text:
Why do electrophillic addition reactions (between hydrogen halides and alkenes) happen?
I understand that the type of solvent (protic or aprotic) may impact the rate of the reaction, but why doesn't the hydrogen just stick with the halide in the first place?
I'm not sure if the reason is just that the hydrogen's electrophilic attraction to the π orbital electrons are just stronger than the hydrogen's electrostatic attraction to the electrons it shared with the halide.
Am I missing something?


text:
Distinguishing among SN2/SN1/E1/E2 for a secondary bromide

a. 2-Bromo-3-phenylbutane reacts with $\ce{NaOMe}$ in methanol. What is the mechanism?
b. This compound then reacted with $\ce{CN-}$ in DMSO. What is the mechanism?

My Attempt
a. I am debating between $\mathrm{E1}$ and $\mathrm{S_N1}$ since it is a polar protic solvent and the carbocation can be stabilized by the tertiary carbon and phenyl group. I am leaning towards $\mathrm{E1}$ since $\ce{OMe-}$ is a strong base. Am I correct?
b. If the answer to the first question was $\mathrm{E1}$, then 2-phenylbut-2-ene would be the product. I don't see how that would react with $\ce{CN-}$ in DMSO. So, would no reaction occur?

I read part b as 2-broom-3-phenylbutane is reacted with cyanide. Otherwise, it doesn't make sense.

Reaction a proceeds via an $\mathrm{E2}$ mechanism. Firstly, elimination reactions are much more favorable in this molecule because of the conjugation of the resulting structure. If the reactant had been 2-bromo-2-phenylbutane, this would have proceeded via an $\mathrm{E1}$ mechanism, as this would delocalize the positive charge around the phenyl ring. There is no such delocalization in the ejection of the $\ce{Br-}$ from 3-bromo-2-phenylbutane though—it would simply be a 2º carbocation. Instead, the delocalization of the negative charge around the phenyl ring resulting form the extraction of the $\ce{H+}$ by $\ce{OMe-}$ is more stabilizing.
$\hspace{1.5cm}$
There are no possible reactions between an alkene and $\ce{CN-}$ in DMSO. If this were a haloalkene, further elimination could occur, and if it were in $\ce{H2O}$, hydrocyanation could occur.
$\hspace{4cm}$

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How can be explained why the specific heat of ice is twice as low as liquid water?
To raise the temperature one degree of ice costs about 2000J but for liquid water it is 4000J.
The reason for that is that the freedoms of vibrations rotation and translations of ice are lower and the molecules of liquid can move more freely.
But I don't understand that. I would say that when molecules can move more freely it would take less energy to raise it one degree, because heat is actionally movement. But why isn't that through?

I think your statement "the freedoms of vibrations rotation and translations of ice are higher" is not correct $\it{vis-\grave{a}-vis}$ liquid water.

I think you are wright, so I changed it,

I seem to recall this has something to do with how energy can go into some bending modes of the hydrogen bonds in water, but that motion isn't available in ice because the molecules are held in place too rigidly.

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Replacing air with nitrogen in a drum.
I have a 55 gallon drum half ful of liquid. I would like to put a "blanket" of nitrogen over the liquid. How can I replace the air in the drum with nitrogen?

Why not use caron dioxide? Much easier.

Is your liquid hazardous in any way?

No but if it works we might look at using it for other applications. Do you foresee any safety concerns? I was trying to avoid safety issues by using Nitrogen.

Just curious about how you were using it to double check that you're being safe :) and to see if there are specialized parts available.  If you check safety equipment suppliers, you can find pumps and gas regulators for drums.


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Preparation of picric acid
My textbook says "Nowadays picric acid is prepared by treating phenol first with concentrated sulfuric acid which converts it to phenol-2,4-disulfuric acid and then with concentrated nitric acid to get 2,4,6-trinitrophenol"
My question: Why is this a better method than simply treating phenol with concentrated nitric acid to obtain picric acid?
What I think is that maybe because $\ce{NO2}$ is a stronger deactivating group than $\ce{SO3H}$, it is more difficult to add 3 $\ce{NO2}$ to a benzene ring directly.
But we would also have to substitute $\ce{SO3H}$ for $\ce{NO2}$. Wouldn't that be hard?

Guess that's because conc nitric acid is too aggressive and would produce a significant amount of other isomeric nitrophenols.

@IvanNeretin is this a case of ipso substitution? Is it possible at room temperature?

Actually it's completely the opposite. Phenol is so strongly activating that you get far too much substitution and tend to end up with a big mess of high molecular weight solids that don't separate easily and aren't picric acid.

Phenolic group can be oxidised and polymerisation occurs.

How is that opposite? True, phenol is much more activated than (say) toluene, hence nitric acid is too strong an agent for it, which results in "too much substitution". So we switch to $\ce{H2SO4}$ which is a weaker agent. Or one may put it the other way around: so we switch to phenoldisulphuric acid which is less activated. Aren't these just two ways to say the same thing?

Is the textbook you are referring to NCERT Class 12 Part 2 Chapter 11?

@varun Yes. That's THE book ;)

Picric acid can be further nitrated to form tetranitro and pentanitrophenol: https://chemistry.stackexchange.com/a/178068/17368

The benzene ring in phenol is highly activated toward electrophilic substitution and hence attempts to directly nitrate it result in charring and copious evolution of oxides of nitrogen. The reaction is highly exothermic and difficult to control.
To reduce the reactivity, the phenol is first mono-sulfonated ( some of the product which is substituted may also be used). The products are ortho- and para-isomers. The para-isomer is separated and then nitrated. The nitration is comparatively far smoother (easier to handle).
Ipso substitution of $\ce{SO3}$ groups occur.

Links: Wikipedia, Research paper
Link: (For ipso substitution): PDF, Wikipedia (see subsection: Ipso substitution)

but how do the nitro groups replace the sulphonic acid ones? I believe desulphonylation requires excess of water hence dilute acidic medium but in the book it's written conc. HNO3. Please help.

@RameshAgarwal Look up IPSO Substitution. I believe the link is attached in this answer.

The
OH
  group of phenol is so activating that the nitration mixture rapidly overheats even with dilute nitric acid. The result is oxidation with the evolution of copious amounts of nitrogen oxides.
Instead, you heat the phenol with sulfuric acid to get mostly phenol-2,4-disulfonic acid.The sulfonic acid groups are so electron-withdrawing that nitration can proceed more slowly and at a controllable rate.
The first product is probably 6-nitrophenol-2,4-disulfonic acid.
In subsequent steps, the nitronium ion attacks at the 2- and 4-positions, displacing the sulfonic acid groups and forming 2,4,6-trinitrophenol (picric acid).

Here is the small scale manufacturing process we are giving to our students in their first organic chemistry laboratory, for their synthetizing and destroying picric acid.
Take $1$ mL concentrated nitric acid $\ce{HNO3}$ ($65$%) in a $\pu{20 mL}$ test tube. Add with care, and in several portions, a tiny amount of phenol (maximum $0.02$ g). The reaction is highly exothermic ! Wait two minutes, then cool down the mixture. Add enough water to nearly fill the test tube (about $20$ mL). The picric acid is an oil, that soon gets solid after one minute or two. Throw the liquid away. Half fill the tube with $\pu{10 mL}$ water. Heat to $\pu{100°C}$ : the picric acid gets dissolved. Wait : nice yellow needles are produced in the tube by slow cooling. It is picric acid, which can be filtered in case of necessity. But don't do it presently. Decant the aqueous solution, and keep the solid in the tube
As picric acid is an explosive stuff when dry, don't dry it and don't save it. Instead, add $\pu{1 mL}$ water to the whole sample of picric acid from the preceding experiment. Add $\ce{1 mL NaOH 1 M}$. This produces a yellow solution of sodium picrate. Add one grain $\ce{Na2S}$. Heat for $2$ minutes. The solution darkens and becomes red-orange. The picric acid has been reduced into picramic acid ($\ce{2-amino-4.6-dinitrophenol}$), which is no more an explosive substance. $\ce{Na2S}$ has been oxidized into $\ce{Na2S2O3}$.

This doesn't answer the question.It just provides a fairly safe small-scale alternative the suggested process in the question.

Picramic acid certainly is explosive.

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Positive vs. negative heat of solution confusion
Given that I have absolutely zero formal (or informal) education in chemistry (sorry about that), I'm struggling to get my head around what appears to be repeated confusions of the term "heat of solution" with regards positive vs. negative. As a specific example, the Wikipedia entry on Erythritol states:

Erythritol has a strong cooling effect (endothermic, or positive heat of solution) when it dissolves in water

However, many other sources, including a number I found via Google Books suggest that Erythritol has a negative heat of solution. For example, this table extract from Food Science, an Ecological Approach:

Erythritol: Noncaloric (0.2 kcal/g), high digestive tolerance, zero glycemic index, does not promote tooth decay, good heat and acid stability, high negative heat of solution.

This is frustrating as I'm trying to find something that has the opposite heat of solution to counteract erythritol's cooling effect, but since many sources seem to contradict each other, I'm struggling to find what I'm looking for.
Which is correct? Or am I using ambiguous terminology?

The heat of solution (aka enthalpy of the solution and $\Delta\mathrm{H}_{\text{sol}}$) is negative if heat is released. The reason is that energy is leaving the system. The reason you got two different answers is that the sources you referenced are describing two different things. The $\Delta\mathrm{H}_{\text{sol}}$ is positive (endothermic) when dissolved in water, but Erythritol is commonly combined with Inulin which produces an exothermic reaction (cancelling out the cooling effect of the Erythritol).

Thanks - I did wonder if it was something like this, because I did stumble across a number of conflicting reports as to the '+/- heat of solution' of erythritol, but none of them seemed to make it abundantly clear if they were talking about the solute or the solvent.


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Liquid that looks like smoke when bottled?
I am making a gift that will consist of a liquid in a sealed bottle. I want the liquid to have a magical appearance. Something like the appearance of smoke - varying opacity from almost clear to almost opaque but never completely mixing.
I am looking for a liquid or combination of liquids with the following properties:

Looks like smoke
Will not become homogeneously opaque over a relatively long period of time (months)
Non-toxic and non-reactive in a household setting (in case it is spilled accidentally)
Relatively
easily found / made assuming access to decently stocked pharmacies,
grocery and hardware stores
Can be coloured would be a bonus

Play nethack and look for "a smoky potion" ;-) To be serious again, it is unclear what do you expect this to look like? In my mind, it would have some movement involved.

Welcome to Chemistry.SE! If I understand, you want something that has two mixed liquids that don't separate, give a "smoky" look and neither separate or become completely opaque. Cool idea, but sounds tricky even without your other criteria.

It seems to me that rheoscopic fluids may be a close fit to the requirements.

Yes, rheoscopic fluids is very close to what I am looking for. Thank you!

So you want to make something that looks like smoke. What makes smoke look the way it does, anyway?
Smoke is composed of extremely fine particles, often below $\mathrm{1\ µm = 0.001\ mm}$ in diameter (check here for comparisons with other particulate materials). These particles are large enough to scatter light, which allows them to be seen, but small enough to be extremely light and stay aloft with even very weak currents in air or liquids.
One interesting and very beautiful consequence of these combined properties is that they can clearly reveal fluid flows, which is what gives smoke its wispy, delicate and complex appearance. That's what fluid flows are like all the time, we just almost never have a direct view of it.
So fundamentally your request is quite simple; all you need is a common, non-toxic source of very fine particulate. Fortunately there are good ways to achieve this in practice.
One of the most popular seems to be using fine mica dust. These non-toxic silicate minerals can be ground to form tiny, almost perfectly smooth crystalline sheets. When the particles align properly, light is reflected into a viewer's eyes giving a shiny, glittery appearance. If the mica dust is suspending in a liquid, then the many millions of particles twinkle momentarily as they tumble chaotically under the action of currents in the fluid.
Liquid mixtures containing similar materials are called rheoscopic fluids. There are many many sources online for ready-made fluids, or you can make your own from mica powder or other fine solids. Suspensions in water eventually settle, but last for a long time without degradation, needing only a shake to awaken. With a myriad of water-soluble dyes at your disposal, you can adjust the fluid to the desired colour.
Though rheoscopic fluids based on water and mica are non-toxic and quite safe, that doesn't mean they pose no hazard whatsoever. Very fine particulate can be an irritant if inhaled or exposed to eyes and mucous membranes (especially a problem if handing the dry powder). Just take a bit of care and you can have fun too! Try putting the fluid on a warm surface and see the convection currents!

Not really sure what you're looking for, but it sounds like you want to make something with liquids that have different densities and do not mix well.
You can find these DIY projects by searching for "ocean in a bottle" or "galaxy in a bottle."  You can make them by adding food coloring, water and mineral oil to a bottle.  For a "magical" appearance, you can add bath products that contain a pearlescent shine, like shampoo, or add glitter.

This is really more appropriate as a comment rather than an answer. It doesn't really provide an answer to the question.


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Can the first ionization energy for a specific element be calculated?
I'm taking a first-year Inorganic Chemistry course in college, and stumbled upon this problem:

For $\ce{Mn}$ and $\ce{P}$, calculate the first ionization energy and explain the difference between the obtained values.

Now, my course materials (written by our professor) apparently say that the first ionization energy can be computed using this formula. I say apparently because the "course materials" are poorly written PowerPoint slides with loads of unexplained terminology and abbreviations, so I may be misunderstanding.
$$
\text{IE} = \frac{Z_\mathrm{eff}^2}{n_\mathrm{eff}^2} \cdot 13.6~\mathrm{eV}
$$
where $Z_\mathrm{eff}$ is the atomic number minus the differentiating electron's shielding constant (calculated using Slater's rules) and $n_\mathrm{eff}$ is pulled out of a table of values (see here).
Unfortunately, it only seems to produce (somewhat) numerically correct results for $\ce{H}$ and $\ce{Li}$, and stops being useful for comparing them at $Z=5$. I wrote an implementation of the algorithm in Mathematica and compared the results it produces to the first ionization energies returned by ElementData (suitably converted from molar to regular), and it's a very rough approximation.
Much too rough, in fact, to be usable for the problem above: it returns $47.38$ instead of $7.43$ for $\ce{Mn}$ and $45.71$ instead of $10.48$ for $\ce{P}$. It sorts them the wrong way around, in other words.
My question is, what am I missing? As far as I can tell from Wikipedia, there's no way to simply calculate ionization energies - they have to be measured. Is there some obvious detail I haven't spotted? Another way to interpret the problem text? Some elementary knowledge I don't have?

The formula I know uses $(Z_\mathrm{eff}/n)^2$. The idea of using either an effective nuclear charge, $Z_\mathrm{eff}$, or an effective principal quantum number $n_\mathrm{eff}$, is that you want to account for the electron-electron repulsions. However, if you use both of them, you're accounting for e-e repulsions twice. Can you try and run your script again just using the unadjusted integer values for $n$?

Actually, scrap that. The difference is way too much - a factor of 4 to 9 - so I don't feel like what I said would make a difference. Wikipedia does also write that Slater was attempting to find one-electron wavefunctions that had both $Z$ and $n$ adjusted. Maybe check your code, make sure it's outputting the correct values of $Z_\mathrm{eff}$. I would help you do it but I'm not familiar with Mathematica syntax at all, sorry.

@orthocresol: tried it anyway - changing from $n_{eff}$ to just $n$ actually makes the numeric differences worse.

Qualitatively, that doesn't seem like it should be the case - since $n_\mathrm{eff} < n$, if you divide by $n^2$ you should get a smaller value for the ionisation energy, which is what you want

@orthocresol: my bad, it was just Mathematica redering the graph differently for some reason. It does give smaller values, but not too much smaller, and of course only for atoms that have the differentiating electron in $n>3$.

The answer would be a tentative yes, but...
You can use quantum chemistry to calculate the ionization energies, but certainly not some simple formula like you presented in the question.

@tschoppi: I see. Is there any chance you could give an overview of the process in an answer?

I'm not really a quantum chemistry geek, so no, I cannot. I simply did some empirical science and looked at the numbers. I hope this can help you.

I tried to find out empirically whether your formula makes sense. The data I used was effective nuclear charges from this Wikipedia page (always the minimal value, Pt was discarded), and ionization energies already gathered for the answer to this question on this site. The principal quantum number was also manually added to the data set.
I explored the data in many ways and tried to find a simple expression which could give an estimate for where the ionization energy should lie.
First, I looked at what your formula would give:

It is quite obvious that your formula describes two atoms perfectly and is absolutely terrible for the overwhelming majority of elements in the periodic table.
I present to you the best guess so far, based on simple manual adjustments and some exploration of the data:
$$ E_\text{ion} / \mathrm{eV} = 6 \left(\frac{Z_\text{eff}}{n^{1.5}} \right)^2 + 4$$
Below is the graphical representation:

The error will still be large for many atoms, I estimate the median error to be about 2 eV. But if you want to use a simple formula to calculate the first ionization barrier, this would be the one I would use.
The data exploration was done in Python and is recorded in this iPython notebook. Feel free to have a look at it and even come up with a better model.


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Platinum vs Silver as Electrode for carbonic acid electrolyte
Is silver a better cathode electrode with carbonic acid in a galvanic cell or is a platinum electrode better with carbonic acid ? The anode electrode is copper with copper sulfate electrolyte, and the salt bridge is sodium chloride.

sorry it should say platinum based electrode

Platinum based? What do you mean?

an electrode made from platinum. A platinum electrode


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Melting point-general chemistry
This one has stumped me for a few days.  What is the specific name of the general chemistry principle that is responsible for the difference in melting points of pure and impure solids.  I have exhausted all other resources available and a precise principle name eludes me.

related http://chemistry.stackexchange.com/questions/30/why-do-impurities-lower-the-melting-point-of-an-isolated-substance

So far as i can remember there isn't a specific name like Charles's Law for gases. It is just known as "melting point depression." https://en.wikipedia.org/wiki/Melting-point_depression

Eutectic mixtures?

This is discussed pretty thoroughly in materials engineering textbooks when melting and freezing of (mostly) metals in general is considered. A good reference is Gaskell’s “Introduction to thermodynamics of materials”. But you will find helpful explnations to fill in the gaps of what uou know by reading up on undercooling in Wikipedia etc. Feezing and melting are related.
And if you read my answer to highest melting point materials, I mention some related issues.
One issue is that you cannot consider materials as “pure” or “impure”: a pure material of any amount large enough to test is in practice almost impossible to make. In fact any real world material is impure once you consider a large enough scale: which is on the order of anything bigger than (say) millions of atoms. You cannot actually make a “pure” material of appreciable size, but can only approach it once you get larger than a relatively small size of sample. In addition, a lot of materials have only been made as thin coatings, or as individual small crystals. And their behaviour is not the same as larger amounts of what you might think is the same material. A single crystal behaves very differently from a polycrystalline mass, and the surface of any material is very different from its interior, so a thin film is dominated by surface effects, as are very fine particles.
So you are looking at degrees of impurity, in terms of orders of magnitude. And the general effect of purity level is not linear but more like exponential in terms of its effect on material behaviour.
A related concept is to look at the theoretical strength of a material. From a theoretical point, a pure and defect free material would be incredibly strong, but in practice this is impossible to achieve, you can only approach it.


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Why do isomeric bis-ethylenediamine cobalt complexes react differently?
A compound $\ce{Co(en)2(NO2)2Cl}$ has been prepared in a number of isomeric forms.One form undergoes no reaction with $\ce{AgNO3}$ or $\ce{(en)}$ and is optically inactive.
A second form reacts with $\ce{AgNO3}$ but not with $\ce{(en)}$ and is optically inactive.
A third from is optically active and reacts with both $\ce{AgNO3}$ and $\ce{en}$. Identify each of these isomeric forms. Name and sketch each of the structures.
My attempt:
I understood how to use the information of reaction with $\ce{AgNO3}$ and optical activity but  couldn't understand what does reaction with $\ce{en}$ represent (I know $\ce{en }$ means ethylene diamine).
In solution, it is given that, if the compound reacts with en it means monodentate ligands are cis to each other,if not it is trans to each other.
I couldn't understand why it is so. please help.

If the monodentate ligands are cis to each other, then one en may substitute them both, which it does quite willingly. Otherwise it can't.

why is it so? how does being cis or trans affect chemical properties?after all stereoisomers have smae chemical properties.

Simple geometry. Ethylenediamine occupies two positions in the coordination sphere, and these two positions must be cis. Trans positions won't do, they are just too far apart.

sir,please explain in detail,what does configuration has to do with reaction with ethylene diammine?and and why should ethylene diammine in molecule be trans to each other for external ethylene diammine to react with the compound.?

I never said that ethylenediamine should (or should not) be trans to each other. This does not even make sense. Likewise, my words don't make sense to you. Sorry, I don't think I can help.


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Why does the acetyl group drop the pH?
I have a molecule that is 8% of the R groups are replaced with Acetyl groups.  When I dissolve it in water it drops the pH.  I am trying to understand why.

Because the electrons in the double bond C=O help to stabilize the CH3 group after it has released an H+ (it is similar in this way to what happens with carboxylic acids), the Acetyl groups are acid. Hence the pH drop.


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How to calculate the dissociation constant of a weak acid from the titration with a strong base?

Problem:
  A solution of an unknown weak acid of unknown concentration was titrated with a solution of a strong base of unknown concentration. During the titration, the pH after adding $\pu{2.00 mL}$ of the base was 6.912. An additional $\pu{14.00 mL}$ of the base was required to reach the equivalence point. Calculate the Ka of the weak acid.

I have decimated six sheets of paper trying different equations. Combining the two ICE tables you can make for each equation, I can still get no fewer than three unknowns. I'm sure I'm overlooking a super-simple solution but I cannot  think of it.
These images are what I have done so far.

It would be useful if you include at least some of the work you've done in detail than generally which you have done here. This is flagged as a homework question but I vote to leave open for now.

This is a strong base weak acid titration. If the weak acid is of the form $\ce{HA}$, then at equivalence point, all the acid has been converted to its conjugate base.
Total volume of base required for equivalence point is $\mathrm{16~ml}$. Let the molarity of base used be $M$. Then, number of milli moles of base for equivalence point is $M\times 16$. This is also the number of milli moles of acid taken.
If only $\mathrm{2~ml}$ of base was used, there will be some acid and some salt of acid and base (A buffer, since the acid is weak and the base is strong).
$$pH=pk_a+\log\frac{M\times 2}{M\times 16-M\times 2}=pk_a+\log\frac{2}{14}$$
From this you will get $$6.912=pk_a-0.845$$
$$pk_a=7.757$$


text:
Comparing two oxidizing agents using electron configuration

Which oxidizing agent from the pair $\ce{K_2IrCl_6/K_2PtCl_6}$ is stronger? Explain your answer using their electron configurations.

We write the electron configurations for $\ce{Ir/Ir^{4+}}$ and $\ce{Pt/Pt^{+4}}$:
$$\begin{align}
\ce{Ir:\ [Xe] 4f^{14}\ 5d^{7}\ 6s^2 \ &\implies \ Ir^{4+}:\ [Xe] 4f^{14}\ 5d^{5}}\\
\\
\ce{Pt:\ [Xe] 4f^{14}\ 5d^{8}\ 6s^2 \ &\implies \ Pt^{4+}:\ [Xe] 4f^{14}\ 5d^{6}}
\end{align}$$
From those, we notice that $\ce{Ir^{4+}}$ has all 5d orbitals half-filled, whereas $\ce{Pt^{4+}}$ has 4 half-filled 5d orbitals and one filled entirely (which is a less stable configuration);
so it's much easier for $\ce{Pt^{4+}}$ to get reduced than it is for $\ce{Ir^{4+}}$.
Therefore, the strongest oxidizing agent is $\ce{K_2PtCl_6}$. What's wrong with my solution?

Read wikipedia's article on crystal field theory. If it won't be sufficient, come again.


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How to synthesise 2-aminoethyl benzene

Starting with benzene, describe the synthesis of 2-phenylethan-1-amine providing all the reagents and conditions.

I am not really great at organic chemistry and my knowledge is pretty limited.
I am currently thinking of forming toluene by Grignard reagent and then reacting with $\ce{Cl2}$ to form benzyl chloride. Then react with sodium cyanide to form benzyl cyanide and then react it with $\ce{HCl}$ and $\ce{H2O}$ to form an amide. Then reduce it with $\ce{LiAlH4}$ to form the compound.
I am not sure if that is correct as it seems a bit complicated for the synthesis of a seemingly simple compound. Also I am not sure on the exact conditions for the chlorination step.
I was also thinking of acylation with ehtanoyl chloride followed by Wolff-Kishner reduction to form ethyl benzene. However I am not sure on how to from the $\ce{NH2}$ group.

Several thoughts:

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

Also, from ethylbenzene, it's very hard to introduce that amine group cleanly. You effectively just have a compound with C-H groups and you have to somehow introduce some kind of functional group into the non-benzylic position. Try to think backwards instead of forward!
The amine you want can be obtained by direct reduction of a nitrile either by simple $\ce{LiAlH4}$ reduction or hydrogenation:

That arrow essentially means "can be made from".
Of course, that's easy to make using a nucleophilic substitution reaction with $\ce{CN-}$ and benzyl chloride (for example):

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

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

I don't like the alkylation step though.


text:
How to calculate amount of nonanoic acid needed to prepare sodium nonanoate salt
I have this procedure for making sodium heptanoate, but I need to adapt it to make sodium nonanoate.
The protocol is as following:

Dissolve $\ce{NaOH}$ in pure ethanol
After all $\ce{NaOH}$ has dissolved, add acid
Precipitates will form, dry the mixture in the oven
Dilute the salt to a working solution

Here is the chemical equation I wrote: (please correct me if I am wrong)
for sodium heptanoate preparation:
$$\ce{NaOH + C2H5OH -> C2H5ONa + H2O}$$
$$\ce{C7H14O2 + C2H5ONa -> C7H13NaO2 + C2H5OH}$$
for sodium nonanoate preparation:
$$\ce{NaOH + C2H5OH -> C2H5ONa + H2O}$$
$$\ce{C9H18O2 + C2H5ONa -> C9H17NaO2 + C2H5OH}$$
Question:

How can I calculate the amount (in ml) of nonanoic acid I need to get 1L of solution?
Does the amount of $\ce{NaOH}$ dissolved in absolute ethanol matter? Because I have found related protocol using 40 g $\ce{NaOH}$ and some using 60 g $\ce{NaOH}$

Please help me to understand step by step, since I am not a chemistry student. I am a biology student and I need to prepare the salt for my experiments.
I have included the picture of the protocol for your reference. *I know the answer is 180ml for heptanoic acid preparation, but how do I get to the answer? I need to know the calculation steps, so I can calculate the ml of nonanoic acid I need to make sodium nonanoate.

As a curiosity, why not just follow the second procedure that you have posted and see how much you get. You have broken down the procedure pretty well. If you do not get enough sodium heptanoate for your purposes, do it again.

Also, the 40 g of sodium hydroxide is to produce a different salt from a different carboxylic acid. The amount of NaOH was chosen to balance the stoichiometric equivalents of acid.

I wanted to know how to calculate the amount of acid I need because I need to use a different acid for my experiment.
I need to make sodium nonanoate using the same protocol, but I dont know how to calculate the amount of nonanoate acid i need...

Okay, then! Edit your question so that is clear. Start with something like, "I have this procedure for making sodium heptanoate, but I need to adapt it to make sodium nonanoate."

Thank you Ben Norris, I have edited the question. I hope it is clear now so that someone can teach me how to do the calculation.

According to the first procedure,  approximately 1 mol of  valeric acid  reacts with 1 mol of NaOH. The reactions are quantitative. In the second procedure, 1.5 mol of NaOH reacts with 1.3 mol of hepatanoic acid. i.e they use a little excess of NaOH in the second approach. The key issue here is to know the molar mass and the density of the nonaoic acid to determine the approximate volume you need to add.


text:
IUPAC Name of n-propyl benzene?
What is the IUPAC name of n-propylbenzene? Is it propylbenzene or 1-phenylpropane, i.e. should the ring or the chain be considered the substituent?


text:
Stoichiometry of alkane-perchlorate reaction

What are the products of an alkane and a perchlorate reaction?
e.g., making the alkane methane and the perchlorate anhydrous for simplicity:

My effort:
$$\ce{Mg(ClO4)2 + 4CH4 -> MgCl2 + 4CO2}$$
leaves 16H free so obviously incorrect, but shouldn't a salt and a gas be part of the products? I guess that the anhydrous form would make this explosive.

I think that there is some $\ce{H2O}$ as product:
$$\ce{2Mg(ClO4)2 + 4CH4 -> 2MgCl2 + 4CO2 + 8H2O}%edit$$


text:
What is the correct discharge reaction in nickel-cadmium cell?

In school textbook

$$\ce{Cd +2Ni(OH)3->CdO + 2Ni(OH)2 + H2O}$$

Wikipedia

$$\ce{2NiO(OH) + Cd + 2H2O->2Ni(OH)2 + Cd(OH)2}$$

Chemwiki

$$\ce{Cd +NiO2 + 2H2O->Cd(OH)2 + 2Ni(OH)2}$$
Which reaction should I follow?

The ChemWiki page got updated on 22nd Jan 2017, an year after this question was posted, and the reaction in question has since been updated. The reaction on the ChemWiki page is now identical with that on the Wikipedia page. This question now has no significance whatsoever but for the sake of the time and effort of the answerer, I am keeping it undeleted

If you're studying a course, follow the textbook.  There will be multiple possible cell configurations for a "Ni-Cd" cell, depending on their configuration they may have different properties such as how the voltage varies during discharge.  Your "chemwiki" equation doesn't balance, check this and this

Note that the three equations are different only in whether the nickel and cadmium oxides are hydrated (and thus where water appears in the equation). Also, the Chemwiki equation is not balanced because nickel oxide is $\ce{Ni2O3}$ and not $\ce{NiO2}$.
The correct equation for the ChemWiki version is:
$$\ce{Ni2O3 + Cd + 3H2O -> 2Ni(OH)2 + Cd(OH)2}$$
Many metal oxides react with water to form hydroxides. This kind of behavior is probably most frequently taught with the alkali metals and alkaline earth metals:
$$\begin{align}
\ce{2NaO + H2O} &\ce{-> 2NaOH}\\
\ce{CaO + H2O} &\ce{-> Ca(OH)2}\end{align}
$$
However, other metals undergo these kinds of reactions. For example, the chemistry of rust involves the equilibrium between iron oxides and iron oxide hydroxides:
$$\begin{align}
\ce{FeO + H2O} &\ce{<=> Fe(OH)2}\\
\ce{Fe2O3 + 3H2O} &\ce{<=> 2Fe(OH)3}\end{align}
$$

Nickel

The three ionic nickel species in the reactants are different only on the degree of hydration:
$$\begin{align}
\ce{Ni2O3 + H2O} &\ce{<=> 2NiO(OH)}\\
\ce{NiO(OH) + H2O} &\ce{<=> Ni(OH)3}\end{align}
$$

Cadium

The two ionic cadmium species in the products differ only by degree of hydration:
$$
\ce{CdO2 + H2O<=> Cd(OH)2}
$$


text:
Does an atom want to give or take electrons when it has a shell half full?
For example, carbon has 4 electrons in the outer shell and so its half full. Does it want to give electrons in a bond or take electrons, and if any atom has a half full electron shell does it want to take or give electrons for a covalent bond?
I'm new to chemistry so this question may be simple.

When it comes to elements like carbon and silicon, which have 4 valence shells, the maximum possibility is sharing of electrons to form covalent bonds, like in $\ce{CO2}$ or $\ce{CH4}$. However in carbides, like calcium carbide or tungsten carbide, it may gain electrons.

Thanks also
If an atom like Boron has only 3 electrons in the outer shells, will it give up its electrons for a bond? and vice versa for nitrogen where it would take electrons for a bond?

Atoms with 1,2,3 electrons lose electrons to form cations(positively charged) while atoms with 5,6,7 electrons gain electrons to form anions(negatively charged)  (or share as discussed)    So boron will give 3 electrons and Nitrogen will accept electrons...So yeah u're right

Except when they don't. There are a whole range of cations with positive charges on nitrogen whereas boron cations are quite rare. Much more common are electron deficient covalent boron compounds.

Boron cations are rare due to much less electropositivity( refer to periodic trends in the 2nd period of the periodic table) And when it comes to positive charge, are you referring to ionization due to polarity?

Boron forms anions, but as clusters or macromolecular structures rather than single atoms. See here.


text:
What is the qualitative definition of enthalpy in beginner terms?
So in my coursebook it says enthalpy is a property that accounts for the increase in internal energy and the accompanying work done by the system and that its equal to the sum of the internal energy $U$ and product of pressure $P$ and volume $V$. Also on the net, mostly on the internet it says:

A thermodynamic quantity equivalent to the total heat content of a system. It is equal to the internal energy of the system plus the product of pressure and volume.

And where it explains in other words I don't quite understand, e.g. as on Wikipedia. So please I would like anyone of you to explain (in beginner terms) that what is the meaning of enthalpy and heat content as used in the definition above.

I got this from Wikipedia:

The $U$ term can be interpreted as the energy required to create the system, and the $pV$ term as the energy that would be required to "make room" for the system if the pressure of the environment remained constant. When a system, for example, $n$ moles of a gas of volume $V$ at pressure $p$ and temperature $T$, is created or brought to its present state from absolute zero, energy must be supplied equal to its internal energy $U$ plus $pV$, where $pV$ is the work done in pushing against the ambient (atmospheric) pressure.

I think it answers my question. Enthalpy is just a property of system we defined essentially to make it easy to calculate the heat of the reaction as Chester Miller says, but its definition is the energy present as particular kinetic and potential energies in addition to the energy needed for the system to exist in its present condition by doing work against the surroundings.

In my judgment, it's not worth spending much of your valuable time trying to assign physical significance to the physical property we call Enthalpy.  Enthalpy is just a convenient parameter to work with when we are doing certain types of thermodynamic analysis.  One situation which it is particularly useful for is in applying the first law to open systems in which material is entering and/or leaving the system.  Another is where changes in a closed system are occurring at constant pressure.


text:
Redox Reactions with buckyballs
I have read that buckyballs are excellent electron acceptors, C60 in particular can accept up to 6 electrons.
However I do not understand why they have this ability and how they can accept a large number of electrons. I would appreciate if anyone could elaborate on this topic.

It's only 1e for 10 C atoms nothing special

you should look into chemistry of aromatic hydrocarbon anion-radicals and molecular orbital theory.

Triply degenerate LUMO, that's why.


text:
Using electron affinity to calculate lattice energy
I am trying to calculate the lattice energy of $\ce{CaBr2}$, I am uncertain of how to incorporate the electron affinity value provided for $\ce{Br2}$ into my calculation.
Since electron affinity is the negative of the energy change that occurs when an electron is accepted by an atom in a gaseous state in the form of an ion, will I have to take the negative value of electron affinity and double it?
At this point in the Born-Haber Cycle $\ce{Br2}$ is already dissociated, and since calcium looses two electrons while forming $\ce{CaBr2}$, would it be logical for the total energy change to be a (negative) double of the electron affinity of bromine ?


text:
Fluorine Preparation:How can this be explained on basis of Lewis Acid-Base Reaction?
How can this be explained on basis of Lewis Acid-Base Reaction ?
$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$
References:
https://en.wikipedia.org/wiki/Fluorine#Industrial
Google Books reference (click me)

$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF3 + F2↑}$
Occurs in two stages. First there is a (Lewis) acid-base reaction
$\ce{2 K2MnF6 + 4 SbF5 → 4 KSbF6 + 2 MnF4}$
where the Lewis acid $\ce{SbF5}$ accepts a lone pair from a fluorine in the manganese complex, and the adduct then falls apart to leave $\ce{MnF4}$. This is thermally unstable at ambient temperatures (Housecroft and Sharpe) and falls apart in the redox reaction
$\ce{2 MnF4 → 2MnF3 +F2}$

This is actually a double Lewis acid-base reaction. One Lewis acid-base complex is broken down while another is formed.
The starting compound potassium hexafluoridomanganate(IV) or $\ce{K2[MnF6]}$, features a complex hexafluoridomanganate(IV) ion which is a Lewis acid-base adduct. Upon adding $\ce{[SbF5]}$, we suddenly have two Lewis acids — $\ce{[MnF4]}$ and $\ce{[SbF5]}$ — both competing for the Lewis bases $\ce{F-}$. It turns out that $\ce{[SbF5]}$ is the stronger Lewis acid, i.e. the acid-base equilibrium is shifted to the product side.
$$\ce{[MnF6]^2- + 2 [SbF5] <=>> [MnF4] + 2 [SbF6]-}\tag{1}$$
Thus, the more stable Lewis acid-base adduct has been formed. Manganese(IV) fluoride is not stable with respect to degradation, i.e. the redox decomposition to form manganese(III) fluoride and fluorine is favourable.
$$\ce{2 MnF4 -> 2 MnF3 + F2 ^}\tag{2}$$


text:
How to name the tautomers of hexane-2,4-dione correctly according to IUPAC rules?
Today I was considering to answer the question "Which tautomer of hexane-2,4-dione is more stable?" I wanted to run a few calculations and while preparing a directory structure I wanted to make sure that I used correct and unique names for the compounds. I am citing the picture in the named question:

I was going to derive the names of the structures from the parent compound hexane-2,4-dione. Hence I would have named compound 1 (Z)-2-hydroxyhex-2-en-4-one and 2 (Z)-4-hydroxyhex-3en-2-one.
When I recreated the above picture in ChemBioDraw and used it to autogenerate the name, the result for 1 was different: (Z)-5-hydroxyhex-4-en-3-one. See below.

I am no expert on nomenclature and I do not have the blue book handy. Usually I rely on the names generated through software. Most of the times they are accurate enough to name a compound unambiguously and that is often what I am going for.
I understood from this answer, P-14.3.5 Lowest set of locants, that I should use the 4,2,2 set over the 5,3,2.
Is my thinking correct and the auto-generated names are wrong, or is the software correct and my thinking is wrong?

You numbered all locants together in a series and you found the lowest set of locants according to Subsection P-14.3.5 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) since the locant set ‘2,2,4’ is lower than ‘3,4,5’.

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

This approach would be correct if all locants refer to detachable prefixes (like ‘2-hydroxy’). However, in the given compounds, the locants refer to prefixes (‘hydroxy’), endings (‘ene’), and suffixes (‘one’). The assignment of low locants to such different features is stipulated in Subsection P-14.4 as follows:

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(e) saturation/unsaturation:
  (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
  (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(…)

Therefore, the resulting name is (4Z)-5-hydroxyhex-4-en-3-one (not (2Z)-2-hydroxyhex-2-en-4-one) since the locant ‘3-one’ for the principal characteristic groups is lower than ‘4-one’.

You will assign lowest locant to the group with highest priority. Here priority order is
Keto>hydroxy>alkene. So use that numbering which gives lowest locant to keto group. So I think the names given by software are correct.

I would really appreciate if you could back that up with the official sources, hence the reference request.

Ok. Sure. I will definitely try to provide a good reference. That's the logic I have learnt and it also matches with software.

One of the reference is 'Organic Chemistry' by John E. Mcmurry. There is an article in the book stepwise describing nomenclature of polyfunctional compounds. It also gives a priority order for common functional groups


text:
What chemicals used everyday will cause an explosive reaction if exposed into outer space?
What chemicals used everyday will cause an explosive reaction if exposed into outer space?

Well, you're asking for a list, and we don't do lists here. And that whether demanding an answer is acceptable is debatable.

I figure that if something is going to explode, it's much more likely to do so under our oxygenated atmosphere than in vacuum.

One example for an everyday chemical that most probably will cause an explosion if exposed into outer space is liquid water at temperatures significantly above 300 K when hold in an inappropriate container.  This phenomenon is called a BLEVE.  But be aware that this is not an chemical reaction.


text:
Resultant temperature of mixture
**Question: ** There are 2 flasks of equal volume, connected by a narrow tube of negligible volume, filled with nitrogen gas. When both are immersed in boiling water, the gas pressure inside the system is 0.5 atm. Calculate the pressure of the system when one of the flasks is immersed in ice water mixture, keeping the other in boiling water.
Problem: There are two ways that I am going to describe that I can think of to solve the question, but they give different answers.
Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry. As both the flasks are identical and at 100 and 0 so it should be 50.
Calculating the answer with this assumption and the formula
P1/V1=P2/V2
The answer comes out to be about 0.432 atm
Way 2: The sum of the moles of gas in both the flasks would remain same.
And using the ideal gas equation to get n = number of moles = PV/RT
After equating the values. The pressure come out to be about 0.422 atm
I was bothered with the 0.1 change. So I took the liberty to calculate the temperature from way 2, and it came out to be about 41.82 degree Celsius.
Note : calculations were done in Kelvin itself.
So we assumed that temperature would be 50 but way 2 contradicts that. I could not find a fault in either of the ways so could not understand what temperature would it really be.
Attempt to solve: is it something about Gibbs Paradox? I was Googleing and that came, but I couldn't understand...
Please help with proper explanation on why the other way is wrong.

I see no reason to believe that the gas temperature in both flasks will be the same.

@IvanNeretin why not? Wouldn't it exchange the gas until it reaches equilibrium in temperature

It would, if the flasks were not heated and cooled. The way we have it, though, there would be a constant stream of heat from boiling water into one flask, and from another flask to icy water. Would the effects of those be totally negated by the heat exchange between the flasks though the tube (narrow tube, mind you)?

@IvanNeretin ahh I am afraid I do not completely understand you. This is a high school question, and the reason that they say that the tube is narrow is I think needed to neglect the volume. But you might be right, so could you write more simple explanation?

I think the tube is there just to equate the pressure; as for the temperature, it will be $0^\circ\rm C$ and $100^\circ\rm C$, respectively. Oh, and this has nothing to do with Gibbs paradox, BTW.

In the first condition when both flasks are boiling, the temperature of the gases in both flasks are the same (100 °C).  Using the ideal gas law on the system that comprises both flasks, where $n$ is the number of moles initially in each flask and $V$ is the volume of one flask.
$$ P_{init} = \frac{(2 n) R\;T_{boil}}{2 V} = \mathrm{0.5~atm}$$
$$ 2n = \frac{P_{init}(2 V)}{R\;T_{boil}} $$
Now if Flask #2 is immersed in ice water, we should assume that the pressure in the system will be constant but that the flasks will have different temperature.  For the pressure to be constant at different temperatures, material will have to flow from the hot flask to the cold flask; now, $n_1$ and $n_2$, the moles of gas in flask #1 and flask #2, will not be the same, but since no gas leaves the system comprising both flasks, then we know $n_1 + n_2 = 2n$
$$ n_1 = \frac{P_{final}V}{R\;T_{boil}} $$
$$ n_2 = \frac{P_{final}V}{R\;T_{ice}} $$
$$ n_1 + n_2 = 2n$$
Substituting:
$$ \frac{P_{final}V}{R\;T_{boil}} + \frac{P_{final}V}{R\;T_{ice}} = \frac{P_{init}(2 V)}{R\;T_{boil}}$$
$$ P_{final}\frac{V}{R}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}V}{R\;T_{boil}}$$
$$ P_{final}\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)= 2 \frac{P_{init}}{T_{boil}}$$
$$ P_{final} = 2 P_{init}\frac{\frac{1}{T_{boil}}}{\left(\frac{1}{T_{boil}} + \frac{1}{T_{ice}} \right)} = 2 P_{init} \frac{\frac{1}{T_{boil}}}{\frac{T_{ice} + T_{boil}}{T_{ice}{T_{boil}}}}$$
$$ P_{final} = 2 P_{init} \frac{T_{ice}}{T_{ice} + T_{boil}}  $$
This answer makes sense in that (i) if the temperature $T_{ice}$ is actually the same as $T_{boil}$, then the final pressure would be the same as the initial pressure, and (ii) the answer doesn't depend on the volume or moles of gas in the flasks, just the temperatures and the initial pressure.

Way 1: Net Temperature of the system should be 50 degree Celsius because there is no heat loss and there is symmetry.

This way is incorrect because a flask immersed in boiling water ($T=$100°C) is unlikely to have a temperature of 50 °C.  Similarly for a flask immersed in ice.

Way 2: The sum of the moles of gas in both the flasks would remain same.

This is the correct method.  It does require assuming that the rate of heat transfer from the boiling / ice baths to the flasks is much much greater than the rate of heat transfer between flasks via diffusion through the tube.  However since the tube is small, this assumption is probably OK.

The gas law is applicable when both elements are at same temperature but it isnt the case here so your way 2 is right

No, I am doing that for the system. And the temperature in the system. Is the same :/


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de-yellowing lego
I have a huge lot of yellowed Lego's, I want to try to de-yellow them.
I have searched around the internet and this method seems the most promising:
http://www.retr0bright.com/
I have a few questions about this :
 - What is happening at the surface of the ABS plastic ?
 - My knowledge on this subject is very limited, but would the process go faster when you perform the process in a ultrasonic bath ?
A bit of unrelated question :
- Is it possible to combine the de-yellowing with filling any scratches gaps in the bricks ? It seems not to me because you have to have some filler in the solvent or break the bonds in the lattice, i guess ?
any help, information is welcome, I'm like to learn new stuff !!
I can only understand it a little bit from a nanotech point of view.

According to the retrobright website the process involves the use of activated hydrogen peroxide and possible UV light. This won't work as well in an ultrasonic bath as the bath will keep out most of the light.
Yellowing is thought to be caused by some degradation reaction of some of the additives in the ABS plastic (probably brominated flame retardants, though it could be something else). What is probably happening is that the yellow products of that degradation are being destroyed by activated peroxide. The UV light probably helps this reaction along a little faster, possibly by activating the degradation products (a yellow colour is often a sign that there is a strong absorption of light from the molecules in the near UV).
If you follow one of the recipes on their site (with their safety precautions) it looks as though it should work for Lego.
Filling scratches, though, will probably not be easy as this is not a chemical problem but a physical one. The only effective way would probably involve remelting the ABS and recasting it which seems like overkill (especially since you would need original Lego moulds). Any chemical that will dissolve the ABS is more likely to destroy it than to selectively reshape the surface.

Also, hydrogen peroxide threatment may lead to significant brittling of the plastic.

What does activated in activated hydrogen peroxide mean?

@wuschi hydrogen peroxide doesn't work well when cold. Certain chemicals can enhance the bleaching power (the TARP used here is a common additive used to enhance peroxide bleaching in low temperature laundry). Basically TARP reacts with peroxide to give a more active intermediate.

@matt_black thanks for your awesome answer, if i do analysis with the spectroscope i could see whether bromine is the cause ? or i could perform a afm sweep to determine different materials ? I know there is another method to determine the material properties (forgot the name).  if i have more questions about selective reshaping the surface, should i open a new question ?

@pwghost Optical/UV spectroscopy will certainly tell you whether the treatment is working but it might be hard to verify what the actual coloured contaminants are. Other methods might be better. It would be a very good experiment to do as it would contribute useful results to science if you can identify the degradation products.

@matt_black thanks for your comment, what kind of analysis method are you proposing ? maybe i can get access to the equipment to do some experiments ?


text:
Looking for information about possible ingredients of a fertilizer
I have found it in a local store, "20.20.0 - Kompoze" is written on the pack. In the pack, there are yellowish greenish irregular formed little balls. What is this, do you know ? Please help and excuse my poor english.

well , maybe you should have had googled it or spoke to the retailer about it . i dont know why you have no but its constituents are 20% nitrogen (N) and 20% phosphorus (P2O5). Afterall there must be NP written above 20-20-0 and these both must must be written above the KOMPOZE and this all will be enclosed in a red disc.and below all this will be a trademark of toros.


text:
What is an entropic trap?
Can someone explain the concept of entropic traps, or provide good sources to look up? The term is not present in any chemistry books I own, and I cannot find online sources explaining it.

Can you give some context of where you came across the term?

I think this should address your query:

The separation of macromolecules such as polymers and DNA by means of electrophoresis, gel permeation chromatography or filtration exploits size-dependent differences in the time it takes for the molecules to migrate through a random porous network. Transport through the gel matrices, which usually consist of full swollen crosslinked polymers depends on the relative size of the macromolecule compared with the pore radius. Sufficiently small molecules are thought to adopt an approximately spherical conformation when diffusing through the gel matrix, whereas larger ones are forced to migrate in a snake-like fashion. Molecules of intermediate size, however, can get temporarily trapped in the largest pores of the matrix, where the molecule can extend and thus maximize its conformational entropy. This 'entropic trapping' is thought to increase the dependence of diffusion rate on molecular size. Here we report the direct experimental verification of this phenomenon. Bragg diffraction from a hydrogel containing a periodic array of monodisperse water voids confirms that polymers of different weights partition between the hydrogel matrix and the water voids according to the predictions of the entropic trapping theory. Our approach might also lead to the design of improved separation media based on entropic trapping. – Source

I don't know if you needed a textbook definition, but thought you'll get a more elaborate view from this.

This is a schematics of an entropic trap device(ref: Anal. Chem. 2002, 74, 394-401)

So basically the energy of a molecule is lowered because the conformational entropy is maximized? In the case above, macromolecules are allowed to "extend" (i.e. stretch out, occupy a larger volume) in suitable pores, which then becomes a favoured state for the molecule ("trapped" in a local minima on its potential energy surface). But why is it trapped? Is the minima deep? Can it not easily escape? Could you explain how the entropic trapping works for chemiluminescence of 1,2-dioxetane? (see http://pubs.acs.org/doi/abs/10.1021/jp074063g)


text:
Energy required to break salt bridges in peptides and proteins
Is there a way to calculate the energy required to break a salt bridge in a peptide or on the exposed surface of a protein in aqueous solution?  Is there any useful literature on this topic?  I'm looking to disrupt salt bridge interactions, and was wondering whether it would be feasible to do this by heating the sample, or whether infeasibly high temperatures would be required!
Thanks in advance for the help!


text:
Is there a chemical treatment for pipes to reduce or eliminate leaching of lead into tap water?
Given events in Flint, Michigan, where a switch to acidic river water in the city water supply resulted in leaching of lead into tap water from the insides of pipes, I wonder -- has anyone has heard of a chemical treatment for pipes that would reduce or eliminate further leaching of lead into the water?

I found this http://cen.acs.org/articles/93/web/2015/03/Lead-Dioxide-Coating-Pipes-Help.html

The best way to avoid leaching of lead into water is to treat the water not the pipes. Leaching is highly dependent on what is in the water and it is is far easier to fix the water than to treat every pipe. Though, of course, the best solution is to completely replace the lead pipes with plastics or copper.

In some cases the Langlier Saturation Index can be adjusted so a small carbonate scale barely forms on a pipe to prevent leaching.  The opposite may have happened to the index making the water corrosive when the water supply was switched.
This adjustment must be done by qualified professionals because if this index is too high the pipes will clog.


text:
Importance of Overall Reaction Order
All my texts make it a point that in a differential rate law, the sum of the exponents on the reactant concentrations add up to the overall reaction order. However, I have yet to see any use in this. So you have two reactants, each undergoing a first-order reaction, and the overall reaction is second-order. So what? We analyze reactions for being first-order, second-order, and zero-order (at my level anyway), but that hasn't to do with the overall reaction order—just the reaction order pertaining to each individual reactant.
So my question is simple: what is the point of the overall reaction order? Why is it important?

In a situation where a reaction mixture is diluted by half by adding water, the rate of reaction for an overall 2nd order reaction will be a quarter, whether the rate = k[A]^2 or rate =k[A][B].  If you knew that overall a reaction was second order, but also that it was first order wrt [A], you can deduce it must also first order wrt [B] without an explicit experiment.

It changes the mathematical form of the solution for the relationship between concentration and time, particularly for A reacting by itself compared to A + A reacting.

@ChesterMiller How so?

Suppose that for a first order rate relation you have $\frac{dA}{dt}=-kA$, while for a 2nd order rate relation, you have  $\frac{dA}{dt}=-kA^2$.  Do you think the time dependent changes in concentration will be the same for both?

This answer is mostly a compilation of the excellent comments by Spontification and Chet Miller.
Differential rate law
The overall reaction order tells you what happens to the rate when you change all concentrations by the same factor at the same time. For example, if you decrease all concentrations by a factor 2 and the overall order is 1, the reaction will be slower by a factor of 2. If the order is 2, slower by a factor of $2^2 = 4$, if the order is 3, slower by a factor of $2^3 = 8$ and so on (overall reaction orders higher than 3 are unusual for elementary reactions). Dilution by a factor 2 would be one way of lowering concentrations; however, not all species will decrease by a factor of 2 (solvent, hydronium concentration in a buffer, any other species buffered by a fast equilibrium).
Integrated rate law
Integrated rate laws describe how the rate changes (decreases) over time. All species concentrations that are part of the differential rate law will change over time unless they are catalysts. If one species concentration is very high compared to others (e.g. the solvent) then its change will be negligible. So the integrated rate law depends on more than the differential rate law.
However, if all of the species in the differential rate law are reactants, and these reactants are present at stoichiometric ratios, the overall reaction order is directly related to the integrated rate law (e.g. overall second order reaction gives second-order integrated rate law). If not, you can get, for example, a pseudo-first order reaction even though the overall order of reaction is two.


text:
How to predict precipitate upon evaporation from calcium ions and either chloride, carbonate, or nitrate counterions?

An aqueous solution containing $\ce{Ca^2+}$, $\ce{Cl^-}$, $\ce{CO3^2-}$, and $\ce{NO3^-}$ is allowed to evaporate. Which compound will precipitate first?

The correct answer is $\ce{CaCO3}$. I had guessed that, and I know nitrates are soluble, so that wouldn't precipitate, but why the $\ce{CO3^2-}$ and not the $\ce{Cl-}$? Is it because of the equal and opposite charges between them? Does electronegativity play a role?

Because calcium chloride is fairly well soluble, too. In fact, $\ce{CaCO3}$ would precipitate right away, not waiting for the solution to evaporate. There is no simple rule; it has nothing to do with electronegativity or charges.

Oh! Duh! Its a halide! I can't believe I missed that

This question is testing your knowledge of general solubility trends, which largely need to be memorized. Here is a fairly comprehensive list, and there are many others scattered through the internet.
You are correct that nearly all nitrates are soluble, so that would be a bad guess.
The halides are generally pretty soluble. There are more exceptions than for nitrates, but as a rule halide salts tend to be soluble.
Carbonates however are frequently insoluble. The Group I cations tend to be an exception, but not the Group II cations, like $\ce{Ca^2+}$.
So, the general trend in solubility for the three anions in your question is $\ce{NO3- > Cl- > CO3^2-}$, so you would expect $\ce{CaCO3}$ to precipitate first.
We can verify this against the known solubilities of each of the salts at room temperature:
$\begin{array}{rr}
\pu{\ce{Ca(NO3)2}}: & \pu{1212 g/L }\\
\pu{\ce{CaCl2}}: & \pu{75g/L}\\
\pu{\ce{CaCO3}}: & \pu{0.01g/L}
\end{array}$
So, the quantitative solubility data confirms the original prediction and $\ce{CaCO3}$ will clearly precipitate first.

Frankly I hate problems like this. The problem didn't specify that equal amounts of the anions were present...

@MaxW, I guess one aspect of this problem that diminishes that to a fair degree is the ~3-5 orders of magnitude difference in solubility between $\ce{CaCO3}$ and the others. I think the fact that you can only dissolve 10 mg of $\ce{CaCO3}$  in a liter of water makes this a pretty impractical physical experiment though. I mean, you start off with a 100 ml of solution and you're basically just left with dirty glassware!


text:
Is activation energy temperature-independent?
I know that activation energy for a reaction is the extra energy given to the reactants to reach the threshold energy so that they can collide and react. But then, why is it said that the activation energy does not depend on the temperature?
If the temperature is higher, the reactant molecules have a higher kinetic energy, and thus the energy difference between the threshold energy and the energy of the molecules will decrease, that is, the activation energy will decrease.

The activation energy $E_\mathrm a = RT^2(\mathrm d \ln k /\mathrm dT)$ [IUPAC Gold Book] is not actually temperature independent, except for within the simplest models. The temperature dependence is usually small, but not zero. This can be seen from, for example, transition state theory analysis. (For various other reasons it may in fact be large, e.g. if there is a significant quantum tunnelling contribution to the rate.)

Here is an example of non-Arrhenius behavior: https://pubs.acs.org/doi/pdf/10.1021/j100310a026

There is an answer here https://chemistry.stackexchange.com/questions/139196/why-does-activation-energy-not-change-with-temperature-change/139222#139222

There is a simple (some might say simplistic) way to get an intuition about this and it involves thinking at the molecular or atomic level rather than about the bulk properties of the reaction (the thermodynamic view).
For a simple reaction where molecule A has to bang into molecule B to create molecule C, the reaction will only happen if the amount of energy involved in the collision is large enough. The amount of energy involved depends on the nature of the reaction which is determined by the electronic structures of A and B. Within limits, those electronic structures don't vary with temperature and a given reaction will only occur when there is enough energy in the collision for the reaction to happen.
What changes with temperature isn't the energy required for the reaction to occur, it is the number of molecules that have enough energy to overcome the barrier. At higher temperatures more molecules have that energy, but the amount of energy required doesn't change. The amount of energy require to initiate the reaction is roughly equivalent to the activation energy. It can be calculated by looking at the way the overall rate of the reaction changes with temperature. So the rate of reaction changes with temperature but not the barrier to the reaction happening.
This view is a little simplified but it gives the right intuition about the problem.

I disagree with your 3rd para. "At higher temperatures more molecules have that energy, but the amount of energy required doesn't change". By amount of energy are you referring to activation energy or activation free energy? If you mean activation energy it's fine, but if you mean activation free energy than it's not right. You are only considering electronic contribution, but at high temperature you also have high vibrational, translational, and rotational contribution. They can change the activation barrier dramatically.

@mamun I did say I was simplifying. And I did mean activation energy. At high temperatures and for some types of reaction this picture won't be right, but it explains the key idea well enough for an introduction and creates some basis for discussion the complications, I hope.

"Within limits, those electronic structures don't vary" You should elaborate more on 'within limits'.

@Mockingbird In extremis if you add enough energy you turn everything into a plasma and every molecule disintegrates. Before that some molecules will fall apart anyway if enough thermal energy is added. But we don't normally think about, for example, the kinetics of bimolecular reactions under such extreme circumstances.

As mentioned above, it is best to first assume that activation energy is temperature independent. The number of molecules that have sufficient energy to do the reaction, can now be calculated via the Boltzmann distribution. The textbook "Physical Organic Photochemistry" from Anslyn & Dougherty has a very good picture to visualize this.

Note that Eq. 7.17 is the Arrhenius equation, and Eq. 7.20 is the Boltzmann distribution.
Another thing mentioned is the dependance of Activation energy $E_a$ on temperature $T$. This is best illustrated by Figure 7.6 of the Physical Organic Photochemistry-book; it shows that the reaction coordinate diagram (shown on the right) must actually be seen as an energy surface (shown on the left).

At higher temperatures, there is more vibrational/rotational/translational energy, in total there are 3N degrees of freedom. Only one of the vibrations will lead to an effective reaction; e.g. the bond of interest to be broken. This is why the length of that bond is chosen as a reaction coordinate.
Thermal energy, however, will be distributed over all degrees of freedom, according to the Maxwell–Boltzmann distributions. The point is that, at a higher temperature, there will be more paths over the transition state than just the path that walks in the valley of the energy surface.
Back to the question. Does activation energy depend on temperature? Well, the minimum activation energy, or enthalpy of activation ($\Delta H^\ddagger$) does not change, but the effective activation energy ($\langle E_a \rangle$, or actually simply $E_a$), which is essential for the macroscopic property of a rate constant, does increase, because paths other than the minimum require slightly higher activation energy.
According to page 368 of "Anslyn and Dougherty - Modern Physical Organic Chemistry", the effective activation energy  ($E_a$) can be connected to enthalpy of activation ($\Delta H^\ddagger$) via an equation. For a single-step unimolecular or bimolecular reaction, the following equation holds:
$$ \text{Activation energy: } E_a = \Delta H^\ddagger + RT \qquad (eq. 1)$$
According to the internet, it may be generalized for single-step multimolecular reaction as follows:
$$ \text{Activation energy: } E_a = \Delta H^\ddagger + (1-\Delta n^\ddagger)RT \qquad (eq. 2)$$
where $\Delta n^\ddagger$ is the change in the number of molecules changed in going from the reactants to the transition state. For example, a bimolecular reaction in the gas phase would have $\Delta n^\ddagger = -1$. In most other cases, however, $\Delta n^\ddagger = 0$, so this equation becomes equal to the one given in the book of Anslyn and Dougherty.
Let's take a look at what the consequence of this $E_a$-temperature-dependence is. Let's plug it into the Arrhenius-equation!
Before I start, I want to make one final change to activation-ennergy-equation. For convenience, I define $n = (1-\Delta n^\ddagger)$, so that upcomming equations appear less scary. Note that $n$ is generally 1 for reactions in the liquid-phase, but $n$ may be 2 in the case of reactions in the gas-phase.
$$ \text{Activation energy: } E_a = \Delta H^\ddagger + nRT \qquad (eq. 3)$$
Below, I start with the original Arrhenius equation, and then substitute $E_a$ with eq. 3. I then try to simplify.
$$ \begin{align}
\text{Arrhenius: } k_1 = \!\! \qquad A \!\! \qquad & e^{\frac{-E_a }{R T}}\\
\text{Arrhenius: } k_1 = \!\! \qquad A \!\! \qquad & e^{\frac{-(\Delta H^\ddagger + nRT)}{R T}}\\
   ... = k_1 = \!\! \qquad A \!\! \qquad & e^{\frac{-nR T}{R T}} \times e^{\frac{-\Delta H^\ddagger}{R T}}\\
   ... = k_1 = \!\! \qquad \frac{A}{e^n} \!\! \qquad & \times e^{\frac{-\Delta H^\ddagger}{R T}} \propto e^{\frac{-E_a }{R T}}
 \end{align}$$
Now, because $A$ is the "pre-exponential Arrhenius constant", I think we can make our life easier by including the $e^{-n}$ term in this constant. For as far as I know, changing $n$ would mean changing reaction mechanism, which is (I suppose) not so easy in practice, and generally not done in kinetics studies. Keeping this in mind, we may state the following:
$$k_1 = \frac{A_{old}}{e^n} \times e^{\frac{-\Delta H^\ddagger}{R T}} = A_{new} \times e^{\frac{-\Delta H^\ddagger}{R T}}$$
According to Arrhenius equation, we must conclude that temperature dependence of activation energy is irrelevant on the kinetics of the reaction, because it just changes the pre-exponential factor with a non-temperature-dependent constant.
If you try to determine the value of activation energy or reaction Gibbs energy, via kinetics, however, Arrhenius is far from ideal, because $E_a = - n R T \ln(\frac{k_1}{A})$, and if you go deeper on Wikipedia, you'll find out that the pre-exponential factor $A$ may also depend on temperature and other fudge factors.
I think that, if you want to determine information on the energy of a reaction, it is better to use the Eyring equation, because it has no fudge factors in it. I'll give it below, just for completeness, but I'll stop before I'll go off topic.
$$ \begin{align}
\text{Arrhenius: } k_1 = \!\! \qquad A \!\! \qquad & e^{-\frac{E_a }{R T}}\\
\text{Arrhenius: } k_1 = \,\, \frac{k_\mathrm{B}T}{h} \,\, \mathrm{e}^{-{\frac  {\Delta G^\ddagger}{RT}}}
 \end{align}$$
$$ k={\frac  {k_{{\mathrm  {B}}}T}{h}}{\mathrm  {e}}^{{-{\frac  {\Delta G^{\ddagger }}{RT}}}} $$

This started out strong, but I don't really get what you're going for with the last section (with the equations).

Activation energy= Energy of transition state - Energy of Reactants. Of which temperature energy of reactants in considered here?

I don't think I understand your second diagram. Can you elaborate more about it?

@orthocresol, I made a start in converting the Arrhenius equation into the Eyering-equation. I didn't want to become off-topic, so I stopped at the point at which e^(temperature dependent activation energy) became e^(temperature independent enthalpy of activation). In the end, I think I prefer to put all the temperature dependent factors into entropy.

Activation energy ($E_\mathrm a$) is temperature independent, and activation free energy ($G_\mathrm a$) is temperature dependent. If we analyze Erying's transition state theory, or the statistical mechanical treatment of free energy it will become obvious. For a generic reaction
$$\ce{A + B <=> \{AB\}^\ddagger -> P}$$
we have
$$\begin{align}
k_\mathrm{forward} &= \frac{k_\mathrm{B}T}{h}\cdot \frac{\widetilde{q}_{\ce{AB^\ddagger}}(T)}{q_\ce{A}(T)q_\ce{B}(T)}\cdot \exp\left(-\frac{E_\mathrm a}{k_\mathrm{B}T}\right) \\[6 pt]
&= \frac{k_\mathrm{B}T}{h}\cdot\exp\left(-\frac{G_\mathrm a}{k_\mathrm{B}T}\right)
\end{align}$$
So here we have two different contribution to free energy. One is entropic or temperature dependent contribution that comes from vibrational, rotational, and translational partition function, and temperature independent contribution from electronic partition function. Electronic contribution or activation energy is temperature independent (though all the electronic structure software use elevated temperature to find the electronic energy, but later they all extrapolate it to $\pu{0 K}$). Even zero point energy is also temperature independent. Electronic partition function is temperature independent as long as they are in ground state.

It didn't become as obvious as you think.

My bad. I should have explained it in simple language.

Activation energy has a term that is proportional to temperature according to transition state theory (at least). This relies on the definition of activation energy as the parameter
$$E_\mathrm a = RT^2\left(\frac{\mathrm d \ln k}{\mathrm dT}\right)$$
(see this answer).
For instance, for the bimolecular reaction $\ce{A + B -> P}$, we have
$$E_\mathrm a = \Delta^{\ddagger}H^{\ominus} + nRT$$
where $\Delta^{\ddagger}H^{\ominus}$ is standard enthalpy of activation, and $n = 2$ for a gas-phase reaction and $1$ for a reaction in solution. That said, for many reactions, the activation enthalpy is much larger than $RT$ (which is ~$\pu{2.5 kJ/mol}$ at $\pu{300 K}$) and the activation energy can be approximately regarded as temperature independent.


text:
Why does phosphoric acid mask the colour of iron(III) complex in water?
In the redox titration of iron(III) with permanganate or dichromate, we use phosphoric(V) acid to "mask" the color of iron(III) because it interferes with the end point color change.
What's the chemistry behind this?

When phosphoric acid is added to an aqueous solution of iron (III), the ligands in the yellow complexes $\ce{[Fe(OH)(H2O)5]^2+}$ and $\ce{[Fe(OH)2(H2O)4]+}$ ( $\ce{[Fe(H2O)6]^3+}$ are weakly colored but fairly acidic), are successively substituted by $\ce{H2PO4^-}$ and $\ce{HPO4^2-}$. This yields colorless hydrogenphosphatoiron(III) complexes, like $\ce{[Fe(H2PO4)(H2O)5]^2+}$ and the water-free $\ce{Fe(H2PO4)3}$ complex, depending on the phosphoric acid concentration (source). In less acidic solutions, where $\ce{PO4^3-}$ is present, the colorless bis(phosphato)ferrate(III) $\ce{[Fe(PO4)2]^3-}$ is formed (source).


text:
The proper formula of sucrose: why both forms are correct and how one is transformed into another?
During preparations for my exam I've stumbled upon a tormenting question how it happens that the formula of sucrose is written in two different forms: the vertical and the horizontal ones:

The horizontal form

The vertical form

The vertical one seems quite obvious and logical when you know that the hydroxyl group on the second carbon in β-ᴅ-fructose directs up, but in the horizontal and more popular form the −OH directs as it was alpha form what's not true. How does it happen that both are correct?

Imagine (or assemble, if you can) a ball-and-stick model of sucrose. Now start slowly turning the fructose half around the bond between glucose and the bridging oxygen.

The stacked form is simplified so the C-O-C bond looks linear, but it's actually bent like an elbow as in the horizontal display. Rotate about that bond and one of the rings is turned upside down relative to your viewpoint, but otherwise remains the same.

http://chemistry.stackexchange.com/questions/43600/structure-of-sucrose

Grab the six-membered ring and move the five-membered ring upside-down while taking care of the relative $\ce{OH}$ configurations. The vertical form will transform into the horizontal one and vice-versa.

I want to thank you all for your answers. It helped me a lot.

Aaah, that linked duplicate question looks horrible; this looks much cleaner. Can’t we close the other one that already has a negative score as a dupe of this one?

@Jan It's not a duplicate in the first place.


text:
How to name the compound with the formula Br3C-CHCl2?

Name the compound:

How should I name $\ce{Br3C-CHCl2}$? Is it 1,1,1-tribromo-2,2-dichloroethane or 2,2,2-tribromo-1,1-dichloroethane?

The first name is correct (1,1,1-tribromo...).  Counting the substituents from either end, you would get 1 for the first substituent going from either direction.  As a tie-breaker, look at the next substituent, which would give you 1,1 going from the bromine end, and 1,1 going from the chlorine end.  Since the tie is still not broken, you go to the third substituent.  Going from the bromine end you would have 1,1,1 but 1,1,2 going from the chlorine side.  So, you number the carbon with the bromines number 1, giving you 1,1,1-tribromo-2,2-dichloroethane.
As a side note, if the carbon on the right had three chlorines instead of two, this method wouldn't work (you'd get 1,1,1,2,2,2 numbering from either direction).  In this circumstance you would use alphabetical order to break the tie, with bromo- having higher priority than chloro.  That would give you 1,1,1-tribromo-2,2,2-trichloroethane.  I know you didn't ask this, but it could prove useful to know!


text:
Is alcohol really a solution?
A friend recently sent me the following chemistry joke:

As a lover of puns, I'd like to enjoy it, but I think that it is actually false the way it is phrased.
A solution is defined as: "a homogeneous mixture composed of two or more substances" Alcohol alone would not constitute a mixture of multiple substances and would therefore not be a solution.
But in English the word "alcohol" is also commonly used to refer to many recreational drinks containing ethyl alcohol in addition to other types of molecules. Are there any common examples of alcoholic beverages which are solutions in the scientific sense that would make the pun valid, such as liquor, wine, or beer?

Well, nothing we see around ourselves is pure, and I consider alcohol to be a solution; since the impurities exist. The purest alcohol I've used is 99.6 % pure, which means out of a thousand molecules in the medium, 4 are not ethanol molecules.

Technically, alcohol is the name of a class of organic compounds containing one or several hydroxyl groups.
Colloquially, the term "alcohol" is understood as you have described: A solution (of varying degree of purity) of ethanol and water. Pure ethanol is impossible to create via the traditional method of purification (atmospheric distillation) as the water-ethanol mixture is azeotropic:¹

An azeotrope [...] is a mixture of two or more liquids whose proportions cannot be altered by simple distillation. This happens because, when an azeotrope is boiled, the vapour has the same proportions of constituents as the unboiled mixture.
Source: Wikipedia

This is why a chemist would never think "100% pure ethanol" when you mention "alcohol".
If you check for example this vendor page for the purest ethanol (>99.5%) I could find, you see various substances listed as impurities:

<0.2% Water
<0.1% Methanol
Other organic substances, such as isopropanol and acetone

Concerning your second question, I would argue that many distilled spirits (vodka, whisky, and cognac just to name a few) are solutions of ethanol, water, and various flavour compounds.
Some alcoholic beverages, such as unfiltered beer and wine, usually also contain solid particles and would as such be correctly termed "suspensions".

¹ I wonder now whether it would be possible to produce 100% ethanol from another mixture somehow. The problem is probably that every common starting mixture already has some amount of water. But synthesizing ethanol in, say, an ionic liquid under an inert atmosphere and then distilling it away might be an interesting approach.

It is possible to get water-free alcohol by distillation but the common methods leave nasty traces of things like benzene. Other methods exist to create some spectroscopic grades with very low water content and no benzene.

@matt_black So it's possible to get water-free, but still not >99.9% pure? I heard about that thing with benzene but didn't want to go off on a tangent too much in this answer.

There is an additive-free approach, pressure-swing distillation, that works to cross the azeotrope in some systems, as long as the min-/max-boiling point of the azeotrope is sufficiently sensitive to pressure.

With this argument, there is no pure liquid at all. (I would otherwise challenge you to make one.)

How we define "pure" is always up to us, but we always must reckon with real data.  Tiny amounts of impurities most people would dismissed created this ... https://en.wikipedia.org/wiki/Polywater

Isn't it true that the volume of a solution is equal to the volume of the water in the solution? (The solute only increases the specific weight.) If so, how come we talk, for example, about 40% alcohol by volume in an alcoholic drink?

Tschoppi’s answer is technically correct, since it is nigh impossible to create a pure compound without any contaminants. The highest-grade liquids I can remember seeing on SigmaAldrich’s web pages were something like $99.9999~\%$ purity or something — far enough away from $100~\%$ to call it a solution.
However in practice, chemists (especially non-physical chemists) would call any liquid phase that does not have substantial dissolved particles a ‘pure’ liquid — that would even include ‘just’ $99~\%$ purity. (As a general rule: The more organic or biological the chemist, the less pure ‘pure liquids’ are.) Some may even call the $96~\%$ ethanol (which is often traditionally labelled ethanol absolute — not to be confused with Absolut Vodka) a pure liquid and totally neglect the $4~\%$ water and other stuff in there.
So in a practical view, the questions remain:

What does one think when one hears alcohol; and
would one consider the liquid phase thought of as a solution or a pure liquid.

The first is the difficult one. When in the lab, chemists may colloquially call ethanol alcohol — while some might also call methanol the bad alcohol or something. The word alcohol no longer describes a specific chemical (if it ever did) but rather, as Tschoppi pointed out, a group of chemicals that all contain $\ce{O-H}$ groups (and excluding the chemical if all its $\ce{O-H}$ groups come from acids). So the pun would probably fail in step 1 when we are defining alcohol.
However, if we do assume that a chemist would understand ‘the bottle labelled ethanol abs.’ (or whichever ethanol purity is most available) in a lab setting, then that chemist would most likely say ‘no, this bottle does not contain a solution, only a liquid/solvent/whatever.’ So the pun would fail at step 2.
Outside the lab, by the way, chemists would often talk about alcohol meaning alcoholic beverage — if they use that word in that sense at all. I don’t think any chemist (at least no-one I’ve met so far) would use alcohol outside of the lab to mean ethanol. If that is the case, then yes, the pun would work.
So all things considered, most of the time chemists would reject the pun for being either too simplified or wrong in a lab setting. However, given the correct set of constraints, they would accept it.

This is very tangential, but I've always been curious about what is the purest substance we can make on a macroscopic scale, so I figure I'll share. Turns out it's superfluid helium-3, which is immiscible with helium-4 due to quantum effects (!), is the poorest solvent imaginable, everything else is frozen solid at the temperatures required (millikelvin or less), and mixing entropy contributes almost nothing to mixing free energy so close to absolute zero.

I would argue that these are mixtures, not solutions. A solution requires at least two different components. A solution might look like alcohol (being the solvent) that would take water (the solute) into solution, making $\ce{H+}$ cations and $\ce{OH-}$ anions. That would be an acid and a base, that would immediately turn back to water. That would never happen. The water molecules would just stay together in the first place. The fact is that it is a mixture, not a solution.


text:
What are the specific dangers associated with Grignard reagents?
I would like to know if anyone has had any specific safety-related problems with Grignard reagents.  I have taught the year-long introductory Organic Chemistry sequence for about 15 years and have never had a student get any Grignard Reagent on their skin, but I've always questioned how safe it really is to have students work with these compounds (specifically PhMgBr).  I have never allowed my students to work with n-butyllithium (we don't even have any in our stockroom), so it seems an odd contradiction that I have much less reluctance to have them work with Grignard reagents.
The dangers of organolithium reagents, particularly t-BuLi and -- to a lesser extent -- n-BuLi are well documented.  It seems to me that the danger of these reagents largely arises from the flammability of the solvent (hexane) in which the compounds are dissolved as opposed to their extreme reactivity as bases.  The organolithium reagent as it reacts with moisture provides the energy needed to ignite the solvent. Since Grignard reagents are themselves extremely basic and dissolved in highly-flammable ethers, it would seem that the dangers of physical contact with a Grignard reagent would be equal to if not greater than those of using an organolithium reagent.  Just out of curiosity, has anyone actually had experience with any accidents and/or injuries involving Grignard reagents?

There is this incident that involves t-BuLi. Yes I know it's not a Grignard but if you read the article, which goes in a lot of technical detail, you might agree that it could have very well happened with a Grignard reagent and the main issue was not the high flammability of t-BuLi but low safety standards in the lab. On the other hand I believe that no organic chemist should be awarded a degree if he never performed a Grignard reaction as an undergraduate in the lab.

K_P I'm aware of the t-BuLi incident you mentioned.  This is actually one of the things which got me worried about using Grignard reagents.  I agree that it seems an essential experiment, though.  I've done it every year I taught the lab.

I think that an undergraduate experiment of standard preparation of a Grignard reagent is perfectly safe and gives the student the experience of performing a "serious" experiment. On the other hand if the experiment involves handling of a stock solution of a Grignard reagent requiring inert atmosphere, syringes, cannulas, etc then it wouldn't be suitable for undergraduates.

You could ask about what are the dangers, but asking about personal experiences isn't good idea here imo as questions should be about specific stuff and number of accidents described could be big.

It would be nicer if you restrict the scope of your question to speak about only one/two Grignard reagent(s).

@Ϻ.Λ.Ʀ. it wouldn't make a difference

I have to disagree your position entirely.
Every chemical has associated hazards, with even the most unassuming chemicals having the potential to cause severe harm if not fatality.
However, the actual risk to the chemist is often severely minimised by taking sensible precautions, understanding the hazards, and working under a controlled set of conditions.
The whole purpose of organic chemistry education, especially the purpose of organic chemistry labs is to teach students how they should safely handle chemicals, not to tell them that they're too risky.

While I appreciate your point (yet do not agree with it), it's not at all what I'm asking about.  I want to know if anyone has specifically encountered any kind of injury resulting from Grignard reagents mainly as an issue of personal curiosity.  However, to your point, are you saying you would allow students in a second-year course (1st year of o-chem, after a year of g-chem) to work with potassium cyanide or lithium aluminum hydride?  I'm talking about a class of 24 students each working independently.  Even with the most careful supervision that's not a risk I would be willing to take.

Perhaps we can continue this conversation in chat, as I think its worth exploring, but the stack exchange model doesn't encourage opinion based questions and answer.


text:
Amine Nomenclature – When to use "N"
I'm currenty wondering what the correct naming scheme for the following molecule would be:
Is this called "2,N,N-trimethylpropan-1-amine" or "2-methyl-N,N-dimethylpropan-1-amine". In other words, do you group the substituents attached to the N with the other substituents.

most likely it would be called isobutyldimethylamine, actually.

The name is N,N,2-trimethylpropan-1-amine.  I was looking this same type of question up in the IUPAC rules last week but couldn't find a definitive answer.  However, my naming software said N comes before a numbered locant.  As to your question, all substituents of the same type are included, regardless of whether they are on the chain or on a nitrogen atom.

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the ‘N’ is a locant just as the ‘2’:

P-14.3 LOCANTS
P-14.3.1 Types of locants
Traditional types of locants are arabic numbers, for example, 1, 2, 3; primed locants, for example, 1′, 1‴, 2″; locants including a lower case Roman letter, for example, 3a, 3b; italicized Roman letters, for example, O, N, P; Greek letters, for example, ‘α-, β-, γ-’ and compound locants, for example, ‘1(10)’ and ‘5(17)’.

Note that ‘N’ is lower than ‘2’:

P-14.3.5 Lowest set of locants
The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
  (…)
  Italic capital and lower-case letter locants are lower than Greek letter locants, which, in turn, are lower than numerals.

Identical simple substituents are indicated by multiplicative prefixes, such as ‘di’, ‘tri’, etc. The current IUPAC recommendations do not contain any rule that would stipulate that such substitutents should be grouped. By way of comparison, similar examples given in the Blue Book include names such as N,N,2-trimethyl-3-{4-methyl-3-[2-methyl-3-(methylamino)-3-oxopropyl]phenyl}propanamide.
Therefore, the preferred IUPAC name (PIN) for the compound given in the question is N,N,2-trimethylpropan-1-amine.

Thanks for the specific IUPAC citation.  I was searching the older rules and couldn't find anything specific about ordering of non-numerical locants.