Where did amino acids get their one-letter codes? Some amino acids have a one-letter code that's just the first letter of the name of the amino acid. This makes sense and obviously, since there is more than one amino acid that begins with the same letter, other letters had to be used.

But why were these other letters chosen? For instance, is there a reason 'W' specifically was chosen for tryptophan (other than the fact that 'T' was taken)?

Tryptophan

For instance, is there a reason 'W' specifically was chosen for tryptophan (other than the fact that 'T' was taken)?

Once you have assigned the other 19 amino acids, there are only 7 letters of the alphabet left: B, J, O, U, W, X, and Z. (Certainly not a nice Scrabble hand to have!)
If one wants to use a letter found within the name of the amino acid, the only available letter would be O. However, the usage of U and O was historically discouraged because these letters could be easily confused with other letters (U with V; O with G, Q, C, D, and the number 0).
It turns out that the choice was made because W is a very fat letter and was reminiscent of the indole ring system present in tryptophan (the only amino acid to contain a bicyclic system). (See below for source.)

What happened to the other six letters, then?
All 26 letters of the alphabet now find use as a one-letter code for amino acids or various combinations thereof.
After their discoveries, selenocysteine and pyrrolysine (the latter only found in bacteria) were assigned U and O respectively. Furthermore, B is used to represent aspartic acid OR asparagine; J is used to represent leucine OR isoleucine; Z is used to represent glutamic acid OR glutamine; and X is used to represent an unknown amino acid.
I believe B and Z find use because, in protein sequencing, acid hydrolysis is often used to break peptide bonds. This has the undesirable side-effect of hydrolysing the amide groups in asparagine/glutamine, leading to the formation of aspartic/glutamic acids, which means that one cannot tell exactly which amino acid it was at the start. J is used in NMR spectroscopy where isoleucine and leucine are difficult to distinguish.

Why did they choose the letters they did?
As far as I am aware the usage of one-letter symbols is adopted by both IUPAC and IUB (since 1991, IUBMB) in their joint 1983 recommendations on "Nomenclature and Symbolism for Amino Acids and Peptides".1 ankit7540's concise summary of the historical development already mentioned these recommendations.
In particular, Section 3AA-21.2 "The Code Symbols" has a description of why the letters were chosen. This document is probably the most authoritative stance on the matter. The rationale is mostly in line with Jan's answer:

Initial letters of the names of the amino acids were chosen where there was no ambiguity. There are six such cases: cysteine, histidine. isoleucine, methionine, serine, and valine. All the other amino acids share the initial letters A, G, L, P or T, so arbitrary assignments were made. These letters were assigned to the most frequently occurring and structurally most simple of the amino acids with these initials, alanine (A), glycine (G), leucine (L), proline (P) and threonine (T).
Other assignments were made on the basis of associations that might be helpful in remembering the code, e.g. the phonetic associations of F for phenylalanine and R for arginine. For tryptophan the double ring of the molecule is associated with the bulky letter W. The letters N and Q were assigned to asparagine and glutamine respectively; D and E to aspartic and glutamic acids respectively. K and Y were chosen for the two remaining amino acids, lysine and tyrosine, because, of the few remaining letters, they were close alphabetically to the initial letters of the names. U and O were avoided because U is easily confused with V in handwritten material, and O with G, Q, C and D in imperfect computer print-outs, and also with zero. J was avoided because it is absent from several languages.
Two other symbols are often necessary for partly determined sequences, so B was assigned to aspartic acid or asparagine when these have not been distinguished; Z was similarly assigned to glutamic acid or glutamine. X means that the identity of an amino acid is undetermined, or that the amino acid is atypical.

One can only hypothesise what they meant by "associations that might be helpful in remembering the code" in the case of N/Q/D/E. My best guess is:

D and E were possibly chosen for aspartic and glutamic acids because they were the only consecutive pair of letters left, emphasising their chemical similarity. Aspartic acid is shorter than glutamic acid by one methylene group (CH2), so it gets the earlier letter D.
Glutamine sounds like Q-tamine. If you don't think it sounds similar, repeat it 50 times until you do.
AsparagiNe was assigned N.

Reference

IUPAC-IUB Joint Commission on Biochemical Nomenclature. Nomenclature and Symbolism for Amino Acids and Peptides: Recommendations 1983. FEBS J. 1984, 138 (1), 9–37. DOI: 10.1111/j.1432-1033.1984.tb07877.x. A HTML version (perhaps more user-friendly) can be found at this address.


Grignard alkoxide intermediate as starting material for ether or ester synthesis I'm having a Grignard reaction between a ketone or aldehyde with RMgX in mind.
The average workup uses water to hydrolize the intermediate alkoxide. But could this intermediate be used without isolation to prepare an ether or ester? For example by simple addition of RX, tosyl chloride, acetyl chloride ...

I don't know if the compoud is reactive enough, however. But it is an alkoxide and you probably wouldn't even need additional base for e.g. HCl scavenging. Maybe someone here knows more about the reactivity or maybe there are some literature examples?

Alkoxide, unless hydrolysed, acts as a very good base and is even used for the synthsis of ethers. However the alkoxide obtained in this case will have a yield that is quite low since the O- anion forms a salt as OMgX, hindering its ability to attack the alkyl group R-X. Thus, it is generally avoided.


Why is a sodium chloride molecule stable? The ionization energy of sodium is $\pu{5.139eV}$. This is the energy absorbed when a neutral sodium atom is stripped of its outermost electron. The electron affinity of chlorine is $\pu{3.62eV}$. This is the energy released when an electron is added to a neutral chlorine atom to form the chlorine anion.
Now suppose we have a system containing exactly one neutral sodium atom and exactly one neutral chlorine atom infinitely far apart so that the systems potential energy is zero. If we wish to make a $\ce{NaCl}$ molecule from our system, we need to put $\pu{45.139eV}$ of energy into the system to ionize the sodium atom. Afterwards, we place the electron we have just pulled off the sodium atom onto the chlorine atom upon which we get back only $\pu{3.62eV}$ of energy. So the total energy of the system has increased by $\Delta E =\pu{5.139eV -3.62eV= 1.519eV}$.
If the total energy of the system has increased after performing the previous supposition, then why does the $\ce{NaCl}$ ionic bond occur? Surely it is less stable than having the two neutral atoms separate from each other? My understanding up till now has been that molecules form when the total energy of the system is lower than the total energy of the sum of its constituents?
Is this all because an individual $\ce{NaCl}$ bond is not stable on its own but it is stable when its within a crystal lattice?
Any help on this issue would be most appreciated!

What you have calculated is the energy required to make two separated ions out of two separated $\ce{Na}$ ad $\ce{Cl}$ atoms. When they are separated, these ions do not make a molecule or a crystal of salt. Separated ions attract one another and are able to release a huge amount of Coulomb energy when approaching each other.
Anyhow your calculation should still take into account the fact that chlorine atoms do not actually exist. They must be produced by breaking a $\ce{Cl2}$ molecule. And this requires some energy. You may also introduce the sublimation energy of sodium atoms


Samsung's S-Plasma Ion - scientific basis Has anyone come across a scientific study of the S-Plasma Ions that Samsung claims will purify air from microbes, allergens and free radicals? I would prefer publicly available research findings.

The following is a quote from http://www.samsungspi.com:

  The Samsung SPi generates Hydrogen atom ($\ce{H+}$) and Oxygen ion ($\ce{O_2-}$) which reacts together to form Hydroperoxyl radicals which the positively charged biological contaminants then reacts with three hydrogen elements of its protein structure before forming water vapor. Also, the Hydrogen atom ($\ce{H}$) will combine with $\ce{OH}$-radical to form a water vapor and neutralize harmful $\ce{OH}$-radical.

It also matches the description found on the product itself. See http://winniekepala.com/2011/03/ionizer-review-samsung-virus-doctor/ .

Here's a quote from the video on the same site:

  $\ce{H+}$ is born with the help of that plasma in this plus electrode. $\ce{H+}$ is hydrogen with a positive electric charge. Through the air, the $\ce{H+}$ moves to the minus electrode which generates electrons. In here, electron meets $\ce{H+}$ and active hydrogen like me are born.... The electron that is generated in minus electrode in the SPi ion device where [I] am born is naturally combined with oxygen. That is oxygen ion ($\ce{O2-}$). (Active hydrogen and oxygen ion together transform into $\ce{(H-O-O)-}$). $\ce{HOO}$($\ce{HOO-}$) ion envelops the positively charged airborne bio-contaminants and react with three hydrogen constituted of protein components, forming water vapor. This modifies the surface protein or breaks the cell membrane to remove bio-contaminants. $\ce{HOO}$($\ce{HOO-}$) ion reacts only with single cells such as virus or germ but does not react with the human body... $\ce{OH}$ radical is an active oxygen species and one of the highly reactive and strong oxidants. It can be harmful to human's body if exposed for a long time. (Active hydrogen combines with OH radical to form water vapor).

(It has numerous grammatical mistakes which I just copied verbatim).

I would like to know what are the various compounds and radicals mentioned in the explanation, and how the reactions above take place. What is the plasma in the positive electrode, and what is the potential difference between the electrodes that allows the generation of the active hydrogen and oxygen ions? What are the real chemical states of the active hydrogen and the oxygen ions? How do they combine to form the $\ce{HOO-}$ ion? What is the $\ce{HOO-}$ ion and how is it stable enough to go around the room? How does the active hydrogen combine with the $\ce{OH}$ radical?

if you can take a look at the main web site, then it seems they make $\ce{HO2}$ and not $\ce{HO2-}$. This a peroxy radical which exists at trace levels in the ambient atmosphere, usually from reactions of light with oxides of nitrogen and volatile organic compounds. Its really reactive and I don't know for sure how long it will last in the air of the room. It will however react with $\ce{OH}$ radical and make water and $\ce{O2}$. $\ce{HO2 + OH -> O2 + H2O}$. $\ce{HO2}$ will also react with $\ce{NO}$ to form $\ce{NO2}$ and then depending if there is sunlight could make ozone. One of the issues for the machine will be that $\ce{HO2}$ will need to made at some point and therefore one of the first molecules it will encounter will be itself and therefore make $\ce{H2O2}$ (hydrogen peroxide). I think getting one into a lab and testing it would be a start.


Optical activity of 3‐chlorobutan‐2‐ol In the picture that I have uploaded, I have read that the given compound is chiral.

But on drawing the sawhorse structures (for better visualisation), I feel like the compound is symmetric about a plane that could bisect both OH and Cl.

What is it that I'm thinking incorrectly?

Originally, both of us, andselisk and I, thought you are taking about a plain which goes through all four groups ($\ce{H-H-Cl-OH}$ squire). It is a forbidden plane when it come to symmetry of a compound, if it had been the case. Fortunately it is not the case. We later realized after Mithoron's comment that the plane you are talking about actually goes through $\ce{C^*-C^*-Cl-OH}$ squire instead (asterisks indicate chiral $\ce{C}$s). That is an allowed plane for symmetry concern. Yet again, you are wrong about it's been a plane of symmetry. For your convenience, I draw all four possible isomers of 3‐chlorobutan‐2‐ol and relevant sawhorse structures showing the plane of your concern:

I tried to put the plane as it is a mirror on a plane where $\ce{C^*-C^*-Cl-OH}$ squire overlap. The particular stereoisomer representing both Fischer projection and sawhorse structure (with your 'plane of symmetry') is $(2R,3S)$-3‐chlorobutan‐2‐ol, so I drew that particular isomer first. Now look at my sawhorse structure below that particular structure. The $\ce{C^*-C^*-Cl-OH}$ squire is on the plane so that those groups and atoms do not count on symmetry. The only countable groups are two methyl groups and two hydrogen residues. If you consider the plane as a mirror (which dissects each $\ce{H-C^*-CH3}$ angle), you would see both $\ce{CH3}$-groups are in the front of the mirror and hydrogen residues are inside of the mirror (resembling mirror images). As a result, we can conclude that the mirror is not a plane of symmetry because each $\ce{H}$ atom can't be the mirror image of adjacent $\ce{CH3}$ group. Therefore, both carbon atoms marked with $\color{red}{red}$ $\color{green}{green}$ circles are chiral centers. As a result, there are 4 stereoisomers for 3‐chlorobutan‐2‐ol: $(2R,3S)$-, $(2S,3R)$-, $(2R,3R)$-, and $(2S,3S)$- where first two and last two isomers are enanthiomers of each other (as indicated in the diagram).


Liquid floating on top of water due to density or insolubility? When a liquid is floating on top of water, how can one tell whether it is because of its insolubility in water or because it has lower density than water?

Both are contributing factors.

The insolubility ensures that it does not mix with water.

The lower density of the liquid on top ensures that it stays on top of the water instead of below the water.


Given solubility of sodium carbonate, find mass of hydrate sodium carbonate dissolved? I am working on a problem from my scholarship exam practice (MEXT). I am not quite sure if I got it right. The question was provided with multiple choices (a) to (f) below:

  The solubility of sodium carbonate in $\pu{100 g}$ water is $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$. How many grams of the hydrate $\ce{Na2CO3∙10H2O}$ can be dissolved in $\pu{100 g}$ of water at $\pu{22 ^{\circ}C}$? (a) $\pu{0.556g}$; (b) $\pu{0.762g}$; (c) $\pu{9.27g}$; (d) $\pu{67.5g}$; (e) $\pu{81.7g}$; (f) $\pu{117g}$

I honestly do not know how to solve this problem. My best guess is to find the moles of sodium carbonate that can be dissolved in water first (from $\pu{25.0 g}$). And use that moles equate to the moles of hydrate form (1:1). Then find the mass of hydrate form by multiplying its molecular weight and my answer would be $\pu{67.5g}$. There is no answer key provided so I would like to hear other opinions and get some advice.

To answer this kind of problem, it'd be better starting with elimination approach. It is given the solubility of sodium carbonate in $\pu{100 g}$ of water as $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$.  By that, you know, solubility of its hydrate should be more than $\pu{25.0 g}$ at $\pu{22 ^{\circ}C}$ because the mass of hydrate is including additional mass of water as well (as your calculation shows). Therefore, you can eliminate answers (a), (b), and (c) directly, because they are less than 25 and are not correct. Your calculations shows it must be greater than $\pu{67.5 g}$ (therefore, answer (d) is incorrect as well). It should be greater than $\pu{67.5 g}$, because you forgot to include additional amount of salt, which would dissolve in 10 equivalents of water in that added salt: Molar mass of $\ce{Na2CO3}$ is $\pu{106 gmol^{-1}}$ while Molar mass of $\ce{Na2CO3.10H2O}$ is $\pu{286 gmol^{-1}}$. This means, each $\pu{mol}$ of hydrate has $\pu{186 g}$ or $\pu{186 mL}$ of water. That is a lot and we need to consider that as well in this question. Now we have only two answers to be considered: (e) and (f).

To do the calculations, suppose $x\; \pu{g}$ of $\ce{Na2CO3.10H2O}$ would dissolve in $\pu{100 g}$ of water at $\pu{22 ^{\circ}C}$. Thus,
$$\text{amount of }\ce{Na2CO3}\text{ in the solution} = \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
$$\text{and, amount of }\ce{H2O}\text{ in the solution} = \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
From given data for anhydrous salt, we can conclude that upon saturation:
$$\text{The amount of }\ce{H2O}\text{ in the solution} = 4\times \text{amount of }\ce{Na2CO3}\text{ in the solution}$$

$$\therefore \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g} = 4 \times \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$
Solve for $x$:
$$\left(4 \times 106 - 180\right)x = 28600 \; \text{and therfore, } x=\frac{28600}{244}=117.2$$
Therefore your answer is (f) ($\pu{117 g}$).


Other metabolic reactions aside, what causes beta oxidation to stop? To my understanding beta oxidation is when a fatty acid has its carbons plucked off in groups of 2 (acetyl groups) into acetly-coa which we can dump into the TCA.

At what point would this process stop ? lets say we have a short chain fatty acid like butanoate (4c) would it still transfer an acetyl group leaving an acetate molecule ?

Different enzymes specialize to work on different size ranges. If the fatty acid is too short, it might not bind to enzymes that work on large fatty acids. An enzyme capable of binding short fatty acids might have most binding determinants targeting the CoA moiety instead of the fatty acid itself.

  At what point would this process stop ? lets say we have a short chain fatty acid like butanoate (4c) would it still transfer an acetyl group leaving an acetate molecule ?

It goes all the way as long as the CoA is attached.


Conservation rule in closed systems In a closed system the reacting species interconvert but do not leave the system. Therefore the amounts of different species in the system would be bound by a conservation rule.

  A simple example:
  $$\ce{A -> B}$$

Here the conservation rule would be: $\ce{A + B = constant}$, which says that at any point of time during the reaction the sum of the amount of both the species would be constant.

  A more complex example
  $$\ce{2A -> B}$$

Here, the conservation rule would be: $\ce{A + 2B = constant}$.

Let's take a more complex system:

  $$\begin{align}
\ce{A &-> B} \\
\ce{2B &-> C} \\
\ce{C &-> D + E} \\
\ce{2A + C &-> F}
\end{align}$$

From empirical analysis I can deduce that the conservation rule would be: $$\ce{A + B + 2C + D + E + 4F = constant}$$

However, I am not able to deduce a mathematical equation that gives these co-efficients. I assume that the stoichiometry matrix may give a clue but I am not sure. The stoichiometry matrix for the above system is:

$$\mathbf{S}=\begin{bmatrix}
-1 & 0 & 0 & -2 \\
1 & -2 & 0 & 0 \\
0 & 1 & -1 & -1 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$$

This is relatively easier to figure out empirically as there is only one starting reactant (A). For reaction systems with multiple starting reactants (as shown below), the deduction of conservation rule(s) becomes even more difficult.

$$\ce{aA + bB -> cC + dD}$$
The problem seems quite elementary but I am not able to solve it. Am I missing something obvious?

With 6 components and 4 reactions, you must have two conservation rules (that is, if the reactions are independent, which they are).

Mathematically, we find ourselves in a 6D space (as there are 6 components), we move in a 4D subspace defined by the 4 reactions, and we need a basis for the 2D subspace normal to it, which is precisely where we can't go. The basis vectors will be our conservation laws.

You may use any CAS or your own hands to solve the problem. In Wolfram Mathematica, this is done as follows:

NullSpace[{
{1, -1, 0, 0, 0, 0},
{0, 2, -1, 0, 0, 0},
{0, 0, 1, -1, -1, 0},
{2, 0, 1, 0, 0, -1}
}]

As expected, the answer includes two vectors. One is $\{1, 1, 2, 2, 0, 4\}$. Another is $\{0, 0, 0, -1, 1, 0\}$, which translates to $\rm [D] - [E] = const$ ($0$ if there were no initial amounts), which is obviously true, since both are produced in the same reaction and never spent. That's your second conservation law.

BTW, the sum of the two vectors corresponds to the conservation law you already know. A basis can be chosen in a multitude of ways, after all.

That's what you do in any problem of this sort. That's linear algebra; it is about as simple and robust as a cleaving axe. It will never run out of gas, or freeze because its usage license has expired, or refuse to chop the incompatible firewood. It just works.

So you prepare the stoichiometry matrix (which you already did; I just transposed it), find the normal subspace, translate the basis back to your compounds. That would be it. The notion of time is irrelevant. The notion of differential equations is irrelevant. The notion of limiting reactants is irrelevant. It is just vectors and spaces. They do not even know you apply them to chemistry, nor do they need to.

So it goes.


How can aqueous KOH give alcohol by SN2 of alkyl halide? I ask this because SN2 requires a strong nucleophile, and I have read $\ce{OH-}$ isn't a strong nucleophile in a polar protic solvent like water. So, how can $\ce{OH-}$ replace, say $\ce{I-}$ from $\ce{CH3I}$ (methyl iodide)?

When you are considering a particular SN2 reaction, you have to consider the nucleophilicity as well as the nature of the leaving group of both the groups.

Here as you mentioned, $\ce{OH-}$ and $\ce{I-}$  (and other halide ions, except $\ce{F-}$, which is a very bad nucleophile) have comparable nucleophilicity in protic solvents. But, as for their leaving nature, $\ce{I-}$ (as well as other halides) is far more better leaving group than $\ce{OH-}$, because we know weak bases are better leaving groups. That is why the reaction is feasible.


How to find a chemical by its formula? There are (obviously) a lot of chemical compounds. Wikipedia is able to translate names to molecular formulas for a lot of them.

But is there a way to enter a molecular formula and find the chemical's name?

Coming from the computer science background, I realize that there should be some canonical textual representation of the molecular formulas for such search to be possible. So is there a language that can represent a full formula, including the structure? I could only find some latex-based languages.

This answer is meant to address your second question about computerized representations of molecule structure since comments have already addressed your first question.

To truly represent the structure, mol and sdf formats are common: https://en.wikipedia.org/wiki/Chemical_table_file

Common string representations include SMILES and InChI. An issue with smiles is that there are multiple valid ways to represent a given molecule, which means a simple string match is not a good test for sameness. Canonical SMILES and InChI attempt to address this issue via additional rules. InChI additionally provides a standard way to convert the sometimes very long InChI strings into 27 character hashed version called InChIKey, which are better for searching.

Example Acetoin
SMILES string: CC(C(=O)C)O
InChI string: InChI=1S/C4H8O2/c1-3(5)4(2)6/h3,5H,1-2H3
InChIKey string: ROWKJAVDOGWPAT-UHFFFAOYSA-N


Determining the orbitals that overlap to form a sigma bond In the chapter on molecular orbitals, my chemistry book mentions three types of sigma bonds: those formed by s-s ($\ce{H2}$), s-p ($\ce{HCl}$), p-p ($\ce{Cl2}$) hybrid-s ($\ce{CH4}$) and hybrid-hybrid ($\ce{C2H6}$) overlap. How is it possible to predict the type of atomic orbital overlap by which a specific σ bond is formed?

The basic answer is go by what is available.

Hydrogen does not have any other occupied orbitals than s-orbitals. Hence it must use an s-orbital for bonding. This goes both directions.

Carbon requires four bonds. In order to achieve this, it must hybridise in some way, otherwise, an electron pair would be in an s-orbital and an empty p-orbital would be unavailable for bonding. Hence carbon uses hybrid orbitals in both directions.

Chlorine could hybridise, but it doesn’t have to fulfill its bonding needs. In general, an electron pair in an s-orbital is more stable than one in a p-orbital because it is ‘closer to the nucleus’. (That layman’s argument is bad but gets us a long way.) So chlorine will happily use its p-orbitals for bonding.
This goes both ways, by the way, so a carbon-chlorine bond ($\ce{C-Cl}$) is a hybrid-p sigma bond.

Finally, let’s consider oxygen and nitrogen. They would love to just bond with their p-orbitals, but even if hydrogen is the bonding partner that leads to too much steric strain. Therefore, they hybridise just enough to alleviate the steric stress while still maintaining a maximum s-character for a/the lone pair. So those would love to form bonds with their p-orbitals but typically can’t (so they must use hybrid orbitals). Phosphorus and sulfur are large enough to generally allow bonding only with p-orbitals.


What is the reason that ice melts with increasing pressure at constant temperature (revelation)? I've seen many explanations regarding this topic where they explain it in the sense of the anomalous expansion of water, but I actually want an answer where I can understand the different forces at play in the molecular level.
I've seen a paper where it is said that when the pressure is applied the h-bonds contract thus making them stronger but this also makes the covalent bond weaker which results in higher probability of the hydrogen to get separated from one of the two oxygen atoms which in turn decreases the energy required to break the bonds therefore lowering the melting point of ice.
Is this explanation correct?

The potential energy between the O atoms in the H bond has the form of an anharmonic potential an example of the shape is the Morse potential or a Lennard-Jones 6-12. The energy is zero at large O-O separation and moves to a minimum as this distance is reduced (the point at which a the H bond exists with the H atom between the two O atoms, typically 0.15-0.25 nm). As the O-O distance is further reduced the potential rises again to zero then positive potential energy. The H bond thus forms in a lop-sided potential well, with the H atoms between the two O atoms. (I imagine the effect on the covalent bonds would be less as they are stronger and shorter than the H bond.)
Suppose that the distance between the O-O atoms is reduced then the potential rises and becomes closer to zero or repulsive (positive) potential. Suppose that the pressure causes this to happen just a little and is such that the H bond energy is now only slightly below zero. At this point the thermal energy (as molecular vibrations) could in places be enough to increase the energy above zero and so break the H bond, at which point the solid would start to melt. The structure beginning to break down could further collapse rather like a building collapses by bending and twisting if it structural girders are removed.


Order of second ionization energy for cations Why is the correct order of second ionisation potential of $\ce{Li , Be , B , C }$ such as: $\ce{Be} <\ce C <\ce B <\ce{Li} $?
I understand that $\ce{Li+}$ has a stable inert gas electronic configuration and $\ce{B+}$ has one fully filled s orbital. Hence these will have a high value of ionisation potential. But on what basis do we then arrange the ionisation potentials of carbon and beryllium? What is the general steps of rules to solve such kinds of questions?

You are correct about $\ce{Li^+}$: the second electron to be extracted would be one of the two extremely tightly bound $\text{1s}$ electrons, which explains the very high second ionisation energy of that ion.

As for the comparison $\ce{Be^+}$ to $\ce{C^+}$, the main effect here is effective nuclear charge, $Z+1$, which in the case of $\ce{Be^+}$ is $5$ and in the case of $\ce{C^+}$ is $7$.

Equivalent electrons (same or similar orbitals) are more tightly bound to the nucleus, hence harder to extract, for higher $Z+1$. This results in a trend of higher second ionisation energies for higher $Z$, across a period.

This is a general trend across each period.

  What is the general steps of rules to solve such kinds of questions?

Chemguide is a good place to start.


What do GGN, AAP, TCN, CAP, CCN, TGQ, etc. mean in DNA analysis? What do the letters Y, Q, P and N mean in three-letter combinations involving DNA and/or RNA?

I know, of course, about the bases A, C, G, T and U but.....

Especially in articles (and forensic tv shows) I see, connected to three-letter amino acid codes, letter combos like GTN, GGN, etc.....

It is not always possible to obtain an absolutely precise result from automated DNA sequencers, mostly due to too much background noise or two very close peaks which the computer has difficulty reading.

An example of a very noisy four-color chromatogram from a sequencer (sanger):

As you would notice, there are several ambiguous spots and the computer algorithm can be confused or mis-call the wrong nucleotide. Often humans are called in to resolve such ambiguities. It's a fairly common practice.

Sometimes the ambiguity goes beyond a solid conclusion, like when two or more peaks have almost the same intensity.

In that case it is better to explicity label those ambiguous nucleotides as either, say A or B, rather than report a possibly incorrect sequence. Thus was born the incompletely specified nucleic acids.

Here's where IUPAC comes in: the single letter code recommendations.

The "N" in GGN, TCN and CCN would mean the third codon was completely ambiguous and could be any one of tbe four bases. Y would mean pyrimidine (C, U or T).

This would satisfy your query with the exception of the letters P and Q, which I assume were wrongly picked up, or the code was for proteins (amino acid) sequence single letter code.


Valence Bond theory, VSEPR theory and predicting the shapes of the molecules I read about the above mentioned theories then used them to predict the molecular geometry. However, I found that the shape/geometry of a molecule obtained from both the theories was same. For example, water has bent/angular shape according to VSEPR theory as well as Valence Bond Theory. So I wonder if there's any difference between the two theories in predicting molecular geometry.

Do these two theories ever yield different results?

VSEPR theory is a purely observational model, based on empirical knowledge. Many of the "rules" were added to fit the theory to what has been observed. It has quite a lot limitations and may only provide an initial guess for the structure of a molecule.

According to VSEPR, the structure of water is bent. While we would expect $\angle^\mathrm{theo}_\mathrm{bond}(\ce{HOH})=109.5^\circ$, it is indeed smaller $\angle^\mathrm{expt}_\mathrm{bond}(\ce{HOH})=104.5^\circ$. This discrepancy is explained with the reasoning that lone pairs require more space. This is wrong, and a caveat added to make the theory fit.
It would also predict the lone pairs of water to be equivalent. A statement which has experimentally been disproved. Quoting myself:

  One popular example is water. In many texts the central oxygen is described as having (approximately) four sp3 orbitals, which essentially makes the lone pairs equivalent. That view, however, is not in agreement with the photoelectron spectrum, which clearly shows that the lone pairs are not equivalent. At this point I'd like to refer you to Michael Laing's article: "No rabbit ears on water. The structure of the water molecule: What should we tell the students?" (J. Chem. Educ. 1987, 64 (2), 124.).

The theory sometimes predicts correct structures, but that has no theoretical foundation. It is basically educated guessing.
When it comes to more complicated molecules, for example hyper-coordinated compounds like $\ce{IF7}$, most explanations (the involvement of d orbitals) break completely down and have been disproved many times. However, just by minimising the inter-ligand distances, it yield approximately the correct shape. This is a nice coincidence.

Valence Bond theory is on the other hand a highly sophisticated and complex approach, deeply rooted in theory. If applied at a modern, computational level, it will always yield the correct molecular geometry for the correct reasons (within its other approximations).

While we can approximately use a piece of paper and a pencil to do these calculations for simple molecules, it really is an approach at formulating a wave function and approximately solving the electronic Schrödinger equation. That requires a lot of maths and integral solving. It is not so much a theory that can be done on-the-go.
As it is such an extensive and accurate theory, it is complementary to molecular orbital theory. That means that they will yield the same results at their respective accurate (infinite) expansions.

For the example of water, VB theory would predict a bent structure. It would also predict non-equivalent lone-pairs, which agrees nicely with the photoelectron spectrum.

In conclusion; A sad fact is that both theories, VSEPR and VB, are usually (more than just often) taught completely wrong at below university levels (sometimes even there) and will introduce many artefacts into young scientists brains. Your statement, that you "used" valence bond theory is a reminder of that.
These theories could not be more different from each other. In most cases they predict structure on completely different grounds. It is basically a coincidence if they predict the same thing.

Take home message: VSEPR theory is a rule of thumb, VB theory is an accurate description of physics (within its approximations).


Gas of independent molecules $$\begin{align}
p &= kT\left(\frac{\partial \ln Q}{\partial V}\right)_T \\
&= \frac{kT}{Q}\left(\frac{\partial Q}{\partial V}\right)_T \\
&= \frac{NkT}{q}\left(\frac{\partial q}{\partial V}\right)_T \\
&= \frac{NkT\Lambda^3}{V}\cdot\frac{1}{\Lambda^3} \\
&= \frac{NkT}{V} \\
&= \frac{nRT}{V}
\end{align}$$

Looking at the first equation in this series I'm really confused how the $\partial \ln Q/\partial V$ has been simplified to $\partial Q / \partial V$.

My hunch is that it's something to do with the derivative of $\ln Q$ being $1/Q$ however I'm not sure how this ties in.

This is more of a maths question, but I can give a quick answer.

As you said,

$$\frac{\mathrm{d}}{\mathrm{d}Q}\ln Q = \frac{1}{Q}$$

You could write this as

$$\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} = \frac{1}{Q}$$

So, by the chain rule,

$$\begin{align}
\frac{\partial (\ln Q)}{\partial V} &= \frac{\partial Q}{\partial V}\cdot\frac{\mathrm{d}(\ln Q)}{\mathrm{d}Q} \\
&= \frac{1}{Q}\frac{\partial Q}{\partial V}
\end{align}$$

If you are studying physical chemistry, it is a good idea to remember this trick. It appears everywhere.


Do changing opt=modredundant to opt in Gaussian makes geometry optimization not to take into account frozen angles? I am a newbie to Gaussian and just generated an input for the geometry optimization for some molecules with multi ring system. However, in the article that was a reference for those calculations, some of the dihedral angles were freezed, so I did the same with adding the needed dihedrals to constrain with lines D X X X X F.
However, someone superior told me that I shouldn't use the opt=modredundant, but opt instead. Does it mean that I shouldn't constrain angles then? Or should I just manually change it to opt and the program will still understand that the angles are constrained?


Meaning of "mono-N-demethylation" I know what demethylation is.

I have no idea what mono- means, the N is nitrogen of course, I know that from the line structure image in front of me if I could not guess it - and after quite a bit of google searching I gave up and came here.

Chemistry is not my field but out of curiosity and for no practical purpose whatsoever I started learning a few things outside my field, and I came across this tiny little naming issue. I thought this is so simple I could easily find something, but apparently it is even MORE simple than I thought - meaning everybody uses the term but nobody bothers to explain it. :-(

Mono-N-demethylation means the removal of a single (mono) methyl group from a nitrogen(N)-containing functional group. For example, a reaction which transforms the alkylated amine group $\ce{-NHCH3}$ into $\ce{-NH2}$.


Difference between oxygens in a carboxylic acid? I've been reading about MO theory and am wondering if the traditional model of carboxylic acids is correct. Usually, how it's presented is that the carbonyl carbon is doubly bonded to one oxygen and singly bonded to a hydroxyl group. This would mean that one oxygen is $sp^3$ hybridization and the other is $sp^2$ hybridized, and the hydrogen is attracted to both (one through a covalent bond and the other through a hydrogen bond).

My book (Ansyln and Doughtery) says that

  Hybridization isn't just limited to $sp^2$ and $sp^3$ and that bonding isn't localized between only two atoms but between the molecule as a whole. (paraphrased)

How does this affect carboxylic acids? Is there actually clearly defined '$sp^2$' and '$sp^3$' oxygens, or are they chemically equivalent? It seems like carboxylic acids react as alcohols (see: thionyl chloride). But does some sort of spontaneous symmetry breaking take place when a reactant approaches?

I can't find anything detailed/reputable about this on the internet.

The question What is the hybridization of the carbonyl oxygen in a carboxylic acid? does not address my concern.  I get what the classical hybridization of the carbonyl oxygen in a carboxylic acid is. I'm just wondering if there's a better non-classical description where the oxygen atoms are equivalent until approached by another molecule or something of the sort.

First of all I would like to put my usual disclaimer about hybridisation: It is just a mathematical concept, nothing too real and it should always taken with a grain of salt. In the linked post I have already discussed certain aspects of that. Furthermore it has been discussed on this site in various occasions.
Going a bit further, especially for delocalised systems, the hybridisation picture starts to break down. In acetic acid (or any other carbon acids) this is not yet a big problem and we can still argue a lot with the involved concept. But I hope the following explanations will demonstrate some of this break down.

Now let's get back to the problem at hand. Here is the structure of acetic acid in the gas phase at 0 K, calculated with Gaussian09 at the DF-BP86/def2-SVP level of theory.

As you can see the carbon oxygen bonds are not equal at all. In the Lewis picture, we would say there is a single bond between carbon 2 and oxygen 1, which is a somewhat valid approximation. On the other hand we would say that between carbon 2 and oxygen 2 there is a double bond, which is also a somewhat valid approximation. In the molecular orbital picture we have three π-type orbitals (10, 13, 15). Therefore we have partial double bond character in the carboxyl unit due to delocalistion. Note that there is also an hyperconjugation effect involved. The main difference, however lies in the the molecular orbital 15, which is bonding with respect to C2-O2 and antibonding with respect to C2-O1. This orbital is basically the key, why the Lewis structure is a valid approximation.

Now we can unitary transform these orbitals to localised orbitals, therefore representing the bonding structure with hybrid orbitals. I am using the Natural Bond Orbital theory for this. The following summary gives the bonds of the carboxyl subunit, the numbers in brackets refer to the orbital number in the picture. Also note, that since these orbitals are no eigenfunctions of the Hamiltonian and they do not have an energy eigenvalue, hence there is no canonical order.

-----------------------------------------------------------
#  (Occupancy)  (Type)    (orbital)  || Contributions below
-----------------------------------------------------------
1. (1.99578)    BD ( 1)   C2 - O1   [6]
            ( 30.48%) C(2) s( 26.60%)  p 2.75( 73.10%)
            ( 69.52%) O(1) s( 35.03%)  p 1.85( 64.91%)

2. (1.99728)    BD ( 1)   C2 - O2   [5]
            ( 33.43%) C2   s( 33.28%)  p 2.00( 66.51%)
            ( 66.57%) O2   s( 42.68%)  p 1.34( 57.24%)

3. (1.99005)    BD ( 2)   C2 - O2   [14]
            ( 31.97%) C2   s(  0.00%)  p 1.00( 99.64%)
            ( 68.03%) O2   s(  0.00%)  p 1.00( 99.87%)

4. (1.98518)    BD ( 1)   O1 - H4   [7]
            ( 74.58%) O1   s( 19.52%)  p 4.12( 80.41%)
            ( 25.42%) H4   s( 99.60%)

5. (1.80107)    BD ( 2)   O1 - H4   [15]
            ( 99.93%) O1   s(  0.00%)  p 1.00( 99.93%)
            (  0.07%) H4   s(  0.00%)  p 1.00(100.00%)

6. (1.97613)    LP ( 1)   O1        [10]
                           s( 45.45%)  p 1.20( 54.52%)

7. (1.97395)    LP ( 1)   O2        [8]
                           s( 57.37%)p 0.74( 42.62%)

8. (1.83220)    LP ( 2)   O2        [16]
                           s(  0.00%)p 1.00( 99.91%)
------------------------------------------------------------

The total compliance with the Lewis bonding picture is 98.3%. Note the artefact in line 5, which should be a lone pair of oxygen.

From here we can see, that O1 is best described with an sp1.85 contribution towards carbon, an sp4.12 contribution towards hydrogen,
an sp1.20 and a p lone pair.
The O2 is best described with an sp2 contribution towards carbon, an π orbital towards carbon, an sp0.74 and a p lone pair.

Concerning the intramolecular hydrogen bond, one can analyse the electron density with the Quantum Theory of Atoms in Molecules (QTAIM). This shows that there is no bonding interaction between H4 and O2.

(Laplacian distribution, solid blue lines indicate charge depletion $\nabla^2\rho<0$, dashed blue lines indicate charge accumulation $\nabla^2\rho>0$, red spheres are bond critical points, black lines are bond paths, dark red lines are zero flux surfaces. Hydrogens H2 and H3 omitted for clarity.)

One could argue that there is an electrostatic interaction, causing the slightly smaller (105°) CHO angle, but since the H4-O2 distance is only slightly shorter than the sum of the van der Waals radii (H = 1.20 A, O = 1.52 A) in vacuum this might already be a stretch.

In solution there are many more factors to be considered, i.e. dimerisation, protonation/ deprotonation, intermolecular hydrogen bonds, and possibly many more.


Purification of water with ferric chloride Ferric chloride ($\ce{FeCl3}$) is a chemical used in water treatment. If ferric chloride is added to a water sample containing calcium bicarbonate, would the $\mathrm{pH}$ of the solution decrease or increase?

If I'd state that the pH drops, would you know why?

If you dissolve $\ce{FeCl3}$ in water, a hexaqua complex is formed. These complexes are known as cation acids:

$$\ce{[Fe(H2O)6]^{3+} + H2O -> [Fe(OH)(H2O)5]^{3+} + H3O+}$$


Is it possible to have accumulation of electron density in the inter-nuclear region even if the overlap integral is zero? Recap:
Lately, I've been studying Hückel theory where I learned about the approximation of neglecting all overlaps and making overlap matrix an identity matrix that is $S_{AB}= 0\;_; \;S_{AA}= 1$ or $S_{ij}= \delta _{ij}\;.$ He then replaced the off-diagonal Hamiltonian elements with parameter viz. $H_{AB}= \beta_{AB}\;_,$ where $\beta_{AB}$ is a negative quantity.
Question then arose if the overlap is zero as is evident from $S_{AB}$ being zero, how can there be a non-zero value of $H_{AB}$. It was known to me that $H_{AB}$ is the

contribution to the energy due to the accumulation of electron density where the two atomic orbitals overlap, including, for instance, the Coulombic attraction between the extra accumulation of electron density and both nuclei.

So, it is contradictory to simultaneously have the condition $H_{AB}\ne 0$ with $S_{AB}= 0.$
I asked this querying for the same; there I got a nice answer from Brian where he showed that even if $S_{AB}=0$; it need not be necessary to have $H_{AB}= 0$; the graphs he provided in his answer prove this otherwise a counterintuitive fact.

Present thinking & associated query:
Brian's answer confirms the fact that $S_{AB}= 0$ and $H_{AB}\ne 0$ can go with each other. But doesn't that mean there is no need to have a net overlapping of AOs between the internuclear region in order to have accumulation of electron density in the internuclear region?
For example, while using Simple Hückel Theory for ethene, Atkins put all $S= 0$ as usual per the approximation and then wrote:

[...] Finally, as explained in the derivation, he also supposed that the terms $H_{AB}$ and $H_{BA} ,$ which represent the energy of the interaction of the two nuclei with the accumulation of electron density in the internuclear region .......

Really, if he writes this after considering $S_{AB}= 0\;_,$ doesn't that mean there can be accumulation of electron in the internuclear region even if the the participating AOs are orthogonal and there is no net overlapping? He is considering the case of ethene $\pi$-network which means there is electron density that leads to $\pi$-bonding even if $S=0$.
So, is it right to conclude that there can be accumulation of electron density in the internuclear region even if there is no net overlap in the atomic orbitals?

If you define overlap as given by the overlap integral,

$
S_{ij} = \langle \psi_{i} | \psi_{j} \rangle
$

than you can have an accumulation of electron density in a specific part in space even though $S_{ij} = 0$.

Of course, if the complete wave functions of the two AOs are completely non-overlapping in all space, for instance because the centers of the two AOs are seperated by a long distance (say more than 5 A), then there can be no accumulation of electron density anywhere. This is of course logical. There is hardly any accumulation of electron density between two atoms that are quite far apart.

An interesting scenario exists where two wave functions have opposite sign and an overlap of these wave functions actually gives a decrease of electron density. This is termed deconstructive interference opposite to constructive interference when there is an accumulation of electron density.


Balloon from float to sink? Everytime I walk by the balloon stall, I know that they use helium to make the balloons float in the air. But my problem is why when I leave the balloon aside, the balloon sinks to the bottom. Is there anything happen to the  helium inside?

When the balloon is fully inflated with helium its density (mass/volume) is less than that of the air around it, allowing it to float. Over time, some of the helium will "leak" out of the balloon. As the helium is released the mass of the balloon decreases slightly, but the volume decreases at a much faster rate, resulting in a balloon with a density greater than that of the air around it, which is why it falls to the ground.


Where can I find the positions of the infrared band for phosphorus compounds in general? Specifically sodium hexametaphosphate Specifically I am looking for the IR of the sodium hexametaphosphate compound.

Specific for your compound you find at least two public IR spectra, either in KBr:

(source)

or from one of the larger chemical vendors as .pdf

(source, Sigma-Aldrich, product number 305553, CAS-RN 68915-31-1)

More generally speaking for phosphates, you equally find some related entries in NIST webbook chemistry here, too.

For IR-spectra in general, beside the SDBS with focus on organic compounds and sources already mentioned, there are three big multi-volume collections: Aldrich (equally one large about NMR) which is why frequently chemicals in their online catalogue have a link to either IR or/and NMR spectrum, too. The Sadtler collection is equally broad and diverse, organizes the entries by field of application (e.g., pharmaceuticals, forensics).  At least in part Sadtler is electronically available via BioRad / KnowItAll, too.  Finally Hummel, with emphazis on polymers / plastics.  Because many products of chemical synthesis are characterized by IR spectroscopy, too, you may find them indexed in databases like Reaxys (Elsevier) and Scifinder (American Chemical Society), too.


Compound with two functional group Compound X has 3 carbon atoms in one molecule. It contains a primary alcohol and a carboxylic acid functional group.

Draw the displayed formula of Compound X.

Help! Condensed formula is fine for me!

Draw a chain of three carbon atoms.
Make one of them $\ce{COOH}$. You have two carbon atoms to choose from.
Attach $\ce{OH}$ to one of the remaining carbon atoms to make it a primary alcohol. You have one carbon atom to choose from.


Protein purification with Cobalt Cobalt exhibits a more specific interaction with histidine tags, resulting in less nonspecific interaction than nickel. For this reason, cobalt is the preferred divalent cation for purifying His-tagged proteins when high purity is a primary concern.

I read this last week, and I'm wondering, why is cobalt like that? What about it causes that?

(Source of quote: https://www.lifetechnologies.com/us/en/home/life-science/protein-biology/protein-biology-learning-center/protein-biology-resource-library/pierce-protein-methods/his-tagged-proteins-production-purification.html)

Your question is about immobilized metal affinity chromatography (IMAC), often used to separate a protein expressed in the E. coli bacterium from other proteins made by it. The protein of interest is expressed with a histidine-containing affinity tag which binds to immobilized nickel or cobalt at neutral or basic pH. Immobilization is achieved by complexing to NTA (nitrilotriacetic acid) or IDA (iminodiacetic acid). For a discussion of the two ligands, see e.g. here.

There are a few proteins from E. coli (listed in this paper) that are known to bind along with the protein of interest. According to a post written by Damien Soghoian in 2004, nickel-NTA tends to bind stronger to histidine-tags while cobalt-NTA tends to bind more weakly, but this does not affect specificity (e.g. it would be across the board, not making the separation easier or harder).


Geometrical Isomerism (Cis-Trans) in trans-2-fluoro-3-methylpent-2-ene trans-2-fluoro-3-methylpent-2-ene this compound in also known as (Z)2-fluoro-3-methylpent-2-ene.
I have given my possible structure in the image but it is wrong, Can anyone explain this ?

The structure you drew

is ‒ according to the CIP-rules ‒ (E)-configured.  These rules are applied on each side of the double bond; namely

C of the ethyl group has a higher priority than C of the methyl group
F has a higher priority than C
C(ethyl group) opposes F, as seen across the double bond

This is not in contradiction that the two methyl groups are opposing each other, are in (trans)-relationship with each other.

In instances with only two subsitutents around a double bond, like (E)-2-pentene, you may use (E), derived from German "entgegen", literally "opposing", as synonym to (cis).  For the opposite instance, (Z), derived from German "zusammen", like "together", to describe a (trans)-configuration.

In instances with three or four substitutions around a double bond, and instances of several conjugated double bonds, follow IUPAC and adhere to the (E/Z) notation; which is absolute and not ‒ as (cis/trans) ‒ relative and potentially ambiguous.  See the example E/Z notation, for example. (Or, as one teacher told me, "stay on the E-Z pass".)


Oxidation and reduction in organic chemistry Why is it that in order to make an carboxylic acid from an alcohol you oxidise it twice...

ie alcohol + 2[O] ---> carboxylic acid

...but to get back to the alcohol you reduce it four times?

carboxylic acid + 4[H] ---> alcohol

My understanding was that oxidation is the opposite of reduction, and in organic chemistry [O] means addition of an oxygen atom or removal of a hydrogen atom, with [H] meaning the opposite.

Addition of oxygen and hydrogen can be useful to predict oxidation and reduction in most of the cases but this criteria fails to classify a reaction as oxidation or reduction when there are no oxygen or hydrogen atoms involved. This method is very primitive.

A better perspective :

Oxidation : Increase in oxidation state (by loss of electrons) of central atom.

Reduction : Decrease in oxidation state (by gain of electrons) of central atom.

If you already know how to find the oxidation state of the central atom (in your case carbon), then I hope you will be able to get your answer now.

The answer to the question why oxidation of alcohol to carboxylic acid takes 2 oxidation reactions, lies in the reaction mechanism. These mechanisms cannot be challenged, they are what they are. We can only observe and provide a acceptable explanation to why they happen like the way they do.


Nucleophilic substitution in polar protic/aprotic solvents I am confused as to why nucleophilic substitution reactions occur. Does a stronger nucleophile substitute a weaker nucleophile in nucleophilic substitution reaction?
For example, I reasoned that reaction
$$\ce{R-Br + KF ->[DMF] R-F}$$
is correct because $\ce{F-}$ is a stronger nucleophile than $\ce{Br-}$ in polar aprotic solvent.
Therefore,
$$\ce{R-Br + KF ->[Polar protic solvent]}$$
should give no reaction because $\ce{F-}$ is a weaker nucleophile than $\ce{Br-}$ in a polar protic solvent. Is my reasoning correct?

It depends whether the reaction progresses through SN1 or SN2. In both, the departing of a leaving group is involved in the rate determining step, but only in SN2 is the nucleophile included in the rate determining step. In the following mechanisms, both of the first steps are rate determining.
(Image from https://www.quora.com/Why-steric-hindrance-doesnt-affect-Sn1-reaction)

It is important to remember that a good leaving group is defined by poor basicity, which can be quantified by a low pKa of the conjugate acid of the leaving group. Intuitively, this means the conjugate acid can lose a proton and the resulting base - our leaving group - will be reluctant to re-react with the proton. Or in our case, re-react with the electrophilic carbon. This generally means that good leaving groups are large, low charge density species; good examples are neutral molecules or even charged species like iodide ions.
This idea of strong and weak nucleophiles is situational. Hard-soft acid-base theory becomes a very useful reference in a lot of situations. It generally states that low charge density 'soft' nucleophiles and electrophiles have an affinity for eachother, and high charge density 'hard' nucleophiles also have an affinity for eachother. You can read more about exactly why this happens here:
https://en.wikipedia.org/wiki/HSAB_theory
Generally, we consider a saturated carbon a 'soft' electrophile, which will react very well with low charge density nucleophiles like iodide ions. What this means, is that in SN2 reactions, species that are considered good leaving groups are also good nucleophiles. For instance, you may predict - correctly so - that iodide ions are good leaving groups as well as good nucleophiles for SN2; this is why KI can be used to catalyse SN2 reactions by acting as both (usually better than your actual nucleophile and leaving group)!
(Image from Clayden Organic Chemistry)

These reactions can end up reasonably reversible, and can be driven to one side by modifying conditions. For instance, the reaction below (from Clayden organic chemistry) exploits the insolubility of a byproduct, effectively removing it from the equilibrium, shifting it to the right:
(Image from Clayden Organic Chemistry)

In SN1, nucleophilicity is irrelevant, since the nucleophile doesn't appear in the rate determining step, but I'd imagine the equilibrium position can be roughly estimated by relative leaving group abilities.
In answer to your specific example, fluoride would actually be a poorer SN2 nucleophile than you might think, worse than the bromide leaving group would be. The polar protic solvent will even further inhibit the nucleophilicity of the fluoride due to hydrogen bonding, slowing it down dramatically (so your assumption is accurate for kinetic reasons); and once it has been substituted, the fluoride ion is not a particularly good leaving group. In DMF, you don't have the hydrogen bonding issue, and due to poor solvation of anionic species in DMF, the fluoride is made slightly more reactive than it would be 'on its own'. And leaving group ability is also reduced due to poor solvation. This seems like it would be clear-cut; the fluorine will replace the bromine, right? However the bromide ion is a good nucleophile in this situation, and has the potential to kick off the fluoride in either solvent.
As you can see, there are a lot of factors to juggle here, and the actual degree to which the reaction progresses is specific to the exact conditions of the reaction. At a first glance, it is very difficult to predict, and is not necessarily as simple as the 'better' nucleophile replacing the 'worse'.


How to calculate the volume of a sodium hydroxide solution necessary to add to water, to raise the pH to a certain value? What volume of $\pu{0.300N}\, \ce{NaOH}$ would be needed to bring $\pu{16.0 mL}$ of distilled water that has a $\mathrm{pH}$ of $7.00$ to a new $\mathrm{pH}$ of $12.42$?

This is my attempt at a solution:
$$\ce{NaOH + H2O-> Na+ + OH- + H2O}$$
\begin{align}
\mathrm{pH}(\text{initial}) &= 7.00 \\
\mathrm{pH}(\text{final}) &= 12.42 \\
\mathrm{pH} &= -\log{\ce{[H+]}} \\
\ce{[H+]}_{\text{intial}} &= 10^{-7} \\
&= \pu{10^{-7} ions} \\ \hline
V_1&= 0.016L \\
C_1 &= (0.300)(10^{-7})= 3\cdot10^{-8} N\\
\mathrm{pOH} &= 14-12.42  \\
[\ce{H+}]&= 10^{-1.58} \\
&=\pu{ 0.02630 ions} \\
C_2 &= (0.300)(0.02630) \\
&=7.89\cdot10^{-3} N \\
C_1V_1&=C_2V_2 \\
(3\cdot10^{-8})(0.016)&= (7.89\cdot 10^{-3})(V_2) \\
V_2&= 6.08\cdot10^{-8} L \\
\end{align}

$$\ce{NaOH + H2O-> Na+  +OH- + H2O}$$
\begin{align}
\mathrm{pH}_1 = 7.00 &\implies\mathrm{pOH}_1 = 7 \\
\mathrm{pH}_2 = 12.42 &\implies\mathrm{pOH}_2 = 1.58
\end{align}
so $[\ce{OH-}]_1 = 10^{-7}$ and $[\ce{OH-}]_2 = 10^{-1.58}$
So, initially, you have $16^{-3}\times 10^{-7}$ mol of $\ce{OH-}$ in your water. Your $\ce{NaOH}$ solution is $\pu{0.3 M}$ so you'll bring $\pu{0.3\times 10^{-3} mol}$ each $\mathrm{mL}$.
Then, the resulting concentration will be $$C = \frac{n_\text{initial} + (0.3\times 10^{-3})\times x}{V_\text{initial} + x\times \pu{1mL}}$$
You know $V_\text{initial}$ to be $\pu{16 mL}$
You know $n_\text{initial}$ to be $\pu{1.6\times 10^{-9} mol}$
And you know the target concentration to be $10^{-1.58}$
I let you do the solving for $x$ where $x$ is the volume of $\ce{NaOH}$ solution to be added ;) be careful with the units.


What does this example mean? Slow cooling of 40% Sn alloy from 800°C to 600°C: L → L and γ → L, γ, and ε → L and ε It's in a phase diagram topic. It's about $\ce{Cu}$-$\ce{Sn}$ phase diagram.

And it has a question of:

Write the schematic diagram describing the following scenarios and identifies the regions and the phases when Slow cooling of $13.1\%$ $\ce{Sn}$ alloy from $\pu{1000^\circ C}$ to $\pu{300^\circ C}$ happens.

I have this schematic diagram that describes the following scenarios and identifies the regions and the phases at $40 \%\,\ce{Sn}$ alloy from $\pu{800 °C}$ "$\ce{L -> L}$ and $\ce{γ -> L}$, γ, and $\ce{ε  -> L}$ and ε" as an example but it's confusing to follow. However, it's just an example and it doesn't follow the $\ce{Cu}-\ce{Sn}$ phase diagram anyway.
But does the first one mean $\ce{L -> L +  γ}$?
And L , y are in different phases that manages to fall both at $40 \%\,\ce{Sn}$ alloy from $\pu{800 °C}?
Also am I correct to assume that the final one means $\ce{ε -> L + ε}$?
This is the closest image that I can give based on my assumption:

Development of microstructure in eutectic alloys (II)
At compositions between the room temperature solubility limit and the maximum solid solubility at the eutectic temperature, $\beta$ phase nucleates as the $\alpha$ solid solubility is exceeded upon crossing the solvus line

[MSE 2009: Introduction to Materials Science; Chapter 9: Phase Diagrams; Page 28]

Here's the PDF file for more information.
"$\ce{L -> L}$ and $\ce{γ -> L}$, γ, and $\ce{ε  -> L}$ and ε" is a schematic diagram for the slow cooling which means it is an equilibrium cooling Like this image given below:

Equilibrium Cooling in a $\ce{Cu-Ni}$ Binary System
Phase Diagram: $\ce{Cu-Ni}$ System.
System is:
--binary: i.e 2 components: $\ce{Cu}$ and $\ce{Ni}$.
--isomorphous: i.e., complete solubility of one component in another; a phase field extends from $0$ to $100\,\mathrm{wt}\%$
Consider $C_o = 35\,\mathrm{wt\%}$ $\ce{Ni}$

Another one is:

Ex: Cooling of $\ce{Cu-Ni}$ Alloy
Phase diagram: $\ce{Cu-Ni}$ system
Consider microstructural changes that accompany the cooling of a $C_o = \pu{35 wt\%}\, \ce{Ni}$ alloy

Cored vs Equilibrium Structures
$C_o$ changes as we solidify

These images are from this PPT file.

Consider the 40 at% (Sn) composition at 800 °C: it's a liquid.
When you cool it down at around 650 °C you enter a two-phase field: the solid gamma forms (Cu-enriched) and the liquid enriches in Sn. As you further cool down the gamma phase will continue to form, but with progressively higher content of Sn
You cool down at 640 °C and you enter a new different two-phase field: the liquid partially reacts with the gamma phase forming the solid epsilon phase. If you further cool down, the ε phase will continue to form from the liquid, but its composition will progressively enrich in Cu.
Just below 415 °C you enter into another two-phase field: the liquid will solidify into ε and η phase.
That's what occurs by cooling down to 300 °C, but if you further would cool down below 189 °C you will enter another two-phase field; in this case the ε and η' phases will form.
The amounts of the different phases at the different $T$ are obtained by applying the lever rule.
Take care, I described what happened from the thermodynamic point of view (that is what is represented in a phase diagram); the kinetics can remarkably change the things.


Molecular orbital (MO) diagram for N2 and N2^- I have been taught that the MO diagram is different for molecules with 14 or less electrons than the one used for molecules with 15 or more electrons.

For $\ce{N2}$ the orbitals in increasing energy are:

$$\sigma_\mathrm{1s} < \sigma^*_\mathrm{1s} < \sigma_\mathrm{2s} < \sigma^*_\mathrm{2s} < \pi_{\mathrm{2p}_x}, \pi_{\mathrm{2p}_y} < \sigma_{\mathrm{2p}_z} < \pi^*_{\mathrm{2p}_x}, \pi^*_{\mathrm{2p}_y} < \sigma^*_{\mathrm{2p}_z}$$

because it has 14 electrons.

For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as that of $\ce{N2}$ or not? Why?

And because this is a good chance to ask, why is there this difference in orbital energies anyway?

I have been taught that the MO diagram is different for molecules with
  14 or less electrons than the one used for molecules with 15 or more
  electrons.

This is (partly) wrong because the change in the order of $\mathrm{\sigma_{2p_{z}}}$ and $\mathrm{\pi_{2p_{xy}}}$ MOs to the left of $\ce{N2}$ is not directly related to the number of electrons. Rather, it is rationalized by a successive decrease of the $\mathrm{s}$-$\mathrm{p}$ interaction moving from $\ce{Li2}$ to $\ce{F2}$. The $\mathrm{s}$-$\mathrm{p}$ interaction is the bonding interaction between the $\mathrm{2s}$ orbital of one atom and the $\mathrm{2p_{z}}$ orbital of another atom which (among other things) increases the energy of the $\mathrm{\sigma_{2p_{z}}}$ MO relative to the hypothetical case without $\mathrm{s}$-$\mathrm{p}$ interaction.

Now the difference in energy between the $\mathrm{2s}$ and $\mathrm{2p_{z}}$ AOs increases from $\ce{Li2}$ to $\ce{F2}$ due to increasing nuclear charge and poor screening of the $\mathrm{2s}$ electrons by electrons in the $\mathrm{2p}$ subshell. As a result, the rising effect of $\mathrm{s}$-$\mathrm{p}$ interaction on $\mathrm{\sigma_{2p_{z}}}$ MO is getting less and less prominent, so that eventually to the right of $\ce{N2}$ $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\pi_{2p_{x,y}}}$ MO.

Image courtesy of the UC Davis ChemWiki.

Now, back to the question.

  For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as
  that of $\ce{N2}$ (because it is actually an ionized molecule of $\ce{N2}$) or not?
  Why?

The short answer is: we could not tell it using the primitive molecular orbital theory introduced in the general chemistry courses. In exact same way we could not tell why $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ and not to the left of, say, $\ce{C2}$. All this is simply because the primitive molecular orbital theory does not explain the things, rather it rationalizes them (for differences between explanation and rationalization, please, read here).

But then how do we know in the first place that $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ so that we can start to rationalize it? Well, we know that from a more advanced molecular orbital theory which is based on the laws of quantum mechanics. And the same advanced theory should be applied for the case of $\ce{N2-}$ to get the MO picture of this chemical species.

Now note that even in this advanced molecular orbital theory a bunch of approximations is introduced, and the answer in general depends on at which level of theory calculations are done. The simplest picture would be the so-called frozen-orbital approximation (in the Hartree-Fock picture) in which MOs of an anion (as well as that of a cation) are assumed to be the same as for the corresponding neutral molecule. But this approximate picture might be far away from the reality.

Update

The question in fact does not make a lot of sense since $\ce{N2-}$ is known to be electronically unstable, i.e. its energetically less favourable then the system of $\ce{N2}$ plus a free electron. There exist few metastable resonance states of $\ce{N2-}$ that correspond to attachment of the scattered electron to excited valence states1, but that's a different story which is beyond the scope of the question.

1) Krauss, M., Neumann, D., Valence Resonance States of N2, JResNatlStandSecA, 77A, 4, 1973. PDF.


Is the term "carboreduction" interchangeable with "carbothermic reduction"? I am currently studying reactions involving the reduction of metal oxides by carbon at high temperature to produce refractory carbides (such as $\ce{ZrC}$) following equations such as:

$$
\ce{ZrO2 + 3C -> ZrC + 2CO (g)}
$$

I have come across both the following terms for this type of a reaction carbothermic reduction and carboreduction; they appear to be used completely interchangeably. I would like to make sure (as I have not yet found a source to say otherwise) that this is the case in chemistry terminology and not that one may be applied to an equation as above and the other to a similar reaction where the product is actually the pure reduced metal, such as that for iron ore smelting:

$$
\ce{2Fe2O3 + 3C -> 4Fe + 3CO2}
$$

The term "carboreduction" has an IUPAC definition:

  Process in which a metal oxide is reduced in the presence of carbon or a carbon-containing compound.

So technically the term does not require heating and isn't limited to elemental carbon.

Carbothermic definitely implies heating.


C=O and C=C antibonding orbitals Why is the pi star orbital of the C=C bond higher in energy than the antibonding pi orbital of the C=O bond?

I thought it was the other way around; it's relatively easy to add something to a C=C alkene bond (i.e. halogenation) but halogenation with diatomic halogens doesn't occur with C=O bonds.

It's relatively easy to add an electrophilic species, such as X-X, to an alkene. As @Philipp points out, the alkene is operating through the (filled) pi orbital.

It's relatively easy to add a nucleophilic species to a carbonyl. In that case, the carbonyl operates through its (unfilled) pi* orbital.


In the graph provided, which of the substances is the most efficient in acting as an antacid? If every dot represents the addition of 0.1 g of the respective base. I suppose that the best would be the one that reaches the highest pH, in the shortest time, for example Calcium Carbonate, but if that is true, why are other antacids used that reach lower pH?

Normally, the efficacy of a drug is determined by the area under the curve of the entire time course of the drug in the body.  The graphs here only show the initial rate of the change in pH but not the entire time course after the peak pH is reached.  I assumed that these results were generated by an acid-base titration experiment in the laboratory and not generated in vivo.  Therefore I cannot answer which antacid is more effective and/or clinically safe.

Although calcium carbonate shows a rapid initial rate of pH neutralization, it may not be suitable for all patients.  Too rapid a pH rise in the digestive tract may cause nausea and vomiting and may trigger an “acid rebound”, meaning that the body tries to compensate by secreting more acid to maintain the normal gastric pH for food digestion, making the heartburn even more unbearable.

Some patients take calcium carbonate as a source of calcium, but one also needs vitamin D in order for the calcium to be absorbed.  Besides, calcium carbonate yields carbon dioxide gas with acid which is still acidic, and bloating.  For patients with compromised parathyroid function, too much calcium in the blood (hypercalcemia) after the bones are already saturated with calcium may start to deposit in the heart and in the kidneys to form calcium oxalate kidney stones (different from uric acid stone).  Therefore calcium antacid may not be the best antacid for all patients.

If calcium antacid is not tolerable, one can use aluminum hydroxide (Aludrox®) which is a good alternative.  However aluminum hydroxide can cause constipation.  Magnesium hydroxide (milk of magnesia) is another good alternative, but magnesium hydroxide can cause diarrhea because magnesium is a smooth muscle relaxant.  Therefore the product Maalox® or Mylanta® which combines aluminum hydroxide with magnesium hydroxide will be easier on the stomach.  Also, there are products that contain simethicone (which is an anti-flatulent) as an extra ingredient to help with bloating.

The choices for an antacid are out there, make a good, sound one.


Can one figure out an optical absorption spectrum from EPR data? I'm wondering if it's worthwhile for me as an optical spectroscopist to read up a bit on EPR. Do EPR signals reveal where, in terms of wavelength, features in the optical spectrum can be expected to appear? I know it tells me
lifetimes of states, but that's about as far as I have dug into it. I have some EPR data on a substance and I wish to figure out where to look for it's absorption with UV/Vis/NIR means.

No, it shouldn't be possible, based on where these spectroscopies lie in the electromagnetic spectrum:

The most common EPR experiments occur at 9.5 GHz (X band), which is in the microwave region. Even the highest-frequency experiments just barely break into the far infrared. Assuming you're doing an X band experiment, and that your system has a transition very close to the free electron $g$-values, 9.5 GHz corresponds to ~0.3 cm$^{-1}$.

For UV-vis, taking the longest wavelength of light in the visible region as 800 nm gives 12,500 cm$^{-1}$. This can be interpreted in two ways:

The EPR experiment needs ~42,000x more energy to probe even the lowest possible electronically-excited state.
Even the UV-vis transitions with the lowest possible energy will not show any fine structure, like in the sodium lines or benzene vibronic coupling. The energy scales need to be much closer. In the case of the spin-orbit-induced splitting of the sodium spectral lines, the unperturbed gap is 2.1 eV and gap between the split states is 0.0021 eV; this is 3 orders of magnitude between the energies. In the case of the benzene UV spectrum, the gap between peaks is ~10 nm, with the peak center at ~245 nm; this is 1-2 orders of magnitude:

$\hspace{20ex}$

Our EPR/UV-vis example has 5 orders of magnitude difference between the two energy scales. There is (hyper)fine structure in EPR:

which here, for DPPH in toluene is ~0.002 Tesla, or 0.35 GHz, so 1 order of magnitude.


Proper dash to use for negative charge What type of dash does one use when indicating something like F$^{-}$? I assumed it was a minus, since you indicate positive charge with a +, but there is one professor I know who insists that it should be typeset with an endash. Is there a style manual on chemical typesetting I can consult.

[…] one professor I know who insists that it should be typeset with an en dash.

I disagree! A standard minus sign should be

on the same height as the horizontal line of a plus sign;
as long as an equal sign.

The en dash is

lower than the horizontal bar of the plus sign;
slightly longer than the equal sign.

Unicode has dedicated characters for minus and superscript minus

standard minus: U+2212: −
superscript minus: U+207B: Cl⁻

See also:
The situation, and how to realize the correct typography using $\mathrm\LaTeX$ has been described in the German article Feinheiten bei wissenschaftlichen Publikationen – Mikrotypographie-Regeln, Teil II by Marion Neubauer in Die TeXnische Kommödie, 1997, 1, 25-44 (PDF).


Oxidation of formate ion by peroxydisulfate in water solution: Why are there "strange" kinetics? As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is:
$r=k\ce{[HCOO^{−}]}^{1/2} \ce{[S2O8^{2−}]}$.
The first reaction is very slow, so most of the peroxydisulfate is consumed in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.

WHAT does this mean?
So, why is the rate constant NOT according to the slowest step?

Why it is not just $k\ce{[S2O8^{2-}]}$?

The rate of the third reaction affects the rate of the first reaction, i.e. they are competitive reactions and in that case the rds method can't be applied.

The rate of the first reaction is $r=-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t$, but the change of the concentration of the peroxydisulfate depends on the third reaction, too:
$$-\mathrm{d}\ce{[S_2O_8^{2-}]}/\mathrm{d}t=k_1\ce{[S_2O_8^{2-}]} + k_3 \ce{[CO_2^-]}\ce{[S_2O_8^{2-}]}$$


How to formally calculate the surface density of layered materials such as graphene? Typically density is defined in units of mass per volume. In the case of graphene it is mass per area, i.e. surface density. What would be the correct way to calculate the surface density (mass per unit area) of graphene, for example?

Would it be correct to multiply the density of graphite by the van der Waals gap of graphite?

The C-C length in graphene is l = 0.142nm and the area of a hexagon can be calculated with the formula:

$ A = \frac{3\sqrt{3}}{2} l^2 =  0.0523nm^2$

In each hexagon, there are 2 full carbon atoms(1/3*6) so the surface density of one single layer is:

$ S_d = \frac{2*massCarbon}{A}  =  \frac{2*1.994 × 10^{-26}Kg}{0.0523× 10^{-18}m^2} = 76.26 × 10^{-8} Kg/m^2 = 7.63× 10^{-8} g/cm^2$

If you are considering 2, 3, etc. layers than the surface density it's twice, three times etc. the surface density of the single layer.

Additional note: The distance between layers is h = 0.335 nm and therefore its density can be calculated as:

$ d = \frac{S_d}{h}  =  \frac{7.63 × 10^{-8} g/cm^2 }{0.335× 10^{-7}cm} = 2.28 g/cm^3  $

This is very close to the experimental value I have found online says that the density of graphene is $2.267 g/cm^3$


Why do colligative properties depend only on number of solute particles? Colligative properties depend solely on the number of even though the interactive forces are different for different solute-solvent pairs. So why is the dependence only on the number of solute?

It's a lie. Colligative properties do depend on the chemical nature of the solute and solvent - their interaction. The trick is: an ideal solution, in which the solute-solute intermolecular forces and solute-solvent interactions are more or less of the same character, has the property that colligative properties are dictated by composition given only in terms of amount of moles. However, any sufficiently diluted solution will behave somehow ideally, that is where the statement goes: for diluted solutions, colligative properties depend only on the number of moles of solute.


d-Menthol vs dl-menthol: Does an enantiomer and its racemic mixture have different melting points? I am checking out Wikipedia and it shows a different melting point (m.p.) for l-menthol $(\pu{42 ^\circ C}$ to $\pu{45 ^\circ C})$ vs dl-menthol $(\pu{36 ^\circ C}$ to $\pu{38 ^\circ C}).$

How is this possible? d-menthol and l-menthol being enantiomers, shouldn't they have identical physical properties including the m.p.?

  The two enantiomers of a molecule possess the same physical properties (e.g., melting point, boiling point, polarity etc.) and so behave identically to each other.

If so, why would the racemic mixture have any different m.p.? I.e. you can't just physically  mix two pure materials with the same m.p. and end up with a material with a different m.p., right?

I checked out Common Fragrance and Flavor Materials by Surburg and Panten [1, p. 55] just to be sure and the data agrees. The m.p. of the d-menthol and dl-menthol again seem to be different:

  Other physical constants of commercially available levorotatory and racemic
  menthols are: (−)-menthol [2216-51-5], mp $\pu{43 °C},$ $n_\mathrm{D}^{20}~1.4600,$ $[α]_\mathrm{D}^{20}~–50^\circ;$ (±)-
  menthol [15356-70-4], mp $\pu{38 °C},$ $n_\mathrm{D}^{20}~1.4615.$

References

Surburg, H.; Panten, J. Common Fragrance and Flavor Materials: Preparation, Properties and Uses, 5th, Completely Revised and Enlarged edition; Wiley-VCH: Weinheim, 2006. ISBN 978-3-527-31315-0.

If solid menthol were just amorphous, then you would expect the racemate and each enantiomer to exhibit identical melting points. However, if you read further on the Wikipedia page you reference, you'll see that racemic vs. enantiopure menthol crystallizes differently:

  The two crystal forms for racemic menthol have melting points of 28 °C and 38 °C. Pure (−)-menthol has four crystal forms, of which the most stable is the α form, the familiar broad needles.

(This information is trustworthy and corroborated in the primary literature; here is one example)

The melting point of a crystalline solid has to do with how well the molecules can pack together. In other words, on a molecular level, there is an orientation dependence. The better the molecular packing, the stronger the intermolecular forces, and the higher the melting point - since it takes more energy to separate the molecules. This is why the enantiopure menthol has a higher melting point.


Are there any examples of nuclear spin isomers having consequences for chemical reactivity? Ortho- and parahydrogen are two forms of the $\ce{H2}$ molecule that are distinguished by their pairing or antipairing of nuclear spins, giving rise to metastable singlet (ortho-) and triplet (para-) states. This constitutes an example of nuclear spin isomerism.

Whilst the thermal properties of these two species are different, my understanding is that their chemical properties are essentially identical. Are there any examples of this kind of isomerism having nontrivial implications for chemical reactivity?

Yes!

Papers in the field of preparation or selection of nuclear spin isomers have introductions that justify the work by potential applications and some of them include chemical reactions, either under that name or under the more descriptive terminology of “reactive collisions”. While no specific cases are cited as examples, the potential exist but only in limited situations such as bimolecular reaction in molecular beams.

From what I have seen, the canonical reference is “Detailed symmetry selection rules for reactive collisions”, Mol. Phys. 1977, 34, 477–504. This is the theoretical paper laying out the foundations of the effect of nuclear spin isomerism on reactions. I don't have access to the full text, but here is the abstract:

It confirms what I suggested in my earlier comment: the different nuclear spin isomers have different rovibrational states, and accounting for effect changes the selection rules indicating which collisions are reactive or not.

One specific example which I found is “Observation of Ortho-Para H3+ Selection Rules in Plasma Chemistry”, Phys. Rev. Lett. 1997, 78, 3844–3847:

  Large dependences of the intensity ratios on the parent gases have been observed, indicating the spin memory of protons even after chemical reactions. The results clearly demonstrate the existence of selection rules on nuclear spin modifications in chemical reactions theoretically anticipated by M. Quack.

Their results are obtained on the formation of H3+ ions from a H2 plasma.


Why do power requirement figures for water electrolysis on the web seem to be wrong? I need to calculate the power needed to electrolyse 1 mole of water for a project.
We will either use Polymer Electrolyte Membrane Electrolysis or Alkaline Water Electrolysis, they operate on 1.75V-2.20V and 1.8V-2.4V respectively. And each has a maximum efficiency of about 69%. So, I have done further calculations assuming that the potential difference between the electrodes is 1.8V and the efficiency is 70%. As per my requirements, I have also considered that I have 8 minutes for conducting electrolysis.

Here are the calculations :

The reaction : 2H2O 2H2 + O2
Half Reactions:
Reduction at cathode: 2 H+(aq) + 2e−  H2(g)
Oxidation at anode: 2 H2O(l)  O2(g) + 4 H+(aq) + 4e−

1 mole of hydrogen gas will be produced from 1 mole of water ie 18 g of water or 18 cc of water.
Time available for electrolysis = 8 minutes ie 480 seconds
Number of moles of electrons = 2
Charge = 2 x 96,485 C/F = 192,970 C
Current = Charge / Time = 192,970 / 480 = 400 Amperes (approx.)

Considering 70% efficiency, Current required = 570 Amperes
Potential Difference = 1.8V (common to both PEM electrolysis and Alkaline water electrolysis)

So, Power Required = 1026 Watts ≈ 1 kW (for 8 minutes per mole)

I took reference of this webpage to learn how to calculate it.

However, then I came across this figure on the internet -
per mole water splitting requires at least 241.8 kJ of energy
Source: The first answer to this question.

So I am a little confused. I guess, this 241KJ energy figure came out because the person might be considering 1.23V potential difference (which is considered ideal), and 100% efficiency. I can't say anything because he has stated, that the values of enthalpy of formation (which he has used) are experimental, but hasn't provided the voltage and neither other conditions for the reaction.  He has just used the values of enthalpy of formation and not considered the efficiency of the method in use, in fact, that figure seems to be so general for all methods. As, PEM Electrolysis and Alkaline Water Electrolysis are the most efficient methods and both don't operate on that "ideal" 1.23V potential difference. Also, I think my reasoning is correct because if I redo the calculations considering 1.23V and 100% efficiency, the answer comes out to be the same as his.

Is my reasoning correct? and if not what is the reason for having different answers.

They said it right, "... requires at least 241.8 kJ ..." . You just found that one can simply calculate the molar formation enthalpy from the cell voltage, and vice versa, those two informations are redundant.

Of course you cannot run the electrolysis at the minimum voltage, because it will be extremely slow! You would just balance the electrochemical potential of the cell (created by the reverse reaction as soon as you make a small amount of hydogen and oxygen).

To make any actual electrolysis, you need a higher voltage, and the difference is converted into heat at the internal ohmic resistance of your cell. The minimum additional voltage is called the cell overvoltage, below that you don't get electrolysis. It is necessary to be able to break through the polarised electrolyte layers at the electrodes. So you cannot get towards 100% efficiency by running very slow.


Hoffman's Isocyanide test: How does addition of concentrated HCl degrade the isocyanide formed? This question is regarding a laboratory procedure followed for Hoffman's Isocyanide test (also known as the "Carbylamine test")

I've been taught that the isocyanide formed in this reaction is highly toxic and must therefore be destroyed in situ.

I'm told that one method of carrying this out is to cool the test tube and carefully add an excess of concentrated $HCl$.

Though it wasn't mentioned, I'd like to know how exactly the addition of "excess concentrated $HCl$" results in the degradation of the isocyanide formed.

I've provided the reaction-mechanism for a primary amine subjected to the Hoffman Isocyanide test below;

Isocyanide is not as toxic as its isomer cyanide and can be tolerated. The purpose of its removal is due to its disagreeable odor which is not at all tolerable.

  While some isocyanides (e.g., cyclohexyl isocyanide) are toxic, others
  "exhibit no appreciable toxicity for mammals". Referring to ethyl
  isocyanide, toxicological studies in the 1960s at Bayer showed that
  "oral and subcutaneous doses of 500-5000 mg/kg can be tolerated by
  mice".(Wikipedia)

Isocyanide is removed on the basis of its property of acidic hydrolysis.

  Isocyanides are stable to strong base (they are often made under
  strongly basic conditions), but they are sensitive to acid. In the
  presence of aqueous acid, isocyanides hydrolyse to the corresponding
  formamides:

  $$\ce{RNC + H2O → RN(H)C(O)H}$$

  This reaction is used to destroy odorous isocyanide mixtures. Some
  isocyanides can polymerize in the presence of Lewis and Bronsted
  acids.

But the story does not here. The formamides are still pretty unstable and further degrades to form amines and formic acid in the presence of acidic medium.(Here)

  Isocyanides are hydrolyzed by dilute mineral acids to give primary
  amine and formic acid (Isocyanides are not hydrolyzed by bases).
  Isocyanides are hydrolysed only by acids and not alkalies because the
  negative charge present on carbon atom in isocyanides initially
  attracts electrophiles ($\ce{H+}$) but repels nucleophiles ($\ce{OH–}$). When a
  proton gets attracted to negatively charged carbon atom, the tendency
  of this carbon atom to attract a nucleophile increases due to presence
  of positive charge on the N atom and hydrolysis is facilitated as
  shown below:

  Mechanism:

Here is a research paper for a reference.


Voltage applied when charging a galvanic cell A working galvanic cell is constructed using an iron electrode in an iron(II) nitrate solution and a
  silver electrode in a silver(I) nitrate solution. When the cell runs down, it can be charged up again.
  Which of the following scenarios will occur during the charging process?

  $\ce{[Eº(Ag+(aq)/Ag(s)) = +0.80 V and Eº(Fe^2+(aq)/Fe(s) = –0.41 V]}$

  A. A minimum of 1.21 volts is applied to the cell and electrons move from the silver electrode to the iron electrode.

  B. A minimum of 1.21 volts is applied to the cell and electrons move from the iron electrode to the
  silver electrode.

  C. A minimum of 0.39 volts is applied to the cell and electrons move from the silver electrode to
  the iron electrode.

  D. A minimum of 0.39 volts is applied to the cell and electrons move from the iron electrode to the
  silver electrode.

  E. A minimum of 2.01 volts is applied to the cell and electrons move from the silver electrode to
  the iron electrode.

I know reduction of silver is spontaneous and oxidation of iron is spontaneous from the values given, so reduction occurs at the silver electrode and oxidation occurs at the iron electrode, so electrons flow from iron to silver.

In the reverse reaction, electrons will now flow from silver to iron. However, I'm unsure about the stoichiometry of the electron transfer occuring during charging:

$\ce{2Ag -> 2Ag+ + 2e-       -1.60V}$

$\ce{Fe^2+ + 2e- -> Fe    -0.41V}$

I multiplied the oxidation of silver by 2, as two electrons must be supplied for iron to reduce, hence getting -2.01V, and answer E. However, the solutions state A, which suggests the oxidation of silver was no multiplied by 2. Why is this?

Voltage is how much push is needed to move the electron, and is the same whether for one or one trillion electrons. Electric charge, measured in Coulombs, is what changes when more electrons are moved.

Your reasoning that moving twice the charge doubles something is correct: instead of ~1.6 × 10−19 Coulomb for one charge, it's twice that, ~3.2 × 10−19 Coulomb for two of them (no discount for large charges, but I hope this kept your interest).


Under what circumstances is working in the chemistry lab hazardous to health in the long term? I would like to know when working in the Chemistry lab would be hazardous to health in the long term. Because as everyone who worked in the chemistry lab knows that they would be handling many compounds that are harmful to human health. Many of those reagents for example especially organic ones are carcinogenic. While some are toxic to humans. Even with safety precaution, we would inevitably be exposed to those chemicals when working with them such as inhaling them. So actually I wonder whether it is safe to work in the chemistry lab especially with so many carcinogens. Im afraid of getting cancer due to this.
(I'm already halfway through my undergrad chemistry course.)

May I start with a disclaimer saying that whatever long term studies can be done, this will always be a subjective question. All I can do here is offer a few comparisons and examples to show that you shouldn't (in my opinion) be worried.

There answer to the question "Am I more likely to come into contact with carcinogenic/hazardous substances than a regular person?" is quite likely yes. At undergraduate level you will be using a range of strong, harmful, poisonous etc. substances that you could not get outside of a lab environment or at school.

However, the answer to the question "Am I more likely to suffer poor health as a result of working in a lab?" is more difficult.

As you correctly state in your question, there are always many safety procedures in place when you undertake experiments which have been carefully thought, require the scientists to have some minimum level of training in the techniques (perhaps learning with less dangerous substances) out and always err on the side of safe rather than sorry. I myself have been working with a new nanotube based material recently and though unlikely to be carcinogenic, all experiments were to be carried out using gloves in a fume cupboard with extraction as the material hasn't been fully classified or tested yet (again: "safe rather than sorry").

Of course, there are unfortunate cases (for example Marie Curie), but in this day and age the risk assessments are so extensive and always erring on the side of caution that the chance of something going terribly wrong are no worse, I feel, than something catastrophic happening in many other walks of life (e.g. mountaineering accident, plane crash, shark attack).

One perhaps reassuring and rather amusing case is that you are far less likely to succumb to terminal cancer if you work in the nuclear industry by virtue of the extensive risk assessments, precautions and health checks. There is so much safety in place at a reactor, that exposure is minimal (indeed, the background radiation from radon in Cornwall is higher than that next to a nuclear power plant!!) and the industry provides so many health checks for workers that and cancer (radiation induced or natural) is often picked up at a very early, curable stage. Earlier than if you simply went to a doctor when you began to feel ill as a regular person.

Therefore depending on the policy of your employer, you could in fact be at a lower risk of dying from cancer, and I see no need for you to worry for your health in the lab!!

As mentioned when I started, this is a subjective issue, but I really see no need for worry.

Best,

Jack


How to memorize 7 crystal systems? Is there any trick to memorize the angles and sides of 7 crystal systems of crystalline solids?
1. Cubic
2. Tetragonal
3. Orthorhombic
4. Monoclinic
5. Hexagonal
6. Rhombohedral or Trigonal
7. Triclinic

The first step is understanding what you have.  The seven crystal systems are derived from decreasing symmetry (lifting degeneracies) of the cubic system by altering lengths and angles.  Example: Cube volume is $\ce{a^3}$.  Parallelepiped volume with all symmetries explicit is

http://en.wikipedia.org/wiki/Parallelepiped

http://www.rockhounds.com/rockshop/xtal/index.shtml
Good reading about crystal systems and their expression.
http://en.wikipedia.org/wiki/Space_group
As usual, nomenclature is a mess.
http://en.wikipedia.org/wiki/Hermann%E2%80%93Mauguin_notation
That's a winner.

Crystallographers often translate the unit cell within the lattice to make its contents  "nice." The rhombohedral system is particularly nasty for choosing the unit cell itself.  65 Sohnke space groups of 230 overall contain chiral formula units.  Racemic screw axes make that interesting, ${2_1}$, ${4_2}$, ${6_3}$.  Within the 65, 11 enantiomorphic space groups - themselves chiral without regard to unit cell contents - allow the same substance to crystallize in two mirror-image space groups.


Maltose Nomenclature Maltose has a 1,4 glycosidic linkage between two molecules of α-D-Glucopyranose.
My textbook suggests the final name of maltose as

  4-O-(α-D-Glucopyranosyl)-D-glucopyranose

I wanted to know what does "4-O-" stands for ?

I dont think so it provides information about the atoms which are linked and moreover feel that Wikipedia's nomenclature (not IUPAC) :

  α-D-Glucopyranosyl-(1→4)-β-D-glucopyranose

more detailized .

Maltose is 4-O-α-D-glucopyranosyl-D-glucose.  The "O" in "4-O" indicates the attachment of atom of the substituent.  The substituent in this case is "-α-D-glucopyranosyl".  It is attached to the oxygen atom that is attached to carbon 4 in the base molecule.  The base molecule in this case is "-D-glucose".  "4-α-D-glucopyranosyl-D-glucose" could be interpreted as a different structure where the α-D-glucopyranosyl substituent is bound directly to carbon atom number 4.

The "base molecule" of "-D-glucose" still has an unsubstituted anomeric OH group.  Thus, in aqueous solution, maltose will be an equilibrium mixture of at least these three isomers:

4-O-α-D-glucopyranosyl-α-D-glucopyranose
4-O-α-D-glucopyranosyl-β-D-glucopyranose
4-O-α-D-glucopyranosyl-D-glucose (open chain on the right-most "glucose" moiety in the name)

Thus, it doesn't really make sense to disambiguate α vs. β vs. open-chain forms of the "right-most" glucose ring in maltose, unless you need to refer to a particular conformation of maltose that matters for a certain reaction or condition.  However, the "left" ring is locked (in maltose) to an α-D-glucopyranosyl conformation because it does not have an anomeric OH group and rapid equilibration in water does not take place.


Synthesis of diclofenac I found a synthesis of diclofenac, and the last step with the addition of hydrazine and $\ce{KOH}$ seems interesting.

However, I couldn't find the mechanism behind this reaction, therefore I proposed one forming $\ce{N2}$.

Could you help me find the right mechanism regarding this type of reaction or correct mine?

This reaction is a Wolff-Kishner reduction [Organic Chemistry Portal, Wikipedia] which accomplishes the exhaustive reduction from a ketone to an alkyl group. Usually Wolff-Kishner requires forcing conditions, but I think this substrate could react under more mild conditions. The mechanism explains why.

The mechanism begins with the condensation of hydrazine onto a ketone (details not shown) to give a hydrazone. Under basic conditions, this hydrazone is deprotonated at nitrogen to give an anionic intermediate. In this case, the negative charge can be delocalized onto oxygen, resulting in an enolate structure. Typically, the negative charge is only shared between a nitrogen and carbon, so this substrate gives a particularly stable intermediate. Protonation of the enolate at carbon gives the first C-H bond necessary to form the product. A second deprotonation at nitrogen gives a similar flow of electrons to form another enolate structure, this time with cleavage of the C-N bond and release of nitrogen gas. Another C-protonation gives the lactam precursor to diclofenac. Cleavage of the amide with hydroxide (details not shown) gives the target.