Chemical processes and chemical reactions difference Whats the difference between chemical processes and chemical reactions?
Ive heard that the chemical industries only uses chemical processes to produce new substances, but isnt that also what happens when a chemical reaction occurs? I couldn't find a lot on this topic online except for the wikipedia aricles that only made me more confused.

According to IUPAC:

Chemical reaction
A process that results in the interconversion of chemical species.
Chemical reactions may be elementary reactions or stepwise reactions
(It should be noted that this definition includes experimentally
observable interconversions of conformers.) Detectable chemical
reactions normally involve sets of molecular entities as indicated by
this definition, but it is often conceptually convenient to use the
term also for changes involving single molecular entities (i.e.
'microscopic chemical events').

A chemical reaction is a kind of process in which interconversion of chemical species takes place. Because of this, we can say that a chemical reaction is a chemical process.
Chemical process can be a broad term, which describe any process that is commonly studied by chemists because they are closely related to chemical reactions. It is somewhat diffuse, as boundaries of the field of chemistry.
Industries use a variety of procedures to obtain substances. Chemical processes/reactions take place. If your interest is realising what is done in industries, do not get stuck in these subtleties. Most of the times they are not intentionally distinguished, and their usage is customary in the field.


Isonitrile to nitrile conversion If we are given $\ce{CH3NC}$ and told to convert it into $\ce{CH3CN}$ by doing some chemical reactions.
My attempt
According to my textbook there are no chemical reaction of isocyanide conpounds.  I tried reduction of isocyanide giving a secondary amine. Then I don't know how to convert a secondary amine to cyanide compound.
Hints are too appreciated.
Edit:
I have got a solution but I don't know whether the reaction are spontaneous.

I have written these reaction on the basis of theoretical knowledge I don't know whether they are spontaneous or not. Correct me if I am wrong somewhere.

Just heat it up: Isonitriles can be thermally rearranged to nitriles.

$$\ce{R-NC ->[\Delta] R-CN}$$

Please have a look at some references, such as:

Michael J. S. Dewar, M. C. Kohn, Ground states of $\sigma$-bonded molecules. XVI. Rearrangement of methyl isocyanide to acetonitrile, J. Am. Chem. Soc., 1972, 94, 2704-2706 DOI

Michael Meier, Barbara Mueller, Christoph Ruechardt, The isonitrile-nitrile rearrangement. A reaction without a structure-reactivity relationship, J. Org. Chem., 1987, 52, 648-652
DOI

UPDATE

As far as the sequence outlined in the question is concerned:

The synthesis of methyl formamide from methyl isonitrile is pretty much the opposite of how things are done in real life. Isonitriles, used in the Passerini and the Ugi reaction, have shown to be available by dehydration of formamides, e.g. using $\ce{POCl3}$. For an example with cyclohexyl isonitrile, see Org. Synth., 1961, 41, 13.
N-Methyl formamide is far from being unstable! Release of methyl amine, which, by the way, has a BP around -6 °C might be possible under alkaline conditions.
Making methanol from methylamine, again, is a rather pointless mind game. In real life, methylamine is produced by reacting methanol with ammonia at hight temperatures in the presence of acidic catalysts.
You definitely want to make methyl chloride in the lab! With a BP around -24 °C it's very easy to handle, and the GHS classification with H351 (Suspected of causing cancer) and H373 (May cause damage to organs through prolonged or repeated exposure), this is a real gem. Yes, I'm kidding! Don't do it!
For a nucleophilic substitution with cyanide, the high volatility of methyl chloride again probably doesn't make your life easier, but just shorter.

In summary, my suggestion is simple: Forget about the proposed synthesis! It is laborious, difficult to handle in the lab, and an economical nightmare.


What is the Difference between Menthol Crystal and Menthol Powder? I want to buy menthol powder from a chemical supplier but he is selling menthol crystal at half price of menthol powder (he says manufacturing process is the difference) so I am really confused what should I do. Can I buy menthol crystal and just powder them to get the powder form I want it or are crystal and powder menthol totally different things altogether?

Unless there is something bizarre going on (which should be mentioned more or less clear on the website/in the catalogue because no chemist would expect it), menthol will be menthol as there is only one compound defined as menthol. This compound is chiral, so I would expect a stereodescriptor. In its absence, I would not be sure whether I am dealing with a racemic sample (more likely in the case of a chemical supplier aimed at practicing chemists) or a natural, chiral sample (more likely in the case of a supplier selling to the general public).

Chiral menthol crystallises easily and nicely, and because crystals are easily separated from contaminants, they are often the preferred form of isolation of compounds for sale; especially since the crystals themselves very rarely contain impurities. While a powder can be the better choice for several purposes (e.g., weighing small sample amounts into a flask), it may not be as simple to produce. In the case of menthol, I expect the powder to derive from powdering the crystals. This is an extra step and thus costs more money.

Depending on how important the quality of the powder is to you, it may make more sense to buy the powder directly or to buy crystals and grind them to powder yourself. However, as I said above, I am very sure that they contain the same chemical in a comparable purity.


How does the presence of various NMR active nuclei in a compound affect coupling and signal multiplicity? I am basically wondering how the presence of various NMR active nuclei present in a organic molecules affect the shift values and coupling states normally seen in a molecule lacking these nuclei.

Can NMR active metallic moieties in organometallic compounds such as magnesium 25 in a Grignard reagents couple with the spin of protons in H-NMR or carbon 13 atoms in C-NMR? Do the peaks show up in the same range and how are the shift values affected?

In a structure containing many active NMR nuclei (such as a structure containing appropriate isotopes of chlorine, phosphorus, and nitrogen) is special decoupling software or supplemental instrumentation needed to interpret these spectra or are there other techniques to readily discern structure?

I'd like to expand on the other answer a bit (at least on the question of coupling).  An NMR active nuclei X can couple with (and detectably split peaks of) another NMR active nuclei Y if X is both not too rare (as discussed in the other answer) AND does not have an extremely short relaxation time.

I'll focus on the common elements that an organic chemist might work with:  $\ce{H}$, $\ce{C}$, $\ce{O}$, $\ce{N}$, $\ce{S}$, $\ce{F}$, $\ce{Cl}$, $\ce{Br}$, $\ce{I}$, $\ce{P}$, and toss in $\ce{Mg}$ (as mentioned in the question), $\ce{Li}$ and $\ce{Si}$.

"The good"

NMR active isotopes with high natural abundance and not extremely short relaxation times: in our list $\ce{^1H}$, $\ce{^19F}$, $\ce{^31P}$, and, marginally, $\ce{^29Si}$ and $\ce{^6Li}$.
These all are decent NMR nuclei and abundant, so they definitely split peaks of nearby NMR active nuclei.

$\ce{^1H}$:  99.98% natural abundance, spin 1/2 (so 0 quadrupolar moment), T1
around 10 seconds for the standard.
$\ce{^19F}$:  100% natural abundance, spin 1/2, sharp peaks (thus
relaxation times not too short).
$\ce{^29Si}$:  A bit marginal.  Around 5% natural abundance (so the 'side
peaks' caused by splitting come from only 5% of the molecules and
thus will be small, called satellites), spin 1/2.
$\ce{^6Li}$:
Around 8% natural abundance, spin 1, but is an honorary spin 1/2 as
it has a very small quadrupolar moment.  Relaxation times are long.

These nuclei will split nearby NMR active nuclei (not many bonds away).  This is either a bug or a feature.  The splitting pattern can tell you a lot about the molecule along with other advantages you can get from internuclear coupling (I have a colleague who uses this all the time on fluorocarbons).  If you don't want the effect, though, you will have to use a decoupling pulse on these nuclei.

For instance, in the text "Organic Structures from 2D NMR Spectra" (Field, Li, Magill) the examples include many uncoupled experiments (Note:  Y{X} means Y is observed while uncoupling X): $\ce{^13C}\{\ce{^1H}\}$ (often), $\ce{^19F}\{\ce{^1H}\}$, $\ce{^13C}\{\ce{^1H, ^19F}\}$, $\ce{^1H}\{\ce{^31P}\}$, $\ce{^31P}\{\ce{^1H}\}$.

"The bad"

There are nuclei that, if enriched, would split nearby peaks (having not-too-short relaxation times), but have a low natural abundance.  These can be analyzed quite well using NMR, and may have peaks split by other active nuclei, but won't usually cause visible splitting in other nuclei.  Out of our list:  $\ce{^13C}$, $\ce{^15N}$.

$\ce{^13C}$:  Only 1% natural abundance, spin 1/2.  While it is often
examined either alone or in conjunction with other active nuclei, it
won't cause anything but the tiniest satellites in other nuclei.
$\ce{^15N}$:  Even worse.  Spin 1/2 but only 0.4% natural abundance.

"The ugly"

These might or might not have high natural abundance, but they have short relaxation times.  When relaxation times are very short, the nuclei is, in effect, self-decoupling.  Nuclei with very short relaxation times have, though the Heisenberg Uncertainty Principal, broad lines and thus often are pretty hard to work with in any NMR context.  They certainly don't show visible splitting of the signal of nearby nuclei.  From our list, we have:  $\ce{^14N}$, $\ce{^33S}$, $\ce{^35Cl}$, $\ce{^37Cl}$, $\ce{^79Br}$, $\ce{^81Br}$, $\ce{^127I}$, $\ce{^17O}$ and $\ce{^25Mg}$.  At least you don't have to use decoupling pulses on these nuclei!

$\ce{^14N}$:  Too bad.  Even though it's 99.6% abundant (the rest is $\ce{^15N}$), it
has spin 1 and a reasonably large quadrupolar moment, thus a short
relaxation time (often around 20 ms).  It might broaden the peaks of
nearby nuclei, but won't show coupling in any but the most symmetric
environments.
$\ce{^33S}$:  Ouch.  0.8% natural abundance, spin 3/2, high quadrupolar
moment, only NMR active sulfur nuclei.
$\ce{^35Cl}$, $\ce{^37Cl}$:  Between these, 100% abundant.  Both are spin 3/2 with
VERY large quadrupolar moments and short relaxation times.
$\ce{^79Br}$, $\ce{^81Br}$:  Same as the sad fate of chlorine.
$\ce{^25Mg}$:  Only active NMR nuclei of Mg, 6% abundance, spin 5/2, VERY
large quadrupolar moment.
$\ce{^17O}$:  The only NMR active oxygen nuclei.  Very rare (0.04% natural
abundance), spin 5/2 with a fairly large quadrupolar moment.
$\ce{^127I}$:  100% natural abundance, but VERY large quadrupolar moment.
Maybe also prone to CSA relaxation (I'm not sure).

COMMENTS on short relaxation in nuclei

Nuclei with very short relaxation times often get these via two relaxation processes.  One is quadrupolar relaxation (only for spin > 1/2) and chemical shift anisotropy (CSA) relaxation.

For high spin (>1/2) nuclei, relaxation can occur because charge on the nuclei is not symmetric (thus a quadrupolar moment) so that energy can be exchanged with an environment that is not highly symmetric.  The effectiveness of this type of relaxation rises as the square of the quadrupolar moments times the square of the applied magnetic field.  Spin 1/2 nuclei have quadrpolar moment zero.  Quadrupolar moments range all the way from values such as 0.0028 ($\ce{^2H}$), -0.0008 ($\ce{^6Li}$) to values such as 0.31 ($\ce{^79Br}$).

CSA:  Also caused by tumbling in an asymmetric environment.  Larger for nuclei with large shift ranges, proportional to gyromagnetic ratio squared times magnetic field strength squared.  Not really predominant in protons or any carbon but Quaternary carbon.

These relaxation mechanisms don't work if the environment of the nuclei is very (especially tetrahedral or octahedral) symmetric, in which case relaxation times can be long.

The perverse thing about these relaxation mechanisms is that they get worse (more decoupling) with higher field strength.  Usually anyone working with NMR dreams night and day of getting a higher field spectrometer.  However, if you are interested in a nuclei prone to CSA relaxation (say, $\ce{^199Hg}$ or $\ce{^195Pt}$, which are spin 1/2 and not prone to quadrupolar relaxation), and you want to see splitting of peaks (coupling) in nearby nuclei, you might have to move to a LOWER field NMR.

References

A good site with lots of information on NMR active nuclei is the NMR periodic table.


Why bath sonicators don't come with adjustable power setting? All the bath sonicators I saw online don't have a power level setting.
(Probe sonicators, on the other hand, do have a power setting.)
Why is that?

All the bath sonicators I saw online don't have a power level setting.

This depends on the product lines.

I've mostly used BANDELIN SONOREX SUPER ultrasonic baths, and indeed, the most simple models of this line just have a (mechanical) timer.

However, models of the SONOREX DIGIPLUS line do come with power control.

In the past, the ultrasonic baths were mostly used for cleaning purposes, while the probes, which typically had a much higher power, were used for sample homogenisation and similar purposes. Here, power regulation was apparently more important to avoid degradation of the same due to local overheating.

As usual, exceptions prove the rule:

I've used ultrasonic baths to support oxidations with $\ce{MnO2}$ or PCC on silica
Cleaning of modern (composite) materials seemingly requires power control, e.g. to avoid ablation


Anomalous oxidation states of Transition Metals I have noted down the available positive oxidation states of the first
  row of transition elements (on the Periodic Table) from the respective Wikipedia articles of
  the elements.

  $\ce{Sc} - 3, 2,1$

  $\ce{Ti} - 4, 3, 2, 1 $

  $\ce{V} - 5, 4, 3, 2, 1$

  $\ce{Cr} - 6, 5, 4, 3, 2, 1$

  $\ce{Mn} - 7, 6, 5, 4, 3, 2, 1$

  $\ce{Fe} - 6,5,4,3,2,1 $

  $\ce{Co} - 5,4,3,2,1 $

  $\ce{Ni}-  4,3,2,1$

  $\ce{Cu} - 4,3,2,1$

  $\ce{Zn} - 2,1$

Why does $\ce{Fe}$ show not show $+7$ or $+8$ oxidation state?
After $\ce{Fe}$ from $\ce{Co}$ why is there a decrease in number of possible    oxidation states?
Why don't $\ce{Fe,Co,Ni,Cu,Zn}$ show higher oxidation states though they have valence electrons left in $3d$ subshell?

Regarding the maximum oxidation state achievable, there are two competing trends. Both have to do with ionisation energies.

In going to a higher oxidation state, the additional IEs are always balanced against the release in energy in the formation of stronger ionic/covalent bonds. However, if the additional IE is too large, then the favourable bonding will not be sufficient to bring out the higher oxidation state. A simple example is sodium, which cannot adopt the +2 oxidation state because its second IE is simply too large, even though the lattice energy of a hypothetical $\ce{NaCl2}$ crystal is larger than that of $\ce{NaCl}$.

Firstly, the maximum oxidation state is limited by the number of valence electrons available. This is exactly analogous to the case of sodium; therefore, manganese does not exhibit the +8 oxidation state, because its eighth IE involves ionisation from the 3p subshell, much lower in energy than either the 3d or 4s. This therefore accounts for the increase in the maximum oxidation state going from Sc to Mn.

Secondly, the ionisation energies generally increase going across the 3d block, because of an increase in effective nuclear charge. This is the factor which leads to the decrease in the maximum oxidation state after Mn. For example, the seventh IE of Fe is larger than that of Mn; therefore, Fe does not exhibit the +7 oxidation state.

You can look up the IEs in any relevant textbook. However, note that since the +2 oxidation state is very common in the later half of the 3d block, it is usually the third and higher IEs that are more important in this discussion. So, looking at the trend of first/second IEs won't get you anywhere.


Is carbon suboxide planar? I was asked to comment on the planarity of the carbon suboxide molecule. I thought that the Lewis structure of the molecule looks somewhat like an Allene and must have similar bonding. The two terminal pi bonds must be out of plane and hence the molecule would be non planar. But the answer given in my book states that it is planar.
I couldn’t find a satisfactory description of its structure anywhere. The molecule would be linear, but that doesn’t imply that the molecule should be planar.

When we talk about a molecule being linear or planar, we only consider the positions of the nuclei and ignore the electron orbitals. Sure, π-bonds might be out of plane, but that does not count. If it would be otherwise, we would have to consider that any atom except hydrogen has p-orbitals, and some of those are out of plane, and so no molecule ever would have the right to be called planar.

So yes, $\ce{C3O2}$ is kinda like an allene, with all its carbons having sp-hybrid orbitals, which inevitably makes the molecule linear and hence planar.

One might point out that the molecule is relatively easy to bend, but that's another story.


Which reaction is useful to prepare odd number of hydrocarbons? Which reaction is useful to prepare odd number of hydrocarbons:

Wurtz reaction
Kolbe electrolysis
Corey–House synthesis
Oxymercuration

Wurtz reaction produces even products except if we take mixtures. Kolbe ones give even ones too. Oxymercuration merely involves addition, so I think it's Corey–House.

But how does one justify that exactly? Is it because of $\ce{Cu(CH3)2Li}?$

The Corey–House reaction constitutes a general and high yielding method for the joining of two alkyl groups or an alkyl group and an aryl group.

Step one involves preparation of the Gilman reagent from an alkyl halide.

$$\ce{R–X + 2Li -> RLi + Li+X–}$$

Second step, a lithium dialkylcuprate (Gilman reagent) is prepared from the alkyllithium by treatment with cuprous iodide (CuI).

$$\ce{2RLi + CuI -> Li+[R–Cu–R]– + LiI}$$

The third step ,involvs reaction between lithium dialkylcuprate (Gilman reagent) and a second alkyl halide ,resulting in the formation of a C–C bond between the two organic fragments:

$$\ce{Li+[R–Cu–R]– + R'–X -> R–R' + "RCu" + LiX}$$

Example 1

In the following example, gem dihalides react with $\ce{Li+[CH_3–Cu–CH_3]– }$ to give the final product. You could read about it in wikipedia

Example 2

source

Reference

1.https://en.wikipedia.org/wiki/Corey%E2%80%93House_synthesis

2.https://www.synarchive.com/named-reactions/corey-posner-whitesides-house-reaction


What would be the major products of this reaction? Obviously, A would be 3-Bromoo 3-methyl cyclo hexene:

as the stable free radical intermediate would be formed.
But B is what I am having problems with. I read that $\ce{LiAlH4}$ forms alkenes with three degree halides. So my guess was
2-Methyl-1,3-cyclohexadiene:

But when I checked my book, it said B would just be the Bromine getting substituted by Deuterium in (A)
(I'm sorry I don't know any site that allows making chemical structures with isotopes included so I had to describe it like this.)
Can someone please help me out with this?

I will give this my best shot. There doesn't appear that there is any conclusive evidence for this reaction since some sources say elimination is favoured while others say substitution is the major product. However, for this reaction, the difference likely resides in the fact that it is an allylic bromide. This is extremely favourable for substitution reactions because the intermediate will be very stable. Just because of that, substitution may be favoured over elimination and result in the major product being Br substituted for a deuterium atom, just as your book mentioned.


Smoke cobweb formations? One day, my mom decides to light the house on fire, and after we put the fire out, all of the cobwebs in the house are black. Obviously, this is because of the smoke. Later, some renovations guy tells her something about how the cobwebs aren't really cobwebs and when a house fills with smoke, there is some weird chemical formation that makes a black cobweb-like-substance that just happens to show up right where cobwebs would be. Basically, the guy told my mom that those black cobwebs were chemical formations from smoke instead of cobwebs that became black after being in smoke. My mom will not believe me when I say that the renovations guy is wrong and they were obviously cobwebs. I ask her "What happened to the cobwebs we already had?" and "Why would a renovations guy know that it is a smoke formation if there are cobwebs in every house?"

I can't prove to my mom that she's wrong, so I'm asking a bunch of chemistry people to confirm that smoke doesn't stick to walls and form cobwebs.

Awkward. It just so happens that the renovations guy is right. The term you need to google is "smokeweb".

The ionisation process of burning (particularly) synthetic materials like plastics, rubbers and polymers, creates charged smoke particles which can attract towards certain surfaces, and attract to each other creating chains, and ultimately web-like formations. They form under certain conditions called wet smoke, which involves low heat, or smoldering type environments. Smoke webs are typically sticky and pungent and will smudge easily and show up in particular areas where there is limited air flows, typically corners and edges of ceilings, in curtain folds, cabinets etc.

Spider webs will burn up very quickly in a real fire.


Anti-Markovnikov Effect We have seen that $\ce{HBr}$ shows an anti-Markovnikov effect when it reacts with alkenes in the presence of peroxides. Why does only $\ce{HBr}$ shows this effect? How about the other hydrogen halides $\ce{HX}~(\ce{X} = \ce{F}, \ce{Cl}, \ce{I})$?

Main reason is behind is the reaction mechanism involved. $\ce{HBr}$ or other halides usually react with alkenes through an electrophilic addition. Due to the weak $\ce{O-O}$ bond, peroxides act as a source of free radicals. Thus in the presence of peroxides, a free radical chain pathway is followed.

The other hydrogen halides are reluctant to this reaction, because the chain reaction can't proceed due to energy requirements for the generation of the corresponding halogen free radicals.


How does beta decay increases the stabilization of the nuclei? How does beta emission by a nuclei increases the general stability of it? I surely am not a specialist or anything, but this just don't make sense to me, since the objective of radiation is to release energy, but I can't figure out exactly how it's done.

The loss of energy from the electron getting expelled, in addition to its kinetic energy, roughly gives half a MeV (source). Besides, as the strong force acts similarly on protons and neutrons, I guess it won't change considerably in piratical effects. Thus, in my conception, at least, the eletrostatic repulsion from the proton added should overlap the energy who got rid in the process.

Hence, I am confused. I wish to understand how it plays a role in nucleus stabilization.

While this is nuclear physics, the relative number and relative stability of isotopes is of some relevance to chemistry. Further, I spend more time than I'd like perhaps at the intersection of materials science, chemistry, and nuclear physics, so...

There are several good on-line resources to get information on nuclear energy levels and the relative stability of nuclei. These include:
The Nuclear Data Evaluation Project at Triangle Universities Nuclear Laboratory (TUNL), and the Evaluated Nuclear Data Structure File data base at the National Nuclear Data Center hosted at Brookhaven National Laboratory. From these you can get more information than you might want with regard to what is known about nuclear energy levels and stability of the elements.

One cut at looking at stability is the so-called isobar diagram. This compares the relative nuclear energy levels of all nuclei with the same $A$, where $A = P + N$, that is, the sum of the number of protons and neutrons in the nucleus. Since $\beta \pm$ converts between protons and neutrons, keeping $A$ constant, this is the best vantage point for the purpose at hand.

Borrowing from TUNL one finds for $A=14$:

So, we see 4 different nuclei shown on the isobar diagram. The lowest energy (most tightly bound) nuclei is $^{14}$N. To the left is $^{14}$C, which you likely know is unstable and $\beta -$ decays to $^{14}$N, used for carbon dating. To the right is $^{14}$O and you will note that it $\beta +$ decays to $^{14}$N. Neither $^{14}$C or $^{14}$O has as much binding energy as $^{14}$N due to the fact that protons and neutrons are not actually identical in the nucleus. So, the ground state of $^{14}$C or $^{14}$O is about 2.4 MeV higher than $^{14}$N.

High and off to the left you can barely see $^{14}$B, 24.71 MeV above $^{14}$N. It desperately would like to be $^{14}$N, but can't get there directly through the available $\alpha$, $\beta$ or $\gamma$ decay channels. So, it happily $\beta -$ decays to $^{14}$C with a half-life of 16 ms, which will then ultimately decay down to $^{14}$N.

Now, your misconception is that you don't know nuclear physics (and I don't mean that negatively - you applied a fairly simple model, did not get the expected answer, so you asked a question here - GREAT!). But, in the nucleus you have Coulomb repulsion, the strong force, and the weak force, and it all has to be treated quantum mechanically. Taken all together one can determine the binding energy (total or per nucleon) of nuclei, the excited nuclear energy levels, and possible decay paths. It is great fun, but is a different world to explore.


Why does beryllium lose two electrons when it’s orbitals are already full? I have a rudimentary understanding of orbitals, as in what they are, the shapes (l, I think) and of the principles.

So I was doing the electronic configuration for beryllium, it has 4 electrons. So 1s2, 2s2. Why does berylium lose two electrons to complete its “inner duplet”? I thought that atoms want to gain, lose or share electrons so their orbitals are complete and thus don’t react.

It’s orbitals are complete, is it something to do with the promotion of an electron to a higher energy level, but won’t that be temporary change? Or perhaps there is some fundamental misunderstanding on my part.

(I’m in my IGCSE, this isn’t in our syllabus, just trying to understand, so I hope my ignorance is forgiven)

Edit: I am not familiar with hybridisation but I don’t think it applies to this, unless you are giving a bonding example)

Electrons exist in the outer spaces of elements, away from the nucleus. Orbitals are mathematical descriptions of places in these outer spaces where electrons seem to be when they are there; they can be mathematically manipulated and reorganized but are a way to help you imagine what is really going on.

Manipulating the electrons is done experimentally, not mathematically. E.g., an element can be ionized and the energy required can be measured. Beryllium can suffer the loss of up to four electrons: the first requires 9.3 eV, the second requires 18.2 eV, the third requires 153.9 eV, and the fourth requires 217.7 eV. The energy required to remove the first two electrons is in the range of other metals that lose two electrons (Mg 7.6, 15.0; Ca 6.1, 11.9 eV). So Be can lose 2 electrons with a relatively small amount of energy (which will be supplied by something like empty orbitals and ionic or covalent bonding energy to other atoms). Losing those last two is a big jump and essentially doesn't happen in chemistry (it happens in physics and mathematics!).

The empty orbitals on Be are quite high in energy, i.e., if you stuck an electron into a p orbital on Be, it would be floating so far out from the nucleus that it would be looking for a better place. The idea of orbitals gives you a mental picture of what might happen - whether it really happens or not depends on whether that mathematical picture is true and is the most stable (lowest energy) condition available.

Completing shells or orbitals is a qualitative way of looking at the issue, but numerically, ionization potentials gives you the data you need to develop the picture, and the orbitals (as calculated for the hydrogen atom) are a reasonable way to begin.


How does localized bonding theory and hybridisation work? So I understand molecular orbitals and how to do VSEPR models, but I seem to be struggling with understanding localized bonding theory and how to do hybridizations.  After drawing the initial lewis structure then molecular orbital, I do not know where to go in terms of whether something has a $\pi$ or $\sigma$ bond.
Also, how do I know when an atom has a carbon attached as an extra sigma bond or not? I understand how to know if it is $\mathrm{sp}$, $\mathrm{sp}^2$, etc. but I do not know what that means.

First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\text{sp}^2$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning. It's just like vector cross product in Physics (for example the torque on a rotating wheel acts along the axis, i mean what does that even mean physically).

Secondly regarding how to know if an atom has a $\pi$ or a $\sigma$ bond: observing many structures(Lewis structures) and recognizing certain trends is the way out.

Some rules of the thumb that will help you draw structures are as follows:
1. Identify the central atom as the one that can make the most bonds.
2. Hydrogen makes only one bond. If Oxygen is present in the same compound, the Hydrogen is     generally attached to the Oxygen forming a $\ce{O-H}$ group.
3. Second period elements cannot expand their octets(i.e have more than 8 electrons in valence shell)
4. Mostly second period elements can make $\pi$ bonds.(That really solves part of your query). Although in some rare cases third period elements may also make $\pi$  bonds.
5. Never neglect the lone pairs(or non-bonding electrons) of the central atom.

Remember that these are only thumb rules and may be violated sometimes.


Reaction of chloroform, resorcinol and strong base A common reaction used to identify resorcinol (benzene-1,3-diol) in the laboratory is to react it with chloroform in presence of 30% sodium hydroxide (NaOH). A red coloured solution (or precipitate) is produced. I have searched many books and websites, but none of them give any clue about the actual reaction, its products, far less the reaction mechanism.

The reaction is as follows:

$$\ce{Resorcinol + Chloroform + 30\% NaOH -> Red Colour}$$

So any ideas about what reaction is taking place?

Well, turns out to be, the reaction proceeds to product a red colored species.

  Reaction with chloroform in the alkaline medium

  Dissolve 0.1 g in 1 ml of water R, add 1 ml of strong sodium hydroxide
  solution R and 0.1 ml of chloroform R, heat and allow to cool. An
  intense, deep-red colour develops which becomes pale yellow on the addition of a slight excess of hydrochloric acid R.


Solubility of Aluminium Hydroxide in Sodium Hydroxide Solution Aluminium hydroxide is not soluble in water at all, however I was doing some reading and read that it does in fact dissolve in a solution of sodium hydroxide. This just seemed a bit strange to me, so I was wondering if anyone could explain why this is the case.

That's because $\ce{Al(OH)3}$ is amphoteric in nature, it acts as both acid and base. When you drop a chunk of $\ce{Al(OH)3}$ into a solution of $\ce{NaOH}$, you would get a soluble salt, which is sodium aluminate, and some extra water.

$$\ce{Al(OH)3 + NaOH -> Na[Al(OH)4]}$$


Why is platinum used as a catalyst in fuel cell electrodes (and elsewhere)? I have rather basic knowledge of chemistry (at the level of a first year college class). But maybe somebody can still clarify with some simple words: What makes platinum a good catalyst for the electrodes of fuel cells? And why can't other materials be used?

Imagine that a fuel cell is like electrolysis - the other way around.

In the electrolysis of water, you'd apply a certain potential (a voltage) to the electrodes and observe hydrogen and oxygen formed at the electrodes.

In theory, you need a particular potential to run the reaction. In reality - and this depends on phenomena on the surface of the electrode material - you often need more.

With platinum, however, this overpotential for hydrogen is zero.

This means that you have no losses, neither in electrolysis, nor in the fuel cell.

This might be oversimplified but I hope it's not completely wrong ;)


Definitions of autocatalysis Is there a formal definition or algorithm that can take a list of reactions and tell me whether it contains a set of species that can produce itself autocatalytically?

It is clear that the following reaction is autocatalytic:
$$\ce{
A + B \to 2 B
}$$
B appears on both sides of the equation, so it's a catalysis reaction, and there is an extra B produced on the right, so it catalyses its own production from A.

But the following reaction scheme can also be considered to contain autocatalysis:
$$
\ce{B + C \to D + E\\
A + D \to 2C}
$$
If this system starts out with plenty of A and B, together with one molecule of C, then the C molecule can react with a B to produce a D, which can then react with an A to produce two C's — whereas, without the initial "seed" molecule, no reactions will take place. (I'm neglecting reverse reactions here of course, but hopefully it gets the point across.)  The autocataltic cycle can easily be seen in the following diagram, where circles represent species and boxes represent reactions:

Another non-obvious example of autocatalysis is a reaction scheme of this form:
$$\ce{
A + C \to C + D\\
B + D \to D + C + E}
$$
Here, C and D each catalyse the production of one another from the precursors A and B. I'm aware of Kauffman's definition of reflexively autocatalytic sets, which applies to schemes of this form, but would miss an autocatalytic cycle of the kind shown above, since it doesn't consist of single-step catalysis reactions. So I'm looking for a definition of autocatalysis that would include both cases, as well as (hopefully) other, more complex schemes that I haven't yet thought of.

I can't tell if the modifier "formal" applies only to "definition" or to "algorithm" also, in the first sentence of your question's body. In case any algorithm would do: this looks like a pretty standard instance of the cycle-detection problem that one learns in undergraduate computer science.

If you only want to detect the presence of cycles, then the algorithm is essentially just to attempt a topological sorting of the graph; if the attempt fails, you have a cycle, and I guess that would mean that you have some sort of catalysis going on. (I didn't major in chemistry; please yell at me if that's non sequitur.)

Okay, so mere cycles are not enough - it seems you want cycles that have "gain". How about a tortoise and hare and have these multiply a state variable by the ratio of outgoing to incoming edges to reaction nodes (denoted by squares in your diagram)? If the hare catches up to the tortoise, that implies a cycle, hence catalysis; then if you compare their state, if the hare has accumulated a larger "gain" than the tortoise, I would say that that shows the presence of autocatalysis?

I'm not sure how this works if there are multiple, linked cycles. You could have one cycle in your graph that's autocatalytic, and another cycle linked to it that "uses up" its catalyst (something of a contradiction). For example:

B + C -> D + E
A + D + H -> 2C + F
2F + G -> H

Is that still an autocatalytic system?


Lithium compounds with good solubility in organic solvents? I want to dissolve enriched $\ce{^6Li}$ into a liquid scintillator (made almost purely of di-isopropyl-naphthalene, mixed isomers, with a tiny bit of fluors and wavelength shifters) for enhanced neutron detection and background rejection. My group's previous attempts at this have been to use Lithium-6 Chloride with surfactants, but this negatively affects the attenuation length of the scintillation light passing through the scintillator (since the droplets of the microemulsion are essentially giant scattering surfaces).

Having had some O-Chem and P-Chem as an undergrad I know some organolithium compounds are soluble in organic solvents, but those are eat the container walls nasty. I know that lithium bromide has some limited solubility in organic solvents, but I need to get to at least 0.1%, preferably 0.5% atom doping/fraction, so I need something that dissolves well. I know that lithium perchlorate is soluble in many organic solvents, but the explosive nature of the mixtures it makes with those solvents kills that idea.

This leads to my question:

Are there any lithium compounds that are not caustic, nor explosive, but are soluble in greasy organic liquids (like di-isopropyl-naphthalene)?

Under inert conditions you may try dissolving pure metallic lithium. Lithium forms naphtalenide salt , I don't see why di-isopropyl-naphthalene should be any different.

Another option would be use of 12-crawn-4, which make a lot of salts lipo-soluble. Presence of TMEDA (tetramethylethilendiamine) should affect solubility of many lithium salts as well, though I would focus on salts like benzoate or perchlorate.


Calculating the electrons an atom wants to gain/lose to reach a noble gas I'm trying to understand ions.

From what I understand, an ion is when the atom gains or loses electrons. More electrons means it is negatively charged (anion). Less electrons means it is positively charged (cation).

My problem is with the following premise:

  Many atoms gain/lose electrons with the hope of having the same number of electrons as the closest noble gas in the periodic table.

The book uses Potassium as an example. Looking at my table, the closest noble gas to Potassium is Argon, which has $18$ electrons. Therefore Potassium wants to lose one electron...

$$\ce{K^+}$$

Nice, I get that. But then I noticed something odd about all the examples in my book: all the elements they picked are elements conveniently one or two steps away from a noble gas. They never picked an element in the middle of the table like, say, iron.

Iron has $26$ electrons. Its closest noble gas is Argon, which has $18$ electrons. So iron wants to lose $8$ electrons (becoming a cation):

$$\ce{Fe^{8+}}$$

Is this correct? My book literally avoids picking elements in the middle of the table, so I feel there's something weird about this.

Also, I noticed that in every example, the book mentions the Group of the element (Potassium is group 1A, etc...). How is that even relevant? All you have to do is count the number of electrons, no?

The book's explanation about a noble gas configuration is somewhat accurate, but fairly incomplete.

The elements on the right and on the left of the periodic table (the alkali (earth) metals, the halogens, the chalcogens (the group that starts with Oxygen) and the pnictogens (Nitrogen group)) have electron configurations that make it somewhat easier to lose/gain electrons to "look" like noble gasses (this is why alkali metals and halogens are so reactive: the destabilization from the charge is very minimal compared to the stability of the electron configuration).

However, as you have observed, the book goes to some effort to avoid talking about transition metals. There is a reason for that.

Noble gas configurations are a subset of the stable configurations of electrons. In reality, what's actually being aimed for is an element with no incomplete electron shells.

What is an electron shell? The actual quantum mechanical definition may be a bit more complicated than you need, but for your purposes, it suffices to say that it's how electrons will tend to be distributed around a nucleus. Note that this is extremely distinct from saying that the electron "orbits" the nucleus.

If you look at a periodic table, there are four general regions to look at:

The labels "s", "p", "d", and "f" all refer to the various electron orbitals, and the blocks all indicate what the outermost (or "valence") shell is.

So while elements in the s and p blocks can lose a small number of electrons get complete shells (which will resemble the noble gases), transition metals are a bit more complicated. And by a bit more complicated, I mean that there are very rough rules in place that have almost more exceptions than they have actual examples of following the rule.

But in very general terms, a transition metal will lose electrons. If they are in the first half of the transition metals in their row, they can lose a number of electrons less than or equal to the number of d electrons plus the number of s electrons. This maximum number of electrons lost will make them resemble the noble gasses in configuration, but it's only appreciably stable when the transition metal is in the first half of transition metals in its group ($\ce{Mn^{7+}}$ is well documented in ionic compounds, but $\ce{Fe^{8+}}$ doesn't exist a lot because it has too many electrons to try to lose to comfortably fall into a noble gas configuration). But transition metals can lose electrons up to this point. The issue is, most transition metals can adopt multiple charge configurations for various reasons, so generally, when writing an ionic compound with a transition metal in words, you need to specify the charge of the transition metal (actually, its oxidation state, but in an ionic compound that's mostly splitting hairs). So you can have "Iron (II) Oxide", which is a black powder and "Iron (III) Oxide" which is rust. These compounds will have different chemistries because of the charge on the transition metal.

Suffice to say, your textbook is avoiding transition metals because transition metals are weird.


Is consumption of chlorine, like fluorine, harmful to bones? Is consumption of chlorine harmful to bones ?

I know fluorine consumed with water can cause fluorosis to bones and joints.

Can chlorine,  a similar halogen like fluorine, also cause damage to bones and bone joints?

The OP uses the terms chlorine and fluorine. In the context of the question, fluorine doesn't make sense and it is assumed here that the the OP means fluoride. In the case of chlorine, it is likely, though not certain, that the OP means chloride. This answer will address the question primarily in terms of fluoride and chloride, with a note about chlorine.

Bones are largely composed of calcium compounds (mainly hydroxyapetite, $\ce{Ca5(PO4)3(OH)}$). The calcium salts of bone are very slightly soluble, and thus they are in an equilibrium between the solid form and the dissolved aqueous form. Because calcium fluoride is extremely insoluble, fluoride can displace the naturally occurring anions to form water-insoluble calcium fluoride. This results in weaker, more easily fractured bones, which is the condition known as skeletal fluorosis.

Calcium chloride is about five orders of magnitude more water soluble than is calcium fluoride, so an analogous displacement of the naturally occuring anions in bone by chloride does not take place. Thus the ingestion of chloride has no negative effect on bone material. The same is true of chlorine, in the case of chlorinated water, as there is no mechanism by which chlorine can be ingested (or inhaled), metabolized and then interfere with bone composition.


When an enzyme is diluted with water, what is slowing down the rate of reaction? I'm specifically talking about the decomposition of hydrogen peroxide by bovine catalase. The number of catalase molecule obviously doesn't change, so there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, and doesn't change the pH or temperature, then shouldn't the rate of reaction remain the same regardless of concentration, since every environmental conditions & the number of catalase molecule in the solution before dilution is the same after it is diluted?

[…] there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, […]

Q: Why is dating between a fixed number of guys and girls more difficult on a large square packed with nuns than in a small club (without the nuns)? ;)

A: It's more difficult to bump into each other since the girls a more far away and there are legions of nuns in between!

You are right in that the absolute amount of catalase molecules does not change upon dilution, but the concentration does!

Reactions are about collisions of the reactants and in the case of dilutions, there are more collisions with innocent bystanders than with an other reactant.


What makes substances explosive Can someone explain clearly to me why some things are explosive?

I could say explosives are things that blow up, but this would not a helpful answer.

It is also not that helpful to describe explosives as compounds with highly exothermic decomposition. This characteristic alone does not define an explosive.

What matters is rate of the reaction in the bulk compound. This rate defines two key types of explosive. If the reaction, once started, proceeds through the bulk compound faster than the speed of sound we call it detonation. If the reaction is slower than the speed of sound we describe it as deflagration.

This distinction matters because of the different uses explosives can be put to. Propellants, such as the explosives used to drive bullets, tend to deflagrate (as you want a relatively slow, controlled reaction to drive the shell out of the barrel without blowing up the barrel). But if you want something to really go bang and create a mess (as the military do) or if you want something to break up rock (as quarriers and miners do) then you want something that will detonate.

The structures of some well-known explosives are shown below:

Most well known real world explosive compounds are made from these compounds plus some added materials to make them easier to handle. C-4, a well known military explosive is mostly RDX with plasticisers to make it easy to mould like plasticine. The favourite terrorist plastic explosive is Semtex, similar to C-4 but made from a mixture of PETN and RDX. TNT is often used in high-explosive military shells (partially because it can be melted by steam and poured into shell casings). The first modern explosive, dynamite, is based on nitroglycerine mixed with diatomaceous earth (which makes it much easier to handle than the dangerously shock-sensitive pure compound).

The chemical reactions typically involve the production of CO2, H2O and, sometimes, N2.

Another characteristic that matters is how easy the reaction is to set off. Some compounds are very shock sensitive like mercury fulminate or nitroglycerin. This is only useful in detonators (in small quantities) or for chemists who have a death wish. Many bulk explosives require a detonator to trigger an explosion, making them much safer to handle and transport. C-4 in small quantities will actually burn slowly (and was sometimes used by soldiers in the Vietnam war to warm rations).


internal alkyne stability to strong alkali Internal alkynes are relatively electron rich and are not usually prone to attack by strong (nucleophilic) bases such as hydroxide or alkoxide.
However, this isn't always the case.  A very nice overview of the most common reactions of alkynes says:

The sp-hybrid carbon atoms of the triple-bond render alkynes more electrophilic than similarly substituted alkenes. As a result, alkynes sometimes undergo addition reactions initiated by bonding to a nucleophile. This mode of reaction, illustrated below, is generally not displayed by alkenes, unless the double-bond is activated by electronegative substituents, e.g. $\ce{F2C=CF2}$, or by conjugation with an electron withdrawing group.
$\ce{HC≡CH   +   KOC2H5    —>[in C2H5OH at 150 ºC]   H2C=CH-OC2H5}$

I'm interested in more information about this and similar reactions, particular for internal (not acetylenic) alkynes.
Unfortunately, the linked page doesn't provide a reference for this reaction, and searching in Google Scholar for "internal alkyne" "potassium hydroxide" doesn't provide useful results.
Thus my questions are:

Does hydroxide ever react similarly with internal alkynes, at any temperature?  If so, under what conditions?
How long does the ethoxide reaction indicated above (at 150 °C) take?  Days?  Hours?  Minutes?  Would it be significantly different for an internal alkyne rather than acetylene?
In the case of a hydroxide reaction, would the ultimate product be a ketone (a.k.a. base catalyzed hydration)?

According to A Textbook of Organic Chemistry by V. K. Ahluwalia and Madhuri Goyal

  1-Butyne may be isomerised to a more stable 2-butyne in presence of alcoholic potassium hydroxide and the reverse of this reaction may occur with sodamide

(see 17.2.1 of Best Synthetic Methods: Acetylenes, Allenes and Cumulenes for reaction conditions in the 2-alkyne to 1-alkyne direction, actually many relevant reactions are in this book)

So in the present of the strong base $\ce{NaNH2}$ the internal alkyne $\ce{H3CC#CCH3}$ rearranges to $\ce{HC#CCH2CH3}$

My thinking is that with just hydroxide the more stable 2-butyne is favored, while if an even-stronger base is used, that fact that the acetylenic-H can deprotonate pulls the equilibrium toward 1-butyne.

Additionally, see 17.2.11 Isomerisation of 1,4 bis(alkoxy)-2-butynes to the corresponding allenes, which is catalyzed by t-BuOK.

Add see table 17.1 "Base catalyzed isomerisations of acetylenic compounds" about 10 different internal acetylene reactions are listed in the table.


Mathematical definition of a symmetry operator I had a hard time understanding what it means to apply a symmetry operator to a function, so I wondered if there is a formal way to define this? As far as I understand, applying a symmetry operation to a function means the following:
$$ \hat Rf(x) = f(\hat Rx) $$
The result has then to be the same function with a coefficient in front of it. Otherwise it would not be symmetric in regard to the symmetry operator $\hat{R}$.
Is this correct and can it be shown that this is how to apply symmetry operators to functions ?

That's not entirely correct. You're making at least an extra assumption here. You're assuming that $\hat{R}$ is a function from $D\mapsto D$ where $D$ is the domain of $f$, but there's no reason why $f$ must operate as a function on $D\mapsto D$. It might just as well be $f: D\mapsto \mathbb{R}$.
I think what you're really asking for is that a symmetry element $\hat{R}$ such that $\hat{R}: D\mapsto D$, and $\forall x\in D$, $f(x) = f(\hat{R}x) = (f\circ \hat{R})x$.
Then for example, for $f(x) = (\cos x, \cos x)$, the transformation $\hat{R}$ mapping $x \rightarrow -x$ would would a valid symmetry element.


How to drain engine oil by understanding oil viscocity issues I asked this on the stackexchange automotive site but didn't get much response perhaps because it entails some chemistry knowledge. I could have asked without mentioning the automotive details however I think it puts the question into context making it easier to give a good answer.

When you drain an engine of oil in some engines about 12.5% of old oil remains because it coats engine parts, sits in galleys, and some is at the bottom of the sump container(below the drain plug) and won't come out unless you tilt the car which is quite impractical.

The general approach is to remove the older oil by dilution\further oil changes. As an example if you have a 4 liter capacity and an oil change allows you to fill 3.5l new oil, this means 0.5l\12.5% of old oil remains and so a further oil change will reduce the initial 12.5% to 1.5% and a further oil change to 0.19%.

However from my understanding this won't work when you mix oils of different viscosity because oils of different viscosity do not mix when they settle at cold\low temperature(which is where it will be at point of drain). But they do mix when they are at hot\engine temperature.  Is this correct?
Do I need to be using the same viscosity oil to ensure dilution happens at point of drain e.g. 5w30 needs to be mixed with 5w30 but adding 5w40 to 5w30 won't get the dilution effect? What would happen if you added 10w40 to 5w30, would you get dilution at point of drain?

Also how do the oils settle in practice? would the thinner oil sit on top of the thicker oil?  Therefore if you are trying to dilute with a thinner oil the thicker oil may settle below the drain plug and sit below thinner oils in galleys so it may not dilute\get drained with successive oil changes?
Suppose I had a container with 0.5 liter of 5w30 and 3.5 liters of 5w40 would the latter sit below the former? What if you had 5w30 mixed with 10w40?

Also can factors besides viscosity e.g. additives in oil or any other factors affect dilution?

My issue is that I have 0.5 liters of black and thick looking oil in my engine(which I think is affecting performance) and I don't really know the viscosity of the oil they put in.  I don't want to do countless and costly oil changes trying to get it out so approaching this in an educated way may help.

Thanks.

If you drain the oil when the engine is hot or at least warm, it is much less viscous and will drain pretty efficiently. Oils of different viscosity or even different density will surely mix well while the engine is running, so if you still want to do a flush, be sure to run the engine for a while after you add the new oil and before you do your second oil change.


What is the reason for more stable keto form to tautomerise to a less stable enol form? In the mechanism of SeO2 I have seen that a more stable keto form first converts to a less stable enol form and then yields a diketone. I have seen many other reaction where the reactant converts to an unstable form first and then reacts. My doubt is why does the reactant converts itself rather than just simply not react ?

We can think of it an an equilibrium that is disturbed by another, irreversible reaction.

Think of a more familiar reaction.  When you try to dissolve magnesium hydroxide in water you get only a little solubility; the dissolved magnesium and hydroxide ions are only a minor species at equilibrium.  But when hydrochloric acid is added the acid reacts selectively with the hydroxide ions, neutralizing them; in response more magnesium hydroxide keeps dissolving into the water (to form magnesium and hydroxide ions, the latter then reacting with more acid) until you have a salt solution containing the accumulated magnesium ions and the chloride from the acid.  Reaction continues until you've dissolved all of the magnesium hydroxide or neutralized all of the acid.

Similarly, the enol form is only a minor species in the equilibrium, but when you add the selenium dioxide it reacts selectively with the enol form.  This draws more of the carbonyl compound into the enol form in the same way that the acid in the example above draws more magnesium hydroxide into its dissolved form.  Thereby the reaction keeps going until you run out of one reactant or another.


Is it rigorous enough to change a DFT functional after geometry optimization to calculate excited state properties? I have done the geometry optimization with the B3LYP functional, however, there is a need to calculate the excited state energy, so the CAM-B3LYP functional should be better suited for that. Do I need to reoptimize the geometry to use CAM-B3LYP or are these functionals simillar enough so no reoptimization is needed?

What is sometimes done is use the geometry and thermochemical corrections (i.e. the results of the frequency calculation) of method X, but use the electronic energy (SCF energy) of another method Y. This is generally noted Y//X. Since the geometries and thermochemical corrections typically are not much influenced by the quality of the method (beyond a certain point), this scheme gives good result. The electronic energy can however benefit from a larger basis set or a better computational method (e.g. MP2 with DFT geometries).
I am assuming this is what you have seen in the literature, as it is relatively common, especially for large molecules like organometallic complexes. I am not too familiar with excited states, but the same thing likely applies.
However, you must always use the exact same method for the geometry optimisation and the frequency calculation. The frequency calculation uses the curvature of the potential energy surface (PES) to calculate approximate corrections. If you are not at an extremum (minimum for equilibrium structures or saddle point for transition state), the results are meaningless. Changing the method will move you off the extremum sufficiently and your frequency calculation will likely have several imaginary frequencies.


IUPAC nomenclature of organic compound I strongly feel that the answer should be option (1) as the ester functional group has a greater preference over the double bond and hence it should get a lower locant.
But the source of the question says that the answer is option (2). My question is why the double bond gets a lower locant than ester in this case?
As far as I know the lowest set out locants rule is applicable for substituents but when the comparison is between functional groups the preference order plays a greater role. Please let me know if I am wrong.

As already explained in a related answer, the most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic 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 principal characteristic group is the carboxylic acid. Therefore, the lowest locant is assigned to the $\ce{-COOH}$ group in accordance with Rule (c).

Next, a low locant is assgined to the double bond according to Rule (e). Therefore, the name of the structure without any further substituents is cyclopent-2-ene-1-carboxylic acid.

Finally, a low locant is assgined to the remaining substituent to Rule (f).

Thus, the correct name is 4-(acetyloxy)cyclopent-2-ene-1-carboxylic acid.


Comparison of evaporation rate of liquids Evaporation: Evaporation
is a type of vaporization that occurs on the surface of a liquid as it
changes into the gas phase. When a molecule near the surface absorbs
enough energy to overcome the vapour pressure, it will escape and
enter the surrounding air as a gas.
Boiling point: The
boiling point of a substance is the temperature at which the vapour
pressure of a liquid equals the pressure surrounding the liquid and
the liquid changes into a vapour.
Volatility:
Volatility is a material quality which describes how readily a
substance vapourizes.
Flash Point: The flash
point of a volatile material is the lowest temperature at which its
vapours ignite if given an ignition source.
Autoignition
temperature:
The autoignition temperature or kindling point of a substance is the
lowest temperature in which it spontaneously ignites in a normal
atmosphere without an external source of ignition, such as a flame or
spark.
Fire point: The fire point
of a fuel is the lowest temperature at which the vapour of that fuel
will continue to burn for at least five seconds after ignition by an
open flame of standard dimension.
Smoke point: The smoke
point, also referred to as the burning point, is the temperature at
which an oil or fat begins to produce a continuous bluish smoke that
becomes clearly visible, dependent upon specific and defined
conditions.

How can we compare the evaporation rate for water, alcohol, petrol and kerosene oil?
If we consider intermolecular forces as the criterion, then the order of evaporation would be:
water < alcohol < kerosene oil < petrol
But part of intermolecular forces still act in vapour phase...
What should be the criteria to compare evaporation rate of liquids at a given temperature?

Evaporation rates are used to compare solvents for different reasons and under various conditions. One condition would be under vacuum in a rotary evaporator, to remove a low-boiling solvent to assist crystallization of a solute. Another condition would be in a paint, where you desire a higher boiler to allow enough time to spread the paint without brush strokes.
A practical way to compare evaporation rates of solvents is to put a small amount into a dish and weigh it as it evaporates.  Pure compounds will have a fairly sharp disappearance and therefore a sharp time to evaporate; mixtures like kerosene or gasoline will evaporate fast at first, then slower and slower. This time, or the rate for 90% evaporation, can be compared with another solvent under the same conditions.
Evaporation rate is a commonly listed on Safety Data Sheets, and in other tables, generally compared to butyl acetate or ether. One table is at https://www.dispersetech.com/content/11-properties-of-common-solvents.
Many tables of solvent characteristics do not include evaporation rate, but do show vapor pressure at room temperature, e.g., https://sites.google.com/site/miller00828/in/solvent-polarity-table. Vapor pressure at some temperature can be a good indicator of evaporation rate, but a very strong factor in evaporation is wind, or blowing across the solvent surface. Water is particularly complicated because of relative humidity, but it and wind and temperature have been investigated for reservoirs: https://agupubs.onlinelibrary.wiley.com/doi/pdf/10.1029/97WR02405.
Chemical distributors often list all the properties of all the solvents they deal with - and sometimes of materials they don't deal with.


Why hyperconjugation can't stabilize carbanion? If Resonance can stabilize both carbo cation and anion then why can't hyperconjugation stabilize carbanion ? Since both Resonance and Hyperconjugation are pretty much the same, i.e both involve delocalisation or diffusion of electrons( at least in model) ?

In carbanion the carbon carrying negative charge contains one $\mathrm{sp^3}$ hybrid orbital which contains a pair of electrons and therefore there is no possibility that it will undergo bonding interaction with the neighbouring $\ce{C-H}$ bond, due to repulsion between the electrons.


Why does AsF3 boil at 60.4 Celsius and AsF5 at -52.8? I don't understand why the boiling point of $\ce{AsF3}$ is at 60.4 degrees Celsius and $\ce{AsF5}$ boils at -52.8 degrees Celsius. I understand that it has to do with the bondings, but I can't see what would cause such difference?

This has to do with the dipole moment of those compounds. As the difference in electronegativity between As (2.18) and F (3.98) is very large, those bonds are highly polarized.

Using the VSEPR theory, $\ce{AsF3}$ is an $\ce{AX3E}$ molecule. As such, it has a pyramidal structure. Therefore, there is a separation of charge between $\ce{F^\delta-}$ and $\ce{As^\delta+}$ in the molecule, leading to a dipole moment. This implies dipole-dipole interactions between those molecules, which greatly increases the boiling temperature.

However, $\ce{AsF5}$ is an $\ce{AX5}$ molecule and has a trigonal bipyramidal structure. The centers of charges coincide and the net dipole moment is null. There is no dipole-dipole interaction between the molecules and the boiling point is much lower.


food grade diatomaceous earth Is all unaltered diatomaceous earth (DE) food grade?

I am aware that all have not spent the money to have it certified and labeled as food grade, but I'm thinking that is just a way for the government to get more money and also have more control. All countries do not have an FDA, and therefore have no need to certify it, and have used it for thousands of years.

I just want to know the answer without somebody being afraid to say it is still food grade, labeled or not. I know the DE exposed to extreme heat is not safe and is for pool filters.

I very much doubt that all diatomaceous earth is food grade.  For some reasons why, consider the Wikipedia page for diatomaceous earth: "Kieselgur" was discovered when folks in the 1800s were trying to dig a well.  Since knowledge of hygiene was some decades away, it's not implausible that those people may have been trying to dig a well that happened to be near an outhouse.  I'd doubt that diatomaceous earth recovered from a site near an outhouse would be food grade.

Also, Wikipedia lists "salt-water-derived" diatomaceous earth and specifically mentions that it is:

not suitable for human consumption
not effective as an insecticide
often calcined, which converts the safer amorphous silica present in most natural diatomaceous earths into a far more dangerous crystalline silica.  This can lead to silicosis if long-term exposure to airborne microparticles of crystalline silica occurs.

Thus, any diatomaceous earth that contains lots of salts or which has been calcined is unlikely to be food grade.


How well can triglycerides be resolved using TLC? The only part of triglycerides that significantly interacts with a TLC plate are the ester moieties which are present in all triglycerides. Other than that, they do not differ much, except in mass (dependent of the chain lenght of FAs) and shape (saturation degree of FAs). How much do these differences alter the interactions causing the resolution?

If you work with normal phase TLC, you may lower the polarity of your mobile phase, and instead of mixing ethyl acetate with petrol ether, you may give diethyl ether / petrol ether (like 5:95, v/v) a try.

Some of the TLC manufactures even provide databases, pointing towards publications were compounds of interest were successfully separated.  Often, these references refer to a separation on TLC-scale.  CAMAG's CCBS Database is one noteworthy example.


What does A. R. in A. R. sodium nitrite mean? I'm reading about a reaction from a 1970s organic chemistry book. It references A. R. sodium nitrite. What does the A. R. mean?
I see from online that "Ar" means aromatic ring. Is that the same thing?
But I've also seen "Sodium Nitrite, Granular, AR®" with a registered trademark, implying it might be a specific brand of Sodium Nitrite, but that didn't make sense. https://www.chemical.net/sodium-nitrite-acs
Are there different types of sodium nitrite?

I don't know if I have completely answered, but here's my go:
AR : (ANALYTICAL REAGENT) : Reagents essential for analytical purpose and research work having high purity. If reagent meets the requirements of the American Chemical Society Committee on Analytical Reagents, it will be denoted as an AR (ACS) reagent.
Also see:

https://www.researchlab.in/?product-details/grade-of-purity/
https://www.avantorsciences.com/pages/en/grade-definitions#:~:text=AR%E2%80%94The%20standard%20Macron%20Fine,an%20AR%20(ACS)%20reagent.

EDIT:And note that it isn't a Analytical ring


Why doesn't a neutral solution at any temperature have pH 7? I am given 4 statements to choose from and find which is correct about a neutral solution:

$K_\mathrm{w} = \pu{1.00\times10^-14 mol^2 dm^{-6}}$
The solution contains only $\ce{H2O}$
$\ce{[H+] = [OH-]}$
$\ce{pH =7}$

I can see that the first statement is wrong as that is only the case at $\pu{25 ^\circ C}$.

The second statement is also incorrect - it can be when $\ce{H+}$ ions and $\ce{OH-}$ are equal in concentration.
So I can see that option 3 is correct.

However, I am confused why option 4 is wrong.
Why is option 4 wrong?

As the temperature changes the value of $K_\mathrm{w}$ changes(as you wrote) and then the concentration of $\ce{H^+}$ and $\ce{OH^-}$ also changes and now the concentration of $\ce{H^+}$ will not remain $10^{-7}$ and hence the neutral point shifts. Therefore, at any temperature the $\ce{pH}$ of neutral point is not 7.


Are there computer searches for new theoretical minerals in a similar vein to Folding@Home or SETI@Home? As a surprise gift to a friend I would like to discover a new theoretical mineral and have it named after them. I have some understanding that not all minerals have been discovered, and some minerals exist in theory but have not been found on earth.

Is there a computer search that looks for new minerals? Would it be difficult to write one, if one does not exist?

By Ross Snider's request, I have moved my comment into an answer - whilst it does not address computational searches for minerals, it touches on the complexities of actually getting a mineral officially named:

The International Mineralogical Association Commission on New Minerals, Nomenclature and Classification seems to require a physical specimen for consideration for official naming of a new mineral. Under current rules, anthropogenic minerals are also excluded, so if you came up with a hypothetical mineral it seems you would have to also find a naturally occurring specimen to get it named.

To add to this, I am not aware of any software for systematically searching for plausible minerals, however computational packages for the simulation of crystalline structures certainly exist. Determining the electronic structure of a crystal - which is what is really required to make conclusions about geometry, stability and so on - generally involves the use of plane wave QM codes such as VASP, CASTEP or ABINIT, the latter of which is free. The feasibility of mineralogical calculations using these methods is strongly dependent on how crystalline the mineral is and how ideal it is (from a perspective of not being heavily defected or substituted) as they work within the constraints of a periodic unit cell or supercell.


Are there chiral carbons in PVC polymers? PVC polymers are formed from the monomer CH2=CHCl.
Below picture is a section of PVC molecule.
Is there any chiral carbon centre in this polymer molecule. If yes, then how?

I have a source of CAIE examination paper implying carbon bonded to Cl group is chiral, but as in the picture the group on both sides of that carbon is same.
(Source, Qno:38)

Yes, all polymer chains of the form - (CHR - CH2 -) are in principle chiral, the problem is that in the left and right side of the backbone in  BBL - CHCl - BBR is in effect indistinguishable.

Even in a PVC molecule with perfect tacticity, only the end groups (they usually are different) make the whole thing chiral, and they are too far away. In a structure with a defect tacticity, the next defects in the chain create chirality at a C atom, but they (usually) occur with the same probability distribution on both sides of the local carbon atom and the whole molecule.

That's why in polymers, optical activity is not observed, instead, the relative occurrence of pairs, triplets, etc. of neighbouring C* atoms with identical stereochemistry can be specified (and detected via NMR). That way, you can differentiate e.g. chains which go RRRRRSRRRRRSRRRRRS or RRRRRSSSSSSRRRRRRSSSSS, which will show remarkably different mechanical properties due to their crystalline structure.

A different thing are of course polymers with chiral side groups, e.g. proteins. A solution of those can (or rather does, unless the side group is racemic) show optical activity.


Why do things want to lower their energy? Like if we consider a container of gas molecules as a system, Why does it want to lower its energy?

I have read that always more spontaneous events are favored. So Gibb's Free energy should have a negative value i.e ΔG=ΔH-TΔS<0. So Entropy Contributes towards more spontaneity. But Randomness contributes towards more entropy and having more Internal energy causes more random motion.
So overall more energy should contribute towards more Spontaneity which is favored in nature. But we observe that things tend to lower their energy. Please Explain.

Edit: Well I guess I might have found the answer -
When a system lowers its energy, ΔH becomes negative ( due to it being exothermic) so at moderate temperatures, the entropy contributes to lower spontaneity but ΔH is more negative so ΔG becomes more negative(more spontaneous). While at high temperatures, endothermic processes are favored (ΔH positive)
Thermodynamics is so beautiful!

Maybe my shortest answer ever:

Energy is loosely a measure of how disturbed a system it is from its natural, grounded, non-excited state, and it is from here that the "less energy, more stable" thing comes from. Your question reverse-engineers that.


Long lasting dry ice I have a chemistry question but I'm not sure which subtopic it'd go under, so I'm asking it here. I'd really appreciate any help I can get, if you have an answer, or if you can tell me where I can find the answer. I should also mention I don't know anything about chemistry, so I'm sorry if my question sounds stupid :-)

I was wondering if it's possible for dry ice (the industrial ones or those that are sold at stores) to stay cold/frozen in a rather warm/hot room temperature for around 2-4 hours. Let's say blocks of 3x3x4 inches. I just want to know if that's feasible or not.

Put them in a container. It could be open at the top, but should be enclosed in the bottom.

The dry ice will insulate itself with a layer of cold air/CO2. As long as they are not under direct heating illumination or the convection of air in the room is high, this should be sufficient for a suitable mass of dry ice.

We routinely keep dry ice in "open" unpowered freezers at my lab (with no refrigeration) for the duration of the day. If your room is very humid you can get buildup of ice / condensed water, so some insulation around the container might be prudent.


Determining the half-reaction equations for extraordinary redox reaction? I have the following question stated in the science  teacher exam:

  Write the half-reaction equations for  the following reaction  by using oxidation numbers:
  $$\ce{Na2SO3  +  Na2CrO4  -> Na2SO4 + Cr(OH)3 }$$

I am having some difficulty determining the half-reaction equations because the question doesn’t indicate either to the states of the species in the reaction nor to the medium of the reaction.

By determining oxidation numbers it is possible to see what is being oxidised and what is being reduced. I have indicated oxidation numbers below in parentheses:

$$
\ce{
\overset{\color{black}{(+1)}}{Na2}
\overset{\color{red}{(+4)}}{S}
\overset{\color{black}{(-2)}}{O_3}
+ \overset{\color{black}{(+1)}}{Na2}\overset{\color{green}{(+6)}}{Cr}\overset{\color{black}{(-2)}}{O_4}
->
\overset{\color{black}{(+1)}}{Na2}
\overset{\color{red}{(+6)}}{S}
\overset{\color{black}{(-2)}}{O4}   + \overset{\color{green}{(+3)}}{Cr}(\overset{\color{black}{(-2)}}{\ce{O}}\overset{\color{black}{(+1)}}{\ce{H}})3}
$$

By supposing the reaction takes place in an acidic medium, I failed to answer the question.

But, by supposing the reaction takes place in the basic medium, I conclude to following  two half-reactions:

Oxidation half-reaction:
$$\ce{2OH-  + Na2SO3  -> Na2SO4  + H2O + 2e-}$$
Reduction  half- reaction:
$$\ce{3e- + 4H2O  + Na2CrO4  -> Cr(OH)3 +2Na+  + 5OH-}$$
The  completely balanced equation:
$$\ce{5H2O  + 3Na2SO3  +2Na2CrO4 -> 3Na2SO4 + 2Cr(OH)3 + 4NaOH}$$

I am not sure about this solution. Are there, and where if there are, mistakes in the solution? What is the correct solution?

You have rightfully assumed the reaction takes place in either basic or neutral medium, otherwise $\ce{Cr(OH)3}$ wouldn't be the product.
This automatically suggests $\ce{H+}$ are not used for balancing half reactions.
Your solution is fine for the provided reaction given that chromium(III) hydroxide stays precipitated.
Here I've basically formatted your solution in a form I'm more familiar with:

\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ \overset{+3}{Cr}(OH)3 + 5 OH-} &|\cdot 2 \tag{red1}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox1}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 5 H2O &→ 2 Cr(OH)3 + 3 SO4^2- + 4 OH-} \tag{redox1}
\end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 5 H2O → 2 Cr(OH)3 + 3 Na2SO4 + 4 NaOH} \tag{R1}$$

This implies that reaction begins in neutral medium; however there is remaining sodium hydroxide ($\ce{4 NaOH}$ on the right), so the pH will increase over time.
This will result in further dissolving of $\ce{Cr(OH)3}$ in $\ce{NaOH}$ yielding $\ce{Na3[Cr(OH)6]}$ complex, which is likely the final product.
Alternatively, addition of a base to this mix would also accelerate formation of chromium(III) hydroxocomplex.
With this in mind, I'd say the more appropriate way to write the reaction is

\begin{align}
\ce{\overset{+6}{Cr}O4^2- + 4 H2O + 3 e- &→ [\overset{+3}{Cr}(OH)6]^3- + 2 OH-} &|\cdot 2 \tag{red2}\\
\ce{\overset{+4}{S}O3^2- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} &|\cdot 3 \tag{ox2}\\
\hline
\ce{2 CrO4^2- + 3 SO3^2- + 2 OH- + 5 H2O &→ 2 [Cr(OH)6]^3- + 3 SO4^2-} \tag{redox2}
\end{align}

Resulting full redox equation:

$$\ce{2 Na2CrO4 + 3 Na2SO3 + 2 NaOH + 5 H2O → 2 Na3[Cr(OH)6] + 3Na2SO4} \tag{R2}$$


Can isopropyl alcohol vapor damage polymers? Since isopropyl alcohol is found in many cleaning products, can the vapor emanating from the isopropyl alcohol corrode, stiffen or react with polymers?

For example, if I use isopropyl alcohol to clean my desk, will its vapor cause any damage to the mousepad, cables and synthetic rubber?

I used 70% isopropyl alcohol to clean my mousepad before, and it ruined the surface by hardening the polymer fibers. I wonder if the vapor can cause the same thing.

Thanks

I will use your example of the mouse pad to answer your more general question.

As you implied in your question, the mouse pad will be exposed to some isopropyl alcohol vapor if it is used to clean in the vicinity of the pad. As the vapor will quickly disperse to negligible levels, a simple solution could be to move the mouse pad (or other rubber-based or otherwise isopropyl alcohol sensitive item) to the other side of the room while cleaning the area around the computer. Once the area is dry and the smell is gone, there should be negligible isopropyl alcohol vapor remaining and you can put the mouse pad back. For the reasons discussed below, even this much effort is likely overkill and you can probably safely just clean around the pad and other items with isopropyl alcohol.

The key reason that isopropyl alcohol vapors are much less damaging to the rubber component of the mouse pad is simply how much more of the compound comes into contact with the rubber when in direct contact with the liquid phase vs secondary exposure via the vapor phase. Depending on a number of factors (contact time with the liquid, temperature of the surfaces that are cleaned with the isopropyl alcohol, proximity and quantity of liquid used in vicinity of mouse pad, etc.) the mouse pad could be exposed to a few orders of magnitude more isopropyl alcohol when in contact with the solution than when merely exposed to vapors from cleaning some area in the vicinity of the mouse pad. This means that the exposure of the sensitive items to the isopropyl alcohol vapor is very small and not likely to have any noticeable effects on the exposed items like the mouse pad.


Dissociation of different compounds? What would be the dissociation of Hydrosulfuric acid in the presence of water?

I was thinking that Hydrosulfuric acid or $\ce{H2S}_{(aq)}$ would form $\ce{2 H+}$ and Sulfur ions.

With this logic, I keep getting the wrong answer :(

The right answer of this question is it would not dissociate at all.

Shall appreciate some help on explaining this. Thanks!

$\ce{H2S}$ has a mostly covalent character. You can decide for it by the crude way textbooks give: Calculate the difference in electronegativity to decide whether it's a non-polar covalent compound, a polar one or has ionic characters.
For the dissociation part (which is usually endothermic overall) it'll need more "ionic character". And it does dissociate in water, resulting in a weak acid named  sulfhydric acid. (The spelling varies with citable sources) Your textbook likes to think about it as not dissociable, only because it's a weak acid.
Addendum:
The "ionic character" plays its role of a rule a good one as explaining acid 'strength'. However, it acts almost flawlessly in the same period, but has exceptions when it gets to groups. As Dave mentions, $\ce{HF}$ is weaker than for instance, $\ce{HCl}$ and that's because the atomic radius changes more dramatically than electronegativity difference. (See here ) But, in the $\ce{H2S}$ molecule, this exception hasn't occurred. Dave pointed out a good example of when the rule isn't working that well, but as I said, we can ignore the other factors in this very case.


Is the hybridization only related to atomic orbitals but not to the molecular orbitals? What is Hybridization?

  (Spelling: Hybridisation (British English) / Hybridization (American English))
  The hybridization is a concept that describes atomic orbitals. In other words: Hybridization is an approach to model hybrid orbitals mathematically. The geometry of a molecule determines the hybridization and not vice versa.

IUPAC Gold Book's definition of hybridization:

  Linear combination of atomic orbitals on an atom. Hybrid orbitals are
  often used in organic chemistry to describe the bonding molecules
  containing tetrahedral ($sp^3$), trigonal ($sp^2$) and digonal ($sp$) atoms.

For a certain molecule structure we can allocate the atomic orbitals of one atom to hybrid orbitals which are also localized on this atom and which cannot be redistributed over the molecule to over atoms, e.g. a hydrogen can only have a s orbital but no sp hybrid orbitals because it has no p atomic orbitals.

The allocation can be done using the following relations:

$\ce{1s + 3p ->4sp^3}$
$\ce{1s + 2p + p ->3sp^2 + p}$
$\ce{1s + 1p + 2p -> 2sp + 2p}$

We say: "The atom X is best described by a $\ce{s^\text{x}p^yd^z...}$ hybridization."

Combining two orbitals / two hybrid orbitals / one hybrid orbital (e.g. $\ce{sp^2}$) with a orbital (e.g. $\ce{s}$) we obtain bonds. $\sigma$-bonds with $\ce{sp}$ orbitals (ethyne) of the atomic centers have lower energy than $\sigma$-bonds with $\ce{sp^2}$ orbitals (ethylene), lower in energy means: they are better acceptors.

What is my question now?

I'm thinking if we can also describe a bond (a molecular orbital) with a hybridization, e.g. if we can combine a $\ce{sp^2}$ hybridized carbon with a $s$ atomic orbital of hydrogen to make two $\ce{sp}$ molecular orbitals (a bonding $\sigma$ orbital with $sp$ character and an antibonding $\sigma^\star$ orbital with $\ce{sp}$ character; just be saying for a $\ce{C-H}$ bond: $\ce{s} + \ce{sp^2} = 2 \ce{sp}$, what's probably wrong).

I guess this question has to be answered with No, because we cannot say a bond has hybridization.

The question might arise why I'm thinking of the hybridization of molecular orbitals.

Over several years I've heard statements like this:
The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character (50%) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) (33%) that's why ethyne has a lower $\mathrm{p}K_\mathrm{a}$ than ethylene (25 versus 44 at standard conditions in water as solvent) because a bond with higher s character can better stabilize a negative charge, because the s orbitals are penetrating deeper into the shell as the p orbitals do.

Again, I think hybridization is only a good model for atomic orbitals (=localized on atom) not for linear combinations of atomic orbitals (=bonds).

Can we really say the $\sigma$ bond in ethylene has $sp^2$ character?

Hybridisation is a concept devised by man to ease explanation of chemical observations. It is not a 1:1 equivalent of anything seen in molecules but rather a way of rationalising what one has seen — one that is as mathematically correct as molecular orbitals (that span the entire molecule) are.

There are two fundamentally different approaches on how to describe ‘bonds’. The first one is the molecular orbital approach. It considers electrons that can freely move around the entire molecule, in a so-called potential energy space: the potential energy derives from the (negatively charged) electron being close to (a series of positively charged) nuclei which attract each other. As you may remember from your very first chemistry classes ever, the mass of even a proton is $2000$ times that of an electron, so a carbon nucleus is some $24000$ times heavier than an electron. According to Newton, the nuclei have a lot more inertia than the electrons do, so it is reasonable to assume that whichever fixed positions the nuclei take, all electrons can then assume favourable movement patterns.

Quantum mechanics, and further quantum chemistry, give us methods to calculate what movement patterns these electrons might assume — although technically the equation is not solveable, because it also must contain electron-electron interactions, there are good approximations out there. The wave functions that one can derive from all this information correspond to molecular orbitals you might have seen before: They abide by the molecule’s symmetry (will always fit into one irreduceable representation of the molecule’s symmetry point group) but extend over the entire molecule (even the orbitals representing atomic core electrons can be thought of being delocalised somewhat).

Using the quantum mechanical method of calculation (operators) you get an energy for each molecular orbital. Thereby you have ordered your molecular orbitals by energy (approximately equalling the number of nodal planes) — this gives you HOMO and LUMO that you might have heard of before.

To reduce this back into ‘bonds’, we need to use mathematical tricks. Simply spoken: All wave functions can be treated as vectors, and vectors can be linear combinated to give other vectors — meaning we take $0.5$ times orbital $\Psi_1$, $0.01$ times orbital $\Psi_2$ and so on for each molecular orbital, add them up, divide them by a normalisation factor, and we get a new orbital that has a different shape (and a new energy). If we do this in a sensible way, we can manage to reduce molecular orbitals to those that only ‘span’ across one bond — because we have cancelled out all wave functional contributions from other parts of the molecule. This would correspond to your typical $\sigma$ and $\pi$ type bonding, nonbonding and anti-bonding orbitals.

The second method is the hybridisation approach. This time, you take the general shape of the molecule and consider each atom by itself. For each atom, you can assume a hybridisation based on its geometry: If there are three other atoms in a triangular shape around it, all sharing the same plane with the atom in question, we can assume it to be sp² hybridised.

Once we have done this, we need to take a look at neighbouring atoms to create ‘bonds’ again. Usually, there will be two hybrid orbitals ‘pointing’ at each other, ready to create a bond. Using the same technique we have to create the hybrid orbitals, we can mix these two hybrid orbitals to create another orbital. This orbital is not considered to be of a new sp-hybrid type, but rather gets a new type of name: It can be $\sigma$, $\pi$ or $\delta$ (to name the most common ones). These Greek letters closely resemble s, p and d, and the bond orbitals in question share a common symmetric trait with the corresponding atomic orbitals: $\sigma$ orbitals can be rotated freely around the bond axis, $\pi$ orbitals have one and $\delta$ orbitals two nodal planes along the bond axis (the bond axis is completely part of the nodal plane).

If we did this correctly, we will arrive at orbitals of similar shape and energy as with the above method, only we are much, much faster, which is why this method is often taught in schools.

Onto the question of how do the $\ce{C-C}$ (and $\ce{C-H}$) bonds differ in $\ce{C2H4}$ and $\ce{C2H2}$:
You just saw how we decided what orbitals we want to combine to form a bond. In $\ce{C2H4}$, we are combining a carbon’s sp² orbital with a hydrogen’s s-orbital to create a $\sigma$ bond. For simplicity, we can label this bond $\sigma_{\ce{sp^2-s}}$. Similarly for $\ce{C2H2}$, only that this one has to be labelled $\sigma_{\ce{sp-s}}$. You see that you created bonds with different hybrid orbitals and therefore the bonds will have different ‘shapes’ and energies. This is what is meant with

  The $\ce{C#C}$ bond of ethyne ($\ce{HCCH}$) has higher s character ($50\,\%$) than the $\ce{C=C}$ bond of ethylene ($\ce{H2CCH2}$) ($33\,\%$)

Technically, this is short for:

  The hybrid atomic orbitals of the carbons of ethyne have a higher s character ($50\,\%$) than those of ethene ($33\,\%$), so the $\sigma_{\ce{C-H}}$ bonds will retain that property somewhat.

Long story short: No, we cannot say that a $\sigma$ bond has sp² or sp³ character; we can (and should) only say that the atomic orbital of one side of that bond has this character.

And: It is wrong to assume $\ce{sp^2 + s} = \ce{2 sp}$, because that would recentre a p-orbital from the carbon to the hydrogen.


Change in heat of photosynthesis Consider a plant that uses the energy it receives from the sun to do work (to undergo photosynthesis) to produce glucose. The minimum amount of work this plant can do to produce 1 mole of glucose is equal to the change in Gibbs free energy that goes into producing one mole of glucose. I calculated the change in Gibbs free energy that goes into producing one mole of glucose to be $\Delta G =2878.94\ \mathrm{kJ}$. Therefore the plant does $2878.94\ \mathrm{kJ}$ of work. Also if we calculate the change in enthalpy for the same process we get, $\Delta H = 2803.04\ \mathrm{kJ}$. Therefore the heat from photosynthesis under these conditions is, $Q=\Delta H - \Delta W = -69.9\ \mathrm{kJ}$.

This confuses me because it implies that photosynthesis under these conditions gives off heat. How can this be?

Generally speaking (without need to refer to your particular problem), the sign of the free energy change and thus the magnitude of maximum possible work does not depend on $\Delta H$ or $\Delta S$ alone, it depends on the combination:
$$w_{\mathrm{max},\text{non-}pV} = \Delta G = \Delta H - T \Delta S.\tag{1}$$
This equation describes the maximum amount of non-$pV$ work that can be obtained from the system or minimum amount that needs to be done if the process is performed reversibly at constant $T$ and $p.$ In addition, when work other than mechanical $(pV)$ is performed, the value of $\Delta H$ is no longer equal to the heat evolved, in fact then $\Delta H = q_p + w_{\mathrm{max},\text{non-}pV}.$ Finally, it remains that at constant $T$ the entropy change of the surroundings is related to the amount of heat evolved as
$$\Delta S_\mathrm{surr} = \frac{-q_p}{T}.\tag{2}$$
This allows the above to be rewritten in terms of entropies of the system and the surroundings (at constant $T$ and $P):$
$$w_{\mathrm{max},\text{non-}pV} =  w_{\mathrm{max},\text{non-}pV} + q_p - T \Delta S = w_{\mathrm{max},\text{non-}pV} -T(\Delta S_\mathrm{surr} + \Delta S_\mathrm{sys}) $$
Removing the cancelling terms we conclude that
$$\Delta S_\mathrm{surr} + \Delta S_\mathrm{sys} = 0\tag{3}$$
that is, the entropy of the universe remains constant, which is consistent with reversibility.
That clears up the question of how it is possible for a process to be formally endothermic (in the sense that $\Delta H > 0$) and yet heat is released $(q < 0).$ This does not however explain why the net process of glucose generation by photosynthesis, assumed ideal (amongst other things reversible) generates heat.
The answer is that the released heat is responsible for an increase in the entropy of the surroundings which offsets a reduction in the entropy of the system when glucose is generated, which can be shown by writing
$$\Delta S_\mathrm{sys} = -\Delta S_\mathrm{surr} = \frac{q_p}{T} < 0\tag{4}$$
Without that increase in the entropy of the surroundings life would not be possible.


What's with the 70% ethanol? Is there a reason for all the bottles being 70% ethanol, or is it convention.

Primarily it is used as an anti-microbial agent. There is lots of data out there showing that the efficacy of ethanol at killing bacteria peaks at the 70% mixture. As they are making tons of it (literally) for the microbiologists, it's easier to just scale production then set up a new production facility for chemists to use at an arbitrary concentration.


Removing potassium chloride crystals from a mineral composed of manganese dioxides How can I remove crystals of potassium chloride from a mineral composed of manganese dioxides?
Hey, i'm a high school student trying to do a good cryptomelane sample to use in my scientific project.
I was able to develop the cryptomelane, but after some necessary processes, the mineral was crusted with crystals of potassium chloride, what would be the best process to remove this salt?

$\ce{MnO2}$ is insoluble in water -- I would expect with a good deal of confidence that the mineral cryptomelane is also insoluble. $\ce{KCl}$ by contrast is highly soluble.

I would just treat the sample with cold water and filter out the dissolved $\ce{KCl}$ (e.g. with a Buchner funnel and filter paper). Then dry with heat.

I would choose a fine-grain filter paper though because, though insoluble, $\ce{MnO2}$ particles could still be small enough to filter through course paper, like coffee filters (from experience).


Does the carboxylate group show a positive or a negative inductive effect? In the benzoate ion, does the carboxylate group show a $-I$ or a $+I$ effect?

I learnt that the carboxylate group is a $-I$ group, but according to a website the carboxylate group in benzoate shows a $+I$ effect.

So, which one is correct? Does the carboxylate group show a positive or a negative inductive effect?

Please neglect mesomeric effect when answering.

The carboxy group is electron-withdrawing both by inductive and by mesomeric mechanisms. I have drawn two depictions in the scheme below. The upper one is how the mesomeric effect of the carboxy group can be explained. This effect is only relevant if the $\ce{COOH}$ fragment is in the same plane as the ring.

In the case of 2,6-di(tert-butyl)benzoic acid, the bulky tert-butyl groups prevent the mesomerism depicted above. The carboxy group is perpendicular to the ring. Yet, this group still exercises an electron-withdrawing inductive effect. It can be explained by a mesomeric structure of the $\ce{C=O}$ double bond with charge separation:

$$\ce{C=O <-> C^+-O^-}$$

This positive charge is, naturally, electron-withdrawing and can be used to explain the $-I$ effect of carboxy groups. No matter which mesomeric depiction you choose, you can never get a negative charge on the carboxy carbon, thus it will always remain partially positive. That also makes sense considering it is bonded to two electron-withdrawing, strongly electronegative oxygen atoms.


Risks of burning melamine formaldehyde adhesives Background- I make biochar for gardening, and small scale carbon capture and storage. Most of my feedstock is clean, unpainted, untreated, end of life timber, but there is a small proportion of engineered particle board- MDF, chipboard, plywood and composite blocks containing, I believe, melamine formaldehyde binding resin. No halogens from PVC, no metals from CCA treated timber or old lead paint.  I want to establish what pollutants may be produced- I would in a worst case scenario be burning less than 50kg of the suspect material per hour.

All the charring happens in TLUDs- Top Lit Up Draft burners where all the pyrolysis gases are ducted through the hottest zone of the burn- round about 800-900C. I'm currently working to improve residence times and temperature gradients- carbon particulates are a known issue.

Does any of the above sound like it'll produce anything nasty from the carbon, hydrogen, oxygen and nitrogen going into the mix? PAHs? Anything else above and beyond nitrogen oxides and carbon particulates from clean wood?

To put it another way, do I err in favour of creating a pollution factory, or in favour of condemning a potential feedstock to be a disposal nuisance?

Thanks for any replies- I'd like to lay this to rest.

I am by no means an expert in the pyrolysis of different woods by any means. But after spending some time reading through some papers for you, I found one which burns MDF in a slightly different burner to you (around 450-600C) and the ash produced contained formaldehyde at 1681 mg/kg.

doi:10.1016/j.enconman.2014.12.032 - incase you want a read yourself.

This document goes through different methods in producing biochars and says that providing the temperature of pyrolysis is above 500C, it shouldn't contain any PAH's or Dioxins within the biochar and if it does, it'll be several times lower than required clean up set by the Model Toxic Control Act.

Everything else I have read seems to not have specific analysis on what exactly is left within the biochar. There are plenty of organics released as gases during the process specifically above 500C which you could argue should no longer be in the biochar. Have a look around at papers, I know it's not a definitive answer but it might be a start for you.