How to balance the reaction between ethanol and chlorine using oxidation number method? Balance the following chemical equation using oxidation number method:
$$\ce{C2H5OH + Cl2 -> CHCl3 + HCO2- + Cl- + H2O}~\text{(In basic medium)}$$

Chlorine in $\ce{Cl2}$ is in oxidation state $0$ and it is changing from $0$ to $-1$ in both $\ce{CHCl3}$ and $\ce{Cl-}.$
$\ce{C2H5OH}$ has two carbons and both of them have different oxidation states: one has $-3,$ the other has $-1,$ and they are changing to $+2$ in both the products $\ce{CHCl3}$ and $\ce{HCO2-}.$
So, it is definitely a redox reaction and thus I want to use oxidation number method to balance this equation.
I am finding it hard to balance because of multiple products of $\ce{C}$ and $\ce{Cl}$ in the same equation.

The key  is to remembering that  $n \cdot m = m \cdot n$.
It seems obvious, but you would not believe how much struggles people suffer when trying to apply this trivial equation to chemical reactions.
In this special case, The total change of oxidation numbers is zero, otherwise, it would mean we have lost some electrons somewhere.
$$\sum{n_i \cdot \Delta ON_i } = 0$$.
The stoichiometric coefficients are in the reciprocal ratio to ( absolute value of ) the oxidation number change ratio.
So the sum of the oxidation number change of carbons in 1 ethanol molecule is $( (+2 - -3) + ( +2 - -1) ) = +5 + +3 = +8$
The sum of the oxidation number change of chlorine in 1 chlorine molecule is $2 \cdot (-1 - 0) = -2$
We have then:
$$n_{\ce{C2H5OH}} \cdot 8 + n_{\ce{Cl2}} \cdot -2 = 0$$
$$n_{\ce{Cl2}} = 4 \cdot  n_{\ce{C2H5OH}} $$
Note that you are missing negative ions of the left side for charge balance, also overall mass balance is missing.

Enumeration of stoichiometric coefficients have also a general, analytical mathematical solution, but personally, I never used it, as it would take much longer time than just enumerating it on the fly by the chemist's heart:

You are solving a set of linear algebraic equations for multiple variables.
Each variable is a stoichiometric coefficient for a reagent or a product.
Each equation represents some kind of balance.
1 equation for the atom count balance for each of the elements
1 equation for the charge balance, if applies.
In most cases there will remain 1 degree of freedom, so 1 stoichiometric coefficient must be chosen and some others will depend on the choice.
E.g. there can be $\ce{n H2 + n F2 -> 2n HF}$, so we choose $\ce{n=1}$ for $\ce{H2}$  and rest will follow.


Structure of Sodium Naphthalenide This question is a sort of follow-up to my earlier question.

I know about a method of solubilising potassium where the potassium atoms are "trapped" between layers of graphite, but I think that sodium naphthalenide exists only in solution (if a contradiction is found, please inform).

In the solution phase, say in THF, I would expect some THF molecules co-ordinated to the sodium, and an $\eta^4$ complex with naphthalene, involving one ring. An $\eta^{10}$ complex seems possible, involving both the rings, as the negative charge is distributed further, but I haven't read about any $\eta^{10}$ complex. An $\eta^1$ complex might be possible,  but the distribution of negative charge will be lesser.

I expect sometehing similar to the tetrahedral form of butyllithium.

What is the structure of sodium naphthalenide?

TL;DR: Naphthalene anion radical won't tolerate a presence of unshielded alkali metal cation nor in solution, neither in crystal. Reduction of naphthalene is possible strictly under air- and moisture free conditions by alkali-metals in the presence of coordinating solvents (cryptands). In this case solvent-separated cations don't form coordination bonds with naphthalenide radical.
On the other hand, further reduced (dianion) naphthalene readily forms various compounds not only with alkali metals, but also with lanthanides, typically $\eta^4$-complexes.

Various studies on stabilization of radical anions, especially in solid state, show that complexation (preferably with polar aprotic solvent or polydentate ligands) of counter-cations is essential:
$$\ce{C_nH_m + [M^0]_{\infty} + $x$ Solv + $n$ L <=> [M+(Solv_x|L_n)](C_nH_m^{.-})},$$
where $\ce{[M^0]_{\infty}}$ - any metal in oxidation state $0$ as a bulk reactant, $\ce{Solv}$ and $\ce{L}$ - solvent and ligand, correspondingly.
If cation is optimally solvated (complexated/sterically protected), radical counter anions $\ce{C_nH_m^{.-}}$ with extended $\pi$ systems are preferred because they lack extensively charged centers which favor contact ion formation (1). For example, crystals of bis(diglyme-O,O',O'')-sodium naphthalenide radical
$\ce{[Na+(diglume)2](C10H8^{.-})}$ ([Na]) (1) has been isolated based on a principle of increasing enthalpy (calcd. based on crystal structures):
$$\Delta H (\ce{[Na+(THF)6]}) = \pu{-587 kJ mol^{-1}}$$
$$\Delta H (\ce{[Na+(DME)3]}) = \pu{-671 kJ mol^{-1}}$$
$$\Delta H (\ce{[Na+(diglume)2]}) = \pu{-677.3 kJ mol^{-1}}$$
Under certain conditions, the reduction of the naphthalene can proceed further to the naphthalene dianion, $\ce{C10H8^{2-}}$. For example, a complex with radical ion $\ce{[Li+(TMEDA)2](C10H8^{.-})}$ ([Li]) can reversibly be transformed into naphthalene dianion-complex $\ce{[Li+(TMEDA)]2(C10H8^{2-})}$ (2):

Of course, bigger cations require better shielding, commonly resulting in a usage of bulkier solvents or supramolecular assemblies, such as ([2,2,2]-cryptand)-potassium naphthalenide radical, $\ce{[K+(crypt-222)](C10H8^{.-})}$ ([K]) (3), in comparison to previously mentioned [Li] and [Na] naphthalenides (in the following crystal structures sphere-packing model is applied to metal atoms, and the rest is represented with sticks for simplicity):

It is also worth mentioning that naphthalene dianion forms numerous complexes with rare earth elements (4), e.g. with [Y] (see picture): $\ce{CpYC10H8(DME)}$, ($\eta^5$-Cyclopentadienyl)-(1,2-dimethoxyethane)-($2\eta^1:\eta^2(2\sigma,\pi)$-naphthalenide)-yttrium(III) with a $\ce{C6}$ ring being bent by $26.1^\circ$:

$\ce{La}$ and $\ce{Eu}$ complexes with bridged $\mu-\eta^4:\eta^4$-naphthalene are also obtained ($\ce{[LaI2(THF)3]2(C10H8)}$, $\ce{[EuI(DME)]2(C10H8)}$).
(1) Bock, H.; Arad, C.; Näther, C.; Havlas, Z. J. Chem. Soc., Chem. Commun. 1995, 23, 2393–2394. DOI: 10.1039/C39950002393
(2) Melero, C.; Guijarro, A.; Yus, M. Dalton Trans. 2009, 0 (8), 1286–1289. DOI: 10.1039/B821119C
(3) Rosokha, S. V.; Kochi, J. K. J. Org. Chem. 2006, 71 (25), 9357–9365. DOI: 10.1021/jo061695a
(4) Protchenko, A. V.; Zakharov, L. N.; Fukin, G. K.; Struchkov, V. T.; Bochkareva, M. N. Russ Chem Bull 1996, 45 (4), 950–953. DOI: 10.1007/BF01431330


Why do different substances containing saturated hydrocarbons burns with different flame? LPG and Candle wax both contain saturated hydrocarbon, yet LPG gives a clean white flame but a candle gives yellow flame with lots of smoke. Why so?
I have read some answers on web which says that it is due to the insufficient supply of oxygen, so I tried burning candle beside LPG ( so there is no difference in supply of oxygen for both candle and LPG ) , but  still candle gives a yellow flame. Why?

LPG is propane, butane or a mixture of both. Paraffin candle wax is $(CH_2)_n$

typically $C_{31}H_{64}$ (other, nonparaffin, waxes are also used in candles, like stearin, beeswax, etc.)
However, LPG is a gas from the start and can mix with air (i.e., oxygen) before it begins to burn, while candle wax has to be heated by the flame before it can begin to burn, so the outer periphery of candle wax vapor (the part of the vaporized candle not yet heated up to ignition) is very small - a dark zone surrounding the wick. The possibility of contact with oxygen is limited to this area.
You will notice a blue zone around the wick also - perhaps obscuring the dark zone closer to the wick. This is a hydrogen-rich flame; under ideal conditions, you can burn off all the hydrogen and leave the carbon (put a glass into the flame: it will get coated with black soot). But if you don't cool the flame with a glass, the carbon which would otherwise deposit will burn yellow.
So, "insufficient supply of oxygen" is an OK answer, but maybe a better one is "scanty mixing of candle wax vapor with oxygen". I suspect that if you boiled candle wax rapidly in a closed container with a vent, and ignited the gases coming out of the vent, they would burn much more like LPG and less yellow.
The colors are dependent not only on line spectra of the atoms, but also on black body radiation from solid carbon. In the candle flame, by the time the carbon get access to oxygen, its temperature is not as high as in a LPG flame, and the possibility of solid carbon particles burning increases the amount of yellowish light emitted. Kind of like charcoal burning, but on a much smaller scale.


Structure of benzene How does molecular orbital or valence bond theory explain the delocalised structure of benzene? I.e. when the carbon forms 3 bonds with other carbons/hydrogen, what happens to the remaining hybridised orbital?

The short answer is that it's not hybridized.

In benzene, the carbon atoms are sp2 hybridized. That means that each carbon's set of atomic orbitals is three sp2 orbitals and one p orbital. The sp2 orbitals are utilized to make sigma bonds to the adjacent carbons and hydrogens. The unhybridized p orbital contributes to the pi system.


Why are some smells so hard to get rid of? What may seem to be a silly question could actually have roots in some core chemistry principles. The other day, I was cooking fish when I noticed that I'd forgotten to use a fork, and touched the fish with my bare hands. When I tried to get rid of the fishy smell, it just wouldn't go away.

Why are some smells so hard to get rid of?

My Hypothesis:

I think that the phrase "hard to get rid of" really means "resistant to washing," because that's what we're actually thinking of when we say you can't get rid of a smell. And when we say "wash" we really mean "attempt to dissolve in water using soap emulsification".

I'd conclude that "a smell being hard to get rid of" is caused by chemicals that are not soluble in water, and for some reason, possibly because the molecules are too large to fit in a stable emulsive bubble, refuse to be emulsified into aqueous suspension by soaps. But that's just my theory.

Food flavors and odors are comprised of complex mixtures of hundreds of chemicals, which act as stimuli to the human sensory organs, for example, human nose. The odor of food may be attributed to the odor of a single chemical or an integrated response by nose to a mixture of chemicals. This reference to a single chemical or a mixture of chemicals allows trained analysts to develop a common terminology in describing odor, one of which called odor thresholds. Threshold is defined as the degree of stimulation of a nerve or nerve center that just produces a response. If the concentration of the odorous chemical (usually given in parts per million or ppm) is below this point, it will be judged not odorous. Hence, accepted definition of threshold is the minimum
identifiable odor (MIO) or recognition threshold. MIO of some chemicals is very high while some is extremely low. The latter chemicals are the ones you said,  "hard to get rid of." For example, the oder of dimethylformamide is described as 'fishy' and pungent, but you won't smell it if its concentration in air (or on your skin) is below $\pu{100 ppm}$. On the other hand, tar-like, pungent smell of p-cresol can be detected if its concentration in air (or on your skin) is greater than $\pu{0.001 ppm}$.
Now considering the fish smell, it was described short and sweet in following website from American Society for Nutrition: Read
Accordingly, it says:

Water in the open ocean is about 3% salt by weight, but the optimal levels of dissolved minerals inside an animal cell is less than 1%. In order to maintain fluid balance, ocean creatures must fill their cells with amino acids and amines to counter the saltiness of seawater. Ocean fish tend to rely on trimethylamine oxide (TMAO) for this purpose.
The problem is that when fish are killed, bacteria and fish enzymes convert TMAO into trimethylamine (TMA), which gives off the characteristic “fishy” odor.

Now, it is nice to know MIO of trimethylamine (TMA) is $\pu{0.00021 ppm}$ or $\pu{0.21 ppb}$ (parts per billion). That's why it is so "hard to get rid of" fishy smell whenever you touch or cook dead fish. But the good news is TMA is a organic base, which would react with any acid to form water soluble ammonium salts. Citric acid from lemon (or lime) or acetic acid from vinegar would do the job.


Rust forming and being washed off by water I have noticed that, when exposed to water for a few days, the steel in my wagon turns into rust. After several more weeks, the rust that formed is dissolved in the water, and once I pour the water out there isn't any rust where there previously was rust.

This is unusual because iron oxides and iron hydroxides (the components of rust) don't normally dissolve in water, and so the water just keeps being reduced and the iron keeps being oxidized.

How can it apparently happen such that steel turns to rust, then back to steel with no rust, in the presence of water if rust does not normally dissolve in water?

This happens in a cycle like this:

steel + water = rust -> rust + water = steel + rusty water

Rust appears because iron is oxidized in presence of oxygen (it is a oxydo-reduction reaction, meaning an electron transfer occurs between species), and water helps this phenomenon occur. Iron loses 3 electrons to the favour of the oxygen.

In presence of water, oxygen can oxidize iron by the mean of hydroxyl group. It leads to:

$$\ce{2Fe(OH)_3 \leftrightharpoons Fe_2O_3\cdot 3H_2O}$$

where water is not the oxidizer, rather it is a mediator for oxidation.

Hydroxides have low solubility, so water can only dissolve a small fraction of it.

On the other hand, poured water can abrade rust (via mechanical stress) and remove it of from the bulk metal, leaving the surface clean for producing more rust. This is how huge rods of metal can vanish in a few months in corrosive environmental conditions (such as wet air or water).

Preventing rust requires the use of specific alloys and surface finishing. You can also add a galvanic electrode (generally made of $\ce{Zn}$, a so-called sacrificial anode) to protect submerged metallic structures such as the hull of a boat.

Therefore, in your example water is both a mediator of oxidation and mechanical abrasion agent.


Strontium is more common than sulfur? I was really surpised to see that strontium is apparently more common than sulfur in the earth's crust, since it occurs in what I thought were pretty rare minerals.

Sulfur which seems to be everywhere such as in volcanoes, in the atmosphere as sulfur dioxide, in coal deposits, and is even found elementally in large lenses.

Sulfur seems to be ubiquitous. For example, fool's gold, iron pyrite is iron sulfide and it can be found in practically any stream bed. I would be hard pressed to find strontium in my local stream beds. And, of course, this sulfide is only one of many commonly sulfides.

Is strontium really more common than sulfur? It seems implausible. What is the basis for this claim?

The primary reason for this and other seemingly implausible features of element distribution is this: some elements are concentrated, and some are scattered. Strontium is much like calcium (which is way more abundant) and can hide as a tiny admixture in its minerals, while sulfur is pretty much on its own.

Just think of the rare earth elements. Their name is in fact a misnomer; they are not rare at all, just very scattered and thus difficult to come by, hence the relatively late discovery. On the other hand, silver and gold, which are thousands times less abundant, were known since antiquity (or probably before that).

Also, you see yourself that there are several columns with the same name "Abundance in crust", and the data for the same element vary quite significantly between the columns. This is due to the fact that finding an element's content in the crust is not easy at all. Wait, don't they say explicitly that the numbers are but estimates?


Reaction of alcohols with PCl5 and PCl3 The reaction of alcohols $\ce{ ROH }$ with $\ce{ PCl5 }$ and $\ce{ PCl3 }$ yields an alkyl halide $\ce{RCl}.$ With $\ce{ PCl5 }$, the reaction is quite simple leading to the formation of $\ce{ RCl }$ and $\ce{ POCl3 }.$

But with $\ce{ PCl3 }$  a problem arises. Since $\ce{ PCl3 }$ has both a lone pair and vacant $\ce{3d}$ orbitals it can act both as a Lewis base and a Lewis acid.

In the first figure, $\ce{ PCl3 }$ accepts a lone pair showing its acidic character and expelling $\ce{Cl-}$ out. Now this is where I am getting confused. On one hand we have  a Lewis base $\ce{ PCl3 }$ and on the other we have the "expelled" $\ce{Cl-}$? Now in this case why does $\ce{PCl3}$ take away the proton and not $\ce{Cl-}$?

And in the next step, why does  $\ce{Cl-}$ attacks the carbon not the  $\ce{H+}$?

Remember your general chemistry. In an acid-base reaction, the equilibrium favors the side of the reaction with the weaker acid/base pair. Strong bases have weak conjugate acids. Strong acids have weak conjugate bases. Weak bases have strong conjugate acids. Weak acids have strong conjugate bases. Consider the following:

$$\ce{HA + B- <=> A- + HB}$$

If $\ce{HA}$ is a stronger acid than $\ce{HB}$, then $\ce{A-}$ must be a weaker base than $\ce{B-}$.

Here is an example with real compounds:

$$\ce{HCl + NH3 <=> Cl- + NH4+}$$

Even though the right side of the reaction contains ions instead of neutral molecules, the right side is favored because $\ce{HCl}$ is a strong acid and $\ce{NH4+}$ is a weak acid.

When we study organic chemistry, we are given some rules of thumb to help us compare relative acidity:

$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is more electronegative than $\ce{Z}$ and they are in the same period.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y}$ is larger than $\ce{Z}$ and they are in the same group.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more resonance stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more inductive stabilization than $\ce{Z-}$ and if $\ce{Y-}$ and $\ce{Z-}$ are otherwise similar.
$\ce{Y-H}$ is more acidic than $\ce{Z-H}$ if $\ce{Y-}$ has more $\ce{s}$-character than $\ce{Z-}$ and $\ce{Y-}$ and if $\ce{Z-}$ are otherwise similar.
$\ce{YH2+}$ is always more acidic than $\ce{YH}$, though it is hard to compare $\ce{YH2+}$ and $\ce{YZ}$ or $\ce{YH}$ and $\ce{YZ2+}$.

With these rules in hand, it is sometimes challenging to remember that we also have an experimental measure of acid strength, and that there are only a limited number of "strong" acids (those acids which are stronger than $\ce{HSolvent+}$).

The $K_\mathrm{a}$ of $\ce{HCl}$ is not easily determinable, since it is more acidic than most protonated solvents. However, the Evans pKa table and other sources often estimate it at -7 or -8, with only $\ce{HBr}$, $\ce{HI}$, and the various "superacids" including such things as $\ce{HSbF6}$ being stronger.

The Evans table lists the $\mathrm{p}K_\mathrm{a}$ of various protonated phosphines. For example

$\ce{CH3PH3+}$ is 2.7 in DMSO ($\ce{HCl}$ is 1.8 in DMSO).
$\ce{Et3PH+}$ is 9.1 in DMSO

Would we expect $\ce{ROPCl2H+}$, which is the intermediate in your reaction, to be more or less acidic than the two reference phosphonium ions above? Both $\ce{RO}$ amd $\ce{Cl}$ are electron withdrawing by induction.

$\ce{Cl-}$ maybe could attack $\ce{H}$ and not $\ce{C}$, but the $\ce{HCl}$ that forms would be so acidic in comparison to everything else in the reaction that something else would take that proton away again, regenerating $\ce{Cl-}$. Once $\ce{RCl}$ forms, there is no other good nucleophile present that is capable of displacing the chloride group.


Why is CaCl2 called calcium chloride? Doing a first year chem class.
Just read through the molecular naming of compounds and now I'm confused as to why $\ce{CaCl2}$ is called calcium chloride and not calcium dichloride?

The current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book) reads as follows:

(…)
The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., …). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. (…)

So far, the systematic name of $\ce{CaCl2}$ should be calcium dichloride.
However,

Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound (…).

Therefore, the customary name calcium chloride is correct.


Spontaneous reaction between chromium(III) and aluminum I'm trying to rationalize this problem. It's been a while since I've done redox problems. So here goes, and any input would be great. Basically - am I right?

Here is what happens when you store chromium ion in an aluminum container:

$\ce{Cr^3+ + Al -> Cr + Al^3+ ~~~~~~E^{o}_{cell}=}$ ${+ 0.94}$$\ce{V}$

$\ce{Cr^3+}$: oxidizing agent

$\ce{Al}$: reducing agent

$\ce{Cr}$: conjugate reducing agent

$\ce{Al^3+}$: conjugate oxidizing agent

This reaction is spontaneous because we are going from strong to weak (just like in a spontaneous acid-base reaction).

So if we are going from strong to weak, all these statements are true:

$\ce{Cr^3+}$ is a stronger oxidizing agent than $\ce{Al^3+}$.

$\ce{Al}$ is a stronger reducing agent than $\ce{Cr}$.

Yes, your rationalization here is correct. Chromium ions will oxidize aluminum container and deplete chromium while contaminating the solution with aluminum ions. It's exactly the same process, as for instance, copper nitrate with an iron paper clip experiment that we have all seen in a chemistry class at some point. But here you have an aluminum container instead of a paper clip :)


Table Salt vs Kosher Salt difference Does kosher salt has a different structure than regular table salt or it is an issue of crystals size?
Why some recipes want us to use one instead of another?

There is no chemical difference between table salt and kosher salt - both are sodium chloride. Kosher salt just has bigger crystals. However, some table salt brands are iodized, meaning that iodide salts have been added to help us maintain a healthy dietary intake of the essential nutrient iodine. Kosher salt is never iodized.

Sea salt is a different matter. Since sea salt is obtained from sea water, it may have other minerals in it, which perhaps alters the taste.

Kosher salt is used in some recipes as a matter of taste - the bigger salt granules take longer do dissolve, and so there may still be little pockets of salty crunchiness. Kosher salt is also used when this property is necessary, for example in the curing of some meats. For more on the differences in culinary use between the various types of salt, you should consult Seasoned Advice.


Resonance in carbocation derived from allenes During a reaction if addition on one of the $\pi$ bonds of an allene (containing even number of $\pi$ bonds like $\ce {H2C=C=CH2}$) takes place then a carbocation is formed. To decide the position of the positive charge, we look at the stability of the carbocation. Example:
$$\ce {H2C=C=CH2 ->[\ce{H3O+}]} \quad?$$

In the above reaction the proton ($\small{\ce{H+}}$) attacks the electron rich $\pi$ bond (say the left one) and can form one of the following carbocation:
$$\ce{H2C+\bond{-}CH=CH2}\quad\quad \text{or }\quad\quad \ce{CH3\bond{-}C+=CH2}$$
$$ (\text{I})\quad\qquad\qquad\qquad (\text{II})$$

Now the major product (acetone) is formed as a result of the carbocation on the right (II), which apparently is less stable than the one on the left (I) as the positive charge in I is in resonance/conjugation with the $\pi$ bond. But our teacher said that since in allenes the overlapping p-orbitals (say $\ce{p_z}$) of one $\pi$ bond are perpendicular to the overlapping p-orbitals of the other (say $\ce{p_y}$; which makes $\ce{p_x}$ in the direction of internuclear axis $x$) and hence the positive charge cannot resonate with the $\pi$ bond (formed by overlap of $\ce{p_z}$) as the positive charge is effectively an empty p-orbital (the $\ce{p_y}$) and hence the carbocation (II) is relatively more stable.

But once addition has taken place the $\ce{C_1-C_2}$ $\pi$ bond no longer exists and the $\sigma$ bond is free to rotate so the empty p-orbital/positive charge can freely rotate and become coplanar with the p-orbitals of the $\pi$ bond. But why does this not happen?
The reason cannot be steric hindrance as the two groups on $\ce{C_1}$ are the smallest, $\ce{H}$.

The answer to your question might need a few clarifying points. My answer will mostly focus on the protonation (and deprotonation) of allene based on gas-phase experiments and theoretical calculations. A few references that point to elements that will be used concerning the allyl and 2-propenyl cations are as follow : G. van der Rest et al. Eur. Mass Spectrom. 3 323 (1997) and K. Raghavashari et al. J. Am. Chem. Soc. 103 5649 (1981).

Relative stability of the allyl and 2-propenyl cations

Both articles make it clear that the allyl cation (I) is the most stable structure for the $\ce{C_{3}H_{5}^{+}}$ carbocations. It is around $\pu{21 kJ mol-1}$ more stable than its 2-propenyl (II) isomer. The other possible isomers are all higher in energy.

Protonation of allene

According to the first reference above, the protonation of allene on an extremal carbon to lead to the allyl cation requires to pass a high energy barrier, whereas no high barrier is present to lead to the 2-propenyl cation. Figure 4 gives the structure of the interaction complex between $\ce{H_{3}O^{+}}$ and allene, showing clearly that the incoming proton is initially in interaction with the $\pi$ orbital on one side of the allene. Their is obviously no conjugation contribution between this protonation structure and the other $\pi$ orbital, since the proton arrival plane is perpendicular to the other $\pi$ orbital.

From this position, formation of the 2-propenyl cation (II) requires only a minimal energy barrier, requiring only a bending of the $\ce{C-C-C}$ bond and bending of the $\ce{CH_{2}}$ group. Once this is done, the $\ce{CH_{3}}$ group can rotate freely, but one should notice that it does not involve the $\ce{sp^{2}}$ vacant orbital located on the central carbon. This orbital will always remain orthogonal to the remaining $\pi$ orbital.

From this position, one could also consider direct protonation of the central carbon, followed by a rotation of the $\ce{CH_{2}}$ group since no conjugation is present at this point. But here, one should consider that stabilization by conjugation can only proceed after formation of a (non stabilized) primary carbocation. These species are much less stable than secondary carbocations (on the order of $\pu{75 kJ mol-1}$), thus leading to the presence of a large barrier for the direct formation of an allyl cation.

A last pathway that one should consider is a 1,2 hydride transfer between (II) and (I). Although hydride transfers are generally easy between carbocations, this transfer proceeds through a three-center system which can be considered a proton attached to a $\pi$ orbital. Thus, this leads us back to the initial problem, which is that no conjugation can serve as a driving force leading to the allyl cation. And only once the unstable primary carbocation is formed can a rotation of the $\ce{CH_{2}}$ group lead to formation of the allyl cation.


Reduce sizing of molecule - Oxygen Can the size of molecule oxygen reduce smaller? If yes, how is that possible? Is it related to proton and electron surrounding the nucleus?

Surprisingly enough, the answer is yes, though it would be extremely difficult in practice to reduce the size of a molecule as large as $\ce{O_2}$ by the method I will describe here.

First, I must be clear that "shrinking" as seen in many science-fiction stories is completely impossible. Size is not some arbitrarily determined parameter which you can change with a ray gun!

Though the concept of size is a little fuzzy at atomic/subatomic scales, we know it is the valence electrons who are responsible for determining the distance scales in atoms and molecules (since they're the farthest from the nuclei). After solving the Schrödinger equation for a system, one finds that the electrons are concentrated in regions around the nuclei called orbitals. A meaningful definition of size for these orbitals implies that they are around 100 picometres or more across, conveniently described as twice the Bohr radius.

Usually, this would be the limit for any atom or molecule. However, nature has a rather curious quirk. We have discovered that all atoms in the Universe involved in chemistry are made of electrons, protons and neutrons (the latter two being themselves made of different combinations of up quarks and down quarks). However, there are more subatomic particles out there, and there is nothing to stop (exotic) atoms being made out of these other subatomic particles, as long as they are combined in the right way.

Physics doesn't yet understand why, but the fundamental subatomic particles in the Universe come in three generations. That is, there are groups of particles which behave exactly the same (have the same spin, electric charge, among other parameters), but differ only in their mass. In particular, electrons are just one type in a group of three types of particles called leptons. The other types are called muons and taus (or tauons). Again, all leptons share exactly the same properties, except for mass; the muon is approximately 200 times heavier than the electron, while the tau is around 3500 times heavier. We don't naturally see atoms containing these heavier cousins of the electron because they decay very quickly and are hence much rarer. Even though the muon is one of the most stable subatomic particles, its mean lifetime is a mere $2.2\ \mu s$, and the tau is much more unstable still. In the case of the muon, however, it is long-lived enough to produce it artificially and get some chemistry done.

If one were to mix muons with normal atomic nuclei in an appropriate chamber, then for a fleeting moment it would be possible to make atoms and molecules where all the electrons are replaced with muons. The resulting atoms/molecules would have muon orbitals instead of electron orbitals. By solving the Schrödinger equation for muons instead of electrons (changing $m_e$ for $m_{\mu}\approx200\ m_e$), the Bohr radius becomes around 200 times smaller and as a direct consequence the muon orbitals are smaller by approximately the same factor. Thus, an oxygen atom where all electrons have been replaced by muons will be about 1/200th the size, but will display almost exactly the same chemistry; two such exotic atoms would be expected to combine exactly as normal oxygen atoms do, creating an $\ce{O_2}$ molecule with sixteen muons instead of electrons, with an expected bond length of around $0.6\ pm$ instead of the usual $121\ pm$.

This may all seem theoretical, but muonic atoms have already been made in the laboratory before. Given the difficulties in producing and manipulating muons in their short lifespan, the only atoms produced as of yet seem to be muonic hydrogen (where the single electron is replaced by a muon) and quite recently muonic helium where only one of two electrons was replaced with a muon. In the latter case, the mix of electrons and muons makes the atom behave chemically quite differently from regular helium or doubly-muonic helium.


Why does ice freeze the way it does? I was just recently looking into some of the details of the different crystal structures that ice can freeze in. I was aware that many of the unusual properties of ice $I_h$ (the structure of “everyday ice”) are a result of the proton disorder necessary for the hydrogen-bonding arrangement it adopts.

What I also found is there are other arrangements of ice which are proton ordered and have nearly identical density to ice $I_h$. Now, I know density has no predictive value on what crystal structure will be taken (rather crystal structure can be used to predict density), but it seems unusual to me that there is such massive divide between ice $I_h$ and the rest of the ice srystal structures.

Is the entropic gain of proton disorder an explanation why ice freezes the way it does? Also, is there a large energy minimization associated with the hexagonal lattice that ice $I_h$ takes on? Doesn't it seem like a proton-ordered arrangement would be preferable over a proton-disordered arrangement? Or is it all much more complicated than that? It just seems like there should be a crystal structure of ice which is comparable to ice $I_h$ in energy, stability, etc. and yet there isn't.

I suppose a true answer to this question, because I admit it's rather broad, would deal with why the hexagonal lattice is so much more preferable for water than some other lattice arrangement. By dealing strictly with water, rather than crystals in general, I believe an answer to this question is more within reach.

Some of what you are asking are still open questions… and the few data published on the topic are often controversial. But let's try to get some of the facts that appear to be accepted rather broadly:

There is a proton-ordered ice structure, called ice XI that is structurally equivalent to our everyday hexagonal proton-disordered ice (aka ice Ih). In the commonly accepted version of the ice phases diagram:

it can be seen that it is the most stable phase of ice at ambient pressure and low temperature. (Although obtaining this phase in practice can be difficult, and is done by doping the ice crystals with dilute KOH.)

Thus, the proton ordering is energetically favored over proton disorder, while proton disorder is entropically favored. Following simple combinatorial arguments [G. P. Johari, J. Chem. Phys., 109, 9543 (1998)], one can estimate the entropy difference at $R \log(3/2) \approx 3.37$ J/mol/K. From the transition temperature of $T_0 = 72$ K, one can then work out the enthalpy difference at $\Delta H = -T \Delta S \approx  -0.243$ kJ/mol. We can note that both terms are very small.
The hexagonal arrangement is not the only one accessible: there is a metastable cubic ice ($I_c$) phase accessible at ambient pressure. The enthalpy differences between cubic and hexagonal ices are even smaller than with ice XI, of the order of 40 J/mol.
There are, finally, numerous other phases of ice — some experimental, some hypothetical — at positive or negative pressures. Since their boundaries are almost horizontal on the phase diagram, it implies that these transitions are dominantly density-driven.
There are also amorphous ices!

To your final question: why ice prefers the hexagonal net? Well, in order to form low-energy crystalline structures, water must adopt a tetrahedral / four-fold coordination, with each molecule surrounded by four nearest neighbors: two on the donating end of hydrogen bonds, two on the receiving end. So the question is: how many periodic structures (crystals) are mathematically possible with this arrangement? They are called four-connected nets, and there are an infinite number of those. However, it happens that among those with relatively low density and “simplicity” (i.e. relatively few inequivalent vertices), most are of high symmetry, including hexagonal.

There is actually an interesting parallel between ice structures and another category of solids having four-connected nets: silicas.


Does aqueous component contribute to the equilibrium constant? Consider an equilibrium reaction with an aqueous solution $\ce{A(aq)}$ in equilibrium phase with gas:
$$\ce{A(aq) + B(g) <=>[$K_\mathrm{eq}$] C(g)}$$
Would the equilibrium constant have aqueous solution $\ce{A(aq)}$ contributing to it?

In my opinion I think it would. Can anyone please correct me? Also, please help me  find literature which explains equilibrium constant in details.

If you are talking about a species dissolved in a aqueous solution then yes it would feature in the constant .
However if you are referring to the solvent itself then no it wouldn't feature in the constant this is because the concentration of the solvent is constant , as well as the fact that it is present on both sides of the equation since both reactants and products are dissolved in the same solvent
If you are referring to a partial pressure equillibrium constant , then a hybrid one will have to be created that has the partial pressure of the gases and the concentration of species that are not gases
Using a Kc value would mean simply taking the concentration of both the gases and the solutes , to end up with a constant , however solvents are excluded because the concentration of solvents is constant and is thereby thought as being absorbed in the eq constant itself
Derivation of the equilibrium constant requires probability and is quite lengthy so maybe you can check the video on khan academy that discusses the derivation


Which temperature is meant in the Gibbs free energy equation? In the equation for Gibbs free energy change, $\Delta G = \Delta H - T\Delta S$, does $T$ refer to the temperature of the system or the surroundings?

I know we have to calculate Gibbs free energy of the system but the criterion for spontaneity says $\Delta S_\mathrm{total}$ should be greater than zero. When we relate it to Gibbs free energy to show that Gibbs free energy change should be always negative, we keep both system and surrounding temperature same with pressure also constant. Then how can the process occur? Wikipedia says that it is chemical potential that undergoes changes there but what about Gibbs free energy change?

From Wikipedia:

  Thus, Gibbs free energy is most useful for thermochemical processes at constant temperature and pressure: both isothermal and isobaric. Such processes don't move on a P—V diagram, such as phase change of a pure substance, which takes place at the saturation pressure and temperature. Chemical reactions, however, do undergo changes in chemical potential, which is a state function. Thus, thermodynamic processes are not confined to the two dimensional P—V diagram. There is a third dimension for n, the quantity of gas

If you write $\Delta G = \Delta H -T\Delta S$, you implicitly admit that temperature is constant and is the same for a non-isolated system and its environment.

You basically assume that the system and the environment are in thermal equilibrium, at a given temperature $T$. Thermal equilibrium means that when system loses heat, environment absorbs it: this must occur equally in both directions, otherwise a temperature gradient would settle between system and environment. We could describe thermal equilibrium as the result of "a zero balance of rate of heat transfer between system and environment". That's what goes on during a phase transition (e.g. solid to liquid, where entropy of the system increases due to heat given by the environment, whereas temperature stays constant).

Now, applying the second law of thermodynamics to an ensemble system + environment, undergoing an infinitesimal transformation, we write:

$$\mathrm dS_\text{syst} + \mathrm dS_\text{env} \geq 0 \tag1$$

As said above, the entropy of the environment changes if there's a  reversible heat exchange with the system:

$$\mathrm dS_\text{env} = \frac{\delta Q_{env}}{T}=-\frac{\delta Q_{syst}}{T} \tag2$$

Merging eq. $\text{(1)}$ and $\text{(2)}$, you write:

$$\mathrm dS_\text{syst} - \frac{\delta Q_{syst}}{T} \geq 0 \tag3$$

Now the first law allows you to write:

$$\delta Q - P\mathrm dV_\text{syst}=\mathrm dE_\text{syst} \tag4$$

and then

$$T\mathrm dS_\text{syst} - \left(\mathrm dE_\text{syst}+P\mathrm dV_\text{syst}\right) \geq 0  \tag5$$

Now (and this is crucial), assuming both pressure and temperature constant, you end up with

$$\mathrm d{\left(E_\text{syst}+PV_\text{syst}-TS_\text{syst}\right)_{T,P}} \leq 0 \tag6$$

The expression within the brackets is a state function of the only system and corresponds to the free energy. Quoting Wikipedia:

  A state function describes the equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state quantities because they describe quantitatively an equilibrium state of a thermodynamic system, irrespective of how the system arrived in that state.

Considering that $E+PV=H$, we write:

$$\left(\mathrm dG_\text{syst}\right)_{T,P} = \left(\mathrm dH_\text{syst}-T\mathrm dS_\text{syst}\right)_{T,P} \leq 0 \tag7$$

Now, while equation $\text{(1)}$ is much more general, equation $\text{(7)}$ applies only to the situation where the work is expansion or contraction and for constant temperature and pression. Nonetheless, since for most situations of chemical interest, those conditions are satisfied, eq $\text{(7)}$ maintains a well deserved relevance and is very convenient to establish whether a process is spontaneous or not, within the limitation of constant temperature and pressure.

A chemical reaction occurs down to the point where $\left(\Delta G\right)_{T, P}=0$.

For a general reaction:

$$\ce{$a$A + $b$B <=> $l$L + $m$M}$$

For constant temperature and pressure:

$$\left(\Delta G\right)_{T, P}=l\mu_{\ce L} + m\mu_{\ce M} - a\mu_{\ce A} - b\mu_{\ce B}  \tag8$$

being $\mu$ the chemical potential for each of the species involved.
The chemical potential relates to activity, according to:

$$\mu=\mu^\circ+RT\ln a \tag9$$

Equation $\text{(8)}$ and $\text{(9)}$ shows that free energy varies with the composition (expressed as activity) of our system: that's exactly what happens during a chemical reaction.

Proceeding further, you will end up with Van't Hoff isotherm:

$$\left(\Delta G\right)_{T, P}=\left(\Delta G^\circ\right)_{T}+RT\ln\frac{a_{\ce L}^{l}a_{\ce M}^{m}}{a_{\ce A}^{a}a_{\ce B}^{b}}$$

Activities vary until a chemical equilibrium is established, so that $\left(\Delta G\right)_{T, P}=0$.


Dialkylation reactions In the synthesis of ethyl 2-benzyl-2-methyl-3-oxobutanoate, is the nucleophilic substituion of a benzyl (chloride) group an Sn1 reaction?

The experiment is conducted in ethanol (protic) and is under reflux, but most textbooks have it as an Sn2?

Will it then be a mixture of the two?

Both mechanisms will probably be present but $S_N2$ will dominate because the enolate anion is a strong nucleophile. $E2$ will not occur because there are no $\beta$ hydrogens available on the benzyl chloride. Despite the relative stability of the benzyl cation, especially in protic solvents, the strong nature of the enolate nucleophile and the lack of an $E2$ pathway means that $S_N2$ will probably dominate.

For a comparison of alkyl halide reactivities with different nucleophiles, see this link.


Experiment to show that air contains about 20% oxygen I remember that in my primary classed I had an experiment in science to show air contains 20% oxygen. The experiment involved taking a trough of water, lighting a candle in the middle of the trough and inverting a glass jar over the candle so that air supply for the candle is cut of. When candle extinguishes, water level in the glass jar rises. I have seen this happen.

My question is why should the water raise in the jar, since burning of candle should produce an equivalent amount of carbon dioxide, so the molar amount of gas in jar is constant? Also the heat from the candle should expand the gasses in the jar, water level should go down I think.

One problem with using combustion reactions to determine the percent of oxygen in air is the additional products, namely $\ce{CO2}$ and water vapor.  If performing a water displacement reaction, these gases add to the uncertainty and imprecision of this method.

Water displacement by a combustion reaction can, however, be used if an appropriate combustion reaction is chosen.  A more reasonable experiment to determine the percent oxygen in air is to use the oxidation of iron $$\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$$ The benefits of this reaction are (a) only one product is formed and (b) the product is a solid so it will not influence the water displacement in a significant way.

This experiment has been reported in the Journal of Chemical Education and a brief search of the web found an adaptation of this experiment if you don't have access to the journal.  Briefly, one measures the volume of a test tube, adds a known amount of steel wool into the tube and inverts the tube in a beaker of water.  The oxidation of the iron is catalyzed by first rinsing it in a dilute acid (vinegar) solution.  Knowing the density of iron one can calculate the volume of the test tube occupied by the solid and then determine the volume of air before and after the reaction is completed.

There are a number of possible errors that can be introduced into this experiment, but unlike the combustion of a hydrocarbon, these errors are much more manageable for an undergraduate or high school chemistry experiment.  Most significantly, the reaction occurs near room temperature so the temperature change of the gas can be ignored.


What is intermolecular distance in water? I am wondering what is an average intermolecular distance in liquid water (say under normal pressure and room temperature).
I need just order of magnitude. A reference would be helpful.

Not all data are provided in explicit form. They are often not frequently needed and can be often deduced from other data with using of high school knowledge.
You can calculate the volume occupied by 1 water molecule ( in average ) from the water molar mass $\pu{M}$, water density $\pu{\rho}$ and the Avogadro constant $N_\mathrm{A}$ as
$$V=\dfrac{M}{\rho \cdot N_\mathrm{A}} $$
Then you can calculate the side length $a$ of a cube of this volume:
$$a={\left( \dfrac{M}{\rho \cdot N_\mathrm{A}}  \right)}^{1/3}  $$
As a rough number, is is equal to
$a  \approx { \left( \dfrac{\pu{0.018 kgmol^-1}}{\pu{1000 kgm^-3} \cdot \pu{6.022e23 mol^-1}}  \right)}^{1/3}  \approx \pu{0.31 nm}$
what can be taken as the typical "center to center" distance of water molecules within the order-like rough precision.
The more precise average distance would have to calculate with the size and shape of the water molecule ( $\ce{O-H}$ bond length about 0.096 nm, the bond angle 104.5 degrees) and to make the integral average over the directions. But it would be extremely complicated, considering also difficulties with molecule borders and hydrogen bonds. In some sense can be said the distance is zero or even "negative", as the molecules form bonds ands overlaps.


Determining the reduction potential of a half cell reaction at a given pH $E^\circ_\ce{Cu^2+|Cu} = \pu{0.34 V}$.
  What will be reduction potential at $\mathrm{pH} = 14$ for the same couple?
  Given that $K_\mathrm{sp}$ of $\ce{Cu(OH)2}$ is $10^{-19}$.

My Attempt

I think that this problem might be solved by using the Nernst equation

$$E = E^\circ - \frac{0.0591}{n}\log_{10} K,$$

but I don't know how do I apply that.
We have the $E^\circ$ and the number of electrons participating (which is equal to 2), but I don't know how do I determine the equilibrium constant.

Any help or hint would be appreciated.

The K you refer to relates to the overall cell reaction and should be Q, the reaction quotient. This is only equal to K when the potential difference between the two 1/2 cells has fallen to zero.

This question does not refer to a full cell, but to a 1/2 cell. The form of The Nernst Equation you must use, assuming we are at 298K, is:

$\rm E=E^\circ+\dfrac{0.0591}{z}log\dfrac{[oxidised\,\, form]}{[reduced\,\, form]}$

Where z is the number of moles of electrons transferred which, in this case = 2.

This becomes:

$\rm E=E^\circ+\dfrac{0.0591}{2}log[Cu^{2+}]$

The 1/2 cell in question is:

$\rm Cu_{(aq)}^{2+}+2e\rightleftharpoons Cu_{(s)}\,\,\,\,\,\,\,\,\,\,E^\circ=+0.34\,V$

At such a high pH the copper(II) ions will precipitate.

$\rm pH=14\,\,\,so\,\,\,pOH=0$

So $\rm c[OH^-]=1\,\,mol/l$

This will precipitate out the copper(II) ions as the hydroxide.

So we have:

$\rm Cu(OH)_{2(s)}\rightleftharpoons Cu_{(aq)}^{2+}+2OH_{(aq)}^-$

$\rm K_{sp}=[Cu^{2+}][OH^-]^2=10^{-19}$

$[\ce{OH-}]$ is 1 (pH is 14), so $[\ce{Cu^2+}]$ is equal to $\pu{e-19}$.

As you can see this is a very low concentration. Applying Le Chatelier's Principle to the 1/2 cell couple we would predict that reducing the concentration of Cu(II) will drive the position of equilibrium to the left. This would push out electrons and make the electrode potential of the 1/2 cell more negative.

Putting in the numbers into the Nernst Equation:

$\rm E = 0.34+\dfrac{0.0591}{2}\log \pu{e-19}$

$\rm E = 0.34- 1.12\,\,\,\,\,V$

$\rm E = -0.78\,\,\,\,V$

As predicted, it has been reduced considerably.


Is it possible to create an acid anhydride that's stable in water and if so what conditions must its structure satisfy? I ask this question because if acid anhydrides are always unstable in water then my interest, as a student of pharmacology, would be greatly diminished. Acid anhydrides are, in my books, at least, the result of a dehydratation reaction between two carboxylic acids and hence their structures look like this:

Where R1 and R2 are organic substituents.

This answer is years later than the original question, but I add it for anyone who come across it now or in the future...
All acid anhydride and acid chlorides are thermodynamically unstable in water.  They will eventually hydrolyze, but that time could be seconds, minutes, hours, days, or years. Moreover, in the absence of a catalyst, the rate of hydrolysis (kinetics) of any acid anhydride or any acid chloride (carboxylic, phosphoric, sulfonic, etc.) is strongly temperature dependent (hotter = faster, colder = slower, as you would expect).
For example, the classic hydrolysis of acetic anhydride is quite slow until temperatures exceed 55—60 °C.  Similarly, sulfonyl chloride hydrolysis is very slow until temperatures exceed about 70 °C.  This is in part driven by solubility - the substrate needs to be in the same phase as the water for fast rates.  Phosphoric anhydrides such as ATP or ADP can be extremely stable in water at physiological temperatures (37 °C).  The presence of an protic acid, or particularly  a nucleophilic catalyst (hydroxide, amines, thiols), can change the kinetics dramatically.  Thus, you have to pay attention to the temperature, the phasing/solubility, and the identities/properties of the other components in your reaction mixture.


Naming a 4 branched cyclic compound If this compound exists, what will be its name?

Can it be called bicyclo[9.2.2.2] hepta-decane ?

Your selection of the main ring and the bridges that results in the bridge lengths ‘[9.2.2.2]’ and your counting of the total number of skeletal atoms that results in the name ‘heptadecane’ are correct.

For your proposed name ‘bicyclo[9.2.2.2]heptadecane’, you obviously followed the nomenclature of similar bicyclic systems, such as bicyclo[9.2.2]pentadecane.

This is not a bad idea since the nomenclature of polycyclic hydrocarbon ring systems is actually based on the nomenclature of bicyclic systems.

However, the structure given in the question is not a bicyclic ring system because it includes one additional secondary bridge; therefore, the structure actually is a tricyclic system.

The additional steps for the naming of tricyclic hydrocarbons are described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) as follows.

  P-23.2.5 Naming and numbering tricyclic alicyclic hydrocarbons

  P-23.2.5.1 Tricyclic hydrocarbons having an independent secondary bridge are named on the basis of a bicyclic system (…). Rings not described by the bicyclic system are defined by citing the number of atoms in the independent secondary bridge as an arabic number. The locants of the two attachment points of the independent secondary bridge to the main ring are cited as a pair of superscript arabic numbers (lower number is cited first) separated by a comma.

  The name of the tricyclic system is then constructed by citing:

  (a) the prefix ‘tricyclo’, in place of ‘bicyclo’, indicating the presence of three rings in the polyalicyclic system;

  (b) numbers indicating the bridge lengths, starting with the two branches of the main ring (…), followed by the main bridge (…), and the secondary bridge (with superscript locants separated by commas indicating its points of attachment to the main ring), all separated by full stops and placed in brackets,

  (c) the name of the acyclic hydrocarbon having the same total number of skeletal atoms.

Therefore, the correct name for the structure given in the question is tricyclo[9.2.2.21,11]heptadecane.


does it make any difference writing NO2 instead of O2N can you write NO2 in all three?

The Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008), include recommendations for contracted labels in structure diagrams. In particular

  GR-2.3.1 General guidelines
  Atom labels representing more than one non-hydrogen atom—also sometimes known as “contracted” labels—rely on the fact that many elements have consistent and well-understood bonding patterns. (…)

  GR-2.3.3 Orientation of symbols within contracted labels
  Contracted atom labels attached to only one bond should be read outwards from that bond, usually from left to right if the bond is on the left of the label. If the bond is instead attached to the right of the label, the label will normally be read from right to left, but ambiguities can result. Accordingly, contracted labels with a bond on the right should be avoided except for simple cases, usually limited to relatively small labels containing four or fewer combined element symbols and abbreviations. (…)

  GR-2.3.4 Interpretation of contracted labels
  In general, contracted labels are interpreted to fill as many valences as possible, as quickly as possible. (…)

Contracted labels for nitro groups cannot be interpreted with a simple application of valence rules, but also need some implicit charges to be added. It is equally possible to avoid entirely the issue of depiction and instead display the functional group using the $\ce{NO2}$ abbreviation. According to Table III in the Graphical Representation Standards, the left-to-right form of the label is $\ce{-NO2}$ and the corresponding right-to-left form is $\ce{O2N-{}}$.


Same set of carboxylate reagents yielding two different products with an alkylide I recently came across two contrasting products for the same set of reagents - carboxylic acid and alkyllithium, and wish to understand why exactly any difference exists between the two.

Reaction 1

Reaction 2

I understand that in steps A, B, and E, the alkylide (hydride in E) is acting as a base and abstracting the acidic proton. I'm fine with that.

However, compare step C and F. Both involve the same reagents (carboxylate and an alkylide). In step C, nucleophilic addition of the ethylide has taken place on the carbonyl group, which is a reasonable step according to me. However, in step F, an ester has been formed instead. I am unable to grasp how an ester can even form in this condition (considering both ethylide and carboxylate are negatively charged), and even if it can form, why wasn't it formed in C as well?

Source: MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; Q64 and 27 in Alcohol, Ethers and Epoxides

Reaction 2 is wrong. Reaction 1 is correct.

Later in the book (q87), another question is asked with the same set of reagents (carboxylic acid and methyllithium). Reaction 1 has been used in that as well.

Reaction 2 is wrong because, as ralk912 noted above, charge conservation has to be followed. If methyl lithium reacts with a negatively charged species, the product has to be negatively charged as well. If the ester is neutral, then lithium must be negatively charged, and would associate with the $\ce{Li+}$ formed in step E to form $\ce{Li2}$, which is only observed in gas phase.

Not to forget that there simply isn't any mechanism wherein a carboxylate would react with an alkylide to form an ester. It does not simply occur. Reaction 2 is wrong.

Waylander has suggested a good reason for a misprint i.e. the reagent $\ce{CH3Li}$ might have been intended to be $\ce{CH3I}$ instead. Then, it would make sense for the carboxylate anion to attack an alkyl halide, and displace an iodide ion (which could easily attach with the $\ce{Li+}$ ion to form lithium iodide) forming the methyl ester.

user55119 has confirmed that the product of the step F is cyclohexylmethyl ketone, and that it is obtained by reaction 1 only. The book containing this is "Margret Jorgensen in Organic Reactions."


Is the %V/V composition defined for homogeneous mixtures? Why do major websites such as NASA and Wikipedia discuss the composition of the earth's atmosphere in terms of percentage by volume instead of percentage by amount of substance? Isn't the percentage composition by volume undefined since all gases in the atmosphere occupy 100% volume of it?

Let's be clear on what percent by volume means.  If I say component $X$ is 25% by volume of a mixture of $X$ and $Y$, with a total volume of 1 L, it does not mean there is some 250 mL portion of the mixture that is solely occupied by $X$.  That wouldn't be a mixture.  It doesn't even mean that if I mentally subdivide that 1 L mixture into tiny identical cubes, one for each atom or molecule, then 1/4 of those cubes are occupied by atoms or molecules of $X$.

Instead, all it means is that I can prepare 1 L of the mixture by mixing some volume $V_X$ of $X$ and 3 times that volume ($V_Y = 3 V_X$) of Y.  There's not even any guarantee that $V_X = \pu{250 mL}$, since sometimes volumes are non-additive, that is, $V_X + V_Y \neq \pu{1 L}$.

Now as it happens, for ideal gases, which the atmosphere is close to being, you can mentally subdivide the volume into tiny little equal-sized cubes, one for each atom or molecule in the mixture, and on average there would be one atom/molecule of $X$ in each fourth cube if $X$ is 25% by volume of the mixture.  (I emphasize "on average" because of course since the gas atoms or molecules are continuously moving around randomly, they will bunch up and spread out momentarily all the time, so at any given instant there may be zero or many more than one $X$ atom/molecule per tiny cube.)  Real mixtures in the liquid state often have at least some nonideality, e.g. even in a dilute solution of $\ce{NaCl}$ in water, you will find "structure" in the solution, with the $\ce{Na+}$ cations and $\ce{Cl-}$ anions surrounded by a fairly fixed arrangement of $\ce{H2O}$ molecules, almost like a tiny piece of clathrate, so a mental arrangement of tiny boxes is an even less accurate atomic-scale description of the solution.

In short, the percent by volume description of a mixture is only a way to characterize the amount of material that goes into it, it is not intended as any kind of implication of what the  mixture looks like at the atomic scale.

As for why NASA reports the composition of the atmosphere by percent by volume: probably because it's closest to the actual experiments done to measure it.  You would typically measure the composition of a gas mixture by physically separating it (e.g. by lowering the temperature until each gas liquefied) and then measuring the volume of each component.  You certainly could from those measurements easily calculate a percent by moles, but – why?  You introduce a calculational step between the measurement and what you report, and scientists tend to prefer getting original data, right what comes from the instrument, if possible, as a way of avoiding even the smallest risk of some error introduced in calculation.


Is this gas in thermodynamic equilibrium? Please consider the following graph of molar volume as a function of pressure for a fixed volume.

Can we say whether this system under thermodynamic equilibrium or not?

Update Thinking about this some more, the molar density vs pressure graph may be useful too.

Is this gas in thermodynamic equilibrium?

I don't think "equilibrium" is a good word to use in your question.  We have no way of knowing if the gas is at equilibrium from the data you present.  A highly non-ideal gas that was exploding might give the same relationship.  A gas undergoing a rapid unimolecular reaction wouldn't really be at a true chemical equilibrium but could behave as an ideal gas from a pressure / temperature standpoint.

Did you really mean to ask if the gas was ideal?  An ideal gas follows $PV = nRT$, which can be re-written in several ways:

molar density $\frac{n}{V}$ as a function of pressure $P$: $$\frac{n}{V} = \frac{1}{RT}\;P$$, which is the equation of a straight line through the origin (bottom graph).
molar volume $\frac{V}{n}$ as a function of pressure $P$:  $$\frac{V}{n} = RT \frac{1}{P}$$, which is the equation of a hyperbola (top graph).

Your data is consistent with the gas being ideal (and isothermal).


Solvents for monoalkylations? Just beginning to grasp this. Betaphenylethylamine to n-methyl-b-pea. Can I use water instead of DMSOesque solvent in regards to cesium carbonate/hydroxide monohydrate? Also are molecular sieves necessary? As well as 1-bromobutane?

Bottle + water + pea + cesium carbonate/hydroxide = NMPEA?

Your phenethylamine and bromobutane/methyl iodide are not going to be soluble in water but will dissolve in each other and react. If you use aq base and stir, possibly with gentle heating for bromobutane, then you will get your product. No need for sieves or drying agents. This is essentially a phase-transfer reaction but without a second solvent. As with any attempted amine alkylation, the danger is over reaction to give the quat. salt. If you find over reaction to be a significant problem then consider reductive amination.


Heavy-oxygen water production It is easy to find information about heavy-hydrogen water production, but not about heavy-oxygen water. How is heavy-oxygen water commercially produced (for both heavier isotopes)?

As far as I know, while fractional distillation would be a suitable technique to obtain O-18 water (yet, in very low yelds and in a way that is hard to control), the most used method nowadays is the membrane distillation technique: basically, a semipermeable and hydrophobic membrane is employed during the distillation (tipically made in PTFE), and the isotopes are separated by their ability to permeate through the membrane: lighter isotopes (O-16 water) tend to flow through the membrane, while heavier isotopes (O-18 water) are trapped in the apparatus and recovered.

You can find many detailed descriptions online, for instance:

http://www.ichtj.waw.pl/ichtj/library/Production%20of%20stable%20isotopes%20by%20membrane%20method.pdf

https://www.google.com/patents/US5057225


Vacuum Distilled Medicine If I were to buy/build a large still, that was air tight, and could hold a vacuum, and filled the reservoir with vegetable glycerin (which has a boiling point higher than the majority or desired organic compounds) and I made the atmospheric pressure inside less than normal atmospheric pressure, could I distill chemicals out of organic matter without altering them? Such as distilling willow bark to get an aspirin containing solution, or distill mint leaves without changing the chemical composition of the medicinal compounds in the plant/organic matter?

This could be a cool project. It would enable us to explore further into natural medicine. But would this procedure work?

In general the answer would be no.

While there are many distillable compounds you could extract this way (not sure how many medicinal ones, though) there are many more that would decompose or simply not be easily extracted.

This is why there are many other ways chemists commonly extract things from mixture or from plants. For example, if you wanted to extract caffeine from coffee or from coffee beans, you could use solvent-based extraction (methylene chloride works well, for example, and has been used in industrial processes to create decaffeinated coffee). There is no single easy solution, though, and often much trial and error will be required to find the right solvent to differentially extract what you want.


What does "correct orientation" in a chemical reaction mean conceptually? My chemistry teacher was talking about how there are two requirements for a reaction to occur:

The reactants must have enough energy; and
they must also have the correct orientation for a reaction to occur.

What does it mean for molecules to have the "correct orientation", perhaps using the gas-phase reaction as an example?

$$\ce{2H2 + O2 -> 2H2O}$$

What does it mean for molecules to have the "correct orientation", perhaps using the gas-phase reaction as an example?
  $$\ce{2H2 + O2 -> 2H2O}$$

This is a good example, but a complicated one. It is a chain reaction with three steps that have radical species as reactants and products (chain propagation), and one step that has radical species as reactants but not as products (chain termination). If you add up reactions 1 through 4, you will get the net reaction.

$$\ce{ O + H2 -> OH + H [1]}$$
$$\ce{ H2 + OH -> H2O + H [2]}$$
$$\ce{ O2 + H -> O + OH [3]}$$
$$\ce{ H + OH -> H2O [4]}$$

In order for the chain reaction to start, you need an initiation reaction starting with stable molecules making radicals. This is a slow reaction requiring a lot of energy, and there is some debate what it is. Here are two suggestions discussed in the literature.

$$\ce{ O2 + H2 -> OH + OH [5]}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ce{ O2 + H2 -> H2O + O [6]}$$

These two reactions require different orientations of the dioxygen and the dihydrogen molecules when they collide. This is because different bonds are broken and made when the different products are made. The picture below shows a "good" orientation for reaction [5] on the left and for reaction [6] on the right (The circles show the reactant molecules, and the dashed lines show the bonds that have to be made to get the products).

With the orientation on the left, it is easier to make two OH radicals and more difficult to make a water (with an O atom remaining). With the orientation on the right, it is easier to make a water molecule (with an O atom remaining) and much more difficult to make two OH radicals. Depending what the products are, some orientations upon collision are productive (lead to product) and some orientations are non-productive (no reaction occurs).

There is a third suggestion for the initiation step in the literature, and it requires an orientation for the collision different from the two orientations shown above. Can you figure out the best orientation for reaction 7?

$$\ce{ H2 + O2 -> H + HO2 [7]}$$

Source for reaction mechanism: NASA report


Can someone intuitively explain the reason for the units of entropy (J/K )? If entropy is just a measure of the "randomness" or "chaos" of a system, then why does it have specific units. What is the reason behind the units of entropy being J/K?

The units of of energy over temperature (e.g. J/K) used for entropy in the thermodynamic definition follow from a historical association with heat transfer under temperature gradients, in other words, the definitions of temperature and entropy are intertwined, with entropy being the more fundamental property. Entropy can be thought of as a potential and temperature (or its inverse, rather) as a generalized force associated with displacements along energy dimensions in the entropy potential. The (inverse) temperature is a measure of the effect of changes in the amount of energy on the entropy of the system, as the following definition suggests:

$$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)_V$$

However if you start from a statistical mechanical definition of the entropy,

$$S=k_\mathrm{B}\log\Omega$$

then it is easy to see that a unitless definition might be just as well suited:

$$S_{\mathrm{unitless}}=\log\Omega$$

However, in the absence of the Boltzmann constant you need some other way to relate entropy and energy (e.g. heat transfer) in our conventional system of units. You can of course subsume $k_\mathrm{B}$ into the definition of a new temperature scale:

$$\frac{1}{T_\mathrm{new}}=\left(\frac{\partial S_\mathrm{unitless}}{\partial U}\right)_V$$

where the old and new temperature scales are related as $T_\mathrm{new}=k_\mathrm{B}T$. The new temperature scale then has units of energy.


Electrolysis of Water: Overvoltage? I've heard from a few sources that over-voltage in a electrolysis cell for water will cause a greater heat buildup with no yield. Is this true, and if it is, why is it true? From my electronics background, voltage can be though of as electric potential, or thinking of it as water in a pipe for an analogy it is the strength of the substance flowing through the pipe, or as the strength of each electron, so why does it cause extra heating with no appreciable/no yield?

To first order, the power dissipated in an electrochemical cell is equal to that of an electrical circuit: $P=IV$. Thus, as the overpotential necessary to achieve a given current throughput increases, the power consumption increases also.

In the fluid flow analogy, think of an elevated overpotential as a throttling orifice inserted into one of the pipes.  In order to keep the water moving through the pipe at the same volume flow rate, you have to increase the pressure drop across the system.  A greater pressure drop to achieve the same flow rate requires a pump with a greater power rating.


Would concentration of a compound in equilibrium increase even if it's not favored by a volume decrease? So say I have a reaction where there are x moles of gas on the reactant side, and y moles of gas on the product side, where x>y. The volume decreases, favoring the products side. The textbook I am using says that the equilibrium concentration of the reactants would decrease, but wouldn't it end up increasing due to the lowered volume?

Suppose you have the following generic reaction:

$\ce{xA_{(g)} <=> yB_{(g)}}$

With the condition, $x > y$. The reaction quotient, written in terms of concentration would be:

$Q = \frac{(\frac{n_B}{V})^y}{(\frac{n_A}{V})^x}$

Where $n_A$ and $n_B$ are moles of the respective species and $V$ is volume. With some algebraic manipulation, this can be rewritten as follows (note here I'm explicitly treating $Q$ as dimensionless, q.v., the relevant Wikipedia page on equilibrium constant units, which applies to the reaction quotient as well):

$Q = \frac{n_B^y}{n_A^x} \cdot V^{x-y}$

Written in that form, it should be apparent that, if $x > y$, then $x - y > 0$, and therefore $Q$, the ratio of products to reactants, varies directly with $V$. In other words, as volume grows, $Q$ grows, and as volume decreases, $Q$ decreases. Suppose that, prior to some volume change, the reaction is at equilibrium, i.e., $Q = K$. If volume instantaneously begins to increase, $Q$ becomes instantaneously greater than $K$, and therefore the rate of the reverse reaction becomes greater than the rate of the forward reaction, favoring generation of $A$ until equilibrium is reestablished. Conversely, if volume instantaneously begins to decrease, the opposite happens and production of $B$ is favored. This is precisely what one would expect based on an intuitive understanding of Le Chatelier's principle: if volume decreases, equilibrium shifts to favor production of fewer moles of gas to counteract the increasing pressure; if volume increases, equilibrium shifts to favor production of more moles of gas to counteract the decrease in pressure.

Edit: I misread part of your question initially. In short, if the volume decreases, generation of product $B$ becomes favored, so the concentration of reactant $A$ must decrease relative to $B$ (since $x$ moles of $A$ are consumed in the production of $y$ moles of $B$, and the reaction quotient approaches the equilibrium constant over time).

Just to clarify this further, it's true that, in an absolute sense, if volume decreases in an instantaneous fashion, then the concentrations of the reactant and product, considered separately, would both have to increase. However, in relative terms, the concentration of $B$ must increase relative to $A$ as the reaction proceeds. The ratio between the two must grow to compensate for the diminished volume, otherwise $Q$ will remain perpetually smaller than $K$, when in fact $Q$ always converges to $K$, for a given set of conditions, over time.


I have a sodium silicate sealant in my car and need to find a way to soften it Poured Blue devil sealant and found that it clogged my engine. Can someone who knows about chemistry and cars maybe give me any ideas on how to solve this issue?

I do not know much about engines, but from a chemical point of view, the clog that formed in your engine is most likely a hydrated silica. This compound is very insoluble and unreactive - there are only few things that can dissolve silica, e.g., hydrofluoric acid or a hot concentrated solution of sodium/potassium hydroxide, but I wouldn't pour any of these to an engine. Therefore, it seems that your only option is to dismantle the engine and try to remove the clog mechanically.


Acids generating nascent gas atoms When sulfuric acid  reacts with copper, then $\ce{H2SO4}$ gives nascent oxygen $\ce{[O]}$. The reaction is:
$$\ce{H2SO4 -> H2O + SO2 + [O]}$$

This nascent oxygen reacts with copper to create $\ce{CuO}$.

Therefore my question is that when copper reacts with $\ce{HCl}$ then why doesn’t $\ce{HCl}$ give nascent chlorine $\ce{[Cl]}$ so that it could react with copper and create copper(II) chloride $\ce{CuCl2}$?

When concentrated sulfuric acid is heated, then it decomposes via the reaction
$$\ce{H2SO4 -> H2O + SO3}$$

I tried to determine under what circumstances the nascent oxygen (a free single atom of oxygen) is formed which would react with copper to create $\ce{CuO}$ but couldn't find the reaction in any modern source. I did find the proposed reaction in Essentials of Modern Chemistry
 By Charles Elwood Dull (1918) but I'm guessing that the reaction to give nascent oxygen is just wrong.

The gist here is that you can't piece reactions together from balanced equations but you must also define the conditions under which the reaction takes place. At a deeper level not only can the chemical equation be specified but also the mechanism of the reaction can be detailed.

Hydrochloric acid, $\ce{HCl}$, is typically in aqueous solution. Although copper won't react with pure hydrochloric acid, it can react with dissolved oxygen in hydrochloric acid.
$$\ce{Cu (s) + 2HCl (aq) + 1/2O2 (aq) -> CuCl2 (aq) + H2O}$$

Oxygen can also be formed in chlorine water:
$$\begin{align}
  \ce{Cl2 (aq) &-> HCl + HOCl}\\
  \ce{HOCl &-> HCl + 1/2O2}
\end{align}$$

But the concept of nacent oxygen from this reaction is wrong.


How is the Maxwell-Boltzmann curve obtained? Is it purely qualitative and determined by experiment?

Or is there some function which defines the familiar graph?

(Image source: Wikipedia)

There is, indeed, a function that plots the graph.

There are two common types of graph, one with kinetic energy as the $x$-axis, and one with speed. For the most part, they are qualitatively the same, although there is a slight difference in curvature near the origin. The speed function is much more commonly used, because one can more easily extract useful information from it, such as mean velocities, or mean free paths. It is:

$$f(v) = (4\pi v^2) \cdot \left(\frac{m}{2\pi kT}\right)^{3/2} \cdot \exp{\left(-\frac{mv^2}{2kT}\right)}$$

where $m$ refers to the mass of the particle, $k$ Boltzmann's constant, and $T$ the temperature.

As for the derivation, it is rather long and perhaps you were not really asking for it. If you are interested, I would recommend looking it up in a physical chemistry textbook, for example: Atkins' Physical Chemistry, 10th ed., section 1B(b), or Levine's Physical Chemistry, 6th ed., section 14.4. I am partial towards Levine because he gives a much more detailed explanation than Atkins, who is always very brief.


Elimination in α,β-dibromohydrocinnamic acid The above is a picture of α,β-dibromohydrocinnamic acid (2,3-dibromo-3-phenylpropanoic acid). In the lab today I carried out an elimination reaction on this molecule using potassium carbonate giving the carbonate anion as a base. I thought about the mechanism and I initially thought that the base would attack the hydrogen on the leftmost carbon atom (out of the two with the bromine atom on it) as I thought that this would be the most acidic proton of those two carbon atoms because of resonance available with the aromatic ring. However, the proton on the carboxylate group is surely the most acidic. How does this effect things in terms of the mechanism and the product?

Dibromohydrocinnamic acid can eliminate in a number of different ways as shown in the following diagram. I'm not sure what conditions lead to cinnamic acid, but I know it can happen.  You have recently been asking questions about E1 and E2 eliminations, so my guess is that you are running a well-known lab experiment that generates cis- and trans-bromostyrene and allows the student to study the effect of reaction conditions on the mechanism of an elimination reaction.

Usually, the experiment starts with the bromination of trans-cinnamic acid.  The bromination is run under conditions that lead to trans-addition of bromine so that only the erythro (see here for diagrams and mechanism) or (S,R) [and its mirror image (R,S) enantiomer] dibromocinnamic acid diastereomer is formed.  The other dibromohydrocinnamic acid diastereomer, the threo isomer [(R,R) and its (S,S) enantiomer], which can only be formed by cis-addition of bromine is not produced.

If you build a model or draw a Newman projection it will make it easier to understand what follows.

Using the erythro [(S,R) and (R,S)] diastereomer obtained from the above bromination, running the elimination reaction in a solvent like acetone (a solvent that won't stabilize a carbonium ion as much as protic solvents like water or methanol) you drive the mechanism towards the E2 elimination side.  With your model or drawing, put the carboxylate group (the base has removed the acidic carboxylic acid proton to produce the corresponding carboxylate anion) and the bromine on the next carbon in an anti-periplanar arrangement (you've discussed anti-periplanar in your earlier questions on the E2 mechanism) and see how in a one-step, concerted elimination of both carbon dioxide and bromide you wind up with the cis-bromostyrene isomer.

If you ran the reaction in a protic solvent (like water or methanol) that stabilizes carbonium ions, you push the reaction towards the E1 mechanism involving loss of the bromine adjacent to the benzene ring to produce a stabilized benzylic carbonium ion.  The stable carbocation ion has an appreciable lifetime and will rotate so 1) the soon-to-be-departing carboxylate group is lined up with the carbocation p-orbital on the adjacent carbon and 2) that the benzene ring and remaining (bulky) bromine are as far apart as possible (the lowest energy conformation).  Elimination of carbon dioxide from this conformation produces trans-bromostyrene.

Typically, the student is given several test tubes and the reaction is run under a variety of conditions where solvent and reagents are changed and then the amounts of cis- and trans-bromostyrene are measured by nmr.  From this information the student can see how solvent and reagent can shift the mechanism.  See this link for a bit more detail on the actual experiment along with some good Newman projections.

BTW, the lab should smell nice since the trans-bromostyrene produces a hyacinth-like odor.


Non polar molecular and attractive forces How come when a non polar molecule increases in size, the attractive forces between the molecules of a substance also increase?

I thought that when a molecule increases in size, the atomic radius increases, which creates a bigger space between particles; thus, the force of attraction would be weaker.

But, why does the force of attraction increase when a non polar molecule increases in size?

There is more than one factor that matters in the forces between molecules. The tradeoff between these different factors is complex and can't be reduced to a single simple equation or factor. So some of the factors become more important because of molecular or atomic size counterbalancing the factor of distance between the entities.

The collective force that cause molecules and atoms to attract are often referred to as van der Waal's force (though some sources use a more narrow definition that doesn't include all of the components below). The key non-bonding forces involved are:

forces between permanent dipoles
forces between permanent dipoles and induced dipoles
forces between instantaneously induced dipoles caused by quantum fluctuations

In non-polar molecules it is the third of these that dominates. Crudely stated the force arises because of uneven distribution of the electron cloud in a molecule or atom that happens as a result of quantum fluctuations in the cloud. An uneven distribution of electron density is the equivalent of a dipole moment. These dipole moments create induced dipoles in neighbouring atoms and can interact with other quantum fluctuations in the neighbouring atoms. These create small electrostatic forces between the atoms or molecules.

The reason why larger atoms or molecules seem to have larger forces is two fold. Larger molecules, crudely, have larger electron clouds and more ways to interact (shape matters too and large flat molecules "stick" together more than irregular lumpy ones: compare the melting points of benzene and toluene).

In larger atoms the explanation is simpler. Larger electron clouds are less tightly held by there nuclei than those in smaller atoms. Crudely, larger, floppier clouds can have more scope for the quantum fluctuations than tighter-held smaller clouds (also they have a larger area and that maters to the total force). So the size effect increases the strength of the force rather than reducing it as might be expected from simpler arguments.

I'm aware that this is a slightly superficial explanation, but a better one would probably involve enough quantum mechanics to make the average reader's head explode. I think i've captured the key intuitions that matter.


How do I create azeotrope mixture? I want to decrease the boiling point of a compound. So I need to create an azeotrope mixture from two or three compounds. Where can I find out how make it?

Boiling point can be decrased by lowering pressure over liquide. If you insist on the azeotrop preparation, find azeotrop diagram such as:

or data table such as: wiki and determine the composition that must be achieved to obtain the azeotropic mixture. Then mix it accordingly.


Why do acids mix in water faster than bases? I have $100\ \mathrm{L}$ of water in a tank with an mixer inside and want to know how fast the tank mixes at a given rpm-setting.
I am adding $50\ \mathrm{mL}$ of $2\ \mathrm{N}\ \ce{HCl}$ and see how long the $\ce{pH}$ goes to down to an equilibrium.
Then I add $50\ \mathrm{mL}$ of $2\ \mathrm{N}\ \ce{NaOH}$ to the solution and wait for the $\ce{pH}$ to come to an equilibrium $\ce{pH}$.
I feel like the 2 times should be the same, but acid mixes much faster than base in all cases. Why?

The time for $\ce{pH}$ to change from $7$ to $4$ is always less than time for $\ce{pH}$ to change from $4$ to $7$ in the same tank at the same rpm.

Any theories regarding why?

You cannot compare the rate of first mixing acid with water with the rate of mixing the base with the diluted acid. You should compare the rate of mixing the acid with water with the rate of mixing the base with water instead.

The reason is that adding the base to the diluted acid will neutralize portions of the base during the mixing process resulting in lower concentration gradients that make the mixing slower.

A simple gedankenexperiment will show this. Consider a cylindrical tank with water of volume 1 with a mixing rod along the axis. Now put an amount of 1 acid into the tank along its axis. When the acid has made half the way to the wall of the tank the concentration in the inner volume is 4. The gradient is 4.

Now think of adding an amount of 1 base to the completely mixed diluted acid. When the base has made half the way to the wall of the tank, 1/4 of the base has already been neutralized. The concentration in the inner volume is 3. The gradient is only 3 compared to 4 before.


what is the structure of bamboo kun? There are claims that the bamboo fiber has inherent antibacterial properties due to "bamboo kun".
However I could not find anywhere what is the chemical composition of this "bamboo kun" and also no evidence to substantiate the antibacterial claims.

Does anybody know if bamboo kun is a myth or it really has a known chemical structure?

Edit1:

check out wikipedia: https://en.wikipedia.org/wiki/Bamboo_textile

  Bamboo also contains a substance called bamboo-kun – an antimicrobial agent that gives the plant a natural resistance to pest and fungi infestation

google books: https://www.google.com/search?tbm=bks&q=bamboo+kun

Edit2:

Also I could not find this Japan Textile Inspection Association which lots of sites making the antibacterial claim above are citing as the source of antibacterial validation (although there is a Japan Textile Federation that has nothing to do with the above claims):

http://www.bambrotex.com/second/bc_nab.htm
https://www.textilemates.com/bamboo-fibre/
https://highlineorganix.com/pages/why-bamboo
https://www.weber-textilien.ch/bademantel/50-210-bambou-bademantel-mit-schallkragen.html
https://web.archive.org/web/20111005122758/http://www.fashionandearth.com/us/bamboo-clothing-facts.html

It is not implausible that bamboo should have antibacterial properties, after all plants generate lots of protective chemicals. However, the use of this particular word (kun or kunh) by a number of commercial textile websites seems like a promotional gimmick without a strong scientific basis.

Claims of an intrinsic antimicrobial property of processed bamboo fibers are likely exaggerated, as the excerpt below (1) suggests:

  3.3. Antibacterial Properties

  The antibacterial properties of the control sample (OB) and the
  composites (ZB, GB, ZGB) were qualitatively investigated by the
  bacteriostatic circle method against one Gram-negative (E. coli) and
  one Gram-positive (B. subtilis), and the results were shown in Figure
  6. Figure 6A displayed the antibacterial activities of bamboo samples against E. coli. It can be observed that OB had little antibacterial
  activity without an inhibition zone.

This is not to say that bamboo lacks antimicrobial properties. Plants employ a myriad of chemical substances as sources of physical and chemical protection. Extracts of bamboo (2) apparently contain bactericidal compounds, although the concentration in processed fibers appears insufficient to grant these significant antibacterial properties.

It's also worth adding that the above refer to specific species of bamboo and there may be significant variety in antibacterial properties across species.

In conclusion it is difficult if not impossible to answer the OPs question as it is unclear what "kun" or "kunh" refers to, it may just be a catchphrase to describe unknown substances granting antibacterial properties to raw bamboo.

References

Junyi Zhang, Bo Zhang, Xiufang Chen, Bingbing Mi, Penglian Wei, Benhua Fei, and Xindong Mu. Antimicrobial Bamboo Materials Functionalized with ZnO and Graphene Oxide Nanocomposites. Materials (Basel). 2017 Mar; 10(3): 239.
doi: 10.3390/ma10030239
The origin of the antibacterial property of bamboo. T. Afrin , T. Tsuzuki , R.K. Kanwar & X. Wang.  The Journal of The Textile Institute. 2012; 103(8): 844-8491. doi: 10.1080/00405000.2011.614742


Alphabetization rule in case of consecutively placed substituents This is the compound:

I am stuck on assigning the locants to the given halo substituents. We know that all halogens are of equal priority. So, using first point of difference, we get either an $\ce{I \rightarrow Br \rightarrow Cl}$ path or a $\ce{Cl \rightarrow Br \rightarrow I}$ path. I am confused as to which of the two paths I should choose. I could choose alphabetical order, but the bromine in between is ruining the alphabetical order of C-I

Is there any new theory that I am missing?

Rule P-45.2.2 in the 2013 IUPAC Recommendations reads:

  The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than 'hydro/dehydro') to the parent structure.

The term "lowest set of locants" is defined in P-14.3.5; see Loong's answer here. As such the correct set of locants is 1,2,3, not 1,2,5 as was proposed in another answer since "1,2,3" is lower than "1,2,5". (The other answer has since been deleted.) As far as I can tell you have correctly identified this. However, the ambiguity between these two possible numbering schemes is not resolved by P-45.2.2:

In order to decide one first needs to generate the corresponding names for both numbering schemes. Regardless of locant numbering, the three substituents are listed in alphanumerical order (rule P-14.5). Scheme A would give the name "2-bromo-3-chloro-1-iodocyclopentane", whereas Scheme B would give the name "2-bromo-1-chloro-3-iodocyclopentane", since "bromo", "chloro", and "iodo" come in that order alphabetically.

Turning to the next rule P-45.2.3:

  The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than 'hydro/dehydro' prefixes) in their order of citation in the name.

Since Scheme A gives a name with the locants cited in the order 2,3,1, whereas Scheme B has the locants cited in the order 2,1,3, Scheme B is used to generate the preferred IUPAC name 2-bromo-1-chloro-3-iodocyclopentane.

As for what is missing in your train of thought: From what I understand of your question, you seem to be thinking that the prefixes must be ordered in increasing locant number i.e. the name generated is "1-X-2-Y-3-Zcyclopentane", and that the locant set must then be chosen such that X, Y, and Z come in alphabetical order. As you already noticed with this example, that is not quite a foolproof rule. The correct way to do it is the other way round: first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants.


Is styrofoam soluble in gasoline or paint thinner? Why doesn't gasoline or paint thinner melt through Styrofoam when I try it yet there are videos of others melting Styrofoam in gasoline and paint thinner? Is there something I am doing wrong? This is the link to the paint thinner, I used.
http://www.kleanstrip.com/uploads/products/Paint-Thinner-gallon-metal-GKPT94002-900.jpg

For gasoline, I used gasoline from a jerrycan which was sitting for a couple of months.

Please help me out by pointing out where and how I might have made a mistake. No excessive chemistry terminologies please. I am merely a fourth grader ( guardian ) trying to work this out.

Thank you,
Guardian of fourth grader

Unless you're trying to make napalm, don't use gasoline.

Use acetone. It dissolves egg cartons within just a few minutes, and you can then take it out and reshape it (use gloves!). Set it out to dry and once the acetone has evaporated, it should be a (crude) solid.

Just ensure whatever container you're using to hold the acetone isn't polystyrene. Check to make sure recycling code 6 isn't on the container. I had a pretty good laugh at my teacher one day who poured acetone into a red solo cup and it went running all over the table.


Do epoxide rings react with bases? We've learned in class that epoxide ring openings can be catalyized with acids, and I seem to recall either the professor or a classmate mentioning that they also react with bases, but some quick internet searching didn't turn up anything useful.  Do epoxide rings react with bases, and if so, what is the mechanism?

(Dis)claimer: this is just for my own curiosity/learning, and not being asked as a homework/test question (yet).

Do epoxied rings react with bases

Yes, they can be opened, e.g. with alkoxides.

  what is the mechanism?

In principle, this is a $S_N2$ reaction, with the typical approach of the nucleophile from the back side. The difference to a non-epoxide case is that removal of the leaving group (bond breaking, ring opening) is facilitated by the ring strain.

But the general rules for the reactivity of alkyl-substituted centres in $S_N2$ nevertheless apply: the higher the substitution, the lower the rate. This is attributed to the steric hindrance of the respective centre.

Note that the regioselectivity is different under acidic conditions!


Salt Bridge for Microbial Fuel Cell: Full or Partly Full? I'm making a microbial fuel cell out of benthic mud and salt water in containers. My salt bridge will be made of agar and table salt solidified in PVC pipes. Does the salt bridge connecting the anode (benthic mud) and cathode (salt water) have to be full of the agar-salt solution, or can it be partly full?

Does the salt bridge connecting the anode (benthic mud) and cathode
  (salt water) have to be full of the agar-salt solution, or can it be
  partly full

There should be no air gaps or cracks so that the circuit does not break anywhere. Take a narrow PVC pipe if your concern is using too much of agar-salt solution.


Can 1 gram hyaluronic acid really hold 6L of water? According to this study (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3970829/), they make an (uncited) claim about hyaluronic acid: "One of its most important properties is that it can attach and hold large amounts of moisture; approximately 6L of water in just lg ."

Now, the MW of HA is $776.651$ g/mol, so there are about $\pu{0.00128865979}$ moles of HA in a gram. And about $\pu{7.757732e20}$ molecules.

Using google, 6L of water has $\pu{2.007e26}$ molecules.

So this is essentially telling me that roughly 1 million water molecules are able to hydrogen bond with 1 molecule of HA...

Firstly, is this fact true? And if so, how is this possible??

Various figures can be found “on the internet”. E.g. ‘Hyaluronic Hydro Gel Has 1000 Times Water Holding Capacity Perfect for Hair to Heel’ (commercial link intentionally omitted).

Gel liquid phase absorption ability measure seems to be a ‘swelling capacity’ (ratio).

For hyaluronic acid (HA), it can be increased by additional crosslinking. In one paper from 1992, usual swelling ratio was found to be in the order of magnitude 1 to 10, the maximum figure was 120 (for crosslinking ratio much higher than 2.1; 25–30 for the ratio 2.1).[1]

In another paper, another guys are using biphasic nanoparticles (some composite of crosslinked and non-crosslinked HA), they state they have achieved for two different particle types ‘… swelling ratio 1270, 1000 %, respectively’[2] (note the percent sign). Which is, in both cases, approximately, not 1000× capacity, but rather 10× capacity, I'd say.

However, there are polymers with the claimed property – superabsorbents:[3]

  3.27 superabsorbent polymer

  Polymer that can absorb and retain extremely large amounts of a liquid relative to its own mass.

  Note 1: The liquid absorbed can be water or an organic liquid.
  Note 2: The swelling ratio of a superabsorbent polymer can reach the order of 1000:1.
  Note 3: Superabsorbent polymers for water are frequently polyelectrolytes.

References:

Shah, Chirag B., and Stanley M. Barnett. "Swelling behavior of hyaluronic acid gels." Journal of applied polymer science 45.2 (1992): 293-298.
Chun, Cheolbyong, et al. "Effect of molecular weight of hyaluronic acid (HA) on viscoelasticity and particle texturing feel of HA dermal biphasic fillers." Biomaterials research 20.1 (2016): 24.
Compendium of Polymer Terminology and Nomenclature (IUPAC Recommendations 2008)


How to rationalise with MO theory that CO is a two-electron donor through carbon? A question I am looking at is as follows:

  $\ce{CO}$ is isoelectronic with $\ce{N2}$. Sketch MO diagrams for $\ce{CO}$ and $\ce{N2}$. Point out key differences between the diagrams and use the diagram to explain why $\ce{CO}$ acts as a two-electron donor through carbon rather than through oxygen.

Understandably, the key difference between these molecules is that $\ce{CO}$ is heteronuclear, and thus will have differences in energy between the molecular orbital and the atoms.

But I can't explain why $\ce{CO}$ is a two-electron donor using MO theory, even though I can with Lewis?

Does anyone have any thoughts?

In the above MO diagram, the 5σ is the HOMO. But it is closest in energy to oxygen's 2p orbitals, so why is it centered on carbon?

The electrons in the frontier orbital(s) play a special role for the chemical reactivity. In $\ce{CO}$, the HOMO is the $ 5 \sigma $ orbital (ref: your diagram), and has mainly $\ce{C}$ character. The two electrons in it act like a lone pair on the carbon.

The predominant $\ce{C}$ character of the HOMO accounts for the reactivity via carbon ($\sigma$ donor

Side note: $\ce{CO}$ is also a $\pi ^*$ acceptor due to the LUMO (which is also predominantly $\ce{C}$ in character.

To answer your comment:

  So overall, how could I tell exclusively from the MO diagram that HOMO electrons are centered on C and not on O?

Don't think you can. The contribution to MOs is determind by the coefficients in the linear combination. You should see Martin's calculated values in the answer link he provided.