text:
When to use iso and sec while naming compounds in alkyl halides?
I am having confusion on when to use iso- or sec- while naming the alkyl halides?

Preferably, you don’t use “iso” and “sec” at all when naming alkyl halides for two reasons:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the prefixes “isobutyl” and “sec-butyl” etc. are no longer recommended (“isopropyl” is still retained, but only for general nomenclature).
Preferred IUPAC names of alkyl halides are formed using substitutive nomenclature, i.e. using the prefixes “bromo”, “chloro”, “fluoro”, and “iodo” (e.g. iodomethane instead of methyl iodide).

If you cannot avoid using functional class nomenclature with traditional prefixes,

the “sec”-compounds are the simplest isomers that contain a secondary functional group, i.e. the halogen is bonded to the second carbon of the chain, e.g.
“sec-butyl chloride” is 2-chlorobutane,
“sec-pentyl chloride” is 2-chloropentane, etc.;
the “iso”-compounds are the simplest isomers that are branched at the end of the chain, i.e. all carbons except one form a continuous chain, e.g.
“isobutyl chloride” is 1-chloro-2-methylpropane,
“isopentyl chloride” is 1-chloro-3-methylbutane, etc.

http://www.chem.ucla.edu/~harding/IGOC/C/common_name.html According to this link,sec shoulld only be used for 4 member chains. Is this always the case?

iso- is used when the all carbons form a straight chain except one methyl at "2nd Carbon", like 2-methyl-R (where R refers to straight chain alkane)

And, sec- is used when any functional group is linked to a secondary carbon.

Then why is the name for iso-octane 2,2,4 trimethyl pentane rather than 2 methylheptane?

look up isoctane it will show up as  2,2,4 trimethylheptane... is that incorrect?

I am sorry, actually, 2-methyl-heptane should be the isooctane. But, since 2,2,4-trimethylpentane is the most important isomer (by application) of octane. Therefore, traditionally it is known as isooctane. Note: All other iso- compounds follows the general rule of 2-methyl-R. After all, exception is a common term in Chemistry ;)


text:
Very elementary question about valence electrons of ionic compound
In a university entrance question, I found this question:

What is the number of valence electrons of $\ce{Mg(OH)2}$? Answer = 24

I know that Mg has 2 valence electrons and O has 6. But I don't know what I must do for a complete ionic compound. If I add them up I get 2+6+6+1+1 = 16, but the answer of this question was 24. Is the answer right? If yes, then how to get it?

Your answer (16) looks right to me. True, this is an ionic compound; also, it is Saturday today. Both facts are equally irrelevant to the question. How 24 came about is a mystery to me.

You get 24 if you say Mg2+ has 8 valence electrons (its electron configuration is 1s22s22p6) instead of zero.
In that case, there are 8 in Mg2+, and 8 in OH-  (6 from O, 1 from H, and and extra 1 because of the -1 charge).  Therefore there are $8 + 2 \times 8 = 24$ per formula unit made of one Mg2+ and two OH-.
Some people do argue ionized Group 1A and 2A metals (above Period 2) have 8 valence electrons instead of zero, because technically there are indeed 8 electrons in the outermost electron shell and they are isoelectronic with the noble gas at the end of the previous row (e.g. in your case Mg2+ has the same electron configuration as Ne).
But I think most chemists would prefer to say Group 1A and 2A cations have zero valence electrons, that is, that the valence shell in these ions is the shell corresponding to their Period that used to have the outermost electrons.  In that case, the answer to your question is that there are zero valence electrons in Mg2+ and 8 in OH-, for a total of 16 in Mg(OH)2.  Note that this is slightly different from your reasoning, however.
An argument for saying Mg2+ has zero valence electrons is that if we draw the Lewis dot structure for the formula unit, I don't know anyone who would put any dots around the Mg.  On the other hand, an argument for saying that it has 8 is that we frequently argue that Mg forms the Mg2+ ion because that gives it a complete octet in the outer shell.  We rarely present the octet rule as requiring 0, 2 or 8 electrons in the valence shell.
I don't think the correct answer can be found in any mathematically rigorous sense, it depends on your preference in labeling, what you want to call "valence" electrons, and there are nontrivial arguments both ways.

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text:
Heat of Combustion

Methanol ($\ce{CH3OH}$) is regarded by many chemists as a possible liquid fuel for the future. The combustion of methanol occurs according to the equation:
$$\ce{2CH3OH(g) + 3O2(g) -> 2CO2(g) + 4H2O(l)}\quad\Delta H=-1506\ \mathrm{kJ\ mol^{-1}}$$
(a) Determine the heat of combustion of methanol, in $\mathrm{kJ\ g^{-1}}$.
(b) If $0.750\ \mathrm{mol}$ of methanol is mixed with $0.750\ \mathrm{mol}$ of oxygen and the mixture is ignited, how much energy will be released?

I have some working for this question, but not sure if it is right, because there is two moles of methanol in the equation.
for 18a)
$$1506\ \mathrm{kJ}\space\text{for}\space1\ \mathrm{mol}$$
$$1=\frac{m}{M}$$
$$1=\frac{m}{32.04}$$
$$m=32.04\ \mathrm g$$
$$\therefore1506\ \mathrm{kJ}\space\text{for}\space32.04\ \mathrm g$$
$$\therefore47\ \mathrm{kJ}\space\text{for}\space1\ \mathrm g$$
$$\therefore\space\text{heat of combustion}=47\ \mathrm{kJ\ g^{-1}}$$
for 18b)
must use $0.75\ \mathrm{mol}$ of oxygen
$$\therefore0.75\times\frac{2}{3}=0.5\ \mathrm{mol}\space\text{of methanol}$$
$$1506\ \mathrm{kJ}\space\text{for}\space1\ \mathrm{mol}$$
$$1506\times0.5\space\text{for}\space0.5\ \mathrm{mol}$$
$$753\ \mathrm{kJ}\space\text{for}\space0.5\ \mathrm{mol}$$
Therefore, $753\ \mathrm{kJ}$ is released for $0.75\ \mathrm{mol}$ of oxygen and $0.75\ \mathrm{mol}$ of methanol. Only $0.5\ \mathrm{mol}$ of methanol can be used.
Not entirely sure if any of this working is correct, any help is appreciated.

Part (a) is correct. In part (b), you have correctly identified the limiting reactant as oxygen, and hence only 0.5 moles of methanol will be combusted. Since 2 moles of methanol yields 1506 kJ, 0.5 moles would yield 376.5 kJ or 377 kJ (to 3 sig.figs.)

@Dr.J. Part (a) is not correct. The given enthalpy is for 2 mol of methanol. The molar enthalpy of combustion of liquid (!) methanol at 25 °C actually is 726.1 kJ/mol.

Correct answer for (a) should be 23.5 kJ/g as pointed out by Loong. Initial comment was made in haste.

Part A
First, the molar mass (MM) of methanol must be calculated. Sum the atomic masses given on the periodic table when multiplied by the instances of each atom in methanol.
$$
\begin{split}
\text{MM}_{\ce{CH3OH}} & = 1 \times 12.01~\mathrm{g} + 4 \times 1.01~\mathrm{g} + 1 \times 16.00~\mathrm{g} \\
& = 32.05~\mathrm{g}
\end{split}
$$
Employ dimensional analysis to change the units of the given quantity.
$$
\frac{-1506~\mathrm{kJ}}{1~\mathrm{mol_{rxn}}} \times \frac{1~\mathrm{mol_{rxn}}}{2~\mathrm{mol}_{\ce{CH3OH}}} \times \frac{1~\mathrm{mol}_{\ce{CH3OH}}}{32.05~\mathrm{g}} = -23.49~\frac{\mathrm{kJ}}{\mathrm{g}_{\ce{CH3OH}}}
$$
The heat of combustion of methanol in $\mathrm{kJ}~\mathrm{g^{-1}}$ is -23.49.
Part B
Identify this as a limiting reagent problem. When given equimolar quantities of both reactants, the limiting reagent is the reactant with the greater coefficient in the balanced chemical equation ($\ce{O2 (g)}$ in this instance). Mathematically, determine the number of moles of $\ce{O2 (g)}$ needed to fully combust $0.750~\mathrm{mol}~\ce{CH3OH}$. If this number exceeds the provided moles of $\ce{O2 (g)}$, then $\ce{O2 (g)}$ is the limiting reagent.
$$
\frac{0.750~\mathrm{mol}_{\ce{CH3OH}}}{1} \times \frac{3~\mathrm{mol}_{\ce{O2}}}{2~\mathrm{mol}_{\ce{CH3OH}}} = 1.13~\mathrm{mol}_{\ce{O2}}
$$
$$
1.13~\mathrm{mol}_{\ce{O2}} > 0.750~\mathrm{mol}_{\ce{O2}}
$$
Therefore, $\ce{O2 (g)}$ is the limiting reagent.
Knowing $\ce{O2 (g)}$ is the limiting reagent, employ dimensional analysis to find the quantity of energy released.
$$
\frac{0.750~\mathrm{mol}_{\ce{O2}}}{1} \times \frac{1~\mathrm{mol_{rxn}}}{3~\mathrm{mol}_{\ce{O2}}} \times \frac{1506~\mathrm{kJ}}{1~\mathrm{mol_{rxn}}} = 377~\mathrm{kJ}
$$
377 kJ of energy is released by the given reaction in the listed scenario.

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text:
Existence of tesserane and it's properties
Could there exist an allotrope of carbon, $\ce{C16}$ where the overall shape of the molecule is a three dimensional representation of a tesseract? And what are its (predicted if unknown/unsynthesised) properties if it does? I imagine this molecule would be highly strained, but I don't know if it shares a stability comparative to cubane's.
The only predicted properties I could find were chemicalize's calculations that it has a high solubility in water, of approximately 1200 mg/ml, and a polarizability of $\ce{17.83 A^3}$ (which has little meaning to me as I don't know typical values of polarizability).

I doubt such a compound is possible. It would seem to require making a much larger version of cubane that could fit cubane inside of it. Any arrangement large enough for that to occur would be too large for the outer carbon vertices to be bound to each other.

Could alkyne moieties make up the outside?


text:
Proof of the chemical potential formula
My textbook as well as Wikipedia both state the formula $\mu_A = \mu_A^0 + RT\ln\{A\}$ which gives the chemical potential $\mu_A$ of a chemical species $A$ as function of the standard chemical potential $\mu_A^0$ and the activity (relative to standard conditions) $\{A\}$.
This formula can also be used to derive the Gibbs free energy formula as $\Delta_rG_{T, p} = \Delta_rG_{T, p}^0 + RT \ln Q_r$ where $Q_r$ is the reaction coefficient.
Both formulas appear to be widely used in chemistry, but I have not found a decent explanation for why they are correct. The texts I have seen did not provide sources for the claim and omitted explanation.
Is this an empirical formula, or is it a consequence of statistical mechanics? I would like to find a source with a clear proof or explanation for the origin of this equation. I know that some will define the activity precisely using this equation, but then the question becomes why the activity so-defined corresponds to concentration for ideal solutions. That is, why in an ideal solution, we have $\mu_A = \mu_A^0 + RT\ln c_A$ where $c_A$ is concentration of $A$ relative to standard concentration.

My calculus and thermodynamics are too rusty to help. An answer explaining simplifications/assuming to move between the formulas would be nice besides the math itself.

In this Wiki article on the Nernst equation the initial section relates it to Gibbs free energy and then to your first equation in the the "Bolzman factors" section. https://en.wikipedia.org/wiki/Nernst_equation. So the connection between the two is reasonably apparent.

Many thermodynamics textbooks, including Atkins PhysChem derives this for ideal solutions.

The equations that relate chemical potential, standard chemical potential, and activity (or fugacity) are indeed direct definitions, or an algebraic manipulation away from the definition. If concentration is used by itself instead of activity, the equation may be known from experiment, is plausible from statistical mechanics, or again simply the definition of an ideal system. The standard chemical potential and activity are so defined that for a pure substance (or infinitely dilute mixture) activity reduces to some type of concentration depending on a chosen basis.

@Greg I had not seen Atkins' book. Based on a glance at the table of contents, it looks quite thorough!

@LinearChristmas yes, it appears like some will use that formula for concentration as a definition of ideal solution, and define activities to obey the same formula for real solutions too. I think I've also seen definitions of ideal solution in terms of the lack of intermolecular interactions, so it would be nice to find a derivation of this formula from this assumption. The answers below seem to be starting from the ideal gas assumption, which I guess might be equivalent?

From page 201 of the Atkin physchem book (https://www.rnlkwc.ac.in/pdf/study-material/chemistry/Peter_Atkins__Julio_de_Paula__Physical_Chemistry__1_.pdf). In Atkin's derivation all comes down to this relashionship: µJ = µJ(std) + RT ln aJ (eq.5.25). This lattest equation is derived from the ideal gas law and the assumption that the solute satisfies Henry’s law (amount of dissolved gas in a liquid is directly proportional to its partial pressure). I do not think this is satisfactory as it assumes that the solute can be both in solution and in the gas phase (immagine an ion Fe3+ for example).

(a) The idea of potentials is familiar from mechanical and electrical systems. In an electric field the work required to move a charge $q$ from one location with potential $\theta_1$ to one with $\theta_2$ is $q(\theta_2-\theta_1)$.  This expression has the form of a constant factor ($q$) times the change in potential. In a gravitational field the potential at height $h$ is $gh$  and so the work needed to lift a mass $m$ (the constant factor) from $h_1 \to h_2$ is $mg(h_2-h_1)= mg\Delta h$. We also know that these systems naturally move to position of lowest potential energy to reach equilibrium: stones do drop to the ground.
In a chemical system we can choose the mole number $n$ as the constant factor and then the molar free energy is the chemical potential which is usually given the symbol $\mu$.
The chemical potential controls mass equilibrium just as temperature controls thermal equilibrium. If there is a gradient of chemical potential then mass flows (or molecules rearrange). At equilibrium the chemical potential of all parts are equal. Similarly  energy flows in a temperature gradient until equilibrium is achieved and the temperature is uniform throughout. In this sense the chemical potential is considered as providing the 'force' to drive chemical systems to equilibrium.

(b) If we wish to transfer $n$ moles of a gas from a state with molar free energy $G_1$ to one of $G_2$ then the work done (other than expansion) is the change in free energy $n(G_2-G_1)$. Suppose now that we isothermally transfer $n$ moles of an ideal gas from a vessel with pressure $p_1$  to one with pressure $p_2$  the work involved is $nRT\ln(p_2/p_1)$ which is also $n(G_2-G_1)$ so that $(G_2-G_1)=RT\ln(p_2/p_1)$.
Usually we relate the pressures to one at a standard state, say 1 atm. then letting $G_1 \to G^\mathrm{o} $ and $G_2 \to G$ then $G= G^\mathrm{o}+RT\ln(p)$ where $p$ is $p/(1\text{atm})$ which is a dimensionless ratio. As the molar free energy is the chemical potential,
$$\mu=\mu^{\mathrm{o}}+RT\ln(p)$$
(The equation $(G_2-G_1)=RT\ln(p_2/p_1)$  can also be derived by starting with $dG=Vdp-SdT$, at constant temperature ($dT=0$) and for an ideal gas by using the gas law substitute for $V$, and integrating from $p_1\to p_2$.)

(c) More formally the chemical potential can be defined via the work done in reversibly compressing a gas composed of $i$ species with mole numbers $n_i$. If the force $F$ is applied to a piston that moves a distance $dx$ then
$$Fdx= -pdV +\sum_i \mu_i dn_i$$
where $\mu_i$ is the chemical potential of species $i$ and is defined by this equation. This means that the chemical potential is the (reversible) rate of change of internal energy with mole number while keeping other variables ($S,V$) constant, thus since $dU=TdS-pdV+\sum_i \mu_i dn_i$, where $U$ is the internal energy, then $\displaystyle \mu_i= \left(\frac{\partial U}{\partial n_i}\right)_{T,S,n_j}$. (The subscript $n_j$ means that other mole number are held constant.)
It is more common, however, do define the chemical potential in terms of the Gibbs free energy and where $T,p$ are held constant.
For a pure substance your textbook will derive the equation
$$dG=Vdp-SdT$$
but if there is a mixture the number of moles can vary then it is necessary to account for  energy changes due to this by adding a term to the energy and doing this produces
$$dG=Vdp-SdT+\sum_i \mu_i dn_i$$
which is sometimes referred to as the fundamental equation of chemical thermodynamics, and then the chemical potential is defined as
$$\displaystyle \mu_i= \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_j}$$
[discussion in (a) and (b) based on arguments in Lewis & Randall, 'Thermodynamics' publ McGraw-Hill.]

The Sackur-Tetrode equation can be written as $$S^{Trans}=R\ln \left( \frac{e}{C\Lambda^3} \right)$$ where $C$ is the concentration and $\Lambda$ the thermal wavelength. You can rewrite this as $$S^{Trans}=R\ln \left( \frac{e}{\Lambda^3 C^\circ\frac{C}{C^\circ}} \right) = R\ln \left( \frac{e}{C^\circ\Lambda^3} \right)-R\ln\left( \frac{C}{C^\circ} \right) = S^{\circ,Trans}-R\ln\left( \frac{C}{C^\circ} \right)$$ where $C^\circ$ is an arbitrary standard concentration. When you combine it with enthalpy and the other entropy terms you get $$G = G^{\circ}+RT\ln\left( \frac{C}{C^\circ} \right) $$


text:
What reactions will occur at the anodes and electodes of two iron electrolytic cells?
What reactions will occur at the anodes and electrodes of two iron electrolytic cells: both with iron metal electrodes, one with an iron(II) sulphate electrolyte, the other with an iron(III) sulphate electrolyte?
I set up an experiment with both types of cells connected in series, which yielded the following qualitative data:

Both cathodes saw vigorous bubbling--to me implying hydrogen gas evolution.
Also, there appeared to be a darkish iron plating on the cathode, implying two reactions occurring here.
Some bubbling at anodes.
Large quantity of dark green precipitate formed in Iron(II) electrolyte, I would assume $\ce{Fe(H2O)4(OH)2}$ precipitate.
A deepening orange in iron(III) solution and a relatively small quantity of orange precipitate, I would assume $\ce{Fe(H2O)3(OH)3}$ precipitate.
pH of electrolytes remained constant (measured with a pH probe).
Conductivity of iron(III) solution remained constant, whilst conductivity of iron(II) solution--which was almost 20 times higher--steadily fell.

It seems to me that similar things are happening at the respective cathodes of the iron(II) and iron(III) cells: both hydrogen gas and iron metal are evolving. The hydrogen gas evolution momentarily increases pH, pushing equilibrium of iron complexes towards the precipitates--thus keeping pH constant.
I'm less sure about the anodes. But I think iron is entering solution as Fe(II) and Fe(III), respectively, with some oxygen evolution.


text:
Why do we multiply two wavefunctions in valence bond theory instead of adding them?
The combined wavefunction of a hydrogen molecule in valence bond theory is the product of two wavefunctions, while in molecular orbital theory it is the sum of two wavefunctions.

Actually, the MO theory wavefunction can also be written as a product. While the individual molecular orbitals are just weighted sums of the atomic orbitals, the wavefunction can, to a first approximation, be written as a Slater determinant of the occupied orbitals.
For your example of the hydrogen molecule, this would mean taking the antisymmetrized product of the occupied $\sigma$ spin orbitals. (Note, if you haven't encountered this before, spin orbitals are a spatial and spin description of a single electron, rather than the commonly encountered spatial orbitals which describe two electrons of opposite spin.


text:
Why does molten salt turn black?
Lately I have been seeing quite a few videos on YouTube about molten salt, specifically table salt (NaCl). One of the things that interests me about it is that regardless of what melting technique or type of crucible, the salt always ends up black afterwards.
Is the issue that the salt is highly reactive and is pulling chemicals out of the crucible regardless of what the crucible is made of? Wouldn't that mean that different crucibles would impart different colorations on the end result? That is the speculation I've seen on the comments of many videos (for example), but nothing definitive.
Or is it perhaps an optical change due to the change of the crystal structure? Or maybe something in the salt itself oxidizing or reacting with something in the air?

It may not be black.  It may only look that way by reflecting/refracting light away from the camera.

@OscarLanzi That's part of what I was asking regarding possible optical changes. Large blocks of naturally-occurring NaCl are translucent white but it also generally doesn't form via melting in nature.

referencing a few of these videos would be helpful.

@A.K. I had linked to one of them; unfortunately most of them don't show a good view of the result or are built as clickbait or are using very dirty crucibles or are using "light salt" or the like.

Part of the color change might be due to creation of F-centers, crystal defects where there are electrons filling the place of an atom. A corresponding H-center (for halogen interstitial) may also be created for each F-center.
That said, such color changes tend to be rather pale, and as you state, $\ce{NaCl}$ heated to melting in an iron crucible may appear dark or even black. This is likely due to dissolution and oxidation of the crucible. The molten salt attacks the metal at grain boundaries and also dissolves protective oxide coatings, allowing oxygen from the air to attack the underlying metal. Even in a closed system, such as a molten (mixed) salt thermal energy storage system, corrosion is a problem.
The color change could also be due to trace impurities in the "table salt". For example, sodium or potassium ferrocyanide, $\ce{Na4[Fe(CN)6]·10H2O}$, is intentionally added to keep the salt free-flowing despite humidity changes. This might react at high temperatures to form a deeply-colored compound.
To find out what is causing the discoloration, you might try heating a small sample of high purity $\ce{NaCl}$ in a ceramic or platinum (a bit pricey) crucible, and you could analyze heated table salt for dissolved iron compounds.
BTW, F-centers and other crystal defects can be created intentionally with optical lasers and x-rays. If memory serves, there was an article in Scientific American's Amateur Scientist about creating F-centers using an electrostatic generator.

Quite logical , solid state in actikn


text:
How does H2SO4 dissociate?
Apologies for this extremely basic question, I'm just beginning with Chemistry so please don't be too harsh on me.
My book says that sulfuric acid, $\ce{H2SO4}$, dissociates in its ions following this reaction: $$\ce{H2SO4 -> H2^+ + SO4^{2-}}$$
My question is, why can't the dissociation reaction happen like this:
$$\ce{H2SO4 -> 2H^+ +SO4^{2-}}$$
I know hydrogen is a diatomic gas, but here I don't know if H will dissociate as a gas or as a liquid (since $\ce{H2SO4}$ is a liquid, not a gas).
I'm trying to learn, thank you for your understanding and your time.

It can and does happen as you suggested. Your book is wrong. Hydrogen the diatomic gas is simply not here.

Thanks, but then how do I know when I will have $H_2^+$ and when $2H^+$?

(The book was written by my teacher, I suppose he made a mistake in this exercise)

I would agree that $\ce{H2^+}$ is not present. The overall reaction is the dissociation of both hydrogen ions, but I'd suggest that the dissociations happen one at a time. Both dissociations would be very fast, but not instantaneous.

@Jose On your current level of theory, this is pretty simple: you always have $\ce{2H+}$ and never $\ce{H2+}$. You might want to ask this question again, say, after a year.

https://chemistry.stackexchange.com/questions/67235/can-common-ion-effect-suppress-the-dissociation-of-water-molecules-in-acidulat

https://chemistry.stackexchange.com/questions/65502/why-do-diacidic-and-triacidic-alkalis-dissociate-in-one-step

https://chemistry.stackexchange.com/questions/23698/how-do-the-strong-acids-dissociate

$\ce{H2SO4}$ is one of common strong acids, meaning that $\ce{K_{a(1)}}$ is large and that its dissociation even in moderately concentrated aqueous solutions is almost complete.
Arrhenius dissociation:
$$\ce{H2SO4 <=> H+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$
Brønsted-Lowry Dissociation:
$$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$
This accounts for the vast majority of protons donated by the acid. However, since it is diprotic, you may want to take into account the second dissociation, which is technically weak but has a larger $\ce{K_a}$ than many weak acids.
Arrhenius 2nd Dissociation:
$$\ce{HSO4- <=> H+ + {SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}=1.2\times10^{-2}$$
Brønsted-Lowry 2nd Dissociation:
$$\ce{HSO4- + H2O <=> H3O+ +{SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}= 1.2\times10^{-2}$$
This second dissociation may need to be taken into account for some calculations, but it is negligible in concentrated solutions.

There might be only 6 strong acids mentioned in your book, but it's by no means total number. Also this Also this Arrhenius/Bronsted division is kinda silly IMO. Both H+ and H3O+ are only symbolical and don't truly reflect hydration of proton.

@Mithoron My teacher defined “strong” acids as those with a “large” Ka (as in too big to be measured). Do you know of a list of the rest?

There is no list as their number is limitless. Add -SO3H group to one of millions organic groups and you have strong acid, voila!

@Mithoron Good to know! I have not taken organic chemistry yet, so I was not aware of this. In the future, you should try to find a better way of critiquing than a downvote and a reprimand. Perhaps an edit to the post in question and a comment explaining it? Just a thought — and I will edit this post to reflect your insight.

Sulfonic acids are just an example. There's also a lot of inorganic acids, just less known, and their number is also probably limitless.


text:
Find product of the oxidation of 2-methylnaphthalene with chromium trioxide

This is a question from an organic chemistry book.
I did not understand the solution given in that book.
Here's the solution given in that book,

I know that CrO3 oxidizes methyl group of toluene, but here
Why are alpha positions more reactive and alpha positions of which functional group have they mentioned?

Fun fact: this synthesis is/was the commercial method used by BASF to make the product, menadione, which is often found as a supplement in animal feeds (patent DE092952709). Apparently the use of stoichiometric chromium is worth it.
Why are alpha positions more reactive? There are two potential questions here, I'll try to answer both.
1) Why does methylnaphthalene react differently from toluene?
The naphthalene ring is generally more reactive than benzene because reaction intermediates do not fully destroy aromaticity (see below image, left)
2) Why is the alpha position more reactive than the methyl group?
This question of reactivity is tricky because other oxidants and reaction conditions will indeed oxidize 2-methylnaphthalene into 2-naphthoic acid. In general, electron-rich naphthalenes are able to oxidize into naphthoquinones, even just by exposure to $\ce{O2}$. If we assume (big assumption!!!) that these reactions to go through radical mechanisms, then we are either oxidizing the naphthalene pi system (left) or abstracting an H* (right). There's various arguments to be made about which pathway is more stable based on the extent of conjugation, and obviously that will depend on the reagent and solvents in use as well.

Alpha positions of which functional group have they mentioned?
The alpha nomenclature they use here is actually referring to the two different positions in naphthalene:


text:
Reaction involving opening of epoxides
The following question was asked in an exam today:

I chose the option (a), but the correct answer given was (b). Following is the explanation why I chose (b):

Step 1 (Reaction with $\ce{PBr_3}$): $\ce{Br}$ substitutes $\ce{OH}$. Hence, the product is $\ce{CH(CH_3)_2Br}$
Step 2 (Reaction with $\ce{Mg}$ in the presence of dry ether): $\ce{Mg}$ gets inserted between $\ce{CH(CH_3)_2}$ and $\ce{Br}$. Hence, the product is $\ce{CH(CH_3)_2MgBr}$
Step 3 (Opening of epoxide and attacking of grignard reagent): First, the $\ce{CH(CH_3)_2MgBr}$ will break into $\ce{CH(CH_3)_2^-}$ and $\ce{MgBr^+}$. Then, $\ce{MgBr^+}$ will attack $\ce{O}$ of the epoxide. After that, $\ce{O}$ will take its electron away from 2nd $\ce{C}$ (referring to the catalyst used) due to the greater stability of secondary carbocation.
Further, $\ce{CH(CH_3)_2^-}$ will attack the carbocation and hence form the product $\ce{CH(CH_3)_2-CH(CH_3)-CH_2-OMgBr}$
Final step (Hydrolysis): On hydrolysis, $\ce{CH(CH_3)_2-CH(CH_3)-CH_2-OH}$ will be our product. Phew!

Where I am wrong and what should be the correct mechanism?

Steric hindrance is the major factor

The carbocation stability argument isn't so valid under basic conditions as the extent to which the C-O bond is elongated in the TS is less significant. Sterics tend to dominate

@NotEvans., How do you know that the medium is basic?

Grignard reagents are pretty basic...

As you are dealing with Grignard reagents, the carbocation argument is inadequate since the reaction is in basic media (carbocations are brønsted acids, so they would be easily deprotonated to produce an alkene). In fact, this mechanism is concerted (everything happens simultaneously, not in 2 distinct steps such as in the formation of a carbocation followed by a nucleophilic attack).

The Grignard reagent could attack either the methylene $\ce{(CH_2)}$ carbon or the methylidyne $\ce{(CH(CH_3))}$ carbon. The methylene carbon is attacked preferentially because its substituents are smaller than the methylidyne ones (2 hydrogens against one hydrogen and a methyl group). So steric hindrance is the key factor for predicting the product correctly in this question, not carbocation stability.
ps.: I drew a specific enantiomer just for visualization, but in this case, the absolute configuration doesn't matter.

When you are using Grignard reagent for opening of epoxides, the Grignard reagent will attack less sterically hindered carbon(i.e. SN2 attack will take place). This is because when we use Grignard reagent, the reaction takes place in basic medium. Therefore in step 3 it will attack the 1st carbon instead of the 2nd one.

Grignard reagents are generally bulky in nature therefore they choose to attack the less sterically hindered carbon which in this  case is the methylene carbon .
Furthermore you should remember that presence of acidic media will destroy the Grignard reagent . The carbocation  stability is not considered only because it is a bronsted lowry acid.


text:
Formula of Manganese (IV) Ion
When determining the formula of manganese (IV) oxide, since it’s made up of $\ce{Mn^4+}$ and $\ce{O^2-}$ ions, why is the formula $\ce{MnO2}$ and not $\ce{Mn2O4}$?
Any help would be greatly appreciated

Both are right, but MnO2 is simpler.

Good question, be careful what do you mean is it solid MnO2 or manganese(VI) oxide in the gas form. I have no idea if manganese (VI) oxide can form a gas but I know that chromium(VI) oxide in the gas state is in fact Cr3O9 rather than CrO3

For ionic compounds there is really no such thing as a "molecule" of the compound. In the solid state the ions exist in a 3D crystalline structure and there is no unique way to associate a particular $\ce{Mn^{4+}}$ ion with a particular pair of $\ce{O^{2-}}$ ions. Therefore the simplest integer ratio is used as the "molecular formula."

Quite important misconception here is assumption of presence of discrete ions in MnO2. It's not a good approximation as it's primarily covalent.

When writing an ionic formula, the simplest ratio is the 'ionic formula.'
Any sample of manganese (IV) oxide will have many, many, many manganese and oxygen ions. The 'criss cross' method of translating the ion charge down to the opposite subscript gives you a starting point. For an ionic substance, you then reduce this ratio to the lowest whole number ratio.
To get a proper formula, the ionic charges have to neutralize, so $\ce{1 Mn^4+}$ ion neutralizes $\ce{2 O^2-}$ ions.
This does not work for molecular formulae, which have a structure based on atoms connected with covalent bonds.


text:
Balancing Redox reaction : with 2 species being oxidized
I have a redox reaction I found from a chemistry book from the 1970's.
Two chemical species are oxidized, the Chromium and Iodine.
$\ce{CrI3 + KOH + Cl2 -> K2CrO4 + KIO4 +KCl + H2O}$
Hope someone can help on how to balance this.

Possible duplicate of What is the simplest approach to balance a complex reaction equation?

Not that this was all that complex, really.  Whatever ...

Your best bet is to consider the entire reactant $\ce{CrI3}$ as the reducing agent.
The chromium (III) iodide starts out as that compound and gets oxidized to a combination of chromium (VI) (in the chromate) and iodine (VII) (in the periodate) with a ratio of chromium to iodine matching the original compound:
$\ce{CrI3 -> Cr^{VI} + 3 I^{VII}}$
Clearly the originally neutral reactant has been oxidized to atoms having a combined oxidation state of $+27$, so we balance this reaction with $27$ electrons:
$\ce{CrI3 -> Cr^{VI} + 3 I^{VII} + 27 e^-}$
Each chlorine molecule takes up two of the electrons so the redox ratio must be $\ce{2 CrI3 : 27 Cl2}$.  Put those in and balance the remaining compounds as you normally would.

Hi Oscar, you are correct i believe. I have the answer for this and the lone Cl(2) has mole coefficient of 27. Thanks for these details. I am trying to do this via oxidation number method, so not doing the explicit electron counting. So you are assuming the chromium and Iodine are one compound and it is neutral, but it breaks apart into two things that are charged and are treated together in terms of adding their charges.

Yup, that's how it works.  In effect the relevant "oxidation number" in tbe oxidation half reaction is the combination of chromium and iodine (which adds up to 0 before but +27 after the reaction), not either element alone.

I have done disproportionation reactions, its kind of similar, but this one is making my head spin a bit.


text:
Is there a more direct way to show this identity for the projection operator?
The following question I jotted down from an online quantum chemistry pdf.

Show that
$$i\hbar \frac{\partial P_j}{\partial t}= [H,P_j]$$

in which $P_j$ is the projection operator and $H$ is the Hamiltonian. We are given the usual time-dependent form of Schrodinger's equation.
What I did: I supposed that this was an application of Ehrenfest's theorem because it involved a commutator and derivatives. $P_j$ does not explicitly involve time and so Ehrenfest's theorem reduces to$^*$
$$\frac{d}{dt}\langle\psi|P_j|\psi\rangle=\frac{1}{i\hbar}\langle\psi|[H,P]|\psi\rangle $$
But assuming I think w.l.o.g. that $\psi$ is normalized we can pre-multiplyby $|\psi\rangle$ to get (using $|\psi\rangle\langle\psi|=1$)
$$i\hbar\frac{d}{dt}P_j|\psi\rangle~= [H,P_j]|\psi\rangle$$
If this works (I don't assume that), my question is whether there is a direct way to do this, using the definition of $P_j$, and if so whether it would not also be an appeal to this theorem?
$^*$ Using eq. (9) from this note.

Before I go into how this result can be derived, I want to point out a mistake in your attempt. $|\psi\rangle\langle\psi|\neq1$, but rather $\sum_i|i\rangle\langle i|=1$, where the sum of is over a complete set of states. Basically, this result only holds when you sum over all the basis vectors that span the space you are studying, not just for any single vector in the space. Its easiest to understand this in analogy to more conventional vectors. Assuming a column vector $v$ is normalized, than $v^\dagger v=1$, which is the inner/dot product of the vector with itself. This is analogous to $\langle\psi|\psi\rangle=1$ in bra-ket notation. If we reverse the order of the product, $vv^\dagger=\mathbf A$ where $\mathbf A$ is some matrix. This operation is called an outer product and is akin to $|\psi\rangle\langle\psi|$. As it turns out, for a complete orthonormal basis, the sum of these outer products has to equal the identity matrix $\mathbf1$. We can make sense of this by recognizing that the sum is a sum of projectors onto all the basis vectors, which in QM might be all of the possible eigenstates of the system. Your wavefunction has to be some linear combination of these basis vectors, so if your project your wavefunction onto all the eigenstates and then sum those projections, you should get your original wavefunction back, hence the sum of the all the projectors is the identity matrix.
For the actual derivation, the result proceeds directly from the time dependent Schrodinger equation (TDSE) $$i\hbar\frac{\partial\psi}{\partial t}=H\psi\to i\hbar\frac{\partial|{j}\rangle}{\partial t}=H|j\rangle$$
where in the second half of the equation, I have written the equation for the state $j$ that the projection operator references. Writing the projection operator explicitly as $P_j=|j\rangle\langle j|$, we can rewrite your original
expression as $$i\hbar\frac{\partial(|{j}\rangle\langle j|)}{\partial t}=i\hbar\bigg[\frac{\partial|{j}\rangle}{\partial t}\langle j|+|j\rangle\frac{\partial\langle{j}|}{\partial t}\bigg]$$ where the equality is due to the product rule. The first term contains $i\hbar\frac{\partial|{j}\rangle}{\partial t}$ which we know from the TDSE equals $H|j\rangle$, so the first term becomes $H|j\rangle\langle j|=HP_j$. To find the other term, we take the complex conjugate transpose of the TDSE to obtain $-i\hbar\frac{\partial\langle j|}{\partial t}=\langle j|H$, which leads to the second term being $-P_jH$ and thus $$i\hbar\frac{\partial P_j}{dt}=[H,P_j]$$

I realized later that $|x><x|u>$~$|u>$ (proportional) and not =1 if we are not running over the basis set.  Not so minor! I see the product rule step. The last step is not obvious to me but when I convince myself I will accept this answer, which was very helpful!

Just out of curiosity, is there any sense to be made of $|x><x|A|x>$ in this context? If $|x><x|=k|x>$ for constant k we get (?) $k|x>A|x>.$ Is it well defined?

@daniel I don't think that 2nd expression has any meaning because I'm not confident $|x><x|=k<x|$ makes sense. However your first equation is just the vector for state x multiplied by the expectation value of A for that state, so that would just be the vector times some constant.

Here is a question that did not occur to me earlier. Where you write the original expression as a product, which makes sense formally, why is the l.h.s. not equal to the first term on the r.h.s.?

@daniel Do you mean $|\psi><\psi|\neq1$? I will add a little about that point.

No that was clear, thanks. Just the part: $ih\partial |j><j|/\partial t = ih[\partial |j>/\partial t <j|...]$...it seems that the LHS =this first term. The equality after "We can re-write your original expression as..."

@daniel that's just due to the product rule $\frac{d(f(t)g(t))}{dt}=f'(t)g(t)+f(t)g'(t)$


text:
Help on alternative terpene names and naming schemes
I am a software developer and currently parsing data involving terpene profiles of plant samples. The data in question uses different ways of naming the same terpenes in-between samples and since this is not my field I would like to ask:

if there is an online index showing all possible names for a specific terpene or
if there is a software library for normalizing those names or
if somebody could simply tell me which ones of the attached list are the same terpene and what the recommended naming scheme would be

The list of terpenes I am concerned with is as follows:

3-Carene
α-Bisabolol
α-Humulene
α-Pinene
α-Terpinene
β-Caryophyllene
β-Myrcene
β-Ocimene
β-Pinene
Camphene
Caryophyllene
Caryophyllene Oxide
CaryophylleneOxide
cis-Nerolidol
δ-Limonene
DLimonene
Eucalyptol
Geraniol
Guaiol
Humulene
Isopulegol
Linalool
Ocimene
PCymene
p-Cymene
Terpinolene
trans-Nerolidol
TransNerolidol 1
TransNerolidol 2
trans-Ocimene
γ-Terpinene

(Some of those names are obvious duplicates of others, but I just wanted to give you the full, unaltered list here)
If anyone could point me in the right direction I would be very grateful!

Chemical naming is a disaster! For example just look at the 25 different names (non-codes) for Humulene on pubchem: https://pubchem.ncbi.nlm.nih.gov/compound/5281520#section=Synonyms
It is very common for databases and publications to use non-standard names and thus it is a real problem in the field of chemistry.

Chemical nomenclature is officially defined by IUPAC (International Union of Pure and Applied Chemistry). You can find most IUPAC names for these terpenic compounds by searching their popular names on Wikipedia, they are quite long and confusing terms though, and can only be interpreted if you have a certain knowledge in organic/inorganic chemistry.
What you are probably more interested in are databases with hundreds of thousands of compounds with unique identifying numbers that are unequivocally assigned to a specific substance. Two of the largest are CAS Index (https://en.wikipedia.org/wiki/CAS_Registry_Number) and ChemSpider (https://en.wikipedia.org/wiki/ChemSpider). I believe working with numbers will facilitate your work a lot, so I would suggest that you look for one of these databases (but be careful, each database uses a different numbering system so you should pick one and stick to it).

Don't forget Beilstein: https://en.m.wikipedia.org/wiki/Beilstein_database


text:
Is there a single or double bond between S and O in propanethial S-oxide?
I am doing some research on propanethial S-oxide, a chemical produced by onions. I am trying to find out whether they have a single or double bond between the sulfur and oxygen atom. I have seen both from different sources, simply search it up on Google Images. Is this because there is resonance going on? Also is this considered a sulfinyl functional group (I am guessing no because it cannot be on the end of a chain)?
In addition, is the C double-bond S part in this molecule considered a thial functional group? A Google search indicates that the atom bonded to the sulfur must be a hydrogen for it to be a thial. If not, what type of functional group (if any) exists between the C, S, or O?

The $\ce{S-O}$ bond in sulfoxides is best considered as a single covalent bond with positive charge on $\ce{S}$ and negative on oxygen. The oxygen is significantly nucleophilic (see for example the Swern oxidation mechanism 1) and the sulfur will stabilise an anion on the alpha carbon.
$\ce{C=S}$ has significant enol content and is better considered $\ce{HS-C=CR}$. The $\ce{S}$ is significantly nucleophilic and is easily oxidised to the ene-disulfide $\ce{RC=C-S-S-C=CR}$.

Or, you could say the bonding is complicated and not easily described as a simple single or double bond. Simple models don't always do a good job of describing complex bonds.


text:
What is the mechanism for the oxidative cleavage of diketones via permanganate under basic/hot conditions?

 The products of the reaction are carboxylates.
What exactly would cause the alkyne to cleave?

In J. Org. Chem. 1979, 44 (15), 2726–2730, the authors Lee and Chang have investigated the permanganate oxidation of alkynes to α-diketones. In addition, they investigated the oxidative cleavage of several α-diketones to carboxylic acids. In the case of hexadecane-8,9-dione, permanganate oxidation under at least three sets of conditions provided n-octanoic acid 8 (major) and n-heptanoic acid 7 (minor). They suggested that the oxidation of the dione 1 leads only to octanoic acid 8 via intermediate 4 while oxidation of enol 2 affords 2-oxononanoic acid 6 and ultimately octanoic acid 8 via oxidative decarboxylation. In addition, oxidation of heptanal 5 affords heptanoic acid 7. The three solvent systems employed were acetone/aqueous HOAc, aqueous H2SO4 and water. Starting material was recovered in each experiment. Tetradecane-7,8-dione showed similar behavior.
For the mechanism of permanganate oxidation of alkynes to α-diketones, see the accompanying post by @Raul Luciano.

Pinging with @ in posts does nothing. In comments it's quite limited. In chat is more useful, still sometimes "superping" from mod would be needed to actually ping someone. Just saying ;)

@Mithoron(ping?): I was trying to draw the attention of Raul Luciano. So how do I accomplish this goal ... in simple terms.

You'd need to comment under his answer. BTW this time @.... wasn't necessary as there was no other users' comments under this post.  Probably it wouldn't work otherwise as you didn't put space after user name and before "(" Here's related post https://chemistry.meta.stackexchange.com/questions/3889/pinging-in-comment-to-suggest-answering-question

Alkynes react just as alkenes in the oxidative cleavage with $\ce{KMnO_4-}$ (the mechanism is the same). The difference is that since there is a triple bond, $\ce{MnO_4-}$ doesn't break the molecule apart like it does with alkenes (only if you use high temperature and high base/$\ce{MnO_4-}$ concentration), instead it forms a diketone. What causes the alkyne to cleave is the oxidant ion $\ce{MnO_4-}$ that is reduced to $\ce{MnO_2}$.

This fails to answer the question asked.


text:
Why some reaction have no transition state?
All stationary point are local minimum, if any parameters changed the energy should be increase. There is no way for a stationary point that energy go straight down to reach another.

I was told that radical reaction have no transition state, but why? What kind of reaction also do not have transition state?
Is there any way to confirm there is no transition state using quantum chemical calculation method?

Radicals are very reactive. They just rapidly collide in the environment where chemical reaction are occuring they just want to one electron by forming bonds even at the price of breaking other bonds.
Also they are so reactive that they can't bear a state where the bond is breaking and forming at the same state like that in transition state.

There is nothing special in having an activation barrier between reactants and products although, obviously, very many reactions do. But some reactions, such as electron transfer, are observed experimentally to have no barrier which happens when the product potential crosses that of the reactant at its minimum (i.e. between the 'normal' and 'inverted' regions).  In fact  electron transfer is an interesting case as it is possible to control the barrier height by changing the redox potential. Some atom diatom reactions such as H+F2 also have very very small barriers.  Whether you want to call the crossing point a transition state when it has a zero barrier is a moot point. The behaviour is more like water going over a waterfall, so limited by the transport of the wavepacket along the reaction path, rather than sometimes being reflected at the barrier and sometimes crossing.


text:
Extraction of citric acid from lemon
Now i am extracting citric acid from lemon juice and i have a problem in adding sulphuric acid to calcium citrate because i don't know how to calculate the proper amount sulphuric acid to add to calcium citrate. Could anyone show me how to calculate the amount of sulphuric acid? And if anyone of you know the procedure for this extraction; please share me. Thank you!

When you add calcium hydroxide to the sodium citrate solution, you end up forming tricalcium dicitrate

In order to neutralize 1 mol of tricalcium dicitrate and form 2 mol of citric acid you need 3 mol of H2SO4 (each mol of tricalcium dicitrate have 6 mol of deprotonated oxygens and each mol of sulphuric acid provides 2 mol of H+).
So the balanced equation should be:
$$\ce{Ca3(C6H5O7)2 + 3H2SO4 <=> 2C7H8O7 + 3CaSO4-}$$
So, the proportion of tricalcium dicitrate to sulphuric acid is 1:3.
To fully neutralize the citrate to citric acid, for each mol of citrate you need 3 mol of sulphuric acid, or mathematically:
$$\ce{3n_{\ce{Ca3(C6H5O7)2}} = n_{\ce{H2SO4}}}$$
You know the molecular weight of both and the mass of tricalcium dicitrate that you have, now you can calculate how much sulphuric acid you need from the equation above.
There are several videos on youtube showing how to extract citric acid from fruits, I particularly like this one: https://www.youtube.com/watch?v=FMtayizdFiw


text:
How to calculate number of each type of atom in a residue in PyMOL?
I have a 3D model of PDB 4UDC (dexamethasone bound to human glucocorticoid receptor), and I'd like to calculate how many atoms of each element are in the dexamethasone molecule (which has the residue name DEX) in this model, just to make sure it is correct (as I have added hydrogen atoms with h_add and if I missed a double bond it will be off). So how do I determine how many atoms of each element are present in this residue?
The sort of output I'd be looking for, from whatever answer you provide, would be something like:
C22H29FO5
if I got the bond order and everything else right. Just so you know, as I would dislike it if you thought I meant if we were to count all the atoms up altogether (which would be $22+29+1+5=57$).

Why not just look at the pdb in an editor and search for DEX. Then cut this out and paste etc.

Not super elegant, but this should work:
select resn dex and symbol C

Pymol will respond,
Selector: selection "sele" defined with 22 atoms.

Repeat for H, F, O.

Well it's a nice and simple method, it's just I'd like to be able to write a program that would represent it C[No]H[No]F[No]O[No] (where No is the number of each and square brackets, [], can be omitted). Is that possible or will have I just have to deal with this output format?

I'm sure it can be done with a few lines of Python, but not by me.

Download the data and using say, Visual Studio Code (which is free) locate DEX in the coordinates by scrolling down. If you open the edit to this post the text is formatted properly)
HETATM 2172  C1  DEX A1778       1.049  40.118  -7.466  1.00 33.40           C
HETATM 2173  C2  DEX A1778       0.936  40.967  -6.455  1.00 33.43           C
HETATM 2174  C3  DEX A1778      -0.392  41.247  -5.876  1.00 35.10           C
HETATM 2175  C4  DEX A1778      -1.589  40.609  -6.463  1.00 32.73           C
HETATM 2176  C5  DEX A1778      -1.450  39.719  -7.454  1.00 34.13           C
HETATM 2177  C6  DEX A1778      -2.723  39.089  -8.030  1.00 33.20           C
HETATM 2178  C7  DEX A1778      -2.784  39.329  -9.551  1.00 34.15           C
HETATM 2179  C8  DEX A1778      -1.457  38.973 -10.271  1.00 33.89           C
HETATM 2180  C9  DEX A1778      -0.254  39.730  -9.617  1.00 34.09           C
HETATM 2181  C10 DEX A1778      -0.096  39.369  -8.094  1.00 32.77           C
HETATM 2182  C11 DEX A1778       1.064  39.565 -10.432  1.00 34.52           C
HETATM 2183  C12 DEX A1778       0.895  39.783 -11.960  1.00 33.48           C
HETATM 2184  C13 DEX A1778      -0.244  38.953 -12.593  1.00 34.24           C
HETATM 2185  C14 DEX A1778      -1.548  39.301 -11.772  1.00 33.89           C
HETATM 2186  C15 DEX A1778      -2.675  38.685 -12.622  1.00 35.44           C
HETATM 2187  C16 DEX A1778      -2.182  38.918 -14.084  1.00 35.12           C
HETATM 2188  C17 DEX A1778      -0.695  39.386 -14.019  1.00 35.80           C
HETATM 2189  C18 DEX A1778       0.119  37.433 -12.584  1.00 32.13           C
HETATM 2190  C19 DEX A1778       0.242  37.886  -7.780  1.00 25.76           C
HETATM 2191  C20 DEX A1778       0.187  38.810 -15.131  1.00 35.76           C
HETATM 2192  C21 DEX A1778       1.535  39.421 -15.434  1.00 35.52           C
HETATM 2193  C22 DEX A1778      -3.098  39.926 -14.771  1.00 34.29           C
HETATM 2194  F1  DEX A1778      -0.577  41.091  -9.664  1.00 36.57           F
HETATM 2195  O1  DEX A1778      -0.478  41.974  -4.905  1.00 39.49           O
HETATM 2196  O2  DEX A1778       1.689  38.298 -10.213  1.00 35.91           O
HETATM 2197  O3  DEX A1778      -0.681  40.815 -14.192  1.00 38.27           O
HETATM 2198  O4  DEX A1778      -0.157  37.869 -15.807  1.00 37.29           O
HETATM 2199  O5  DEX A1778       2.356  38.523 -16.169  1.00 35.99           O

@pentavalentcarbon Many thanks for edit, I wasn't sure how to format.

Say you have selected the dex residue, and named the selection 'Dex'.
You could've done it using
select Dex, resn dex

Than you can get the count of each atom using this:
from collections import Counter;
counts = Counter([atom.symbol for atom in cmd.get_model('Dex').atom])
print(counts)

This will print
Counter({'C': 22, 'H': 29, 'F': 1, 'O': 5})

to get the format you asked for:
from collections import Counter;
counts = Counter([atom.symbol for atom in cmd.get_model('Dex').atom])
elements = [atom + str(count) for atom, count in counts.iteritems()]
print(''.join(elements))

This will print
C22H29F1O5

Notice that After the F there is a '1' instead of blank.
I could change that if you'd like with a bit more code.


text:
Why are signals broadened in polarography?
In the attached polarogram from Heyvrovsky's 1959 Nobel lecture, transitions from one plateau to the next are smoothened out rather than sharp and abrupt. Why is that?
I understand that currents were small in this experiment, so metal ion concentrations should not change significantly.
I would have expected the current to jump from one level to the next as soon as the necessary voltage is reached. What broadens these transitions?
https://www.nobelprize.org/nobel_prizes/chemistry/laureates/1959/heyrovsky-lecture.pdf

Related question: Polarography is less popular nowadays than it used to be, but can anyone point me to contemporary applications / results that I could show my students?

Before actually reaching the reduction potential, reduction of the metal ions happens only in very tiny amounts, but eventually becomes measurable as the equilibrium constant approaches 1 from below, (e.g. 0.001 → 0.01 → 0.1 → 0.5 → ...). I discuss something similar in this question.


text:
Working out pH of solution when excess NaOH is added

In a pH titration, $30\text{cm}^3$ of $\ce{NaOH}$ is added to $\pu{20 cm3}$ $\ce{CH3COOH}$. The concentrations of both solutions was $\pu{0.5 mol dm-3}$. $K_\mathrm{a} = \pu{1.7 x 10^{-5}}$ for $\ce{CH3COOH}$. Calculate the pH of the solution after all the $\ce{NaOH}$ is added.

Clearly, $\pu{10 cm3}~\ce{NaOH}$ is in excess. So moles $\ce{NaOH}$ in excess $= \pu{0.005 mol}$. Then in the mark scheme, it says $$\left[\frac{\ce{NaOH}}{\ce{OH^-}}\right]=0.005 \times \frac{1000}{50} = \pu{0.1 mol dm-3}$$ and I don't understand where this part has come from.

The excess NaOH is, indeed 0.005 moles. The total final volume is 30 ml + 20 ml = 50 ml = 0.05 L. Thus, the final concentration is 0.005 moles/0.05 L = 0.1 M


text:
Are we capable of deliberatly creating DNA from inorganic compounds?
Good day
The  Miller–Urey experiment showed that organic compound can naturally form from inorganic substances. What i wish to know is this: Could we as humans deliberately create a DNA molecule starting with only inorganic substances?

Yes. You can make very many different types of organic compounds from carbon monoxide and hydrogen gas (syngas) . You then need, say, N2,  NO2,  NO or nitrates and phosphates to have all the elements you need. Not an easy thing to do though. No doubt the synthetic chemists can suggest very many other ways.

Yes we could.
It wouldn't be worth the effort since all the building blocks are available from biological sources, but if you really wanted to you could have a chemist prepare nucleotides from inorganic substances.
Commercial labs routinely build long strands of DNA from individual nucleotides. You can tell them what sequence you need and they will make it for you. There's a Wikipedia article on how it works.


text:
Selectivity in aldol condensation of 7-oxo(6,6-²H₂)octanal
I am supposed to find the product of the following reaction:

I know that six-membered rings are more stable than five-membered rings, but the C-D bonds are stronger, so it confuses me where will the attack happen.

C-D are only marginally stronger, the reaction is reversible and 6-member rings are very much favoured over 8-member.

There are two factors to take into consideration.

The acidity of the deprontonated $\alpha-\ce{H}$.
The electrophilicity of the carbonyl being attacked.

If 1 dominates, then the reaction is under kinetic control. While if 2 dominates then the reaction is under thermodynamic control. (Detailed arguments are omitted here.) Since addition to carbonyls is generally highly reversible, the reaction should generally be under theromodynamic control and yield 1-(2-hydroxycyclohexyl)ethan-1-one:

There are existing experimental results on aldol condensation of 7-oxo-octanal.
Ghobril et al. (Eur. J. Org. Chem., 2008: 4104-4108) used TBD as base and obtained the thermodynamic product, while Pidathala et al. (Angew. Chem. Int. Ed., 42: 2785-2788) used proline and obtained the kinetic product. Ghobril et al. also pointed out that

When unsymmetrical ketoaldehydes such as 3 are submitted to intramolecular aldol reactions, the ketone usually acts as the CH-acidic component, whereas the aldehyde plays the role of the carbonyl-active counterpart to afford regioisomer 4. This regiochemical outcome is also favoured when thermodynamic conditions are used.

In both cases, it would form a 6-membered ring (I can't see where you spotted a 5-membered ring product). The alpha hydrogen of the aldehyde (pKa around 17) is more acidic than the alpha deuterium of the ketone (pKa around 20) since neighboring alkyl groups tend to destabilize carbanions by donating electron density. Furthermore, as you mentioned, C-D bonds are slightly stronger than C-H so it is safe to say that the enolate will form on the alpha carbon at the aldehyde side. (I'm sorry I wrote the carbanion attacks the carbonyl, to be more correct I should've written the enol double bond acting as the nucleophile, not the carbanion. For practical purposes, it makes no difference though.)

Information on the acidity of alpha hydrogens: http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch21/ch21-2.html

Both enolates will reversibly form and the selectivity of the reaction should not be determined by which enolate forms but rather by thermodynamic stability of the products. In this case it also seems highly likely that a subsequent E1cb reaction will furnish the enone instead of the hydroxyketone.

I thought about continuing to the enone formation but the conditions were not specified and I wasn't sure which product they were hoping to obtain.


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Dependence of mass transfer coefficient on solubility
Does solubility of the solute affect the value of the mass transfer coefficient?
The book I am following mentions that the mass transfer coefficient is a characteristic of the constituent and its environment, and solubility is a property of the environment.

No.
In all but highly exotic situations, diffusive and convective fluxes of solutes are linear in the solution-phase concentration of the species:
$$
\begin{align}
J &= -D {\mathrm dC\over \mathrm dx}  & \mathrm{(diffusion)} \\
N_c  &= k_c\left(C_\infty - C_\mathrm{surface} \right)  \quad & \mathrm{(convection)}
\end{align}
$$
Whatever the dissolved concentration $C$ (or concentration gradient ${\mathrm dC\over \mathrm dx}$) of the species is, that is the value you use in both of the above equations.  Whether that concentration (gradient) arises from a species with low, intermediate, or high solubility doesn't matter at all. I suspect your book is using the term environment to refer to things like the geometry, temperature, and fluid-flow characteristics of the system.
The main way I can think of where the solubility might affect the overall mass transport achievable in the system is in its effect on the maximum possible dissolved solute concentration at the surface.  If solubility limitations constrain $C_\mathrm{surface}$ to a low value, then the magnitude of $N_c$ will necessarily be low.  The low solubility is typically assumed not to appreciably change the value of $k_c$, however.


text:
Terminology for describing nuclei participating in metallic bonds
I've read many times that in metals (for example sodium metal), what constitutes the crystal lattice is an arrange of atoms, since no electron is given or lost, they are just shared between all the atoms within the structure.
In other texts though, it is written that metallic bonding is comprised of cations "surrounded" by free electrons since electrons are not held to the nucleus as in a single metal atom in the gas phase. From a formal point of view, which of the two statements is correct? Does it make a difference to consider cations or atoms when describing metallic bonding?
Which is the more appropriate terminology?

Well, the solid state physics is more complex than either...

I know that it is complicated, I just want a clarification.

At an introductory level, the two really aren’t that different. An atom sharing electrons isn’t conceptually that different from cations surrounded by the free electrons.

https://chemistry.stackexchange.com/a/8082/9961

Possible duplicate of Clarification for metallic bond

I don't see the answer to my question there. I want to know if, when referring to metals, I should say: "atoms in the metal lattice" or "cations in the metal lattice". It's more of a formality than chemistry itself. Which is more adequate.

At least in solid state physics and materials science we would say that atoms sit on lattice sites. Unless, particularly for a polar compound, you want to distinguish between anions and cations. But for a metal, just say atoms...

Also, though there is a "sea" of free valence electrons, they pretty much stay in position around the atoms. If you apply an electric current across a metal, for every electron added, at the negative side, one pops out by the positive. The actual drift velocity of the electrons is quite slow, though the wavefront travels at the speed of light (in the metal).

Both your statements mean the same thing, however it is convenient to think of metals as kernel + sea of electrons, as it greatly simplifies visualization, conceptualization and thus calculations. From a purely numeric viewpoint, i prefer the 'sea of free electrons' model or that cations are surrounded by free electrons.


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Sargassum seaweed affects porcelain?
We are inundated with Sargassum seaweed here in the Virgin Islands right now. It has a pungent sulphur odor as it decomposes. Coincidentally (or NOT) the porcelain tubs in the condos I work at are all turning brown. Some are not used at all and still this stain is appearing and it appears to be beneath the surface, or at least you can't FEEL it, it feels smooth. It "cleans" in contact with bleach, but reappears later.
What chemical reaction could be occurring and is it perhaps sargasso related or just coincidence?

Interesting question, but it doesn't seem to me that there is enough information to give an answer. I'll point out that the stain can't be "undersurface" since bleach cleans it. // Question - Where is the source of fresh water? Desalination or groundwater?

Desalination. And thank you, the surface must be porous, or course, but to the touch, it is smooth and the stain detectable by touch alone.

All that can be said is that there are many desalination processes. I'd assume that distillation is not used since it would be the most energy consuming. Now the desalination would lower mineral concentrations (eg NaCl) to be fit for human consumption and also be treated for bacterial contamination. So I'd expect the water to be fit for human use, but it wouldn't be "pure water" like you'd get from distillation. What might be different in the water now would be a complex research project. // All in all, you could be right that there is a difference...

No water has touched the tub in my office. It is not used, so the water isn't the issue. It seems to be coming from the air and the only environmental change we are aware of is the sargasso grass that is deteriorating in the marina and the beach nearby.

Interesting! Dimethyl sulfide (DMS), $\ce{H3C-S-CH3}$, structurally, is given off by some marine algae. It's volatile, with a boiling point ~37 °C: just about normal body temperature. It's odor is as you describe, sulfury, resembling cooked cabbage.
As a wild guess, could your condo be air-conditioned or the porcelain fixtures be cold? The maximum sea water temperature at British (and U.S.) V. I. is ~29 °C in June, and it's likely some DMS would evaporate at that temperature, perhaps condensing on cold fixtures and serving as a nutrient for archaea and bacteria? Try warming a section of the tub, perhaps with a waterproof heating pad, to see if that prevents condensation and subsequent discoloration.
DMS itself should be a clear, colorless liquid at room temperature, but bacterial slime, particularly iron-sulfur (which turns to sulphur in BVI) metabolizing critters, produce a reddish-brown deposit. And underneath that porcelain is probably a cast-iron tub. So if bleach blasts away that brown, it might be debris from a biofilm.
If that is the case, then at last be consoled that you've got some of the same bacteria and archaea as found in Yellowstone National Park.

Thank you!  Yes, the discoloration is eliminated with chlorine beach, but it reappears overnight. It is not detectable by touch and doesn't rub off. The tubs ARE ceramic over cast iron and the surface glaze may have been compromised by cleaning compounds, exposing the metal microscopically (?) but not visibly. Is the archaea and bacteria harmful to humans? It is not possible to rub away this stain so it does not SEEM to be on the surface. Is that the nature of a biofilm?

No, these archaea and bacteria are not harmful, AFAIK, though in water pipes they can cause clogs. To remove iron stains, there are commercial products such as Iron Out and Red-B-Gone. These have to sit for a while on embedded stains in porcelain. Though more effective at first, avoid cleaners with HF, hydrofluoric acid, which is toxic and dissolves the porcelain glaze.

Thank you for your help. It has helped us assuage the  our owners concerns and certainly helped me in formulating an intelligent response to their inquiries!


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Naming a hexasubstituted benzene ring

In naming the attached compound, is there a reason it is termed:

3,5-dibromo-1-chloro-4-ethyl-2,6- dimethylbenzene

as opposed to:

1,3-dibromo-5-chloro-2-ethyl-4,6-dimethyl benzene?

Shouldn't bromine be assigned the smaller numbers, given that it comes first alphabetically?

It is not named "3,5-dibromo-1-chloro-4-ethyl-2,6-dimethylbenzene". Why do you think so?

Related: What is the numbering of substituted benzene derivatives according to IUPAC nomenclature?

The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

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

The compound that is given has no principal characteristic group that would be expressed as a suffix. Thus, Rule (c) is not relevant in this case.
No locants are assigned to the delocalized electrons of the benzene ring. Thus, also Rule (e) is not relevant.
Usually, Rule (f) is important for benzene rings with multiple substituents. In this case, however, all six substitutable positions are completely substituted; i.e. the locant set (all considered together in a series of increasing numerical order) is always ‘1,2,3,4,5,6’ – no matter how we turn it.
Finally, Rule (g) stipulates that the lowest locant is assigned to the substituent that is cited first as a prefix in the name. Therefore, the correct name for the compound that is given in the question is 1,3-dibromo-5-chloro-2-ethyl-4,6-dimethylbenzene.


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Why is the rate of a reaction always positive and why?
Why rate of reaction can never be negative? And why rate is an experimental derived expression?

Otherwise you could not call it a reaction.

No mystery. Chemists write equations so that reactants (on the left) yield products (on the right). So:
$$\ce{H2O -> H^+ + OH^-}$$
and
$$\ce{H^+ + OH^- -> H2O}$$
The reactions are read from left to right and reactivity is defined as positive from left to right - this means the reactivity for the reaction we want to look at is always positive. (great clarification from user Stian Yttervik.

I'd include the sentence "The reactions are read from left to right and reactivity is defined as positive from left to right - this means the reactivity for the reaction we want to look at is always positive"

In a first order reaction such as a cis-trans isomerization or decay of an excited state the rate is
$$\displaystyle\frac{\mathrm d[\ce{C}]}{\mathrm dt} = -k[\ce{C}],\tag{1}$$
so is negative, but the rate constant $k$ is always positive. The negative sign is because the initial species $\ce{C}$ decays away in time. Integrating this expression leads to exponential kinetics. If $\ce{C}$ decays into a product $\ce{P},$ then
$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = +k[\ce{C}].\tag{2}$$
The rate constant is determined experimentally as it us usually impossible to calculate what it will be. Similarly, the actual mechanism of a reaction is not always what the stoichiometry would indicate, in fact it is usually not and so has to be determined experimentally.


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Does activation energy depend on steric factors?
I thought that activation energy depends on steric factors.
By Arrhenius equation it must depend on steric factors since $A$ depends on steric factors. $$k=A\cdot\mathrm e^{-\frac{E_\mathrm a}{RT}}$$ (for bimolecular reaction)
But my book says it doesn’t.
Does it depend on steric factors or not and why?

The Arrhenius model of a rate rate constant is very basic so has to be modified in many cases. One of these is when steric factors (generally undefined exactly) are used as an empirical  multiplier to A and used as a rationale to fit experimental data.


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Is neutron a compressed form of hydrogen atom?
As I know:

$E_\mathrm{p} = \pu{938.272 MeV}, \quad
    E_\mathrm{e} = \pu{0.511 MeV}, \quad
    E_\mathrm{n} = \pu{939.565 MeV}$
$E_\mathrm{n} − (E_\mathrm{p} + E_\mathrm{e}) = \pu{0.782 MeV}$
$M_\mathrm{p} = \pu{1.6727 x 10^-24 g}, \quad
    M_\mathrm{e} = \pu{9.110E-28 g}, \quad
    M_\mathrm{n} = \pu{1.6750E-24 g}$

Furthermore, following reaction also happens:
$\ce{p+ + e− -> n + ν}$
So i was wondering if this is a combination reaction at atomic scale and if the neutron is really a highly compressed form of hydrogen atom? As mass difference is negligible and higher energy of neutron can be regarded as the moving energies of electron(i.e. rotational, kinetic and others) are locked down into potential energy and charge of electron and proton neutralizing eachother?

While technically true, it does not make much sense to think of it this way. Look at everything around you (people, buildings, ground, vegetation, etc.) Now compress it with the same force. Everything will turn into neutrons. Are neutrons really a highly compressed form of all that?

No, it seems not as the neutron can decay into a proton an electron and an anti-neutrino, and the proton can decay into a neutron a positron and an electron neutrino.  Thus there are extra particles involved than just the proton and electron.

General thoughts along those lines were pretty common between the time of the identification of deuterium and the discovery of the neutron. But, nuclear physics makes you look at the quark constituents and that takes you into a completely different space, far from chemistry. So, the neutron is not a compressed form of hydrogen.


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Why do some chemical reactions require many steps?
I posted the following question in Physics SE and was advised to transfer it to Chemistry SE.

I studied physics in college ten years ago and I recently started to learn biochemistry. I enjoy finding out that some familiar concepts in physics play important roles in biochemistry such as entropy and the Gibbs free energy.
For example, as a (ex-)student of physics, I am happy to know that the Gibbs free energy determines the directions of chemical reactions. I feel this is a good example where a sort of fundamental law of physics determines how a phenomenon looks like.
However, I still can not understand why the chemical reactions in a body need to be so complex. Many chemical systems consist of more than several steps to achieve their purposes. According to Wikipedia, glycolysis takes ten steps through its process. Why are so many steps necessary?
I tried to find out a physical law that prohibit the glycolysis process from being achieved in one or two steps, but I could not find an answer. I would like to know (or discuss) whether there is a physical law that makes chemical systems so complex (many steps required).
My assumption is that some physical law prohibit the existence of an enzyme that realizes a one-step process of glycolysis.

There is no such law. It is just that you need to carefully rearrange a lot of delicate things (chemical bonds). Can you, say, print a book in one hour? I don't think so. What physical law prohibits that?

@IvanNeretin My question is about why many steps are required. So, if there is no physical law which prohibit that a single step acheive the same production, a researcher might discover a way to realize this single step. Can you agree with this? In my intuition,  some chemical systems fundamentally require taking more than several steps. Probably, this intuition of mine comes from the fact that most of the systems  in the real world take many steps to acheve its purpose. If no physical law prohibit this, theoretically, printing a book can be also achieved by just a single step.

My intuition is opposite to yours.

Do you think any chemical system can be theoretically  achieved in a single step?

Do you think any book can be printed in one hour?

Is it not just a matter of where you start? If with the elements, say carbon, oxygen etc then even to make a small and specific molecule rathe than just  any type of molecules several steps are needed as the reaction has to be directed towards your product.

@IvanNeretin I joined just so I could comment,,, I work at a printing company, and many times we print multiple books (and finish them) within the span of an hour. I had my thesis printed (6 copies) and binded within 45 minutes. :)

@SynerCoder Great! Likewise, sometimes a surprisingly complex chemical transformation can be achieved in one step.

Indeed, it's is just not an easy thing to do ;)

If you look at the whole of glycolysis - citric acid cycle - respiratory chain, in the end C6H12O6 + 6 O2 → 6 CO2 + 6 H2O, we have a lower complexity reaction path to compare: burning glucose in oxygen atmosphere. So what the complex and delicate biochemical reaction chain achieves is to harvest the energy in portions that are physiologically usable.

There is no fundamental law preventing simple chemical reactions: things are complex because of the combinatorial complexity of chemical compounds
The complexity of many chemical reactions is a byproduct of the fact that there is a very, very large variety of possible chemicals. Much of that complexity happens because of the almost infinite way even some simple elements can be combined together to give complicated structures (carbon being the archetypal example). Theoretically, for example (theoretical because not all of the examples can exist in 3D space) there are 366,319 ways to build different alkane compounds from just 20 carbon atoms and hydrogen atoms (see this question here and this entry in the Encyclopaedia of integer sequences). And this number drastically understates the real complexity as it ignores mirror images and more complicated ways of joining the carbon atoms together (like in rings, for example). The complexity just gets more mind boggling if you start adding other elements to the mix.
No physical law prevent us making any possible compound in one step. But the sheer complexity of the end products makes simple ways to reach many of them extraordinarily unlikely from the laws of probability alone, never mind the specific ways chemical components can be easily joined up to make more complex things.
Here is a simple analogy. Let's say you want to assemble a Lego model of the Star Wars Death Star weapon. There are 4,016 pieces of lego that have to be assembled in the right combination and the right order. There is no physical law that says you couldn't somehow do that in a single step. But no sane person's intuition would assume that this was easy or likely. It isn't physical law that prevents one step assembly: it is combinatorial complexity. Chemistry is, do I really need to say this, more complicated than Lego: not least because atoms can be joined up in many more complex ways than the simple, standard-sized physical pins that join Lego bricks together.
Both nature and synthetic chemists have explored many ways to achieve particular end products from simpler building blocks. Sometimes new chemical Death Star equivalents (like the geometrically beautiful hydrocarbon dodecahedrane, which, incidentally, has 20 carbons but isn't counted in the list of 20 carbon alkanes) are made only after long sequences of reactions. The original synthesis of dodecahedrane took 29 steps but others found better, higher yielding, routes that took only 20. Many important drugs are first synthesised in long sequences of reactions but are later found to be available via much shorter routes (there is nothing like the economics of manufacturing cost to encourage creativity).
So the reason many chemical reactions take multiple steps isn't physical laws but probability theory. There are just too many possible chemicals and too many ways to combine things for single step routes to most given products to be likely to work. Doing one thing at a time (just like you would if building the Lego Death Star) is the way to get what you want.

Thank you very much! If I understand you correctly, you mean that there are significant number of routes for, say, Glucose to two molecules of pyruvic acid by the combinatorial complexity. The step size of each route is various. When one route is selected randomly, the probability of $step_size = 1$ is very few. That's why almost all complex chemical reaction takes more than one step. Is my understanding correct?

This and the other answer touches on the fact there is, in fact, a fundamental law: thermodynamics. To get, say, 25 atoms in the correct arrangement would require a colossal decrease in entropy. Individual steps lowers the entropic penalty, allowing reactions to take place.

@rkjt50r983 Yes, I think your intuition is getting the right idea.

The Lego death star is a fine example: It is geometrically, mathematically impossible to build a machine that assembles it in one step, because there simply isn't enough space around it. Making things worse: The machine needs to be made of Lego parts, too. :/

@awjlogan I missed this when the answer was first posted, but the thermodynamic view focussing on entropy is, essentially, the same as the probabilistic view in my answer.

I do not believe there is a fundamental law which prohibits complex reactions from taking place in a single step — it’s just extremely improbable.
Collision Theory
Gasses
This is especially relevant in gasses, but I will relate it to glycolysis later. Kinetic-Molecular Theory simplifies gasses to dimensionless points moving in constant, random, straight-line motion and colliding 100% elastically with each other. While none of this is exactly true, it is a good model.
For gases to chemically react, molecules must collide with the proper orientation and enough energy. Let’s look at the following reaction.
$$\ce{CH4 + Cl2 -> CH3Cl + HCl}$$
While it is theoretically possible for these particles to collide, it would most likely confuse the system rather than actually make the desired products. Let me propose a mechanism.
$$\ce{Cl2 -> 2Cl}$$
$$\ce{CH4 + Cl -> CH3+ + HCl}$$
$$\ce{CH3+ + Cl2 -> CH3Cl + Cl}$$
$$\ce{2Cl -> Cl2}$$
By using two intermediates (unstable substances created during a reaction which quickly rereact) and four steps, I have broken a complex reaction into a series of unimolecular and bimolecular collisions (favored by probability). Furthermore, it is far easier for the molecules in these steps to collide with a proper orientation. Let’s look at the second step. Methane has a tetrahedral electron-pair geometry, and when a monoatomic chlorine collides with enough energy 180 degrees opposite a hydrogen, the electron clouds may overlap, simultaneously forming a $\ce{C-Cl}$ bond and breaking a $\ce{C-H}$ bond.

Google Images
Glycolysis
Oversimplifying, but glycolysis turns a glucose molecule into two G3P molecules and two ATP. If methane is complicated with its five atoms, glucose is far more with its twenty-four. Not only would it be nearly impossible for one reaction to split this sturdy sugar ring but to rework the products as well into their forms that are comparable with the Krebs cycle, etc. would take insane luck. Instead, a carefully controlled process makes the essential reaction reliable.
Furthermore, enzymes are ‘invented’ by random mutations, so an enzyme to carry out this process may be feasible, but evolution would probably not invent it. And even if it did, it would probably not confer that much of a survival advantage and would pass out of the gene pool.
Hope this helps!

(+1) While I like matt_black's answer a lot, I think this gets more to the point OP is confused about. Yes, there are a lot of things that could form if many atoms collided at once, but the bigger issue is that it becomes astronomically more unlikely to get 3, 4, ... things to collide in the proper orientation to produce anything at all and so most reactions proceed in steps involving a single collision. Even a very intricately designed enzyme would have little change of producing benzene from 6 carbon and 6 hydrogen atoms, it simply requires too much coordination.

Not to mention, each enzyme in glycolysis has a specific function, and the metabolic intermedates often have functions by themselves. For example 2,3-BPG is "only" an intermedate in glycolysis, but it has vital functions on its own.

Authors note: While there are some good answers already, I wish to help you understand by explaining in a different way. I do agree with the other posts that there is no physical or chemical law to prevent a different, more straight-forward process.
Reason behind the glycolysis process
The reason why this process is as it is, is efficiency towards reaching the goal. And the goal is not to break down glucose to smaller molecules. The goal is to store energy in a carrier that can move through the body and is compatible with other biological processes.
The three bold words are key here. The body needs energy to perform various tasks such as muscle contractions (breathing, heart beat), cell growth, fighting bacteria and many more. It’s not handy to always generate the energy needed at the place where it is needed. Instead, we have energy carriers (most importantly ATP) that is produced in certain parts of our bodies and then distributed via the blood.
Energy in the body
Before I continue, you need to understand a bit about Gibbs free energy. As you mentioned, it determines the most energy-efficient way for a process from beginning state to end state. However, if you provide energy, the process can go the reverse direction just fine. So looking at Gibbs free energy only shows the process that is most likely to spontaneously occur under normal circumstances, but not in all circumstances.
Second in-between background info is that the energy in the body is transported using adenosine triphosphate (ATP) and adenosine diphosphate (ADP). Adding a phosphate group to ADP (which then becomes ATP) costs energy that can later be extracted by the reverse process.
Third is that energy availability in the body is limited. We have two major sources of energy: ATP and body heat. An ATP molecule will always provide a specific amount of energy, while body heat can provide from 0 up until a certain limit, depending on body temperature (this maximum is lower than ATP’s energy). Any process that needs more energy than ATP can provide will have to be broken down into separate smaller steps.
Back to glycolysis
With this background information in mind, we can explain the reason behind the (complex) glucolysis process better. From a Gibbs free energy point of view, we don’t need to go from high energy glucose to low-energy pyruvate as fast as possible. Instead, we need to do this in a way that has the most steps that provide the exact amount of energy needed to transform ADP into ATP.
As you can see in the image caption on the glucolysis page you linked, we need 1 glucose + 2 ATP, to generate 4 ATP. Why is the initial ATP needed? This is to get the specific break-down chain that allows 2*2 steps of energy extracting throughout the process. We need the initial energy investment to allow for in-between steps to happen, chemically speaking. Without this investment, you will not be able to form the intermediary molecules needed to give enough energy in order to store it in $\ce{ADP\bond{->}ATP}$.
Comparison to nuclear fusion/fission
I normally don’t like to make comparisons to unrelated subjects, but I think this one fits well enough to mention and hope you will understand it better with your physicist background. In nuclear fission and fusion, you determine possible nuclear decays and fusions by looking at the available energy and energy levels of an atom. And if we ask your original question here, we get the same answers as in chemistry.

Is there anything preventing 6 hydrogens atoms fusing into a carbon atom?
Is there anything preventing U-235 into splitting into 20 different atoms in a single step?

To the first: no, but it’s very unlikely that 6 atoms meet at the exact same time and with the correct amount of energy. And even if they did, carbon is not a stable atom without neutrons, so where do they come from? You need multiple steps to go from hydrogen to carbon…
To the second: No, nothing prevents this. But splitting of atoms goes via a set of strict rules regarding the stability and energy of the atoms and their radiation products. The start and end point may be clear, but there are almost always multiple in-between steps (example: decay chain of thorium. In the same way, chemistry has many rules for reactions and atom/electron rearrangements within a molecule, limiting how molecules can break apart or combine.
The part where this comparison goes wrong is that biology does not always lean towards the most energy-efficient solutions. Sometimes nature takes a complex, unefficient route for a different purpose, as in glycolysis.

The chemical reactions that you learn about in chemistry classes are designed by humans. Although these can be sometimes quite complicated, there is a strong bias toward designing reactions that can be rationalized by the human brain. The reaction networks found by evolution are not constrained by what humans can understand and, therefore, can appear more complex.
In fact, evolution in some cases may favor complex reactions, because they can be more efficient. Have a look at the Krebs cycle (also known as the citric acid cycle), which converts most of the energy used by your body from sugars to a more usable form (as ATP, where the energy is stored in phosphate linkages). The RCSB Protein Data Bank has a good description (description of Krebs cycle with the enzyme structures) of the steps involved. The cycle is more energy efficient than other simpler options for converting energy from oxidation of acetate to ATP (paper by Krebs on the efficiency of the cycle).

Another thing to ponder is how these processes came to be. Evolution is the progenitor of basically all biochemical processes. Evolution doesn't [necessarily] care about efficiency, it cares about being effective (ie. creature stays alive). Sure, I can cut through someone's yard to get across the neighborhood, but taking the sidewalk keeps the neighbors from releasing the hounds on me.
Then there's energy. Everything in chemistry boils down to being energetically favored. You seem to understand that with Gibbs Free Energy. While not necessarily the case with glycolysis, but some chemicals will keep their bonds the same because it's so stable, unless a lot of energy is thrown at it. That's where proteins come in. The proteins bend or twist molecules so that the activation energy is much smaller and certain reactions are favored. (A tetrahedral carbon will not like having bond angles significantly less than 109.5 degrees so a change in bonding is likely to occur to return to that energy stability).

Many of the specific sentences of this post are sensible and factual, but I'm afraid I don't see how this answers the question posted.

It's a generic answer to the title question. Will it get best answer? No, the previous answers definitely fit the bill better. However, it does explain why there isn't one singular reaction and provides some context to help the OP with rest of biochem.

OP asked specifically about biochemistry. "Nothing in biology makes sense except in the light of evolution", so this answer adds an important perspective that is missing in the others. It could explain in more detail how unnecessary complexity would evolve, though. When there's a simple and a complex way to do it and a species has already evolved 90% of the complex way for other reasons then that's what you will end up with.


text:
Is the pressure in an HPLC system the same throughout?
I'm not sure if I'm articulating myself well enough, but here's my problem: I have a current HPLC system, where we typically have backpressure readouts of around 15 MPa. We are looking to get a new HPLC, and the rep said its DAD detector has a maximum pressure of 6 MPa. My questions are:

Is the pressure at the pump outlet the same as the pressure at the column inlet and outlet?
What pressure is actually being shown on the system readout when looking at the pump? (I'm assuming just the pressure at the pump outlet)
Is the pressure at the detector usually significantly less than at the pump outlet?
Assuming everything transfers from one system to our new one, and we end up with the same types of backpressures, will I have issues with this new detector?

I figured this out. The pressure displayed is between pump and column inlet. There's a pressure gradient from the column inlet to the outlet, so the pressure reaching the detector is significantly less than the readout.


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Why is the conjugate of a strong oxidizing agent a weak reducing agent?
I am a grade 12 student. While I was reviewing some electrochemistry, I had the following question. Take this reduction half-reaction:
$$\ce{F2 (g) + 2e- <=> 2F- (aq) + $\pu{2.87V}$}$$
I know that $\ce{F2 (g)}$ is a strong oxidizing agent, but why does this make $\ce{F- {(aq)}}$ a weak reducing agent? I thought that since the electrical potential of the reduction half-reaction:
$$\ce{2 F- (aq) <=> F2 (g) + 2 e-}$$
was $\pu{-2.87 V}$. Therefore, should $\ce{F- (aq)}$ not be a strong reducing agent?

You're always looking for a positive potential as an indicator of spontaneity. $-2.87\ \mathrm{V}$ is not strong...

It is not the magnitude of the standard cell potential that indicates whether or not the cell reaction will be spontaneous, but the sign. In the same way a reaction is only spontaneous if $\Delta G < 0$, a substance only oxidizes/reduces if the corresponding cell potential is positive.


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Hydrogen bonding in chloral hydrate
Normally, chlorine doesn't form hydrogen bonds because despite its electronegativity, the size of the atom is such that its electron density is too low to form hydrogen bonds.
However, chlorine forms hydrogen bonds in chloral hydrate (2,2,2-trichloroethane-1,1-diol). What could be the possible reason for this?

I am far from to be a specialist of organo-chemistry. However I remember learning that geminal diols are particularly unstable. So I would say maybe in this case the hydrogen atoms exhibit "a smaller electronic density" (the electrons are diffused in the whole molecular orbitals) then even the small electronic density of the chlorine atoms would be enough to get a very weak hydrogen bond.

Possible duplicate of Is hydrogen bonding generally defined to include only three period two elements?

In other words, your starting premise is false - Cl hasn't got much problem with hydrogen bonding. It's just weak. For some reason weak hydrogen bonds often aren't considered at all.

How is that a duplicate if chloral hydrate includes a $\ce{Cl\bond{...}H}$ hydrogen bond? Or is this a misconception?

@pentavalentcarbon it is misconception to be debunked. I couldn't find straight up dupe, but there as many posts about it.

@Mithoron Then that should be part of an answer. Closing a question because it contains a misconception is not right; closing should be indicating a question should be improved.

@pentavalentcarbon There is no problem with Cl in hydrogen bonds and this misconception was thoroughly debunked already.

As noted in the comments, the fallacy is the notion that only a select few atoms can form hydrogen bonds with protic hydrogen.  In reality, the strongest hydrogen bonds involve nitrogen, oxygen, or fluorine, but most nonmetals can form such bonds if they have an electron pair to donate to the bond.  Hydrogen bonding with chlorine is seen in the $\ce{HCl_2^-}$[1], which is analogous to the more familiar bifluoride ion, as well as chlorine-bearing organic compounds[2].
References
1.
Harry F. Herbrandson, Richard T. Dickerson Jr., and Julius Weinstein, "
Cite this: J. AThe Bichloride Ion", J. Am. Chem. Soc. 1954, 76, 15, 4046. https://doi.org/10.1021/ja01644a066
2.
R. Banerjee, G. R. Desiraju, R. Mondal, J. A. Howard. "Organic chlorine as a hydrogen-bridge acceptor: evidence for the existence of intramolecular O--H...Cl--C interactions in some gem-alkynols." Chemistry 2004 Jul 19; 10(14):3373-83. https://doi.org/10.1002/chem.200400003. PMID: 15252783.

In chemistry, generally most of the notable exceptions are associated with the thermodynamics or the kinetics of the reaction.
In the particular example  of chloral hydrate, chlorine tends to form hydrogen bonds with hydrogen (even though usually their hydrogen bonds don't exist) majorly due to the thermodynamic changes that occur in the process. By forming such "pseudo-rings" (because H-Bonds aren't completely a "bond", rather only very strong attraction) the structure of chloral hydrate gains excessive stability as 5 & 6 membered rings are very stable. This pseudo-ring formation is called "Chelation Effect".
So, basically that extra stability is the driving force for chlorine to form H-Bonds and the process, as a whole, becomes thermodynamically feasible (spontaneous).

Most chemistry is associated with thermodynamics!

Hydrogen bonds are bonds!  They involve the formation of molecular orbitals that produce a bonding overlap.