Why is fluorine a oxidising agent? An oxidizing agent pulls the electron cloud of the substance being oxidized towards itself, for example:
$$\ce{F2 + 2 X- -> 2F- + X2}\qquad (\ce{X} = \ce{Cl}, \ce{Br}, \ce{I})$$
We also know that fluorine has very high charge/mass ratio as it is very small in size and its electron gain enthalpy is know to be very high, i.e gain of electron is favourable thermodynamically.
Also, many sources state that:

Fluorine has low bond dissociation enthalpy and high hydration enthalpy due to which fluorine is a good oxidizing agent.
Fluorine is achieving octet, so electron gain is favorable.
I found another source stating that bond dissociation enthalpy is the major driving factor.

My questions(along with reasons why fluorine shouldn't be oxidising agent)

What has hydration enthalpy and bond dissociation enthalpy got to do with oxidizing power?
(As we know that lithium is the best reducing agent, and it also has high hydration enthalpy but we can’t say it is an oxidising agent.)
What am I missing here? What is the reason for such high standard electrode potential for fluorine?
Why wouldn't the electron cloud of fluorine atom repel the electron cloud of substance being oxidized? It seems paradoxical that the fluorine pulls electron cloud towards itself so easily.

Even in the post Why does chlorine have a higher electron affinity than fluorine? it is mentioned that electron cloud is repelled.
Is there any better explanation other than "octet is being achieved", such that it resolves the repulsion issue, to explain why fluorine is a good oxidising agent?

If we consider

$\ce{F2(g) -> 2 F(g)}$ with $\Delta H_\mathrm{F, bond}$

$\ce{2 F(g) + 2 e- -> 2 F-(g)}$ with $\Delta H_\mathrm{F, aff}$

$\ce{2 F-(g)  ->[H2O] 2 F-(aq)}$ with $\Delta H_\mathrm{F, hydr}$

and respectively

$\ce{Cl2(g) -> 2 Cl(g)}$ with $\Delta H_\mathrm{Cl, bond}$

$\ce{2 Cl(g) + 2 e- -> 2 Cl-(g)}$ with $\Delta H_\mathrm{Cl, aff}$

$\ce{2 Cl-(g)  ->[H2O] Cl-(aq)}$ with $\Delta H_\mathrm{Cl, hydr}$

Then overall reaction enthalpy for $\ce{ F2(g) + 2 Cl-(aq) -> Cl2(g) + 2 F-(aq)}$ is

$$\Delta H_\mathrm{r} = \left({\Delta H_\mathrm{F, bond} - \Delta H_\mathrm{Cl, bond}}\right) + \left({\Delta H_\mathrm{F, aff} - \Delta H_\mathrm{Cl, aff}}\right) + \left({\Delta H_\mathrm{F, hydr} - \Delta H_\mathrm{Cl, hydr}}\right)$$

The differences between fluorine and chlorine enthalpies of atomization(bond breaking ), electron affinity and ion hydration lead to the overall reaction enthalpy, that is pushing reaction toward chloride oxidation.

The truth is, that the equilibrium is driven by Gibbs energy, not enthalpy,

$$\Delta G^{\circ}_\mathrm{r}=\Delta H^{\circ}_\mathrm{r} - T \cdot \Delta S^{\circ}_\mathrm{r}=-RT\ln{K}$$

but for the rough reasoning you get the picture. Futhermore, the entropy change will not be major factor, due analogical structures.


Why are 720 nm LEDs so uncommon? Disclaimer: Potentially more of a physics question, but this area is muddy.

I'm trying to find 720 nm LEDs but they're proving to be very uncommon. In fact, there seems to be a gap between 660 nm up to ~730 nm. I've only been able to find one supplier of 720 nm LEDs (data sheet here). These LEDs also tend to be very low powered compared to their counterparts in the nearby range. For example, a typical 690 nm LED outputs 690 mW, whereas the 720 nm LED only ouputs 170 mW, and that's the strongest one I could find (which is no longer carried).

My gut feeling says this has to do with the material being used in the LED (in this case GaAlAs). Perhaps it's very difficult to dope the material in such a way that it emits 720 nm light? I've already noticed that these LEDs are prone to burning out. Is the material just not very good for this purpose? Hopefully someone with more insight into this topic could shed some light (heh) on the situation.

I suspect the answer is that AlGaAs is an alloy with a bandgap between that of AlAs - 2.16 (which is indirect) and GaAs - 1.42 eV (which is direct). If there is enough aluminum, the resulting semiconductor is an indirect bandgap semiconductor, which means that to emit a photon, it must be coupled with a lattice phonon, greatly reducing the probability of photon emission, which would make a AlGaAs LED of short enough wavelength rather inefficient at emitting light, instead producing a lot of heat rather than light.


Why does a reaction take time? A reaction in nature is, as we know, generally an interaction between the electron clouds of different atoms and nuclei. Why does this takes time?

Firstly, some reactions have a very low probability, secondly, the probability depends on temperature (activation energy!), and thirdly, diffusion is not exacly a fast process, and takes time even in a gas phase. Once the reaction progresses, the remaining educt particles have an increasingly hard time finding each other even if you thoroughly mixed everything at the beginning.


Bonding in a bridged metal carbonyl Can someone please explain the bonding in a bridged metal carbonyl (like diiron nonacarbonyl)?

I cannot understand what kind of bonds (sigma/pi) exist between the bridging carbonyls and the metals.

Also why aren't the metals charged in such a case?
What is the bond order in such a case?

I don't like to reinvent wheel -- if you go to Wikipedia's  Metal carbonyl and follow just two 1st literature position, you'll find your answer concerning bridging.

For MO calculations and a general scheme applicable to systems featuring “dative” MM bonds and non-compensating, semi-bridging carbonyl ligands, see: R. D. Barr, T. B. Marder, A. G.
Orpen, I. D.Williams, J. Chem. Soc. Chem. Commun. 1984, 112–114


Why is the sneaky method of smothering a grease fire more effective? In this demonstration, a grease fire is temporarily extinguished if you put the cover on the pan, but reignites as soon as you take the cover off.  However, if you slowly slide the cover across the pan, then the fire stays extinguished.

Why does sliding the cover on make such a big difference?

By smothering the fire your starving it of oxygen. Without oxygen it can't burn.

If you cover the fire and take the cover off quick it reignites because it quickly interacts with the oxygen again because the heat of the oil is still around the same temperature as when it caught on fire.

If you slowly slide of the lid your increasing the time you've starved the fire of oxygen and also allowed the grease to cool slightly so when the oxygen comes back in contact with the grease at a slow rate the grease won't reignite.

The best way to deal with a fire is to prevent it from happening in the first place , make sure you know the boiling point and smoke point of what your using and use a thermometer .


When naming alcohols based on alkanes, when do you keep the "e" at the end of the alkane's name? For example, you would write pentane-1,2-diol but pentan-2-ol, correct? Is the difference simply the number of hydroxyl groups?

The names pentane-1,2-diol and pentan-2-ol are correct.
Systematic elision of the terminal letter ‘e’ or other vowels appears on several occasions in the IUPAC nomenclature.
A brief rule for simple alcohols is mentioned early in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-15.1.7.2.1 Alcohols are named by attaching the suffix ‘ol’ to the name of the parent hydride, with elision of the final letter ‘e’ in the parent hydride, if present. When alone in the structure, the characteristic group(s) must receive the lowest locant(s) possible, which is (are) cited immediately in front of the suffix (see P-14.3.2).

For example: pentane → pentan-2-ol
Extensive general rules for the elision of vowels can be found in Subsection P-16.7:

P-16.7 ELISION OF VOWELS
P-16.7.1 Vowels are systematically elided as follows:
(a) the terminal letter ‘e’ in names of parent hydrides or endings ‘ene’ and ‘yne’ when followed by a suffix or ‘en’ ending beginning with ‘a’, ‘e’, ‘i’, ‘o’, ‘u’, or ‘y’;
(…)

Clearly, this rule also applies to the suffix ‘ol’ for alcohols.
A similar rule is mentioned later in Chapter P-6 concerning the applications of general principles to specific classes of compounds:

P-60.2 PRESENTATION OF NAMES.
(…)
(a) names formed substitutively using suffixes follow the general method described in P-15.1. Substitutive names are formed by adding a suffix such as ‘al’, ‘ol’, ‘yl’, ‘carbaldehyde’, ‘carboxylic acid’, etc., to the name of a parent hydride, with elision of the final letter ‘e’ of the parent hydride, if any, before ‘a’, ‘i’, ‘o’, ‘u’, and ‘y’;

Clarification concerning multiplying prefixes and numerals can be found in Section P-59 concerning name construction:

P-59.1.9 Characteristic groups
(…)
If, and only if, the complete suffix (that is, the suffix plus its multiplying prefixes, if any, as described in Appendix 1) begins with a vowel, a terminal letter ‘e’ (if any) of the parent hydride name is elided. For example, ethanol (not ethaneol). Elision or retention of the terminal letter ‘e’ is independent of the presence of numerals between it and the following letter, for example, propan-2-ol (not propane-2-ol).
(…)

Since the complete suffix (the suffix plus its multiplying prefixes) of pentane-1,2-diol is ‘diol’ and thus begins with ‘d’ and not with a vowel, the terminal letter ‘e’ of the parent hydride pentane name is not elided.


Why does mercury have low melting and boiling points? Many metals have relatively high melting and boiling points, but mercury has relatively low melting and boiling points. What are the possible reasons for this?

Mercury is different!  It is not as reactive as its neighbors in the Periodic Table, it doesn't conduct heat and electricity as well as other metals, and it is a liquid unlike other metals.  The electronic structure of mercury is $\ce{[Xe] 4f^{14} 5d^{10} 6s^2}$, so the first thing we notice is that all of its orbitals are full, there are no unpaired electrons in mercury.  This helps explain mercury's reluctance to react with other materials or form bonds with other mercury atoms.  Most metal atoms share their outer electrons with other metal atoms, in fact, all of these "shared" metal electrons exist as a diffuse "sea" of electrons.  It is the extended nature of this sharing that makes metals good conductors of heat and electricity.  The fact that the metal electrons are shared provides a strong bonding interaction between metal atoms that gives metals their solid structure.  None of this applies to mercury.  With its filled shell electron configuration it is very reluctant to form bonds even with other mercury atoms.

One additional, interesting effect makes mercury even more reluctant to share its electrons.  Special relativity suggests that for larger nuclei (and mercury with atomic number 80 is large enough for the effect to apply) electrons will begin to travel closer to the speed of light.  In turn, their mass will increase and as mass increases, orbital radius decreases.  The electron is closer to the nucleus, more attracted to the nucleus and less available for bonding.  Here is a nice reference that explains this relativistic effect and how it applies to mercury in plain English and a bit more detail.

These two effects, filled outer electron shell and contraction of the outer orbital closer to the nucleus, combine to make mercury reluctant to form bonds, even with other mercury atoms.  It is this lack of electron bonding between mercury atoms that makes it melt and boil at such low temperatures.


Why does alkyl rearrangement happen in the Hofmann rearrangement? In the mechanism of Hofmann rearrangement, one step involves migration of an R−group from carbon to nitrogen.

This reaction proceeds even if the migratory aptitude of the R-group is poor (like in the case of ethanamide). What motivates the migration of the R-group and makes the reaction proceed?

The ‘motivation’ is that migration is the only way in which the reaction proceeds in a productive manner towards the product. From seeing the mechanism alone, you have no idea if the reaction of a particular substrate is rapid or slow, if it requires higher or lower temperatures etc. Only if you start taking these into account, you can tell whether the compound in question reacts rapidly or not.

I would not be surprised to see vastly different reaction times between compounds, depending on whether the $\ce{R}$ group easily migrates or not.


Large difference in boiling points of tetrafluoromethane and tetrachloromethane Why is carbon tetrachloride $\ce{CCl4}$ is seen to posses liquid state (b.p. $\pu{76.72 °C}),$ whereas carbon tetrafluoride $\ce{CF4}$ is in gaseous state at room temperature (b.p. $\pu{−127.8 °C})?$

The vapour pressure can be estimated by equating the rate of leaving the liquid surface with that returning to it from the vapour. This leads to an expression of the form $\displaystyle p\sim n_{l}RTe^{-\Delta E/RT}$ where $\Delta E$ is the energy needed to remove a molecule from the liquid to the vapour phase and $n_{l}$ is the number density (number / unit volume) of the liquid.

The energy $\Delta E$ will clearly depend on the type of molecules involved. CCl$_4$ has large easily polarisable chlorine atoms. This means that these electrons are easily influenced by nearby atoms and this leads to an increased intermolecular interaction compared to CF$_4$ because the F atoms, having fewer electrons, and ones close to the nucleus, are only slightly polarisable. The increased $\Delta E$ (CCl$_4$ vs CF$_4$) then leads to a lower vapour pressure for CCl$_4$ and hence a higher boiling point which is defined at the temperature when the vapour pressure reaches one bar, or one atm. if you prefer.

(Sometimes the polarisable interaction is called induced-dipole, induced dipole or London or in general van der Waals interaction.)


Assessing fire hazard of compounds Following up on a somewhat too broad question about amateur chemist safety, this will be the first to address some of the issues discussed there.

I often use the NOAA MSDS database (material safety data sheet) to investigate the chemicals I'm working with (they also have a nice reactivity predictor). As an example, let us examine the compound cyclohexylamine (I chose this as leading example for these followup question, because of the availability of extensive research in my country)

sometimes used as a plasticizer or catalyst. As an amateur I would carefully go through the MSDS and notice for example that it is flammable, caustic compound, scoring a 3 on flammability and health in the NFPA 704 hazard diamond.

As an amateur hobby chemist without all comfort of a basic lab (aside from safety goggles, gloves, lab coat, fire blanket), how would I judge if I can safely handle this compound with regard to its flammability, aside from conforming to the Protective Clothing section?

For example, I could imagine that flammable liquids with high vapor pressure (11 mm Hg for cyclohexylamine according to NOAA, I guess this is measured at a standard temperature as vapor pressure varies with temperature) or low auto ignition points (560 ° F) or low boiling points (274.1 °F) are unsafe for amateurs. Are there empirical (or theoretical) recommendation considering vapor pressure and ignition points, i.e., starting at what value will these factors become too dangerous for the compound to be handled in a relatively amateurish lab.

(I understand that the answers will be partly subjective, as for all answers regarding safety, but please try to give an objective as possible answer, preferably referring to external sources)

I guess I don't completely agree with the premise of your question. Gasoline is highly flammable, even potentially explosive, yet "amateurs" deal with it constantly. The key is that it is stored and used properly, from the pump to your gas tank. It becomes less safe if you pump it into a gas can, but gas cans also have certain requirements for storage safety.

So I think the issue may be what sorts of flammables would an amateur have a high chance of storing or using inappropriately. For example, any pyrophoric compound would be a good candidate to avoid, as most require a way to exclude air and moisture, which often requires specialist equipment.

Solvents with low flash points and high vapor pressure should also be treated with respect, but rubbing alcohol (isopropanol) or acetone might qualify for that designation. Storage in airtight fireproof containers in cool and dry places is often sufficient. Better yet, use a flammables cabinet. This simple step is something an amateur might skip for cost, but it may be the most sensible precaution. I have flammables cabinets, and I use them for ordinary hardware solvents along with experimental chemicals.

Finally, work safely! No flames, as they say in organic chemistry. Use hotplates or heating mantles rather than burners. If you must use flames for some reason, put away all flammables and keep them a safe distance. Pay careful attention to any spills and attend to them immediately before resuming work. And perhaps key for amateurs, small quantities! No need for an amateur to have gallons of flammables on hand. Even professional labs are trying to reduce quantities in the name of green chemistry.

EDIT: My go-to source for using and disposing of hazardous chemicals is Hazardous Laboratory Chemicals Disposal Guide by Margaret-Ann Armour. In addition to disposal techniques, each chemical has an extensive incompatibility section about what combinations can make a chemical even more hazardous, and all these entries have literature references. I use this book literally every time I do an experiment. It deals with much more than flammables, but flammables are definitely present.


Is there a connection between M1V1=M2V2 and Boyle’s Law? Recently in Chemistry 101 we learned about $M_1V_1=M_2V_2$, where $M$ is the molarity of the two mixtures and $V$ is their respective volumes.

The first thing that popped into my mind when I saw this formula is, “Hey, that reminds me of Boyle’s Law from high school physics - $P_1V_1=P_2V_2$ - the relationship between the volumes of a gas at different pressures.”

Is it just a coincidence that these two formulae look the same, or is there a connection between the pressure variable in Boyle’s Law and the molarity variable in the first equation? Are there other formulae which are structured similarly which are connected?

Yes, they are related. The first comes directly from the conservation of number of moles of the solute in a dilution,

$$n_1 = n_2 $$

Since $n = MV$,

$$M_1 V_1 = M_2 V_2$$

The second is related to the conservation of the total number of moles in an isothermal compression or expansion,

$$n_1 = n_2$$

Using the ideal gas law $n = pV/RT$,

$$\frac{P_1 V_1}{R T} = \frac{P_2 V_2}{R T}$$

Or $P_1 V_1 = P_2 V_2$.

Again, both are related to the conservation of moles (which by the way, isn't an universal law, since it's broken in trivial situations like the event of a chemical reaction), although the first one refers to solute only and the second to total number of moles.


Why would a polar protic solvent favor the addition of HBr to an alkene? I understand that a polar solvent will help dissociate the $\ce{HBr}$ into ions.
But why a protic solvent?
Why would we want a solvent that would encage the $\ce{Br}$ and $\ce{H}$ ions?
Why is the aprotic solvent not favored? My textbook does not discuss this and I searched online, with no result.

why a protic solvent?

Protic solvents tend to have higher dielectric constants than typical, unreactive, non-protic solvents.  Look at the following table.
\begin{array}{|c|c|c|}\hline
\text{Solvent} & \text{Dielectric constant}    \\ \hline
\text{methanol} & 32.6    \\ \hline
\text{ethanol} & 24.6   \\ \hline
\text{methylene chloride} & 9.08 \\ \hline
\text{chlorobenzene} & 5.69\\ \hline
\text{chloroform} & 4.81    \\ \hline
\text{diethyl ether} & 4.27    \\ \hline
\end{array}
The higher dielectric constant stabilizes ions which means that 1) $\ce{HBr}$ will tend to be more dissociated and 2) (as Dissenter noted) the resultant carbocation intermediate will be more stabilized thereby lowering the energy of the transition state leading to it.


What is the region of this wavelength (462 nm)? My lecturer said 462 nm is infrared region. I think the wavelength is visible light. Which region is the correct one? I need confirmation about this one.

Light with a wavelength of 462nm is in the visible light region (blue light visible by the human eye).

For certain, your lecturer was wrong that the light is in the infrared region, this occurs at much higher wavelengths closer to 700nm. More possibly he meant to say ultraviolet, which occurs somewhere below 400nm (shorter wavelengths), though even this isn't technically correct.

The Wikipedia article gives a fairly good overview of the various regions, along with a schematic describing where they come:


Determine the position of an equilibrium process For the exothermic reaction of nitrogen gas with iodine solid to produce nitrogen triodide gas. Write the balanced equation including heat and select the overall direction of shift in the reaction.

I have balanced the equation: $\ce{N2 (g) + 3 I2 (g) <=> 2 NI3}$
Is this equation correct?

I am confused on how to find the overall direction of shift. Should I use the Le Chatelier's principle? If so, then I think the overall shift is a "shift to the left"/reverse reaction. Am I correct?

You are right, you can use Le Chatelier's principle to predict the shift of the equilibrium when temperature or pressure is in- or decreased. It is stated that the reaction is exothermic, and this tells you on which side of the reaction equation the heat goes. And as both products and educts include gases, it would also be easy to predict how the equilibrium shifts when pressure is varied. I presume that the question, in its current form, is asking for the shifts that happen in all possible combinations of in- or decreased pressure or temperature.


Why is iron used in the body? I'm writing a report on the spectrophotometric determination of iron. In my introduction I want to briefly explain the importance of iron in the human body.

Wikipedia states that "Iron is essential to life due to its unusual flexibility to serve as both an electron donor and acceptor"

1.) So is iron the only element that can do that? If not, what is it about iron that makes it the only element found in hemoglobin?

2.) What is it about being able to gain and lose electrons make it essential to life? I think it has something to do with oxygen being able to bind with iron. But there are so many elements that can bind with iron! What is special about iron?

1.) So is iron the only element that can do that? If not, what is it about iron that makes it the only element found in hemoglobin?

No, iron is not the only metal that will fit inside the porphyrin structure. But its ability to be an electron acceptor and donor makes it very useful for oxygen and carbon dioxide to bind to the hemoglobin. Since iron is also very abundant, it was easy to incorporate it and utilize its properties.

Other macrocyclic structures include chlorophyll with magnesium in the center and vitamin B12 (cobalamin) with cobalt in the center.

  2.) What is it about being able to gain and lose electrons make it essential to life? I think it has something to do with oxygen being able to bind with iron. ... What is special about iron?

Iron is abundant. The iron in the hemoglobin can weakly bind both oxygen and carbon dioxide.

You can find more information on the Wikipedia page.


Reaction between NaOH and CO2 So I wanted to know what the reaction between sodium hydroxide and carbon dioxide can be, and upon research I got 2 answers. The first one is

$$\ce{CO2 + NaOH(aq) -> NaHCO3(aq)}$$

and the second one is a two-step reaction, first with water then sodium hydroxide:

$$\ce{CO2 + H2O(l) -> H2CO3(aq)}, \\
\ce{H2CO3(aq) + NaOH(aq) -> NaHCO3(aq) + H2O(l)}$$

So my question is which one of these answers is right?

To me, I do not understand why the 2nd option takes into account water because really concentrated $\ce{NaOH}$ would/should be written just as $\ce{NaOH}$, which to me makes the 1st option the right one. Am I wrong?

No both are right.

If you add the latter equations, then you will get the former.

If sodium hydroxide solution was highly concentrated like you said, then both equations are wrong since sodium hydroxide solution will strip away the second hydronium as well forming carbonates.

All state notations with (aq) implies that it is dissolved in water. If it was pure NaOH, it would be in a solid state under room temperatures. It is also noteworthy to know that NaOH must be contained in a dry environment as it will spontaneously absorb water from the air.


What (simple) household chemicals react with oxygen? I'm interested in making a small "preservation chamber" that I can basically use to store objects without the presence of oxygen. I have a resealable stainless steel canister and the plan is pretty much just to put whatever object I want to be preserved inside, together with a chemical/object/thing that absorbs the oxygen from the air inside the can, protecting the other item. I don't have access to anything fancy, so I need something is/can be made from household materials that will react with/contain oxygen.

Here's the catch: It CANNOT be a candle. The container has a plexiglass cover that will melt if it's exposed to the heat of a flame. (I tried it before; it worked, but the plastic almost cracked. I don't dare to do it again.)

You don't have to reinvent the wheel.
Food industry already uses quite simple oxygen scavengers for decades, and one of the most affordable formulations include iron powder mixed with sodium chloride.

Typically, an oxygen scavenger is packaged in small packets (sometimes labeled as "oxygen absorber" or "antioxidant") that are added with the product inside the pact and sealed.
The absorber itself consists of iron powder and an inert absorbent (sodium chloride, activated carbon, food-grade clay or a composite such as nylon 6/nanoclay).
When placed inside a package with a product intended for storage, it begins to actively absorb oxygen and completely removes it within 12–96 hours.
The drawback of using Fe is that it promotes rapid lipid oxidation, if it's leaked and gets in contact with the product.
However, since you are already using stainless steel container, going with iron-based quencher seems like an optimal choice to me.

More historical background and interesting tech details can be found in Wiley's Encyclopedia of Packaging Technology [1, p. 845]:

  […] The iron-based oxygen scavengers are the most prevalent and they are examined
  first.

  The iron technology is based on rapid oxidation of iron powders to prevent headspace oxygen from reacting with foodstuff components or to reduce oxygen permeation rates through container walls. Reduced iron powders (zero-valent iron $\ce{Fe^0}$) are highly reactive and possess a higher capacity to react with oxygen than partially oxidized ferrous iron ($\ce{Fe^2+}$) compounds. Large exposed surface area of iron powders per unit weight, which increases with reduction in mean particle size and especially when using highly porous "sponge" iron morphologies, promotes simultaneous oxidation at multiple reactive sites, resulting in faster rates of oxygen scavenging. Presence of electrolytes and an aqueous interface to facilitate electron transfer during oxidation is known to accelerate oxidation reactions of transition metals. This is exempliﬁed by rapid rusting of ferrous metals and steels exposed to road salt and salted seawater, in contrast with slow rates of corrosion in the dry climates. Ferrous and ferric iron oxides and hydroxides produced by oxidative reactions are basic. As a result, acidic environments promote oxygen scavenging by shifting equilibrium to iron oxides. At low pH values, oxidation of reduced iron proceeds in humid environments even in the absence of oxygen, by deoxygenizing water molecules and resulting in hydrogen gas formation.

  Since low pH values and aqueous interfaces require the presence of water, water vapor diffusion and variation in relative humidity of the environment can be used to control activation of oxygen scavenging capacity of iron powders. The speciﬁc formulations are designed to trigger the activation at desired RH levels, control the reaction rate, achieve the full oxidative potential of the iron powder, maintain compatibility with various products, and ensure food safety. The reaction formula for complete oxidation of reduced iron to eventually form ferric oxide trihydrate complex (commonly known as rust) is

  $$\ce{4 Fe^0 + 3 O2 + 6 H2O → 4 Fe(OH)3 → 2 (Fe2O3* 3 H2O)}$$

  Many transition metals are recognized as being sufficiently oxidative to become candidates for an oxygen-scavenger formulation; however, iron does offer unique advantages that have driven the industry to general use of this medium. Iron has a relatively high afﬁnity to combine with oxygen on a per unit weight basis compared to most alternatives. Complete oxidation of 1 g of reduced iron removes 300 cm³ of oxygen at STP conditions. Elemental candidates that exceed it have drawbacks such as the odor problems with the use of sulfur and the propensity of aluminum to form an almost impermeable oxidized skin layer that limits further oxidation. Iron rust, on the other hand, ﬂakes off as it forms and it is highly permeable to further duffusion of water and oxygen. Of great importance is the food safety of iron powders, which are recognized as nutrients important for healthy blood and commonly used as food enrichments in milled flour, breads, breakfast cereals, and baby foods. The same cannot be said about copper, zinc, aluminum, and many other transition metals. The abundance and relatively low cost of producing elemental iron are also important, especially in comparison to choices such as palladium and platinum metals and catalysts based on them. Then there is the ready ability to manipulate reactive capability of iron powder formulations to adapt to a wide variety of applications for both the rate of oxygen scavenging and the activation by the moisture in a package headspace. One of the few drawbacks is that the oxidation of iron is a temperature-dependent reaction and normally slows dramatically as the temperature approaches freezing.

  Electrolytes and acidifying agents in a scavenging formulation are commonly supplied by metallic salts […]. The practical considerations of food safety have dictated that common table salt ($\ce{NaCl}$) is often the best choice for the electrolyte, while various metal and organic salts of strong acids are often used to increase the acidity of the formulation. […] When the packaged product does not have a required water activity to fully activate an iron-based scavenger and provide the desired reaction rate, premoisturized silicas, zeolites, and activated carbons are used as self-contained water carriers in various proprietary formulations to provide a reliable activation.

References

The Wiley Encyclopedia of Packaging Technology, 3rd ed.; Yam, K. L., Ed.; John Wiley & Sons: Hoboken, N.J, 2009. ISBN 978-0-470-08704-6.


How to calculate the quantum density of states? I've been thinking recently about Fermi's Golden Rule. Sparing the details of how to get Fermi's golden rule (you'll find like five derivations if you google it), the final result is,
$$\frac{\partial P}{\partial t}=\frac{2\pi}{\hbar}|\langle f|H'|i\rangle|^2\rho(E_f)$$which says that the probability per unit time, to first order in the perturbation from $H_1$, is a constant. Hence, this quantity tells you the transition rate. Notice, however, that it depends on the density of final states. Notice, if I'm not mistaken, that this is not for a specific state, but this is more accurately the density of all states above the original states (let's say we're in the ground state and ignore stimulated emission). I understand how to calculate the density of states for a classical, and hence continuous system. It's just an integral in phase-space.

Things become more complicated in a molecule, however, because the states are quantized. So, how do I go about doing this calculation of the density of discrete states? A really great answer would sketch out how one might do this calculation for a quantum harmonic oscillator.

I've seen a couple things online which do this kind of calculation for the motion of a particle in an infinite potential well on a solid. You basically just use the relationship between energy and the wavenumber, which is easy because you know the solutions to the Schrodinger equation for all states ahead of time.

If you really do want to calculate the density of states then you can try direct counting to get the total number of states N and then calculate the density of states $\rho =[N(E+dE)-N(E)]/dE$. This, of course, will only work if you have a few vibrational normal modes as all the combinations of quanta have to be included, e.g. in water with 3 modes, quanta (in the order of increasing energy) are: $(0,0,0), (0,0,1),(0,0,2),(0,1,0), (1,0,0),(0,0,3),(0,1,1)$, etc.  But for a molecule with just a few more atoms, it is quite probable you will get $10^6$ vibrational states /wavenumber at an excess energy of, say,  $5000 \pu{~cm^{-1}}$.

In the field of unimolecular reactions (RRKM models etc.) there has been much effort in calculating total and density of states. There is the Beyer-Swinehart algorithm; Beyer, T., & Swinehart, D. F. 1973, Commun. Assoc. Comput. Machinery, 16, 379 and Stein S & Rabinovitch B. J. Chem. Phys. v58, 2438, 1973. Whitten and Rabinovitch have also modified the classical sum of states to give good agreement with the direct counting (quantum) method. (J. Chem. Phys. 41, 1883, 1964, and 48, 1427 1968). Their modification $^{[1]}$ gives the total sum of states as:

$$ N(E)=\frac{(E+\alpha E_{zpe})^s}{s!\prod_{i=1}^s h\nu_i}$$

where s is the number of (harmonic) oscillators, $E_{zpe}$ is the zero point energy for the i th oscillator,

$$\alpha = 1-\beta w(E')$$

$$\beta = \frac{s-1}{s}\frac{<\nu^2>}{<\nu>^2}$$

and w depends on the range of $E'=E/E_{zpe}$. When $0.1 \lt E' \lt 1; w(E')=(5E'+2.73\sqrt(E')+3.51)^{-1}$, and when $1\lt E' \lt 8; w(E')=\exp(-2.4191(E')^{1/4})$.
The equation in N can be differentiated to find the density of states.

If, however, you want to calculate a rate constant using the Fermi Golden rule it is better not to start with the density of states separated out, but to calculate the Franck-Condon weighted density of states.

The rate constant has the Golden Rule form
$$ k= \frac{2\pi}{\hbar}|<av''|H|bv'>|^2\delta(E_{av'}-E_{bv''})$$

and then this is modified depending on what actual situation is to be analysed, e.g. thermally averaged over the Boltzmann distribution if the molecules are in solution.

In this case the best place to start is probably with the papers

(a)  Heller, Freed & Gelbart, J. Chem. Phys. 56, 2309 (1972),

(b)  Bixon & Jortner, J. Chem. Phys. 48, 715-726, 1968 and

(c) S. H. Lin; J. Chem. Phys. 58, 5760 (1973).

Each deal with non-adiabatic interactions causing radiationless transitions in molecules, i.e. from one vibronic level into a continuum of states, such as singlet to triplet intersystem crossing. There is a summary of different methods in Eyring, Lin & Lin 'Basic Chemical Kinetics' ch 7.

$^{[1]}$ equations given in Steinfeld, Francisco & Hase, 'Chemical Kinetics & Dynamics'. 2nd ed. p333.


Symmetry C3V For MeCl Methyl chloride belongs to the $C_{3\mathrm{v}}$ point group. This is because it has a $C_3$ axis down the $\ce{Cl-C}$ bond.

However what I do not understand is how it has three vertical mirror planes? It has three planes down each of the $\ce{C-H}$ bonds, but don't these bisect a bond making them dihedral planes? Although I'm wrong I can't fathom why it isn't a $C_{3\mathrm{d}}$ point group. I have used a molecular viewer to visualise it but I still get the vertical mirror planes bisecting a bond...

You've correctly identified the $C_3$ axis that contains the $\ce{C-Cl}$ bond.  Methyl chloride also has 3 planes of symmetry that contain the $C_3$ axis and each of the $\ce{C-H}$ bonds.  Since these planes contain the $C_3$ axis they are referred to as $\sigma_{v}$ planes of symmetry.  There are no other symmetry elements in methyl chloride so it belongs to point group $C_{3v}$ (n $\sigma v$ planes plus a $C_n$ axis).

There is no $C_{nd}$ point group.  There is a $D_{nd}$ point group which contains a $C_n$ axis, n $\sigma_{v}$ planes (so far methyl chloride fits the description), but also n $C_2$ axis.  Methyl chloride does not contain any $C_2$ axes, so it does not belong to the $D_{nd}$ point group.


What is the color and structure of CrO3? What is the structure and color of $\ce{CrO3}$?

Also, is it possible to produce $\ce{CrO3}$ when dichromate is heated with sulphuric acid?

According to its Wikipedia page:

Color

$\ce{CrO3}$ is a dark-purple solid under anhydrous conditions, that turns bright orange when wet.

Production

$\ce{CrO3}$ is generated, as you said, by the following reaction:

$$\ce{H2SO4 + Na2Cr2O7 -> 2 CrO3 + Na2SO4 + H2O}$$

Structure

Its monomeric structure is predicted to be pyramidal like so:

Its crystalline structure is:

Each chromium center shares two oxygen centers with neighbors, but two oxygen atoms are not shared, giving the overall stoichiometric ratio of 1:3.


How does a group 15 oxide react with water? Given the following reactants:

  $\ce{P2O5}$ and $\ce{H2O}$

How does one determine what product the two will form in a reaction?

I instantly tried to treat the problem as a double displacement reaction. However, the compounds that I got by doing so were absurd. Do double replacement reactions only occur with ionic compounds (e.g., the cations of two ions switch)?

The two elements listed above form $\ce{H3PO4}$. How do I even go about determining that? What indicates that such a product forms from those two reactants?

The reaction between the two substance you have mentioned above in synthesis
it is an acid-base reaction .
and about the other question you asked above i can say that we have reactions where two ionic substances take part in oxidation reduction reactions instead of double replacement.


Why does color of halogen deepens and the boiling point decreases down the group? Why does the color of the halogens deepens as we go down the group? also why there is a change in physical state of halogens down the group?

\begin{array}{c|c}
\mathbf{Gas} & \mathbf{Color} \\\hline
\text{fluorine} & \text{very pale yellow (almost invisible to naked eye)}\\
\text{chlorine} & \text{yellow-green} \\
\text{bromine} & \text{red-brown}\\
\text{iodine} & \text{violet-black}\end{array}

As for the BP:
The attraction of covalent molecules is due 'dispersion' interaction. It is a quite complex effect, commonly described as interaction of 'instantaneous dipoles', i.e. interaction of dipoles formed by nucleus and electrons around it. Its efficiency generally increases with number of electrons (due to sheer numbers) and size of electron shell. Since outer electron shells vastly grow in size in fluorine-iodine row, the efficiency of dispersion interactions grow. In fact, close halogen-halogen contacts for heavier halogens (bromine and iodine) are very common in crystal structures of organic compounds.
As for color. As a guess, I'd say that the active transition is lone pair orbital $\rightarrow$ $\sigma$-antibonding orbital (see any introduction to MO theory for the terminology). The energy of this transition should lower from fluorine to iodine because of great decrease in bonding energy and increase of size of atoms. According to color wheel, the absorption in violet spectrum (with highest energy of photons in visible spectrum) should give yellowish apparent color (case of fluorine and chlorine), red apparent color to adsorption in lower-energy and violet color for adsorption at even lower-energy red spectrum.


Decomposition products of PET On Wikipedia, it says that polyethylene terephthalate (PET) decomposes at temperatures above 350°C.
What are these decomposition products? I read further that acetaldehyde is produced, but is that the only decomposition product?

Also, what compounds are produced when PET is burned? ($\ce{+O2->?}$)

This is a partial answer to your query.  According to the paper Degradation and Recyclability
of Poly (Ethylene Terephthalate)
 (Venkatachalam, 2012), there are a few degradation products, from the article:

  Apart from aldehyde, $\ce{CO}$, $\ce{CO2}$, ethylene, benzene, biphenyl are also identified as
  degradation products

(The article also shows diagrms of the chemical processes of PTE degradation).

an important aspect to note is experiments/observations have

  shown that rate of thermal degradation in the closed system is
  about three times higher than that in open system.

In regards to burning of PET, the products can vary according to the page EPB 433- Health and Environmental Effects of Burning Waste
Plastics
 (Saskatchewan Ministry of Environment) stating that

  the composition of byproducts of plastic combustion as to their type and concentration depends
  on the combustion temperature and the flame residence time.

According to the Peiyu Plastics page (a reference from Wikipedia), if complete combustion occurs, then the products would be $\ce{CO2}$ and $\ce{H2O}$, incomplete combustion could produce soot and carbon monoxide.

Further reference:

Here is the MSDS for PET

A study of thermal decomposition and combustion products
of disposable polyethylene terephthalate (PET) plastic using
the High Resolution Fourier Transform Infrared
Spectroscopy, Selected Ion flow Tube Mass Spectrometry
and Gas Chromatography Mass Spectrometry  (Sonova et al. 2008).


Carbon Free energy vs Temperature Diagram I've recently read that the Carbon takes a unique position as a reducing agent , and i think it has to do with the very special form of its Ellingham diagram  , but i am still not able to make a connection , So what makes exaclty the carbon a good reducing agent ?

Here is the link to a set of Ellingham Diagrams

At high temperatures, smelting and carbon reductions have several practical challenges some of which are energy efficiency, formation of metal carbides, equipment concerns etc. Therefore, to assess the reducing ability of any reducing agent, we must work within practical limits. Let us select a temperature of $1600\ \mathrm{K}$ as our upper bound beyond (which is still quite high) which, practical concerns outweigh any benefit of having a good reductant. At this temperature $\Delta_\text fG(\ce{CO2})$ is approx $-394.39\ \mathrm{kJ/mol}$ and $\Delta_\text fG(\ce{CO})$ is $-504\ \mathrm{kJ/mol}$. These values are high enough and can already reduce several oxides of metals like $\ce{Co, Fe, Cu, Pb, Ni, Bi}$ and $\ce{MnO2, CrO2}$ etc (subject to practical constraints). This gives it a good reducing character, better than $\ce{H2}$ or ammonia.

Another thing to notice, is that, although the variation of $\Delta_\text fG$ for carbon dioxide is flat, for carbon monoxide it slopes downward and hence with increasing temperature, it can reduce many more oxides further expanding its capacity to reduce metal oxides. $\ce{V2O5, VO2, SnO2, WO3, Mn3O4, MoO3, GeO2}$ can be reduced by carbon. (All this just implies thermodynamic favourability; it can still be unfavourable kinetically, or due to practical restraints.) Hence, the negative sloping $\Delta_\text fG$ for $\ce{CO}$ makes carbon a very good reducing agent, and it can reduce a huge variety of metal oxides, within a considerable temperature limit.


Is the standard enthalpy of formation always non-positive? If the standard enthalpy of formation is defined as

  the energy change when 1 mole of a substance is formed from its elements in their standard states

and forming chemical bonds is an exothermic process, then does this mean all standard enthalpies of formation are non-positive?

Why do some compounds such as $\ce{NO}$, $\ce{NO2}$ have positive standard enthalpies of formation? Is this because to form $\ce{NO}$, we must first break the bonds in $\ce{N2}$ and $\ce{O2}$ (endothermic) to form the bond in $\ce{NO}$ (exothermic, smaller energy produced than the energy required to break the bonds of original standard state elements)?

You'v answered the question yourself.

In general, any enthalpy of reaction is the measure of the energy remainder after breaking the bonds (positive) and what energy we got back from the formation of the bonds (negative).

The net sum can be positive or negative, and it depends on what the reactants and products were. As you correctly said yourself, $\ce{NO}$ happens to have a positive standard enthalpy of formation because, as experimentally determined, more energy went into breaking the bonds of reactants than what we got back from forming the products.

So,

  Is the standard enthalpy of formation always non-positive?

The answer is no.


EMF measured while verification of Nernst equation keeps on increasing I was doing this experiment to verify Nernst equation in my school's lab. I had two beakers one with $\ce{ZnSO4}$ and the other with $\ce{CuSO4}$ both having a concentration of $1~\mathrm{M}$. I had a saltbridge having $4~\%$ Agar in $1~\mathrm{M}\ \ce{KCl}$. The temperature was $21.3\ \mathrm{^\circ C}$ Now the problem was that when I connected a multimeter across the electrodes it showed an EMF of $0.8\ \mathrm{V}$ initially and it went on increasing to $1.001~\mathrm{V}$. The increase was rapid initially and became slower and slower. The EMF still kept on increasing but with a very slow pace.

By what I have studied the EMF I get initially should be the maximum one and then go on decreasing as the equilibrium is attained.

So why was the EMF in my experiment increasing? Is it because I had not used wool at the ends of the salt bridge? But the Agar was already a thick gel and was not flowing out of the U-tube.

I asked my teacher and she said that there might be some problem with the multimeter. Does any one of you know the reason for the increasing EMF?

You should arrive to a plateau value, the EMF can't increase perpetually. The ideal potential difference is around 1.1 V given the experimental arrange, this is also known as the Daniell Cell. I think the low initial value you get is because of the time response of the system, until it reaches a steady state.
There are basically two types of connections:

i) Electrode/Solution: Some specie B transfers an electron from the solution to electrode surface (by receiving or by giving). The electron transfer rate is very high, specially on batteries, so this can not limit the whole process.

ii) Ion migration through solution: This step is the limiting process. The ions are the charge carriers in bulk solution, which move much slower compared with electrons in a conductive metal. A way to overcome this is using strong acid solutions, as sulfuric acid in the Lead-Acid Battery used on cars.

The response you have will depend on the time of measure (immediately after closing the connection, 5 seconds after, etc.) and the time interval of such measure. It will depend also on the initial concentrations of the solutions (high or low). If the concentration is low, the ionic strength of solutions is weak and few ions are to carry the charge; adding table salt (NaCl) is a possible solution. The ionic bridge may be highly conductive but, if the water permeation is also limited, you have an extra limitation there, which will be higher as the phase thickness increases (ie using too much agar, or long bridge). A solution may be, using an U-Tube with NaCl 4 M and covered in the ends with commercial water-purifier filters well sealed to the glass.

A delay on response could be due to the hydration time of the agar gel, which can take a while. Once water has permeated the agar, the ions begin to move faster, until steady state is reached.


Do amines or alcohols have stronger intermolecular hydrogen bonds? Which has stronger hydrogen bonding, $\ce{CH3OH}$ or $\ce{CH3NH2}$

I think it comes down to which has more dominance; number of hydrogens, number of lone pairs, or electronegativity.

The boiling point of non-ionic compounds are highly depend on their H-bonding abilities. For example, boiling point of water (molar mass: $\pu{18.02 g/mol}$) is $\pu{100 ^\circ C}$ at $\pu{1 atm}$ while that of ethanol (molar mass: $\pu{46.07 g/mol}$) is $\pu{78.4 ^\circ C}$ at $\pu{1 atm}$, even though ethanol is heavier and have more other intermolecular attractions, excluding H-bonding, such as van der Waals interactions than water. This difference in boiling points displays clear dominamce of H-bonding (HB) on boiling point over other forces on these non-ionic compounds.

Therefore, since molar masses of methanol and methylamine is very close ($\pu{32.04 g/mol}$ for $\ce{CH3OH}$ versus $\pu{31.05 g/mol}$ for $\ce{CH3NH2}$), comparition of their boiling point would be excellent indicator of the comparison of their H-bonding strength: Boiling point of $\ce{CH3OH}$ is $\pu{64.7 ^\circ C}$ at $\pu{1 atm}$ while that of $\ce{CH3NH2}$  is $\pu{-6 ^\circ C}$ at $\pu{1 atm}$. Therefore, it is safe one to say that $\ce{CH3OH}$  has stronger H-bonding capabilities than that of $\ce{CH3NH2}$ even though $\ce{CH3NH2}$ has extra $\ce{H}$ compared to only one in $\ce{CH3OH}$. The best comparison measures of these two types of hydrogen bonding (HB) are the $\Delta H_\mathrm{HB}$ values in Wikipedia (Hydrogen Bonding):

  $\Delta H_\mathrm{HB} = \pu{21 kJ/mol}$, listed for $\ce{O−H···:O}$ in water-water.
  $\Delta H_\mathrm{HB} = \pu{13 kJ/mol}$, listed for $\ce{N−H···:N}$ in ammonia-ammonia.


Are hydroxides soluble in each other? Let's say I have molten potassium hydroxide and then I add some Calcium hydroxide. I imagine, since there are hydroxide groups floating around, it would be energetically neutral for the calcium ions to bind to a "new" hydroxide group, therefore they should be floating freely with the hydroxide groups, becoming solvated. I can't find any empirical data; is my reasoning correct and I can dissolve a hydroxide in any other molten hydroxide?

Probably not so much calcium hydroxide, but alkali metal hydroxides mixing with each other in the liquid phase.  The pairs $\ce{LiOH-NaOH}$ and $\ce{KOH-NaOH}$ are not only fully miscible, they have been studied as fuel cell electrolytes.  The picture below, from the referenced article, shows results for the a range of electrolyte compositions that imply the extensive miscibility.  The fuel cell itself is based on carbon oxidation to carbon dioxide.


Sample: How to calculate mass of nanopaticles with given density and Vol % I'm trying to calculate the wt% and the mass of the nanoparticles in a ferrofluid solution.

Attached is data from the supplier:

**EMG 605 Specifications and Physical Properties**
- Appearance:   Black-brown fluid
- Carrier Liquid:   Water
- Nominal Particle Diameter:    10 nm

- Saturation Magnetization (Ms): 220 Gauss or 22 mT (SI)
- Viscosity @27°C:  <5 cP or <5 mPa·s
- Density @25°C:    1.18 g/cc or 1.18x10^3 kg/m3 (SI)
- Initial Magnetic Susceptibility:  0.24    or 3.02 (SI)
- :Magnetic Particle Concentration  3.9 % vol.
- pH:   9-10
- Nature of Surfactant: Cationic

Defining the following

Density of solution: $\rho = \pu{1.18 g/cc}$
Density of water: $\rho_w$
Density of nanoparticles: $\rho_{np}$
% (v/v) n.p.s: $p_v=3.9$%
% (w/w) n.p.s: $p_{wt}$

Then solve for $\rho_{np}$ in the following equation (you'll need to look up missing value of water density):

$$\rho = \frac{p_v}{100}\rho_{np} + \frac{100-p_v}{100}\rho_{h2o}$$

With $\rho_{np}$ you can compute $p_{wt}$

$$p_{wt} = 100\frac{p_{v}\rho_{np}}{p_{v}\rho_{np}+(100-p_{v})\rho_{w}}$$

As noted in the comments, the resulting density of the particles ignores the effect of the added cationic surfactant. Also, the solvent density is assumed that of water, free (not particle bound) detergent or other additives assumed to be negligible.


Stereoelectronic effects in torquoselectivity of 4π electrocyclic opening Thermal 4π electrocyclic ring openings of cyclobutenes to form butadienes are known to proceed in a conrotatory manner, consistent with the Woodward–Hoffmann rules. However, there are two different conrotatory modes which can lead to isomeric products if the cyclobutene is suitably unsymmetrically substituted. Torquoselectivity refers to selectivity arising due to the preference of one conrotatory mode over the other.

The Houk group has reported some examples of counterintuitive 4π ring openings1,2 (simply considering steric interactions, the larger group generally wants to move outwards). In both cases only one isomer of the product is formed.

Why is this selectivity seen?

References

1. Houk, K. N.; Spellmeyer, D. C.; Jefford, C. W.; Rimbault, C. G.; Wang, Y.; Miller, R. D. Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds. J. Org. Chem. 1988, 53 (9), 2125–2127. DOI: 10.1021/jo00244a058.

2. Rudolf, K.; Spellmeyer, D. C.; Houk, K. N. Prediction and experimental verification of the stereoselective electrocyclization of 3-formylcyclobutene. J. Org. Chem. 1987, 52 (16), 3708–3710. DOI: 10.1021/jo00392a047.

The torquoselectivity of the above reactions is dictated by stereoelectronic effects, which (in this case) are significant enough to take precedence over steric effects. In general, electron-donating substituents (e.g. OMe, OTMS, ...) prefer to move outwards, and electron-accepting substituents (e.g. CHO) prefer to move inwards.

These can be understood by considering the frontier orbitals in the transition state of the conrotatory opening, as explained by the Houk group in earlier publications1,2 (the reactions given above were, in fact, carried out in order to provide experimental verification of their theoretical predictions). A more accurate orbital correlation diagram for the reaction is given below (modified from ref 2 to include diagrams of the relevant MOs):

A brief explanation is as follows. As the reaction proceeds, the C–C σ bond in the reactant is stretched, and consequently the σ* orbital is initially strongly destabilised (its energy increases rapidly). At a certain point near the transition state, though, it becomes sufficiently close in energy to the π* orbital such that orbital mixing can occur; this mixing leads to stabilisation and eventual "evolution" into the HOMO of butadiene.

Therefore, in the transition state, the HOMO is mostly σ-like, and the LUMO σ*-like; this was also verified computationally (details in ref 2). These frontier orbitals are of the correct symmetry to interact with pi-type orbitals of substituents. For donor substituents such as OMe, these orbitals are filled (lone pair on oxygen); for acceptor substituents such as CHO, these orbitals are empty (low-lying C=O π*).

The dashed orbitals in the following diagram represent the positions of these orbitals in the TS for both cases of outward and inward rotation. In the case of outward rotation, the substituent orbital can only interact with the lobe on C–3 which is in close spatial proximity; however, in the case of inward rotation, the substituent orbital can interact with the lobes on both C–3 and C–4.

In general, the interaction of a filled orbital and an unfilled orbital leads to stabilisation, whereas the interaction of two filled orbitals leads to destabilisation. The torquoselectivity is determined by maximisation of the former and minimisation of the latter. If the interaction with the lobe of C–4 leads to destabilisation, then inward rotation will be disfavoured, and vice versa.

For a donor substituent:

Inward rotation decreases the stabilising interaction between the donor orbital and the LUMO of the TS, as the lobe on C–4 is of the opposite phase as the lobe on the donor orbital, which leads to poorer overlap between the two orbitals.
Inward rotation increases the destabilising interaction between the donor orbital and the HOMO of the TS, as the lobe on C–4 is of the same phase as the lobe on the donor orbital, which leads to stronger overlap.

Based on these reasons, donor substituents generally prefer to rotate outwards. In the case of OH, this preference was calculated to be a whopping 16.4 kcal/mol.3

For acceptor substituents, the interaction with the LUMO is not relevant as the interaction of two unfilled orbitals does not lead to any stabilisation. However, inward rotation maximises the stabilising interaction with the HOMO, since this is enhanced by overlap with the lobe on C–4. Hence, acceptor substituents generally prefer to rotate inwards. This preference, however, is markedly smaller than in the case of donor substituents - perhaps because there is only one factor that biases the course of the reaction (as opposed to two in the case of donor substituents). Experimentally there are fewer examples of torquoselectivity with electron-accepting substituents, and for the second reaction above, the preference for the inward rotation of the formyl group was calculated to be a much more modest 4.5 kcal/mol.4

Interestingly enough, for the cyano group (-CN), a preference of 4.3 kcal/mol was calculated - for outward rotation.4 It turns out that the donor π orbitals are more important than the acceptor π* orbitals, which are relatively high in energy. The formyl group only rotates inwards because it has an exceptionally low-lying π* orbital!

References

1. Kirmse, W.; Rondan, N. G.; Houk, K. N. Stereoselective substituent effects on conrotatory electrocyclic reactions of cyclobutenes. J. Am. Chem. Soc. 1984, 106 (25), 7989–7991. DOI: 10.1021/ja00337a067.

2. Rondan, N. G.; Houk, K. N. Theory of stereoselection in conrotatory electrocyclic reactions of substituted cyclobutenes. J. Am. Chem. Soc. 1985, 107 (7), 2099–2111. DOI: 10.1021/ja00293a046.

3. Houk, K. N.; Spellmeyer, D. C.; Jefford, C. W.; Rimbault, C. G.; Wang, Y.; Miller, R. D. Electronic control of the stereoselectivities of electrocyclic reactions of cyclobutenes against incredible steric odds. J. Org. Chem. 1988, 53 (9), 2125–2127. DOI: 10.1021/jo00244a058.

4. Rudolf, K.; Spellmeyer, D. C.; Houk, K. N. Prediction and experimental verification of the stereoselective electrocyclization of 3-formylcyclobutene. J. Org. Chem. 1987, 52 (16), 3708–3710. DOI: 10.1021/jo00392a047.


What safety measures should I take when operating a backyard aluminum furnace I have made a DIY furnaces out of cement. Before I do any smelting, I would just like to take any suggestions in regards to any safety measures I should take.

I plan to only melt aluminum cans and small parts. I then plan to pour them onto a flat plane to make aluminum plates. Later I plan to make casts.

My furnace is very similar to the one pictured below.

I plan to connect a hair drier to the hole (with an extension metal rod) in order to increase air flow. It will operate by burning charcoal brickets.

Set this up in an area free of leaves and debris.

I recommend wearing safety gear: heat resistant goggles and fire-resistance clothing and gloves that will protect you from heat, especially when pouring.

Get a couple practice runs in pouring before using it so you get a feel for the best angle to hold in and learn what NOT to do. Even better, you could assemble something that would do the pouring for you. (I'm thinking of something along the lines of this, but it doesn't have to be as elaborate.)

You may also want to scrape the paint off the cans before putting them in. Otherwise, you'll  get a lot of crud from the paint in your molten aluminum.


Euclidean distance between atoms using RDKit I'm trying to find the Euclidean distance between two atoms in the molecule with SMILES representation O=CC1OC12CC1OC12 using the rdkit package. Looking online, I have converged to the following code.

import numpy as np
from rdkit import Chem

mol = Chem.MolFromSmiles('O=CC1OC12CC1OC12')

conf = mol.GetConformer()
at1Coords = np.array(conf.GetAtomPosition(bond_i.GetBeginAtomIdx()))
at2Coords = np.array(conf.GetAtomPosition(bond_i.GetEndAtomIdx()))
print(np.linalg.norm(at2Coords - at1Coords))

But I get the error

conf = mol.GetConformer()
ValueError: Bad Conformer Id

I was wondering how to fix the issue. Thanks!

The issue was that I had to embed the molecule to have a 3D conformation. So I added following line:

AllChem.EmbedMolecule(mol)

Feel free to answer with better suggestions and alternatives.


Are there substances that cannot undergo combustion? Are there substances that cannot undergo combustion? If so, what are they?

It depends a little on your definition of combustion.

Most combustion we see involves things reacting with the oxygen in air. Most combustion reactions are a self-sustaining exothermic reaction of some substance with oxygen gas. But atmospheric oxygen is relatively dilute (only about 21% of air is oxygen) and this limits what will burn sustainably in air (also, at slightly lower concentrations many commonly flammable things won't stay lit). The requirement for things to be combustible in air is, crudely, a reaction path with oxygen that releases energy.

Examples include many types of compound. Hydrogen reacts with oxygen to give water; hydrocarbons react to give carbon dioxide and water. Many metals react to give metal oxides (eg magnesium, spectacularly when lit; iron, only when hot and finely divided).

Many things won't burn in air for two reasons: there are activation barriers to a potential oxidation meaning a potential reaction can't be sustained (aluminium has a strong oxide layer on its surface protecting the reactive metal from further oxidation, for example); other compounds are already highly oxidised an no further oxidation can happen (eg silicon dioxide, a major ingredient of sand and glass).

But there are many more things that will combust in pure oxygen as Apollo 1 discovered, tragically. And as many demonstrations show spectacularly (see demo here). But highly oxidised things like silica can't be oxidised further and won't burn not matter how concentrated the oxygen is.

But oxygen isn't the only gas that can support exothermic reactions. Rather more things will burn in fluorine, for example. Even relatively unreactive substances can support combustion if there is an available reaction path that releases enough energy and a reaction can get started. One of the best known examples is that magnesium will burn in carbon dioxide (the reaction producing magnesium oxide releases energy) and even nitrogen (producing magnesium nitride). See the explanation here and the video here.

So, to return to the original question, are there any substances that won't combust? Whatever the definition of combustion is the answer is still yes. Noble gases, for example, though some can be coerced into reacting under extreme circumstances, never combust or support combustion.

If we restrict our definition to combust in air, many substances won't combust though many of them might do so under different circumstances.


PCl3 vs SCl2 boiling point Why does sulfur dichloride have a lower boiling point than phosphorus trichloride? Is it to do with the higher number of electrons causing stronger id-id interactions in PCl3, or do we need to compare pd-pd interactions also?

I think you're looking down the wrong path to explain this. Boiling point can often be explained by looking at the interactions between molecules such as hydrogen bonding, dipole-dipole interactions, and london forces, as well as overall molecular mass. I would explain this phenomenon by looking at two data points for each molecule. The dipole of $\ce{SCl2}$ is 0.4 Debye and the dipole of $\ce{PCl3}$ is approximately 0.780 Debye.$\ce{^{1,2}}$ Based on  this alone, I would expect $\ce{PCl3}$ to have a higher boiling point. If you also look at the mass of each molecule, 102.97 g/mol for $\ce{SCl2}$ and 137.33 g/mol $\ce{PCl3}$, it becomes pretty clear which substance one should predict to have the higher boiling point. It's $\ce{PCl3}$.

Ref 1. http://openmopac.net/PM7_accuracy/data_molecules/sulfur%20dichloride_jmol.html

Ref 2. https://cccbdb.nist.gov/diplistx.asp


The molecular orbitals of lithium hydride So my understanding of molecular orbitals is as follows: the molecular orbitals of $\ce{LiH}$ can be thought of as being formed from the interaction/overlap of two atomic orbitals, one from lithium and one from hydrogen in this case. I also understand that the atomic orbitals have to be quite close in energy for any significant interaction to take place.

The occupied atomic orbitals of lithium are 1s and 2s, with energies of around -66 eV and $\pu{-5.3eV}$ respectively. The occupied atomic orbital of hydrogen is 1s, with an energy of -13.6 eV. I believe that it's safe to assume that the 1s orbital of lithium is far too low in energy to interact with the 1s orbital from hydrogen, meaning that only the interactions are between the 2s orbital of lithium and the 1s orbital of hydrogen, producing a bonding MO and an anti-bonding MO.

Wikipedia states that "Bonding MOs are lower in energy than the atomic orbitals that combine to produce them." In other words, the 2$\sigma$ bonding MO should have an energy less than -13.6 eV, which is the energy of the 1s hydrogen orbital. However, computer  calculations of the MOs of LiH show that this bonding MO has an energy of -8eV. which would seem to contradict the statement from Wikipedia.

So assuming there is no fault in my reasoning so far, we can safely say that this Wikipedia statement is only generally true, and false in this particular case. So why is this?

I was trying to rationalise this as follows: generally, the reason why bonding MOs are lower in energy than the atomic orbitals used to create them is because bonding MOs concentrate electron density between the two nuclei, resulting in the electrons being attracted by both nuclei (as opposed to just one), leading to the MO being lower in energy than the original AOs. However, in LiH, due to the energy of the 1s AO on hydrogen being significantly lower than that of the 2s AO on lithium, the 1s AO will be the major contributor to the bonding MO, meaning that the electron density in the bonding MO will be significantly closer to hydrogen's nucleus than lithium's nucleus.

Now, originally, there is one electron each in the 1s hydrogen AO and 2s lithium AO, and when they come together, two things change: Firstly, there is electron-electron repulsion between these two electrons. Secondly, the electrons are now bound by two nuclei rather than 1. However, due to the high polarity of the bond, this second change is minimal, such that the first change (introduction of repulsion) actually outweighs the second change, meaning that the resulting bonding MO is actually higher in energy than the hydrogen 1s AO.

Is my rationalization of this accurate/correct? If not, please suggest alternative explanations/ improvements to my thought process because I'm frankly not fully convinced by myself.

We may want to consider also the energy change in the lithium 1s orbital.  Formation of the molecular orbitals also changes the energy level of the core orbitals even though these do not participate appreciably in the bonding.

This is because what I call "reverse shielding" occurs.  In a lithium atom in the ground state, there are of course two 1s electrons and one 2s electron.  We commonly suppose that the 1s electrons, packed close to the nucleus, partially shield the 2s electron from the nuclear charge and raise the energy level of the latter.  But, because the probability distributions of the 1s and 2s clouds overlap, there is a small probability that the 2s electron is instead shielding one or both of the 1s electrons.  The effect on the 1s wavefunctions is small, due to the small probability, but the effect on lithium 1s energy can still register in electron volts because the 1s wavefunction is crowded into a region of strong Coulonb potential.

Then, when the hydrogen atom is added and the bonding orbital forms, what had been the 2s electrons are drawn away from the lithium atom and thus the reverse shielding of the lithium 1s electrons is diminished.  Therefore, the lithium 1s orbital is lowered in energy, possibly enabling a stable molecule even if the bonding molecular orbital is higher than the isolated hydrogen 1s orbital as the OP suggests.


Why do we still use the 'P' subscript in a thermodynamic derivative such as (dH/dT)P? Consider the following thermodynamic derivative:

$$
\left(\frac{\partial H}{\partial T}\right)_P
$$

Taking the partial derivative w.r.t. $T$ means that all other variables are held constant. Why do we still write it with a subscript like $P$ or $V$?

In thermodynamics, the number of degrees of freedom available to describe a system is given by Gibbs' phase rule:

$$
F = C - P + 2
$$

Here, $C$ is the number of components in the system, $P$ is the number of phases in the system, and $F$ is the number of resulting degrees of freedom.

Consider a pure substance in a closed system.  One can name a wide variety of thermodynamic parameters to describe its state: temperature, pressure, specific volume, internal energy, enthalpy, entropy, etc. However, Gibbs' phase rule says that we can only independently specify a subset of these parameters before we run out of degrees of freedom:

$$
F = 1 - 1 + 2 = 2
$$

So, in any mathematical description of (in your example) the enthalpy, $H$, it can only properly be a function of two independent variables:

$$
H = H\left(X_1, X_2\right)
$$

But: which two variables?

Simply writing the partial derivative $\partial H \over \partial T$ indicates that temperature is to be one of the two available degrees of freedom:

$$
H = H\left(T, X_2\right)
$$

Now, though – what is $X_2$? Well, including the subscript on the differential is a way of explicitly indicating what that other independent variable is:

$$
\left({\partial H \over \partial T}\right)_P \quad \Longrightarrow
\quad H = H\left(T, P\right)
$$

As porphyrin rightly notes in a comment, this applies to multivariate calculus generally, not just the calculus of thermodynamics.


How to know if something will be a good oxidizing/ reducing agent? (Besides the activity of the metal) One of first redox reaction I was taught in school, is probably single replacement. It is easy to predict cause all you need is a table of metal activity series.

But then there are more complicated or less well-known agents. Like permanganate ion, which is used commonly in determining the purity of iron.

For agents like this, how do you know if it is strong enough to reduce or oxidize the other elements?

(Is there also a table for this?)

The relative strengths of oxidising and reducing agents can be inferred from their standard electrode potentials. The strongest oxidizing agents are shown from the standard electrode table.


How does a shock, such as the snap of the disk in a reusable hand warmer, trigger the re-crystalization of a supersaturated solution? Reusable handwarmers are made of something dissolved nearly to its saturation point in hot water (Sodium Acetate in the case of HotSnapz) at a high temperature, and then cooled to a temperature at which some would normally be insoluble, but remains dissolved due to a lack of a site to grow a crystal from. When the crystal finally grows, triggered by popping a little disk, the process is exothermic and it releases heat.
What I'm wondering is: how does popping the little metal disk inside the hand warmer allow nucleation to begin?
EDIT: I got a bit more information, but nothing definitive. What I’ve heard is that the metal disk may either be initiating nucleation by releasing tiny metal particles when it snaps or by snapping quickly enough to cause cavitation. The first seems unlikely, because if bits of metal were released and functioned as nucleation sites, the warmer wouldn’t be reusable - the crystals could just reform on those metal bits. Cavitation seems more plausible, though I still don’t understand how that would allow crystallization to begin.
EDIT 2: I did some experimenting, and a whack also starts nucleation, not just the disc snapping. So, any shock can cause the sodium an acetate ions to come out of supersaturated solution. How?

Edit: Contribution of enthalpy and bond formation
Edit 2: Changed the description of entropy
After thinking for sometime, I realized that I neglected to mention three things:

The Driving force of precipitation reactions depends on the charge and size of the ions which corresponds directly to the magnitude and order of the hydration shells formed upon dissolution of the salt.

The enthalpy as a driving force of the precipitation of the salt. (Remember the Gibbs Free energy equation has two terms, entropy and enthalpy)

Additionally, the formation of ionic bonds can be lower in energy (more stable) than the electrostatic interactions present in solution. The release of energy during salt formation is transferred to the surroundings as heat. This would be a negative enthalpy (heat flowing out of the system) and if large enough, would overcome any decreases in entropy stemming from the formation of highly order crystalline structures and drive the precipitation.

In this case it is probably the enthalpy term that is driving the reaction, since you are putting a ton of heat into the system to dissolve the salt, once the solution cools, there is a lot of potential energy that is stored in the position of the atoms in the system. This is then released as heat upon the formation of the crystal structures.
Entropy changes in precipitation reactions:
In order to make a supersaturated solution, you have to add energy to form hydration shells around the ions (entropy in the system is decreasing). The thing to remember is that these hydration shells decrease the dispersal of the ions relative to that in the crystalline structure for some salts. When you heat and mix the solution with excess solute, you are providing sufficient energy to form hydration shells. When the solution cools (Slowly), the hydration shells remain intact, however, the system is unstable and applying enough energy to the solution will disrupt the hydration shells and crystallization of the salt will occur. The large amount of heat that is released is coming from the increase in the micro-states available for the ions (gain in entropy) in removing the hydration shells around the ions. The hydration shells are less dispersed than the crystalline structures of the salt and that is why entropy increases during some crystallization reactions, making this process exothermic.
Here is why the disc snapping is important:
Even though this process is very exothermic, it still requires activation energy. The mechanical force you are referring to (i.e. snapping the disc or whacking the heating pad) is doing just that by simply transferring energy to the molecules, through vibrations, which provides the activation energy for crystallization.
Just as walking can cause energy transfers from the mechanical forces of your legs to the molecules of the floor (i.e. through friction) and increase the temperature of the surface, some of the force applied on the disc transfers to the solution and if it is enough, will overcome the activation energy barrier of the process and crystallization will begin.


Isn't the change in entropy zero for a reversible process? I'm calculating $\Delta G$, $\Delta H$ and $\Delta S$ when an ideal gas expands isothermally and reversible at $20$°C from  $5$ MPa to $1$ MPa. I used the relation that:
$$\left(\frac{\partial G}{\partial P}\right)_T=V$$
This relation can be shown from the defintion $G=H-TS$. Then I integrated G so that:
$$\Delta G=\int_{P_1}^{P_2}VdP$$
We have an ideal gas $=>V=\frac{RT}{P}$. Which gives us that:
$$\Delta G=\int_{P_1}^{P_2}\frac{RT}{P}dP=RT\ln\frac{P_2}{P_1}$$
If you calculate this you'll get that $\Delta G\approx3.9$ kJ/mol. This was correct.
However I then tried to calculate $\Delta H$. Since $dH=dQ+VdP$ and $dQ=TdS$:
$$dH=TdS+VdP$$
After this we know that we have a reversible process which according to me means that $dS=0$. Which means that $dH=VdP=dG=>\Delta H=\Delta G$. However the answer was that $\Delta S= 13.38$ J/(Kmol) and $\Delta H=0$. How is this possible? Isn't $dS=0$ for a reversible process?

As mentioned in the comments, there is an entropy change that occurs in the system, however, when combined with the entropy change of the surroundings, is zero.
When solving for the change of entropy in the system, you can use the equation below,
$ \Delta S = R \ln(\frac{P_1}{P_2})$
converting MPa to atm and plugging in, we get
$ \Delta S = 8.314 \ln(\frac{49.3}{9.87})$
which yields an entropy change that is equal to the answer you provide.
The above equation can be rearranged to have different variables instead of pressure. Here is a link for a ChemLibretexts article on entropy and reversible processes if you're interested.


Nomenclature of an alkyne with bromo and hydroxy substituents Is the IUPAC name of above compound 7,7-dibromo oct-5-yn-4-ol or 2,2 dibromo oct-3-yn-5-ol ?
I understand that the former name is given keeping in mind -OH as the functional group but by giving the latter name we will have the least sum of locants (ie, following lowest sum rule) .
When should the lowest sum rule be dropped ?
Thank You in advance.

In this case, the relevant criteria for the numbering are:

lower locants for suffixes (here: ‘ol’)
lower locants for multiple bonds (here: ‘yne’)
lower locants for prefixes (here: ‘bromo’)

The corresponding subsection 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:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (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;

  (…)

Therefore, the correct name for the compound given in the question is ‘7,7-dibromooct-5-yn-4-ol’ (not ‘2,2-dibromooct-3-yn-5-ol’) since the locant ‘4’ for the suffix ‘ol’ is lower than ‘5’.


Partial Aromaticity and Anti-aromaticity The following resonance structures are possible for the given compound. (a) promotes both aromaticity and Anti-aromaticity whereas (b) promotes aromaticity. In the resonance hybrid, (b) contributes more as we can see. Will the resonance hybrid be called aromatic or anti-aromatic?

From a classification point of view,  there is no reason for not calling the molecule aromatic.

First, saying a molecule is aromatic doesn't imply that  aromaticity fully extends over it. A trivial example is, just to say, cyclohexylbenzene, or whatever aromatic molecule of your choice.

Now, let us assume the two rings being completely independent so to exclude any delocalisation. The phenyl moiety will be just aromatic while the Huckel antiaromatic cyclobutadiene ring will simply adopt a rectangular shape, so effectively avoiding the antiaromatic unstable configuration.

Again, there wouldn't be reason for not calling the molecule aromatic.

This is often a sort of misunderstanding when antiaromaticity is discussed. It is just a criterion to decide if full delocalisation within planar rings is attained. If the answer is no, because the system is in fact antiaromatic, then there is nothing actually specific to attain.

In short: while a ring that is Huckel-aromatic does actually tends to be aromatic, the Huckel-antiaromaticity is simply a relative high energy situation that a system will always try to avoid.*

Finally let us consider the actual case. The resonance structures you wrote indicate indeed a certain character of aromaticity in the cyclobutadiene moiety and a partial double character of the interring bond. This alone, let aside the above considerations, definitely tell us that the molecule is aromatic.

For more insights I've found the following papers, analyzing delocalisation in phenylcyclobutadiene experimentally (enthalpy of hydrogenation and EPR) and theoretically.

https://pubs.acs.org/doi/abs/10.1021/ja0525511

https://www.academia.edu/25340037/Aromatic_pathways_in_conjugated_rings_connected_by_single_bonds

As a very quick resume enthalpy of hydrogenation point to a destabilising effect of the Huckel-antiaromatic moiety, while electron spin resonance and calculation show overall aromatic rings current and an interring C-C bond length of about 1.44 Å, about midway from a single to a double C-C bond length.

*note that chemists are sometimes able to constrain structures in otherwise highly unfavourable systems. I won't be surprised if exotic beasts forcing an inner ring into antiaromaticity were synthesised. I might even have seen a related Q here in SE.


R & S Stereochemistry of an unsaturated ring system Using the R- & S- nomenclature assign a configurational symbol to decalin 4-methyl (above the plane), 4-hydroxy (below the plane), 7-isopropyl (above the plane), 1,9 diene. (Numbering the carbons in decalin ring-system: ring-junction carbons are 1 (top), 6 (bottom))

Carbon 7—holding the isopropyl group: I think this is S (isopropyl $ > \ce{C-C=C} > \ce{CH2} > \ce{CH}$).
Carbon 4—holding $\ce{CH3}$ & $\ce{OH}$: I think this is S ($\ce{OH}> \ce{CH3} > \ce{CH2=CH2}$).

Is my reasoning correct?

I don’t really know how you arrived at your reasoning, but it seems to fail as soon as double bonds are involved. If you encounter a double bond, it must be expanded with ghost atoms to two single bonds in the following manner:

$$\ce{(R2)C=C(R'2) -> (R2)C(C)-C(C)(R'2)}$$

The carbon atoms in bracket are ghost atoms not connected to anything else. These ghost atoms are treated just like normal atoms when determining priorities. You can read more on Wikipedia, which also includes a better graphic example. (But since your case is rather simple, we can do without.)

Let’s first determine the tertiary alcohol’s carbon’s absolute configuration:

$\ce{O}$ from the hydroxy group.
$\ce{C}$ (along the ring to the left)

$\ce{C}$ (the quarternary carbon)

$\ce{C}$
$\ce{C}$ ← This is the first point of difference.
$\ce{(C)}$ — this is the double bond’s ghost atom.

$\ce{H}$
$\ce{H}$

$\ce{C}$ (along the ring upwards)

$\ce{C}$

$\ce{C}$
$\ce{H}$ ← This is the first point of difference.
$\ce{H}$

$\ce{H}$
$\ce{H}$

$\ce{C}$ from the methyl group.

$\ce{H}$
$\ce{H}$
$\ce{H}$

Thus, the priorities are $\ce{OH} > \ce{CH2C_{(q)}} > \ce{CH2CH2R} > \ce{CH3}$, because the quarternary carbon that one of the two $\ce{CH2}$ groups is bonded to beats the secondary carbon at the top of the right-hand ring. That makes that carbon’s absolute configuration (S).

Similarly, for the carbon attached to iso-propyl:

$\ce{C}$ (the quarternary carbon)

$\ce{C}$
$\ce{C}$
$\ce{C}$

$\ce{C}$ (the iso-propyl group)

$\ce{C}$
$\ce{C}$
$\ce{H}$

$\ce{C}$ (the secondary carbon)

$\ce{C}$
$\ce{H}$
$\ce{H}$

$\ce{H}$

As you can see, we only need go down to the second level to find a difference: $\ce{C(C3)} > \ce{C(C2H)} > \ce{C(CH2)} > \ce{H}$ Thus, that carbon’s absolute configuration is (R).


Dissolving Organic Tissues Watching some movie/TV this question came to mind, purely theoretical of course, if one wanted to dissolve a human corpse, like getting rid of a body after killing someone, is it better to use an acid or a base? In the movies (as well the popular perception for some reason) they always use acid. The movie I was watching they had a bathtub full of acid in which they throw in the body. But I also remember my chem teacher telling me (a long long time ago) that if she had a choice of an acid or a base in her eye, she would take acid any day because a base will dissolve organic tissue faster. So assuming that the acid and base have the same level of strength (say a pH of 1 and 13 respectively) which is better for getting rid of a corpse?

Thanks!

I would say that acids are better.

Why? Bases and acids dissolve flesh pretty easily (I'm thinking of $\ce{NaOH}$ and $\ce{HCl}$ here, weaker acids/bases -- not so much). Most of our body is made up of proteins, which are overall pretty much neutral (generally). Both acids and bases can dissolve protein.

On the other hand, there is one major component of our body which is easier to dissolve with acid. Bones. The major component of bone is $\ce{CaCO3}$. The reaction $\ce{2HCl +CaCO3 -> H2O + CO2 ^ + CaCl2}$ is much more efficient/faster than $\ce{2NaOH + CaCO3 -> Na2CO3 + Ca(OH)2}$. Why is this? Well, for one, in the acid reaction, a gas is being released. This speeds up the reaction, by Le Châtelier's principle.

Apparently, NaOH is not able to dissolve bone completely, you are left with a fragile shell. Either way, it will be much slower than $\ce{HCl}$ while dissolving bone.

Here's an interesting article (no claims on validity) of why lime shouldn't be used to dissolve a corpse.

While (as far as I can tell) acids may be the better way to dissolve a body, I feel that filmmakers use acids because the term is known to most of the general public (unlike "bases", which isn't)


How does the Wurtz reaction produce alkynes? Wikipedia mentions that alkynes can be produced from geminal dihalides through the Wurtz reaction. However, I am unsure of the reaction mechanism for this reaction. I have not been able to find any sources which show the mechanism for this reaction. Does anyone have any idea how the reaction is likely to occur?

@tox123: You have generated two radicals on the same carbon by a two-electron reductive α-elimination (1 --> 2). So far, so good. In other words, you have created a carbene (2). Carbenes can rearrange to alkenes, NOT alkynes. Think of the two orbitals on the carbenoid carbon as one bearing two-electrons (negative) and the other one vacant (positive) (3). When a vicinal hydride migrates to the vacant orbital, one has a negative charge on one carbon and a positive charge on the vicinal carbon (4). Structures 4 and 5 are identical. If dichloride 1 is treated with base, one gets a double elimination to an alkyne. There is no change in oxidation state. Therefore, dichloride 1 cannot give an alkyne via a 2-electron reduction.


Distillation of diethyl ether and tetrahydrofuran In the lab, I'm working in over summer, there is a set of solvent stills.

Most of them use $\ce{CaH2}$ as a drying agent, however ether and THF use Na and K respectively in the presence of benzophenone.

What I don't understand is why the ether still uses Na, whilst the THF uses K. Equally, I'm not sure what the purpose of the benzophenone is — it makes the stills blue, but none of the other solvents (DCM, toluene, triethylamine) have this property.

Sodium and potassium are both drying agents. Being alkali metals they react (often vigorously) with water.

Sodium wire is often used in diethyl ether stills, whilst potassium is used in tetrahydrofuran stills. In general, potassium is more reactive, and hence a better drying agent.

The reason different desiccants are used is due to temperature. THF boils at a higher temperature, meaning a lump of potassium metal can be used in the still ( the lump of K melts once the THF reaches reflux ). Diethyl ether boils much lower, meaning potassium would not melt, for this reason sodium is often used in the form of a wire (giving a much larger surface area).

In respect to the use of benzophenone (or other indicators such), it generally indicates the presence (or absence) of oxygen in the solution. In the past, it was thought that the purple (blue is actually not great) colour was related to the moisture content, however we now know that the colour is related to the oxygen content. For this reason, using the indicator allows us to determine how oxygen-free the solvent is. A purple THF/ether still in which metal is still present indicates that the solvent is both dry and oxygen free,


Is the concentration of negative ions always equal to that of positive ions? Is there any way in which a solution can have more negative charge than positive charge, or vice versa? For example, in the context of acid-base chemistry, is there any scenario in which $[\ce{H+}]\ne[\ce{A-}]$? I have never heard of a solution with a non-neutral overall charge, but is it possible in some extreme scenario?

The only point that $[\ce{H+}]=[\ce{A-}]$ is when the $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ of a given acid, so indeed in most instances $[\ce{H+}] \neq [\ce{A-}]$. In addition to $\ce{A-}$, the solution also contains negatively charged $\ce{OH-}$ that balances the positive charge of the $\ce{H+}$ ions. If the acid is titrated with a strong base (e.g. $\ce{NaOH}$), the positively charged $\ce{Na+}$ ions balance the negative charge of $\ce{OH-}$ and $\ce{A-}$.

To answer the question in your title though, yes, charges are always balanced. To quote another good post on this topic:

  While it is true that solutions are not always locally electroneutral, that non-neutrality is small, and does not extend to larger scale. Thus, it does not impact chemical concentrations: you cannot separate positive and negative charges in a solution at a macroscopic level.


How do manufacturer remove oxygen from metal during melting? How do manufacturers remove oxygen from metal during melting? I melt aluminium with nichrome wires and during melting it, it forms lots of slag (aluminium oxide) and I have to remove it manually.

They use flux, that's how. That is, an inert substance which melts easier than the metal and floats on top, preventing it from oxidation in the air. Common fluxes for aluminum seem to be based on eutectic mixtures of various chlorides (say, $\ce{NaCl + KCl}$).

Covering your crucible with a lid does not offer complete protection, and working in inert atmosphere is much more costly (though sometimes they do that, too).


kinetic description of an autocatalytic system I have previously asked this question at physics SE here and it was suggested to post it here.
I am looking at the system:
$2X + Y \Leftrightarrow 3X $
$A \rightarrow Y$
$X \rightarrow B$
The description in terms of differential equations is this:
$\frac{dx}{dt} = x^{2}y - x$
$\frac{dy}{dt} = a - x^{2}y$
What is puzzling me is that the fact that 3X are produced is not captured in these equations.
Why is the description not something like
$\frac{dx}{dt} = 3x^{2}y - x - x^{3}$ ?
What fundamental thing am I missing? Help is greatly appreciated.
The source for this is "mathematical models in biology (1987), p 357" by Leah Edelstein-Keshet.
(also, I have a hard time finding relevant tags for this, so please feel free to change)

Let's first assign rate constants to each reaction for completeness:

$\ce{2X + Y <=>[k_1][k_{-1}] 3X}$

$\ce{A ->[k_2] Y}$

$\ce{X->[k_3] B}$

Then let's consider what's happening overall. A is converting to Y, Y is converting to X, and X is converting to B:
$\ce{A ->[k_2] Y <=>[k_1X^2][k_{-1}X^2] X ->[k_3] B}$
Y and X may be in equilibrium, and X catalyzes the X-Y interconversion.
Since X and Y are intermediates, I'm guessing that you are writing the rate equations for them in order to do a steady-state approximation?
Regardless of the purpose, we can see that X is created only in the reaction $\ce{Y -> X}$, which has $\frac{d[\ce{X}]}{dt}=k_1[\ce{X}]^2[\ce{Y}]$. However, it is consumed in two possible reactions, the back reaction $\ce{X -> Y}$ and the forward reaction $\ce{X-> B}$. So the consumption rate is $-\frac{d[\ce{X}]}{dt}=k_{-1}[\ce{X}]^3+k_3[\ce{X}]$. Combining the two equations, we find that
$$\frac{d[\ce{X}]}{dt}=k_1[\ce{X}]^2[\ce{Y}]-k_{-1}[\ce{X}]^3-k_3[\ce{X}].$$
This equation is similar to yours, but complete with rate constants and including the extra term for the reverse reaction. If that reaction is negligible, that term would drop out and give an equation like yours.
For Y, you can use the same process to again get a slightly longer equation than what you have written, because it includes an extra term for formation of Y from X in the back reaction. Clearly the derivation of the equations you provided assumes that only the forward reactions are significant.


Driving force in distillation If the pressure is lowered in a distillation column, then the equilibrium curve will move further away from the diagonal (bulge out more) in an equilibrium diagram. This leads to a decrease in ideal step stages required to reach the separation requirement.
My question is why does the ideal step stages decrease? Is it only because of how the stages are drawn; meaning if the equilibrium curve is more bulged less horizontal and vertical stages are required. Or has it to do with the driving force increasing as pressure is lowered?
All help is appreciated. Thank you!

In an x-y diagram, the "bulging out" is demonstrating the increase in relative volatility of the components; the number of stages is a strong function of relative volatility, so an increase in relative volatility leads to less stages.