Why should ether O be named as OXA in a bicyclo ring when it is a substituent?

The suggested name 3-methoxy-6-methyl-7-oxabicyclo[2.2.1]hept-5-en-2-one is correct. Indeed, the ketone is the principal characteristic group in this case; however, this is not the reason why the other oxygen is named as ‘oxa’. Already the unsubstituted parent structure is named 7-oxabicyclo[2.2.1]hept-2-ene.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), such bicyclic heterocycles are named using skeletal replacement (‘a’) nomenclature (a system in which a heteroatom replaces skeletal atoms of a parent hydride).

  P-31.1.4.4 Bi- and polycyclic von Baeyer heterocycles named by skeletal replacement (‘a’) nomenclature. In heterocyclic compounds formed using skeletal replacement (‘a’) nomenclature, low locants are assigned to heteroatoms, in accord with the fixed numbering of the system, then to unsaturated sites.

Nondetachable prefixes, called replacement (‘a’) prefixes representing the element being introduced, are used to designate the replacing skeletal atoms with their standard bonding number, for example ‘aza’ (N), ‘oxa’ (O), or ‘thia’ (S).


Which is easier, the nitration of toluene or the nitration of mesitylene (1,3,5-trimethylbenzene)?

The kinetics of aromatic nitration depend on the nitrating method, which also bear directly or indirectly on the subject of nitration. For example, aromatic nitration displays second-order kinetics in sulphuric acid (when used in nitrating mixture), and first-order kinetics in nitric acid. Aromatic nitration also displays first-order kinetics in either nitromethane or acetic acid (when used as a solvent) with nitric acid in constant excess. It exhibits zeroth-order kinetics for sufficiently reactive aromatic compounds, and first-order kinetics for sufficiently unreactive compounds used for nitration (Ref.1). The reference predicts that:

  The numerical results vary with the conditions, but are not very sensitive to them; so that the following statements may be taken as approximately true over a considerable range of temperatures and nitrating media. Toluene is attacked 24 times faster than benzene, fluoro- and iodo-benzene 6 times more slowly than benzene, chloro- and bromo-benzene 30 times more slowly than benzene, and ethyl benzoate 300 times more slowly than benzene. Qualitatively we know that benzenesulphonic acid is attacked still more slowly, and nitrobenzene even more slowly. Thus the relative effects of substituents in activating or deactivating the benzene nucleus towards this type of attack is expressed by the following sequence:
  $$\ce{Me > (H) > F, I > Cl, Br > CO2Et > SO3H > NO2}$$

Thus, it is fair to conclude that the nitration of mesitylene is faster than that of toluene regardless of the steric effects enforced by neighboring 1,3-dimethyl functions. That conclusion is confirmed by the relative rates listed in Ref.1 & 2 where it is the case in all conditions considered (obs. rates are at $\pu{25 ^{\circ}C}$):

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\textrm{Compound} & 7.5\% \: \textrm{aq.} & 15\% \: \textrm{aq.}  & 61.01\% \: \textrm{aq.}  & 68.3\% \: \textrm{aq.}  & \textrm{Estimated}  \\
\textrm{nitrated} & \textrm{Sulfolane} & \ce{CH3NO2} & \ce{HClO4} & \ce{H2SO4} & \textrm{Rel. Rate}  \\ \hline
 \textrm{Benzene} & 1 & 1 & 1  & 1 & 1 \\ \hline
\textrm{Toluene} & 20 & 25 & 19  & 17 & 23 \\ \hline
\textrm{p-Xylene} & 114 & 130 & 84 & 38 & 50 \\ \hline
\textrm{m-Xylene} & 100 & 146 & - & 38 & 400 \\ \hline
\textrm{Mesitylene} & 350 & 400 & 78  & 36 & 16000 \\ \hline
\end{array}
$$
Estimated relative rates are from reference 3 (the authors have calculated from the observed value for toluene and the known isomer ratios by assuming the additivity principle).

References:

E. D. Hughes, C. K. Ingold, R. I. Reed, “493. Kinetics and mechanism of aromatic nitration. Part II. Nitration by the nitronium ion, $\ce{NO2^+}$, derived from nitric acid,” J. Chem. Soc. 1950, 2400–2440 (DOI: 10.1039/JR9500002400).
J. G. Hoggett, R. B. Moodie, J. R. Penton, K. chofield, In Nitration and aromatic reactivity; Cambridge University Press: London, United Kingdom, 1971, “Chapter 3: Nitrating Systems: B. Inert organic solvents,” pp. 32–49.
R. G. Coombes, R. B. Moodie, K. chofield, “Electrophilic aromatic substitution. Part I. The nitration of some reactive aromatic compounds in concentrated sulphuric and perchloric acids,” Journal of the Chemical Society B: Physical Organic 1968, 800–804 (DOI: 10.1039/J29680000800).


Best oxidizing and reducing agents: Na, Zn^2+, Ba, Ba^2+, and Ag?

It is a good idea to recap the the definitions of oxidation and reduction when approaching this kind of question, and also what it means to be an oxidising or a reducing agent:

Oxidising is defined as the loss of electrons, reduction as the gain.
An oxidising agent is good at oxidising other compounds while being reduced itself; and vice-versa.

Take another quick look at your list; all you have are neutral elements or the most common cations of said elements. If you want to oxidise or reduce, then we would always be going back and forth between either. Therefore, only sodium, barium and silver are potential reducing agents on your list (being oxidiseable themselves) while only zinc(II) and barium(II) can be oxidising agents. Here are the relevant equations:

$$\begin{align}\textbf{Reducing agents:}\\
&&\ce{Na &-> Na+ + e-}\\
&&\ce{Ag &-> Ag+ + e-}\\
&&\ce{Ba &-> Ba^2+ + 2 e-}\\
\textbf{Oxidising agents:}\\
&&\ce{Ba^2+ + 2 e- &-> Ba}\\
&&\ce{Zn^2+ + 2 e- &-> Zn}\end{align}$$

Using your table or our chemical intuition, we can then decide which reduction or oxidation is the easiest. The best oxidising agent without doubt is zinc(II) as it is much more noble than barium (although still not a noble metal). The best reducing agent is a slightly more tricky call without data, but barium(0) being further down in the periodic table allows us to assume it to be stronger. Luckily, this is supported by the data.


What's the difference between precipitate and turbidity?

My understanding is that turbidity is an adjective describing the presence or degree of particulate matter suspended in a liquid (see turbidimetry), whereas a precipitate is a particulate that can result in turbidity.

The turbidity generated by a precipitate is a function of its ability to form a stable suspension. If the precipitate simply drops to the bottom of a container or readily flocculates, the resultant turbidity will be low. The turbidity of a suspension is also going to be a function of the solvent and the nature and concentration of dissolved ions - a dramatic example of this is the rapid clarification of muddy freshwater at river deltas mixing with salt water.

If the word turbidity is being used as a noun, it is probably referring to a light-occluding suspension.


Why don't molecular orbitals change when electrons are added or removed from a molecule?

The molecular orbitals do change when electrons are removed or added. But what does not change is the relative order of these orbitals with respect to other orbitals. If a given orbital has a higher energy than another one, it will remain so if you add one electron or it you remove one electron, whatever the absolute values of these energies.


Calculating heat of combustion via calorimetry

You have 60 g of $\ce{C60}$ (which has a molecular weight of 720.6), so that means you'll need to multiply your answer by 720.6/60 which (cleverly) turns out to be 12.01. Another way of saying it is that you have calculated the heat of combustion for 1/12.01 moles of $\ce{C60}$, so you'll need to multiply your answer by the inverse in order to get the desired (per mole) quantity.

Using your notation, numbers, and noting that a change in one degree Celsius is equivalent to a change in one kelvin:

$$\Delta E_{\rm molar} = -12.01\;\mathrm{mol} \cdot 300\;{\mathrm g} \cdot 4.18\;{\mathrm J} \cdot {\mathrm g}^{-1}\, \cdot {\mathrm K}^{-1}\, \cdot 10\;{\mathrm K} = -150.6054\;\mathrm{kJ}\cdot\mathrm{mol}^{-1}$$


Are heterolytic dissociations barrier-less or do they have an activation energy?

In my theoretical research I have often seen that, indeed, there is no barrier for such reactions. In modeling the kinetics you take the bond dissociation energy as the forward barrier and assume a zero barrier going backward. This will also be true for radical recombination reactions.


Intuition for why effective nuclear charge increases down a column

Initially, let's take the electrons as probability density fields instead of classical particles.

The increase on the number of protons in the nucleus increases the electric force on all electrons, but the core electrons suffer higher attraction due to higher proximity and lighter electronic blindage, so their probability density field is forced nearer the nucleus.

Reminding of Electric Force:
$$\mathbf{F}_\text{E} = \mathbf{k}_\text{C} \frac{q_1q_2}{\mathbf{r}^{2}}$$

Consequently, the outer electrons will suffer lower blindage force from these core electrons, parallelly to the nucleus attraction, allowing them to get closer to the nucleus as well.

As effect, of course, the outer electrons also force the inner electrons towards the nucleus, but their probability densities reach a minimum repulsion energy in such a distance that electrons on the same level are forced closer to the nucleus than on the atomic element above, going down a column in the periodic table.


What makes some atoms/compounds reactive and some noble?

Unpaired electrons and how tightly these electrons are bound to their respective nuclei.

  Why do reactions happen at all? What is the fundamental truth behind a reaction?

Because atoms are more stable together than they are apart, and systems will tend towards their lowest energy state.

  But why NaCl?

This is really a matter of ratios, which is the basis of chemistry. Different elements share their electrons in different ways under different circumstances. Some can donate their electrons ($\ce{Na}$, $\ce{Al}$, $\ce{Au}$), some accept electrons ($\ce{Cl}$, $\ce{O}$, $\ce{N}$), and others are somewhere in between ($\ce{C}$, etc.). Sodium and chlorine are very consistent with how they share electrons. Sodium has 1 valence electron, chlorine lacks 1 valence electron, and they combine in a one-to-one ratio. The formula unit of their salt is $\ce{NaCl}$.

When considering the salt of aluminum and chlorine, ask how many valence electrons does each atom have/need. I think you might begin to see the pattern.

$[1]$ Bond energy diagram


You are right about the rarity of amide to ester conversion. Amide to ester conversion has remained a challenging and underdeveloped synthetic transformation. It is often due to the fact that amides are sufficiently stable because of its resonance-stabilized $\ce{C–N}$ group that imparts a planar geometry to amides and as such, esterification is difficult and requires the use of harsh acidic or basic conditions, while employing a large excess of nucleophile (e.g. using alcohol nucleophile as solvent). In contrast, esters are generally reactive under a variety of mild condition and therefore the opposite conversion of ester to amides is quite normal.
The most promised method of achieving amide to ester conversions is Keck’s methylation/hydrolysis although this method is limited to the synthesis of methyl esters and thus researchers has developed some method to expand this conversion's feasibility and efficiency.

Hie et. al. has achieved amide to ester synthesis using nickel complex catalyst. They studied the conversion of various aryl and heteraryl amides. One studied reaction is conversion of benzamide to methyl benzoate using $\pu{10 mol{\%}}$ $\ce{Ni(cod)2}$, $\pu{10 mol{\%}}$ SIPr, 2 equivalents of methanol, and toluene as solvent at $\pu{110 °C}$ for 12 hours. (Ref. 1)

An efficient two step transformation of 1° and 2° amides to methyl and ethyl esters has been developed using trimethyl- and triethyloxonium tetrafluoroborates, and dilute acid. This method was applied to 1-benzyl-4-phenylamino-4-piperidinecarboxamide, a precursor in the synthesis of carfentanil, to produce the methyl ester in $\pu{60{\%}}$ yield and the ethyl ester in $\pu{80{\%}}$ yield. (Ref. 2 and 3)

Conversion of amides to ester has been achieved using $\ce{CeO2}$ catalyst and HY zeolites. A variety of metal oxides as catalyst and ester yields has been discussed. (Ref. 4)

References

Hie, L., Fine Nathel, N., Shah, T. et al. Conversion of amides to
esters by the nickel-catalysed activation of amide C–N bonds. Nature
524, 79–83 (2015). DOI: 10.1038/nature14615 (PDF link)
Amide to ester conversion: a practical route to the carfentanil class of analgetics
Douglass F. Taber and Mohammad Rahimizadeh, The Journal of Organic Chemistry, 1992, 57 (14), 4037-4038, DOI: 10.1021/jo00040a061
Anthony J. Kiessling & Cynthia K. McClure (1997) The Conversion of Amides to Esters with Meerwein'S Reagent. Application to the Synthesis of a Carfentanil Precursor., Synthetic Communications, 27:5, 923-937, DOI: 10.1080/00397919708004212
Conversion of Amides to Esters via Selective Cleavage of Amide C–N Bonds over a CeO2 Catalyst by Shimizu et. al. (PDF)


I can't comment on spreadsheet-like tools.

But I'm frequently using Open Babel, which is free (as in free speech and free beer).

If you're not so much into using command-line tools or using programmes via their bindings to programming languages, a GUI for Open Babel is available too, but I've never used it.

  Generates 2D structure for the entered compounds (based on name or SMILES)

Names might be tough in general, but SMILES (Simplified Molecular Input Line Entry System) can be processed with Open Babel when the identifier is read from a file. Output to large number of chemical formats and image formats is possible.

obabel -ismi stuff.smi -omol -O stuff.mol

would convert a SMILES string to a mol file.

Converting a collection of SMILES to MOL and writing them to a SDF file is not a big thing:

#!/usr/bin/env python

import os
import pybel

# dictionary of SMILES taken from chemspider.com
insecticides = {
    'parathion' : 'CCOP(=S)(OCC)OC1=CC=C(C=C1)[N+](=O)[O-]',
    'disulfaton' : 'CCOP(=S)(OCC)SCCSCC',
    'permethrin' : 'C1(C(C1C(=O)OCC2=CC(=CC=C2)OC3=CC=CC=C3)C=C(Cl)Cl)C',
    'furathiocarb' : 'CCCCOC(=O)N(C)SN(C)C(=O)OC1=CC=CC2=C1OC(C2)(C)C',
    'methomyl' : 'CC(=NOC(=O)NC)SC'
}

# write dict items to multi-component SDF file
outdir = '/home/klaus/workspace/chemistry/stackexchange/similarity/'
outname = 'insecticides.sdf'
outpath = os.path.join(outdir, outname)
outfmt = 'sdf'
datafile = pybel.Outputfile(outfmt, outpath, overwrite = True)

for name, smiles in insecticides.iteritems():
    mol = pybel.readstring('smi', smiles)
    mol.title = name
    datafile.write(mol)

datafile.close()

EDIT

  Easy to search and create customized reports (such as similarity search)

For the substructure search, I used the Drug-likeness NCI subset from ligand.info. The unpacked SDF file (445.2 MB) has 193,323 entries on 14,757,262 lines (wc -l).

From the SDF file, I first generated a binary fastsearch file using

obabel data.sdf -ofs -O data.fs

Searches are based on SMILES and SMARTS patterns.

I thought that 2-aminobenzothiazole might be an interesting fragment to search for and with a bit of fiddling, I wrote the suitable SMARTS pattern: c12ccccc1nc(N)s2

If in doubt about the correct representation, it is always a good idea to convert the pattern to an image file here: obabel -:"c12ccccc1nc(N)s2" -opng -O pattern.png.

The search was performed using:

obabel -ifs data.fs -s"c12ccccc1nc(N)s2" -osvg -O results.svg -xd -xc5 -xA -xC

and generated a SVG table with 233 entries.


Difference between continuously overlapping p orbitals and conjugated pi bonds

Interesting question!

To have continuously overlapping p-orbitals, the p-orbitals must be arranged in a cyclic loop.  The p-orbitals in benzene are continuously overlapping.  The p-orbitals in benzene are also conjugated pi bonds.

image source

In buta-1,3-diene the p-orbitals form conjugated pi bonds, however, they are not continuously overlapping - there are two "end" carbons that are not connected with each other, there is no continuous loop.

image source

Let's take it a step further:

Continuously overlapping p-orbitals are not always aromatic.

Example 1: Cyclobuta-1,3-diene (the square planar structure, $\ce{D_{4h}}$) has 4 continuously overlapping p-orbitals but it is not aromatic, it is antiaromatic.  For a continuously overlapping system of p-orbitals to be aromatic, it must contain 4n+2 pi electrons.

image source

Example 2: Have you heard of the Möbius strip? The mathematical (topological) concept has application in chemistry.  Imagine a large cyclic loop of p-orbitals where each p-orbital is twisted a few degrees from the preceeding p-orbital.  If the loop is large enough, then by the time we come back to the first p-orbital we have twisted the last p-orbital by 180°; there is a phase inversion (node) between the first and last p-orbital.  In a Möbius system the p-orbitals are continuous, but not continuously overlapping!  Möbius systems are aromatic if they contain 4n pi electrons.

image source

Continuous pi systems with an even number (0, 2, 4...) of nodes (cyclobutadiene, benzene, etc) are called Hückel systems in order to differentiate then from Möbius systems where there are an odd number of nodes (1, 3, 5...).

Lest you think the Möbius concept is purely theoretical take a look at these two molecules

image source

Isn't this what makes chemistry fascinating!


how do electromeric / inductive effect theories compare with resonance

The concepts of electromeric and inductive effects are not obsolete.

Only the term "electromeric" is obsolete and has been subsumed in the terms "mesomeric effect" (+M and -M) and "resonance effect" (see iupac source). The term "isomeric effect" is still in use and I learned it not a decade ago along with "mesomeric" and "resonance".

Amusingly, your SE question is my first result in google for "Marxist chemistry". From the little I have just read in this book, Marxist chemistry had apparently more to do with refusal of the concept of resonance than anything else. So... maybe? Maybe not.


Is this asking for the equivalence point?

If you were titrating acetic acid with a base, then the equivalence point would be where as much base has been added (i.e. moles of base vs moles acid) to the solution as there was acetic acid. This is essentially the point where all the acetic acid has been neutralized.

In general the "pKa" is the term used to define the point at which the protonated and unprotonated forms are equal. The pKa is from the negative log of the acid dissociation constant as given by the reaction:
$$\ce{HA <=> H+ + A-}$$
and the equation: $$ K_\mathrm{a} = \dfrac{\ce{[H+] [A- ]}}{\ce{[HA]}}$$

where $\ce{[H+] = [A- ]}$ then:
$$K_\mathrm{a} = \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$

and $$ \ce{pKa} = -\log \ce{Ka} = -\log \dfrac{\ce{[H+]^2}}{\ce{[HA]}}$$


How the equation that describes evolution of alkyl nitrates has been derived?

You have to quote a bit more of the paper for context, namely the reaction scheme. In the paper, it is given as

$$\ce{RH ->[\beta k_\mathrm{A}] RONO2 ->[k_\mathrm{B}] P}$$

The first thing to notice is that the concentration of $\ce{RH}$ as a function of time is given by the first-order integrated rate law

$$[\ce{RH}]_t = [\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)$$

and therefore equation (11) becomes

$$\frac{\mathrm{d}[\ce{RONO2}]}{\mathrm{d}t} = \beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)- k_\mathrm{B}[\ce{RONO2}]$$

This is a ordinary differential equation and there are many ways of solving them (this is hardly the place to discuss ODEs; consult a maths book if you are interested). In this case, the way I did it is to multiply throughout by $\exp(k_\mathrm{B}t)$. I am going to use $c$ to represent $[\ce{RONO2}]$ to simplify the equation a little bit:

$$\begin{align}
\frac{\mathrm{d}c}{\mathrm{d}t} &= \beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)- k_\mathrm{B}c \\
\exp(k_\mathrm{B}t)\frac{\mathrm{d}c}{\mathrm{d}t} + \exp(k_\mathrm{B}t)k_\mathrm{B}c &= \exp(k_\mathrm{B}t)\beta k_\mathrm{A}[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t) \\
&= \beta k_\mathrm{A}[\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t]
\end{align}$$

Now if you consider the derivative of $\exp(k_\mathrm{B}t)c$ with respect to $t$, we have (by the product rule, and bearing in mind that $c$ itself is a function of $t$)

$$\frac{\mathrm{d}}{\mathrm{d}t}[c\exp(k_\mathrm{B}t)] = \frac{\mathrm{d}c}{\mathrm{d}t}\exp(k_\mathrm{B}t) + ck_\mathrm{B}\exp(k_\mathrm{B}t)$$

which is the same as the left-hand side of the equation above. The thing we multiplied by throughout, $\exp(k_\mathrm{B}t)$, is called an integrating factor. So, we can now integrate both sides:

$$\begin{align}
\frac{\mathrm{d}[c\exp(k_\mathrm{B}t)]}{\mathrm{d}t} &= \beta k_\mathrm{A}[\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] \\
\int_0^t \frac{\mathrm{d}[c\exp(k_\mathrm{B}t)]}{\mathrm{d}t}\mathrm{d}t &= \beta k_\mathrm{A}[\ce{RH}]_0 \int_0^t \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] \mathrm{d}t \\
\left[c\exp(k_\mathrm{B}t)\right]_{t=0}^{t} &= \beta k_\mathrm{A}[\ce{RH}]_0 \left(\frac{\exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - 1}{k_\mathrm{B}-\beta k_\mathrm{A}}\right) \\
c_t\exp(k_\mathrm{B}t) - c_0 &= \beta k_\mathrm{A}[\ce{RH}]_0 \left(\frac{\exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - 1}{k_\mathrm{B}-\beta k_\mathrm{A}}\right) \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_0 \exp[(k_\mathrm{B}-\beta k_\mathrm{A})t] - [\ce{RH}]_0) \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_t \exp(k_\mathrm{B}t) - [\ce{RH}]_t\exp(\beta k_\mathrm{A}t)) \\
c_t\exp(k_\mathrm{B}t) &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} ([\ce{RH}]_t \exp(k_\mathrm{B}t) - [\ce{RH}]_t\exp(\beta k_\mathrm{A}t)) + c_0 \\
\frac{c_t}{[\ce{RH}]_t}\exp(k_\mathrm{B}t) &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (\exp(k_\mathrm{B}t) - \exp(\beta k_\mathrm{A}t)) + \frac{c_0}{[\ce{RH}]_t} \\
\frac{c_t}{[\ce{RH}]_t} &= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{\exp(-k_\mathrm{B}t)c_0}{[\ce{RH}]_0\exp(-\beta k_\mathrm{A}t)} \\
&= \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{c_0}{[\ce{RH}]_0}\exp[(\beta k_\mathrm{A} - k_\mathrm{B})t] \\
\end{align}$$

I have shown quite a lot of the intermediate steps, if anything seems puzzling just work through it, it is really just algebraic manipulation.

Replacing $c$ with $[\ce{RONO2}]$ gets you to nearly your desired equation

$$\frac{[\ce{RONO2}]_t}{[\ce{RH}]_t} = \frac{\beta k_\mathrm{A}}{k_\mathrm{B} - \beta k_\mathrm{A}} (1 - \exp[(\beta k_\mathrm{A}-k_\mathrm{B})t]) + \frac{[\ce{RONO2}]_0}{[\ce{RH}]_0}\exp[(\beta k_\mathrm{A} - k_\mathrm{B})t]$$

The only difference in the paper is that $\beta$ was sometimes dropped (i.e. some of the $\beta k_\mathrm{A}$'s simply became $k_\mathrm{A}$). I glanced through the paper and can't find a rationale for it, but I will admit I did not look at it in great detail.


Interpretation of the following mass spectrum

I was on to the wrong structure. My main issue with the carboxylic acid was the fragment m/z=60 that it would produce by McLafferty. I didn't have that, so I thought that left me with t-Bu-COOH, which was the only structure with no $\gamma$-hydrogens. Besides I couldn't explain the m/z=74 peak with an acid. On the other hand I couldn't explain the M-17 peak by ethylpropanoate.

But of course, if the carboxylic acid has a methyl substituent at the $\alpha$-carbon, then that methyl group would also be part of the McLafferty fragment, therefore resulting in m/z=74!

The spectre is of: 2-methyl-butanoic acid

M-17 and M-45:

Alpha cleavage of carbonyl => loss of hydroxyl radical or carboxyl radical

m/z = 74:

Loss of ethene by McLafferty rearrangement

m/z = 45:

$\ce{COOH^+}$ fragment

M-15 and M-29:

Loss of methyl radical or ethyl radical

m/z = 29:

Ethyl carbocation


How to interpret spectroscopic data to obtain the structural formula?

Your calculations of the double bond equivalents seems correct, and there is also another hint towards a $\ce{C=O}$ double bond: the corresponding $\ce{^{13}C}$ chemical shift.

Also, while it says you have $\ce{^1H}$ decoupled $\ce{^{13}C}$ data, this is clearly wrong, because they’re showing you multiplets. The $\ce{^{13}C}$ data is $\ce{^1H}$ coupled data, so you can use the information there to deduce the number of carbons with which number of hydrogens.

A third hint given to you is the amoun of signals: You see signals that add up to six carbons and eight hydrogens, meaning there is no symmetry in your molecule.

The 6.8 hydrogen could, by its chemical shift, be aromatic, but you do not have enough DBE’s for an aromatic ring (you would need at least 4). Do you know other types of protons that give chemical shifts in similar ranges, and are consistent with both your DBE count and the structure element(s) you have already found out?

Further ideas to think of:

You have certain integrals on your hydrogen signals, especially one that stands out (not the $6.8\,\mathrm{ppm}$). What could this integral mean, and how does it affect the rest of the structure?
Make a list of the carbons you have and the amount of hydrogens attached to them.
How can you turn all that into a cyclic compound?
Once you have, you can check if the multiplets of the hydrogen signals can be fitted to that structure. Hint: You seem to have no H–H couplings other than across three bonds. (So no long-range couplings.) And no H–C couplings other than one bond length.

   Right, in this first spoiler tag I’m just gonna state the different signals. Check the second one, if you really want to know the solution.
 The carbon at 209.6 (singlet) is most likely not an aldehyde but a ketone. If it had a proton connected, it would need to be a doublet.
 The singlet at 1.8 is a methyl group, belonging to the carbon at 12.1.
 The triplet at 6.8 needs to be next to a double bond (strong low-field shift). The corresponding carbon needs to be 157.9 (doublet).
 Since it’s a triplet, it couples to two hydrogens. There is no other hydrogen on a double bond, but there is a triplet and a quartet left. One of these two must couple to the proton integral one at 6.8. Luckily they are both integral 2. A little further reasoning tells us, it must be the quartet, because the other triplet (2.4) has to couple to two protons, which have to couple back, therefore no other possibility.
 There is another carbon at a double bond (142.1) and it has no hydrogens attached to it (singlet). It needs to have the methyl group on it.

   The structure I deduced is 2-methylcyclopent-2-en-1-one.


How do you identify reducing / non-reducing sugar by looking at structure?

Many sugars are drawn in the cyclic, closed form where the carbonyl group has been converted to a hemiacetal.

Once you realize that a hemiacetal can equilibrate with a carbonyl (e.g. it is a carbonyl in disguise), identification of reducing sugars becomes easier.

Here is an example from Wikipedia, it is Maltose, the same as your third compound

See the aldehyde in the open isomer drawn on the right; see how it can close to form a hemiacetal on the left.


A lab coat will eventually get dirty, that's why it is a good idea to wear it. I had a lot of students complaining about the condition of their coat and they asked me if they could just wash it in the washing machine. My answer was always no.

As far as I remember - I am a theoretician now - a labcoat costs much less than a pair of trousers and a T-shirt. It should be renewed annually and disposed of according to the norms how you would dispose of an unknown chemical. You never know where the chemicals incorporated in your coat will end up, so dispose of them as safely as you can.

In some countries, it is required by the employer to provide equipment like that and taking care of either cleaning or disposing of it eventually. (In Germany that should be the case.) Wear it as a disposable layer of protection and please - for the safety of the environment and yourself - do never ever put it in a normal washing machine again.

As an additional note, if you wash it in the same washer like your normal clothes, how can you assure that it will not end up in your everyday clothes? I can imagine, that having chromium whatever in my underwear would not make me feel alright.


When is rotation around double bonds to form cis-trans isomers allowed?

In rhodopsin this normally occurs only after the 11 cis in the chromophore (a retinylidene schiff base) is placed into an electronically excited singlet state after absorbing a visible photon. This excited state has one electron promoted from the HOMO orbital into the LUMO (antibonding) orbital of the conjugated $\pi$ system. Most data has been obtained from the protein bactoeriorhodposin, which acts a  proton pump, as it is easier to use and has a similar chromophore to rhodopsin.

In the protein the rhodopsin molecule is somewhat strained by adjacent (tryptophan) and other amino acids. The isomerisation takes about 200fs so is extremely fast; among the fastest reaction ever studied. After that there is a series of at least 6 intermediates that slowly returns the trans molecule to the cis form and these thermal reactions take about a second or so.

The initial reaction is thought to proceed as follows; just after the electronically excited state is formed calculations indicate that the cis bond is rapidly stretched (~20 femtoseconds) and experiments show that a  small barrier exists between the minimum energy of the excited state and the pathway to isomerisation. At or close to this barrier rotation & oscillation of the groups either side of the 'double' bond occur during isomerisation. (Kobayashi et.al Nature, 2001, v414,p531).

Many experiments have ben performed on trans/cis stilbene isomerisation which shows similar behaviour. Trans stilbene is the most stable isomer and can be converted into the cis also by photo excitation. The cis ground state has a relatively low barrier to the trans at room temperature (the opposite is not true) and so reverts to the trans spontaneously. This does not happen in rhodopsin  so the protein has to go through a complex set of reactions.


A key difference between carbon and silicon (and why you usually can't make an all-silicon version of an organic compound where all its carbons are replaced with silicon) is that long chains of silicon atoms are not stable. Another key difference is that a hydrogen attached directly to silicon, as in silane, is typically more reactive--say toward hydrolysis or oxidation--that one attached to a carbon atom. I do not know of any process to directly "silanize" a carbon compound, as stated in your question.


Why is Aniline a poor substrate for Nitration reaction?

In the case of option c, the $\ce{-OCOCH_3}$ group will not be showing $\mathrm{-M}$ effect as the ring is directly attached to the oxygen atom. The group rather shows $\mathrm{+M}$ effect because of the two lone pairs present on the oxygen atom, increasing the electron density in the benzene ring and thus making it a good substrate for a nitration reaction.
On the other hand, as nitration is generally done with mixed acid as the reagent, the $\ce{-NH_2}$ group in aniline takes up a hydrogen atom to become $\ce{-NH_3^+}$,  which is a ring deactivating group.
Therefore, among the options given, aniline is the poorest substrate for nitration reactions.


Yes, this equation of state is original. While I cannot absolutely guarantee that it has never been used or proposed before, I can say that I have not encountered it in any “classical” model — and I teach thermodynamics and statistical physics, so I have seen many textbooks and classical exercices :)

Extending beyond the question of originality, I read your demonstration and what bothers me to some extent is the link you establish between $Z$ (the compression factor) and $U(r)$, the intermolecular potential energy. First, let's state that linking $Z$ to $U$ is indeed a common way to derive equations of state. However, the way you are doing it is lacking a sound basis. The sentence “From these explanations it can be said that $Z-1=U(r)$” (page 1) is the problem:

Yes, if $U(r)>0$ then $Z>1$, and the other way around… but that does not mean that the two quantities are proportional.
In fact, $Z$ is a scalar (function of $P$ and $T$), while $U(r)$ is a function of distance (and not thermodynamic variables). So they cannot be linked directly.

You intuited between $Z$ and $U$, which exists, but it needs to be formalized a bit more carefully. Standard approaches to do so rely on perturbation theory or mean-field method. You can check for examples the classical derivations of the van der Waals equation of state to see how this can be done. But it requires some fundamentals of statistical physics…


How to get equation of state from fugacity

That first equation is supposed to be at constant T.  You can rewrite it as $$RTd\ln{(f/P)}=RT(z-1)\frac{dP}{P}$$or $$\ln{\frac{f}{P}}=bT=\int_0^P{(z-1)\frac{dP'}{P'}}$$So if z behaves like z = 1+aPT, this becomes:
$$bT=aPT$$ or b = aP


At a glance, the values listed in the table in the CHEMKIN database are standard enthalpies of formation at 298 K, $\Delta _fH^○(T)$, and standard entropies at 298 K. Why? If only because CHEMKIN's $H^○(298)$ for the elements in their stable states are presented as equal to zero.

You may have just mixed up different enthalpies. Along with the heats of formation, thermodynamic tables (including NIST JANAF) often present the enthalpy increments (sometimes called simply enthalpies), $H^○(T)-H^○(0)$ or $H^○(T)-H^○(298)$. These are not to be confused with the enthalpies of formation $\Delta _fH^○(T)$.

By definition, the standard enthalpy of formation (at a certain temperature) of a complex substance AB is equal to the enthalpy of a chemical process A + B = AB, where A and B are elements in their most stable modifications (at a certain temperature). By convention, $\Delta _fH^○(T)$ for the elements in their most stable modifications is zero.

The enthalpy increment $H^○(T)-H^○(0)$ is an integral of the heat capacity at constant pressure, $\int\limits_0^T C_p(T)dT$. It is no less important than the heat of formation. For example, knowing $\Delta _fH^○(298)$ for our hypothetical substance AB and all the enthalpies $H^○(T)-H^○(298)$ for A, B and AB, we can calculate $\Delta _fH^○(T)$, an enthalpy of formation at different temperature:

$\Delta _fH^○(T) = \Delta _fH^○(298) + [H^○(T)-H^○(298)]_{AB} - [H^○(T)-H^○(298)]_{A} - [H^○(T)-H^○(298)]_{B}$

Note that similar formulae can also be found in the introduction to NIST JANAF (at least to the 4th ed. of it).


Your small copper-zinc cell may not light up a bulb as often shown in textbooks. Check your ammeter settings, it must show some current. How are you connecting it in series? voltmeter is connected in parallel. Now an interesting question. Imagine you join 12 (e.g., Duracells) in series to generate about 12 Volts. Can one start a car with 12 Duracells? Why do you need a large car lead-acid battery? The answer lies in current. Same potential difference does not mean same amount of current supply. You may think 1 V is close to an ordinary cell, your single copper-zinc cell cannot supply enough current to even light up a bulb. Cell potential is a thermomdynamic concept. Current supply is a kinetic issue-how fast the electrons transfer to or from electrodes. Now you can think of the reason why there is no commercial battery made from zinc and copper. See Daniell cell for an interesting history. Check the size of electrodes in Daniell cells.


What tools can be used to estimate binding energies?

The steps for predicting a binding energy is relatively straightforward in computational chemistry but the process itself can become extremely difficult depending on

The types of molecules involved (size, complexity, etc.)
The desired accuracy of your final answer
The environment that the molecules reside in (gas phase, aqueous phase, etc.)

For very small molecules in the gas phase, the process is easy and routine assuming that electron correlation is trivial.  Aqueous phase environments, complex/large molecules and high accuracy quickly makes the problem more difficult if not impossible.

I have written up a previous answer which addresses how to predict an interaction energy ($E_{\mathrm{int}}$) which is completely different from a binding energy ($E_{\mathrm{bind}}$).  However, much of the concepts will remain the same.  I will try to outline here what in fact is the difference between an interaction energy and a binding energy.  Unfortunately, 'binding energy' is tossed around a lot in the literature and consequently grossly misused.  I hope to clarify the subtle differences between these two terms in my answer to your question.

I will focus my discussion around an elementary system of two water molecules to make this conceptually easier.  In the gas phase, an isolated water molecule will adopt a particular geometry (at 0K and in a vacuum).  We can determine what this minimum energy configuration will be at a particular level of theory by approximating the Schrodinger equation.  The geometry that gives rise to the lowest possible energy for an isolated water molecule will be the geometry that you will observe under those conditions.  Remember that the internal coordinates of water include things like the O-H bonds, and the H-O-H angle.  We don't need to guess at what these parameters will be in the end, solving the Schrodinger equation will determine that.

Now, we know exactly what an isolated water molecule looks like and what the corresponding energy of that water molecule will be.  Both of these components are equally important when it comes to determining a binding energy because (as the name suggests) we are trying to determine an energy.  So the energies that we start off with are crucially important.

We can now consider what happens when two water molecules interact (form a hydrogen bond).  What will be the binding energy of these two water molecules (i.e. what will be the strength of this bond that forms between then?).  It is common to approximate the strength of this bond by simply computing an interaction energy (its computationally easier to determine $E_{\mathrm{int}}$ rather than $E_{\mathrm{bind}}$).  How would we compute an interaction energy?  In this case, you wouldn't need to figure out what geometry or energy an isolated water molecule in its lowest energy configuration.  You would simply need to determine what the geometry of two interacting water molecules (i.e. dimer) would be (and corresponding energy) as well as the energy of each water molecule isolated from each other in the geometry that it adopts in the dimer.

I'll explain.  We can determine the lowest energy configuration of a single water molecule in the gas phase.  This geometry is inevitably different than a water that is not isolated... a water that is interacting with something else such as another water.  If you want to get an interaction energy, you only need...

Energy of the lowest energy configuration of the water dimer
Energy of each water (in the water dimer) isolated from each other but only at the geometry that the water takes on within the dimer.

In this case, you are NOT allowing each isolated water molecule to 'relax' into its lowest energy configuration.  You are simply going to neglect the contributions of this energy relaxation effect to the overall bond strength.  Therefore,

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

where mon1 is the first water molecule in your dimer and mon2 is the second water molecule in your dimer. (opt-dimer means optimized dimer)

However, in the case of a binding energy, you MUST account for the relaxation effects that occur when you break the dimer apart and allow the waters to become isolated from each other (and everything else).  To do this, you simply carry out what I've stated above... you simply optimize each water and then take its corresponding energy.  (Determining an interaction energy means you don't have to do this extra optimization(s)... you would simply do a single point energy on each monomer).  Clearly for a water dimer, you would only do one optimization as each water will become identical when isolated.  For a heterodimer, you would have to optimize each piece separately.

To determine the binding energy, you would use the following equation:

$E_{\text{bind}} =  E_{\text{opt-dimer}} - (E_{\text{opt-mon1}} + E_{\text{opt-mon2}})$

Of course the problem becomes more complex when taking into account basis-set superposition error (you will need to perform counterpoise corrections though there exists a heated debate in the literature about the usefulness of this type of correction).  Also, zero-point vibrational energy corrected binding energies will require that the zero-point energy of the dimer and each monomer will need to be determined (which comes from taking the 2nd derivative of the wavefunction which is computationally demanding).

As far as tools go, you can choose from a number of available quantum software packages.  Gaussian 09 is a popular choice but very expensive.  Q-Chem (paid) has the benefit of having the most beautiful, gorgeous, well-laid out documentation of anything I've ever seen (makes for a good supplemental textbook to boot).  NWChem is opensource.  Google can help you find plenty of options.

A quick note: Basis set superposition error (BSSE) should NOT be referred to as an error as 'error' suggests deviation from the RIGHT answer. BSSE has been shown to actually make  non-corrected answers even WORSE.

One last important note:  When determining differences of energies (such as you do to get an interaction or binding energy), ALL energies being compared must be evaluated at the SAME LEVEL OF THEORY (method/basis set, etc.).


I think it would be nice to consider the charge density. The S- is more stable than N- because it has more electron shells, so the negative charge density is better distributed in the atom. It makes that the N- is more likely to give up electrons and act as a nucleophile


Mechanism of dehydrogenation with nosylate

This is the reaction mechanism that I propose for this reaction. Though it does not show the complete reduction of the 3-nitrobenzene sulfonate, it could be reduced further through the nitroso-hydroxylamine pathway. This means that in total, a single nitrosyl anion could oxidize three molecules of the starting material.

Extraction of a hydrogen by the nitro group. The hydrogen is extremely acidic compared to an ordinary aliphatic $\ce{C-H}$ bond due to the massive conjugation in the resulting ion:

Amide-imidic acid tautomerization:

Resonance structures of the molecule, showing its extensive conjugation, and hence stability:

Radical reaction pathway and formation of the product. Every atom in product is $\mathrm{sp^2}$ hybridized and is conjugated over the entire system. It is much more stable than the starting material, and is the driving force of the entire reaction:


How to explain regioselectivity in nucleophilic aromatic substitution

In the benzyne mechanism, mesomeric effects are not involved as when protons are removed, the electrons are situated in orbitals perpendicular to the pi-system, in non-bonding orbitals (which makes sense since the electrons come from a sigma bond). Therefore the electrons left over are not delocalised, which is why such a strong base is required. This also means regioselectivity is explained by inductive effect, which acts through the sigma bonds. The amide is not involved in any -M stabilisation in this case.
In the first part of the mechanism, the most acidic proton is abstracted, which will be either of the identical protons either side of the Cl, as this provides the most stabilisation by inductive effect. This is correctly shown in your diagram. Deprotonation at the carbons adjacent to the amide is less favourable and leads to no reaction due to lack of leaving group anyway.
The point where the alkyne is attacked is determined by both inductive and steric effects; and optimisation of both simultaneously usually occurs in this mechanism. What I mean by this is that in the major product above, the negative charge is situated as close to two -I substituents as possible, so is stabilised, and at the same time the added NH2 is furthest from the other substituent, reducing steric strain.
In the minor product, the negative charge has less -I stabilisation, and more steric strain since the substituents are now closer together.
Of course this only applies if it progresses through the benzyne mechanism. As @Waylander hinted at, a likely possibility is aromatic nucleophilic addition-elimination since you have an electron withdrawing substituent para to the leaving group. This just so happens to give a product identical to the major product of the benzyne route.


Why does sodium react with water to produce a hydroxide, while zinc produces an oxide?

The overall potential of an electrochemical cell is equal to the reduction potential for the substance being reduced plus the oxidation potential for the substance being oxidized. In other words, $ΔE_{rxn} = E°_{red} + E°_{ox}$. Now, all that we need to do is find a list of reduction potentials. From here, I found the following reduction potentials:

$$\ce{Na+(aq) + e- -> Na(s)}: E°_{red}=-2.71 V $$

$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH^{-}(aq)}: E°_{red}=-0.83 V $$

$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}: E°_{red}=-0.76 V $$

A redox reaction occurs between one species being oxidized (losing electrons) and another being reduced (gaining electrons). Since all of these values are reduction potentials, we can flip reactants and products to get oxidation potentials. The oxidation potential for a reaction is negative one times the reduction potential (due to conservation of energy) and since we possess sodium and zinc metal rather than ions, we want for these reagents to be oxidized. This gives us the following equations with oxidation potentials:

$$\ce{Na(s) -> Na+(aq) + e-}: E°_{ox}=2.71 V $$

$$\ce{Zn(s)->Zn^2+(aq) + 2e-}: E°_{ox}=0.76 V $$

Now, you might be wondering what any of this has to do with your original question. It turns out that a reaction is spontaneous (will occur) when the $ΔE_{rxn}>0$. Calculating $ΔE_{rxn}$ for the oxidation of sodium gives us $ΔE_{rxn} = E°_{red} + E°_{ox}=-0.83V + 2.71V=1.88V$, which is greater than zero, so this reaction will be spontaneous. On the other hand, if we calculate the cell potential for zinc being oxidized and water being reduced, we get a negative number, so this reaction is not spontaneous. In other words, sodium will readily reduce water, but zinc will not. Since zinc cannot directly reduce water, it will react with water to produce zinc hydroxide and hydrogen gas. The zinc will then decompose into zinc oxide and water. The reaction mechanism for this is given here:

$$\ce{Zn + 2 H2O → Zn(OH)2 + H2}$$
$$\ce{Zn(OH)2 → ZnO + H2O }$$

You might be wondering why this reaction occurs. It turns out that the standard electrode is:

$$\ce{2H+(aq) + 2e- -> H2(g) + e-}: E°_{red}=0 V $$

which means that while zinc cannot reduce $\ce{H2O}$, it can reduce $\ce{H+}$, reacting with the protons present due to the natural decomposition of water. After losing two electrons, the positively charged zinc will react with the hydroxide found in solution. This creates zinc hydroxide, which is an insoluble compound. Sodium will remain dissociated from the hydroxide, as sodium hydroxide is water soluble. From here, the decomposition of zinc hydroxide is spontaneous, producing water and zinc oxide.


Why don't alcohols undergo nucleophilic substitution with ammonia?

To build on ssavec's answer: Nucleophilic substitution reactions require two species: a nucleophile (a Lewis Base) and a substrate with a leaving group. Nucleophiles are often generically represented as $\ce{Nu}$ and leavings groups as $\ce{LG}$.

All leaving groups are not created equal. The leaving group ability of a leaving group is defined as the relative rates of a particular analogous series of substitutions. For example, the relative leaving group ability might be based on the following reaction, with rates determined for various leaving groups:

$$\ce{EtOH + PhCH(LG)CH3 -> PhCH(OEt)CH3 + H-LG}$$

This is an $\mathrm{S_N1}$ substitution, so the first (and rate determining) step of the mechanism is loss of the leaving group (and is independent of the nucleophile):

$$\ce{PhCH(LG)CH3 -> PhCH+CH3 + LG-}$$

The relative rates of this reaction are influenced by the stability of the $\ce{LG-}$ anion (see the Hammond Postulate, which proposes that the transition state of an endothermic process resembles the products).

The stability of the $\ce{LG-}$ anion can be predicted from the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of $\ce{LG-}$, which is controlled by the same ability to stabilize a negative charge.

The conjugate acid of $\ce{Cl-}$ is $\ce{HCl}$, which is a strong acid. Using some Hess's Law trickery, the $\mathrm{p}K_\mathrm{a}$ of $\ce{HCl}$ has been predicted to be around $-8$.

The conjugate acid of $\ce{OH-}$ is $\ce{H2O}$, which has a $\mathrm{p}K_\mathrm{a}$ around $+16$.

$\ce{H2O}$ is a much weaker acid than $\ce{HCl}$, which means that the negative charge on  $\ce{OH-}$ is less stabilized than $\ce{Cl-}$. Thus, $\ce{OH-}$ is a poorer leaving group than $\ce{Cl-}$ (by a lot).

Another good discussion of leaving groups (and the $\mathrm{p}K_\mathrm{a}$ data used above) come from this site.

In your example reaction (ammonia + ethanol), the product of the reaction has a better leaving group ($\ce{NH3}$, conjugate base of $\ce{NH4+}$, which has a $\mathrm{p}K_\mathrm{a}$ of $+9.75$) than the $\ce{OH-}$ leaving group in the reactant, so the reaction will also run in reverse, and the equilibrium will strongly favor the reactants.

$$\ce{CH3CH2OH + NH3 <=> CH3CH2}\color{red}{\ce{NH3+}}\ce{+ OH-}\ \ K_\mathrm{a} \ll 1$$

As for your confusion about bond energies...

Bond dissociation energies are defined for homolytic bond cleavage, in which each atom keeps one electron when the bond breaks (see image).

Substitution reactions involve heterolytic bond cleavage, in which one atom gets both electrons:


I'm quite certain about the following, because it's from the related (german) woorkbook to the 4th (german) edition. The beneath numbering is from the linked online version, which should be the 9th english edition.

Your way to the solution is part of subsection 3.9 Properties of the Gibbs energy and subsubsection 3.9 c) The variation of the Gibbs energy with pressure.

I seriously entrust you to read this subsubsection and try to solve it with the given information. If this does not help, the following "hidden" section reveals what you need to solve it.

   You need to insert the equation for the molar volume of an ideal gas $V_m = R~T/p$ into (3-57b) $$G_m(p_f) = G_m(p_i)+\int_{p_i}^{p_f}V_m~\mathrm{d}p \tag{3-57.b}$$ which yields one of the two equations that you need: $$G_m(p_f)=G_m(p_i)+R~T\int_{p_i}^{p_f}\frac{\mathrm{d}p}{p}=G_m(p_i)+R~T\ln\frac{p_f}{p_i}\tag{3-59}$$ The other one is the Boyle's law.

Good luck.


In the gas phase (molecular), SnBr2 is bent with an angle of 95 degrees.

The IUPAC definition of valence is such that there is only one valence for each element.

Valence is independent of the compound being consider.


The NMR spectrum of acetylferrocene is very well known and widely published as an undergraduate teaching resource. It would be useful for reference if you were to provide a copy or image of your spectrum. At least you should describe it including chemical shifts, integration and peak appearances. I suspect that your troubles here come from a misunderstanding of your observations of the actual spectrum.

In order to help you with your analysis, you should make sure you complete an accurate integration of all peaks. This will give you some clue as to how many protons are contributing to each of the peaks you are observing. That is, each of your peaks that you think are part of your doublet contribute two protons. Also, when you suggest you are observing a doublet - what is the splitting of this doublet. I doubt it is of a magnitude that conforms to a typical H-H coupling. It is probably closer to 100Hz. So perhaps what you are seeing is in fact two distinct singlets. Or at least two distinct resonances with poorly defined splitting.

Your analysis of the expected splitting pattern for the Me-Cp ring isn't quite correct. From your drawing, H1 and H4 are chemically equivalent, but magnetically non-equivalent. This is because coupling between H1 and H2 is not the same as the coupling between H4 and H2. The same can be said for H2 and H3; H2 and H3 are chemically equivalent, but magnetically non-equivalent.  This means you have an AA'XX' spin system, which is not a simple first order spectrum in appearance. You can read more about these types of spin systems on one of my favourite web resources, Reich. Each signal actually appears as an apparent triplet.

So, why aren't you seeing any splitting in these peaks? There will be a number of reasons why this could be the case, most of which are perfectly reasonable and are no cause for concern. Sample concentration, purity, magnetic field strength, field homogeneity (shimming), temperature and solvent will all contribute to the appearance of this spectrum, and many published spectra for this molecule have splitting which is barely resolved. Without seeing your spectrum, it is difficult to make further interpretation. However, as a suggestion, you should measure the linewidths of your peaks (the width of the peak at half height). You will probably find that the Cp peaks from Me-Cp are considerably broader than the other peaks, especially than for your residual CHCl3.


Why are amidines so much more basic than amides?

Here is a diagram comparing what happens when we protonate an amide and an amidine.

Let's compare the protonated products in the two reactions.  In the case of the amidine we can draw two important (the fact that they are equivalent resonance structures is why they both are important) resonance structures to describe the protonated product.  In the case of the protonated amide the resonance structure with the double bond to oxygen places a positive charge on oxygen.  Therefor this resonance structure is a less significant contributor than the corresponding structure for the protonated amidine where the positive charge resides on the less electronegative nitrogen.  Since we can draw more significant resonance structures for the amidine case, the protonated amidine is more stable than the protonated amide.  This difference in protonated product stabilities explains why amidines are more basic than amides.


The reaction of phenol with bromine is known as bromination of phenol. Solvent has great influence on the reaction. In different solvents, different products are obtained. The action of bromine on phenol can be explained as:
Reaction with bromine in water:—-

Phenol reacts with bromine water to give 2,4,6-tribromophenol.
In water, ionisation is facilitated. Phenol gets ionised to form phenoxide ion, which is even better ortho,para-directing. Bromine also gets ionised to a larger extent to form a large number of bromonium ions. And Bromine ions are highly stabilized in ionic solvents. So the formation of strong o,p-directing group and stabilization of $\ce{Br+}$ enhances the formation of tribromophenol.
Reaction with bromine in $\ce{CS2}$:—-

Phenol reacts with bromine in a  presence of Carbon disulphide to form a mixture of o-bromophenol and p-bromophenol. Among which p-bromophenol predominates. In $\ce{CS2}$ ionisation is not facilitated that much. As it is a non-polar solvent it doesn't the $\ce{Br+}$ on. Also $\ce{-OH}$ group is moderately o,p-directing. So as usual para-product is more formed.


With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes. On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Note that Rule e takes precedence over Rule f.

For example, a preferred IUPAC name is 3-bromocyclohex-1-ene (not 1-bromocyclohex-2-ene): (This example is given in the Blue Book.)

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

Therefore, the structure given in the question is named as 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.


There is a subsection in March’s Advanced Organic Chemistry [1, pp. 450–451] that provides a straightforward definition and a couple of examples:

10.G.viii. Ambident Substrates
Some substrates (e.g., 1,3-dichlorobutane) can be attacked at two or more positions, and these may be called ambident substrates.
In the example given, there happen to be two leaving groups in the molecule.
Apart from dichlorobutane, and in general, there are two kinds of substrates that are inherently ambident (unless symmetrical).
One of these, the allylic type, has already been discussed (Sec. 10.E).
The other is the epoxy (or the similar aziridine524 or episulfide) substrate.525
Selectivity for one or the other position is usually called regioselectivity.

Substitution of the free epoxide, which generally occurs under basic or neutral
conditions, usually involves an SN2 mechanism.
Since primary substrates undergo SN2 attack more readily than secondary, unsymmetrical epoxides are attacked in neutral or basic solution at the less highly substituted carbon, and stereospeciﬁcally, with inversion at that carbon.
Under acidic conditions, it is the protonated epoxide that undergoes the
reaction. Under these conditions the mechanism can be either SN1 or SN2.
In SN1 mechanisms, which favor tertiary carbons, attack may be expected be at the more highly substituted carbon, and this is indeed the case.
However, even when protonated epoxides react by what is expected to be an SN2 mechanism, attack is usually at the more highly substituted position.526
This result probably indicates signiﬁcant carbocation character at the carbon (ion pairing, for example).
Thus, it is often possible to change the direction of ring opening by changing the conditions from basic to acidic or vice versa.
In the ring opening of 2,3-epoxy alcohols, the presence of $\ce{Ti(O\textit iPr)4}$ increases both the rate and the regioselectivity, favoring attack at C-3 rather than C-2.527
When an epoxide ring is fused to a cyclohexane ring, SN2 ring opening invariably gives diaxial rather than diequatorial ring opening.528
Cyclic sulfates (108), prepared from 1,2-diols, react in the same manner as epoxides, but usually more rapidly:529

524 Chechik, V.O.; Bobylev, V.A. Acta Chem. Scand. B, 1994, 48, 837.
525 Rao, A.S.; Paknikar, S.K.; Kirtane, J.G. Tetrahedron 1983, 39, 2323; Behrens, C.H.; Sharpless, K.B. Aldrichimica Acta 1983, 16, 67; Enikolopiyan, N.S. Pure Appl. Chem. 1976, 48, 317; Dermer, O.C.; Ham, G.E. Ethylenimine and Other Aziridines, Academic Press, NY, 1969, pp. 206–273.
526 Biggs, J.; Chapman, N.B.; Finch, A.F.; Wray, V. J. Chem. Soc. B 1971, 55.
527 Caron M.; Sharpless, K.B. J. Org. Chem. 1985, 50, 1557. See also, Chong, J.M.; Sharpless, K.B. J. Org. Chem. 1985, 50, 1560; Behrens, C.H.; Sharpless, K.B. J. Org. Chem. 1985, 50, 5696.
528 Murphy, D.K.; Alumbaugh, R.L.; Rickborn, B. J. Am. Chem. Soc. 1969, 91, 2649. For a method of overriding this preference, see McKittrick, B.A.; Ganem, B. J. Org. Chem. 1985, 50, 5897.
529 Gao, Y.; Sharpless, K.B. J. Am. Chem. Soc. 1988, 110, 7538; Kim, B.M.; Sharpless, K.B. Tetrahedron Lett. 1989, 30, 655.

Reference

Smith, M. March’s Advanced Organic Chemistry: Reactions, Mechanisms, and Structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013. ISBN 978-0-470-46259-1.


Our teacher started with $\ce{AuCl}$ and wrote down Gold (I) Chloride as the nomenclature. How did he get from $\ce{AuCl}$ to Gold (I)?
There are two possibilities: Gold (I) and Gold (III).

True, but there is only one possibility for chloride ;)
Karl's comment under your question gives the crucial hint:
Start with the part that you know and that is unambiguous!
Let's have a look at some examples:

Chloride is (always) $\ce{Cl-}$; it has a charge of (-1).
If your teacher would have written $\ce{AuCl3}$, you would know that there are 3 chloride ions with a total charge of $3\times(-1)=-3$. Consequently, you would know that you're having $\ce{Au(III)}$ here. The copper (II) cloride example belongs to the same category and if you like, you can try iron(II) bromide, iron(III) chloride and copper(I) iodide.

Karl also mentioned oxygen. You can safely assume one or more $\ce{O^{2-}}$ units in your metal oxides, depending on the oxidation state of the metal. Can you assign the oxidation state in $\ce{Fe2O3}$?

Nitrite, nitrate, sulfite, sulfate are some anions that you should memorize sooner or later. You might know that the names that start with nitr~ refer to nitrogen, the sulf~ refers to sulfur. All these anions contain various number of oxygen atoms.


How can I determine the chemical composition of local rocks and minerals?

Let's talk about rocks. The art and science of getting useful things from rocks is one of the oldest in human civilization. All of the example products you mention (gunpowder, petroleum, silica, and iron) were discovered or developed from experimentation on rocks.

In the ancient world, composition was determined by qualitative analysis, or a series of chemical tests and determination of important physical properties. For example, if a lump of shiny yellow stuff was soft, melted easily, did not lose its lustre in air or water, did not dissolve in vitriol (sulfuric acid) or aqua fortis (nitric acid), but did dissolve in aqua regia (made then by dissolving sal ammoniac (ammonium chloride) in aqua fortis), then we probably have gold.

In the modern age, the field of analytical chemistry was developed to answer the very questions you asked: What is this stuff? What are its constituents? How can we separate them? How do we determine that we successfully separated/purified them?

To continue the rock example, the modern technique of x-ray fluorescence is used to determine the elemental compositions of rocks (and other things, like ceramic fragments found at archaeological sites). In this technique, high frequency x-rays are used to excite the core electrons in the atoms of the sample. When these core electrons relax, they spit out x-rays of characteristic frequency/wavelength that is specific to the element. These emitted x-rays are collectively measured and converted into a spectrum like the one below.

This spectrum indicates that many elements will have more than one spectral line because there is more than one electronic transition that is possible. It also demonstrates the clear resolution of one element from another. After some maths (and better yet a calibration), you can use the heights of these peaks to determine the specific ratios of elements in your sample.

Not everyone is ignorant of the utility of natural objects. Most of us who become scientists do so to study the natural world and learn from it. For example, organic and medicinal chemists do not design new drug molecules from scratch. Rather, they take cues from nature. A folk legend about the curative properties of the leaves of a plant that grows along the riverside might turn in to an effort to extract and analyze the phytochemicals in those leaves, including throwing those compounds at bacteria, viruses, cancer, etc. Any that look promising are studied further. Taxol (paclitaxel) is a successful chemotherapeutic agent that was isolated from the bark of the pacific yew. It is now mass-produced by a semi-synthetic pathway which isolates a late-stage precursor from the needles of the European yew.

If you are specifically interested in rocks, then you should study geoscience.


How to purify sodium chloride table salt that contains sodium ferrocyanide

Don't.  Ferrocyanide is a good way of tying down cyanide ions so they don't do their usual bad things to you.  Trying to purify the salt could release the cyanide ions.


Assumptions made in calculating melting point from thermodynamic data

In real life the melting point is determined by measuring the temperature. The entropy of fusion is usually determined by dividing the enthalpy of fusion by the melting temperature.

$$\Delta S_f = \Delta H_f / T_f$$

Your text book calculation just solves this equation for $T_f$ to recover the temperature that probably had been measured before.

Note that for constant pressure the temperature is constant $T_f$ as long as the phase transition takes place.


Electrophilic substitution of an indole as opposed to benzofuran and benzothiophene

because when electrophile react on 3 position more resonance structures are posible than the 2 position and thus it show stability.Thats why we say that position 3 is more good for indol to attach electrophile.


Here they replace the ferricyanide for bleach, but the description does say that the luminol solution is stable in base for up to three weeks. So I would follow your 2nd way.


Why is concentration used as a proxy for activity instead of mole fraction?

The posted notation seems unorthodox if not incorrect.
For the Gibbs free energy of reaction one usually writes
$$\Delta_r G = \left(\frac{d \Delta G}{d \xi}\right) = \sum_i \nu_i\mu_i^\circ  + RT\sum_i  \nu_i \ln \gamma_i x_i$$
$$ = \Delta _r G^\circ + RT \ln Q$$
where $ \xi$ is the reaction progress, $Q=\sum_i  \nu_i \ln \gamma_i x_i$ is the reaction quotient, and one can equate $\Delta _r \mu= \sum_i  \nu_i\mu_i $ with $\Delta_r G$. For the reaction $A \leftrightharpoons B$ one would write $\Delta \mu = \mu_B - \mu_A$.
Changing concentration units is ok but a set of units usually implies a choice of thermodynamic reference state. The reference state generally defines the limit in which $\gamma \rightarrow 1$ (the limit of ideal behavior). If you decide to express the reaction quotient Q in terms of molar concentration rather than mole fraction but want to retain the same general form of the above equations, then it becomes necessary to modify both Q and $\Delta _r G^\circ$. Some reference states are better choices than others under certain circumstances, and it is not accurate to say that mole fraction is not used, it is, just not always. For solids or solvents a mole fraction unit is often convenient. When dealing with solutions, and particularly with solutes as a minor component, it is generally more convenient to deal in molal and molar units (the prior being favored), in combination with Henry's law. In addition, at small concentrations of solute the mole fraction, molarity and molality are linearly proportional, differing by a multiplicative constant.  For instance, at low concentration we can write the mole fraction of solute as
$$\chi _2 = \frac{n_2}{n_1 + \sum_i n_i} \approx \frac{n_2}{n_1} = \frac{c_2}{c_1} = \frac{c_2}{1000\delta_1/M_1} \approx \frac{M_1}{1000 \delta_1 ^\circ}c_2$$
where 1 and 2 label solvent and solute and $\delta_1 ^\circ$ is the mass density of the pure solvent, $M_1$ its molar mass.


How would you go about adding a Cl group para to a brominated cinnamic acid?

There are a number of conditions that are selective for p chlorination of substituted benzenes. KCl and Oxone in MeCN at rt as described in Synthetic Communications vol 31 (2001) 2021 looks easy to use and should not disturb the other substituents details here


Let’s consider the following cases:

getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ air on one’s skin
getting $1\,\mathrm{mol}$ of $100\,\mathrm{°C}$ water vapour on one’s skin

With the slightly irrealistic assumption that all of these liberate all their thermal energy to the skin while cooling down to $40\,\mathrm{°C}$.

1st case

Assuming isobaric conditions (constant pressure), the heat energy liberated from $100\,\mathrm{°C}$ water that is cooled to $40\,\mathrm{°C}$ can be calculated as follows:

$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 4.5\,\mathrm{kJ}$$

2nd case

A similar formula holds true. However, now we are considering a gas all the way, not a liquid. (And ‘air’ is technically not a single substance, but we can work with it.)

$$\Delta Q = \Delta T \cdot C_p(\mathrm{air}) \cdot n(\mathrm{air})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 29.19 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 1.7\,\mathrm{kJ}$$

3rd case

Unlike the previous cases, we start off with steam, which will condense to water. Therefore, we need to add the heat of vapourisation to the equation.

$$\Delta Q = \Delta T \cdot C_p(\ce{H2O}) \cdot n(\ce{H2O}) + \Delta H_{\mathrm{vap}} \cdot n(\ce{H2O})$$
$$\Delta Q = 60\,\mathrm{K} \cdot 75.327 \frac{\mathrm{J}}{\mathrm{K\,mol}} \cdot 1\,\mathrm{mol} + 40.66 \frac{\mathrm{kJ}}{\mathrm{mol}} \cdot 1\,\mathrm{mol}$$
$$\Delta Q \approx 45.2\,\mathrm{kJ}$$

You will note that the third case delivers a lot more heat energy than the first two cases.

(Part of the simplifications I used means that I grossly overestimated the heat transferred by hot air. Which is also why one can put one’s hands under a hair dryer and be fine with it.)


Contradictory Properties of Mercuric Cyanide

Mercury cyanide behaves like Mercury chloride $\ce{HgCl_2}$. Both are soluble in water, and both do not dissociate in water. These compounds are not salts. The bonds $\ce{Hg-Cl}$ or $\ce{Hg-CN}$ are more covalent than ionic. When the atom Hg is included in compounds it often behaves as a non-metal.

This is due to a relativistic effect. In all elements with atomic number higher than $70$, the outer electrons move at a speed that is not far from c, the velocity of light in vacuum. Of course everybody knows that electrons do not move like planets in the atom. But the relativistic corrections are significant "as if" the electrons were moving. And one of the effect due to relativity is that the length decreases at velocities going near to c. Pekka Pyykko has shown that this entails a contraction of the relativistic electron orbitals. And this contraction is proportional to the sum of the two quantum number n and l.

Mercury has the configuration {$\ce{Xe}$} $\ce{4f^{14} 5d^7 6s^2}$. The three external orbital have a sum n+l equal to $8$ for $4$f, $8$ for $5$d, and $7$ for $6$s. So the electron $4$f and $5$d (with n+l = $8$) are so much contracted that they nearly disappear in the internal core of electrons. It must be nearly the same for the $6$s electrons. And the mercury atom looks a bit like Xenon when seen from outside. That is why it is "nearly" a noble gas : its boiling point is rather low with respect to its neighboring atoms.

All the chemical properties of Mercury are due to its two $6$s electrons which are rather difficult to extract, as they are partly included in the inner core. These electrons "prefer" making covalent bonds rather than ionic bonds, when included in molecules with $\ce{CN}$ group or $\ce{Cl}$ atoms.

Pekka Pyykko, Relativity and the Periodic System of Elements, Accounts of Chemical Research, Vol. 12, No. 8, (1979), p. 276 - 281.


Why do bonding and anti-bonding states split in energy (mathematically)?

You are (essentially) correct and the off-diagonal terms are related to so-called "overlap" between atomic orbitals (AOs). It is common to talk of the AOs "overlapping" to form molecular orbitals (MOs), and as you rightly said, the off-diagonal terms (or "overlap") goes to zero in the limit of infinite bond length. The term "avoided crossing" is not commonly seen in introductory MO theory, but I think it is fair to describe it as such.

Of course there are some slight subtleties. Let's stick to a one-electron system such as $\ce{H2+}$ for now. $\psi_i$ refers to the 1s orbital on hydrogen atom $i$ and $E$ is the energy of a 1s orbital on hydrogen. The minor issue is that $\hat{H} \neq \hat{H}_1 + \hat{H}_2$ (following your notation where $\hat{H}$ is the molecular Hamiltonian and $\hat{H}_i$ is the Hamiltonian for hydrogen atom $i$), because:

$$\begin{align}
\hat{H} &= -\frac{\nabla^2}{2} - \frac{1}{r_1} - \frac{1}{r_2} + \frac{1}{R} \\
\hat{H}_1 &= -\frac{\nabla^2}{2} - \frac{1}{r_1} \\
\hat{H}_2 &= -\frac{\nabla^2}{2} - \frac{1}{r_2} \\
\end{align}$$

(I used atomic units. $r_i$ is the distance between the electron and nucleus $i$, and $R$ is the distance between the two nuclei). So actually, it is more like

$$\hat{H} = \hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}$$

and therefore the first term in the matrix

$$\begin{align}
\mathbf{H}_{11} &= \left<\psi_1\middle|\hat{H}_1 - \frac{1}{r_2} + \frac{1}{R}\middle|\psi_1\right> \\
&= E + \left<\psi_1\middle|-\frac{1}{r_2}\middle|\psi_1\right> + \frac{1}{R} \neq E
\end{align}$$

(since $\langle\psi_1|\hat{H}_1|\psi_1\rangle = E$, and also $\langle\psi_1|(1/R)|\psi_1\rangle = (1/R)\langle\psi_1|\psi_1\rangle = 1/R$).

For more information I refer you to Atkins' Molecular Quantum Mechanics 5th ed., pp 262–266. The terminology can be a bit confusing and also Atkins does not use atomic units (in his text he uses $j_0 = e^2/4\pi\varepsilon_0$ which is simply equal to $1$ in atomic units). Adjusting for this, the final quoted formulae for the matrix elements are:

$$\alpha = \mathbf{H}_{11} = \mathbf{H}_{22} = E - j' + \frac{1}{R}; \qquad j' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_1\right>$$

(as found earlier), and

$$\beta = \mathbf{H}_{12} = \mathbf{H}_{12} = \left(E + \frac{1}{R}\right)S - k'; \qquad S = \langle\psi_1|\psi_2\rangle;\quad k' = \left<\psi_1\middle|\frac{1}{r_2}\middle|\psi_2\right>$$

If you now solve the generalised eigenvalue equation $\mathbf{Hc} = E\mathbf{Sc}$ (the simple $\mathbf{Hc} = E\mathbf{c}$ only works when your basis set is orthonormal, but in this case it is not) you obtain the two eigenvalues

$$E_\pm = E + \frac{1}{R} - \frac{j' \pm k'}{1 \pm S}$$

Note that when the atoms are infinitely far apart both $S$ and $k'$ vanish, so (1) the off-diagonal element $\beta$ vanishes and (2) the "bonding" and "antibonding" orbitals are of the same energy, i.e. $E_+ = E_-$.


What is the state of aggregation (gas, liquid) of oxygen in blood?

Regarding the state of oxygen in blood: It is in solution in the blood plasma (which mostly consists of water), in the form of single molecules. Think of water which you leave exposed to air: carbon dioxide will be captured and dissolved (along with the other gases in air), but these molecules are not gaseous or liquid, but rather "in solution", which is different from the "classical" states.

Back to oxygen: As your reference already states, most of the oxygen in solution will bind to hemoglobin. The actual state of oxygen in that complex has been debated, but it is believed to be reduced by the hemoglobin iron to the superoxide anion, coordinated to Fe$^{3+}$. See Wikipedia on this.

Also, the lungs do not "convert" the atmospheric oxygen to anything, they rather allow, due to their very large surface area, the quick exchange of oxygen/carbon dioxide in solution and in the air.


What are the other ways (if they exist) that atoms can gain or lose electrons that aren't bonding?

My question rather concerns if electrons are ever lost or gained naturally.

Yes, they can!

Here are two examples of natural processes by which atoms can gain or lose electrons "naturally":

Electron Capture

In the natural radioactive decay process known as electron capture the nucleus captures an inner-shell electron, K- or sometimes L-shell. This results in a proton being converted to a neutron, and the emission of a neutrino. The atom has lost an electron, but conveniently a proton as well, so it remains neutral but with a deep vacancy.

Auger Process

If the vacancy is filled by a higher shell electron falling into it this can result in the emission of a photon, or in the emission of yet another electron. That emission is referred to as the Auger process.

See Electron capture; Reaction_details for examples.

 Source


Stability of a carbocation, in part of camphor synthesis

Since mechanistic studies of the camphene/ bornyl acetate system are not available, I will focus on the parent compound norbonyl, and with it explaining the stability of the cation.

For quite some time there has been active debate over the structure of the 2-norbonyl cation. A very brief description of the two sides can be found in the Wikipedia article. A short history of the argument can be found at the homepage of the Royal Society of Chemistry (RSC): "The nonclassical cation: a classic case of conflict" by Mark Peplow.

In 2013 the case was finalized after I. Krossing's group was able to obtain a crystal structure of the compound. Another blog post at the RSC gives some more insight into the matter: "Crystal structure closes classic carbocation case" by Andy Extance. Finally we can speak of the 2-norbonyl cation as a true non-classical cation. The findings are also in good agreement with high-quality calculations. F. Scholz, D. Himmel, F. W. Heinemann, P. v. R. Schleyer, K. Meyer, and I. Krossing, Science, 2013, 341 (6141), 62-64.

This does in principle mean, that at very low temperatures, the 2-norbonyl cation is best described by the following resonance structures (courtesy of wikipedia):

At higher temperature the structure is quite disordered. In solution this means basically, that there is an ultra fast equilibrium (at the rate of molecular vibrations) between several structures. This would also explain the high reactivity of this compound.

Since camphene is a substituted norbornene, similar reactivity can be expected. The reaction is therefore most likely determined by the small steric effects. Since this reaction is highly selective, one can also safely assume, that the thermodynamic favoured product is formed. In this case it is the isobornyl acetate.

I have already mentioned it in the comments, but I am including it here nonetheless. The second row of your image is contains the wrong numeration and therefore depicts the mirror image. Correct is the following depiction:


Why does increasing surface area of the electrode decrease resistance?

Fluid flow is always an excellent analogy for understanding current flow.

Fluid flow rate -> electrical current

Pipe friction, valves and other throttlers -> electrical resistance

Fluid pressure -> voltage.

In this case, the amount of contact between the electrode and the solution is most likely the bottleneck of the circuit so increasing the area of contact should increase the current.  Similar to how cranking the valve on a faucet or outdoor spigot more and more open increases the flow rate of the water coming out.


Dimethylene Cyclohexane to Octalin-1,4-Dione | Hydroboration vs. Ozonolysis Pros Cons

Comparison of the two approaches as far as yield/cost/etc. is void because one of them does not provide the same product as the other. Hydroboration-oxidation of 1,4-dimethylenecyclohexane yields 1,4-bis(hydroxymethyl)cyclohexane with undergoes oxidation to yield 1,4-cyclohexanedicarbaldehyde or 1,4-cyclohexanedicarboxylic acid.

Now, you may be able to take the carbocylic acid, covert to acid halide and then react with $\ce{LiCuMe2}$ to get the ketone, which should undergo Baeyer-Villager oxidation followed by hydrolysis to give the alcohol you want. Or you could covert the diacid to the diamide and subject it to Hofmann degradation to get the amines, which you might be able to convert to the diazo salt with HONO, which may hydrolyze to give the desired diol. Both routes ae significantly more convoluted than ozonolysis, which decreases oveall yield and increases cost.
Now, there are other oxidative cleavage reactions which might be cheaper.


How do you separate microcrystalline cellulose from amino acids in a powder mixture?

I would try to use a 5-10 cm layer of fine SiO2 particles above a slow filter in a Buchner funnel. I'd be surprised if it didn't do the trick.
Consider that the cellulose particles you see after filtration could come from the paper filter as well (I sometimes see such tiny tiny particles floating around after filtration), in which case it may help to submerge the filter in a beaker of hot water before use, and in the worst case, carry out the SiO2 filtration AFTER the filter paper (in a column with a fritted disk for instance, above larger sized SiO2 not to waste the fritted glass with tiny particles).


Explanation of calculating equilibrium constant

Your reasoning is correct. Initially, you saw that there was no iodine in the vessel. After reaching equilibrium, you saw 0.25 mol. Where did this come from?

Answer: Hydrogen iodide broke down to form this iodine and hydrogen. How much hydrogen iodide? Well this is a super simple moles problem.

0.25 moles of iodine was formed, and the mole ratio for $\ce{HI:I_2}$ is $2:1$. For 0.25 moles of iodine to be formed, 0.5 moles of hydrogen iodide must have been used (just look at your balanced equation).

If 0.5 moles of hydrogen iodide was used, how much hydrogen iodide is left? 0.60 moles minus 0.50 moles to leave 0.10 mol.

Then, as you did correctly, you plug it into the $K_c$ formula.

While you have stated that you just want to know how to do it (which you should be able to now), it is useful to know what's going on. Basically, the $K_c$ is a measure of where the equilibrium is. If your $K_c$ value is high, there will be a higher concentration of the products at equilibrium.


Is the adsorption of dissolved iodine onto activated carbon physical adsorption or chemical adsorption?

The mechanism(s) of iodine adsorption into carbons are complex and strongly dependent on each carbon. That is, iodine can be physisorbed in the carbon micro/mesopores, but in carbons with a rich surface chamistry, it can also be chemically adsorbed through charge transfer interactions.
There are many papers about this, but this one shows clear evidence of both mechanisms. (Written by the carbon "connoisseur" Harry Marsh).
They clearly mention:

  Thus, both classical physical adsorption theory and more specific chemical interactions have been proposed to model the adsorption of iodine by carbonaceous materials.

Strong evidence is provided in the paper itself.


Several thoughts:

I can't see how you form toluene with a Grignard - it's certainly possible though
Radical chlorination of toluene with $\ce{Cl2}$ isn't the cleanest reaction.
To reduce the nitrile to the amine, you don't need to go through the amide.

Also, from ethylbenzene, it's very hard to introduce that amine group cleanly. You effectively just have a compound with C-H groups and you have to somehow introduce some kind of functional group into the non-benzylic position. Try to think backwards instead of forward!

The amine you want can be obtained by direct reduction of a nitrile either by simple $\ce{LiAlH4}$ reduction or hydrogenation:

That arrow essentially means "can be made from".

Of course, that's easy to make using a nucleophilic substitution reaction with $\ce{CN-}$ and benzyl chloride (for example):

To get from benzene to benzyl chloride, you could use a Blanc chloromethylation. So your forward synthesis would be:

and you might want to use catalytic $\ce{KI}$ in the nucleophilic substitution step. Of course, for such a simple compound, there are tons of ways to synthesise it. The amine could be made from reduction of an azide, or by a reductive amination, and so on. I strongly doubt my proposed route is the most efficient. Still, I think it should work.

Theoretically, you could use a Friedel-Crafts alkylation and a radical bromination to get to benzyl bromide instead:

I don't like the alkylation step though.


How to round correctly when calculating the average and standard deviation?

I figured it out.

The means are rounded to only two sig figs because its subtraction.
And significant figures follow different rules under subtraction/addition vs multiplication/division.

And its 0.16 for the same reason that the means are only two sig figs.
Because its addition, not multiplication/division.

derp, two hours down the drain


Difference between equivalent conductance and equivalent conductivity

Conductance and conductivity are distinct, although related, magnitudes.

Conductance, $G$, is a measure of the total electric current, $I$, that can pass through a conducting element (such as a wire, a space through which a charged beam flows, or an electrolytic solution) when a given voltage, $V$, is applied through it:

$G = \cfrac{I}{V}$

Correspondingly, the IS unit for conductance is the siemens (S), equivalent to an ampere per volt. Since conductance is the inverse of resistance, a siemens is also the inverse of an ohm.

However, in general, conductance does not only depend on the nature of the material; it also depends on the geometry of the conducting element. For instance, the conductance of copper wires of different length and/or different section will be different. Conductivity, $\sigma$, is defined in order to compare the contribution to conductance of materials, isolating it from shape effects.

In general, for a linear electric flow across a conducting element, conductivity increases with cross-section (as more flow is able to pass through simultaneously) and decreases with length (as more resistance is found). Conductivity is thus defined as

$\sigma =  G \cfrac{l}{A}$

which, ideally, should be identical for all conductors made of the same material (in practice, there are a number of effects that can make the relationship between conductance and geometry non-linear, but for macroscopic ohmic conductors it holds fairly well). This allows direct comparison of the electric properties of materials, effect of doping, temperature, etc.

The IS units of conductivity are therefore siemens per metre (S/m).

Note that unlike conductance, which is an extensive quantity, conductivity is an intensive quantity which doesn't depend on the size of the conducting element. According to 19th century naming conventions, conductivity can also be seen as a specific conductance (in the same way specific gravity is an intensive measure for mass, or specific heat is an intensive measure for heat capacity) and indeed that name was widely used in the early days of both electrochemistry and electromagnetism. While in electromagnetism that standard has changed and currently conductivity is the almost universal name for $\sigma$, in electrochemistry that name lingered and conductivity measurements in solutions were reported as "specific conductance" for longer.

Compound the facts that conductivity measures in electrochemistry are usually made with standard-sized cells in which geometry isn't a factor, and that specific values of conductivity (such as molar conductivity or equivalent conductivity) need an additional qualifier, and these magnitudes were often called "molar conductance" or "equivalent conductance" short for "molar specific conductance" or "equivalent specific conductance". However, you'll find that these magnitudes are always reported in units of conductivity (typically $\mathrm{\mu S/cm}$ or $\mathrm{mS/cm}$).

The use of specific conductance to mean conductivity is now considered obsolete; see for instance the IUPAC Gold Book's page for conductivity.


I think you are getting confused see firstly methyl free radical is a planar with two hydrogens facing each other . The dipole moment vector of the single electron is away from the central carbon
Carbon has Electronegativity value 2.55 while for H it is 2.2 so all the hydrogens have a small positive charge and the carbon has a small negative charge . the direction of the dipole is from negative to positive so the dipole moment vector of the single electron egts cancelled by the resultant of that of the C-H bonds hence it is a non polar structure


Is there a preferred way of naming the resonance hybrid in keto-enol tautomerism?

Nomenclature of Organic Chemistry, IUPAC Recommendations and Preferred names 2013 in the section P-76 DELOCALIZED RADICALS AND IONS mentions only "totally delocalized" ions like cyclopentadiene derived cyclopentadienyl radical, cyclopentadienylium cation, cyclopentadienide anion; benzo[7]annulenylium (azulenium?) and pentadienyl radical.

However, there is a draft having this chapter expanded, allowing more systematic naming of delocalized ions.

  P-76.1.2 Partial delocalization is denoted by the descriptor deloc preceded by the locants indicating the extent of delocalization. This descriptor and its locants are cited immediately before
  the appropriate suffix and enclosed in parentheses.

  cyclopenta-2,4-dien(1,2,3-deloc)ide
  ...

But I am not 100% sure how exactly would that apply to your 3 structure.. let's be brave and creative:

(2Z)-but-2-en-2-(2,3,O-deloc)olate or
(2Z)-but-2-en-2-ol(2,3,O-deloc)ate or if you don't need stereochemistry
3-oxobutan-2-(2,3,O-deloc)ide or maybe even just
butanon(2,3,O-deloc)ide ? .. or
butanon(O,2,3,-deloc)ide ?

However, that draft is from 2004, but this rule did not make it into the current official nomenclature (2013), so for now we can just speculate, or contribute to the upcoming version.


I suspect that you could be forming an aqueous solution of $\ce{(NH4)2PtCl6}$.

I would like to make an observation as a person who has done platinum group metal chemistry, the anionic chloro complexes of PGMs like $\ce{PdCl4^2-}$ and $\ce{PtCl4^2-}$ are very toxic. They can induce a nasty allergy to PGMs. I would suggest that you do not work at home with these substances.

Also in my experience at least with $\ce{PdCl4^2-}$, I have always found that if you want to bring it out of solution then cation methathesis with a potassium salt can bring the substance out of solution as a brown solid. I suspect for the hobby chemist that low sodium salt would be the best reagent for doing this transformation.

For the lab chemist I would suggest using tetramethyl ammonium chloride or even tetraphenyl phosphonium chloride


I cannot guarantee that the following explanation for the proof of the ring size of the principal cyclic form of glucose is due to Haworth. The link I provided in the comment above explains one method. In "Methylation of Pyranoses" methods are illustrated for the permethylation of the hemiacetal form of glucose and its subsequent hydrolysis to the tetramethoxy hemiacetal. Although the 6-membered ring is shown, a priori the ring size is unknown. This is the issue to be determined. continued below
In "Ring Size of Pyranosides" mild HNO3 oxidation of the tetramethoxy hemiacetal affords a keto carboxylic acid (upper left). Continued oxidation of either enol of the keto group leads to oxidative degradation to an  optically-active dimethoxysuccinic acid (red path) and a meso trimethoxyglutaric acid (green path). These two dicarboxylic acids are only derived from a 6-membered ring of glucose.

Source: http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbohydrates44.pdf


Does Gibbs free energy of formation always equal zero for elements in standard state?

No.  Elements occur in different allotropes.

$\Delta H_f^o$ and $\Delta G_f^o$ are defined to be zero at 298K, 1 bar for the lowest energy allotrope, with the exception that the values for white phosphorous are defined to be zero even though it is not the lowest energy allotrope.

The other allotropes do not have zero $\Delta H_f^o$ and $\Delta G_f^o$ even though they are still elements.

  So of course the compounds that have a $\Delta H_f$ of zero are the elements in their standard states but why does that make their $\Delta G_f$ zero?

$\Delta H_f = 0$ does not imply $\Delta G_f = 0$

The two values are independently defined to be zero for a particular allotrope as explained above.

  Does every compound that has a $\Delta H_f$ of zero also have a $\Delta G_f$ of zero?

If a compound (other than by definition) happens to have $\Delta H_f = 0$, then $\Delta G_f$ could be any value, it is not necessarily zero.

  What exactly is $\Delta G_f$

It is the change in Gibbs energy upon forming a compound in its standard state from its constituent elements in their standard states at 298K.


Why does the helium hydride ion have such a large dipole moment?

The dipole moment of an array of charges is defined as
$$
\vec{\mu}=\sum_i \vec{r}_iq_i
$$
where $\vec{r}_i$ is the position of charge $q_i$ in the molecule with respect to the center of mass. In a neutral molecule the sum of the charges is zero and therefore the terms in the sum partially cancel. For ions this is of course not the case and the dipole moment is much larger than for the corresponding neutral molecule (provided that the ion is not completely symmetric such as H$_2^+$).


I just thought that the rearrangement might be occurring during the analysis I.e. The thermal ramp, implying a lower temperature would be better for you. The bp of each compound is too close to separate purely in that way.


Finding rate of reaction from a concentration vs. time graph

Let the reactant be $S$ and the product be $P$.  Let $p$ be the dimenionless product concentration, $\frac{P}{S_0}$.  First order kinetics:

$$\frac{P}{S_0} = p = 1- e^{-k t}$$

Slope of first-order kinetics:

$$\frac{dp}{dt} = k e ^{-kt}$$

Slopes and angles.  One confusing thing about this problem is the definition of $\theta$ is different than what you usually see in math class.  It's the angle with respect to the $y$ axis, not the $x$ axis.  As you note, if the $y$ angle is 30 degrees, then the $x$ angle is 60 degrees, meaning the slope is $\tan~60^{\circ} = \sqrt{3}$.
Units problem.  However there is a big units problem in this question.  The slope is $\sqrt{3}$ what?  Gigamolar per nanosecond?  Zeptomolar per eon?  The angle of the tangent curve will change depending on the units used to plot the data.  If you use nanoseconds as the time unit, the angle of the tangent curve will be very flat, but conversely if eons is the time unit, the the angle will be very sharp.  The question does not explicitly say what units are used in the plot.  And the expression for the derivative of a first-order reaction shows that the slope should have units of concentration per time, but $\sqrt{3}$ is dimensionless.  So there is not really enough information to solve the problem.
Dicey assumptions about units.  Let's assume that the time unit is minutes.  Let's further assume that the $y$ axis plots a dimensionless concentration, i.e., $p$. Then the derivative (slope) is $-k e^{-k t}$ and apparently has a value of $\sqrt{3}~\mathbf{{min}^{-1}}$.

The question is asking for the rate, so you are done.  But there are a lot of weird assumptions in this problem.


Conversion between anisotropic temperature factors and geometric ellipsoids

I found the publication titled Atomic Displacement Parameter Nomenclature which seems to have the details.


Is this a valid method to find percentage of pure ethanol in an impure sample?

What if this unknown material produce part of the energy stated in the problem?

It would be stated in the problem that the amount of heat that the unknown material produced.

  The standard molar enthalpy is measured under standard conditions (1 atm, 25°C). However, the condition inside the bomb calorimeter is unknown.

We are given some information, the final temperature of the system. But you are right in that we do not know the pressure, and the specific heat of the remaining items is difficult to calculate.


Questions like this are not prone to easy answers. But, lets look at what is known about the liquids, then at the various binary diagrams.

Lithium - Melting point 453K, density at melt is 0.512 g/cm$^{3}$, or 0.0737 mol/cm$^{3}$

Sodium - Melting point 371K, density at melt is 0.926 g/cm$^{3}$, or 0.0403 mol/cm$^{3}$

Potassium - Melting point 336K, density at melt is 0.828 g/cm$^{3}$, or 0.0211 mol/cm$^{3}$

Calcium - Melting point 1112K, density at melt is 1.376 g/cm$^{3}$, or 0.0343 mol/cm$^{3}$

Clearly Li has by far the smallest atomic volume in the melt.

The Li-Na binary, as you point out, shows a miscibility gap, tilted towards the Li side, but peaking around 300C. The K-Li system is less well studied, but it also shows a miscibility gap which peaks at perhaps a similar value of 300C.

In sharp contrast, the Ca-Li system exhibits an intermediate compound (Li$_{2}$Ca) and can be described with a 'virtually ideal' liquid phase (C.W. Bale and A.D. Pelton, Bulletin of Alloys Phase Diagrams 8(2) 125-127 (1987)).

Given that the molar density of liquid Ca lies between those of Na and K it is clear that you cannot ascribe the observed miscibility gaps solely to atomic radii.

To further complicate matters, lets keep looking at more binary diagrams. The K-Na diagram shows an intermediate compound, KNa$_{2}$ and no miscibility gap.  KNa$_{2}$ has the same crystal structure as Li$_{2}$Ca (both are MgZn$_{2}$ prototype). The Ca-Na binary shows -- surprise! -- a miscibility gap towards the Na side, peaking at 1200K. So, Ca and Na, even with fairly similar atomic volumes, do not want to mix in the liquid, while Ca and Li are perfectly happy to.

Bottom line - atomic radii do not really mean that much when predicting binary phase diagrams.  Even with comparable radii, you would be surprised at the weirdness than can be found on phase diagrams. As one example, Ag-Cu (both fcc metals) do not form a continuous solid solution, and have a slightly positive enthalpy of mixing in the liquid (not enough to form a miscibility gap).  Yet, Au-Cu (Ag and Au have similar atomic radii) form a fcc continuous solid solution (with intermediate ordered fcc-based phases at lower temperatures), and has a strongly negative enthalpy of mixing in the liquid.

Have fund exploring binary phase diagrams (and ternary, ...), but don't rely on simple heuristics.