text:
Predicting the product in a series of reactions
I am stuck in finding the final product $\ce{C}$ formed in the following series of reactions:

$$\ce{Bu-C#CH ->[NaNH2] A ->[Ph-CHO][H2O] B ->[MnO2] C}$$
  where $\ce{Bu}$ denotes the butyl substituent and $\ce{Ph}$ the phenyl substituent.

I could easily predict product $\ce{A}$ and $\ce{B}$. $\ce{A}$ is simply formed by replacement of hydrogen with sodium and $\ce{B}$ by the meta substitution in benzene ring.

However, I am having trouble predicting the final product. I know that $\ce{MnO2}$ is a mild oxidising agent.

Um, exactly how is your acetylide anion reacting with benzaldehyde? That anion is a nucleophile, not an electrophile, so electrophilic aromatic substitution (which involves the benzene ring reacting with an electrophile) isn't gonna work here. Is there any part of benzaldehyde that might react with nucleophiles?

@orthocresol yes I was mistaken.it should be para nucleophillic aromatic substitution.yes benzaldehyde also has  a replaceable hydrogen with itself but why it should be the major pathway.

@aniline It can't be nucleophilic aromatic substitution either, since there is no leaving group on the aromatic ring... Most reactions are nucleophile / electrophile reactions, so the questions are: what's the most reactive nucleophile and what's the most reactive electrophile. I'll give you the hint that aromatic rings are generally very stable, so that's not likely to react as either nucleophile or electrophile...

One more hint... What type of functional group is MnO2 likely to react with?

And another hint: A will react with $\ce{Ph-CHO}$, but it will react in similar fashion with $\ce{R-CHO}$ (R = alkyl) to give a B-type product. This, in turn, will react with $\ce{MnO2}$ as well.

You seem very sure about your intermediate B; we’ll come back to that in a second.
For now, let’s concentrate on the reagents for your final step.
$$\ce{B ->[MnO2] C}$$
Manganese dioxide is the product you get from heating any manganese salts in the bunsen burner until red-hot; thus, one might call it the most stable or thermodynamic oxide. It seems very unreactive and in fact I only know of one type of reaction that it is able to perform on organic molecules: allylic and benzylic alcohols are oxidised to the corresponding carbonyl compounds while fully saturated (i.e. not allylic) alcohols are untouched.
Since you have that reagent, it might be helpful to go backwards; assuming that the final step is an oxidation of a benzylic or allylic alcohol to a corresponding carbonyl. Thus, $\ce{C}$ would either be a ketone or an aldehyde and $\ce{B}$ would have to be an alcohol. But your proposed structure $\ce{B}$ does not include an alcohol! Therefore, your proposed reaction of the second step is very likely wrong.
$$\ce{A ->[Ph-CHO][H2O] B}$$
Indeed, here you assumed that the anion $\ce{A}$ you proposed reacts with the phenyl ring of benzaldehyde in an electrophilic aromatic substitution. Take a good look at the two emboldened words and consider whether an anion can really be electrophilic. Then ask yourself what anions are in philicity terms. Once you have done that, attempt to locate the position of benzaldehyde with the corresponding electronic properties.
The solution is given in the spoiler below; please try to find it yourself, though; it is a much better exercise than just looking things up.

 Anions are negatively charged as are electrons. Thus, anions are practically never drawn towards negative charges — remember that like and like repulse each other. Instead, anions are drawn to positive charges: they are nucleophilic.

 Nucleophiles are drawn to the most positively charged bit of a molecule they are attacking. In the case of benzaldehyde, the carbon side of the $\ce{C=O}$ bond is most positively charged due to the resonance depiction we can draw below:

 $$\ce{C=O <-> \overset{+}{C}-\overset{-}{O}}$$

 Therefore, the acetylide anion will attack the carbonyl carbon of benzaldehyde nucleophilicly, creating a benzylic alcohol. It is now no secret that this is cleanly oxidised to the corresponding alkynophenone.

 $$\ce{Bu-C#C^- (A) ->[Ph-CHO][H2O] Ph-CHOH-C#C-Bu (B) ->[MnO2] Ph-CO-C#C-Bu (C)}$$


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How do I model non-covalent interactions in Gaussian?
I'm doing some research about the dyes found in photography, and found Organic Chemistry of Photography
by Shinsaku Fujita, which shows the dyes used. I'm looking at doing an excited states and UV-VIS calculation, of course optimized before with a geometry optimization, with some of the dyes presented in the book. Right now, I'm trying to build number one in the image below, but have come upon a bit of a stumbling block.

The problem is the $\ce{(CH2)3SO2O-}$ and $\ce{(C2H5)3NH+}$ part on the rightmost nitrogen, as I'm not sure how to model the interaction between $\ce{(CH2)3SO2O-}$ and  $\ce{(C2H5)3NH+}$ (triethylamine), as both have no more covalent bonds available.
To visualize the problem, I built out two molecules (everything but triethylamine as one, then triethylamine below). As you can see, both are connected with as many covalent bonds as possible.

How can I model the interaction between the two molecules (pictured above) in Gaussian, for UV/Vis calculations?

Ah, I think you are essentially asking - "how can these two molecules be covalently connected?". The answer is : presumably they are not! It seems likely that they form a mixture with some kind of non-covalent interaction. In other words, when you say 'occupied' you mean "each atom is as connected as possible", or similar?

@gilleain That is correct, when I say "occupied" I do mean "as connected as possible." If they are not covalently connected, then I'll have to figure out how to model the dye. (I'm using WebMO and Gaussian, not sure if that's relevant)

I've no idea, sadly! There may be many ways to model mixtures like this - that seems like a good question, so I could suggest that you alter your question to be something like "how do I create a model for a mixture of charged dyes", mentioning the tools you are using.

what's the property you want to know? For some things it might be fine to omit the counter ion.

The quaternary amine appears to be the counterion to the negatively charged dye species.

@DSVA I'm doing a uv/vis curve

So what is the question? Can you clear it up a little?

@Greg How can I model the interaction between (C2H5)3NH+ and the rest of the system in Gaussian, since they're not covalently bonded?

@DavisDude Since it is not clear what do you want to calculate, it is itself a question what non-covalent interactions you need to calculate (if any). I would also question if you have a useful idea about the structure of the solid material, which is most probably amorphous.

@Greg Sorry for the confusion, I've updated the question. Please let me know if there is still anything else that needs clarification.

This would be a very involved problem to solve and very difficult to specifically answer in full (sampling the potential energy surface in various configurations, benchmarking results, choice of model, etc.).  As for computing non-covalent interactions, please see https://chemistry.stackexchange.com/questions/31/do-terminal-amides-participate-in-%CF%80-stacking/38#38 and https://chemistry.stackexchange.com/questions/32714/whats-the-bonding-in-the-diborane6-hydrogen-chloride-dimer/32758#32758.

@DavisDude Are you sure you need the cation for the calculation? What effect do you expect?

As hBy2Py, Greg, and others have pointed out in their comments, the cation is just a counter ion for your benzothiazole-based cyanine dyes.
I suggest to be reasonably lazy and perform the calculations on the cyanine dye only.
Even for calculations in solvents of sufficient polarity (from an experimental photochemical perspective, acetonitrile seems to be a good choice), I would just ignore the cations.

Sorry for the long delay in response. Unfortunately, I am unable to perform the job with the time allotted, but it seems reasonable that it would work

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Before going into stable states ( i.e less energy states ) , do exothermic systems become unstable for a short time on the basis of Kinetic Energy?
In most of the cases the reactions are exothermic because all systems want to reduce their net energy and one way to do that is by loosing the $KE$ of the molecules to the surrounding molecules by collisions.
Now suppose that that the molecules of our system cannot emit any radiation so the only way for them to reduce their net energy is by giving their $KE$ to other molecules which are not a part of the system.
Now if I make water with Hydrogen and Oxygen, it results in an exothermic reaction. My question is that hasn’t the $KE$ of the molecules increased when we have mixed Hydrogen and Oxygen together before they pass their $KE’s$ to the molecules of the air? Or in other words does every exothermic system becomes unstable ( because of the increase in the $KE$ of the molecules and hence in net increase in energy ) before it can pass out it’s energy to the surroundings ? And is it that exothermic systems will tend to increase their energy if they cannot pass on their energy to the surroundings ?

I think you are talking about a transition state.

@Zhe , wikipedia states that transition state is related to potential energy but here I am referring to Kinetic energy! Moreover in a video at youtube she said that the transition state theory assumes that the reactants and products are in equilibrium... . But the the question which I have addressed does not mention that too! So I think that transition state is somewhat related to the distance between the atoms ( potential energy ) but $KE$ is obviously related to the change in that length per unit time.

If we're nitpicking, then the first statement in your question is wrong. Systems don't care to lose energy. The universe just needs to increase its entropy. Based on your comment, it's no longer clear what you're actually asking. What do you mean by "unstable"?

The KE of molecules does not somehow increase as they already have a distribution of energies (Maxwell-Boltzmann) and distribution of potential energy (Boltzmann)  which both vary with temperature. These energies interconvert and occasionally there is enough energy to overcome an activation energy and cross a transition state towards products (and vice versa but with lower chance if reaction is exothermic). (btw. In the TS model of reactions there is equilibrium between reactants and the TS and between products and the TS).

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Rates of Keto-Enol Tautomerism
Are there any databases with rate constants for keto-enol tautomerism in small molecules? I'm looking for rate constants if possible.

They're equilibrium processes so a rate constant isn't really meaningful

I'm asking if there is some kind of reference where I could obtain kinetic data of this type.  It is useful for what I am trying to do.

My point is that almost any carbonyl will exist to some extent in its enol form. You can't really measure a rate because no sooner as it's made, it's equilibrated. It's not a chemical reaction that you can 'start' and then 'quench' once it's done. Perhaps you can [edit] your post with more details/an example of what you're looking for

I'm interested in the times for intramolecular proton exchange in small molecules. I understand what you are saying regarding tautomerization being an equilibrium reaction

You could use standard  perturbation (relaxation) methods such as (small) temperature or pressure jump to study such reactions, for example $\ce{H2O<=>H^+ + OH^-}$ Searching for the technique may be a way to find data.

This is not a database, and I don't know about how small molecules you are looking for, I found that article on Jacs Kinetics and thermodynamics of keto-enol tautomerism of simple carbonyl compounds: an approach based on a kinetic study of halogenation at low halogen concentrations
If you can access the file it will be better for you, if not I let you some matrix of values. I hope it can help ! I give you them in the same the order of the publication.
I added some other publication at the end.

Links
Tautomerization of 2-Acetylcyclohexanone. 1. Characterization of Keto−Enol/Enolate Equilibria and Reaction Rates in Water
Kinetic studies of keto-enol and other tautomeric
equilibria by flash photolysis (a lot of constants in this one)


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Synthesis of 5-chloro-2-ethenyl-1,3-diiodobenzene
The question is basically to figure about the steps for the following transformation.

Since ethene is a electron donating group and weakly ortho para directing so it makes electrophilic substitution difficult.hence we must add bromine water($\ce{HOBr}$) to the double bond.After this if we add $\ce{Cl2}/Fe$ we will get ortho and para positions substituted by chlorine.However I am facing difficulty changing the ortho chlorine atoms to iodine.Please suggest me a way to proceed.Thanks.

A vinyl group is electron donating and is weakly ortho-para directing.

@ron thanks for your comment.I edited the question accordingly.

Masking your double bond  through addition of one bromine atom is a reasonable first step.  Take a look at this earlier answer and consider how sulfonation and the Sandmeyer reaction might prove useful in your synthesis.


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How does heptafluoropropane suppress fire?
How does heptafluoropropane suppress fire?
It says here that it does so by "inhibiting the chain reaction."  What does that mean specifically? Does $\ce{O2}$ attach itself to the heptafluoropropane?

@CurtF. And most burning material is heavier than air, and so tends to burn near the ground -- thus, if you can push the oxygen higher, even if it doesn't escape the room, displacing it may still starve the flames.

Curt F. is probably correct in his estimate that heptafluoropropane extinguishes fire primarily by physical means. According to an overview by Choy and Fong, 'An Introduction to Clean Agents: Heptafluoropropane' (Int. J. on Eng. Performance-Based Fire Codes, vol. 5, nr 4, p. 181$-$184, $2003$),

For heptafluoropropane, the contribution of physical mechanisms to the extinguishment of fires predominates over the chemical mechanism. It suppresses  fires  primarily  by

extracting  heat  from  the flame reaction zone,
reducing the flame temperature below that which is necessary to maintain sufficiently high reaction rates by a combination of heat of vaporization [and] heat capacity.

They add that,

Oxygen depletion plays an important role in reducing flame temperature. The energy absorbed in decomposing the agent by breaking fluorine bond[s] is quite important, particularly with respect to decomposition production formation.

The chemical effect does indeed involve tying up active radical species. As an example, consider the mixture $\ce{H2/O2}$. The primary active species are $\ce{OH^\bullet}$ radicals.
$$\ce{H2 + O2 -> 2OH^\bullet}$$
The chain starts to grow.
$$\ce{OH^{\bullet} + H2 -> H2O + H^\bullet}$$
$$\tag{two reactive sp. instead of one}\ce{H^{\bullet} + O2 -> OH^{\bullet} + O^\bullet}$$
$$\tag{two reactive sp. instead of one}\ce{O^{\bullet} + H2 -> OH^{\bullet} + H^\bullet}$$
The bottom two steps show especially clearly why fires are violent, quick reactions (after incubation period is over). Heptafluoropropane's chemical extinguishing effect arises from

[----] the thermal decomposition of small amounts of heptafluoropropane in the flame which form fluorinated fragments such as $\ce{CF3}$ and $\ce{CF2}$. These will then consume the key combustion chain-propagating species $\ce{H}$ and $\ce{O}$, but to a lesser extent on $\ce{OH}$ radicals. The rates of chain-branching combustion reaction will decrease, the chemical flame is inhibited and the flame propagation is halted. (Choy, Fong)

To increase the chemical inhibition, heptafluoropropane (or HFC-227ea) is sometimes mixed with $\ce{NaHCO3}$.

Flame inhibition by sodium species is believed to be due to chemical scavenging of major radical species (e.g., $\ce{OH}$, $\ce{H}$) in the flame.
source: Skaags, 'Assessment of the Fire Suppression Mechanics for HFC-227ea Combined with $\ce{NaHCO3}$', U.S. Army Research Laboratory, link

One commercial source, unlike Wikipedia, groups pentafluoroethane together with heptafluoropropane.

source: Group of Companies "RealSnabService"

Further reading, more in-depth discussion and related topics:

Luo et al.. 'Effect of Hydrofluorocarbons and Perfluorocabons on Suppression Efficiency of $\ce{CH3I}$'. Process Safety and Environment Protection Research Group. link
Williams, et al.. 'Intermediate Species Profiles in Low Pressure Methane/Oxygen Flames Inhibited by 2-H Heptafluoropropane: Comparison of Experimental Data with Kinetic Modeling'. Annual Conference on Fire Research, $1998$. link
Sheinson. 'Heptafluoropropane With Water Spray Cooling System As A Total Flooding Halon 1301 Replacement: System Implementation Parameters'. Navy Technology Center for Safety and Survivability. link
Grosshandler, et al.. 'Extinction of Hydrofluorocarbon Flames with F/H Ratios of Unity and Greater'. Annual Conference on Fire Research, $1998$. link

This is a much better answer than mine.  I recommend switching the accepted answer to this one.

It's OK, @CurtF. We convey a similar point; that these compounds' primary means of extinguishment are physical rather than chemical in origin. In any case, thanks for your kind comment! Here, have some well-deserved upvotes on your previous answers. :-)

TIL: StackExchange/Chemistry experts are very polite.

That wikipedia page on gaseous fire suppression is not very good. It is very difficult for me to believe that pentafluoroethane has a different mechanism of fire suppression than heptafluoropropane. I suspect that all the inert-gas agents work by lowering oxygen concentration, both by simple dilution and by virtue of their density selectively displacing oxygen at the bottom of a room, where fire is more likely to be (flammable materials, like all materials, are rarely stored on the ceiling).
Rooms aren't usually hermetically sealed so if you suddenly unleash a lot of gas in one, pressure won't build up very much at all. Leaks through doorways, windows, and vents will let all gas leave the room. Thus the oxygen in the room will go down as it is diluted by the new gas. If the vents are high up in the room, oxygen and other less dense gases could be selectively displaced.


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Why is the melting point of lactic acid higher than pyruvic acid
I have a quick question about pyruvic acid and lactic acid, which I think I know the answer too but I'm unsure.

The melting point of lactic acid is said to be higher than that of pyruvic acid despite very similar molecular masses.

Is this because lactic acid has 2 $\ce{OH}$ groups, and pyruvate only has 1, thus there are twice as many hydrogen bonds between lactic acid molcules that must be broken for the substance to melt?

Do you men boil or melt?

@A.K. I know it sounds weird, and it seems way more intuitive to ask about the boiling point, but that literally is what the question asks.

Your analysis is practically correct but not quite — more due to technical reasons. You said:

Is this because lactic acid has 2 $\ce{OH}$ groups, and pyruvate only has 1 […]?

You probably know the structures of both; they both include a $\ce{COOH}$ group which you apparantly counted as a hydroxy group. You should not do that; the carboxy group ($\ce{COOH}$) has different properties than a lone hydroxy group ($\ce{CR2OH}; \ce{R} \ne \ce{OX}$). The most obvious is the $\mathrm{p}K_\mathrm{a}$ value which differs by about $10$ logarithmic units.
So you should say that lactic acid has higher melting and boiling points becuase it contains a (or: one) hydroxy group while pyruvic acid has a carbonyl group in the same position.

So basically the presence of the OH causes there to be hydrogen bonding between lactic acid molecules, which don't exist between pyruvate molecules due the the presence of a Carbonyl group instead.

@frostedcake Well, the carboxy group does hydrogen bond, too; but you can form more hydrogen bonds per molecule with lactic acid than with pyruvic acid.


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"Alcohol intoxication" naming confusion
From Wikipedia: "Alcohol intoxication is the result of alcohol entering the bloodstream faster than it can be metabolized by the liver, which breaks down the ethanol into non-intoxicating byproducts."
Why is it called intoxication if ethanol is broken down to non-intoxicating byproducts? This is so confusing.

Read that sentence again. Intoxication occurs when the liver, which is responsible for detoxification, cannot keep up with the alcohol intake.

@orthocresol I think the reason I don't understand that is I don't know what intoxication really means chemically. Can you explain this a little, to a person who only have basic chemistry knowledge?

If that's your question, vote to migrate to biology.SE.

You're losing the forest in the trees. Ethanol is intoxicating. The metabolized by-products are not.
Note:

"Alcohol intoxication is the result of alcohol entering the bloodstream faster than it can be metabolized by the liver..."

Alcohol can only be removed from the body my metabolism of the liver. This phrase is just stating that if you intake alcohol faster than your liver can destroy it, then the level of alcohol in your blood is rising.
Now the other half:

"...which breaks down the ethanol into non-intoxicating byproducts."

The by-products that the liver made ethanol into are not intoxicating thus only the ethanol contributes to your intoxication.


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How to desalinate seawater at freezing temperature or high pressure?
The deep ocean depths have very high pressure and freezing temperatures. Could one or both of these characteristics be used to desalinate seawater?
Reverse osmosis uses high pressure to separate salt from seawater to make
freshwater. Freezing seawater can also make freshwater. Less energy would
be utilized in both processes.

No, you cannot. You'd still have to raise the newly made freshwater up to sea level. There's no free lunch.

If there is a problem raising the freshwater to the top, why not make ice
cubes from the freezing ocean depth and have them float to the top via a vertical pipe? The high ocean depth pressure would make membrane technology
less energy intensive.

The deepest points in oceans are a bit far away from continental masses and also not near desert countries (Persian Gulf, Sahara and Central Asia)
however there is another idea, to use mechanical pressure forced onto seawater, in large vessels, to squeeze the NaCl molecules from H2O molecules, just so they sink to the bottom.


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Proper definitions of resistivity of conductivity

Is resistivity the same as specific electrical resistance?
And conductivity is same as specific conductance when I am dealing with electrical properties ?

The specific modifier simply means that stated property per a given quantity, usually mass, usually a gram. In other words resistance or conductance per gram of resisting or conducting material.  But the fundamental definition of the properties for the different word forms of resistance and conductance are not changed by this wording.


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Calculate limiting molar conductivity of BaCl2
How can I calculate limiting molar conductivity of BaCl2 in water at 25 degree Celsius?
I have got a formula for calculating limiting molar conductivity .

limiting molar conductivity  =  V+ A+ +
V- A-
V+  =  number of cations per formula unit of the
electrolyte
V-  =  number of anions per formula unit of the
electrolyte
A+ = molar conductivity of the cations
A- = molar conductivity of the anions

Here I figured out, V+ = 1 and V- = 2
How can I solve the question and tell me  if there is any information that is not present above to solve the question.

I'll answer the second part of your question first: you have one equation with three unknowns, so you need the value's of two more unknowns, namely the molar conductivities of Ca++ and Cl-.  They are readily available from the CRC Handbook of Chemistry and Physics.  Every chemist and chemistry prof and library has one.  You could also find those values from online resources, just be sure to check with reliable sources.  An internet search will turn up many.  Once you have those two values you just plug them in to your equation and you are done.  You are correct with your V values.


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etymology of phosphatidates and phosphatidic acid
A phosphatidic acid is an ester of a glycerol, two fatty acids and a phosphoric acid.
Who coined this term, and is the word's "phat" part related to english "fat"?

Apparently, "phosphatidate" is from "phosphate" + "-ide" + "-ate".
The word "phosphatidic acid" is from "phosphate" + "-ide" + "-ic acid".
Phosphate ($\ce{PO4^3-}$) is the anion of phosphoric acid ($\ce{H3PO4}$).
The suffix "-ide" is used in biochemistry to just about anything. It just denotes an element of a group of related compounds.
Now, "-ate" and "-ic acid" denotes the acidic nature of the compounds.

And no, it is not related to the English word "fat".

Phosphatidate can be broken down to "Phosph(orous) + at(e) + id(e) + ate"
-ate means either "having" or "characterised by" and is used in chemistry to denote esp. a salt or ester of an acid.
-ide is a suffix meaning "a compound of " or to denote a group of relate do compounds as in monosaccharide or to indicate that is a binary compound as in Sodium chloride.
-ic also means relating to or characterised by.
Therefore phosphatidate literally means it's a compound containing a moiety characterised by having an ester of phosphorus.

Who coined this term?

According to this site it was 1st used in 1884 by International scientific vocabulary.

Can it be one of "ate" was for phosphoric acid, and other for fatty acid. and ide for glycerol ( like a monosaccharide)?

Yes it could be.


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Why are Bohr's formulae for radius and velocity still in use when Heisenberg's uncertainty principle predicts otherwise?
I learnt about Heisenberg's Uncertainty principle which states that it is not possible to determine the position and momentum of a small particle at the same time.
I was told that it is because in order to be able to see the atoms, we need to make use of radiations, and when we do that, the electrons get excited by those radiations and their position changes. No matter how sophisticated our instruments get, there will be some error. If we use radiation of wavelength
"D" to measure the position of electron, then minimum error in position will be "D"

I didn't really get how would we use radiations to make the atoms visible. Can you please explain that?
This principle clearly states that we cannot be certain about the position and velocity of an electron in an atom, and this must also be true for hydrogen like species. But we still use the formulae given by Bohr's model of atom to find out velocity of electron in nth orbit and its position for hydrogen like species. My teacher says that these formulae give satisfactory results for hydrogen like species. How could that be? Is it because the error is minimum in the case of hydrogen like species?

I am not able to intuitively arrive to an explanation to both these above question.

Are they still in use? I'm not sure about that.

I was told that they give satisfactory results for hydrogen like species.

They are in use (in a way), but their meaning is quite different from that in Bohr's theory.

They're in use insofar as the Bohr model is in use. And the Bohr model does a great job in terms of explaining the emission spectra for hydrogen and motivates some kind of quantization within the atom. That's about it though.

Fair enough, the Bohr radius $a_0$ is still in use and is a pretty important length scale; however its interpretation as "the radius of the first orbit" is quite obsolete though. Why does it work for hydrogenic atoms? That's more like Bohr was trying quite hard to explain the spectrum of the hydrogen atom, and he found a way that worked for it, and it turns out the quantum mechanics of a hydrogenic atom is pretty much the same as that of the hydrogen atom.

$a_0$ is now useful because it combines a bunch of fundamental units together into something with dimensions of length.


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Can there be oxygen dissolved in water in a hydrogen atmosphere?
I'm trying to determine (for a science fiction story) whether a certain type of planet could have a significant amount of oxygen dissolved in water (e. g. produced continuously by some aquatic organism) if its atmosphere was made mostly of hydrogen. Is there a way, or will any oxygen just "pop" out of the water into the atmosphere and react with the hydrogen? I'm assuming the hydrogen pressure would be in the order of 10-100 atm.

Maybe the hydrogen can go into the water to react with the oxygen?

@DHMO Not only can but has to. Bodies of water would be saturated with H2, and even water in organisms would be if they wouldn't actively remove it.

This question is very good as it brings together many disciplines$-$astronomy, chemistry, atmospheric physics. Regretfully, as a consequence it is hard to compentently answer by a member of one field. I look forward to a healthy discussion in the comment section.

Hydrogen
A scientifically sound story ought to be such in most aspects. Specifically,

how does that planet possess hydrogen in high concentrations?

Hydrogen is a molecule with small molecular mass. As such it occupies the high end of Maxwell-Boltzmann speed distribution:
$$f(v)\begin{align}&=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right)
\end{align}.$$
$\ce{H2}$ would thus escape the atmosphere over time since they often reach escape velocity
$$v_e\approx\sqrt{\frac{2GM_{planet}}{r}}.$$
Possible explanations why this would not have occured include:
1. Relatively young planet
A newly-formed celestial body is very likely to be hot. As such, it will not contain liquid water, definitely not to the extent of forming oceans.
One study by Tian et al. 'A Hydrogen-Rich Early Atmosphere' ($2005$) suggest the possibility of oceans and a concurrent hydrogen mixing ratio of $30$%. However, this claim has been contested.

Tian et al.’s argument that hydrogen escape was inefficient on early Earth assumes that the upper atmosphere would be cold as on Venus, because the atmosphere is composed mostly of CO2. Their claim is controversial because their model depends on gases other than hydrogen to provide the radiative cooling, but their model does not actually include gases other than hydrogen. In particular, the key assumption that the upper atmosphere was cold is not obvious and needs to be addressed quantitatively. [----]
Visconti’s calculations contrast markedly with the cold thermospheres in Tian et al.’s pure hydrogen escape models.
source: Zahne et al. 'Earth’s Earliest Atmospheres'. (2010) link

2. Massive size and a magnetoshpere
Massive size would mean it has not really lost any atmosphere. Unless it is close to a star, in which case hydrodynamic escape would become important. These planets are mostly gas giants. We do not really know what is going on at their center, but it is almost certainly not a liquid ocean.
Without a magnetosphere the planet's atmosphere would be washed away by solar wind, similarly to that of Mars.
3. Hydrogen-producing microorganisms, e.g., a hyperactive, modified Chlamydomonas reinhardtii
This would be your best  bet. I am not qualified to discuss the likelihood of simultaneous thriving of both hydrogen-producing and oxygen-producing organisms. At face value$-$and this is speculation$-$it is probably not viable. Especially with so high pressures of hydrogen (high solubility).
What if?
Whether the reaction between $\ce{O2}$ and $\ce{H2}$ actually takes place (with considerable speed) depends on numerous variables. The ratio of these gases, temperature, pressure.

Can there be oxygen dissolved in water in a hydrogen atmosphere?

Short answer is yes.

[Is it] a significant amount?

Probably not. There ought to be some equilibrium amount but negligible. Even leaving aside the reaction between hydrogen and oxygen, the oxygen would be used to oxidise minerals. After these are, in a sense, saturated and when aquatic organisms adapt to more oxygen-rich water will the concentration of $\ce{O2}$ grow. It is now that the high hydrogen content becomes an issue.
Extra discussion requires actual calculation with supplied parameters, and the definition of what constitutes a significant amount. I have provided some sources for the kinetics of $\ce{O2/H2}$ mixture if this a route you wish to undertake.

Marcus Ó Conaire et al.. 'A Comprehensive Modeling Study of Hydrogen Oxidation' link
Alekseev et al.. 'The effect of temperature on the adiabatic burning velocities of diluted hydrogen flames: A kinetic study using an updated mechanism' Combustion and Flame, 162, 5, 1884$-$1892. (2015). link
Martin Hersch. 'Hydrogen-Oxygen Chemical Reaction Kinetics in Rocket Engine Combustion'. NASA Technical Note. (1967). link

Rogue planets
In a paper by David J. Stevenson 'Life-sustaining planets in interstellar space' (Nature vol. 400, page 32, 1999, link) he proposes that planets which form sufficiently far from their star or are gravitationally accreted may be sent on a course outside their star system. These planets could retain their hydrogen because the effective temperature would be fairly low: estimated to be about $30 \ \mathrm{K}$. According to Stevenson,

[P]ressure-induced far-infrared opacity of $\ce{H2}$ may prevent these bodies from eliminating internal radioactive heat except by developing an extensive adiabatic (with no loss or gain of heat) convective atmosphere.

Thus surface temperatures may be enough for liquid water to exist. Assuming the planet collided with enough asteroids and comets at a stage where water was retained we arrive at the conditions:

hydrogen-rich atmosphere, Stevenson estimates that the pressure could be $100$ to $10000\ \mathrm{atm}$, so your intuitive estimate was pretty adequeate;

Sidenote: Pressure at the bottom of oceans on Earth is (at the deepest) about $1000\ \mathrm{atm}$. So we would have the same amount of pressure on the surface of the ocean.

liquid oceans.

A possibility for life would be anaerobic chemoautotrophs: they do not need oxygen, and would be able to be their own source of organic molecules. They do require a geothermal energy source. Volcanic activity at the bottom of the ocean floor suffices. Through random genetic mutation these organisms might transition from being obligatory to being facultative anaerobs. Remember: if $\ce{O2}$ is your product, $\ce{CO2}$ is likely the raw material the organism is utilising.
I have no idea how things could proceed from here. We could have a cosy ecosystem in such an ocean (especially near vents) but how and why there would be evolutionary pressure toward oxygen-rich life at other depths... I simply do not know. Only way is to decrease hydrogen content but, well, we need that to keep warm :-). Perhaps a gradual enrichment or replacement of hydrogen by other gases without lowering temperature further is possible? (see suggestion below)
As Stevenson notes,

If life can develop and be sustained without sunlight (but with other energysources, plausibly volcanism or lightning in this instance), these bodies may provide a long-lived, stable environment for life (albeit one where the temperatures slowly decline on a billion-year timescale). The complexity and biomass may be low because the energy source will be small, but it is conceivable that these are the most common sites of life in the Universe.

Suggestion
If you wish for more information, consider follow-up questions on our

biology,
physics,
world building,
astronomy site.

I suggest against asking in all sites at once. Consider taking a concrete point (for example the transition from anaerobic to aerobic on biology.SE). Read their guidelines to be sure it is on-topic, and check for possible duplicates. Link to this question when asking again.
Good luck!

Thanks a lot for the sources! What I was thinking about was (in line with what I saw in pop-sci articles such as this): a large rocky planet (a Super-Earth) not bound to a star which keeps its internal heat thanks to the insulation provided by a dense hydrogen atmosphere.

"A significant amount" would mean "significant enough for there to be an evolutionary pressure toward the use of mollecular oxygen as a source of energy for organisms". How that oxygen could be produced in the first place in the absence of sunlight to split water is another matter.

@pablodf76 Glad I could be of some use! I have added a new subsection addressing your comments.


text:
Why do alkyl chlorides have lower boiling points than the corresponding alkyl bromides and iodides?
To my knowledge, the $\ce{C-Cl}$ bond is stronger than the $\ce{C-Br}$ and the $\ce{C-I}$ bond.
I always thought that the stronger a bond, the harder it is to break so the boiling point would be higher.
If that is the case, why does, among $\ce{RCl}$, $\ce{RBr}$ and $\ce{RI}$ (R = alkyl), $\ce{RI}$ have the lowest boiling point?

Your knowledge of the bond strengths is correct, but your understanding of boiling is flawed.
As orthocresol has mentioned in his comment, boiling is not about breaking the carbon-halogen bonds in your tetrahalomethanes.
Boiling does not mean decomposition!
You only transfer your material from the liquid to the gaseous state. Here, the molecules are still intact. But upon heating to the the boiling point, you have overcome the interactions between the molecules that existed before and held the whole bunch of molecules together as a lquid.
So, why is there the observed trend in the boiling points? Did you have a look at the molecular weights? ;)

What about the boiling points of alkyl halides, $\ce{R-Hal}$?
Let's have a look at the data for haloethanes, $\ce{C2H5-Hal}$:
$$\begin{array}{cr}
\mathrm{Hal} &\mathrm{bp~[°C]}\\
\hline
\ce{F} & -37 \\
\ce{Cl} & 13 \\
\ce{Br} & 38 \\
\ce{I} & 71 \\
\end{array}$$
Again, the trend is similar: Fluoroethane has the lowest, iodoethane has the highest boiling point.
Apparently, compounds with a higher molecular mass have a boiling point.
As a rule of thumb, the heavier the molecule, the more kinetical energy is needed to transfer it to the vapour phase. However, please note that this is a simplified model which might break when strong interactions between molecules, such as hydrogen bonds are involved!

So you mean the boiling points and the bond strengths aren't related at all? and btw I meant to put RCl, RBr, RI, alkyl halides, not CCl4, etc.

I'm sorry I thought of boiling point as bond breaking, I feel so stupid, anyway thank you for clearing that up.

one last thing, could you please explain the trend on boiling points? This is like a completely new thing to me, what factors affect the boiling points of an alkyl halide?


text:
What happens to the water inside the egg when it is boiled?
An egg contains 90% of water. When it is boiled, the water inside it cannot escape as water vapor because of the covering of calcium carbonate shell. After boiling, when the egg becomes solid the water cannot be seen. So what happens to the water inside the egg when it is boiled?

This is an interesting question. The water cannot be used to hydrolyze the proteins since the primary structure is not broken by heating.

Check out the background section here. The water from the egg white and the yolk essentially remains within the egg, but is trapped between the denatured protein chains. The process of denaturation makes the egg solid, but the water within makes it somewhat soft and compressible. See also Compound Interest's post about this subject.

An egg contains 90% of water. When it is boiled, the water inside it cannot escape as water vapor because of the covering of calcium carbonate shell.

That is correct, the water is still in there. Also, you are not really boiling the egg but the water bath it is in. If you place a zip lock bag filled with water into boiling water, the water in the bag will not boil and the bag will not pop open.

After boiling, when the egg becomes solid the water cannot be seen.

Yes, you would expect water at room temperature to be a liquid, but there is no visible liquid in a hard-boiled egg. This is the same when you make jello by adding a small amount of gelatin to a large amount of water (or an agarose or polyacrylamide gel). The wikipedia article on gels states:

By weight, gels are mostly liquid, yet they behave like solids due to a three-dimensional cross-linked network within the liquid.

So the answer to your question is that the water is still there but now behaves like a solid because the egg white protein that was dissolved in it changed conformation and physical properties (from a soluble globule to a sticky extended structure).

When it is globular, there is less surface area available for interaction with the surrounding globular proteins. Due to this, it is liquid. But, when we heat it to such an extent that the thermal energy is enough to overcome the H-bonding and other interaction that holds the globule together, they 'denature' to form an extended linear chain. Now, there will be extensive interaction with other extended chains and hence is it 'kind of' solid. Is my understanding correct?

The water present in the egg is a colloidal solution of mainly proteins in water.  On boiling proteins get coagulate. The coagulate proteins absorb water (solvent and it disappear).

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
How does one apply the knowledge of the direction of a molecular dipole moment?
I was watching Khan academy's video about molecular dipole moments, which explained what they are and how they can be calculated. I also read this article.
Both sources explain the concept similar to common organic chemistry textbooks, but they did not provide any information regarding the uses or applications of a molecular dipole's direction, nor did the google search I conducted later.
So far, two things are clear to me:

A dipole moment is a vector, and therefore has a quantity and a direction.
Larger dipole moment is associated with a more polar molecule. This a good enough reason to find out how large is the dipole moment.

My question is: What possible applications can knowing the direction of a molecular dipole moment have?

The knowledge of the direction of a molecular dipole moment can help us figure out how molecules would react, since regions of positive charge in one molecule can react with regions of negative charge in another molecule. This can also help us identify electrophiles and nucleophiles, since electrophiles have regions of low electron density (positive charge) and nucleophiles have regions of high electron density (negative charge).

Take a look at chemlibretexts. I'll just write an overview.

A dipole is a vector representing the charge distribution in a molecule. It's drawn from the positive-density charge side towards the negative.
Some examples

Phase change

How do we explain $\ce{CO2}$ fusion temperature is $-78\,^\circ$C while $\ce{SO2}$ is $-10\,^\circ$C? Because of its dipole magnitude. In the first one, there's no net dipole, but there is for the second one.
The bigger molecular interactions the more they oppose to temperature-increasings.

Solubility is higher for alcohols than lipids* in water. Again dipole-dipole interaction is stronger (energetically more favourable) than dipole-induced dipole interaction.
CO$_2$ has no electric dipole in the symmetric vibration. According to quantum mechanics, if electric dipole does not change when it absorbes light of a specific energy, it will not be susceptible of a infrared measure. This particular vibrational mode will be (and it is) infrared inactive.

As mentioned in the comments the information can be used for understanding reactivity. It can also tell the direction of an electron transfer. This is useful in materials chemistry to understand and predict where electrons and holes will prefer to go. This is applicable in,  for example,  organic photovoltaics or chlorophyll.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Is hydroperoxyl radical(HO2) toxic to the human body, or even flammable?

I understand that radicals are damaging to cells, due to their high chemical reactivity. Though I do not know whether they are flammable in any level.

Flammability is a property of compounds. A radical can't be flammable, much like it can't be happy or thoughtful.

It is very reactive.  It is an oxidizer rather than a flammable substance.
You can't collect a significant amount in one place because it will react with itself:
$\ce{HO2 + HO2 -> H2O2 + O2 }$


text:
Zero order kinetics
For a given reaction A→ B, the rate constant is 1.2 × 10$^{–2}$ M s$^{–1}$. What is the concentration of B after 20 min., if we start with 10 M of A. It is zero order kinetics.
When I try to solve the question, the equation would be
$$x_i - x_f = kt$$
$$10 M - x_f = kt = (1.2×10^{–2}M s^{–1})(20 min)(60 s/min)$$
So, $$x_f=10M - 14.4 M$$
I do not understand how it can be negative.

Why am I getting negative votes ?

How did you get 14.4 ?

@Shailesh I get it from kt = 1.2x10$^{-2}$ (20x60)

Your math checks out.  B cannot be negative, and indeed since the reaction forms B, it has to be zero.  What's the highest it could go?  Hint: first solve an alternate problem where you want to calculate the time $t$ it takes for A to be completely depleted.  What is the concentration of B at that time, and at all times thereafter?

It is negative because your reactant would be depleted before 20min. When the reaction is reaching the end, the mechanism is no longer 0th order.
Imagine a catalytic reaction mechanism where the rate is limited by the concentration of catalyst. You'd get a rate that is independent from the reactant concentration because the reaction rate is limited by the disassociation process of reactant-catalyst complex whose concentration is the same as the initial concentration of the catalyst.
However, when the reaction is ending, it would be limited by the interaction between the reactant and the catalyst instead, where the reactant would probably take a pseudo-first order mechanism.


text:
Treatment of acid attack by milk
I saw a post on Facebook saying if acid falls on some body part instantly wash it by milk but after a search on google I got this link which says it will actually worsen the condition.
Now I'm confused. Is milk 'good' or 'bad' for acid on the skin?

This seems close to personal medical advice. But, if I spilled a concentrated acid on my skin and only had a glass of milk at hand, I pour the milk on and then go find the emergency shower. No, just why I'd have a glass of milk nearby while doing chemistry is a different question...

Which acid? It's not like there's no difference...

@JonCuster The glas of milk is on the bench, right between the open beaker with fuming nitric acid and the plate with the cookies ;)

"Treatment" of acid burns by pouring milk over it falls into the same category as

drinking milk in the case of intoxications
covering burns with flour or shaving cream

All these "household remedies" are not recommended by modern medical standards and should be avoided.
Jon Custer's tongue-in-cheek comment is right. At a place where you work with concentrated acids, a glas of milk should not be easier to reach than a huge sink with a water tap.
In addition, proper eye protection and acid-resistent gloves are advised!
You will not find the suggestion to use milk in any recent occupational safety first aid guide!
The DGUV Information 204-008 (PDF, German), a first aid handbook for schools and kindergardens explicitely states on p. 106 ff that the proper procedure to treat others with acid burns is:

protect yourself (rubber gloves)
remove all clothes that were in contact and soaked with the acid
wash the affected areas with lots of warm water until the pain has ceased
if necessary, call an ambulance

Ok i got it but just out of curiosity can milk really cause any harm as stated on the link provided in question.


text:
IUPAC name of this compound

Please solve this question .I hope its name is $1-$Tertiary amino cyclopentanone. But it have some mistakes.

Cyclic amides of amino carboxylic acids are called lactams.
According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), lactames can be named by two methods:

P-66.1.5.1 Lactams and lactims
Intramolecular amides of amino carboxylic acids, $\ce{-CO-NH-{}}$, are called ‘lactams’ and their tautomers, $\ce{-C(OH)=N-{}}$, are ‘lactims’. Lactams are named in two ways:
(1) as heterocyclic pseudo ketones;
(2) by substituting ‘lactam’ for the ‘ic acid’ ending of a systematic ‘oic acid’ name for the parent acid without the amino substituent, and inserting a locant designating the position of the amino group between the ‘o’ and ‘lactam’. Lactims are named in the same way, using ‘lactim’ in place of ‘lactam’.
Method (1) generates preferred IUPAC names.

The corresponding heterocyclic parent structure is piperidine.

Thus, the preferred IUPAC name for the lactam according to Method (1) is the pseudo ketone piperidin-2-one.

The systematic name of the parent acid without the amino substituent is pentanoic acid.

Therefore, the lactam can also be named as pentano-5-lactam in accordance with Method (2).


text:
Spinel or salt?
In reaction of $\ce{Al2O3}$ with $\ce{NaOH}$, we get product written as "$\ce{NaAlO2}$", "$\ce{Na2Al2O4}$" or just "$\ce{Na2O.Al2O3}$".
My question is what's the real structure of this compound? Spinel or a salt of aluminate anion ($\ce{AlO2-}$)? Maybe just mixture?

Related: http://chemistry.stackexchange.com/questions/60452/reaction-between-alumina-and-aqueous-sodium-hydroxide-which-ones-right/60463#60463

From wikipedia article of sodium aluminate:

Anhydrous sodium aluminate, $\ce{NaAlO2}$, contains a
  three-dimensional framework of corner linked $\ce{AlO4}$ tetrahedra.
  The hydrated form $\ce{NaAlO2·5/4H2O}$ has layers of $\ce{AlO4}$
  tetrahedra joined into rings and the layers are held together by
  sodium ions and water molecules that hydrogen bond to O atoms in the
  $\ce{AlO4}$ tetrahedra.

The composition of sodium aluminate is basically written as $\ce{NaAlO2, NaAl(OH)4}$ (hydrated), $\ce{Na2O·Al2O3}$ or $\ce{Na2Al2O4}$.


text:
How to perform a total synthesis on paper ? (Think before to make it in lab)
I studied organics chemistry during five years. All those years I just made reactions to made them, the only moments where I had to perform a synthesis on paper was during exams or exercises but all steps were given. So, even if, it was very interesting, there were no really fun to apply all this knowledge to think about the synthesis of a molecule. Obviouly with all these lessons I'm not that bad and I'm able to do a little bit of retrosynthesis.

However I know that total synthesis of natural origin molecules is perform a lot for some decades, we can find a lot of examples here, and sometimes it looks pretty weird to think that human people were able to think on how to make a synthesis in more than twenty steps.
For example, the weirdest synthesis I found, is the synthsis of Azadirachtin "only" perform in $79$ steps. That is abosolutely insane.

My question:

I'm wondering, for every molecules in general, how do chemist do to
  think about the retrosynthesis of such big molecules which contain in
  general a lot of stereocenters which then add more difficulties?
  I mean without to use a computer, is there some rules, such like theorems in mathematics, to think in the best possible way? Because with a bit of luck for $79$ steps at $90$% yield it makes a final yield of $0.024$%...

There are a few "classic" books on total synthesis which have pretty decent introductions that cover this idea of retrosynthesis; you may wish to consider checking them out. Off the top of my head I remember 1) E.J. Corey and X.-M. Cheng, The Logic of Chemical Synthesis; 2) K.C. Nicolaou and E.J. Sorensen, Classics in Total Synthesis

Retrosynthesis, the way to determine which cuts to make and which steps to use in which order (at least in the general picture), is an art. It is not a science or a craft, it is an art. You learn it by doing it and there is only so much you can get taught by others.
Recently, a multi-group seminar I attended started a ‘molecule of the month’ challenge. We were given a molecule and divided into groups. Each group then had about six weeks to come up with a total synthesis plan, find literature precedence or good arguments for all steps and then present it. Our basic strategy was to draw substructures into SciFinder and rely heavily on its reaction search functionality. With more experience and having read more synthetic journal articles, we probably would have known what we would want to go for.
Finally, I feel like adding that sometimes it is not the molecule whose total synthesis you are publishing that piques your interest. There are also those that primarily do method development and then try to find out ‘which molecule could be a good target for the new and super-cool method I just established?’ Thereafter, the synthesis is modelled to fit the desired novel key step. This paragraph is just to show an alternative approach to total synthesis.

I'll add that you might use "extra" steps for purity or yield. So the fewest steps is not necessarily the "best" method.

Yes, overall yield is a very important quantification parameter @MaxW


text:
Naming a keto ester

Find the IUPAC name of the above structure.
This question was asked in my Chemistry exam. I could not figure out whether it is a ketone or an ester.
My attempts:
1.If i consider the whole compound as an ester then part to the left of C-O-C is the acid part and the part to the right is the alcohol part. Thus the IUPAC name is:
Ethyl 4,5-dibromoheptan-2-one-6-oate
2.If the consider the part to the left of C-O-C as ketone and the part to the right as ester then IUPAC name comes out to be:
4,5-dibromopentan-2-one Ethyl Methanoate

Is any of my attempts correct and if not then why not?

Ester group is more important but your first name is still quite thoroughly wrong.

What about the second one?

Related: Nomenclature priorities with ethers, acyl groups and ketones

Second is two molecules, you'd get if you'd manage to cleave it reductivly.

What is the IUPAC name then and why has this question been downvoted?

I think the downvotes are a bit harsh, to be honest.  IUPAC naming gets far more attention than it deserves, and the more complex the molecule, the more useless it is.  "The ester (A)" (accompanied by a diagram labelled A) is far more useful.

According to Section P-41 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the order of seniority of classes in decreasing order of seniority is as follows.

(…)
  9. esters
  (…)
  16. ketones
  (…)

Therefore, the compound that is given in the question is named as an ester.
The names of esters are usually formed by placing the alcoholic component in front of the name of the acid component as a separate word.

P-65.6.1 General methodology
Neutral salts and esters are both named using the name of the anion derived from the name of the acid. Anion names are formed by changing an ‘-ic acid’ ending of an acid name to ‘-ate’ and an ‘-ous acid’ ending of an acid name to ‘-ite’. Then, salts are named using the names of cations, and esters the names of organyl groups, cited as separate words in front of the name of the anion.

Therefore, the name of the parent structure without further substituents is ethyl hexanoate.
This ester is substituted using the usual principles of substitutive nomenclature. A low locant is assigned first to the principal characteristic group (i.e. the ester group). The prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix.
Thus, the complete name for the compound that is given in the question is ethyl 2,3-dibromo-5-oxohexanoate.


text:
Why do Grignard reagents add to propargyl alcohols?
In Phil Baran's lectures on heterocyclic chemistry at Scripps, one of the lecture problems is concerned with this synthesis of Vioxx:

The first step supposedly involves an anti addition of the aryl Grignard reagent to the propargyl alcohol:

and this intermediate is trapped with $\ce{CO2}$ to form a lactone (a butenolide in this case), and mCPBA oxidises the sulfide to a sulfone.
Why does the aryl Grignard add to the alkyne in this case? I would expect it to deprotonate the alcohol, and it does seem to do so, but as far as I know, Grignards don't usually add to alkynes which aren't conjugated to electron-withdrawing groups. Why is this propargyl alcohol special?
On top of that, what is the rationale for the selectivity observed (both regio- and stereoselectivity)?

It looks wrong. It doesn't seem to be made this way https://books.google.pl/books?id=fsKKI82uqK8C&pg=PA13&lpg=PA13&dq=rofecoxib+synthesis&source=bl&ots=IQ2sW9FSkU&sig=-Vg6ESf3XVuis2m5piHCpGSMixw&hl=pl&sa=X&sqi=2&ved=0ahUKEwjCvNaI0KvRAhWC2SwKHcL9AtcQ6AEIMTAG#v=onepage&q=rofecoxib%20synthesis&f=false

@Mithoron I think that this wasn't the one that Merck opted to go for. However, this alternative synthesis was reported in Tetrahedron Lett. 2000, 41 (1), 17–20 (apparently, there is a non-paywall version here as well).

Then the reaction is rather with salt after neutralisation of alcohol.

@orthocresol looks like reference 1 in that paper is the original report of that reaction. Check if they postulated a mechanism?

The addition of Grignards to propargyl alcohols occurs via the alkoxide salt, after deprotonation:[1]

Here:

the reactivity may be explained by the stabilising influence of complexation;
the regioselectivity may be explained by the preference of a five-membered chelate ring over a four-membered one; and
the stereoselectivity arises because syn addition requires a five-membered ring with a trans double bond and is therefore precluded.

In fact, propargylic alcohols also get reduced by $\ce{LiAlH4}$ (Hans Reich's website has a page on it), with the same trans stereochemistry observed.

Reference
(1) Forgione, P.; Fallis, A. G. Magnesium mediated carbometallation of propargyl alcohols: direct routes to dihydroxydienes and enediyne alcohols. Tetrahedron Lett. 2000, 41 (1), 11–15. DOI: 10.1016/S0040-4039(99)01994-2.

Note that the oxygen can withdraw electron drnsity from the methylebe grouo, giving it some catbovation character; this polarized carbon is then conjugated with the end of the triple bond where the nucleophilic part of the Grignard reagent attacks.


text:
how to keep a vessel with liquid in it a certain temperature?
how do i keep a vessel with liquid in it a certain temperature?
in a room and don't have to buy something expensive?
i am taking a biogas vessel and placing them in different temperatures.
PLs help me CHAR

In the construction of DIY biogas reactors, it is quite common to rely on Intermediate Bulk Containers (IBC). These are

affordable
available in different sizes
sturdy, particularly when used with baseplate and frame
stackable

The relevant reactions take place in the absence of oxygen and light.
If you run your reactor outside, which is a good idea, given that the degradation of sulfur-comtaining amino acids yields volatile sulfur-containing compounds, it needs to be insulated.
Typically,

PU foam (from a spray can)
styrofoam panels or
bales of straw

are used for this purpose.
The non-profit Solar C³ities community provides a lot of background information and practical hints, such as a detailed instruction how to build  an IBC biodigester.


text:
Ionic Mechanism for Wurtz Reaction
Instead of going for the normal free radical mechanism for Wurtz reaction, What reasons prevent us from believing in the following ionic mechanism?
\begin{align}
\ce{R-X + 2Na &-> R^--Na+ + NaX}\\
\ce{R-X + R^--Na+ &-> R-R + NaX}\\
\ce{2R-X + 2Na &-> R-R  + 2NaX}
\end{align}
Related but doesn't answer my question here

Maybe the fact that Wurtz reaction occurs even in mediums that do not solvate ions properly (like non-polar ones) ?

That is only a case

The actual mechanism is believed to be strongly dependent on the nature of halide, metal (not only $\ce{Na}$ is used in this reaction), and solvent. The 'ionic' mechanism you have written here is the part of the most accepted mechanism which involves both, an 'ionic' and a radical reaction.
First step. Through the single-electron transfer from $\ce{Na}$ to the halogen atom, an alkyl radical $\ce{R^.}$ and sodium halide is formed:
$$\ce{Na + R-X -> Na+X^- + R^.}$$
Alkyl radical further acts as an electron acceptor from another $\ce{Na}$ atom to form highly nucleophilic intermediate (can be isolated in some cases):
$$\ce{R^. + Na -> R^-Na+}$$
This alkyl sodium intermediate reacts with another alkyl halide in the aliphatic nucleophilic substitution:
$$\ce{R^-Na+ + RX -> R-R + NaX}$$

Actually, nothing prevents us .This mechanism is well accepted but only for 1 degree & 2 degree carboanions . It does not give results in case of 3 degree carboanions because the resulting alkyl sodium would act as a strong base for the R-X and give elimination products


text:
What are the practical differences between the various DMSO oxidations (Swern, etc.)?
Recently, I was reading a paper about a total synthesis of (+)-rubriflordilactone A (the article is open access: Angew. Chem. Int. Ed. 2015, 54 (43), 12618–12621). In two different steps where an alcohol -> aldehyde oxidation was required, the authors favoured using the Parikh–Doering reaction (DMSO activated with pyridine-sulfur trioxide complex):

In practice, is there a reason why the Parikh–Doering might be favoured over the Swern (DMSO activation with oxalyl chloride), or is it simply a case of "we just got better yields with it"?
I checked in Kurti–Czako. No hints were offered, but they also did list two more Swern variants, namely Pfitzner–Moffatt (activation with a carbodiimide) and Albright–Goldman (activation with an acid anhydride). On top of that, as Jan mentioned in a comment, there is also the Corey–Kim oxidation where N-chlorosuccinimide is used to generate the chlorosulfonium salt from DMS (instead of DMSO).
In general: are there benefits or drawbacks that are specific to each variant that would influence a synthetic chemist's decision, and if so, what are they?

Will think about it properly later but in practice I think here's little difference— labs seem to have a preferred method. I have a feeling that the Parikh Doering doesn't have the issue of an unwanted Pummerer that I've seen once or twice with a Swern but practically its of little consequence. DMP all the way!

We often had this in our seminar. Parikh-Doering has $\ce{SO3 . Py}$ as the activating agent meaning that $\ce{Py . HSO4}$ is eliminated and no formal $\ce{HCl}$ is formed that would need to be quenched additionally. Or something along those lines.

I imagine "we just got better yields" is frequent, but the nature of the byproducts could be one rational consideration.

Sometimes it is a matter of scaleability. The pyridine.SO3 conditions are easier to use than handling a lot of oxalyl chloride or TFAA

I'm a theoretician, but could solubility of reagents/substrate be a factor?

I've been out of the loop for a while, but in general, I think for reaction variants, we tend to care about how strong/mild the reagents are, in terms of unwanted side reactions. Also, we care a lot of cleaning up side products for purifications. Too bad I can provide some detailed examples of which conditions fit which properties... :(

A belated comment. Oft times it is simply a matter of  what has worked for the experimentalist in the past or what reagents are readily available in the lab. The discussion is reminiscent of the line in a synthesis lecture when the speaker says, "And the last reagent we tried worked!" Well, duh, are still going to keep trying? ;)

tl;dr:
Swern and Parikh–Doering are by far the most practically useful and widely used oxidations. Whilst both have slight differences, the reality is that most research groups have a preference. This may be because they just have the reagents on hand (distilled oxalyl chloride vs good quality sulfur trioxide–pyridine), or because someone had a bad experience with one over the other.
The Anderson group (in the post you linked) commonly uses the Parikh–Doering conditions, whilst other groups appear to favour the Swern conditions using oxalyl chloride in place of sulfur trioxide–pyridine.

A brief overview of various DMSO oxidations and their drawbacks:
One thing to mention about all of these DMSO based reactions is the formation of stoichiometric dimethylsulfide (stench) as a by-product, which increasingly makes them avoided, especially by process chemists. That said, the reagents are cheap (compared to a Dess–Martin, for example), and the reaction is generally very reliable and amenable to use on moderate scales.  The DMSO and triethylamine must also be distilled for optimal results (which can be a drawback, especially if large scales are needed).
1. Moffatt oxidation
Reagents: DMSO, DCC, PPA.
Reference: Pfitzner, K.; E.; Moffatt, J. G. A New and Selective Oxidation of Alcohols. J. Am. Chem. Soc. 1963, 85 (19), 3027–3028. DOI: 10.1021/ja00902a036.
The Moffatt oxidation was one of the earliest examples of a DMSO-mediated oxidation. Two issues generally exist. Firstly, a strong acid (polyphosphoric acid, TFA, etc.) is required to get the reaction going, and, although the reaction becomes basic as it proceeds, the initial conditions using polyphosphoric acid/similar quite often have unwanted side effects (stripping of protecting groups, from experience). Secondly, and more importantly from a scale point of view is the stoichiometric N,N'-dicyclohexylurea produced as a by-product (similar to the Steglich esterification).
Due to the difficult-to-remove stoichiometric by-products, and the harsher conditions, Moffatt oxidations have largely been forgotten/ignored. Reagents such as EDCI have been introduced, however the high cost relative to oxalyl chloride/sulfur trioxide–pyridine have limited the appeal of Moffatt oxidations.
2. Albright–Goldman oxidation
Reagents: DMSO, acid anhydride
Reference: Albright, J. D.; Goldman, L. Dimethyl Sulfoxide-Acid Anhydride Mixtures. New Reagents for Oxidation of Alcohols. J. Am. Chem. Soc. 1965, 87 (18), 4214–4216. DOI: 10.1021/ja01096a055.
The Albright–Goldman oxidation commonly uses acetic anhydride, and goes at room temperature - although with significant reaction times which can give side reactions time to occur.
The biggest drawback, and reason why the Albright–Goldman is rarely used, is due to a competing Pummerer reaction (which is often significant), along with protection of the alcohol as its acetate ester if unhindered.
Some improvements have been made using TFAA in place of acetic anhydride. However, mixing TFAA and DMSO causes a violent exotherm unless the temperature is carefully and cautiously controlled.
3. Parikh–Doering
Reagents: DMSO, sulfur trioxide–pyridine, triethylamine
Reference: Parikh, J. R.; Doering, W. v. E. Sulfur trioxide in the oxidation of alcohols by dimethyl sulfoxide. J. Am. Chem. Soc. 1967, 89 (21), 5505–5507. DOI: 10.1021/ja00997a067.
Along with the Swern (see below), the Parikh–Doering is one of the most widely used DMSO based oxidations, with many examples in the synthetic literature.
The Parikh–Doering is able to be ran at mild temperatures (0 °C to room temperature), and the by-products are easily removed (crashing out and filtration). One advantage to the Parikh–Doering over the Swern is the fact that sulfur trioxide–pyridine is a bench stable solid, compared to oxalyl chloride which is a liquid that must be distilled and stored in the freezer to prevent degradation.
4. Swern oxidation
Reagents: DMSO, oxalyl chloride, triethylamine
Reference: Omura, K.; Swern, D. Oxidation of alcohols by “activated” dimethyl sulfoxide. a preparative, steric and mechanistic study. Tetrahedron 1978, 34 (11), 1651–1660. DOI: 10.1016/0040-4020(78)80197-5.
The Swern is the newest of the widely used DMSO oxidations, and often represents the best combination of reliability, cost of reagents, and ease of reaction.
Unlike the Parikh–Doering, the initial stages of the reaction must be ran at −78 °C to avoid unwanted decomposition. However, the conditions are equally as mild. One advantage over the Parikh–Doering is that oxalyl chloride decomposes into gaseous by-products, meaning that the only thing needed to be removed is the hydrochloride salt of triethylamine. Swern reactions can often be filtered, concentrated, and telescoped into the subsequent reaction without attrition in yield.

References:
The majority of the chemical information here comes from Comprehensive Organic Synthesis, Vol. 7. A whole chapter is dedicated to DMSO based oxidations and their examples in total synthesis: see Lee, T. V. Oxidation Adjacent to Oxygen of Alcohols by Activated DMSO Methods. DOI: 10.1016/B978-0-08-052349-1.00191-8.


text:
What is the classical partition function for a system of anharmonic oscillators?
This is an exam question that I'm trying to solve. Here's the text:

A system is composed of N localised, but independent one-dimensional
  classical oscillators. Assume that the potential energy for an
  oscillator contains a small anharmonic term
$$ V(x) = \frac{k_0x^2}{2} + \alpha x^4 $$
where $\alpha < x4 << kT$. Write down an expression for the Canonical
  partition function for this system of oscillators. Then, to first
  order in the parameter $\alpha$ ($\alpha > 0$), derive an expression
  for the internal energy and the isochoric heat capacity for this
  system and show that the anharmonic correction tends to reduce the
  energy per oscillator compared to the equipartition result of a
  perfectly harmonic oscillator. Explain also why letting $T \rightarrow
\infty$ in the expression for the heat capacity represents an
  unphysical result.

Attempted solution
The classical canonical partition function, $Q$, in this case will be
$$
Q = \iint \limits_{-\infty}^{\infty} e^{-\left( \frac{k_0}{2}x^2 + \alpha x^4 + \frac{p^2}{2m} \right)\frac{1}{kT}} \text{d}p\text{d}x
$$
Now comes the difficult part. I have tried to find the Maclaurin series of $V(x)$ around $\alpha = 0$ to first order, but this just returns the original expression (weird, huh?). Then I tried to evaluate the integrals as they stand above, but this seems to not be possible:
\begin{align}
Q &= \iint \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} e^{-\frac{p^2}{2mkT}} \text{d}p\text{d}x \\
        &= \int \limits_{-\infty}^{\infty} e^{-\frac{p^2}{2mkT}} \text{d}p     \int \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x
\end{align}
The first integral I can do, but not the second. However, if I use that $a < x^4 << kT$, which logically leads to that $ax^4 << kT$, then I perhaps could simplify the second integral by using that $e^{-\frac{ax^4}{kT}} \approx e^{-0} = 1$. But this totally neglects the anharmonic term, which clearly I should draw some conclusions about later on.
Instead, the trick I think is to do the correct Maclaurin expansion. Should I expand the expression for $Q$? Using
$$
Q \approx Q \rvert_{\alpha = 0} + a\frac{\partial Q}{\partial a}\rvert_{\alpha = 0}
$$
results in
\begin{align}
Q &\approx \iint \limits_{-\infty}^{\infty} e^{-\left( \frac{k_0x^2}{2kT} + \frac{p^2}{2mkT} \right)} \text{d}p\text{d}x + a\left[ \frac{\partial}{\partial a} \left( \iint \limits_{-\infty}^{\infty} e^{-\left( \frac{k_0x^2}{2kT} + \frac{p^2}{2mkT} \right)} \text{d}p\text{d}x \right) \right] \\
&= \int \limits_{-\infty}^{\infty} e^{-\frac{p^2}{2mkT}} \text{d}p     \int \limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} \text{d}x + 0
\end{align}
Since the integrals in the second term do not depend on $a$, the whole thing reduces to $0$, and the anharmonic contribution here is lost. I expected from the start that the anharmonic term would give a contribution to the final result, but it does not when I do it like this.  Anyway, continuing, the integrals that are  left are easily evaluated, and we get a first order approximation to the canonical partition function
\begin{align}
Q &\approx 2kT \sqrt{\frac{m}{k_0 }}
\end{align}
where the square root sort of resembles the angular frequency of the oscillator (assuming that $k_0$ is the force constant).
Now, finally, we can start to actually answer the question, and find an expression for the internal energy $U$
$$
U =kT^2 \left( \frac{\partial \ln Q}{\partial T} \right)_{N,V}
$$
We need the logarithm of $Q$
$$
\ln Q = \ln T + \ln \left( 2k\sqrt{\frac{m}{k_0}}\right)
$$
And then get the expression for $U$
\begin{align}
U &= kT^2 \frac{1}{T} \\
&= kT
\end{align}
Taking the derivative of $U$ with respect to $T$, should give the isochoric heat capacity:
$$
C_V = k
$$
So in this case, the isochoric heat capacity does not depend on $T$ at all, so I can't really evaluate whether the high-temperature limit is meaningful or not. My approach may have been incorrect from the start.

I can draw up an answer to this question later today. I have found the solution.

Here follows a complete mathematical derivation of the expressions for the internal energy and isochoric heat capacity. I am not sure why taking the high-temperature limit is unphysical, but maybe someone else knows.

The potential is
$$
V(x) = \frac{k_0x^2}{2} + \alpha x^4
$$
for one dimensional, localized oscillators. The Hamiltonian then becomes
$$
H(p,x) = \frac{k_0x^2}{2} + \alpha x^4 + \frac{p^2}{2m}
$$
for each particle. Since we have a system of $N$ particles, we sum over all particles to get the full Hamiltonian for the entire system (assuming all oscillators have the same mass)
$$
H(p,x) = \sum\limits_{j=1}^{N} \frac{k_0x_j^2}{2} + \alpha x_j^4 + \frac{p_j^2}{2m}
$$
The general expression for the classical canonical partition function is
\begin{align}
Q_{\text{N,V,T}} &= \frac{1}{N! h^{3N}} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}e^{-H(x,p)/kT} \text{d}x\text{d}p \\
\end{align}
The text says that the oscillators are localized, so we should take away the N! in the expression for Q, since we are dealing with distinguishable particles. Plugging in the Hamiltonian, we then get for the partition function {dropping the subscript from now on}
\begin{align}
Q = \frac{1}{h^{3N}} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty}e^{-\left( \sum\limits_{j=1}^{N} \frac{k_0 x_j^2}{2} + \alpha x_j^4 + \frac{p_j^2}{2m} \right)\frac{1}{kT}} \text{d}x\text{d}p
\end{align}
We know that an exponential where the exponent is a sum we can rewrite into a product of exponentials $e^{a+b+c} = e^ae^be^c$
\begin{align}
Q &= \frac{1}{h^{3N}} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} \prod\limits_{j=1}^{N} e^{-\frac{k_0x_j^2}{2kT}} e^{-\frac{\alpha x_j^4}{kT}} e^{-\frac{p_j^2}{2mkT}} \text{d}x\text{d}p \\
   &= \frac{1}{h^{3N}} \prod\limits_{j=1}^{N} \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x_j^2}{2kT}} e^{-\frac{\alpha x_j^4}{kT}} e^{-\frac{p_j^2}{2mkT}} \text{d}x\text{d}p \\
  &= \frac{1}{h^{3N}} \prod\limits_{j=1}^{N} \int\limits_{-\infty}^{\infty} e^{-\frac{p_j^2}{2mkT}} \text{d}p  \int\limits_{-\infty}^{\infty}  e^{-\frac{k_0x_j^2}{2kT}} e^{-\frac{\alpha x_j^4}{kT}} \text{d}x \\
&= \frac{1}{h^{3N}} \left( \int\limits_{-\infty}^{\infty}e^{-\frac{p^2}{2mkT}} \text{d}p \right)^N \left( \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \right)^N
\end{align}
where in the last step we saw that the integrand will be the same for all values of $j$. The integral over the momenta are easily evaluated; just look up any table of definite exponential integrals to find the expression. We then get
$$
Q = \left( \frac{2mkT\pi}{h^2}\right)^{3N/2} \left( \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \right)^N
$$
The remaining integral is too complicated to evaluate analytically, so we perform a Taylor series to first order around $\alpha = 0$ (first order Maclaurin series in $\alpha$) for the anharmonic exponential. This is justified since, as the text explains, $a > x^4 >> kT$. Doing the expansion:
\begin{align}
e^{-\frac{\alpha x^4}{kT}} &\approx e^0 + a \left( \frac{\text{d}}{\text{d}a} e^{-\frac{\alpha x^4}{kT}} \right)_{a=0} \\
&\approx 1 + a\left( e^{-\frac{\alpha x^4}{kT}} \left( -\frac{x^4}{kT} \right) \right)_{a=0} \\
&\approx 1 - \frac{ax^4}{kT}
\end{align}
We now use this expression instead of the exponential in the integral. Lets just evaluate the integral for now, and then plug in for $Q$ once we have an expression for the integral. Calling the integral $I$, we get
\begin{align}
I &= \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} e^{-\frac{\alpha x^4}{kT}} \text{d}x \\
&= \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} \left( 1 - \frac{ax^4}{kT} \right) \text{d}x \\
&=  \int\limits_{-\infty}^{\infty} e^{-\frac{k_0x^2}{2kT}} \text{d}x - \frac{a}{kT} \int\limits_{-\infty}^{\infty} x^4 e^{-\frac{k_0x^2}{2kT}} \text{d}x \\
&= \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} - \frac{a}{kT} \int\limits_{-\infty}^{\infty} x^4 e^{-\frac{k_0x^2}{2kT}} \text{d}x
\end{align}
While this integral looks intimidating, we can look it up in standard tables. It has the following solution
$$
\int\limits_{0}^{\infty} x^{2n} e^{-cx^2} \text{d}x = \left( \frac{\pi}{c} \right)^{1/2} \frac{1\cdot 3 \cdots (2n-1)}{2^{n+1}c^n}
$$
Since our limits go from $-\infty$ instead of $0$, we multiply with a factor of 2 to get the correct results for the integral. Using this, we obtain
\begin{align}
I = \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} - \frac{2a}{kT} \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} \frac{3}{8} \left( \frac{2kT}{k_0^2} \right)^2
\end{align}
We can now cancel the 8 in the denominator with the $2^2$ and 2 in front of $a$, cancel a factor of $kT$ and then factorize by moving the square root out
\begin{align}
I &= \left( \frac{2kT\pi}{k_0^2} \right)^{1/2} \left(1 - \frac{3akT}{k_0^4} \right)
\end{align}
We can now insert this into our expression for the partition function:
$$
Q = \left( \frac{2mkT\pi}{h^2}\right)^{3N/2} \left( \frac{2kT\pi}{k_0^2} \right)^{N/2} \left(1 - \frac{3akT}{k_0^4} \right)^N
$$
In statistical thermodynamics, all thermodynamic properties are related to the natural logarithm of the partition function, so we now take the logarithm. This is the reason for factorizing the expressions; it just makes taking the logarithm much easier.
$$
\ln Q = \frac{3N}{2} \ln \left( \frac{2mkT\pi}{h^2} \right) + \frac{N}{2}\ln\left( \frac{2kT\pi}{k_0^2} \right) + N\ln \left( 1 - \frac{3akT}{k_0^4} \right)
$$
From here we can start to get expressions for the internal energy and isochoric heat capacity. First, internal energy
\begin{align}
U &= kT^2 \left( \frac{\partial \ln Q}{\partial T} \right)_{N,V} \\
&= kT^2 \left( \frac{3N}{2} \frac{1}{T} + \frac{N}{2} \frac{1}{T} - N \frac{3ak}{k_0^4 - 3akT} \right) \\
&= \frac{3NkT}{2} + \frac{NkT}{2} - \frac{3Nak}{k_0^4 - 3akT}
\end{align}
The isochoric heat capacity is
\begin{align}
C_V &= \left( \frac{\partial U}{\partial T} \right)_{N,V} \\
&= \frac{3Nk}{2} + \frac{Nk}{2} - N\left( \frac{3ak}{k_0^4 - 3akT} \right)^2
\end{align}
The first term in $C_V$ comes from the kinetic energy, and is exactly what the equipartition theorem predicts. The second term is related to the potential energy, while the last term clearly is related to the anharmonic pertubation.
The exam question now asks why taking the high-temperature limit for $C_V$ is unphysical. This I am unable to  answer. Mathematically, the limit seems to diverge, but I'm not sure what kind of answer is expected here. But at least I think the mathematical derivation is correct.

Or does it not diverge? In the high-temperature limit, the anharmonic term approaches zero, does it not?

I couldn't spot anything wrong with your derivation. And based on that final expression, for T going to infinity the last term should go to zero.

The first answer has one problem, you have N oscilators not N particles. So we have $$(\frac{2mkTπ}{h^{2}})^{N}$$ instead of $$(\frac{2mkTπ}{h^{2}})^{\frac{3N}{2}}$$. With this correction your internal energy becomes: $$U=NK_{b}T+\frac{3\alpha NK_{b}T}{3\alpha-\frac{k^{2}}{K_{b}T}} $$. Note that if $\alpha=0$ the energy becomes:
$$U=NK_{b}T$$ which is the internal energy of N harmonic oscillators as expected.

In high temperatures the oscillator tends to occupy larger values of x, therefore the approximation you performed to turn the original partition function into a Gaussian  integral (assuming small x) becomes invalid.
You can also see this in the final result for $C_v$ which unphysically turns negative for high enough T value


text:
Liquid water below freezing temperature
I got in my car this morning and the temperature read 13 F (that's around -10.5 C.) While driving down the road, I noticed a bottle of water in my cup holder was frozen solid as expected.
Another bottle of water (cheap, thin-walled, plastic disposable water bottle) contained liquid water! I shook the bottle, not a single crystal of ice!
Shortly thereafter, I noticed the water became slushy, so it had begun freezing quite quickly!
The interior of the bottle was hardly perfectly smooth given the quality of the bottle and the water was half drunk so it was probably not super pure.
What caused this? Was this supercooling?

You may have just experienced supercooling.
Ordinary crystallization upon reaching the freezing point of a liquid requires nucleation sites - places for crystals to form. In absence of nucleation sites, water can be cooled down to lower temperatures until crystal homogeneous nucleation occurs. In water, this can occur as low as $224.8\ \mathrm{K\ (-48.3\ ^\circ C,\ -55\ ^\circ F)}$.
Nucleation can also be induced by disrupting the system, which happened when you shook the bottle. Even so, the fact that it only became slush may attest to the presence of impurities. A $\mathrm{19\ ^\circ F,\ 10\ ^\circ C}$ freezing point depression is not unreasonable depending on the amount of saliva that might have ended up back in the bottle.


text:
Why must a voltaic cell take place in a conductive solution?
In order for a voltaic cell to function, its electrodes must be placed in a conductive electrolyte solution, most often solutions consisting of their metal ions. While I understand the use of the metal ions (allows for the plating to occur on the cathode), why must the electrolyte be conductive? Thanks!

If it's not conductive, the cell doesn't work - the electrons need a complete circuit to be present in the system in order for current to flow.

The current flow in the solution is not done by electrons per sey. Within the solution ions conduct the current.  Cations would flow in one direction and anions in the other.

The electrolyte must be conductive to ions, not electrons. In fact, if there's an easy path for electrons to move between the two electrodes, the cell will immediately self-discharge. The exact reason for needing ionic-conductivity kind of depends on the cell. For example, in lithium-ion batteries, lithium-ions are generated at the anode and consumed at the cathode (or vice-versa). If ions cannot move freely between the two (through the electrolyte), then you will quickly build up an ion-rich region at one electrode and an ion-deficient region at the other at which point the reaction will stop.
It's not even a requirement that the electrolyte be liquid, as long as it conducts ions. Eg. the sodium-sulfur battery uses a form of solid alumina that conducts sodium ions.


text:
Storage of the Halogens in PTFE Bottles
I'm looking to buy some Iodine for various reactions, and noticed that the person who sells it has it in a glass jar with a plastic lid of unknown type.
I'm looking up chemical resistance charts and it says that Iodine, Chlorine and Bromine both work absolutely fine when placed inside of a PTFE bottle, assuming that the Chlorine is not in a gaseous form.
As for Fluorine, one website states that it attacks the plastic severely, whereas the other states that the PTFE is resistant until around the 100°C mark.
These sites are both here and here.
Now when I go look at people asking the question, "How should I store elemental Iodine?", everyone seems to respond along the lines of keeping it in a glass jar with a ground glass joint or PTFE/Teflon lined cap.
However since my research suggests that PTFE alone is capable of storing Iodine, is this a viable option? I would love to know from past experiences, because no-one seems to recommend PTFE on its own on any site I checked. Of course however I will also be storing said PTFE container inside of another non-reactive material container.
I understand that I'm probably better off just storing it in glass containers but I have PTFE bottles lying around and would like to use them if at all possible.
Since this question revolves around the Halogens themselves, could the same be potentially done for Bromine and Chlorine?

http://chemistry.stackexchange.com/questions/34190/is-there-any-chemical-that-can-destroy-ptfe-or-teflon?s=5|0.9921 http://chemistry.stackexchange.com/questions/6183/does-a-sulfuric-acid-nitric-acid-mixture-dissolve-ptfe-ie-teflon?s=9|0.5349 Teflon is very resistant

Though PTFE is not attacked by the halogens, they may discolor PTFE (particularly iodine, and at elevated temperature), since the plastic is somewhat permeable to halogens (end even to water).

Hmm... I might be better off just buying a glass jar with a Teflon-only lid instead. Thanks for the reference, some good source material for later storage!


text:
Which characteristics of the compound affect the number of theoretical plates in a GC experiment?
Some days ago, I tried for the first time a chromatography machine.
The data I collected are the following:
$$
\begin{array}{lrrr}
\hline
\text{Compound}                 & t_\mathrm{R}/\pu{min} & w/\pu{min} & N     \\
\hline
\text{Benzene}                  & 2.325                 & 0.047      & 39153 \\
\text{Toluene}                  & 2.712                 & 0.067      & 26214 \\
\text{Chlorobenzene}            & 3.690                  & 0.099      & 22228 \\
\textit{p}\text{-Chlorotoluene} & 4.332                 & 0.106      & 26723 \\
\hline
\end{array}
$$
I know the relationship between retention time $t_\mathrm{R},$ peak width $w$ and theoretical plates $N.$
Known the different properties, boiling points, polarities, or shapes of the molecules involved, is there a way to predict which molecule will yield an higher number of theoretical plates?
Which characteristics of the molecules affect the different number of plates for each molecule?

The number of theoretical plates can be calculated as:

$\mathrm{N_t = \frac {H}{HETP}}$
$\mathrm{N_t =}$ number of theoretical plates
  $\mathrm{H =}$ total height (or usually length) of the column [$\mathrm{m}$]
  $\mathrm{HETP =}$ Height Equivalent to a Theoretical Plate [$\mathrm{m}$]

Where $\mathrm{HETP =}$ can be calculated from the Van Deemter equation:

$\mathrm{HETP = A + \frac {B}{u} + (C_s + C_m) * u}$
$\mathrm{A =}$ Eddy-diffusion parameter, related to channeling through a non-ideal packing [$\mathrm{m}$]
  $\mathrm{B =}$ diffusion coefficient of the eluting particles in the longitudinal direction, resulting in dispersion [$\mathrm{m^2 s^{−1}}$]
  $\mathrm{C =}$ Resistance to mass transfer coefficient of the analyte between mobile and stationary phase [$\mathrm{s}$]
  $\mathrm{u =}$ Linear Velocity [$\mathrm{m s^{−1}}$]

So, the characteristics specific to the analyte molecule that determine the number of theoretical plates are B and, unfortunately from the standpoint of developing an ab initio model, C.
Various models have been developed, however.  Below is a list of references to some articles of interest in this field.
[ 1 (abstract)]"The problem of plate modeling: Theoretical and computational results", Computer Methods in Applied Mechanics and Engineering, Volume 100, Issue 2, October 1992, Pages 249–273
[ 2 ]"Plate Models in Chromatography: Analysis and Implication for Scale Up", Chromatograpy,Springer-Verlag
[ 3 (abstract)]"Neural Network Modeling of Structured Packing Height Equivalent to a Theoretical Plate", Ind. Eng. Chem. Res., 2000, 39 (5), pp 1520–1525

I think it may be worth noting explicitly in this answer that the column chemistry is a primary determinant of it HETP and of N.  GC usually uses nonpolar stationary phases such that vapor pressure is a primary determinant of these parameters.  Would the same be true if an e.g. an aqueous stationary phase was used to analyze ammonia gas?


text:
Band gap of Transition Metal Dichalcogenides
What phenomena is responsible for the shift of the nature of the band gap of transition metal dichalcogenides from being indirect band gaps to direct band gap material once it is slimmed down from bulk to its respective monolayers? (I had asked this question in physics stack exchange and nobody bothered with an answer, so I'm here)

Don't repost questions on different sites of SE. You could have flagged it for migration.

I didn't know that such a possibility existed, neither do I have any clue, how it can/could have been done. Plus I hardly had people looking at them, let alone bothering to rack their brains over it. My guess is that experimental work on materials isn't really much of a physicist's thing as per SE

From "The indirect to direct band gap transition in multilayered MoS2 as predicted by screened hybrid density functional theory"[1]:

First, in the bilayer, there is strong splitting in the conduction
  band along the $K$ $\rightarrow$ $\Gamma$ line, which shifts the minimum
  down relative to the $K$ point minimum, leading to an indirect
  transition. Adding additional layers increases this splitting,
  reducing the indirect gap as more layers are added, eventually
  converging toward the calculated bulk value as seen in Figure 1.

In summary, splitting of energy levels is responsible. The atoms in a monolayer are in a different environment than those in the bulk. As a result, the energy levels are split differently. Most of the time, the important splitting occurs in the K point and $\Gamma$ points. In most metal chalcogenides, the valence band of the $K$ point energy decreases in energy as you add more layers. This consistency in behavior is due to the crystal structure of metal chalcogenides. The majority of them form hexagonal structures.
Source:

Applied Physics Letters  99 , 261908 (2011); doi: 10.1063/1.3672219

Interesting!!! Will look more into it.


text:
Why is it impossible to liquefy a gas with only pressure?
Why is it impossible to turn it to liquid by using only pressure after it reaches critical temperature? What happens in the molecular level? If we put enough pressure, the molecules are supposed to come close enough to become liquid. Then why it is impossible?

Why, it is possible with some gases.

With some gases... at some temperatures. You have to be below the critical temperature for the particular gas.


text:
Trends in pi backbonding for main group elements across a period
Recently, I have been re-reading some undergraduate notes and found that there's a period trend in pi backbonding of main group elements (i.e. no transition metals involved) but no explanation were given other than it correlates with the increase in number of nonbonding pairs (lone pairs?):
$\pi$ backbonding strength: Si>P>S>Cl.
While it is well known that the lewis acidity of boron halides are explained by the strength of $\pi$ backbonding which decreases down the group, the same kind of explanation cannot be used for period trends like those above (especially B-P, B-S and B-Si are almost unheard of to check that) as across a period,orbitals in the same principal quantum number will be involved mainly in bonding, thus we should not expect a large difference in the size of the orbitals across a period and hence their extent of overlap due to spatial overlap alone.
In Chemwiki usdavis an explanation of the periodic trend for the same ligand but different metals were provided as an increase in the energy gap of the d, $\pi^*$ orbitals. However since we are not talking about d orbitals here, that is not very relevant
I suspect that since periodic trends have orbital energies in general decreases across the period, that it might be down to the increase in energy gap of the overlapping orbitals in the $\pi$ bonding between a given electron deficient atom and the donor atom of the type listed above. However this explanation has a problem  in that the role of nonbonding pairs are not accounted for in the trend. I then suspect it might be due to an increase in repulsion of the nonbonding pairs, but a repulsion argument seemed to be too simple to explain since the oribtals involve in the same period should have roughly the same size

What is the mechanism or underlying explanation of $\pi$ backbonding periodic trends in main group chemistry?

Pi backbonding to what? and I'm not sure whether this is a valid "trend" if there's no series of compounds to compare.

It never specifies, but suspecting on the context of main group elements (Gp III-VII), the acceptor atom might be an electron deficient lewis acid. B is the most well known electron deficient atom, but there are others such as Al. I don't recall there are stable compounds for any of them except Cl though

Okay. I don't know and am not really inclined towards guessing the context, but what I can offer (don't know if it's relevant) is that the size of the 3p orbitals should decrease going across the period (from Si to Cl) due to increasing effective nuclear charge.

I suppose you mean ligands like phosphine (PH3), hydrogen sulfide (H2S) etc. and not these atoms alone.


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SN1 or SN2 reaction?

Here, this is a 1° allylic halide. If it goes by SN1, the intermediate will be resonance stabilised.
Also, since it is 1° Halide, it can also go via SN2. Which one will be preferred?

Don't forget $S_{\mathrm{N}}2'$...

Sn2 might be preferable because {\ce{CN-}} is a strong nucleophile and they didn't mention of any polar solvent...

This reaction will happen by both SN1 and SN2 mechanism, apart from this SN1' and Sn2' (allylic mechanism ) will also occur. Thus from SN1 and Sn2 one will get primary cyanide and from SN1' and SN2' one will get secondary cyanide .( read organic chemistry by jerry march for the detailed explanation).
I am attaching a screenshot of clayden which explains this beautifully.

Please add the reference for the table and the highlighted block of text.

organic chemistry by clayden - oxford university press - 1st edition -2000 page 424

this reaction does not include any polar solvent so SN2 is prefered over SN1.

Although I haven't found this exact example in the literature, I suspect this reaction is exclusively SN2.
The analysis with respect to the electrophile is good: the allylic leaving group is suitable for both SN1 and SN2. That means that the other conditions will determine which is preferred.
Cyanide is among the best nucleophiles for SN2 reaction. First, it is negatively charged. Second, because it only has two atoms, it is extremely unhindered.
As @CupC_56 suggests, solvent (unlisted) could play a role. However, typically for SN1 reactions, a polar protic solvent acts also as the nucleophile. If one was trying to obtain a nitrile product, it would be a poor choice to run the reaction in the presence of a polar protic solvent that would (1) reduce the nucleophilicity of cyanide and (2) compete with cyanide to react with the electrophile.

Well, Your Question seems  a bit off as the type of solvent has not been mentioned.
Is it:-

Non-polar
Polar (protic or aprotic)

Obviously, SN1 can occur only in Polar protic solvents, and in reactions where the alkyl halide is a conflicting one (reaction can occur both by SN1 and SN2), the mentioning of the said solvent is essential.


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Hydrolysing a compound by repeated methyl and hydride shifts
The question is to predict the major product in the following reaction.

First of all the proton will abstract the bromide ion resulting in the formation of a primary carbocation which being unstable would rearrange.I have the following idea about how the reaction should proceed.

However I have doubts if such repeated methyl and hydride shifts are actually possible.Is the same product possible through a different energetically favourable mechanism?Thanks.

http://chemistry.stackexchange.com/q/42043/16683 same alkyl halide and pretty much a duplicate (I'm not entirely convinced about the answer there though)

The reaction will stop after first methyl rearrangement as it leads to tertiary allylic carbocation which is highly stable, so there will be no further rearrangement, as it does not lead to more stable carbocation intermediate. Thus one will get tertiary allylic alcohol.