text:
What's the most efficient way to find crystal structures of drugs?
I'd like to be able to find the crystal structures of drugs, for example at the moment I want to find the crystal structure of (R)-fluoxetine so that I can use Jmol to create an accurate 3D model of it. This is so I can then upload it to Wikimedia Commons so it may be used alongside Benjah-bmm27's structure of (S)-fluoxetine on the Wikipedia page for fluoxetine. I have tried this database [1] but when I searched for the SMILES of (R)-fluoxetine, CNCCC@HOC2=CC=C(C=C2)C(F)(F)F (from ChemSpider [2]), I obtained no search results.

I think my first question is why you need the crystal structure?
For organic molecules, the definitive resource is the proprietary (and fairly expensive) Cambridge Structure Database.
The Crystallographic Open Database which you linked is fairly new and expanding, but does not have the many, many molecular crystals in the CSD. For example, American Chemical Society journals required submission to the CSD for publication.
My suggestion would be to start with the NIH Chemical Resolver or PubChem. The latter has a 3D computed structure of (R)-fluoxetine, which I'm sure is pretty decent. Jmol should be able to load these directly, or you could save SDF files and open them in Jmol.
I'll point out that Avogadro can also do these tasks, but since I'm the creator, I'm biased.

Avogadro with UFF performs fine for simple cases. Some special cases, including atoms with unusual coordination and/or delocalized structure do not work, one need QC software and good initial guess to get adequate structure in the case.

Consider http://www.rcsb.org/pdb/home/home.do , that contains structures for proteing-drug complexes and http://www.drugbank.ca/ . Also, don't drop plain google searches

Are there any crystal structures of the free base?
The structure of the hydrochloride was published by D. W. Robertson and co-workers in  J. Med. Chem., 1988, 31, 185–189 (DOI).
The structure of 1:1 co-crystals of the hydrochloride with different acids was published by S. L. Childs and co-workers in J. Am. Chem. Soc., 2004, 126, 13335–13342 (DOI).
The CIF of the 1:1 co-crystal of the hydrochloride with benzoic acid is freely available.

authors: Петар Васиљевић


text:
Are radioactive isotopes of silver dangerous?
I know we all think of things like uranium as being radioactive in the nasty radiation poisoning sort of way. But do elements like silver have similar effects in their radioactive isotopes?
I'm an author doing some research for a book I'm working on. Would there be a particular quantity or exposure time that would be necessary to make it toxic?
Say we did know the amount to make it toxic. I know regular silver can cause Argyria which is mostly a cosmetic illness, but would the side effects of the radioactive silver be similar to that of other radioactive materials?

@Kaylyn, that depends on the radioactive decay chain. Turns out the majority of radioactive silver isotopes decay via beta minus decay. The biological effect of radioactive silver is therefore similar as for radioactive isotopes of other elements that emit beta minus radiation. This is more dangerous externally to biological systems than alpha particles emitted by radioactive isotopes of uranium say which are more dangerous once they get into the body since they can be blocked by the skin.

@TheJsyChemist If you wanted to expand on that, it would make a good answer (the other answer covers some of it, but not all)

I incorporated some of the information from the comments into your question so they are more permanent.

Please do not mix up toxicity of silver with "radiation poisoning" or other health hazards coming from radiation. Carcinogenic health hazards work different way than regular poisons, and the type of risk caused by a minimum amount of carcinogen is different from regular toxic chemicals.

A glance through the table of the isotopes in the venerable CRC handbook reveals that the longest lived radioactive isotope of Ag is $Ag^{108m}$, made by neutron capture by $Ag^{107}$.  this has a listed half life of "> 5y", but the capture cross section is only $35\pm5$ barns, pretty small for thermal neutrons.  So, it would be really hard to get a lot of radioactive silver.
Next, all isotopes listed decay by $\beta, \beta^{+}$, EC, or IT (electron capture of internal transitions for the metastable nuclei), with decay energies ranging from 93 keV ($Ag^{107m} \rightarrow Ag^{107}$) to 6MeV ($Ag^{116}$).  Those are ionizing radiation, which would increase your probability of getting cancer.
The 'nasty radiation poisoning' things are actually nasty primarily because of being heavy metals, the radiation comes along for the ride, so to speak.  The normal Uranium isotopes are very long lived, so emit very little radiation per unit time.  Polonium, the scary poison of choice in stories and perhaps by the KGB, in contrast, has a short lifetime (138 days for $Po^{210}$), and spews lots of ~5MeV $\alpha$ particles, heavily damaging tissue (much more so than gammas).
Given the difficulty in making the stuff, assuming you would try $Ag^{108m}$, (small cross sections, can't chemically separate from the stable $Ag^{107}$, etc., your best bet would just be acute heavy metal poisoning from Ag salts (which may or may not be highly toxic). Or, using a silver candlestick to bludgeon the poor victim to death.

Hahaha! Candlesticks may work a bit better. Maybe a dirty candlestick bomb. Everyone's been really helpful, though. Chemistry was my worst subject at university so I really appreciate the time everyone has put in to my random question.

The most important radioactive silver nuclide is Ag-110m (half-life: 249.9 d), which is generated in nuclear reactors.
The effective dose coefficient of Ag-110m for ingestion by adult members of the public is 2.8E−09 Sv/Bq. However, this committed dose is evaluated over 50 years. In order to estimate the deterministic short-term effects of radiation sickness, only the dose that is actually accumulated during a few days is taken into account. For a period of seven days (time after intake), the effective dose coefficient of Ag-110m for ingestion by adults is 1.8E−09 Sv/Bq.
The first symptoms of acute radiation syndrome may occur at a dose of about 1 Sv. Thus, you would need to eat at least 5.6E+08 Bq (or 560 MBq) of Ag-110m in order to find such symptoms within seven days after acute intake.
The corresponding mass of pure Ag-110m amounts to 3.2 µg. However, any real sample of Ag-110m would be “diluted” with a larger amount of stable silver.

The energy of radioactive emission does depend on nucleus decaying. This, however, does not depend only on number of protons in the nucleus, so the fact that silver is, well, silver, does not make its radioactive isotopes less or more dangerous if they somehow enter biological system.
However, it must be noted, that if a radioactive nucleus belongs to element/compound that can deposit/accumulate in human system, its damage potential increases as it stays there longer. This applies to elements/compounds that can deposite in bones (calcium, stroncium, phosphorous) or accumulate by different means (mercury, some other heavy metals, some hydrocarbons. And yes, silver also can accumulate in animals)
Another particulary nasty habit is capability to produce radioactive gases, usually radon. Unlike most common radioactive elements, radon can be inhaled, effectively bypassing skin, that is some (even if little) defense against radioactivity. This is why uranium and torium mines must have especially effective ventilation systems.

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text:
What is the color of elemental silver?
From my lab book it says it is black? I don't understand how we know the color of it is black. Can anyone explain?

Here is the reflectance spectrum for silver

It reflects the entire visible spectrum, absorbing light only in the ultra-violet region.  An ingot of elemental silver will have a silvery-white color.  In fact silver is said to be the "whitest" of all metals.  If you grind the silver up into smaller and smaller particles, the silver will appear progressively grayer or darker.  This is because smaller particles become less effective at reflecting light and produce more scattering.  Eventually when the particles are very small, even less light is reflected to the eye and the material appears black.  So depending upon the particle size, the color of silver can change from white to black.
Also, silver can react with chemicals present in air and tarnish.  It is the reaction of silver with sulfides in the air to produce silver sulfide that turns silver black, but this is through a chemical reaction.

Fun fact: Gold nano particles appear purple-ish for the same reason.

Ah, the glorious Lycurgus Cup.

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text:
Is there a compound where one element has a CCP structure with all voids occupied by others?
My Google-fu hasn't got me anywhere. All results have been quite fuzzy about it, nothing explicit.

Not quite sure what this is asking. If you tried to occupy all the voids, you'd just create new ones. Imagine you pour marbles into a box and try to fill the holes in the structure with more marbles---you'd never catch up with the holes you're creating!


text:
What is the oxidation state of oxygen in Na2O2?
I am asked to determine what is the O.S. of oxygen in $\ce{Na2O2}$
I was under the impression that oxygen had a charge of -2 and I multiplied that charge by 2 giving me a total of -4.
However, the computer program I am using, states that I am wrong. Can someone clarify why?

In the vast majority of cases this is true, but teachers like to assign substances that test the exceptions to rules, so be on the lookout for peroxides and superoxides.

I don't see why does -4 come up, I believe you do understand that in such questions you are usually asked about the average oxidation number per atom of that type found in the given species.

Oxygen can take multiple oxidation states. This particular compound is sodium peroxide.
You're right that usually oxygen has a charge of -2, but in this case, there's no way that each $\ce{Na}$ can have an oxidation state of +2.
So you then work backwards, deciding if it's $\ce{Na+}$ then you have +2 from the sodium, and oxygen must have an average oxidation number of -1 per oxygen atom.
Besides peroxides ($\ce{O2^{2-}}$) there are also superoxides ($\ce{O2^{-}}$).
It's not surprising that these compounds are usually very reactive.

I think this is a terminology difference. I meant to imply that there's no way to balance a -4 charge from two normal oxides with two sodium ions. You'd need +2 on each sodium, which clearly isn't the case.

Do you really mean that "oxygen must have an average oxidation number of +1", rather than -1? I'm trying to understand more clearly the difference between oxidation number and oxidation state, but I must be missing something.

@Cohen_the_Librarian Err yes, thanks. Fixed the typo.

There are "special rules" for determining the formal charge in some situations.  Peroxides fall into one of these special cases, see here (see the section entitled "Oxygen in peroxides").  Each oxygen in a peroxide is assigned a formal charge of -1.  Since sodium peroxide is neutral overall, each sodium atom is assigned an oxidation number of +1.

The structure of peroxides - $\ce{ROOR}$ - implies that the oxygens will generally exhibit -1 oxidation states. Oxidation states are assigned as if all bonding is ionic.
Given that oxygen is the second most electronegative element on the periodic table, we would expect that all the electrons in the $\ce{R-O}$ bond to be "taken" by oxygen (unless $\ce{R=F}$ as in the case of $\ce{FOOF}$).
In the $\ce{O-O}$ bond, however, the electrons are split 50/50 between the two oxygens (since both are of equal atomic electronegativities).
Therefore each oxygen has an oxidation state of -1 in peroxides (usually): each oxygen has 3 bonding electrons and 4 lone pair electrons for a total of 7 electrons, and oxygen by itself only has 6 electrons. 6 minus 7 is -1.

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text:
By what mechanisms can molecules with the same empirical formula be so different?
I understand the difference between empirical and chemical formula. But I always thought that compounds and molecules took their specific shape because it's the lowest energy configuration. How then, can fructose ($\ce{C6H12O6}$) and formaldehyde ($\ce{CH2O}$) both exist?
If for example formaldehyde had overall the smaller energy, why then wouldn't fructose just decompose in 6 $\ce{CH2O}$ molecules? What makes both of those molecules so stable?
I guess this is a very general question of what makes certain elements combine the way they do to form compounds.

What makes both of those molecules so stable?

This is a fundamental question. Let's start by looking at the following diagram; a chemist would call it a "potential energy diagram." The y-axis shows the energy of a molecule as it travels along the reaction coordinate (x-axis; transforms, reacts) from the starting material on the left to the product on the right.

Notice that there are 3 "potential energy wells" (valleys) in the diagram, these reflect molecules that are stable enough to have a finite lifetime.  Notice too that to move from one well to another, an energy barrier must be surmounted.
So to return to your question, a molecule would be "kinetically stable" if it was protected on all sides by high energy barriers.  A molecule would be "thermodynamically stable" if its potential energy well was located at an energy that had a very low value on the y-axis.  A molecule that is very stable in a thermodynamic sense, could still be kinetically unstable (and therefore very reactive) if it was surrounded by low energy barriers.  By the same token, a molecule that is very unstable in a thermodynamic sense (exists at a high energy value on the y-axis) could be kinetically stable if it were surrounded by high energy barriers to reduce the likelihood of further reaction.
Millions of molecules are stable at room temperature, but if you supply enough energy to the system (heat them for example) they can all transform into other compounds.  At room temperature there are high enough energy barriers separating fructose and formaldehyde and they will both be stable in a jar, but under more energetic conditions such transformations might be feasible.

Just a small correction: a chemist would, of course, call the diagram in your post potential energy curve. It is the variable along the abscissa which chemist would call reaction coordinate.

@Wildcat  Good point Wildcat, I'll edit my answer accordingly.  Thanks!

Thanks for this! Very helpful! Now I'm wondering: if you just dump a whole bunch of atoms in a box at a certain temperature, so a couple of C,H and O, is there a deterministic solution to which compound will be formed? Also: do we know enough about chemistry and physics to synthesise any molecule we want if we had all the atoms separated and could add those to a mix at will?

@Moppentapper, answer to your first question is, I think, yes. Roughly speaking, you have to calculate potential energy surface which will presumably have many minima corresponding to different (meta)stable combinations of these atoms and many maxima connecting those minima. Then given a temperature you can estimate the average amount of kinetic energy atoms and forming (meta)stable molecules possess, so that you can predict which maxima will eventually be crossed and which minima will be popoulated.

@Moppentapper, note, howeever, the word eventually, since while we can predict the final state of a system, estimating reaction rates theoretically is in general impossible. See http://chemistry.stackexchange.com/questions/4031/could-you-estimate-the-rate-of-reaction-from-the-chemical-structure-of-the-react/5712#5712

Thanks @Wildcat that was all very useful!! I appreciate the help!

@Moppentapper, your second question sounds for me like it is about molecular assemblers - hypothetical devices that are a popular topic in the area of molecular nanotechnology and science fiction. :D

I guess this is a very general question of what makes certain elements
  combine the way they do to form compounds.

At the fundamental level, I think, a good answer to this question would be the principle of minimum potential energy, which basically states that a system has a tendency to be in a configuration that minimizes its total potential energy. (Please, note the word tendency!)
For the case of molecules it means that if forming a molecule out of atoms lowers the potential energy comparing to the situation when these atoms are infinitely far apart of each other, than there is a natural tendency for this molecule to be formed.
But there is a one important detail that complicates matters: given some set of atoms, there is no guarantee that there exist only one possible arrangement of them which minimizes the potential energy. In fact, the situation is usually the opposite: there are many different minima in the potential energy corresponding to different possible arrangements of the same set atoms. Often we can clearly identify one global minimum (the lowest one) and call all other minima local minima.
Any local minima in potential energy (directly on indirectly, i.e. through other local minima) is connected with the global minimum in a sense that there exist some way for a system to transit from a local local minima to the global one. But on it is way to the global minimum from a local one potential energy first rises and only then falls. It falls to the lower value, but first the system need to overcome a barrier. Thus, we say that a structure that corresponds to a global minimum is stable, while structures that correspond to local minima are metastable in a sense that a system (a molecule) actually may reside in these local minima for quite a long time if the barriers are relatively high. Relatively to what? Basically, to the amount of kinetic energy a molecule posses so that it can be converted to the potential energy to overcome the barrier.
A good analogy may be drawn with a ball resting in a hollow on a slope (image below courtesy of Wikipedia)

If the ball being in a hollow (1) does not have enough kinetic energy to be converted into potential with the aim to overcame the barrier (2) and move on to the lower energy minimum (3) it will stay at (1) for ages. But if the ball has enough kinetic energy it will roll to the lower level. Here (3) might be just another local minimum or the global minimum, it does not matter. If (3) is not the global minimum then this process is just one of the steps on a way to the global minimum.
Now, back to molecules, think about (1) as of 6 molecules of formaldehyde being far apart of each other and about (3) as of 1 molecule of fructose. Both these different combinations of atoms correspond to minima in potential energy, and here I assumed that fructose is lower in potential energy. Note that I could not guarantee that, as well as, I could not guarantee that of of these minima is the global minimum for a system composed of 6 atoms of $\ce{C}$, 12 atoms of $\ce{H}$, and 6 atoms of $\ce{O}$. Apriori we do not know what form is lower in potential energy, so we have to do some calculations, say. But here let me just assume that $\ce{C6H12O6}$ is lower in potential energy than $\ce{6 CH2O}$.
Now recall that (as I noted at the very beginning) the principle of minimum potential energy only tells us that there is a tendency for a system to favour a configuration with the lowest potential energy (i.e. $\ce{C6H12O6}$ out of two choices we discuss). But starting from $\ce{6 CH2O}$ the natural tendency to form $\ce{C6H12O6}$ would not necessarily be realized in reality if the barrier between these configurations is relatively high in a sense already mentioned. So, it might be the case that at, say, room temperature, most of the $\ce{CH2O}$ molecules do not posses enough kinetic energy to be converted to potential energy to overcame the barrier. But you can increase the amount if kinetic energy of $\ce{CH2O}$ molecules by rising the temperature, so that the system will roll to overcome the barrier. Chemical reaction will take place, although, I'm not sure that these two ($\ce{6 CH2O}$ and $\ce{C6H12O6}$ are directly connected). But conceptually, the situation is similar to just pushing the ball from the picture above. :D


text:
Why do different impurities create different colors in diamonds?
Traces of nitrogen make diamonds yellow while traces of boron make them blue.  What accounts for the differences in color for a given element?

If you had googled a little bit you would have found this wonderful resource.

In a yellow diamond, a few carbon atoms per million have been replaced
  by nitrogen atoms, each containing five valence electrons. The
  structure of the diamond crystal does not change significantly, but
  the extra electrons occupy a donor level.
The nitrogen donor level energy in a diamond is large, peaking at
  about 4 eV. With a concentration of a few nitrogen atoms per million,
  instead of a clean "spike" donor level energy, the nitrogen donor
  level energy broadens into a band because of a number of complex
  factors, including thermal vibrations. This broadened donor energy
  band results in the difference between donor and conduction bands
  being as low as 2.2 eV. The most likely transition allows incident
  light with energy of 4 eV per photon to excite electrons from the
  donor level to the conduction band. However, it is possible for
  electrons to be excited to the conduction band with energy of 2.2 eV
  and upwards. This means that blue and violet light are absorbed from
  the full spectrum of light normally transmitted, and the resulting
  color is yellow.

Unlike blue boron-doped diamonds, which conduct electricity,
  nitrogen-doped diamonds remain insulators. This is because nitrogen is
  a deep impurity: a relatively high energy (2.2 eV, as compared with
  0.4 eV for boron) is required to excite electrons from the donor energy level to the conduction band, and only a fraction of the
  available electrons are freed to carry a charge.
An extremely rare green color can result from a higher nitrogen
  content of about 1 atom per 1000 atoms of carbon. At even higher
  nitrogen concentrations, the donor level broadens so that all visible
  light can be absorbed, resulting in a black color.
Synthetic blue diamonds are created by adding boron as an impurity.
  Boron is trivalent, having three valence electrons; where a boron atom
  replaces a carbon atom in the diamond structure there is one fewer
  electron than usual. This missing electron, or hole, creates an
  acceptor energy level above the valence band. The boron acceptor
  energy is only 0.4 eV, so light of any energy can be absorbed during
  excitation. The boron acceptor band is broadened, and the absorption
  tapers off throughout the visible light energies, resulting in
  stronger absorption at the red end of the spectrum. At a level of one
  or a few boron atoms for every million carbon atoms, an attractive
  blue color results. Natural diamonds of this color, such as the Hope
  Diamond, are rare and highly priced.

There are two mechanisms by which diamonds can become colored.  The first mechanism involves impurities.  As the diamond crystal is growing, materials in the atmosphere or soil can become incorporated into the diamond structure.  The bonds that form between carbon atoms adjacent to the impurity and the impurity can absorb photons and result in electronic transitions.  If these transitions happen to occur in the visible region of the spectrum, then any reflected light will not contain the absorbed wavelength and the diamond will no longer appear white (as it does when all wavelengths are reflected).
A second mechanism that can impart color to diamonds involves structural deformation of the diamond crystal.  As the diamond crystal is growing, tectonic forces or radiation can cause the diamond crystal to grow in an irregular manner.  There are no impurities as described above, but a carbon atom may be missing here and there.  These structural anomalies also result in a shift in the frequency of light absorption by the diamond.  Again, if the light absorption shift is into the visible region, the diamond will appear colored.


text:
Why does gallium nitride allow an AESA radar to be more powerful than a radar with elements made of gallium arsenide?
AESA (Actively Electronically Scanned Array) radars use thousands of individual transmit/receive elements to create a radar.  All in service AESAs use gallium arsenide as the material in the t/r elements.  It is said that gallium nitride allows a tenfold increase in the power emitted by these elements?  What allows GaN to emit so much more radio frequency energy than GaAs?

The short answer is that GaN transistors have a much higher breakdown voltage and saturation velocity than GaAs, allowing them to operate at higher voltages at high frequency, handling more power. The large band gap (~2x GaAs) means that GaN still behaves like a semiconductor, even at relatively high temperatures, allowing for higher operating powers.
GaN has really been a breakthrough in semiconductor technology. It's what allows us to have 405 nm violet laser diodes and could potentially replace magnetrons in microwave ovens. NXP has a little document here comparing GaN with a few older technologies that's a quick read.


text:
Heat of combustion for octane
I'm a Mechanical Engineering student and I have been working on a school project but I want to go a bit beyond that and it has led me here. I'm designing an engine cylinder, and I figured I should be able to calculate the heat of combustion of the working fluid.
Please remember that I have no chemistry textbook and all I have been doing up to now is using the internet and some vague memory from high school.
So I wrote the balanced equation,
$$\ce{C8H18 + 12.5\left(O2 + 3.76 N2\right) -> 8 CO2 + 9 H2O + 23.5 N2}$$
using that I was able to compute the air-to-fuel ratio, individual masses for the $1.8\ \mathrm L$ volume, etc. I settled on a compression ratio of 8 which after some thermodynamic computation (if neccessary I'll present them) I got to the following state,
$$
T_2 = 650.9 \; \text{K} \\
P_2 = 1779 \; \text{kPa}
$$
basically those are the conditions prior to the spark. I'm trying to compute the heat of combustion using those properties. Using Mathematica I found out that for Octane the heat of combustion is somewhere along,
$$
\Delta H=5474\;\frac{\text{kJ}}{\text{mol}}
$$
but I assume this is for STP, my interest is to calculate for my conditions at state 2. After some bit of research I noticed one has to use the specific heats of the participants, but the website stated 'find these at the end of the book' – which I don't have one. To be honest, it will probably take me days to figure out this problem on my own using the internet. So does anyone know of a shortcut such as an online database or some public tables that display specific heats (maybe as a function of temp) for octane. Or how would you go about solving my problem?
Remember that I don't have much knowledge regarding advanced chemistry, and most of my interest lies in the field of engineering.

First, nitrogen does not participate in the reaction, so remove it (BTW, it's not balanced). You would still need the same volume of air since oxygen is about 21% of air. Could you tell me how much octane is in the cylinder when it is ignited? Also, is there an excess of oxygen?

balanced equation is 2C8H18+25O2----> 18H2O+16CO2

Remarkably, also this exercise (like many textbooks and lectures) selects the combustion reaction of octane of all the possible combustion reactions of gasoline to show the principle of combustion, although gasoline actually does not contain a high concentration of octane.

Enthalpy of a system is a state function, meaning, it does not depend upon the path you take to reach the end state. What that means in this case is that the change in the enthalpy of combustion will be the heat required to raise the temperature of the combustion material to the new temperature and to increase the pressure. Essentially, think of it as if the combustion is occurring at STP, and you must construct a path from the beginning state to the end state which will pass that combustion at STP. In this case, that would be:

Initial state of reactants -> STP reactants
Heat of combustion at STP
STP products -> final state of products

Your final state will be the same as your initial state, 650.9 K/1779 kPa. The important point is that the heat in the change of the reactants to STP will be different from the change of the products back to the initial state. This difference would be reflected if you found a heat of combustion at some condition other than STP.
Nitrogen is a spectator to the reaction, so you can completely ignore it for the sake of reaction. However, you do need to account for raising it to the final pressure/temperature if either is changing. I'll assume you've got that under control.
Now, the issue comes in at the fact these are somewhat extreme conditions, at least in pressure. To calculate this properly would take a great deal of time and effort. That said, there may be corners you are capable of cutting in the process. The general process you need to follow is:

Take octane and $\ce{O2}$ from 1779 kPa to 1 atm, 650.9 K to $T_{1,\mathrm i}$.
Take octane and $\ce{O2}$ from $T_\mathrm i$ to 25 °C at constant pressure (using $C_p$).
Use the heat of combustion.
Take $\ce{H2O}$ and $\ce{CO2}$ from 1 atm to 1779 kPa, 25 °C to $T_{2,\mathrm i}$.
Take $\ce{H2O}$ and $\ce{CO2}$ from $T_{2,\mathrm i}$ to 650.9 K.

The overall enthalpy change of this process will be the heat of combustion at your conditions. Now, you'll have to decide for yourself about steps 1 and 4, as you need to be considering intermolecular interactions, and thus need to pick an equation of state (not ideal gas). I feel like I am not necessarily knowledgeable enough in this specific field to offer suggestions in that regard. I personally would just pick something like VDW, but there might be a specific standard for combustion processes. Further, only one book that I have access to right now that has the proper temperature-dependent-equation form for $C_p$, and it doesn't have one for octane. These are the ones for the other gases involved:
$\ce{O2}$: $C_p = 25.46 + 1.519 10^{-2} T - 0.715 10^{-5} T^2 + 1.311 10^{-9} T^3$
$\ce{CO2}$: $C_p = 22.243 + 5.977 10^{-2} T - 3.499 10^{-5} T^2 + 7.464 10^{-9} T^3$
$\ce{H2O}$: $C_p = 32.218 + 0.192 10^{-2} T + 1.055 10^{-5} T^2 - 3.953 10^{-9} T^3$
From Chemical and Engineering Thermodynamics (Sandler).
I wish I could help more. Hopefully this explanation at least points you in the right direction.

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text:
Accessible statement on the intuition and usefulness of group theory to chemistry
I am writing a lesson plan for high school students who are studying group theory,
and I am having difficulty in relating the subject matter to the real world.
I have quite a few examples using the Rubik's cube, but I would like to use
some statement that might provide some intuition to students of how group theory
is useful for chemists.
I understand that group theory is used with spectroscopy, but I am not sure how to
relate this back to the theory.
I have seen some good reasonable non rigorous attempts to relate group theory to
the insolubility of quintic polynomials, but I think it would great to have a second
more concrete example, such as chemistry.
I understand this may be a big thing to ask, and would be happy with just a broad brush strokes answer without being too technical or trivial.

Individual molecules rely on point group symmetry - that is, the symmetry about a point. Crystal structures rely on space group symmetry including the translational symmetry in a crystal.
Many molecules, particularly important ones, are highly symmetric. The different symmetry operations (reflections, rotations, etc.) form a finite group.
There are a number of interesting group theory results, but a simple statement comes down to two things:

Understanding Orbitals, Bonding, Vibrations..: Using point groups, any function of the molecule corresponds to a linear combination of irreducible representations. This means I can quickly understand vibrations of a molecule, spectroscopy (as you mentioned) and orbital diagrams and bonding using just pencil and paper.
Faster Calculations: Using symmetry, you can greatly reduce the amount of time required for quantum calculations. For one, you only need to consider the symmetry-unique atoms and bonds. For another, you can show that certain types of integrals and calculations will always be zero and can be completely ignored. This leads to immense speedups.

For example, chemists label orbitals with symmetry labels based on the point group. Using the direct product we can realize that certain interactions will never occur.
Similarly, vibrations have symmetry labels, and we can determine what types of spectroscopy will interact with those vibrations, or whether they are "silent."

THanks for that answer Geoff, that gives some nice insight without getting too technical.

I realize my answer renders this off-topic for Chemistry, but what about general relativity? The fact that there is no "special" point in the universe is made concrete by saying that general relativity is invariant under the group of rigid motions (rotations and translations). That's an interesting real world group on its own.

Thanks for that all coments are helpfull, I know this is a very difficult topic.


text:
Seeking word equations
is there a general word equation so I can learn how to write out the following:

With chlorine, it seems to be: Hydrocarbon + Chlorine ---> ____ + HCL
With Hexane, it seems to be: Hydrocarbon + Br ---> ____ + HBr
I am not sure how to derive the first part of the final reaction (____)
It seems that methane has lost one hydrogen atom.
It seems that hexane has lost one hydrogen atom too.
Does that mean, if it was say Pentane with Chlorine, the first part of the final reaction would be CH4CL (not ch5cl) ? Likewise, would Pentane with Bromine be C6H12BR (and not c6h13br)
It feels like i'm rote-learning a skill, wonder where my knowledge is missing hmm..

In each case an atom of a halogen has replaced one of the hydrogens on the alkane.  That hydrogen sticks to the other halogen atom, forming an acid.  Your missing product is a called a haloalkane or an alkyl halide.  Pentane, C5H12, would turn into bromopentane, C5H11Br.  If this is for an organic chemistry class, your professor should have talked about the mechanism of the reaction, and not just its existence.

I don't understand your question completely, but If you want to understand what's happening in an organic reaction you need to know which type of reaction is in question and you have to look at the mechanism. This is a radical reaction with the following mechanism (excluding the termination step):


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The role of halogens in electrophilic aromatic substitution
I'm trying to understand what halogens do in electrophilic aromatic substitution. Consider the following.
Bromobenzene is introduced to a carbocation. Bromobenzene will likely react, but at a slower rate than unsubstituted benzene, because the bromine is inductively withdrawing. The bromine is also not resonance donating/activating, but can stabilize the intermediate product through resonance donation.
The bromine is an ortho/para director, but not because it donates electron density to these positions. Rather, when bromobenzene attacks a carbocation, the meta positions are formally positive, and through resonance donation, bromine can complete an octet in one of the resonance depictions.

Is this an accurate depiction of what a halogen does to benzene?

The bromine is also not resonance donating/activating

You can't have it both ways.  If the bromine can resonance interact with the intermediate, then it can also resonance interact in the starting bromobenzene.
Let's back up to the starting bromobenzene.

Bromobenzene will likely react, but at a slower rate than plain
  benzene, because the bromine is inductively withdrawing

That's correct, but let's compare further.  The oxygen in phenol is electron withdrawing (oxygen's electronegativity is even greater than bromine's), yet phenol is a very reactive aromatic compound towards electrophilic aromatic substitution (it will decolorize a solution of bromine, it doesn't even need a Lewis acid catalyst).  The difference between bromobenzene and phenol is that the phenolic oxygen can interact strongly via resonance and donate electrons into the aromatic ring, while the bromine in bromobenzene also resonance interacts with the aromatic ring, but much less efficiently (the larger size of bromine's $\ce{4p}$ orbitals makes overlap much less efficient with carbon's 2p orbitals).  Although the interaction is much less efficient, there still is some resonance interaction between the bromine substituent and the ring.  Look at the analogous resonance structures for bromobenzene that you drew for the intermediate, in the contest between inductive deactivation and weak resonance activation, the ortho and para positions are the least deactivated.

When bromobenzene undergoes electrophilic aromatic substitution, the whole ring is deactivated because (unlike phenol, where the resonance effect overwhelms the inductive effect) the weak resonance effect doesn't offset the strong inductive effect.  However, since the bromine substituent does interact with the ortho and para positions by resonance, these two positions are less deactivated that the meta position which cannot interact via resonance with the bromine substituent.

So two competing effects? @ron - induction vs resonance? But still not enough induction to completely deactivate benzene from reacting, unlike with a nitro group (both inductively and resonance deactivating)?

@Dissenter  That is exactly right, enough inductive deactivation to make all positions react slower than benzene, enough resonance activation to differentiate the ortho and para positions from the meta position. Nice analogy with nitrobenzene!

thank you; so any substituted benzene will react with a halonium ion, but not any substituted benzene will react in a Fridel crafts alkylation or acylation?

No, I don't think so.  Just as some substituted aromatics are to deactivated to react in a Friedel-Crafts reaction, I suspect there are also some very deactivated aromatics that won't react with halonium ions.

In a FC reaction if we have an exclusive meta director then the benzene is deactivated for reactions, correct? @ron

@Dissenter Yes, that's correct.


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Is there a general method for working out electron configurations?
For up to period 3, elements' electron configurations are relatively easy to work out until the transition metals begin. Then some elements may have $\ce{4s}$ orbitals which are preferred by electrons over $\ce{3d}$ orbitals. These small rules don't make sense all the time and only cover some elements. Is their any master method for working out the electron configurations of any given element?

No, at least not by hand. The Aufbau Principle is only a guideline but in later elements many different effects are at play which are nearly impossible to consider without doing a thorough calculation. Maybe this question is of interest to you.

There is a semi-empirical method to fill the atomic orbitals: atomic orbitals are filled  in the order of increasing n+l. If the sum n+l is equal for two atomic orbitals, they are filled in order of increasing n.
This gives the following order for filling the orbitals:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, 5g, 6f, 7d, 8p, and 9s


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What is the oxidation state of all the atoms in this compound, W(CH3CN)3(CO)3?
$\ce{W(CH3CN)3(CO)3}$
I'm thinking:
$\ce{CO}$ is neutral so $\ce{O}$ is -2 and $\ce{C}$ +2
$\ce{CN}$ is negative one so $\ce{C}$ is +2 and $\ce{N}$ -3
$\ce{CH3}$ is positive one so $\ce{H}$ is +1 and $\ce{C}$ is -2
$\ce{W}$ is neutral so $\ce{W}$=0
is this right?

Why is the compound a compound? Why is the neutral W there? And what exactly does your class use as the definition of "oxidation state" anyway? What would the oxidation states of molecular oxygen and nitrogen be? (i.e. <O=O> and |N≡N|)

HINT:
Compound is neutral, So the sum of oxidation state of all the element should be ZERO.
In what you have got, is sum of oxidation state of all the element is zero? This way you can check whether you are right or wrong.

I know its an old question but just wanted to make a technical correction .
The final integral values deduced by you are quiet correct but the method of arriving at OS of W looks clumsy .
As Freddy says , the compound is neutral , hence sum of all OS must be zero .
So we get OS of W as zero not by

"W is neutral so W=0"

but rather using the fact that the compound is neutral and the other elements (which are in the form of ligands) form neutral ligands .

This is marginally an answer, but please be cautious that the answer space is for answers only, not for new questions and not for comments.  Once you attain 50 rep, you can comment everywhere on the site.  I mention this because you've had a few instances where you've misused the answer area.

@jonsca the OP's question was whether the answers were correct . I have mentioned in my answer that they are correct but just the methodology needed to be refined . I am sorry if I did something wrong .

Perhaps my choice of the word "marginally" was a poor one. I just meant that you'd had a couple of instances where there had been some confusion, and I wanted to just give you a reminder about it.  I apologize if it came out in a condescending manner.


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Converting from Grams to Molecules
I was doing some problems from the textbook on converting between units using dimensional analysis and I came across this problem.
A vat of Hydrogen Peroxide ($\ce{H2O2}$) contains
455 grams of oxygen atoms. How many
molecules of ($\ce{H2O2}$) are in the vat?
Sorry if I didn`t format the question correctly, I am a still a newbie.
But how would I set this up? I would not like the answer to this question but I would like some input on how would I get how many molecules of hydrogen peroxide from grams of oxygen atoms? Any help would be much appreciated. Thanks in advance!

A vat of Hydrogen Peroxide ($\ce{H2O2}$) contains 455 grams of oxygen atoms.
  How many molecules of ($\ce{H2O2}$) are in the vat?

How many moles of oxygen atoms are in 455 gms of oxygen atoms? Let's call the answer $x$.
How many moles of $\ce{O2}$ are in $x$?  Let's call the answer $y$.
How many moles of $\ce{H2O2}$ could you make from $y$ moles of $\ce{O2}$?  Let's call the answer $z$.
1 mole of any substance contains Avogadro's number of molecules.  How many molecules are in $z$ moles?

Thanks for the response. For #3, is the answer 14.21 moles of Hydrogen Peroxide?

And is the answer to the question 8.56 * 10^24?

@ToxicTechnetium  You got it!


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reaction with Barium Chromate, Potassium nitrate
I'm confused, in the reaction:$$\ce{K_2CrO_{4(aq)} + Ba(NO_3)_{2(aq)}\rightleftharpoons BaCrO_{4(s)} + 2KNO_{3(aq)}}$$
I got a white precipitate in an opaque yellow solution and I was told that the white precipitate was $\ce{BaCrO_4}$ but while searching on Internet I found that its color is yellow, and that the $\ce{KNO_{3(aq)}}$ is white, so my question is what is true? or what causes the color of the final solution?

I obtained barium chromate once, it is indeed bright yellow. I still can believe that fine dust may look pale enough to be confused with white, especially if the lights are inappropriate, but it is unlikely.

maybe there was an interference in our reagents

Certainly $\ce{KNO3}$ is highly soluble.

Any possibility that the potassium chromate is contaminated with a sulfate salt? That would give barium sulfate as a white precipitate.

${\ce KNO_3}$ is definitely not the precipitate, nor should it be responsible for the color of the solution (that was a fine precipitate of your intended product, barium chromate.)  Is it possible that your potassium chromate was made using tap water, or that your glassware wasn't cleaned with distilled water prior to use?  Barium hydroxide and barium carbonate both give white precipitates that could occur easily from contamination.

Actually, after some though. Are you sure that your precipitate is white? Could it be that it is white-yellow, but under more deep yellow solution it looks like white?

If you used distilled water to prepare the solutions of potassium chromate and barium nitrate, it is more likely that you got a yellow precipitate of $\ce{BaCrO4}$ in a yellow solution of excess $\ce{K2CrO4}$. The yellow color originates from $\ce{CrO4^{2-}}$, and a small amount of finely divided precipitate may look white at first sight (namely when the yellow color of the solution is more intense), when it is actually pale-yellow (it can look like this).


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Why is K2CrO4 Potassium chromate?
I was given $\ce{K2CrO4}$ and was asked to name the compound.
Can this be done by:
1) Determining the molecular compounds oxidation state
or
2) Does one need to memorize the polyatomic ions and that is the only way to know what is what?

There are many hows for nomenclature.  #2 above is correct.

Wow! My teacher gave us 24 to remember.

Memorizing a small number (and 24 is a small number) of things is sometimes preferable.

Definitely you need to remember it. Potassium Chromate  is IUPAC name of $\ce{K2CrO4}$. There are other names such as Chromic acid and dipotassium salts. But mostly IUPAC names are accepted.
I don't think you should try to mug up all 24 names instead you should keep getting familiar with them when they come in reactions or somewhere else. After seeing them at few different places you will be remember them all. They will come very often while you are learning chemistry.

"I don't think you should try to mug up all 24 names" We're right around that time in General Chemistry classes where brute force memorization of all the polyatomic ions are necessary.  ;)

@LordStryker In chemistry there are few stuff you need to memorize, which does not include small compounds :)

According to the chemistry exams I write and administer, there are a lot of things you need to memorize.

@LordStryker Just now i am memorizing uses of each element of S-block.

2 - Yes and no.
It is recommended to known common ions. They are, however, derived in uniform fashion from name of the element in latin. For example, for $Fe$ the name is Ferrum so anions are Ferrates. The 'fancy' part of the anion may consist either of naming the oxidation state (and optionally the number of oxygens and other atoms), i.e. (tetraoxo)ferrate(VI) or known suffixes and prefixes
(hypochlorite-chlorite-chlorate-perchlorate, also hydro-, pyro-, orto- and meta- prefixes). Probably the easiest way to remember the second nomenclature would be to learn several (tens of) common ions and assume that isoelectronic ions have same prefixes and suffixed, but are derived from names of respective elements.
In practice, even though the first nomenclature is normative and is understood universally, the second is traditional and more common.
Some oxoions, however, have special names, like dithionite-ion.

authors: Prîñçess Wïñdÿ Bãrøñ


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Do lone pairs on substituents (e.g. in aniline) count towards Hückel's rule?
Why is aniline aromatic? Doesn't it have 8 π electrons including the lone pair on nitrogen, thereby violating Hückel's rule?

The way I see it, there are 6 π electrons from the benzene ring, and an additional 2 from the $\mathrm{sp^2}$hybridized nitrogen.
I am suspecting that I am mis-applying Hückel's rule. I asked an instructor today and the only rationalization he had to offer was that "we just don't count those electrons on the nitrogen" ... which is deeply unsatisfying. He still insisted that aniline is aromatic. Can someone offer an alternate, more through explanation?

It is only the electrons from atoms in the ring that count when applying Huckel's rule. Electrons from substituents on the ring are only cross-conjugated with the aromatic π-system. When you think of it in terms of perturbational molecular orbital theory the substituent electrons take the part of a perturbation for the ring's aromatic system.
The reason for only using the ring electrons traces back to the origin of aromaticity. Take a look at the π-orbital energy levels of benzene (figure taken from p 36 of Fleming, Molecular Orbitals and Organic Chemical Reactions, Reference Edition):

Here, $\alpha$ is the energy of an isolated p-orbital and $\beta$ is the stabilization energy of the π-bond in ethylene. You can see that for benzene you get the best stabilization if you have 6 π electrons ($4n + 2$ for $n=1$) in the system because then you fill the stabilized MOs $\psi_1$, $\psi_2$ and $\psi_3$ which are lower in energy than an isolated p-orbital and gain the maximum stabilization energy possible. If you'd have more electrons in the system you would have to fill the destabilized MOs $\psi^{*}_{4}$, $\psi^{*}_{5}$ and $\psi^{*}_{6}$ which would cost a lot of energy because of the large energy gap between the bonding and antibonding MOs.
Now, what happens if you add a cross-conjugating substituent to the ring. This can be seen in the (Huckel-)MO-diagram of styrene (Fleming, p 71).

The main effects of the substituent are that the degeneracies of some orbitals are lifted and that the orbital energies shift a bit, especially those of HOMO and LUMO are of interest. In the case of styrene the HOMO lies higher in energy and the LUMO lies lower in energy compared to benzene. Now there are 4 stabilized energy levels instead of 3 - any electron that occupies one of those levels will lead to a stabilization of the molecule. But that is nice and well as you get 2 electrons from the substituent which will want to occupy one of the 4 levels and that leaves you with 3 levels that have to be occupied by ring electrons. Thus, your system will be most stable if the π-system of the ring contains 6 electrons, exactly like in benzene. Any more than 6 electrons will lead to destabilization as there is still quite a big HOMO–LUMO gap (even though it is smaller than in benzene).
The argument I made for styrene will remain valid for other molecules with different substituents (like aniline). The MO scheme of aniline will look a little different but the essential features should be the same.

Pedantic time here. The cited orbital drawing is slightly inaccurate missing the minus sign in the energy level of the LUMO. Don't let this happen to your work -- watch your signs!

Hückel's rule (or the $4n+2$ rule) suggests that systems with $4n+2$ π electrons will be stabilized (aromatic), while systems with $4n$ π electrons will be destabilized (antiaromatic).  It was written with planar, monocyclic, π-systems in mind.  Applied without care, it would suggest that cyclooctatetraene with 8 π electrons is antiaromatic, yet the molecule is a stable, ethylenic like, cyclic, polyolefin - because it is non-planar (it exists in a tub shape conformation).

The Hückel rule also has trouble with polycyclic π systems, as we discussed in your question about pyrene.  The rule was not intended to take substituents into account at all.  We can say that furan or pyrrole has aromatic properties because the lone pair on oxygen or nitrogen contributes to the π-system.  But note that in these cases the oxygen or nitrogen is part of the monocyclic ring, not a substituent appended to the ring.
In the case of aniline, the molecule is aromatic. The benzene ring is, in agreement with Hückel's rule, aromatic. The lone pair of electrons on the amino substituent clearly interact with the π system of the benzene ring as evidenced by the reduced basicity of aniline in comparison to ammonia. But the amino group is a substituent attached to the ring and the lone pair on nitrogen is not counted as part of the $4n+2$ system. The same concept applies to phenol, halobenzenes, etc.: these substituents interact with the benzene ring, but their lone pairs are not part of the electron count when Hückel's rule is applied.

I think the argument is as follows: planar, octagonal, cyclooctatetraene is antiaromatic, and hence distorts to a non-planar conformation which reacts as a conjugated diene instead of as an aromatic ring.

I agree with you that in aniline there are 6 π electrons from the benzene ring, and an additional 2 from the lone pair of the amine group. But aniline is still aromatic, because there are only 6 π electrons resonating, i.e. the 6 π electrons in the benzene ring are conjugated in a plane perpendicular to the ring. While the additional lone pair of electron on the nitrogen atom cannot resonate with them, as it is not in that plane perpendicular to the ring.
We have to notice that the nitrogen atom in aniline is not $\mathrm{sp^2}$ hybridized but $\mathrm{sp^3}$.

Inaccurate answer. See here

As Thomas Aquinas declared in the title of one of his most famous essays, "Sic et Non". Many aspects of chemistry feature classifications that are taught as  definitively affirmative or negative, but in reality they show blurred lines of demarcation. Thomas Aquinas wrote his essay on theological questions, but the same answer might apply to such chemical questions as whether certain basic compounds are alkalies -- and whether and how electrons are conjugated into delocalized pi systems.
The formal lone pair on nitrogen either is or is not conjugated into the pi system of aniline. Wikipedia describes the ambiguous, intermediate situation:

The amine in anilines is a slightly pyramidalized molecule, with hybridization of the nitrogen somewhere between sp³ and sp². The nitrogen is described as having high p character. The amino group in aniline is flatter (i.e., it is a "shallower pyramid") than that in an aliphatic amine, owing to conjugation of the lone pair with the aryl substituent. The observed geometry reflects a compromise between two competing factors: 1) stabilization of the N lone pair in an orbital with significant s character favors pyramidalization (orbitals with s character are lower in energy), while 2) delocalization of the N lone pair into the aryl ring favors planarity (a lone pair in a pure p orbital gives the best overlap with the orbitals of the benzene ring π system).[1][2]

Of course, aniline definitely shows aromatic character experimentally, for instance in its proton nmr spectrum, but that does not mean there must be only six conjugated pi electrons confined to the ring. Alternatively, including the nitrogen "lone pair" into the conjugation would put the molecule outside the scope of the $4n+2$ rule because that rule applies only to an exact ring of conjugated pi electrons. Given this limitation on the $4n+2$ rule, aniline could be aromatic whether the nitrogen lone pair is conjugated or not.
Cited References
1.
Alabugin, Igor V. (2016). Stereoelectronic effects : a bridge between structure and reactivity. Chichester, UK. ISBN 978-1-118-90637-8. OCLC 957525299.
2.
Alabugin I. V.; Manoharan, M.; Buck, M.; Clark, R. J. "Substituted Anilines: The Tug-Of-War between Pyramidalization and Resonance Inside and Outside of Crystal Cavities". THEOCHEM, 2007, 813, 21-27. http://dx.doi.org/10.1016/j.theochem.2007.02.016.

The paper for the MMFF94 vs. MMFF94s force fields are interesting. Results suggested some pyramidal geometry / sp3-character to the aniline nitrogen with dynamic inversion. Enough people wanted a planar nitrogen that Halgren published MMFF94s parameters.


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Comparing rate of reaction of acetic acid and sodium bicarbonate at different temperatures
Given the reaction between $\ce{CH_{3}COOH {(l)}}$ and $\ce{NaHCO_3 {(s)}}$, which happens at the same pressure but at different temperatures, is there any way to mathematically compare the rate of reactions?
I thought using the rate equations of order 2, but I use the same concentration of materials. Suppose I could have found the connection between temperature of he material and its concentration, I would be able to find my required rate-indicators.
How to find the connection between temperature and concentration, or how to describe the rate of the reaction more exactly?

Yes, there is, it has been done by others.  Typical acid-base reactions that occur in solution are diffusion controlled reactions. Such reactions do not follow second order kinetics because at virtually every encounter of the acid and base, rapid proton transfer occurs and the reaction is complete. The reaction rate is limited only by how fast the acid and base can diffuse and encounter each other. It is usually easy to identify diffusion controlled reactions because they are extremely dependent upon the rate of stirring or mixing.  They are also very sensitive to the solution viscosity. Rate constants for these reactions are typically $>\pu{10^11 sec^-1}$.  Here is a link to a full paper that provides considerable experimental detail on one way to measure the rate of diffusion controlled acid-base reactions.


text:
Why is the difference in boiling point so much greater between methanol and methane than it is between 1 decanol and decane?
So any data sheet will tell you;
Group 1: methane boils at: -164 deg C
         methanol boils at: 65 deg C

Group 2: decane boils at: 174 deg C
         1-decanol boils at 229 deg C

that being said; why is the difference in boiling point so much greater between methanol and methane than it is between 1 decanol and decane ?
As well is there anything else other than difference in non-polarity and IMF's that contribute to differences between molecules in individual groups (i.e. between methane and methanol; between decane and 1-decanol)

The difference in boiling point, between methanol and methane on one hand and between 1-decanol and decane on the other hand, is primary due to the presence of hydrogen bonds (HB) in methanol and 1-decanol and not in methane and decane. The density of these intermolecular forces (HB) is higher in methanol than in 1-decanol (In 1-decanol, for every ten atoms of carbon there's one possibility to have HB. While in methanol for every atom of carbon there's one  possibility to have HB.


text:
Is the transformation of pi bonds to sigma bonds always energetically favorable?
Is the move from pi bonds to sigma bonds always favorable?
My professor claims so, except in the case of conjugated pi bonds. I can see how pi bonds going to sigma bonds might be favorable; I know that double bonds are actually not twice as strong as single bonds because of poorer overlap ... but are there any exceptions to this guideline that the reduction of pi bonds to sigma bonds is favorable except in the case of conjugated systems?
EDIT: what about in the case of organic chemistry? It certainly seems that hydrogenation (a way of transforming pi bonds to sigma bonds) is generally a favorable move.

I'm wondering if your question should ask "is hydrogenation always energetically favorable?"  Kinetically it is not but thermodynamically it may be.

I think that should be the question too ... but my professor (falsely) equates hydrogenation with transformation of pi to sigma bonds. Hydrogenation is ONE way to transform pi bonds to sigma bonds, but there are other ways for one, and it seems to also be confusing a lot of answerers. @LordStryker.

By the way, in case it wasn't clear, I do like your question.

@LordStryker I also appreciate all the answers here; do you suggest opening a separate question though specifically about the favorability of hydrogenation? As you said there is a thermo and a kinetic component to favorability ... and the guy's just throwing around favorability rather loosely.

@Dissenter I've edited my answer to include some counterexamples from organic chemistry.

@Dissenter "favorable" usually means "final energy is lower than initial" so that is how I am trying to use it in this context.  I've updated my answer to take into account Phillip's comment indicating that with hydrogenation, we are not just considering the pi to sigma bond transformation of the CC bond but ALSO the creation of two single bonds with the hydrogens.  These are two different things.  For more info on hydrogenation and related energetics check out https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/addene2.htm.

@LordStryker and Philipp Let's make the comparison as clean, simple and direct as possible, let's consider isomerization. The heats of formation of 1-hexene and cyclohexane are -10.2 and -29.5 kcal/mole. Doesn't this suggest that a (if you will) carbon-carbon two-membered ring ("double bond) is less stable than a 6-membered ring?

@ron Ah yes but we're accounting for the entire molecule with that energy comparison no?  The individual double-bond itself is lower in energy than its corresponding single bond, however the MOLECULE itself is more stable (i.e. lower in energy) if the double bond is transformed to a single bond.  I think that the initial question was a bit unclear whether we were only considering the actual bond or the entire molecular unit.  Time has sort of clarified that but this is where this whole issue came from.

@Dissenter, could you clarify, as LordStryker phrased it, "whether we were only considering the actual bond or the entire molecular unit." My answer addressed the latter case, what was your intent?

The answer to your question is no.
In terms of organic chemistry, consider the two hydrocarbons, $\ce{H_3C-CH_3}$ and $\ce{HC\equiv CH}$.  The $\ce{C-C}$ bond dissociation energy is 368 kj/mol while the $\ce{C\equiv C}$ triple bond has a bond dissociation energy of 682 kj/mol, nearly twice that of the single bond.
The transformation of this triple to single bond may be thermodynamically favorable via hydrogenation but may not always be the case.  How does temperature affect hydrogenation and can $\Delta H$ be positive under other conditions? Furthermore, if we want to talk about the 'strength' of the CC bond, then clearly the triple bond is stronger.
Phillip brought up a good point in the comments that merits attention.  Hydrogenation contains an appreciable barrier for this process to occur.  When we take into account the formation of the additional single bonds by hydrogen then the energy is lower.  However, if we consider strictly a pi to sigma transformation (between two carbons in this example) and not the formation of other bonds, based on the energetics of the bond we are transforming, the energy will increase.
Reference for bond energies: http://web.chem.ucsb.edu/~zakariangroup/11---bonddissociationenergy.pdf

but isn't the heat of hydrogenation of acetylene still negative?

@Dissenter You asked for 'always'.  Hydrogenation is just one way and we can consider a multitude of environmental conditions.

@Dissenter Yes it is, but that is not your frame of reference. A double bond will always be stronger than a single bond, so going from double to single costs energy in any case. But usually when you reduce a double or triple bond you create new bonds at the expense of them. Only when those new bonds + the reduced bond are stronger than the old double/triple bond then the reaction is favorable.

@Philipp Your comment is spot on.

@LordStryker Thanks, I also could finally think of some organic reactions/compounds where double bonds win out over single bonds and included them in my answer.

LordStryker and @Philipp  "bond has a bond dissociation energy of 682 kj/mol, nearly twice that of the single bond." The point is that it is less than twice.  There is some destabilization (strain energy is one way to describe it) in a C-C double bond compared to a single bond, and even more in a triple bond.  Considering that the OP was addressing (roughly) homopolar, carbon-carbon systems, I think his professor's statement was correct.  Hetero-polar (C-O) situations or breaking bonds without forming new ones aside, can you think of any C-C violations.

@ron For C-C, no. But in my answer I showed that for some other homopolar bonds the bond dissociation energy of a double bond is higher than twice the energy of the single bond. At the bottom of my answer I linked a paper that analyses this relationship for the main group elements quite nicely. It is a little outdated but still useful.

@Philipp  Yes, but it perhaps it is the bonding scheme (hybridization) in these other homopolar cases that differentiates them from the carbon case.

@ron It has to do with the different electronegativities of carbon, oxygen and nitrogen. This entails different relative positions of the s and p valence electron energy levels and so leads to different degrees of s-p-mixing in the bonding orbitals. For 3rd and later row elements some other factors come into play. With them the p-p-$\pi$-interaction is generally weaker due to the higher bond distance. Carbon is indeed a special case. The enormous variety of organic chemistry is largely due
to the fact that C-C $\sigma$- and $\pi$-bonds are competitive with each other.

When thinking about questions along these lines, I always try to remember the bent-bond description of ethylene.  In all aspects related to physical reality (structure, electron distribution, etc.) the bent bond model pictured below on the right is equivalent to the pi model shown on the left (at least within the scope of all measurements made to date).

The "nice" thing about the bent bond model is that it casts ethylene as a highly-strained, two-membered ring.  When wouldn't releasing this strain and forming sigma bonds be an energy lowering process.  Aside from conjugated examples, I think that transformation of a strained, ethylenic system to a saturated, sigma-bonded system would always be energy lowering.
EDIT: Further Thoughts
Instead of analyzing hydrogenations or other reactions where the starting material and product have a different number of atoms, let's consider the following isomerization reaction.

Thermodynamically, cyclohexane is the most stable $\ce{C6H12}$ isomer in this series by a significant margin.  All we've done is convert a C-C pi bond into a C-C sigma bond and convert 3 $\ce{C(sp^2)-H}$ bonds into 3 $\ce{C(sp^3)-H}$ bonds.  Which (at least as I read it) was the OP's premise.  So I would say that, in carbon-hydrogen systems, pi bonds generally prefer to transform into sigma bonds, as pi bonds are inherently destablilized due to poor overlap (pi model) or strain (bent bond model).
That said, I retract what I wrote above that carbon-carbon pi to sigma transformations would "always" be energy lowering.  Consider the photolysis of benzene to produce dewar benzene, prismane and benzvalene.  These conversions are examples where one or more pi bonds have been converted to one or more sigma bonds and the reaction is not energetically (thermodynamically) favorable.

The products are much less stable (due to the strain in the small rings that have been created) than the starting benzene.  Because of these relative stabilities, the forward reaction does not occur thermally (in fact, thermally, each of those 3 products will revert to benzene), but by pumping in excess energy photochemically, the transformation can be brought about.
To summarize, I would maintain that generally, in model, carbon-hydrogen systems, ethylenic linkages are destabilized (due to poor overlap in the pi model, or strain in the bent bond model) and will thermodynamically prefer to exist as thermodynamically more-stable saturated linkages.

"the bent bond model pictured below on the right is equivalent to the pi model shown on the left (at least within the scope of all measurements made to date)." how exactly are the two equivalent?

Bond angles, bond lengths and electron distribution are all unchanged.  We just arbitrarily assign different hybridizations to carbon.

What would the hybridization of carbon be in the bent bond model? @ron

The carbon orbital used in the C-H bond remains around 2.2, but the orbitals used in the two equivalent C-C bonds are around 4.3, instead of sp2 and p

This is certainly not generally true. Have a look here for example: There you find that the bond energy for an $\ce{O-O}$ single bond is 142 kJ/mol while that of an $\ce{O=O}$ double bond is more than thrice as large with 494 kJ/mol. So, a reduction of $\ce{O=O}$ that would produce two $\ce{O-O}$ bonds would not be favorable. The same is true for $\ce{N-N}$ (167 kJ/mol) and $\ce{N=N}$ (418 kJ/mol) bonds. For nitrogen you even have the rare case that the $\ce{N#N}$ triple bond with 942 kJ/mol packs more bond strength than two isolated $\ce{N=N}$ double bonds.
In the context of organic chemistry you often have the situation that a reaction destroys a $\pi$ bond while in the process creating a new $\sigma$ bond. A double bond will always be stronger than a single bond, so going from double to single costs energy in any case. But usually when you reduce a double or triple bond you create new bonds at the expense of them. Only when those new bonds + the reduced bond are stronger than the old double/triple bond then the reaction is favorable. There are reactions where the opposite is the case. Phosphorous is infamous for its affinity to oxygen and very prone to form $\ce{P=O}$ double bonds. In the Wittig reaction (see for example here) you have an oxaphosphetane intermediate with a strained 4-membered ring with 4 $\sigma$ bonds that springs open to form one $\ce{C=C}$ and one $\ce{P=O}$ double bond.
Another case comes to mind. Hemiacetals (or 1,1-geminal diols), having two $\ce{C-O}$ $\sigma$ bonds, are usually unstable and prefer to destroy one of the $\ce{C-O}$ bonds by expelling the $\ce{R-O}$ group (or one of the $\ce{O-H}$ groups) thus forming a $\ce{C=O}$ bond. So, in this case one $\ce{C=O}$ bond beats two $\ce{C-O}$ bonds.
If you want to read more on the matter of the relation between single and multiple bonds concerning the strengths of the $\sigma$ and the $\pi$ interactions have a look at this paper by Kutzelnigg.

what about in the context of just organic chemistry? Hydrogenation certainly seems favorable most of the time.

you should confront my professor with your knowledge. @Philipp

@Dissenter I smell a science battle :)

@Philipp In the words of Manishearth... "FIGHT! FIGHT! FIGHT! FIGHT! FIGHT!" (http://meta.chemistry.stackexchange.com/questions/493/can-site-users-suggest-some-design-specifications-regarding-graduated-site-devel/495#495)

@LordStryker I will be a slaughter unlike any other witnessed in the course of history... only without all the corpses and the bloodshed

@Philipp The NSA appreciates your comment and would like to inform you that their 'list' is now more rigorous. ;)

@LordStryker I always strive to make new friends ;)


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How does a Frost diagram reproduce the solutions to the wave equation?
I came across a Frost diagram for cyclic compounds in my book, and all my book had to offer was that it geometrically reproduces the solutions of the wave equation, and can therefore determine the relative energies of each pi molecular orbital.
How, exactly, does a Frost diagram work? My book was of no use.

I'm really confused. I'm learning about Frost Diagrams as well in class. But they look completely different than what you just posted. Are these the same?

@JohnSnow  I believe it's the same Frost (Arthur Atwater).  He has two Frost diagrams.  The one discussed here is related to organic MO theory; the one you referenced relates to electrochemistry.

Frost developed this mnemonic patterning as an extension of the Hückel ($4n+2$) rule. A Frost diagram is usually applied to all-carbon, monocyclic, π systems.  It allows one to find the number of molecular orbitals in the molecule's π system and their energetic positions. To construct a Frost diagram, proceed as follows:

Draw a circle and inscribe a regular polygon with a vertex located at the bottom of the circle. The polygon has the same shape as the ring you are interested in. For example, if you are interested in benzene, draw a hexagon; for the tropylium ion, draw a heptagon.
Energy-wise, the top and bottom of the circle are defined as $\alpha+2\beta$ and $\alpha-2\beta$ respectively (so the circle has radius $2\beta$); the center of the circle is located at $\alpha$; other points can be interpolated accordingly; the bottom of the circle is at lower energy than the top of the circle.
Wherever a vertex of the polygon touches the circle, that is the energetic location of a molecular orbital.

Using benzene as an example, the lowest MO has energy $\alpha-2\beta$; the HOMO is degenerate (2 MO's) and located at $\alpha-\beta$; the LUMO is also degenerate and located at $\alpha+\beta$.  Any orbital below the center of the circle is bonding, any orbital at the center is non-bonding and any orbital in the top-half of the circle is antibonding.

But why does this work?

It works because it is an extension of Huckel's 4n+2 rule. Since it is related to Huckel's rule, it has its weak points, just like Huckel's rule.

@ron what are the weak points of Huckel's rule?

Huckel's rule only applies to a small, highly-defined class of molecules.  For example, it predicts that cyclooctatetraene is antiaromatic; in fact, the molecule distorts from planarity and is stable.  Frost predicts 2 degenerate non-bonding MOs at 0 beta for cyclobutadiene; in fact, the molecule distorts and the MO's split.

@ron I'd point out that cyclobutadiene basically undergoes a Jahn-Teller like distortion, which eliminates the triplet character and lowers the total energy. It's not that Frost is wrong, just that cyclobutadiene doesn't exist with equal length bonds.

@GeoffHutchison  Thanks Geoff, you said it much better than I did with my cyclooctatetraene-cyclobutadiene analogy.

Frost-Muselin method is simply a mnemonic to the solution for the secular equations of this very small set of per-cis annulenes. There is no physical reason that one "puts the polygon on its point" and draws energy levels at the intersections. And to add to the discussion about how Hueckel theory is generally wrong for ground state of highest symmetry planar "anti-aromatic" species:  they are ground state singlet even before J-T distortion (violations of Hund's rule in molecules)

@EricBrown There is no physical reason that one "puts the polygon on its point" and draws energy levels at the intersections. In which sense do you mean that? The equation for the Hückel energy levels, $e_{j} = \alpha + 2 \beta \cos \frac{2 j \pi}{N}$, gives you that equilateral polygon shape and since the level for $j=0$ is always non-degenerate and the lowest-lying level ($\beta$ is negative) its tip will always be at the bottom. Or do I misunderstand you?

Algebraically, the levels of cyclic polyenes may be derived using simple
Hückel theory (see also: Pi molecular orbitals of polyenes). The general result for the energy of the $j$-th level for a cyclic system containing $N$ atoms is:
$$e_{j} = \alpha + 2 \beta \cos\left(\frac{2j\pi}{N}\right)$$
where $\alpha$ is the energy of each carbon $\mathrm p_{\pi}$ orbital before interaction (Coulomb integral), $\beta$ is the interaction energy between two adjacent $\mathrm p_{\pi}$ orbitals (the resonance integral) and $j= 0, \pm 1, \pm 2,\ldots, \pm \frac{N - 1}{2}, +\frac{N}{2}$ for even $N$, and $j= 0, \pm 1, \pm 2, \ldots, \pm \frac{N - 1}{2}$ for odd $N$. The very simple form of this equation leads to a useful mnemonic for remembering the energy levels of these molecules. Draw a circle of radius $2\beta$ and inscribe an $N$-vertex polygon such that one vertex lies at the bottom position. The points at which the two figures touch define the Hückel energy levels. And that is what is called a Frost diagram.

+1 I hadn't heard of these diagrams until teaching quantum and it was in the book without any derivation. Then I looked up the general result for cyclic polyenes above and it made sense.

@GeoffHutchison I made a small mistake at the range for $j$ which should be correct now. The Frost circle is indeed a pretty neat example for very useful and pictorial (and historical) chemical rules/mnemonics/concepts that while being very easy give a lot of intuition for how chemical compounds will react and what properties they have. I find it very interesting to discover the actual underlying physical theories and approximations that lead to these rules and to see when they are valid and when they should be used with caution.


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Nucleophilicity of bonding vs lone pairs
Does the bonding pair in $\ce{FeCl4^-}$ attack a proton or does a lone pair on the chlorine attack the proton?

My prof has an argument for the first one; the first one involves fewer steps, so it is probably more favorable. My proposed mechanism has three steps of electron pushing versus two, so okay.
However, I can see an argument for the second mechanism - the first one involves a bonding pair acting as a nucleophile rather than a lone pair on chlorine. The bonding pair is stabilized by the chlorine and the iron atom, while the lone pair is stabilized by only the chlorine.
So, which one wins? I'm guessing both mechanisms are operative here.

I don't know the answer to your question, but we can speculate.  A lone pair is more sterically exposed than a bonding pair of electrons and therefor more accessible.  Also a lone pair is generally more nucleophilic because the electrons are concentrated in a smaller region since they are not being shared.  These two arguments would favor the second mechanism.  Finally, when we consider the SN2 reaction where, for example, water is the attacking nucleophile; I've never seen the mechanism drawn where the attacking electrons are from the O-H bond, but rather they always seem to show involvement of the lone pair electrons.  All of these arguments favor the second mechanism.

The example involves an analogous choice of which electron reservoir to tap.  I didn't think of it as an SN2, but isn't it really just a bimolecular, nucleophilic displacement?  It just has H at the center of things rather than carbon.

does the three vs two step argument hold water?

do you mean the "fewer steps are preferred" argument?

yes that's what i meant

Generally, when you're guessing about something, then yes.  But here, especially with electrons that move so quickly across very short distances, I think such an argument would have less significance - just my opinion.

is there a way to verify which mechanism is operative here? Or could it be that both are operative? Could we check by checking the orbitals that the lone pairs and bonding pairs are in? I.e. which one is in the higher HOMO?

The differences between the 2 mechanisms are pretty minor.  I don't think there is an experiment with high enough precision that could be done.  Perhaps a computational study.

just curious; what is your background? you seem to know more than all my profs combined

organic chemist, was a professor, moved to industry; I suspect your professors know more, but would rather spend time on their research than talk with students

last question for now; what orgo book do you recommend?

I haven't compared introductory organic texts in a while, but I like "Organic Chemistry" by M. Jones, Jr.

"Unlike every other organic chemistry book I've looked at, this book does not once use the excuse "...but that is beyond the scope of this course..."" ... sounds like the book for me @ron, thanks!


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How to analytically quantify a chemical's potential to intercalate DNA?
This shows that polyphenols (like quercetin) have the potential to intercalate DNA, but at potency 100x less than that of ethidium bromide. How does one analytically measure/quantify "100x weaker"?
I'm also interested in how this could apply for the compound 9-fluorenol (hydrafinil) - and of the relative potency of 9-fluorenol and quercetin in intercalating DNA.

Rather than linking to a different site, can you quote your source(s) in the question?

There are assays to determine binding affinity. I can't remember any references right now, but our lab uses a thiazole orange displacement assay where we bind DNA with thiazole orange, which becomes fluorescent on binding. We titrate in another compound and measure loss of fluorescence as the thiazole is displaced.

@user137 You should turn that into a proper answer. Looks like you've got the experimental authority on the subject.

Our lab uses a thiazole orange exclusion assay to measure how well our polyintercalating peptides* bind DNA. The assay works by mixing thiazole orange with DNA so it becomes fluorescent. Then we measure fluorescence while titrating in peptide and the loss of flourescence indicates binding. This is because thiazole orange is only fluorescent when intercalated into DNA, and as the competing compound displaces it, the free thiazole no longer fluoresces. Note that this assay is not foolproof, polycationic compounds can condense DNA without intercalating, but this conformation change still forces the thiazole out and reduces the fluorescence, so loss of fluorescence implies binding, not necessarily intercalation. This method works for RNA too, but you'll need more thiazole to see signal because there are usually fewer double helices to intercalate into.
Another method is to measure viscosity, because intercalators make the DNA longer and the solution becomes more viscous, but you either need a very sensitive method to measure the change or a very concentrated DNA solution, and the viscosity can probably also be changed by condensing the DNA.
Another possible method MIGHT be fluorescent anisotropy, if your intercalator is fluorescent. It would spin rapidly outside the DNA, but slow down once it binds the double helix. However, conformational changes in the DNA can change its rotation speed and confound the results.

That paper doesn't really mention the assay, it's just to show off the peptides so you understand what I'm talking about a little more. Those peptides also bind by electrostatic interaction and condense DNA, so it's not a pure measurement of intercalation.