azeotropic behavior.  Fugacity coefficients $\phi_i$ are negligible (except at enormous pressures) a much higher percentage of the time than activity coefficients $\gamma_i$.  This is because liquid phases are often far more dense than vapor phases, meaning that intermolecular forces govern behavior to a much stronger degree than in vapors.


Weight percent is essentially the same as (I arbitrarily picked grams as the mass unit):

$$
\%~\mathrm{w/w~X} = \mathrm{g~X\over g~solution}
$$

In order to get to the $\mathrm{mol\over L}$ units of molarity, you have to convert by multiplying by the solution density, $\rho$, and dividing by the molecular weight of $\mathrm X$, $M_\mathrm{X}$:

$$
\mathrm{g~X\over g~solution} \times \stackrel{\left(1\over M_\mathrm X\right)}{\mathrm{mol~X\over g~X}}
 \times \stackrel{\left(\rho\right)}{\mathrm{g~solution\over L~solution}}
 = \mathrm{mol~X\over L~solution}
$$

One key thing to remember is that $\rho$ is the density of the solution, not of the pure solvent. If you're working with dilute solutions (below $\sim\!100~\mathrm{mM}$ for aqueous) these two densities will be similar, but in concentrated solutions they will diverge.


Determination of structure by 13C and 1H NMR spectroscopy

This is a tricky little problem, given there is an element of symmetry to consider. As mentioned in the comments, once you come up with a proposed structure, and your one is not a bad first attempt, you should go through and test the spectrum against what you would predict from your structure. Unfortunately, in your suggestion, there are 4 non-equivalent $\ce{-CH3}$ environments, and the spectrum only shows two, so we can eliminate that structure as a possibility. But don't be disheartened — that is the process of structure elucidation! Propose a stucture, test it, refine or eliminate it. And then start again.

Looking at the $\ce{^1H}$ spectrum, we can see:

$\ce{-CH3}$
$\ce{-CH2}$
$\ce{-CH3}$
$\ce{-CH}$
$\ce{-CH}$

We can construct the $\ce{-CH2CH3}$ spin system based on coupling, and the chemical shifts of the two downfield protons indicate a double bond (and, using some old tricks of the trade, looks to be cis — the splitting here is much less than about $14~\mathrm{Hz}$, which would be the expected splitting between the outside lines of the methyl triplet). The $\ce{CH3}$ at about $3.3$ is in the general ball park for a methyl ether, I'd say.

That gives us 10 protons, but the molecular formula says $\ce{H20}$ … so we must have symmetry in there somewhere which gives us:

2 $\ce{-CH2CH3}$
2 $\ce{-CH3}$ (probably $\ce{-OCH3}$)
2 $\ce{-CH=CH-{}}$

So, looking at the carbon spectrum:
I count 7 carbon peaks.
From the proton information, we have 5 $\ce{C}$ environments, each appearing twice, which gives a total of 10 carbons. The other 2 carbon environments occur once only in the molecule.

From DEPT experiments (which nobody in the real world should ever run when they can instead run a phase-edited HSQC instead) we can see we have

$\ce{CH3}$ at $9.9$ and $51.4$
$\ce{CH2}$ at $38.0$
$\ce{CH}$ $132.8$ and $133.2$
$\ce{C}$ at $47$ and $110$.

We know (or should know) what the non-quaternary carbon cantres all are. We worked that out from the $\ce{^1H}$ spectrum. So let's concentrate on the quaternaries. The peak at $110$ is interesting, and is typical of a carbon centre attached to 2 oxygens. Many sugars have carbons that come around this area. The other quaternary at about $47$ is typical of an aliphatic quaternary; not overly exciting.

So somehow we need to construct a molecule that has 2 non-equivalent quaternary centres that have a plane of symmetry through them, with the $\ce{-CH2CH3, -OCH3}$ and $\ce{-CH=CH-{}}$ attached. Each of these three spin systems must be isolated, otherwise we would see coupling to the vinylic protons.

As this is clearly a homework question, I'll leave it to you to piece together, and ask further questions if you need, but my answer gives the exact same predicted spectra in ChemDraw as your spectra.


What method do I use to synthesize propyl-2-(2-methylpropyl)-ether and why?

There are two basic choices here if you are trying to do this on a lab scale. The alcohol plus alkene with catalytic acid is not an approach I would consider unless I needed a lot of this. t-BuOH plus acid is a recipe for isobutylene. Sodium or potassium t-butoxide is commercially available and reacted with an excess of iPr-I should give a decent yield of the ether.

The other approach is to use t-Butyl 2,2,2-trichloroacetimidate compound details here - a general reagent for the formation of t-Butyl ethers originally reported by Armstrong et al, Tet. Letts 29 2483 (1988)


Why is substitution of halogens to alkanes a chain reaction?

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


What do numbers in parenthesis after elements respresent?

These are called Miller indices and they identify a particular surface of a crystal.  (You can form different faces by cutting the crystal in different ways.)  When you see something like "Mo(110)" it means "the 110 surface of a crystal of molybdenum."


What do the numbers in the preferred IUPAC name for ascorbic acid mean?

The numbering of (R)-5-[(S)-1,2-dihydroxyethyl]-3,4-dihydroxyfuran-2(5H)-one has four parts:

1) Numbering of the furan ring

2) Numbering of the ethyl side chain

3) Stereodescriptors R and S

4) Italicized element symbols

The italic element symbol H denotes indicated or added hydrogen.


Why is it considered acid rain with pH <5.6?

The $\mathrm{pH}$ of pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{CO2}}= 10^{-3.5}\ \mathrm{atm}$) can be calculated as follows.

$$[\ce{H2CO3^*}]=K_\mathrm H\cdot p_{\ce{CO2}}$$

where $[\ce{H2CO3^*}]$ is the total analytical concentration of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$, and
$K_\mathrm H= 3.39\times10^{-2}\ \mathrm{mol\ l^{-1}\ atm^{-1}}$ is Henry's law constant for $\ce{CO2}$.

$$\begin{align}
\log[\ce{H2CO3^*}]&=\log K_\mathrm H+\log p_{\ce{CO2}}\\
&=-1.5-3.5\\
&=-5.0
\end{align}$$

The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.3$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction
$$\ce{H2O + CO2(aq) <=> H2CO3}$$
and the protolysis of true $\ce{H2CO3}$
$$\ce{H2CO3 <=> H+ + HCO3-}$$
For a weak acid
$$\begin{align}
\log[\ce{H+}]&\approx\frac12\left(\log K_\mathrm a+\log[\ce{H2CO3^*}]\right)\\
&=\frac12\left(-6.3-5.0\right)\\
&=-5.65\\
\mathrm{pH}&=5.65
\end{align}$$

Thus, pure rain in equilibrium with the atmosphere has about $\mathrm{pH}=5.65$. Any acid rain with lower $\mathrm{pH}$ would be caused by additional acids.


Above are the resonance forms of m-cresol after deprotonation. As you can see, the negative charge resonates around the aryl ring, but never resides on the carbon directly bonded to the methyl group at the meta position.

Thus, the hyperconjugative effect from the meta methyl group does not affect the acidity of cresol in this isomer. Notice how the destabilising hyperconjugation effect would destabilise the resonating negative charge, if the methyl group was at the ortho- or para- position.


How does density of alkali metals increase down the group?

$Density=mass/volume$

1) If mass is increasing and volume is decreasing, then density (mass/volume) will increase.

2)If mass is decreasing and volume is increasing simultaneously, then the density (mass/volume) will decrease.

*3)If both mass and volume are increasing, then we need to check which one of them is increasing at a faster rate(since both are contradictory factors)

a) If mass is increasing at a faster rate than volume, then density will increase.

b) If volume is increasing at a faster rate than mass (i.e denominator in mass/volume is increasing making the overall fraction smaller), then density decreases.

You can apply same logic when both volume and mass are decreasing simultaneously.

Generally, we see that in alkali metals the rate of increase of mass is greater than rate of increase of volume, therefore the density increases down the group.

Also, note that there is no definite reason for why mass is increasing at a faster rate than volume. So asking "why" does it happen isn't a great question.


I hate to break it you, but they're hiding the math from you that is really going on behind the scenes.

It is very common in situations where the students are not expected to know calculus, that wherever a derivative would be needed, to replace, for example, the $dx$ with a $\Delta x$. This is because for many simple physical systems, taking the derivative results in simply measuring the average change in a property with respect to some other property.

For instance, consider a system where you know the position of some particle and how this position changes with time. If you want to know the velocity of that particle at any given time, you would take the derivative with respect to time, $$v=\frac{dx}{dt}$$but, in many cases you might be only interested in the average velocity and could thus skip the derivative part and write that $$v=\frac{\Delta x}{\Delta t}=\frac{x_2-x_1}{t_2-t_1}$$

But, and now we're getting to your question, let's take it one step further and say we wanted to know the acceleration of that particle. This would be the derivative of velocity which is also written as the second derivative of position with respect to time. Thus, we would write $$a=\frac{d^2x}{dt^2}$$Notice how similar this looks to your question about $\Delta^2$ if you simply replace the $d$ with $\Delta$.

Again, if we decided we were only interested in the average acceleration, then we would write $$a=\frac{\Delta^2 x}{\Delta t}$$
The reason why they say that $$\frac{\Delta(\frac{\Delta a}{\Delta b})}{\Delta b}=\frac{\Delta^2 a}{\Delta b^2}$$is again because that is true of derivatives. This would be written as $$\frac{d}{dt}(\frac{dx}{dt})=\frac{d^2x}{dt^2}=a$$

So, to summarize, the symbol $\Delta$ is often used in situations where you care about a change from one state to another (as with a change in energy) or when you only care about the average of some quantity.

As to why they are able to write$$\frac{\Delta^2 a}{\Delta b^2}=\frac{a_3-a_2}{\Delta b}$$ I'm not certain about. I would have to know what this dealt with specifically. My guess is that the quantity $a_3$ is somehow related to the second derivative of $a$ with respect to $b$ and they did some algebra behind the scenes.

Hope that helps.


VSEPR theory, chemical bond and quantum mechanics

VSEPR is a simple and generalised approach based (mainly) on empirical observation. It is a great theory in predicting, in a first order approximation, the geometrical shape of molecules. That is a lot harder with other methods. It is of course based on a physical foundation, not only empiricism. You might want to explore electron-domains and the Pauli principle in this context.

There have been a lot of post-rationalisations that cause more harm than good, because they clearly go beyond the limitations of the theory. One of the most miss-taught ones is the involvement of d-orbitals in hybridisation schemes, but that comes from the lack of understanding of some instructors, and its confusion with valence bond theory.
One of the most prominent examples where it (basically) fails is outlined in my answer here: Are the lone pairs in water equivalent?  At this point it should be noted, that the general topology of the electron density is quite well reproduced.

With that in mind, a reconciliation of this empirical approach with quantum theory to understand bonding is dangerous, if not futile. It should be used for providing a reasonable guess for a molecular structure and its principle explanation. Anything beyond that might lead to wrong conclusions. It cannot, in any way, be used to generate a guess for a wave function, because it is not based on it.

You will have to use another method for that. Often confused with VSEPR is the valence bond theory (VBT). It cannot be stressed enough, that the two are completely independent (even at its crudest level). There has been a lot of development in that field, and what is often taught as VBT can only be classified as a very crude approximation. It is in principle an exact theory, but carried out at the theoretical limit is not as easily understood as that what is taught commonly. (Just have a look at resonance, and its misconceptions as are outlined here: What is resonance, and are resonance structures real?)
Another way to describe bonding is molecular orbital theory (MOT), which already comes with the necessary properties of electron delocalisation. Unfortunately, this theory is more difficult to get started in, and does not provide an easy picture to follow. The good thing about this is, that it doesn't get much more complicated at its theoretical limit.
The two approaches are at their respective theoretical limit (VBT decribes electron correlation via resonance structures, MOT need multiple determinants for this) equivalent.

Understanding bonding is not easy and it is by far not without controversy. Even at approximate VBT or MOT there can be many misconceptions, and incorrect deductions, conclusions, rationalisations. For everything in the "grey" areas, there are opinions, interpretations, and opinions about interpretations.
In any and every case one should always be aware of the limitations of the model used. One should also be critical about the found results. One should always expect everything to be a lot more complicated than expected (point in case: CO2).
If you keep all of that in mind: VSEPR is an awesome model system.


I'm assuming you only need to liquefy ice to drain a septic/grey water tank. Calcium chloride (sold as damp rid) will melt ice down to $-56^\circ C$.  Just add about 10 Pounds to the tank wait a few hours (perhaps even drive a little to slosh it) and drain.


SN2 reaction with alkyl bromide and pentanenitrile?

It appears that you are having some difficulty with this synthesis. The suggestion in the Comments that you employ lithium diisopropylamide (LDA) to generate the anion of a nitrile is a wise one. The anion is usually formed a -78oC in tetrahydrofuran (THF). The bromide is added and the reaction mixture is allowed to warm to ambient temperature. There are two possible bonds that can be made in what is termed a retrosynthetic analysis. One by method A and another by method B. The bonds formed are shown in red in structures 2. Method A is better than method B because an SN2 reaction is being conducted on a primary halide, ethyl bromide. Method B requires an SN2 reaction on a secondary halide, 2-bromopropane, a process that is hindered and in competition with an E2 elimination that would produce propene 4 and the starting nitrile 3. Nitrile 2 must be a racemate given its method of formation. It is comprised of equal amounts of the two enantiomers, (R)-2 and (S)-2. By the methods employed here, a single enantiomer cannot be formed. Thus, a "normal bond" (not dashed nor bold) is shown in nitriles 2.


The Wikipedia page for each element is probably the easiest place to start with. If you want some significantly more in-depth data, the USGS publishes free-access mineral commodity summaries on many (but not all) elements, including information about their origin, abundance and applications. Other good sources are books, especially descriptive inorganic chemistry books, such as a few of John Emsley's publications including Nature's Building Blocks and The Elements, as well as many other college-level inorganic chemistry texts.


Well, 2 cannot be equal to three to start off, because a double bond it way shorter than a single bond.

The Answer to this is D, 2 has the shortest bond because it is a double bond, next 4 is a benzene ring, which is a resonance structure, which means that all the carbons technically have partial bonds between each other, or you can imagine the double bonds to be continuously rotating between the different carbons. Lastly 3 and 1 both have normal single bonds, so would have an equal bond length.

Therefore, since the double bond is shortest, the resonance between a single and double is a bit less short, but not as long as a single bond, and a single bond is the longest, the answer it 2(double bond)<4(between double and single bond)<3&1(which are both single bonds)


Is it possible to form an 8 carbon ring using a Diels-Alder reaction?

You can obviously use Diels-Alder Reaction for the synthesis of the desired product but not so directly. But Diels-Alder will give an optimal yield when the ring formed is Six-Membered. After the formation of that Six-Membered product, you have to increase the no. of carbon atoms in the carbon-chain by some way, and the yield of your product will be also better if you no. of steps as less as possible.

I will not suggest you to directly add 1,3-butadiene to acetylene because there will be a chance of aromatisation, and then you cannot do anything, as to break the aromatic rings are highly unfeasible. Instead I can suggest you the following steps:

First you partially reduce the acetylene to ethene, and then perform Diels-Alder reaction with butadiene. You will get cyclohexene.
In the second step, you perform ozonolysis, and then to the resulting compound, perform  Wittig reaction with excess reagents. You will get octa-1,7-diene.
Lastly, you can perform hydroboration, oxidation to get less substituted diol, and then oxidise that to acid by any strong oxidising agent like $\ce{KMnO4}$ or $\ce{K2Cr2O7}$ in slightly acidic medium.

Thus, you will get your desired product. The path of reaction is shown below: