text:
Should bromine water be called a solution?
Bromine water is a reagent which is used to test for unsaturation in organic compound. It is $2.8~\%$ bromine in water. In many places, it is refer to as bromine solution.
But it is observed that adding bromine to water gives hypobromous acid ($\ce{HOBr}$) and hydrobromic acid ($\ce{HBr}$) via a disproportionation reaction.
$$\ce{Br2 + H2O <=>> HOBr + HBr}$$
So, if bromine reacts with water to form a mixture of acids, how can it be called a solution? Components of solution do not react with each other.

Just ignore the reaction (it is not that fast) and make the solution.

Much like with chlorine or iodine, the disproportionation of bromine can occur in pure water, but it is not fast. As you can see, the reaction generates acids, so without any base present the equilibrium is pretty close to the bromine side. You do need bases to disproportion bromine in any meaningful amount — much like chlorine which is generally quenched with $\ce{NaOH}$ solution to generate hypochlorite solutions.
So if you just put bromine and water together, they may react very slightly, but not much, and most of the bromine will sink to the bottom (assuming you used a lot) while the lesser part will dissolve in the water first. Since the concentration of $\ce{Br2}$ is much higher (and we’re talking orders of magnitude) that those of $\ce{Br-}$ and $\ce{OBr-}$, it is best to consider this a bromine solution and not a bromide/hypobromite one.

According to the IUPAC Goldbook a solution is:

A liquid or solid phase containing more than one substance, when for convenience one (or more) substance, which is called the solvent, is treated differently from the other substances, which are called solutes. When, as is often but not necessarily the case, the sum of the mole fractions of solutes is small compared with unity, the solution is called a dilute solution. A superscript attached to the  ∞ symbol for a property of a solution denotes the property in the limit of infinite dilution.

There is no mentioning of that the solute may not react with the solvent. If you investigate a few common solutions, you will also find that it is indeed quite common, that the solvent reacts with the solution. Take for example the dissolution of hydrochloride in water, which is often expressed with the following equation:
$$\ce{HCl (g) + H2O (l) <=> Cl- (aq) + H3O+ (aq)}$$
You can see here, that the proton is transferred onto a water molecule and from one point of view this is a displacement reaction; the chloride is displaced by water.
Another example is the dissolution of acetic acid. It is often described with the following equation:
$$\ce{H3C-COOH (s) + H2O (l) <=> H3C-COO- (aq) + H3O+ (aq)}$$
The same happens here. We hide the fact that the dissolution process includes getting the solvent into a highly flexible network of hydrogen bonds in the addendum of $\ce{(aq)}$.
When you look at common salts that get dissolved, it is often that the metal ion reacts as a weak acid, co-coordinating the solvent molecules. In many cases this is so flexible, like for the alkaline metals, that we cannot describe even the first solvation sphere approximately.
For other salts this is quite possible. Take for example the dissolution of copper sulfate. We often write for short
$$\ce{CuSO4 (s) <=> Cu^2+ (aq) + SO4^2- (aq)}$$
when we actually mean the formation of a complex
$$\ce{CuSO4.5H2O (s) + H2O (l) <=> [Cu(H2O)6]^2+ (aq) + SO4^2- (aq)}.$$
A chemical reaction actually takes place. And since everything is in equilibrium, there are many chemical reactions happening in solutions all the time.
Now the dissolution of bromine in water is no different. In fact the disproportionation
$$\ce{Br2 (l) + 2 H2O <=> BrOH (aq) + Br- (aq) + H3O+ (aq)}$$
is quite necessary to dissolve bromine in the first place. While I do not even slightly agree to the causality of having a dipole means being a polar molecule, I can still see, that the bromine molecule is less polar than water. It is however quite well polarizable; it has a quadrupole moment.[1] This is what also leads to the above equation.
The bromine hydrolysis has been investigated a couple of times.
$$\begin{align}
\ce{Br2 + H2O &<=>[K_1] HBrO + H+ + Br- }&
K_1 &= \ce{\frac{[HBrO][H+][Br- ]}{[Br2]}}
\end{align}$$
The temperature dependency of $K_1$ has been studied by Liebhafsky as early as 1934.[2] In their studies they also considered the effects of the tribromide equilibrium
$$\begin{align}
\ce{Br3^- &<=>[K_3] Br- + Br2}
K_3 &= \ce{\frac{[Br- ][Br2]}{[Br3^- ]}}.
\end{align}$$
And later continue to state that the existence of the pentabromide ion is highly probable.
$$\begin{align}
\ce{Br5^- &<=>[K_5] Br- + 2 Br2}
K_5 &= \ce{\frac{[Br- ][Br2]^2}{[Br5^- ]}}
\end{align}$$
From the equilibrium constants you can see, that the main dissolved species is still $\ce{Br2}$.
$$\begin{array}{lrrrrr}
T/~\mathrm{^\circ C} & 0    & 10   & 25  & 30  & 35  \\\hline
K_1\times10^{-9}     & 0.70 & 1.78 & 5.8 & 8.3 & 11.3\\
\end{array}$$
Conclusion
Solutions do not exist without chemical reactions. There are often multiple different species in solution, that have very low concentration, so for various purposes a valid approximation is to neglect those species' concentrations.

Jacek Bieroń, Pekka Pyykkö, Dage Sundholm, Vladimir Kellö, and Andrzej J. Sadlej; Phys. Rev. A 2001, 64, 052507.
Herman A. Liebhafsky; J. Am. Chem. Soc. 1934, 56 (7), 1500–1505.

Iirc, wasn’t $\ce{CuSO4 . 5 H2O}$ actually $\ce{[Cu(H2O)4(\mu-SO4)2] . H2O}$? If that’s the case, then you should rewrite the corresponding equilibrium where you only discuss the formation of a hexaaquacomplex while you are actually creating a hexaaqua from a diaquabissulfato complex. Or you switch to discussing the dissolution of $\ce{CuSO4}$ (water-free). But anyways, great answer; let me give you $+1$ =3

@Jan I really had no idea. There you have it again, when you think you pick an easy example, you don't. It always depend how deep you went when you learned it. If I edit it now, I'll mess it up since I am only on mobile and not in the office during the next week, so we'll consider your comment as a fine extension and revisit it next year. Thank you for pointing it out though.

" If I edit it now, I'll mess it up since I am only on mobile and not in the office during the next week, so we'll consider your comment as a fine extension and revisit it next year." - It has been almost a year that you posted this answer. You want to revisit the answer for some extension.


text:
Why exception is noted in the solubility of lead salts?
There are many exception in the solubility of lead salts like:

Lead sulfate is insoluble in cold water whereas most of the sulfates are soluble in cold water.
Lead chloride is also insoluble in cold water but is soluble in hot water. Why?
Carbonates are generally insoluble in water but they dissolve in dil. $\ce{HCl}$ except lead carbonate.

What property of lead causes such type of exceptions?

Your question 2 is the answer to your question 3.

Many things are more soluble in [hot solvent of your choice] than in [cold solvent]. Also, 1 is at best a simplification, probably incorrect, though.

Lead sulfate is insoluble in cold water whereas most of the sulfates are soluble in cold water.
Lead chloride is also insoluble in cold water but is soluble in hot water. Why?

Lead ion is a borderline acid in the context of hard-soft acid-base theory. Water is a hard solvent. Hard-hard and soft-soft interactions tend to be favorable; mixed interactions are not as favorable. Remember, solvation entails not only breaking bonds within the salt but also creation of bonds between ions and solvent.
Hot water likely weakens the lead ion-anion bonds.

Carbonates are generally insoluble in water but they dissolve in dil. $\ce{HCl}$ except lead carbonate.

Many carbonates are soluble in acid because the acid can protonate the carbonate portion. Singly protonating carbonate ion creates hydrogen carbonate, or bicarbonate, and bicarbonate salts tend to be soluble, likely because of its reduced negative charge density. Doubly protonating the carbonate portion creates a neutral entity - carbonic acid. This is rather unstable and tends to disproportionate to carbon dioxide in water. A common qualitative test for carbonates is to acidify the (unknown) salt and if bubbling is observed, the salt is likely a carbonate salt.
The interesting thing about lead ion is that many of its salts are insoluble in water. This suggests to me that there is a) an especially strong bond between lead ion and its counterions and/or b) an inability of water to effectively solvate lead ion. The latter is discussed above. The former, I'll leave to someone more familiar with the chemistry of metals.

(source)

I'd quibble a bit about solubility of carbonates in HCl. In general you'd get carbon dioxide gas and the chloride salt of the metal.

http://www.bbc.co.uk/schools/gcsebitesize/science/add_edexcel/ionic_compounds/ionicanalysisrev2.shtml

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text:
How are electronic configurations including hybrid orbitals notated?
I know the general scheme for notating electronic configurations is (with s and p orbitals) [noble gas core]nsel.$n$pel. etc., where $n$ is the principal quantum number and el. is the number of electrons occupying that subshell, e.g. carbon: [He]2s22p2.
Now, how do you include hybrid orbitals? The problem arises because some hybrid orbitals already have superscript numbers (sp2, sp3 …) – how do you differentiate between these superscript numbers and those denoting how many electrons fill that orbital? If I were to write out the electronic configuration for the hybridised carbon atom in $\ce{CO2}$, this would not be a problem: [He]2sp22p2, but what about the carbon atom in $\ce{CH4}$, which has sp3 orbitals? Would you write [He]2sp3, 4 or [He]2sp34 or something else?

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The electronic configuration concept does not really extend itself well to hybrid orbitals, because these are best considered mathematical concepts only. However, the problem is easily solveable:
Always use brackets when dealing with hybrid orbitals.
Even your $\ce{CO2}$ case is potentially misleading: I misread it once as $\mathrm{2s^1\ 2p^2}$. It is best to show what belongs together with: $\mathrm{[He]\ 2(sp)^2\ 2p^2}$ or, for methane: $\mathrm{[He]\ 2(sp^3)^4}$.

What description of bonding is anything more  than a mathematical concept?

@ron True, but turn it around. The orbitals of an atom in ground state are about the closest we can get to quantumphysical reality. I didn’t want to stress that other descriptions of bonds be less mathematical but rather that the ground state is less so.

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text:
Which has maximum entropy
Which state has a higher entropy — the
liquid–gas equilibrium or the gaseous state? I have learned that at equilibrium the entropy is maximised. But I also learned that the gaseous state has the highest disorder (entropy), if the latter is true then why does the equilibrium exist, why doesn’t the liquid transform into the gaseous state at equilibrium completely? What factor is responsible for this?

Enthalpy. It is positive for liquid-gas transition as bonds need to be broken.

Whatever one observes (whether it is a liquid, a gas, or a mixture thereof) is the state of maximum entropy for the particular values of the internal energy and volume the system currently has. As OP noted, this is a consequence of the second law of thermodynamics. According to this law, a system with a fixed internal energy self-adjusts its free parameters in such a way that the entropy is maximized. The free parameter in OP's example is the fraction of molecules in the liquid (or gaseous) phase.
When we say that the entropy of the gaseous phase is larger than that of the liquid phase, we are referring to the specific entropies (i.e., entropy per mole) of these phases when they coexist. When the two phases coexist, they have a common temperature $T$ and a common pressure $P$. Furthermore, their specific Gibbs free energies are the same, that is,
\begin{equation}
0 = \Delta g (T,P) = g_{\mathrm{gas}} (T,P) - g_{\mathrm{liquid}} (T,P) = \Delta u(T,P) + P \Delta v(T,P) - T \Delta s(T,P).
\end{equation}
The gaseous phase has a higher specific internal energy ($\Delta u>0$) and a larger specific volume ($\Delta v>0$) as well as a higher specific entropy ($\Delta s >0$) in such a way that the effects of the three terms ($\Delta u$, $P\Delta v$, and $T\Delta s$) cancel out.
One may still wonder why liquid-gas coexistence is ever possible. As the gaseous phase has a higher specific entropy, why doesn't the system self-adjust to have all its molecules in the gaseous phase? The Answer: What we know for sure is that if the evaporation took place under a constant temperature and pressure, the entropy would increase (because $\Delta s >0$). But this process would also increase the internal energy and the volume ($\Delta u >0$ and $\Delta v >0$). Therefore, to have more molecules in the gaseous phase but stay at the same internal energy and volume, the system has to lower its temperature and raise its pressure. This results in a decrease in the entropy,1 which counteracts its increase that would have happened if the evaporation had taken place at constant temperature and pressure. Therefore, evaporation at fixed internal energy and volume doesn't necessarily lead to the increase of the total entropy. As stated in the first paragraph, if one observes liquid-gas coexistence, this should correspond to the state of maximum entropy for the given values of internal energy and volume.

1 For any (stable) thermodynamics system, the entropy decreases as the temperature is lowered. On the other hand, no such strict rule exists for the relation between the entropy and the pressure. Still, for most systems, entropy decreases as pressure is raised. [Notice the Maxwell relation $(\partial S/\partial P)_{T} = -(\partial V/\partial T)_{P}$, meaning that the entropy decreasing with increasing pressure is equivalent to the volume increasing with increasing temperature.]


text:
If glucose can react with phenyl hydrazine to form osazone why can't it react with 2,4-DNP to give positive test?

If glucose can react with form phenyl hydrazine to form osazone why
  can't it react with 2,4-DNP?

Yes I know about cyclic form of glucose and its equilibrium with straight chain and all …
I read this, this and this Stack Exchange one too Does glucose react with Brady's reagent?
Still, didn't get an answer to my question. So requesting for clarification …

Glucose itself keeps interconverting between alpha beta glucopyranose and the chain form but it is more stable in this ring forms. A reagent should be strong enough to convert cyclic structure to the chain form so as to react with aldehyde group. I think due to electron withdrawing action of the Nitro groups in DNP they are less active/weaker than phenyl hydrazine and so DNP shouldnt convert glucose to chain form and react with the aldehyde. That's my explanation though.Someone might post a better answer.

Glucose converts itself to the chain form in appropriate solvents, doesn’t it?

@Jan I guess. When the reagent /solvent is highly reactive towards aldehyde group, it becomes chain.

Although glucose 1a exists principally in the hemiacetal pyranose forms (α and β), nonetheless it reacts with one equivalent of phenylhydrazine to form the hydrazone 1b via the aldehydic open form. Two additional equivalents serve to form the osazone and the products of reduction of phenylhydrazine, ammonia and aniline.The glucose 2,4-dinitrophenylhydrazone 1c has been reported1 but not the corresponding osazone. In their studies on the mechanism of osazone formation, Shemyakin and coworkers2 utilized p-nitrophenylhydrazine labeled with nitrogen-15 at the terminal nitrogen (the red N in the structures). Although glucose was not examined, the p-nitrophenylhydrazones  of fructose (forms the same osazone as glucose), benzoin and 2-hydroxycylohexanone were studied. The labelled nitrogen was located in the ammonia that was liberated during osazone formation.
Structures 2-7 represent the essentials of the mechanism.  Enolization of the hydrazone 2 produces the enol 3, which, via the zwitterion 4, undergoes oxidation of the carbon framework and reduction of the phenylhydrazine moiety. The transformation 3 --> 4 is essential. The basicity of the electron pair on the nitrogen in the 2,4-dinitrophenyl case is sufficiently reduced by delocalization into the aromatic ring, thereby repressing the proton transfer. This step must be completed before osazone formation can be accomplished.

1: Lloyd, E. A. and Doherty, D. G., J. Am. Chem. Soc., 1952, 74, 4214.
2: Shemyakin, M. M., Maimind, V. I., Ermolaev, K. M. and Bamdas, E. M., Tetrahedron, 1965, 21, 2771.

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text:
Why is sulfur hexafluoride more stable than selenium or tellurium hexafluoride?
I was just reading about p Block elements from two different books. Both books say that $\ce{SF6}$ is extremely stable in 16th group due to steric reasons but my question is why is it more stable than $\ce{SeF6}$ and $\ce{TeF6}$ when $\ce{S}$ is more electronegative and smaller than $\ce{Te}$ and $\ce{Se}$ and so it would bring the electrons in bonding with fluorine closer to the central atom i.e $\ce{S}$ and thus it would bring the electrons closer and there would be more repulsion between the electron pairs.
Maybe this can be explained by Fajans rules that due to such conditions $\ce{SF6}$ has more covalent character than others and thus is more stable. But if this is the reason, where does "steric" come in here?

Related: How does steric hindrance cause sulphur hexafluoride to have almost no reaction chemistry?

Not quite. For example, fluorine is extremely reactive, but absolutely stable by itself.

$\ce{SF6}$ is extremely stable for purely steric reasons, because S is completely blocked by fluorine atoms from all directions, so the reactions starting with an attack on S that otherwise would readily occur (hydrolysis, etc.) never have the chance to occur. This has nothing to do with electronegativity. For another similar example, look at $\ce{CCl4}$.
Now, the word "stable" is commonly used in two different senses. A compound may be stable by itself, i.e. not prone to spontaneous decomposition; in that sense $\ce{SF6}$, $\ce{SeF6}$, and $\ce{TeF6}$ are all totally stable under normal conditions, so there is nothing to compare. On the other hand, we may consider "stability" as inability to react with common reagents, e.g. water. In that sense $\ce{SF6}$ is indeed far more "stable" than the analogs, and that for the steric reasons (see above).
As for the thermal stability, that's another story altogether, so I don't really know how it would turn out. I wouldn't be surprised if $\ce{TeF6}$ would indeed happen to be the most stable of them all.

Ok it does not react, but does it mean it is more stable?

@Quark that's a twisted question. How do you define stability?

Is my  terminology wrong? If I talk about the thermal stability, TeF6 would be the most stable right? but when I say least reactive , I'd say SF6 right?

So what you’re saying is basically $\ce{SF6}$ is not only (relatively) stable but also inert, isn’t it?

I don't think I said anything that broad. "Inert" may mean a variety of things (thermodynamic or kinetic); the same goes for "stable".


text:
Using results of current experiments in silanization, how to achieve a higher contact angle?
I've done several silanization tests with different solutions and here is a brief of my results:
DichloroDimethylSilan: Advancing=90, Receding=56
ChloroTrimethylSilan: Adv=91, Rec=70
OctaDecylTrichloroSilan: Adv=108, Rec=90
TetraEthoxySilan: Adv=90, Rec=44
OctaDecylTrichloroSilan+ChloroTrimethylSilan(50%): Adv=89, Rec=74
I didn't dilute any of the above solutions and since I want to achieve Maximum Hydrophobicity I gave enough time (~20 min) to all the above reaction to become fully completed.
I'm wondering whether its possible to get a contact angle higher than 108 that I have now. Please let me know what changes should I made? (any pre-treatment of glass before silanizing, any other solutions, etc.)

It's always good to clean glass surfaces with oxygen plasma or UV ozone treatment before silianizing, but 108° is pretty good with an alkylsilane.
The best you will be able to do with this type of surface treatment is using fluorosilanes.
Silanization agents are basically made of three parts, a functional group, a linker and a coupling group:

The coupling group is what allows the silane to attach to a surface and is found bound to the silicon atom.
Typical couplers are hydrolyzable groups like alkoxy, acyloxy, or halogen.
These don't generally influence the hydrophobicity of the resulting surface directly, though they will affect how well the silane adheres to a given surface, the manner and speed with which it can be applied, and the deposition density.
Alkoxy and acyloxy groups can be applied from aqueous solutions and are generally less reactive, but halogen groups can be deposited from anhydrous solvents or the vapour face quite quickly.
The linker is typically just a short alkyl chain between the silicon atom and the functional group to reduce steric interactions between adjacent silanes.
The functional group is what has the greatest influence on hydrophobicity.
Many different functional groups can be added, some even conferring hydrophilicity instead, if desired.
In general, longer alkyl chains are more hydrophobic, but long fluoroalkane chains are the most hydrophobic and will also not interact easily with organic solvents either so fluorosilane coated surfaces are often known as omniphobic by virtue of being both hydrophobic and oleophobic.

So if you want the maximum contact angle possible from a simple surface treatment, you're looking for something like 1H,1H,2H,2H-Perfluorododecyltrichlorosilane.
It may also be possible to simply spin or spray coat a fluoropolymer like Teflon AF or Fluoropel, depending on your application and surface. All these types of coatings will give contact angles up to about 120° in the best case and to get any higher will require surface patterning to achieve superhydrophobicity.

I'll add that the theoretical maximum seems to be about 120 degrees. (ref 1) http://www.gelest.com/goods/pdf/Library/advances/HydrophobicityHydrophilicityandSilanes.pdf  (ref 2) http://mysbfiles.stonybrook.edu/~spongkitwito/SP%20CME320.pdf

Could you please have a look on this question as well. Thanks!

If I deep the glass surface into H2SO4 for a few hours and then perform silanization with OctaDecylTrichloroSilan, is it likely to get a higher contact angle?

Not sure how you'd propose to use sulfuric acid. If concentrated sulfuric acid, how would you remove the adhering sulfuric acid? // In general silanization depends on O-H groups for a reaction. If you remove OH groups with concentrated sulfuric acid, then you wouldn't get a reaction. If you wash off the sulfuric acid with water, then you're more or less back where you started. // So all in all I don't think that dipping in sulfuric acid as a pretreatment would be helpful.


text:
Free radicals vs Hund's rule contradiction?
How do we reconcile these two statements?

Formation of electron pairs is energetically favorable.  Therefore, free radicals are particularly reactive
Hund's rule - it takes energy to pair electrons, so orbitals are typically filled singly first

Formation of electron pairs is not all that favorable. It's the formation of bonding orbitals that's behind the reactivity of free radicals.

See the answer here: http://chemistry.stackexchange.com/questions/13513/why-are-radicals-unstable

The first statement in the question is not correct. Formation of electron pairs is not favorable. That would lead to putting two electrons in the same orbital and there is some effect from charge-charge repulsion.
Electrons end up paired in bonding or stabilized orbitals because there is space for a second electron, and the overall stabilization that would result from putting a second electron in is greater than the destabilization of getting that electron from somewhere else.

A free radical is generated by homolytic bond cleavage of a covalent bond and each of the elements have one electron each.This makes them very reactive.Whereas Hund's rule is based on the fact that in electron pairing each orbital should be singly filled.

This doesn't provide an answer to the main question, since the OP is trying to find why the two seem contradictory per the explanation given in the question.


text:
For group 7 elements, are anions a lower energy state than being neutral atoms?
Given that electron affinity is positive, why don't these atoms attract free electrons and "prefer" to exist as anions?  Never made sense to me — I could be missing something obvious.

There are no free electrons, much like there is no such thing as a free lunch. Every electron belongs to some entity, which may be more or less unwilling to share it.

Yes, the electron affinity for group 17 elements (group VII by older IUPAC nomenclature) is generally exothermic, so the uptake of an electron should be favourable.
But as Ivan mentioned in a comment, there is no such thing as a free or lonesome electron. Each electron must first be displaced from another atom — ionised; the corresponding thermodynamic property is the ionisation enthalpy. For every element, the ionisation enthalpy is positive, thus it is always unfavourable to displace an electron from a neutral atom. No exceptions.

isn't the first electron affinity of any element (except perhaps noble gases) exothermic?

@Dissenter No. Manganese, beryllium and nitrogen and possibly others have endothermic EAs

authors: Степан Гениман


text:
Why doesn't CaO decompose into Ca and O2?
Many metal oxides decompose into the free metal and oxygen gas at high temperatures, but why doesn't $\ce{CaO}$ do that? What happens to $\ce{CaO}$ at high temperatures?
This is from problem 2 of the 2011 USNCO local exam:

Oxygen gas can be produced by the decomposition of all
  of the following substances EXCEPT
(A) calcium oxide. (B) hydrogen peroxide.
  (C) mercury(II) oxide. (D) ozone.

The correct answer to this question was (A).

Nothing happens to $\ce{CaO}$ at high temperatures.

Stability, the thermodynamic property, is defined as whether a decomposition reaction is exergonic or endergonic at a certain temperature. The reaction $\ce{CaO -> Ca + 1/2 O2}$ is endergonic across a very large temperature range. But that doesn’t necessarily help you.
In that list of four, we have to clearly molecular compounds and two mostly ionic ones. We need different discussions for both. Basically, the molecules are unstable towards decomposition into $\ce{O2}$ and something ($\ce{H2O}$/$\ce{O2}$) because they contain energetic bonds whose replacement by an $\ce{O=O}$ bond is favourable.
For the ionic ones, we have a number of different discussions we can do. For one, the metallic partner is calcium versus mercury; while the latter is known since ancient times and pretty stable in liquid form the former is not: It will readily react to form salts. For another, the $\ce{Ca^2+}$ ion is rather small and thus a hard Lewis acid according to Pearson’s concept. $\ce{Hg^2+}$ is much larger and thus a much softer acid. Therefore, the adduct of $\ce{Ca^2+}$ and $\ce{O^2-}$ is predicted to be more stable (hard–hard interaction) than that of $\ce{Hg^2+}$ with $\ce{O^2-}$. Finally, we can also consider heats of formation which should (I unfortunately don’t have values) be much more negative in the case of $\ce{CaO}$ than for $\ce{HgO}$ — therefore, the latter can be decomposed with enough heat due to $\Delta G = \Delta H - T \Delta S$ while the former cannot.

I think the chemistry olympiad question was getting at hard-soft acid-base theory mainly.

@Dissenter - if you think you can explain it with HSAB (which looks like it would work), go ahead and add an answer.

@BenNorris HSAB is one of the three explanations given in my answer …

CaO can thermally decompose at the right temperature and partial pressure of oxygen, it just so happens that everything becomes a gas at such temperatures.

In short ... before the decomposition of CaO, it start to melts. That means, melting point of CaO is less than the theoretical decomposition temperature of CaO. So it do not decompose upon high temperature. It melts. Compounds like Na2O , K2O, MgO .... etc also do not decompose due to this reason.
In fact these compounds are thermodynamicaly more stable.
Compound like Silver oxide, oxides of mercury readily decompose as their decompose temperature is less than their melting point. Oxides of elements below Hydrogen in electro-chemical series act like this.
That is these compounds are thermodynamically less stable with respect to above compounds. (that doesn't mean that these compounds are thermodynamically unstable. Only less stable with respect to Na2O , K2O, MgO .... etc.)

Okay, so it melts. What happens if you keep heating it? Do you get oxygen gas eventually?

Yes, I don't see why the physical state would matter. Solid, liquid, gas, so what? The question is whether decomposition is endothermic or exothermic at any given temperature, isn't it?

I'm tempted to downvote you - all compounds decompose given high enough temperature.

A comparison of the heats of formation of the 4 molecules yield the correct result.  At -630 kJ/mol /\ Hf, CaO requires  the least energy to form - and therefore the most energy to decompose - compared to HgO (-90), MgO (-600), and H2O2 (-188).  In other words, CaO is the happiest of the 4 molecules at 'high temperatures'.
CaO will eventually decompose if the temperature is increased sufficiently, but the other 3 molecules will do it first.  I think the term 'high temperatures' is a source of confusion. :)

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text:
Why does bromine add to a double bond instead of a triple bond if both are present?
According to my textbook, the addition of bromine to pent-1-en-4-yne at low temperature yields 4,5-dibromopent-1-yne:

Why does bromination of the double bond occur selectively?

It would be a much more complete question if you'd mention the solvent, and the temperature at which the reaction is taking place. These things can often affect reaction rates a lot.

I'm going to take "more reactive" as in the alkene undergoes bromination at a faster rate than the alkyne, which seems to imply that the activation energy of the transition state is greater.
A Kinetic Perspective
The formation of the bromonium ion involves a cyclic transition state where the the Br-Br bond is attacked by the alkene/alkyne pi-bond, and the pi-bond donates to the sigma* antibonding orbital to form a sigma C-Br bond. In doing so, both carbons in the alkene/alkyne will have their p-orbitals hybridized with the sp/sp2 orbital to form the sp2/sp3 orbital respectively. However, the sp2 bonding orbital will have more ring strain, as it naturally exists with bond angles of 120 degrees away from other sp2 orbitals, while the sp3 orbital is somewhat less strained with bond angles ~109.5 degrees apart. Thus the higher energy transition state will make the alkyne less reactive, and bromination occurs on the alkene first.

In addition, (as was pointed out here : https://www.masterorganicchemistry.com/2013/05/29/alkyne-addition-pathways-the-3-membered-ring-pathway/) the p orbitals on Cl conjugate to the pi-bond on the alkene transition state, resulting in a 4-electron anti-aromatic system! This further increases the instability of the transition state and results in a slower reaction.
TL;DR: There's more ring strain on the alkyne's transition state because the carbons are sp2 hybridized, and its transition state involves the formation of an anti-aromatic ring, so the reaction occurs slower than bromination on the alkene.

Nice summary. I would just add a cautionary note about intermediates vs transition states: in this case, the possible factors which destabilise the intermediate (ring strain and antiaromaticity) also destabilise the corresponding transition state (by the Hammond postulate).

@orthocresol It's tricky to emphasize the instability of the transition state without bringing up the instability of the intermediates, but I agree it's important to avoid confusing the kinetic role of transition state stability vs the thermodynamic role of intermediate stability. Do you know of any examples where an transition state is destabilized but an intermediate isn't, or vice versa?
p.s. gimme a break i dont have a diploma yet :^P

No, I can't think of an example offhand. That wasn't meant as criticism, by the way, I thought it was just worth a mention.

@orthocresol according to this summary, even in OMDM reaction alkenes will be chosen over alkynes right?

A carbon-carbon double bond is about 133 pm in length; a carbon-carbon triple bond is 120 pm long. It would seem that the pi bonds in a double bond are a wee bit more diffuse and might be a wee bit more attractive to a bromonium ion.
The specification of "low temperature" suggests that the selectivity is not very great. It would be interesting to know the ratio of products at different temperatures.

In general, halo addition is an 'Electrophillic' addition, not a 'Nucleophillic' addition and hence it is $\ce{Br+}$ which attacks first on the saturated hydrocarbon.
Now coming to your question,
According to Jerry March (4th edition, Pg748) Advanced Organic Chemistry,

There is a higher concentration of electrons between the
  carbons of the triple bond than in a double bond, and yet triple
  bonds are less subject to electrophilic attack and more subject
  to nucleophilic attack than double bond.

The probable cause, as suggested by Sujith seems a pretty reasonable explanation backed up by the fact that while additions with 'bridged' intermediates react faster with double bonds, additions without such intermediates react almost at the same rate with both entities.

Could this be attributed to the electronegativity difference caused due to the higher s-character?

It probably has a (small) reinforcing effect, but instability of the intermediate still seems far more likely as the major cause of difference to me.


text:
How to do pre-treatment on surface and how to clean the glass surface after silanization?
By Silanization, I want to treat surface of some spherical glass beads. Each bead is supposed to be shot in a particle launcher (like a bullet from a gun). I have some detailed question for the process:
1) Do you think that the hydrophobic coating which is created on the glass bead surface is durable enough that not to be removed or scratched by friction due to SOFT rubbing to the particle launcher tube?
2) Before silanization, should I perform any pre-treatment on glass surface? (e.g. pouting the glass in acid, or heating the glass surface, etc ...)
3) After I take out the glass from pool of silan, the glass surface is partially covered with reactant agents which is hardly cleaned with water. How should I clean the silanized surface after taking out from the pool? (using acetone and rubbing with a napkin is okay, etc)


text:
Can baking soda or citric acid help clean a pot with burned-in milk rice?
The other day I was cooking milk rice and forgot about it so it burned a little on the bottom. This lower layer was very hard to clean. A lot of websites recommend boiling water in the pot with baking soda (Sodium bicarbonate) or citric acid to dissolve the burned residue and make it easier to clean the pot. I tried this but it did not help. I doubt there is even a chemical reaction between these reagents that would help in any way.
Is this another myth or is there scientific basis for given recommendation?

Not sure about acid, but soda does help sometimes, and that's for a reason.

The hotter the water, the more likely it will be to remove burnt-on starch. Not only does $\ce{NaHCO3}$ raise the boiling point of the water, but it also helps saponify fats (actually, "washing soda", $\ce{Na2CO3}$ would be better).
However, you still need to do a lot of scrubbing afterwards to remove the carbonized layer.

1.) I suppose the carbonized layer is by far the hardest part to remove so this is what sodium bicobonate and acid should work on. 2.) Isn't burnt-on-starch the same as carbonized layer. Your answer implies them to be different, if I understand it correctly.

Use either sodium bicarbonate or acid, not both at once. 2. You can't clean the hard bottom layer off until you remove the partially burnt porridge. 3. No, the alkali will not significantly remove pure baked-on carbon -- use abrasive & elbow grease.

Then is there any significant advantage to using NaHCO3 instead of regular dish liquid to solve the fats? If there isn't then the recommendation is non-effective and a myth. Since as you already stated removing the carbonized layer does not become easier with NaHCO3 or acid.

Boiling with liquid detergent will cover the stove in foam.  If you prefer, use detergent for dish washing machines, which is alkaline and does not foam.

Since an abrasive action is required to remove the burnt protein material after a significant amount of baking soda is covered, I suggest using a stiff plastic scraper and pressure to lift layers of the adhered material.  The peeling tensions, shear, of a dry stiff material help with removal.
I found that a 1/8 to 1/4 inch thick application of baking soda with a spray of water without stirring or puddling of water encourage the ions to migrate and bond at the metal interface.
After the application is dried, a spray of water will encourage more ions to combined at the metal interface.
We might view this as a way of drying the burnt material like we might dry meat.  The tension and bonding at the metal surface is releasing and favoring the burnt protein structure.
So the adhesion of the protein to the material surface is altered by changing the properties of the Burt material to create more stress or tension or contraction by shortening the bonds of the protein and changing the bond strength at metal or ceramic surface.
Adding water relieves the stress.  So a dry mechanical scraper acts like a knife lifting encouraging the dry tensile material to pull away from the pan.
Take the wet pan with no liquid in it.
Coat the bottom area with thick Baking Soda
Spray mist the baking soda with water
Let dry for 4 hrs
Spray mist with water again
Let dry over night
Scrape the dry material with a strong force of a hard plastic kitchen scraper blade
Persist with strong force from clean surface across material coating.
This applies to stainless steel counters and stoves

Welcome to ChemSE.  Initially I assumed putting three times the same image into the reply was an accident.  Reading your answer now leads me to the speculation you either were distracted, stressed or/and tired.  If so, stay safe in your doing now, and be welcome to polish your answer in about a day.

Soaking burnt on food with baking soda - I do it all the time!  I think boiling the pot is counter productive.  Just let it sit covered with water and baking soda - lots of baking soda.  It's very cheap.  B
Putting the pot back on the stove just heats the burnt food more potentially making it even harder to remove.  Just soak it - overnight for a serious build up.  You still might have to abrasively remove the last of the residue.
I just cleaned a Pyrex casserole from dinner that had cheese scalloped potatoes burnt all around the top edge.  Soaked it in q larger bowl covered with a water and a generous shake of baking soda.  A few hours later it all came off with very little effort.


text:
Why is tetramethylammonium hydroxide a stronger base than tri- or dimethyl amine?

Of the following, which compound is the most basic??

$\ce{Me3N}$
$\ce{Me2NH}$
$\ce{Me4NOH}$
$\ce{NaOH}$

Between 1 and 2, 1 is more basic as the 3 methyl groups would produce a +I effect resulting in better availability of the lone pair than in option 1.
But the book says it is 3. I don't understand why. I do make out that it is ionic, but how does this fact make it more basic than option 1?

Trying to compare $\ce{NaOH}$ with $\ce{Me4NOH}$ sounds like tricky business because the cations seem like they should solvate rather differently, and that's hard to account for qualitatively. Does anyone have a reliable source for the basicities of both compounds, in any solvent?

Between 1 and 2, 2 is more basic as the 3 methyl groups would produce a +I effect resulting in better availability of the lone pair than in option 2.

I agree that more methyl groups would result in a greater +I effect.

But the book says it is 3. I don't understand why. I do make out that it is ionic, but how does this fact make it more basic than option 1?

The fully methylated ammonium part is not basic. It has no easily labile proton. As a result, tetramethylammonium hydroxide acts as a plain hydroxide salt, like sodium hydroxide. Hydroxide ion is much more basic than ammonia (at least in water).
So we've eliminated the amines with lone pairs as contenders for "most basic" because we've established hydroxide to be much more basic than an amine, regardless of the extent of the +I effect in an amine.
We're left with $\ce{NaOH}$ and tetramethylammonium hydroxide. The only difference between the two is the cation.
$\ce{Na+}$ is electron-withdrawing (electronegative); it has a positive charge. This lessens (stabilizes) the negative charge density on the hydroxide portion, reducing the basicity of the hydroxide ion slightly.
$\ce{(H3C)4N+}$ has a similar effect on the hydroxide ion part. However, tetramethylammonium ion has four inductively donating methyl groups to lessen its electron-withdrawing tendency. I suspect this is what your book is getting at.
Some speculation follows. In addition, $\ce{Na+}$ is likely smaller than $\ce{(H3C)4N+}$, which has a lot of bulk associated with its four methyl groups. This likely allows $\ce{Na+}$ ions to more effectively solvate and thereby stabilize hydroxide ions in solution. This again decreases basicity, but I'm unsure of how much of an effect this has, since the solution molecules will be present in a much greater quantity than the solute ions.

Proper words here are: weaker coordination.

Also case with amines is more complicated.

@Mithoron - can you elaborate on why the case with amines is more complicated?

authors: ص2اسامه المنصوري


text:
Disproportionation of hydrogen peroxide
Hydrogen peroxide is decomposed as follows:
$$\ce{2H2O2 -> 2H2O +O2}$$
This is a disproportionation redox reaction of $\ce{H2O2}$ involving the 2 half reactions
$$\ce{H2O2 -> O2 + 2H^+ + 2e-}$$
$$\ce{H2O2 + 2e- + 2H^+ -> 2H2O}$$
But I noticed semantically that it can also be the sum of the two following half reactions:
$$\ce{2H2O2 -> 2O2 + 4H^+ + 4e-}$$
$$\ce{O2 + 4H^+ + 4e- -> 2H2O}$$
Both reactions are feasible according to the following:
$$\begin{array}{ccc}
\text{Oxidised species} & \text{Reduced species} & E^\circ (\mathrm{V}) \\
\hline
\ce{H2O2} & \ce{H2O} & 1.763 \\
\ce{O2} & \ce{H2O} & 1.23 \\
\ce{O2} & \ce{H2O2} & 0.695 \\
\end{array}$$
Can it happen in the second way?
If yes, can you explain the mechanism in simple terms?

If it is what I think it is, and the fourth equation is just a typo (H2O should be H2O2), then yes, it is formally correct. However, if you just keep H2O2 by itself, without any oxygen, it will still decompose. This isn't captured by your proposed pair of half-equations, which essentially says that you need oxygen gas for the total reaction (i.e. sum of 2 half-reactions) to proceed.

In principle you could find a lot of equations, all leading to the same result. But these are only (half-)equations. Most of the times the reaction mechanism is much more complicated than just these reaction equations.
The decomposition of hydrogenperoxide is still an active field of research in various media with various catalysts.
Unfortunately with the massive overflow of the above, I was only able to find a quite old PhD thesis by Trice Walter Haas (Iowa State University, 1960)
Here is a small line-up of reaction equations that will play a role in the decomposition, at least to that part that I can imagine it. And I am just assuming that there are radicals involved, because there usually are.
\begin{align}
\ce{H2O2 &<=> H+ + HO2- }\\
\ce{HO2- + H2O2 &<=> HO2 + HO + OH-}\\
\ce{HO2 + OH &<=> H2O + O2}\\
\end{align}


text:
Which of the following has higher enol content?

Which of the following has higher enol content?

The enol form of (A) and (B) is:

Both the enols have hydrogen bonding.
If the oxygen were to accept the pi electrns from carbon forming a carbocation like this (a resonance structure):

the compound becomes stabilized due to resonance as well. So isn't (B) the answer?

You think right. It is the question itself that is wrong (too easy). It should have contained a double bond in (A) as well, and that's where things would start getting somewhat interesting.

If there was another double bond in (A), then its resonance structure will be anti aromatic, making it unstable. Is that correct?

Exactly! While (B) is aromatic and feels quite well.

@AdityaDev - Since you are correct, consider expanding on your explanation and posting it as an answer to your own question. If you are able to do so, you are encouraged to answer your own question.

OK. I will post an answer.

The two main factors affecting the stability of the enol forms of (A) and (B) are

Hydrogen bonding
Aromaticity

In the enol form of both compounds, hydrogen bonding is present:

But the resonance structure of enol form of (B) is aromatic. It contains $6$ $\pi$-electrons:

The enol form of (A) lacks this stability. Hence, the enol content of (B) is more than that of (A)


text:
Order of reactivity of carbonyl compounds towards nucleophilic addition

Arrange the following according to reactivity towards nucleophilic addition reaction:

I have already seen this post regarding the reactivity. It helped me to some extent.
$\ce{Cl}$ increases the positive charge on the carbonyl carbon increasing its reactivity.
So (B) > (A) > (the remaining 3)
Further, aldehydes are more reactive than ketones. But I should also consider steric factors. In compounds (C), (D), and (E), compound (D) has highest carbocation stability but also has highest steric hinderance.
Compound (C) has lowest steric hinderance but low carbocation stability as well. How can I arrance (C), (D), and (E)?
To all those who say steric effects dont play a role: This is taken from page 147 of organic chemistry book by Clayden:

The same structural features that favour or disfavour hydrate formation are
  important in determining the reactivity of carbonyl compounds with other
  nucleophiles, whether the reactions are reversible or not. Steric hindrance and
  more alkyl substituents make carbonyl compounds less reactive towards any
  nucleophile; electron-withdrawing groups and small rings make them more reactive

It will be b>a>c>e>d In nucleophillic addition reaction carbocation doesn't form. You Compare only by electrophillic nature and steric hinderance.

Also remember that formaldehyde is the most reactive aldehyde due to least steric hinderance

B and A are not acid chlorides.

Sorry. But the Cl increases positive charge on the carbonyl carbon. I will edit the question

There's no use of carbocation stability if the bulky groups prevent nucleophiles from attacking there, as per molcalc the order should be $C>E\geq D>B\geq A$

You are saying that  (D) has highest carbocation stability, or more correctly, the positive charge on the carbon is least in (D) among (C),(D),(E).That makes it less reactive, and so does the steric factor.They are not opposing each other, both steric and electronic effects show the same trend, so (D) is least reactive considering any of the two.

In both Himanshu Pandey and MS chauhan, different answers are given. According to MS chauhan's answer, formaldehyde comes first. According to Himanshu Pandey, B comes first.

Carbocation doesn't come into the picture at all, when we are discussing about nucleophilic addition reactions. Steric hindrance will also not play a role because the compounds are planar (attack is going to take place on $sp^2$ hybridised carbon). The nucleophile is going to attack only from the top and bottom* as shown:

The only factors that come into picture in the options are the electron withdrawing or releasing nature of the groups attached to the carbonyl carbon. If the group is $e^-$ withdrawing in nature, the deficiency on the carbon atom increases, making it susceptible to nucleophilic attack.
So, the correct order is : B > A > C > E > D
P.S.:* Since the attack is from top and bottom, both of which are equally likely, the resulting product is a racemic mixture (if the carbon on which attack happens is chiral after the attack).

Read this: https://www.google.co.in/url?sa=t&source=web&rct=j&url=http://www.chem.ucla.edu/harding/notes/notes_14D_C%3DOfun01.pdf&ved=0ahUKEwiUjY7fqffJAhWGcI4KHe2aCZsQFggjMAI&usg=AFQjCNEl6cImoq1NdAUGmgBQ2XW6e0Q0XQ&sig2=M1W6iKQqz1vphDbov0hfjw

And read this answer by Greg: http://chemistry.stackexchange.com/questions/7412/order-of-reactivity-of-carbonyl-compounds-to-nucleophilic-addition-reaction

Also, you modify those arrows showing nucleophilic attack since its not exactly perpendicular to the x-z plane. Due to repulsions from the filled $\pi$ orbitals of carbon, the nucleophile approaches at an obtuse angle. You can read about it by searching Burgi-Dunitz angle on Google.

authors: عبد العزيز الزعبي


text:
Calorimeter and bond energy
Why does 1,3-Butadiene has different standard heat of formation value calculated from calorimeter than the one calculated from bond energies?

Its because of the resonance. buta-1,3-diene is resonance stabilized. The Heat of combustion actually will be less than calculated theoretically because some of energy released will be used to destabilize the molecule.


text:
Mixing naphthalene, formaldehyde and sulfuric acid (Marquis reagent + naphthalene)
Excuse me if this a dumb question but it has haunted me for years. Back in my old school days I mixed  naphthalene, formaldehyde and sulfuric acid in a mad attempt to make $\mbox{2-Methoxynaphthalene}$. I can't remember if I heated the mixture or not.
It turned black instantly and my teacher at school couldn't tell me what had happened, even less why. Did I produce nano-carbon? Just soot? Or something else?
Edit:
way back then, I did try the reaction without naphthalene, and also without formaldehyde, with no (noticeable) result. My teacher also repeated the experiment with the considerably more pure chemicals available at the school lab.
Edit2: as of the comment by Loong, what I did was using Marquis reagent on naphthalene. What I found on the net about this was very little and poor. I couldn't even find anything on the color of $\mbox{polymethylnaphthene}$. (Instinct tells me it should be colorless.)

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center.

@Ϻ.Λ.Ʀ. this has nothing to do with my question here, but could you please give a newcomer like me some hint on how to do all those beautiful formulas? I saw your \mbox and that is quite familiar, but most formulas I have looked at by now are pictures, mostly *.png and I think they are not hand-made with some primitive drawing-tool.

Please visit this page, this page and this one on how to format your posts better using the Mathjax syntax our site uses. Regarding the structural formulas, there are certain molecular editors (some online), that you can use for this purpose. I suggest just Googling for "molecular editor" and choose the one you're most comfy with.

@GyroGearloose Your description reads as if you made a Marquis reaction.

@Loong Thank you, so I produced polymethylnaphthalene?

Marquis Reagent (1) is a spot test for alkaloids, first reported in 1896. The original agent was a mixture of 2 drops of 40% formaldehyde and 3 milliliters of concentrated sulfuric acid and was used to detect certain (e.g., opium) alkaloids and distinguish between them. The signature of the alkaloid is both the initial color produced and the sequence of color changes with time. Morphine reacts with Marquis' Reagent to give a purple to violet color. Two molecules of morphine and two of formaldehyde are proposed to condense in the presence of sulfuric acid to the dimeric product which is protonated to the oxoniumcarbenium salt.
The key reaction is protonation of formaldehyde, which attacks an aromatic ring, ortho to the activating phenol group, producing a hydroxymethyl derivative which is immediately protonated and dehydroxylated to give a benzylic cation which attacks another aromatic ring. And it happens again at ortho positions to close a new ring.

Now the ingredients in your old school days reaction (naphthalene, sulfuric acid and formaldehyde) are used to make what was a common concrete admixture. Kao Soap (2), who made Mighty 150 (“Naphthalenesulonic acid-formaldehyde condensate sodium salt”) offered it to my company. We were making a similar material for paint dispersion and promoted our material for concrete strengthening by water reduction.
The condensate is a polymer and is made by first sulfonating napthalene in the beta position with concentrated sulfuric acid. (The alpha sulfonate is the first product, but quickly rearranges to the more stable beta isomer.) Then formalin, 40%, is added, slowly. The formaldehyde protonates, attacks the alpha position of the (still activated) unsulfonated ring (the sulfonated ring is deactivated), forms a hydroxy methyl aryl group, is immediately protonated, dehydrated and attacks another naphthalene sulfonate in the alpha position of the unsulfonated ring.
$\ce{H2SO4 + H2CO -> H2COH+ + HSO4-}$
$\ce{H2COH+ + HO3SNap -> HO3SNapCH2OH + H+ -> HO3SNapCH2+}$
$\ce{HO_3SNapCH_2^+ + HO_3SNap -> HO_3SNapCH_2NapSO_3H}$
This methylene group is the first link of a polymer chain. The other alpha positions (one on each unsulfonated ring) are still open and will be attacked if there is enough formaldehyde. Controlling the amount of formaldehyde added is a method of controlling ultimate molecular weight of the polymer, which affects its dispersing power because of the ability of longer polymer chains to adsorb more strongly and repel other particles without deabsorption of the polymer.
Controlling the rate of formaldehyde addition is key to preventing runaway polymerization. The addition of formaldehyde at near room temperature lowers the temperature of the 500-1000 gallon batch; however, the batch temperature must be raised to cause polymerization, which is exothermic - sulfuric acid generates heat as it hydrates - and the batch viscosity is already very high, so the heat cannot be quickly removed.
There is a possibility of a runaway reaction in your mixture: the naphthalene is not even sulfonated, so both rings are activated (relative to benzene). If you did get 2-methoxynaphthalene as an intermediate, the sulfuric acid would protonate the methoxy oxygen, generating a benzylic cation, and water and heat of hydration, and a polymer with chains coming out of up to 4 alpha positions per naphthalene. Now the typical naphthalene sulfonate condensates (the acids and the sodium salts) are brown (light to dark), but I could imagine a polymer coming out with a black color.
This approach to 2-methoxynaphthalene seems destined to fail, because the protonation of 2-methoxynaphthalene is easier than  protonation of formaldehyde. But organic chemists frequently generate tars.
Ref.1 https://en.wikipedia.org/wiki/Marquis_reagent
Ref.2 https://chemical.kao.com/en/products/B0012344_en/?region=gl


text:
How can the size of atom be the cause of octet expansion?
Lately, I was reading The Lewis Theory of Covalent Bonding by Peter Atkins in Appendix 4 of 'Elements of Physical Chemistry'. There he was talking about expansion of octet . As he wrote:

Many molecules cannot be written in a way that conforms to the octet rule. Those classified as hypervalent molecules requires an expansion of octet. Although it is often stated that octet expansion requires the involvement of d-orbitals, and is therefore confined to Period 3 and subsequent elements, there is good evidence to suggest that octet expansion is a consequence of an atom's size, not its intrinsic orbital structure. [...]

I had always bore in mind that octet expansion was always due to d orbitals.
But size of atoms? How can it be related to octet expansion?

Since the central atom in such compounds is coordinated to more groups than expected, but is not expanding its valence electron shell, the term hypercoordinated is generally preferred over hypervalent.

Your revised version of Atkins’ statement is correct, no hypervalent main group compounds require the involvement of d-orbitals. Rather, all of those molecules formerly explained by d-orbital participation can just as well be explained with 4-electron-3-centre bonds (including resonance structures) and charge separation (e.g. in sulphate; no need for resonance), giving rise to all-octet Lewis structures. In fact, calculations have shown the d-orbital participation to be minimal and neglegible at best.
A much more relevant question is whether more than $x$ substituents actually fit around the central atom. This is especially notable for the difference between the second and third periods: The reason why nitric acid is $\ce{HNO3}$ while phosphoric acid is $\ce{H3PO4}$ is likely due to the much smaller size of nitrogen, not allowing for tetracoordination by oxygens in the $\mathrm{+V}$ oxidation state while the larger phosphorus atom can support four oxygen atoms. If I remember correctly, that was also discussed as a reason why $\ce{NF5}$ hadn’t been found at the time of Klapötke’s article while $\ce{PF5}$ has been known for quite a long time.
It gets more pronounced when you go even lower in the periodic table, i.e. to higher periods. Antimony is known to be coordinated by six oxygens in the $\mathrm{+V}$ oxidation state ($\ce{H7SbO6}$, if you wish, although $\ce{H[Sb(OH)6]}$ is a more accurate description) — something phosphorus could technically do ($\ce{PF6-}$ with small fluorine atoms is known) but usually doesn’t. It is mainly the greater atom size of antimony that allows the coordination of two additional oxygen atoms.


text:
Which of the compounds give the same SN1 and SN2 products?

Which of the following compounds give same SN1 and SN2 product?

I think (b) should be the answer, but the given answer is (c). I think so as it is the only compound where carbon is not chiral. While in others racemic mixture in SN1 and enantiomer in SN2 are formed, in (b) there is no question about this. Please tell me if I got it right.

If B were to go for SN1 mechanism, rearrangement will take place to form a more stable carbocation.

‘where C is not chiral’ … which C?

C means Carbon here.

Yes, I was fully aware that C meant carbon. But you have many carbons.

For starters, none of your products show chirality in any way, so it does not matter whether stereoinformation is lost or whether you convert a racemic mixture into a racemic mixture where every single molecule if you had labelled it pre-reaction now had the opposite stereochemistry — in both cases, both reactants and products are racemic mixtures. So you cannot take the presence or absence of chiraliy to distinguish here.
What we can say, is that every compound can take part in an $\mathrm{S_N2}$ reaction (some faster, some slower) but not every one will easily take part in an $\mathrm{S_N1}$ reaction. Specificly, (a) and (d) would give a secondary carbocation intermediate while (b) would even give a primary carbocation. These cations are very unstable and would immediately undergo Wagner-Meerwein type rearrangements towards more stable — read tertiary or resonance stabilised — carbocations. Or, to quote one of my professors from uni:

Exactly one single unstabilised secondary carbocation exists, and that is the prop-2-ylium cation. (Because it cannot rearrange to a more stable one.)

Thus, only (c) will form a stable carbocation without any rearrangement happening: It would be an allyl cation which is well resonance stabilised. But now we have to dig a step deeper, because while all other reactants will certainly not give the same product in both cases, it could be that the correct answer is ‘none of these’, i.e. that somehow we can distinguish between products — because once we have the allyl cation, we could let the nucleophile attack on carbon 1 or carbon 3. Thankfully, though, both attacks give the same product here, because there are no other asymmetries present in the molecule, so we can transform one into the other by rotation.
And, never forget, the $\mathrm{S_N2}$ reaction could also follow the $\mathrm{S_N2'}$ mechanism — this, too, would switch the identities of the two carbons, so we can truly say that both mechanisms will lead to the same product.

Those compounds in which 1,2 H-Shift is not possible to increase carbocation stability in $\mathrm{S_N1}$ reactions give same $\mathrm{S_N1}$ and $\mathrm{S_N2}$ product. It is because $\mathrm{S_N2}$ is a one step reaction and substitution takes place without rearrangement. Therefore answer would be (C)

I fail to understand this answer.

Can you tell which line you are not able to understand so I can clarify

I can’t understand how the first and second sentence go together.

I think compound c will react either via $\mathrm{S_N1}$ and $\mathrm{S_N2}$ and the product will be the same. Allylic carbon reacts by $\mathrm{S_N1}$ because the stability of carbocation and reacts via $\mathrm{S_N2}$ because its primary carbon.
Hope that helps.

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


text:
What is the chemical redox equation between hydrogen peroxide and melanin?
I am investigating hair bleach, and want to figure out the redox reaction between $\ce{H2O2}$ (hydrogen peroxide) and $\ce{C18H10N2O4}$ (melanin). $\ce{H2O2}$ is a strong oxidizing agent in this redox equation, but I don't know how to write that out. Someone please help?

Your formula for melanin is incorrect. Melanin does not have a clear and concise chemical formula; it is a class of natural polymers that differ in their exact composition. The only thing they have in common is that they derive from the amino acid tyrosine and that they at one point polymerised.
Therefore, unfortunately there cannot be an exact description of what happened between melanin and hydrogen peroxide. You can draw a half-equation for the hydrogen peroxide part which would be:
$$\ce{H2O2 + 2 e- + 2 H+ -> 2 H2O}$$
But what exactly happens in melanin and which groups are oxidised is hard to say. What probably happens is a series of radical reactions that in some way oxidises away the aromatic system and thereby moves the absorpion wavelength out of the visible spectrum. But that’s probably about as much as you can say.


text:
Why did I get these inconsistent experimental results in silanization?
I used a 50% wt mixture of Tetraethylorthosilicate + Kerosene for silanization of some glass plates with different treatment times (from 1 sec to 35 sec). However, the contact angle(CA) that I got does not seem consistent as it does not always increase as I increase the treatment time (when treatment time increases, the CA fluctuates).
What might be the reason? and whether I should change the method or the solution(s)?
P.S. I cleaned the glass with acetone in a sonicator. Then I put the glass into solution and after specific time put it into water to clean the solution from surface and then clean the remaining oil (from kerosene) by rubbing a napkin to the glass plate. Then I measured the CA.

Rubbing is a very inconsistent process, and will introduce a lot of variation. I would suggest using a volatile non-polar solvent such as acetone, hexanes, or ethyl acetate to remove residual kerosene and allowing to air dry.

So, do you mean the main issue that may cause this fluctuations in CA could be this rubbing that I've done with a napkin?(I mean do you think that if I perform the process correctly, I should basically get the lowest CA in 1 sec treatment and the highest CA in 35 sec. Is that right?)

In theory there should be a correlation, but tetraethoxysilane (TEOS) is not a good chemical for silanization, as the ethoxy group can easily hydrate leaving you with a hydrophilic $\ce{Si}-\ce{O}-\ce{H}$ surface.  I did assume everything else you did was the same, but factors such as humidity and exposure time to air can affect results due the the nature of TEOS

I have the following three more solutions available. Based on your above comment that TEOS is not very appropriate for silanization, do you think which solution is the more suitable one  for creating different levels of hydrophobicity? (please also let me know if there are any instructions for silanizing that you think I should know) 1) ChloroTrimethylSilan 2) OctaDecylTrichloroSilan 3) DichloroDimethylSilan

I like chlorotrimethylsilane because it has only one chloro group, and octadecyltrichlorosilane because it is sterically bulky. The latter should work best from the steric, but the former gives complete surface reaction, so I cannot say with certainty.

One other thing to consider. Are you using a consistent amount of water?  This can affect your contact angle.

I'm not sure what you mean by "using water" as I use water only for cleaning purposes. Could you please explain it?

What material are you using for contact angle measurements?

Oh, I see what you mean. Yes, I don't change the water drop volume in all my CA measurements.

I need one CA less than 90 as well and I realized that when I use kerosene in my silan solution, it is hardly cleaned from glass surface. Since each coated particle is only one time in contact to water (and the coating cannot be easily washed), do you think I can use only kerosene as a coating to give a ~60 deg CA to my glass particle that I want to launch them to my water droplets? OR it may not give a consistent CA.


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Formation of a stable hydrate

Question: Which of the following does not form a stable hydrate?

I think they mean nucleophilic addition of water. For compound (C), on reaction with water, the product formed is

The product has lesser bond angle strain than the reactant. Hence, the compound formed is stable. This is because $\ce{sp^2}$ hybridised carbonyl carbon requires $120^{\circ}$ but only gets $60^{\circ}$ in cyclopropanone. But converting it into a diol makes the angle $109^{\circ}$ which reduces strain to a large extent. In fact, cyclobutanone is also reactive to water.
For (B) however, there is not much change in the bond angle strain. The initial angle is already close to the required angle.
What can I say about (A) and (D)?

D forms stable hydrate due to intra molecular H-Bonding

So the answer is B. Its correct.

What can I say about (A) and (D)?

When a carbonyl compound and water are present together, an equilibrium is set up between the carbonyl compound and the corresponding gem-diol. Like any equilibrium, this equilibrium can be shifted to one side or the other by factors that stabilize (or destabilize) one side or the other.
If water is present, poly-vicinal carbonyl compounds usually exist as hydrates. Drawing the resonance structure of a carbonyl compound we see the the $\ce{C=O}$ bond is highly polarized with the carbon terminus being the positive end.  If we have several carbonyls adjacent to one another, then each carbonyl has this same polarization and an electrostatic destabilization results. This is particularly true for the central carbon since there is a partial positive charge on each of the carbons attached to it.

In order to avoid this electrostatic destabilization, such compounds will form a hydrate at the central (the most electrostatically destabilized) carbon. In the hydrated form there is no repulsive partial positive charge on adjacent carbons.
You've already figured out why cyclopropanone exists as a hydrate.  So the correct answer is B, since there is no driving force that would provide any special stabilization of its hydrate.

Through option elimination we can arrive at a conclusion that B is the answer. But how does one explain that B is the right answer?


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Why only one peak is observed in NMR spectrum of H2?
Assuming a downward magnetic field is applied and the nuclear spin of two equivalent protons in $\ce{H2}$ are all in ground state $\downarrow\downarrow$ initially. Would the required energy of one of the protons excited to spin up state $\uparrow\downarrow$ be different from the energy of proton excited form state $\uparrow\downarrow$ to state $\uparrow\uparrow$ ?
And if it's true why only one peak is observed in NMR spectrum?

Coupling between equivalent spins cannot be observed in the NMR spectrum; this is explained quite thoroughly here: https://chemistry.stackexchange.com/q/53478/16683.

For starters, it is incorrect to think of the ground state being substantially more populated in NMR spectroscopy. Because the energy difference between aligned and misaligned nuclear spins in the external magnetic field is very low, we can say that both states are almost equally populated due to thermal excitation alone. For a $500~\mathrm{MHz}$ experiment, the energy difference can be calculated to be $0.21~\mathrm{J \cdot mol^{-1}}$ which corresponds to a population ratio as follows:
$$\frac{n_\mathrm{ex}}{n_\mathrm{g}} = \exp \left(\frac{\Delta E}{RT}\right) = 0.999916$$
Next, consider the extremely high symmetry of the $\ce{H2}$ molecule: It has the point group $D_{\infty\mathrm h}$, one of the most symmetric ones around. Due to this high symmetry, we cannot distinguish between either proton at all — both can be transformed into each other by a rotation as defined by the point group. Thus, the two nuclei are magnetically equivalent — not only must their chemical shift be identical but also the coupling to the other one.
Think of it this way: Assume, that the spin transition of one proton happened at a slightly different frequency than the other’s. That would mean that one proton had to be different from the other in some way (a different transition energy is equivalent to a different chemical shift is equivalent to a different environment). But how would you explain a different environment on one side of the $\ce{H2}$ molecule compared to the other? Correct: You cannot.
Now we could still assume that the energy required to excite a nuclear spin whose neighbour is parallel be different from one whose neighbour is antiparallel. However, no matter how we excite, the transition is always $\ce{parallel <=> antiparallel}$. And since we cannot distinguish between the hydrogens, we also cannot distinguish which one pointed in which direction, so those two energies must be identical. Therefore, we only see one peak and no $^1J_\ce{HH}$ coupling.

Well, if no magnetic field applied, would the energy level of parallel states $\uparrow\uparrow  or \downarrow\downarrow$ be different from the  energy level of antiparallel states $\uparrow\downarrow  or \downarrow\uparrow$? I think there should be a energy different because of the spin-spin coupling, and when magnetic field applied, the two degenerate parallel states should split equally in energy difference. Therefore, the transition energy between antiparallel states and two distinct parallel states should be different. Am I thinking right?

You are correct in that ↓↓ and ↑↑ are different from ↓↑ and ↑↓. However, each pair of them is identical. However, ↑↑ is still identical to ↑↑ and the transition between antiparallel and parallel is identical no matter what the different ones are.

Technically, for two magnetically equivalent spins, the transitions are no longer between αα/αβ/βα/ββ but rather between linear combinations of them. This is explained more thoroughly in the answers to https://chemistry.stackexchange.com/q/53478/16683.

It’s also slightly inaccurate to say that αα and ββ are the same, as the α and β spin states are no longer degenerate in the presence of an external magnetic field. The correct point is that the transition (to αβ, or βα, or more accurately, (2^-1/2)(αβ+βα)) occurs at the same energy.

If by referring to peak you mean a single chemical shift — in low-resolution H-NMR — the number of peaks available on the spectrum would depend on the type of hydrogen environments there are. For example, ethanol ($\ce{C2H6O}$) has three different H environments: $\ce{C\mathbf{H}_3-C\mathbf{H}_2-O\mathbf{H}}$.

(source: rsc-cdn.org)
The H's in the two methyl groups are of different environments because one methyl group has its electrons pulled on stronger by the oxygen in the hydroxide functional group ($\ce{OH}$).
In that case, $\ce{H2}$ only has one hydrogen environment, with both hydrogen atoms in the same environment, resulting in a single chemical shift signal.
However, you could have meant the spin-spin coupling determined by high-resolution H-NMR — the one where the tinier individual peaks within a certain signal are visible. The number of tinier peaks that show up for a particular chemical shift is dependent on the number of hydrogen atoms on a neighboring carbon atom.

(source: alevelchem.com)
In ethanol, the hydrogen environment of the central carbon atom ($\ce{CH3-C\mathbf{H}_2-OH}$) would have a quartet (4 peaked) signal because the neighboring carbon has 3 hydrogen atoms. The number of peaks due to spin-spin coupling is = number of neighboring hydrogen atoms + 1
With respect to $\ce{H2}$, the hydrogen atoms have no neighboring hydrogen atoms — there is no carbon in $\ce{H2}$ — so a singlet or single peak would apeak above the chemical shift for $\ce{H2}$. Even if a compound contained carbon atoms, like the H-NMR standard tetramethylsilane $\ce{Si(CH3)4}$ only a singlet is produced because all hydrogen atoms are in the same environment of a methyl group attached to a silicon atom.

This is quite false. The “n+1 rule” is a very basic simplification which assumes that only three-bond couplings contribute to the multiplet shape. While couplings over three bonds are the most commonly seen ones, it is not correct to imply that they are the only types of coupling. One-bond coupling is quite a real thing, it’s just that we cannot observe it due to the equivalence of the spins.


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Why at constant pressure and temperature Gibbs energy change of a process can be negative?
The following two statements appear to contradict each other:

For a spontaneous reaction at constant temperature $T$ and pressure $p$ the change in Gibbs free energy can be negative $(\mathrm dG < 0).$
Since
$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT,\tag{1}$$
at constant pressure $(\mathrm dp = 0)$ and constant temperature $(\mathrm dT = 0)$ the change in Gibbs free energy is always zero:
$$\mathrm dG = V\cdot 0 - T\cdot 0 = 0.\tag{2}$$

Why does this happen?

Statement 2 is applicable when there is no reaction, otherwise it would be more complicated.

The relation
$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT\tag{1}$$
implies that the Gibbs free energy of the system depends only on the two variables $T$ and $p$, e.g., as in the case of a single-component gas consisting of a fixed number of molecules. For such systems, the macrostate is completely determined if $T$ and $p$ are given. Nothing can change further.
On the other hand, suppose that there is an additional parameter $X$ describing the system and that $X$ is not fixed. Then, according to the second law of thermodynamics, the system, being at constant $T$ and $p$, self-adjusts $X$ in such a way that $G$ is minimized.
In case there exist multiple species that can undergo a chemical reaction, $X$ could be a measure of how far the reaction has proceeded. The condition
$$\frac{\partial G}{\partial X} = 0\tag{2}$$
gives the well-known law of chemical equilibrium.

This is a 6 yr old query. But I will try to answer this question. Equation of dG for electrochemistry (for galvanic cell for example) is dG = -SdT + VdP + EdQ. Where E = electric potential of anode/cathode of the galvanic cell and dQ is the amount of charge reduced from anode or added to the cathode. So for galvanic cell, dG < 0 since it is spontaneous (no required power source to force a reaction). Negative charge like electrons flow from lower potential to higher potential with no power source required. For anode, dG_A = +E_AdQ (since it reduces electrons or "adds positive charge"). For cathode, dG_C = -E_CdQ (since it adds electrons or "adds negative charge"). Adding these two equations:
dG_total = dG_A + dG_C = (E_A - E_C)dQ
Since E_C > E_A, dG_total is negative or dG_total < 0 (spontaneous process).
Another equation of dG is dG = -SdT + VdP + (mu)dN. Where mu = chemical potential and dN is the number of molecules added/reduced to/from the system. Analogous to the electric potential and charge. Hope this clarifies even though this is an old query.

Please visit this page, this page and this one on how to format your posts better with MathJax and Markdown. Also, I'm not sure how this is related to the question.

The entropy change of the system in any process is given by
$$ΔS=\int\frac{\mathrm dq_\mathrm{rev}}{T},\tag{1}$$
the calculation of which assumes a reversible process taking place between the initial and final states. In this reversible process, both the temperature and pressure can vary. Hence while calculating entropy change of the system, which is subsequently used in the calculation of $ΔG$ using $$ΔG = ΔH - TΔS,\tag{2}$$ the assumption that temperature and pressure are constant, i.e. statement 2, becomes invalid.


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What happens to the bleach after mopping?
At home, I clean the tiled floors by mopping with highly diluted bleach.  After mopping, I let the floor dry.  What happens to the acid in the bleach?
Will it crystallize on the floor, stay there invisibly in solid form, and get solved again next time the floor gets wet?  Will it instead evaporate, or turn to fine dust that floats in the air and leaves through the window?  Will it decompose in some sort of chemical reaction?

This depends largely on the acid we’re talking about.

What makes you say there is an acid in bleach? "Pure" bleach, sodium hypochlorite, is highly alkaline - with a pH of 12.6 for a 6% solution source

Do you have the package label of the product?

What happens to the acid in the bleach?

First of all there is no acid in bleach.  In fact bleach uses base, the antithesis of acid. and is essentially chlorine dissolved in a solution of sodium hydroxide.

Will it crystallize on the floor, stay there invisibly in solid form, and get solved again next time the floor gets wet? Will it instead evaporate, or turn to fine dust that floats in the air and leaves through the window? Will it decompose in some sort of chemical reaction?

In the short term residual bleach will remain on the floor as a crystalline film that can re-dissolve and can become dust in the most literal way.  This should not be of concern though, as carbon dioxide in the atmosphere acidifies and reacts with the sodium hypochlorite and sodium chloride, releasing chlorine from the bleach forming sodium carbonate and eventually sodium bicarbonate (baking soda).

Sodium hypochlorite decomposes by a couple of different pathways:
$\ce{NaOCl\leftrightharpoons Na^+ + OCl^-\\
2OCl^- \leftrightharpoons O_2 + 2 Cl^-}$
In that case, the oxygen escapes (if it didn't react with "things to bleach", which is of course how bleach works in the first place), and you are left with ordinary salt.
The other pathway is
$\ce{3 OCl^- -> ClO_3 + 2 Cl^-}$
In that case you get some sodium chlorate, and again some ordinary NaCl salt.
Both NaCl and NaClO3 are crystalline solids that may be blown around like fine dust; some NaOCl may also remain behind. Note that according to wikipedia most commercial bleach solutions contain a small amount of sodium hydroxide to slow down the decomposition along the pathways mentioned above.
Straight dope link


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can we use con HCl zn amalgam to reduce alcohol?
all says that we can't use zn con HCl to reduce alcohol. can anyone explain this with the mechanism or with explanations

I think we can because when reducing aldehyde or ketone, In mechanism alcohol is formed which further reduces to give hydrocarbons.

but some peoples say that alcohol is not formed as an intermediate in this reaction

I have answered with mechanism. Take a look at that

Here is the mechanism for reduction of aldehyde/ketone by clemmenson reduction.

Note that alcohol is formed as intermediate which further reduces. Considering that alcohol can be reduced by Zn/HCl


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What is the mechanism of the Clemmensen reduction?
I have learned that we can reduce a ketone to an alkane using the Clemmensen reduction, where we treat the ketone with amalgamated zinc in the presence of $\ce{HCl}$.
My book didn't mention the mechanisms for this type of reaction and I couldn't find an answer by searching Google.
How does this reaction actually take place? Are there carbocation rearrangements in this reaction?

As with many complex mechanisms, many variants can and indeed have been proposed. The most widely accepted is the one proposed by Rosanti (J. Org. Chem. 1991, 56, 4269. DOI: 10.1021/jo00013a036).
The key steps of the mechanism are given below, crucially showing that the alcohol is not an intermediate (hence why secondary alcohols are not able to undergo a Clemmensen reduction to afford the corresponding alkane).

A single electron reduction of the ketone takes place to give a radical anion. The exact species responsible for this  has been debated, but is thought to either be the zinc chloride radical (as shown) or the mercury chloride radical (not shown)
The radical anion picks up a hydrogen from water (the reaction is carried out in aqueous hydrochloric acid), this generates a hydroxyl radical, but also means that the alkoxide derived from the starting ketone is now simply an anion rather than a radical anion.
Homolytic cleavage of the C–O bond, the oxygen has now been completely removed from the substrate, simply leaving behind a secondary radical which must gain a hydrogen (again from water) in order to provide the desired product..

As shown below, several other steps also operate in order to quench out the various radical intermediates produced during the course of the reaction.

Proposed mechanism of the Clemmensen reduction
Ref: Comprehensive Organic Named Reactions, Wang.

So it means that we can't reduce alcohol with it?

So I want to ask that with is better method for reduction of 5 hydroxyhexanal to hexane 1,2 diol Clemmensen or wolff kishner and why?

@Mohit None. U can use NaBH4 or LiAlH4 for that.


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Simple method to lengthen alkane?
I'm working on a science project which involves the production of methane. I would like to be able to use the methane as fuel in (e.g.) cars (using octane) or gas tanks (using butane and propane) etc. I have found that there are a variety of coupling reactions out there, but a lot of them seem to involve 'aryl Grignard reagents' which are not relevant to me.
So, I'm looking for a relatively simple method to lengthen an alkane. Preferably the reaction should occur below 100 degrees Celsius.

Alkanes are not very reactive. I'd say there is no simple method to do anything at all with an alkane.

I agree. Thankfully, cars don't run on noble gas fuel. That would be a real problem. :D

If you produce methane, can't you modify that step to generate longer alkanes? (https://en.wikipedia.org/wiki/Fischer%E2%80%93Tropsch_process)

I'd love to. Thing is, I'm using methanogens.

The procedure to convert methane to longer hydrocarbons at low temperature is one of the so-called Holy Grails in catalysis (http://pubs.acs.org/doi/abs/10.1021/ar00051a001). The person who is able to invent a process wherein this is possible (at an efficient rate) would be a very good candidate for a Nobel Prize.
Why is it then so difficult to convert methane to longer hydrocarbons? Methane is a highly stable molecule. In order to form a longer hydrocarbon, methane first has to be broken down to smaller CH$_{x}$ fragments. The first dehydrogenation step of methane (to CH$_{3}$) has a very high activation energy and even using a good catalyst, this is not feasible at an appreciable rate under 100 degrees celcius.
Industrially, this process is done by first creating synthesis gas (CO + H$_{2}$) from methane by partial oxidation using pure oxygen (which is cryogenically separated from air). Thereafter, the synthesis gas is compressed to roughly 20 atm of pressure and converted to very long hydrocarbons (waxes) in the Fischer-Tropsch process. In turn, these waxes are cracked to form diesel fuels and lubricants.
I am afraid that there is no easy way of converting methane to longer hydrocarbons. Though please keep trying because basically the whole world would benefit if you succeed! :-)

But make sure the methane come from $\ce{CO2}$ and $\ce{H2}$ in some way, because otherwise it would be terrible for the environment ;)

I fully agree :) And let us generate the H$_{2}$ also in a sustainable manner.

(Hint: Use $\ce{H2}$, that automatically generates $\ce{H2}$ ;))

You sir, are technically correct. The best kind of correct. Thanks for sharing. ;)

Thanks @ifilot! By the way, do you think that halogenation of the methane to create $\ce{CH3Cl}$ and using a coupling reaction (or something like it) to get ethane and $\ce{Cl2}$ would work?

The idea is good, but you still run into the problem that activating $\ce{CH4}$ in order to form $\ce{CH3Cl}$ remains difficult. Furthermore, the reaction requires a certain driving force. In Fischer-Tropsch (see above), the driving force is the formation of water by which side-reactions such as oxygenates are avoided, because water is much more stable (in terms of enthalpy). Here, I am not sure whether $\ce{Cl2}$ and the alkanodimer is so much more stable than the alkanohalide. You might also want to look into Suzuki coupling. (https://en.wikipedia.org/wiki/Suzuki_reaction)

Another important note: There are probably reactions out there that enable you to form longer hydrocarbons from methane by activating methane and then performing one or more coupling steps. An important thing to keep in mind is that, most likely, these reactions are quite costly in terms of time and resources to perform. In the end, you are going to use your product as a transportation fuel. It might just be the case that converting an engine to run on methane is much cheaper than synthesizing a transportation fuel from methane.

You might want to look into the Heck reaction
As ifilot said, you would win a Nobel Prize for this kind of chemistry, and as it happens, a Nobel Prize was awarded for the discovery of this reaction!
This can be used to join together any two carbon chains where one ends in a halogen and the other ends in a hydrogen bonded to a carbon which is part of a double or triple bond.
Here it is in its generalised format (credit Wikipedia):

As you can see, any R group which it attached to a halogen (X) will bond to another R group which ends in a double or triple bond.
The exact mechanism is a little involved so I won't bother talking about it.
So now you need halogenated alkanes and alkenes to proceed.

Look into the halogenation of alkanes. Even though
alkanes are usually unreactive, they can be halogenated in the
presence of ultraviolet light.  Check out this link for more:
http://www.chemguide.co.uk/organicprops/alkanes/halogenation.html
Look into cracking alkanes, which produces alkenes
as a result.
EDIT: There are other reactions that can join carbon chains together such as reactions involving Grignard reagents, the Wittig and the Diels-Alder reaction.


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Is food that burnt in an 18/10 stainless steel pot still edible?
I was cooking Christmas tamales in an 18/10 stainless steel pot and I accidentally let the water evaporate and the pot started burning.  I think it was probably burning for around 5 minutes or less.  There were a lot of tamales in there and I'm wondering if they are safe to eat or if I have to throw them out?

No reason to believe that they would be particularly harmful, but I'd really wonder about the taste. My guess would be that a burnt taste would have permeated all the tamales.

did you eat it ? are you still alive ? :) you can add your own answer

Your stainless steel pot is thoroughly thermally stable well above the temperature of your stove, and there are no volatile compounds present that could migrate out of the pot and into your food. It is (was) perfectly safe to consume the tamales cooked in this pot.

authors: 港豐人民幣找換匯款有限公司


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What does radiofrequency really do in NMR & MRI?
In NMR, the protons’ frequencies are affected by the magnet which makes them precess around its magnetic field so now all protons will have different frequencies according to their environment including shielding and de-shielding effects. As I read, the radio frequency pulses only change the resonance of protons and make their $\omega$ angle  change  ($180^\circ$) to be induced inside the coil and make signals that can be measured.
But if this true why do we have to use different gradient coils and phase encoding gradient coils in MRI, if the protons really have different frequencies?

This is a very highly technical question =O

The radiofrequency pulses and the gradients have completely different purposes in NMR. Radiofrequency pulses are the basic building block of NMR experiments, you use them to manipulate your spin populations and coherences. While each proton has a different frequency depending on its environment, you generally can't (or don't) target individual protons with your pulses. In most experiments you hit either all protons with your pulses, or a specific group of protons.
Gradients have different purposes in NMR and MRI, what they allow you to do is to change the magnetic field depending on the position inside your sample. In NMR this allows you to remove unwanted signals and select specific coherences.
In MRI, the gradients are what makes imaging possible. They allow you to encode the spatial position of your protons into their frequency. In NMR you look at all the molecules in your sample at once, you don't care where your molecules are floating around. In MRI the position is important, as you want to know what the protons at a particular point in the body are doing. Gradients also allow you to only look at a slice of the body and not all of it at once.
I'll skip phase encoding in gradients as it is MRI-specific and I don't know much about it. Each part of your question has whole books written about it, so this is a drastically simplified answer.

So, for MRI by using gradient coil (that is only responsible for changing frequencies of protons )and then give a pulse of RF that related to specific slices of body the protons will go exited and then relax but the relaxed protons will know have the same frequencies or different frequencies ?

The purpose of MRI is to get spatial information from the spins.  It is true that chemical shifts give protons in different chemical environments different frequencies, however, the chemical shifts will not reflect any spatial information.  For example, all of the water in the body has (roughly) the same chemical shift, whereas fat all has (roughly) a different chemical shift than water, but (roughly) the same chemical shift as all other fat in the body.  If you left out the magnetic field gradients and the phase encode pulses, you would simply end up with a stand 1D proton NMR spectrum that had two broad peaks in it (one for water and the other for fat).
The gradient coils artificially spread out the precession frequencies much more than the chemical shifts do (in fact, the chemical shift differences are largely overwhelmed by the gradient coils).  RF pulses with a bandwidth sufficient to excite the frequencies spread out by the gradient coils will have no problem with the relative small difference created by the chemical shifts, so all of the spins are still excited (Note: A 180º pulse would produce no signal.  A 90º pulse would produce the maximum signal, but for signal-to-time reasons, MRI typically uses trains of small angle pulses, often as small as just a few degrees).
However, we can't ignore the chemical shifts entirely.  After-all, we interpret differences in precession frequency as differences in position.  This means that spins with different chemical shifts will produce images that show up at different positions in our final MRI image!  Even worse than simple ghosting depending on the MRI protocol, you may get very bizarre artifacts due to the chemical shift differences... A great many clever protocols have been developed that exploit off-resonant effects to selectively excite only a narrow chemical shift bandwidth or to minimize the resulting artifacts, etc.
I would highly recommend getting a copy of Dwight Nishamura's book Principles of Magnetic Resonance Imaging.  Very readable with lots of insight on what the spins are doing and what the pulse sequences are doing. One of the best introductions to MRI that I've read.  He discusses this issue of chemical shift artifacts about half-way through the book if I remember correctly.


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Difference between Resonance Effect and Mesomeric Effect
While studying electron displacement effects in organic chemistry, I read that the Resonance Effect and Mesomeric Effect are the same. Every source I checked used the heading or opening "The Resonance (or Mesomeric) Effect is..."
However, I was able to find the difference between Resonance Structures and Mesomeric Structures is and found out that the two differ in how they depict conjugated pi bonding. The Mesomeric Structures are all the compounds with all possible arrangements or the pi bonds drawn separately. The Resonance Structure is one diagram that uses dashes and lines to denote the partial double bond characters in the compound.
So my understanding from this difference between the structures is that the effects must only differ in how they are depicted. But I was unable to find a source that could help confirm this. Can someone please confirm or explain it for me?

The IUPAC gold book (IUPAC are the organisation which 'define' standards for things such as chemical nomenclature) describes them as being essentially synonymous. The following definitions are given (link):

Resonance  In the context of chemistry, the term refers to the
representation of the electronic structure of a molecular entity in
terms of contributing structures. Resonance among contributing
structures means that the wavefunction is represented by 'mixing' the
wavefunctions of the contributing structures. The concept is the basis
of the quantum mechanical valence bond methods. The resulting
stabilization is linked to the quantum mechanical concept of
'resonance energy'. The term resonance is also used to refer to the
delocalization phenomenon itself.
Mesomerism Essentially synonymous with resonance. The term is particularly associated with the picture of π-electrons as less localized in an actual molecule than in a Lewis formula. The term is intended to imply that the correct representation of a structure is intermediate between two or more Lewis formulae.

There does seem to have been some historic difference that has largely been reconciled into a single definition however one use of Mesomerism that still gets commonly used is the definition of groups as having a positive (+M) or negative (-M) effect to describe the effect of a group on a conjugated system.

The Mesomeric Structures are all the compounds with all possible arrangements or the pi bonds drawn separately. The Resonance Structure is one diagram that uses dashes and lines to denote the partial double bond characters in the compound.

This statement is not entirely correct.  The 'Mesomeric' structures you talk about are usually refered to as resonance structures/forms. The depiction with the dashed lines and partial bonds is a resonance hybrid.

Mesomerism and resonance are identical, completely fungible, used interchangeably. Hückel-Lewis-Projection Method: A “Weights Watcher” for Mesomeric Structures: Y. Carissan, D. Hagebaum-Reignier, N. Goudard, S. Humbel J. Phys. Chem. A. 2008, 112, 13256–13262

Yes, they're almost synonymous. The only minute difference between them is this:
Mesomerism = donation/withdrawal of electrons
Resonance = delocalization of electrons