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Ease of reaction of hydrogen with alkali and alkaline earth metals

The ease of reaction of group 1 metals with hydrogen decreases as we
go down the group.

But why lithium requires high temperature than sodium, potassium and
rubidium?

What is exactly "ease of reaction"? I think it is related to kinetics (rate) of reaction and not related to thermodynamics of reaction. If so, is there any chemical data table from which I can predict the ease of reaction for group 2 metals with hydrogen?
Also, at what temperature, the ease of reaction should be compared?
The following processes involve in formation of $\ce{MH(s)}$ from $\ce{M(s)}$ and $\ce{H2(g)}$:
$$\ce{M(s)\rightarrow M(g)\rightarrow M+(g)}$$
$$\ce{1/2H2(g)\rightarrow H(g)\rightarrow H-(g)}$$
$$\ce{M+(g) + H-(g)\rightarrow MH(s)}$$
I know that it can be done by measuring and then comparing the formation of products over time, i.e. $\ce{\frac{\Delta [MH]}{\Delta $t$}}$.
References:

Page 297, Concise Inorganic Chemistry, 5E, J. D. Lee (Wiley)

Page 115, Descriptive Inorganic Chemistry, 3E, House (Elsevier)

Page 342, Concise Inorganic Chemistry, 5E, J. D. Lee (Wiley)

Have you expected the same or lower temperature for lithium ? //  Ease: It is rather a linguistic problem. Ease as a verb has 2 listed meanings: 1/ to make or become less severe, difficult, unpleasant, painful, etc.  2/ to move or to make something move slowly and carefully in a particular direction or into a particular position   // so in context of comparing temperatures for the reaction, it looks like 2/  ="the hesitation with which they do so  decreases from lithium to caesium"

@Poutnik I was thinking, if Li reacts at higher temperature than Na, then at a particular temperature, the rate of reaction for Li should be slow, or the ease of reaction should be low for Li. But both are exothermic reactions, so increase of temperature is not favourable. So, at what temperature, the ease of reaction to be compared? I added a sentence in the question. Also, in the first reference it is given that "the hesitation with which they do so, increases from lithium to caesium."

No, the first reference says "ease(supposed ~= hesitation) decreases from lithium to caesium", what is in agreement with 700 deg C for Li.  Lithium has higher ionization energy, therefore also the reaction activation energy, so slower at the same temperature, so higher "ease = hesitation".

@Poutnik I think at first instance, it should take the meaning of "easily" as "which reaction is easier to proceed?" as mentioned by you in 1/ "become less difficult". If it is taking the meaning of "hesitation", then it is very confusing to me, considering English is not my mother-tongue.

I took it from the Oxford dictionary - see the link.  1/ =easily  is in contrary to the mentioned chemical fact 700 vs 400 and IE 2/=hesitate is in agreement with it, so I have voted for 2/. ( I am not a native speaker either so US/UK guys may correct us ).


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The effect of nonbonding electron pairs on ozone polarity
I know that ozone is a polar molecule. And our master has used the existence of nonbonding electron pairs on the central atom to justify this subject.
I want to know how much the nonbonding electron pairs affects the polarity of the molecule? And if it does, why do nonbonding electrons have no effect on the side atoms?
My English is not very good and I apologize if the wrong terms are used in the phrases.

Have you checked ozone page on wikipedia ? // In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. When you ask, it is expected you have thoroughly searched and thought about the topic, providing explicit summary of partial answers/ideas/thoughts you have got until then. Effort not shown may be considered as effort not done and such a question may get closed.
How do I ask a good question.

In fact the nonbonding electron pairs on the end atoms do affect ozone's polarity, and because the end atoms have more nonbonding electrons they are actually the ones at the negative end of the dipole. The central atom with only a single nonbonding pair (and a positive formal charge) is at the positive end. See also this answer.

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Reason for adding TFAA in this synthesis pathway?

I have included a short part from a synthetic pathway from: https://synarchive.com/syn/112
$\ce{Mg3N2}$ reacts with $\ce{MeOH}$ to liberate $\ce{NH3}$ which I believe is used in nucleophilic substitution at the esters and amides in the last step. However, before this, the designers of this reaction pathway added in TFAA, introducing new groups onto the amines. Why is this step necessary? Why can't the researchers just use $\ce{NH3}$ or $\ce{Mg3N2}$ right after $\ce{SOCl2}$ and $\mathrm{h\nu}$?

I wondered about that one too. It seems unneccessary. Would need to read the full paper to see if there is an explanation.

Very interesting. I see that they do that trick in another step, too, the one where they replace I by CN using CuCN. Is it possible that those two NH2 groups on the fused Ph ring impart so much electron density to the system, that whenever you need to do a reaction where a nucleophile attacks, you need to subtract electron density first, in this case by attaching an electron-withdrawing acyl to the NH2 groups? Just a hypothesis though, might be something else entirely.

@user6376297  I note that they use sulfuric acid conditions to remove those trifluoroacetyl groups immediately before a KOH treatment to hydrolyse the methyl ester when (in my opinion) the KOH would easily remove them. A little odd!

I had a look at the paper - much simpler: it was because of solubility. They say that the free bis-amino compound is too insoluble in organic solvents to be purified on silica.

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Mechanism of acid-catalyzed ring opening of a cyclopropane ring
I have a question regarding following synthesis which apparently only needs a sour medium to occur:

What exactly happens as a mechanism? What I thought was: The alcohol group gets protonated and forms a stable carbocation at it's tertiary position. This charge can now wander to the tertiary carbon next to it; but an H-Shift should not be possible. What is the mechanism here?

Ring opens instead of shift.

Because of the carbocation in this position? Will it destabilize the ring because it is sp2?

While acids are indeed sour, the medium is acidic. ;)

I believe the mechanism should be somewhat like this (after the formation of carbocation) -

It would account for the product shown, but I am not sure if it's correct. I feel it's more favourable for a dehydration to occur with the hydrogen from the ring itself (shown on the dash at the top) to form this -

This double bond is conjugated and more substituted, so I should have thought that this would be the major product.

If my mechanism is wrong, please let me know.

Probably elimination is concerted with ring opening, so "less favorable" product is created.

Did the OP's question state that the observed product was the major product? No.

@Mithoron Thanks for the insight. Yes, if it's a single step then to form my proposed product it would need to start making a double bond before the ring was completely broken - which of course would be difficult owing to angle strain.

@user55119 Great point! It shows how reading things closely matters a lot :)


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Mechanism of cyclisation with tin(IV) chloride

I have taken this photo from synarchive: https://synarchive.com/syn/325
I tried reading the research paper this synthesis was found in (Couladouros, E. A.; Vidali, V. P. Chem. Eur. J. 2004, 10 (15), 3822–3835) as well as some of the references of that paper but I could not figure this mechanism out.
Could it just be the Lewis acid attaches to alcohol, causing the alcohol to leave; alkene attacks carbocation, ester attacks the carbocation formed (losing tBu in the process)?

The stereochemistry of this type of cyclization is well-understood. An extensive review may be found here. As you have realized the role of the Lewis acid stannic chloride is to generate the carbocation 1. The vacant p-orbital is aligned in the sigma sense with the π-bond of the double bond to permit cyclization, in the Markovnikov sense, to carbocation 2. There are two possible modes of cyclization of this carbocation to form either the less strained cis-ring juncture or the more strained trans-ring juncture. The transition state leading to the cis-lactone involves a steric interaction with an axial methyl group. The transition state (2 $\rightarrow$ 3) leading to the trans-lactone is less hindered and kinetically controlled. The utilization of the tert-butyl ester is to allow the rapid collapse of cation 3 to the trans-lactone 4, thereby insuring that the ring closure is not reversible. The tert-butyl cation 5 is scavenged by the solvent, nitromethane.

It was the trans fusion that confused me...


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Vapour pressure and boiling
I have come across two definitions for boiling point:

It is the temperature at which a liquid gets vaporised

It is the point where vapour pressure = external pressure

How are they related? How does vapour pressure being equal to external pressure cause a liquid to vaporise?

Do you really think water on a wet floor boils, while the floor is getting dry ? Have you searched about terms "vapour pressure" and "boiling" ? // In contrary to some other Q/A or forum sites, answers on CH SE site are figuratively paid by the user's own effort. When you ask, it is expected you have thoroughly searched and thought about the topic, providing summary of partial answers/ideas you have got until then. Effort not shown may be considered as effort not done and such a question may get closed.
How do I ask a good question.

The first condition is incorrect and the 2nd condition is correct.  To boil,, a liquid must form vapor bubbles beneath the surface, in the bulk of the liquid.  These bubbles will contain pure vapor at the equilibrium vapor pressure.  In order for these bubbles to form, the pressure inside the bubble must be high enough to expand the total volume liquid and bubble below the surface by pushing back the outside atmosphere.  This can only happen if the pressure in the bubbles is equal to the outside atmosphere pressure.

@Poutnik I know about vapour pressure, but these are the definitions given in the book I'm following, hence the question

I have got it, but critical thinking is supposed to be applied on everything. What is written is not necessarily correct, and in this case it is easy to verify without asking. // Def 1 is also about wording evaporates versus gets vaporized ( the former anytime, the latter is the final stage of evaporation where temperature cannot get higher and heat is used for phase change ( for given p and neglecting metastability)


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Action of EtONa/EtOH on compound containing -CN group

In the above question, I know compound A will be prepared by just replacing -Cl with -CN group in the starting compound. I wanted to know what work would be C$_2$H$_5$ONa/EtOH doing in order to make compound B?
I have seen this C$_2$H$_5$ONa/EtOH to perform elimination reactions on alkyl halides forming alkenes but no idea how will it react with benzene containing -CN group.
NOTE : The original compound should be as given below

This looks to me that someone has either drawn up a reaction scheme with an extra methylene group in it, or they haven't looked at the pKa tables. What do other people think?

@Waylander: I agree that the chloride has an extra methylene group given the way the structure is drawn. Otherwise, what purpose would benzaldehyde serve. EtONa/EtOH itself without water is not going to effect hydrolysis.

@Waylander maybe the compound should not contain it. you can proceed without the extra methylene group. I'll edit the question.

@Waylander doesn't it look like the ethoxide will form a carbanion next to the CN group, like what happens in an aldol reaction? I agree, though, that it would form better next to a phenyl group, otherwise the benzene ring has no use

@TRC ethoxide is nowhere near strong enough to deprotonate an unactivated nitrile - hence my comment above about pKa tables

You might want to google "ivanov phenylacetonitrile" and look at the images.

In the event that the phenethyl chloride is meant to be benzyl chloride then this is the solution I believe the question setter is looking for:
A is phenylacetonitrile which is deprotonated by ethoxide and condenses with the benzaldehyde to give benzylic alcohol B. The acid conditions dehydrate B to give the 1,2-diphenyl-cyanoalkene C (or if pushed hard enough the unsaturated acid). The $\ce{SOCl2}$ followed by $\ce{MeNH2}$ forms the chloride from the alcohol which is displaced by the methylamine to give the $\alpha-$N-Methylamino-nitrile.
As shown by the discussions below there is considerable doubt about whether this reaction sequence has any basis in reality.

I didn't understand how A condenses with benzaldehyde to give benzylic alcohol. Can you share a mechanism?

Much like an aldol only the nucleophilic species is the nitrile anion rather than the enoalte.

I'd opt for B being the unsaturated nitrile via loss of water from the initial adduct.

@user55119 in which case, what is the SOCl2 reaction with B?

Good point! Missed that. The problem is, in IMO, conceptual. A Chem. Abst. search of the reaction between PhCH2CN and PhCHO to form the adduct can be initiated with kinetic bases, CeCl3 or Rh(I). There is a plethora of conditions for the condensation to form the unsaturated nitrile. The postulated chloride derived from either the phenyl acetonitrile or the propionitrile is unknown. The latter is a viable product but not the former, where elimination by methylamine is likely. continued..

... I could find no conjugate additions of amines to PhCH=C(Ph)CN. The problem is an exercise in individual reactions ungrounded in reality. What does your experience say?

My experience says that this is a question with one error already, it could very easily have another. I would expect the unsaturated nitrile to be formed as there is some much driving force from the extended conjugation.

Let us continue this discussion in chat.

@Waylander in the textbook answer to B is given Ph(CN)CH - CH(OH)Ph

That is the answer they want but it is not likely what actually happens under the conditions described.

Okay, got the point @user55119

My answer will be somewhat contradicting that of @Waylander, but I believe it's correct since I know where the question has come from and have solved it before.
The first step with $\ce{KCN}$ substitutes $\ce{-Cl}$ group with $\ce{-CN}$ forming 2-phenylethanenitrile which is A
The next step is deprotonation of the benzyl carbon by a strong base, $\ce{EtO-}$ to give a carbanion which subsequently condenses with $\ce{PhCHO}$ by a nucleophilic addition on the carbonyl group. Like an aldol, it completes a dehydration of the $\ce{-OH}$ group to give $\ce{Ph(CN)C=CH-Ph}$ which is B.
The acidic medium and heating is for the hydrolysis of the $\ce{-CN}$ group to form $\ce{-COOH}$. The subequent steps with $\ce{SOCl2}$ and $\ce{MeNH2}$ are standard reactions.
(I should not be surprised if experimental evidence proves my answer to be completely wrong. But the exam this question is meant for is strictly theoretical and requires us to predict products of reactions in the same manner which I have followed. I know Waylander and user55119 are actual chemists whose answers and comments would be far more correct in reality.)

It is unclear from the reaction sequence drawn whetherD is made from C or from B. I took it that D is made from B. If the intention is that it is made from C then you are correct.

@Waylander I see your point now. As I said, I've done this question before, so it didn't cross my mind that the image formatting might make people think that it's going B to D directly. Of course, you're right that it looks ambiguous.

This is just a very poorly written question


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Sn(II) compound stereoselectivity in aldol reactions
In "advanced organic chemistry: part B" fifth edition, page:130, the authors state that chiral proline-like additives to Sn(II) compounds can further increase stereoselectivity of aldol reaction. Then they showed how different TS evolve to the observed product

Wouldn't it be more accurate if the the enolate approached the aldehyde from the aldehyde's Re face not Si. because I find the Si face attack more hindered (at least in the open TS)


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How can steel be so different from iron, even if amount of carbon is small?
Perhaps I am too naive, but I am having a hard time visualizing how even 'high-carbon' iron and steel alloys are maybe, at most, about four percent carbon by mass, which still means only one (smaller) atom in 120 or so is carbon rather than iron...
How?  Is there a diagram that could help me visualize this?  Doesn't this mean that most iron atoms are not even remotely close to, or even indirectly affected by, a carbon atom?

Maybe think of a stack of dinner plates and same thing with a little sand between the plates. Which stack better resists a side shove?

Nice question. All chemist's more or less have studied this stuff at least at elementary level. But I feel we were not or rarely presented to the scenario OP has made clear. Of course the comment by Ed V (beside the answer is not straightforward, to me) put the issue in place. But a nice question indeed.

What is your criterion of calling steel "different" from iron? If you are just talking about steel, it has other metals (nickel, chromium etc.) besides some carbon.

You should really brush up on chemical calculation - molar percent here is much higher then mass percent, not lower.

You might have heard of building relocation. Sure, there are some iron beams involved, but their weight is negligibly small compared to that of the building. How can they make a difference? Well, just like that.

Presumably you mean pure iron when you say "iron". Steels frequently contain less than 0.1 % carbon, not counting stainless steel with 0.01 % carbon.  Steels rarely contain over 1.0% C ; above that you are getting into cast iron.  I suppose it is the net affect of all elements present as they affect the iron response to heat-treatment and/or cold work.

Have you had a look at iron / steel phase diagrams?

Rebar is only a few percent of the mass of reinforced concrete, for an example that's maybe more intuitive.  That few percent makes a huge difference.

One of the key considerations is that much of the interesting mechanical behaviour doesn't occur within the bulk of the material - it occurs at the interfaces between crystals (known as grains). Grain boundary slip, rotation and growth often defines the mechanics - carbon has a tendency to block the motion of grain boundaries, which are one of the key deformation modes.
There are a multitude of reasons, but carbides in the material cause faults in the crystal (which are the key thing that makes metals deform plastically) to slide less easily (reduces dislocation "glide"). This is because carbon forms carbides that cannot be "split" by the dislocation as easily, and have to deform themselves to climb over the carbides, before re-straigtening via dislocation line-tension (energy penalty for being longer than a straight line).
So its a complicated picture already, and this is only scratching the surface of the issue!
I'd highly recommend Dieter's Mechanical metallurgy if you want a more in-depth understanding - its probably mid-undergraduate to post-graduate level though.
Edit: in case it isn't clear, carbon preferentially segregates to grain boundaries, arguably due to the higher open-volume at most boundaries (this is a simplification). Though even in a random solid solution, carbon still has a mild strengthening effect (the same can be seen in Aluminium-copper random solution alloys, vs pure Al)

Do alloying metals also preferentially segregate to grain boundaries, as you describe carbon doing?

Here is a visual tour of steel grains, and where carbon is depending on the temperature history of the material: https://youtu.be/uG35D_euM-0?t=562

And here is a picture of pearlite grains (containing carbon) between ferrite grains (pure iron): https://upload.wikimedia.org/wikipedia/commons/7/77/Pearlite.jpg

@theorist Yes - the classic example is Gallium in Aluminium, but in that case it causes the Aluminium grains to separate!

@theorist  It depends on what the alloy elements are. Generally intentional alloying elements have good chemical affinity for the parent metal so they will be soluble within the grains of it or else form separate intermetallic phases between parent metal phase grains. But unintended elements (impurity traces) elements like As, Sb, etc that do not have solubility in parent metal phases often migrate to grain boundaries where they form separate phases in either discrete globs or grain surrounding films.

Grain boundary sliding is not a principal mode of ambient (< 0.3 Tm) deformation in steels although it is an important creep deformation mode.

Whatever you read on steel deformation do not read works like Dieter. I say this because in the main this book is concerned with mechanical metallurgy - the treatment of metal forming using a continuum mechanics approach. As a chemist you'll know that all the crucial phenomena derive from phenomena at atomic/molecular level. I would advise a perusal of texts by Reed-Hill and Cahn. https://www.amazon.com/Physical-Metallurgy-Principles-Reza-Abbaschian/dp/0495082546   https://www.elsevier.com/books/physical-metallurgy/cahn/978-0-444-89875-3  And build a model of a dislocation as a visual aid.

Elastic deformation properties like stiffness (Young's Modulus) vary very little with alloying element concentration since we are only working on the pre-yield bond strength. A pure iron, a carbon steel and a high alloy steel will only vary by about 15% in stiffness. So elastic properties will, as you suggest, be fairly invariant under small changes in alloy concentration.
But the other mechanical properties - those depending on plastic deformation like yield strength, hardness, toughness - are different in how they are affected by alloy element concentration.
Plastic deformation in metals at ambient temperatures, and below 1/3 of the absolute melting point, is mostly about moving dislocations within the crystal structure. Strengthening therefore means finding effective methods of impeding normal dislocation flow.
Let's look into why alloying helps any metal's plastic yield behaviour.
The word "alloying" may embrace situations where two or more metals exist as a mixture of separate elements. Such mixtures will show modifications of properties roughly representable by a rule of mixtures equation, i.e.
$$ S = v_1 S_1  +  v_2 S_2$$    where $v_1$, $v_2$ are the volume fractions of the respective elements.
But alloying in the usual meaningful sense will only occur when the chemical equilibrium prevailing favors either a solid solution or intermetallic phase (e.g. $\ce{Ni3Al}$, $\ce{Fe3C}$, etc) in preference to existing as a mixture of two separate elements. Solid solution alloying can occur in two ways. It can occur when an atom of the minor component element substitutes for one in the crystal lattice of the major component element. This usually happens when the minor component has comparable atomic mass, e.g. nickel in iron.
Solution alloying can also occur if the minor component has a much smaller atomic mass than that of the major component, e.g. carbon in iron. In this case the minor element atoms fit inside the existing iron lattice, a type of solution known as an interstitial solid solution. Of course, a carbon atom within the body-centered cubic form of iron will expand the lattice around it somewhat. So we have an ambient temperature limit of 0.002% by weight (or about 1 carbon atom per 10,000 iron atoms) on how much carbon is soluble within solid b.c.c. iron, known as ferrite or $\alpha$-iron.
A dislocation within a metallic crystal is a mismatch in the usual ordered 3D-repeating arrangement of ions. It is virtually impossible to crystallize any substance without having lots of dislocations within each crystal. More importantly, since a perfect crystal will have all its bonds yield at once, it is the very existence of dislocations that allow crystalline materials to deform plastically under moderate stress rather than break in a brittle mode at a much higher stress. So the yield strength of any crystalline metallic material is effectively the stress under which its dislocations begin to move. If we want to improve the strength of some element, e.g. iron,  we need to find a way to impede the movement of dislocations in it.

In the vicinity of a dislocation, the lattice is stretched on the side of the missing half-plane of ions and contracted on the side of the extra half-plane. Interstitial solute atoms within the crystal would therefore - given adequate temperature - tend to diffuse and settle on the expanded side of a dislocation zone since here they create minimal lattice strain energy increase. This is what happens with carbon atoms in iron.
Since the chemical interaction between a carbon atom and surrounding iron ions will be of partly covalent nature there will be significantly greater resistance to dislocation flow under stress and thus a higher yield stress.
Concentrations of carbon ~ 1 per 10,000 iron atoms would equate to one carbon atom per cube of iron crystal with an edge length of 22 interatomic spacings (~ 60 Angstroms). So a single carbon atom alone can significantly restrain dislocation loop movement within that volume and that is more than adequate for normal dislocation densities observed in carbon steel.
Another factor affecting dislocation impedance is the other carbon-containing phase in carbon steels, i.e. $\ce{Fe3C}$ or cementite. This phase precipitates simultaneously from $\gamma{}$-iron (austenite) or carbon supersaturated ferrite as it cools. This type of simultaneous solid-phase precipitation is referred to as eutectoid decomposition. Eutectoid pair phases are very finely mixed and often show a fixed relationship between their crystal orientations. In the case of carbon steels, the eutectoid mix consists of very fine alternating lamellae of ferrite and cementite. Being so close any dislocation movement towards the edge of the ferrite will be mechanically blocked by the neighboring cementite lamellae.
So the nominally "small" additions of carbon to pure iron will have two important effects in restraining dislocation advance and therefore strengthening the material:

The interstitial solubility of carbon in the ferrite b.c.c. lattice which locates carbon atoms in the "pipes" of dislocations where their partly covalent bonding with neighboring iron atoms on the extra half-planes requires a greater shear stress to move the dislocations within ferrite grains.

The close lamellar morphology of the ferrite/cementite eutectoid mix resists dislocation movement out of ferrite grains.

You have asked a seemingly simple question.
But the proper answer is long, involved and demands close study of a number of aspects of physical metallurgy. If your work requires a better understanding of this topic, please consult texts like those by Reed-Hill & Abbaschian and Cahn & Haasen at your local university library.

Let's consider that steel may contain $\pu{1\%}$ carbon. I know it is a bit much. Let us express this mass concentration as molar concentration for a sample of $\pu{100 g}$ of steel. Since this sample contains $\pu{99 g}$ of iron, this amount equates to
$$n(\ce{Fe}) = \frac{\pu{99 g}}  { \pu{56 g mol^{-1}} } = \pu{1.768 mol}$$
and
$$n(\ce{C}) = \frac{\pu{1 g}}  { \pu{12 g mol^{-1}} } = \pu{0.083 mol}$$
In iron-carbon alloys, these two elements often combine to form cementite, $\ce{Fe3C}$. It means that $\pu{0.083 mol}$ of $\ce{C}$ is combined with $3 \cdot \pu{0.083 mol} \approx \pu{0.25 mol}$ of $\ce{Fe}$. This $\pu{0.25 mol}$ of iron is not available to act as solvent of carbon containing molecules. So $\pu{100 g}$ of steel with $1$% C contains only $(1.768 - \pu{0.25) mol} = \pu{1.518 mol}$ uncombined iron. The molar proportion of $\ce{Fe3C}$ in steel is $\frac{0.25}{1.768} = \pu{14\%}$. Then $\pu{14\%}$ is a relatively concentrated solution of $\ce{Fe3C}$ in iron taken as the solvent.
This shows that steel is not a diluted solid solution. It is even a moderately concentrated solid solution of carbon containing solute in the solvent iron.

Fe3C isn't molecular https://en.wikipedia.org/wiki/Cementite

This assumes that all the carbon in a 1 wt% C steel will go preferentially into the cementite phase. This is not so. The amount of cementite phase in a steel depends a lot on its carbon content. Steels with low carbon wt% will have less cementite than those with higher wt%. See the phase diagram  https://www.tf.uni-kiel.de/matwis/amat/iss/kap_6/illustr/s6_1_2.html

Besides, cementite is a hard brittle phase which would be of little use for a material requiring both tensile strength and enough ductility to make it tough. Cementite's main benefit is to provide a measure of wear resistance in the final steel. Its existence in a fine eutectoid mixture with $ \alpha $-iron greatly helps this.


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Is absorbance from HPLC UV detector comparable between runs?
Suppose that I have an HPLC with a variable wavelength UV detector, as well as two pure protein samples that differ by a point mutation (otherwise same concentration, extinction coefficient, buffer conditions, etc.). If I run them through the same column sequentially and integrate the A280 peaks after baseline subtraction, should I expect the areas under the curves to be the same? I guess another way to ask the question is whether the absorbance unit is something that can be compared from run to run, or is the elution volume/peak shape the only physically meaningful quantity for comparison?

There is nothing absolute in HPLC related measurements. Take the easiest example first. Think of a separation of a racemate using a chiral column on a UV detector. After injection, you will see two peaks, with different retention times, and equal peak areas. Their absorbance at the peak maximum will be different because both peaks will have different widths. Change the flow rate and now your peak areas change but their peak ratio will still be 50:50.
The same ideas apply in your case. You state that everything is same (conc., molar extinction coefficient etc.). In HPLC, your peak areas will be affected by flow rates only given the rest of the conditions remain identical.

so if I understand you correctly, if the flow rate is the same in two runs then the area should be identical?

Yes, if the concentration and molar extinction coefficients are identical.

Why does flow rate change peak area? I was under impression that same amount of analyte will result in the same peak area. Is it because of the solvent which absorbs at the same wavelength (and its amount changed)? But in that case peak ratios will be different too..

@StanislavBashkyrtsev, Yes, this is the tricky part of HPLC-UV. It is a concentration sensitive detector (another type is mass sensitive detector). So at low flow rate, you would expect that peak areas are larger, because the analyte is spending more time in the detector (recall peak area is integration with respect to time).

But the height of the signal is lower at low flow rate, which should compensate for the longer exposure. Or do you mean that the same portion of analyte is being measured again and again with the low flow rate and therefore each individual molecule contributes to the area more?

@StanislavBashkyrtsev, Let us take the extreme example. Suppose you are running HPLC analysis and the peak is appearing on the detector. Stop the flow for 10 min and start the flow again. What do you think will happen to the peak area? It will be huge, just because the peak area is an integration of signal with respect to time.

@M.Farooq, yep, got it. So it is because the same instances of molecules are measured multiple times. And this effect is bigger if the flow rate is lower and/or sampling rate is higher. Thanks, never thought of it before..

@StanislavBashkyrtsev, Another important point: Sampling rate should not affect peak areas as long as we are above Nyquist frequency.

Ah, right. In case of DAD increasing sampling rate will decrease exposure time (for how long the diode is discharging on each measurement). And MS is destructive, which means the same molecule won't be measure twice anyway. But does this mean that MS peak area does NOT change if flow rate changes? If we stop the flow, MS detector will remove ions from the injected sample until nothing is left, therefore it doesn't depend on time (not entirely at least).

@StanislavBashkyrtsev, You are right about MS, but peak areas still change due to ionization efficiency issues. This is an interesting topic, can you start a separate question on it?


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Grignard-type reaction of 3-bromocyclohexene, or rather recombination?
I have a question regarding a proposed Grignard-type reaction which involves the conversion of 3-bromocyclohexene to 1-(cyclohex-2-en-1-yl)propan-1-ol, in diethylether. As the bromine atom lies in allyl position to the double bond, it should form the [1,1'-bi(cyclohexane)]-2,2'-diene, instead of the alcohol, shouldn't it? See below:

Allyl Magnesium Bromide is commercially available in ethereal solution https://www.fishersci.co.uk/shop/products/allylmagnesium-bromide-1m-solution-diethyl-ether-acroseal-acros-organics-2/p-3771057. It does all the usual Grignard reactions and does not dimerise. I see no reason why the cyclohexenyl analogue should be any different.

Some preparations of allyl magnesium bromide suggest lower temperatures while others suggest 30 mesh magnesium over the traditional magnesium turnings The 30 mesh increases the surface area of the Mg to maximize Grignard formation over coupling to form 1,5-hexadiene.

The dimerization under Grignard conditions is described here: https://www.sciencedirect.com/science/article/pii/S004040200700316X

I am completely ignorant as to why an allylic halide should be more susceptible to coupling with Grignard than a normal alkyl halide (which generally doesn't couple). Is it because the allylic halide is more reactive to substitution reactions?

Like so much organic chemistry the answer is it depends.
As I said in the comments you can buy Allyl Grignard in ether solution, but you can also dimerise it by allowing it to reach room temperature Organic Syntheses prep here.
Assuming the cyclohexenyl Grignard behaves similarly (and I think it will) then you can get the expected Grignard addition product with propanal if you add you bromide to the magnesium suspension slowly, keep the Grignard solution cold and add the aldehyde promptly (basically using standard conditions). Let reach rt and take a long lunch break before adding the aldehyde and you will get dimer.

My understanding is that dimerization occurs during Grignard formation. As bromide is added to the pot, it either reacts with already formed reagent to give dimer or with Mg to give more Grignard. Dimerization would be more prevalent later in the addition rather than in the early stages of addition. Your Org. Syn. link uses rapid addition of allyl chloride which, intentionally,  maximizes biallyl formation.  How does a long lunch break after the reagent is formed lead to dimer? ;)

I was under the impression that the dimerisation is a slow process that occurs at higher temperatures. You may be correct.

Dimerization of the Grignard reagent would require an oxidation.

There is no obvious oxidant present which leads me to ask is this actual dimerisation or nucleophilic displacement of Cl- by allyl anion?

Exactly! Not a true dimerization. Because the halide is a good electrophile, substitution competes with Grignard formation unless steps are taken to minimize substitution.

So likely to be somewhat less favoured in the cyclohexenyl case due to steric factors.

See my link in the Comment to the OP for the cyclohexenyl coupling.

Can only see the abstract, the rest is behind a paywall

I just came across this paper. Allegedly solves the coupling problem. https://www.tandfonline.com/doi/pdf/10.1080/00397919209409258?needAccess=true

@user55119  Paywalled again. I know Keith Jones too


text:
How can the oxidation of TEMPO be selective?
According to the article "TEMPO oxidizes alcohols to aldehydes and ketones", the oxidation of TEMPO (2,2,6,6-tetramethylpiperidine-1-oxyl, 2564-83-2) can be highly selective. I assume it means you can easily choose to oxidize primary alcohol or secondary alcohol by controlling the conditions of reaction. But how to do that exactly? What conditions should be modified and how?

I have not done this reaction, but I think temperature and rate of oxidant addition will both be important factors in controlling selectivity

Thanks for your help!  I did a little research. Turns out PH value is the key. Secondary alcohols are more easily oxidized in a acidic environment than primary alcohols.

@SamLv Suggest Put your finding (e.g., one of the literature reference you found indicating the dependence of the rate of reaction / selectivity of the reaction on pH value; the structure of their molecule submitted to the oxidation, etc.) into an answer.  Other than a comment, answers (typically) stay on chemistry.se.


text:
How to derive formulas for n factor of different types of reactions?
While studying Redox reaction, I came across this term called n-factor which seems to me as a short-cut trick to do the questions quickly. Since I am preparing for a competitive exam so our chemistry teacher gave us a bunch of formulas to find n-factor of different types of reactions as follows :-
Case 1

Reduction /oxidation reactions in which only one atom of molecule
reduce or oxidise.
$$
A_{x}^{+p} B_{y} \rightarrow A_{x}^{+q} B_{y_{1}}
$$
$$
\begin{array}{l}
p>q \Rightarrow \text { Reduction } \\
p<q \Rightarrow \text { oxidation }
\end{array}
$$
$$
n \text {-factor of } A_{x} B_{y}=x|p-q|
$$

Case 2

If more than one atom oxidise or Reduce in molecule.
$$
A_{x}^{+P} B_{y}^{+r} \rightarrow A_{x}^{+q}+B_{y_{1}}^{+\omega}
$$
$\left.\begin{array}{l}p>q \\ r>\omega\end{array}\right\} \Rightarrow$ Reduction
$\left.\begin{array}{l}p<q \\ r<\omega\end{array}\right\} \Rightarrow$ Oxidation

$n$-factor of $A_{x} B_{y}=x|p-q|+y|r-\omega|$

Case 3

Disproportionation reaction
$A_{x}^{+p} B_{y} \rightarrow A_{x_{1}}^{+q}+A_{x_{2}}^{+r}$
$$
\begin{array}{l}
p>q \Rightarrow \text { Reduction } \\
p<r \Rightarrow \text { oxidation }
\end{array}
$$
$n$-factor of $A_{x} B_{y}=\frac{(x|p-q|) \times(x|p-r|)}{x|p-q|+x|p-r|}$

Now what I am really interested in is How to derive these formulas ?
The first and second cases are pretty easy and logical to think of but the third case really bounced off my head and I am not able to derive the third formula. Please help !

Hmm, a simplifying tool should not be more complex than tasks it should simplify.

@Poutnik The third formula may look big at first but it indeed is quite easy to apply. I have tried it on several reactions and it gives the correct answer although the normal method of calculation n-factor by balancing was also quite easy but out teacher told us to learn this formula to be used in case of large and scary chemical equations

What I mean, I have never needed that, as redox equation enumeration is so simple.

It should be simplified as $$
\text{n-factor} =x \cdot \frac{|p-q| \times |p-r|}{|p-q|+|p-r|}
$$

I will try to explain in the third case of disproportionation reaction first after that I believe you would be able to derive the first two cases easily. Let's start
We will begin with the definition of n-factor.

n-factor is the total number of mole of electron transferred for oxidation or reduction of 1 mole of molecule.

Now let's take a general disproportionation chemical reaction as
\begin{array}{c}
A_{x}^{+p} B_{y} \rightarrow A_{x_{1}}^{+q}+A_{x_{2}}^{+r} \\
p>q \Rightarrow \text { Reduction } \\
p<r \Rightarrow \text { Oxidation }
\end{array}
Now we will start balancing this reaction as follows:
Split the reaction in oxidation half and reduction half.
Reduction half
$$
A_{x}^{+p} B_{y} \rightarrow A_{x_{1}}^{+q}
$$
$$
n_{1}=x|p-q|
$$
Oxidation half
$$
A_{x}^{+p} B_{y} \rightarrow A_{x_{2}}^{+r}
$$
$$
n_{2}=x|p-r|
$$
Here $n_{1}$ and $n_{2}$ are n-factor of the the respective half reactions (i.e. the number of mole of electron transferred per mole molecule $A_{x}^{+p} B_{y}$).
Now, in order to balance the equation we must equate the number of electrons transferred in oxidation half and reduction half and then add these half reactions.
Easiest way to equate the electons transferred is as follows:
Reduction half
$$
n_{2} A_{x}^{+p} B_{y} \rightarrow n_{2} A_{x_{1}}^{+q}
$$
Oxidation half
$$
n_{1} A_{x}^{+p} B_{y} \rightarrow n_{1} A_{x_{2}}^{+r}
$$
In the above half reaction the number of electrons transferred = $n_{1} n_{2} $
Now we add these two reactions to get the final balanced reaction as
$$
(n_{1} + n_{2}) A_{x}^{+p} B_{y} \rightarrow n_{1} A_{x_{2}}^{+r} + n_{2} A_{x_{1}}^{+q}
$$
Now in the above reaction we know that the number of electrons transferred is $n_{1} n_{2} $ for $(n_{1} + n_{2})$ mole of  $A_{x}^{+p} B_{y} $.
Therefore,
$$
\text {n-factor }=\frac{n_{1} n_{2}}{n_{1}+n_{2}}
$$
And if we expand this by putting the formulas of $n_{1}$ & $ n_{2} $ we get,
$$
\text{n-factor} =\frac{(x|p-q|) \times(x|p-r|)}{x|p-q|+x|p-r|}
$$


text:
Dry ice sublimation rate in water
As I got some a finger sized of dry ice in water, it started to bubble vigorously as expected and gave off a lot of smoke. But after 3 minutes or so, the rate of sublimation seems to slow down a lot, and in only bubbles from one spot of the ice, and the rest of the dry ice seems to be not sublimating anymore. I dropped the dry ice into tap water.
So what causes the decrease in rate of the sublimation? Would drinking/distilled water cause different results from tap water?
Thanks!

I guess it got covered with water ice.

Exactly so - water ice insulates the dry ice.

I remember preference of warm/hot water for visual fog ( not smoke ) effects.


text:
Comparing Rate of Hydrolysis of Acyl halides

In this particular question, my answer is (D) as steric hinderance is minimum there, but answer (B) was given due to the electron withdrawing (-M) effect of the nitro group.
Isn't steric hinderance the dominating factor in nucleophilic addition type reactions? Is it that a planar phenyl ring doesn't offer that much steric hinderance?

@Jay Yes knew it , but problem was that steric hinderance should've been more dominating factor than this ?

Both are 1° carbon but your view can't be ignored.

The high reactivity of acyl chlorides towards nucleophilic attack is due to the highly polarised situation of the carbon-oxygen/chlorine bonds i.e.

As you have mentioned due to the phenyl ring being planar, the steric hindrance does not overpower the polarising thing.
Now on comparing the reactions -NO$_2$ group's -M effect will be the most successful to make the carbon-oxygen/chlorine bonds more polar as compared to methyl's +I effect.

Yes , seems right

I disagree. This paper compares the rates of alcoholysis of various acid chlorides and it is found that acetyl chloride has a much higher rate than p-nitro benzoyl chloride.

@RobinSingh the paper is not explicitly showing any comparison between p-nitro benzoyl chloride and acetyl chloride. It does talks about ethyl p-nitrobenzoate which is a different compound.

@DivyanshVerma It does lol. Check Table I and II on page 4

I don't have access to the whole pdf


text:
What is the correct molecular, total ionic, and net ionic reaction equation of reaction between magnesium nitrate and sodium chromate?
From the question, I think that the reactants are $$\ce{Mg(NO3)2(aq) + Na2CrO4(aq)}$$
But the problem is, I am confused about the result of the reaction. Is it $$\ce{MgCrO4}$$ and/or $$\ce{NaNO3}$$?
How is the state of the product? In my opinion, there should be at least $$\ce{NaNO3(aq)}$$ as the product, but what about the magnesium chromate? Is it really magnesium chromate or it is something else?

Please put the question (which you ask in the title) also in the body of the post along with a citation where it is taken from.

There is no solvated molecule of either of 4 ionic compounds. All interactions ( or lack of ) happen on hydrated ionic level, including eventual precipitation or formation of ionic pairs.
All four salts are soluble, including magnesium chromate with the solubility $\pu{137 g/100 mL}$ at $\pu{20^{\circ}C}$ (solubility table).
When  dissolved, salts form independent hydrated ions:
$\ce{Mg(NO3)2(s) + Na2CrO4(s) ->[H2O]  Mg^2+(aq) + 2 NO3-(aq) + 2 Na+(aq) +CrO4^2-(aq)}$
If the solution is being evaporated, the salt with lowest solubility starts precipitating the first, with  the repective ions recombining into the solid, stripping off their hydration cover.
The same solution is formed, if we start with equivalent amounts of the other two salts:
$\ce{2 NaNO3(s) + MgCrO4(s) ->[H2O]  Mg^2+(aq) + 2 NO3-(aq) + 2 Na+(aq) +CrO4^2-(aq)}$
Ions have no memory which ions they were paired with in the solid state.

This question shows up a couple times in a google search, e.g.

Source: https://faculty.uscupstate.edu/cbender/Web%20page%20folder%20enmass/CHM111L/NIE%20answers%20worksheet%20update%20Su2013.pdf

or

Source: https://www.fusd1.org/site/handlers/filedownload.ashx?moduleinstanceid=10419&dataid=21159&FileName=Net%20Ionic%20Equations%20worksheet.pdf

However, it seems that magnesium chromate is soluble in water, so there is no precipitation reaction (see Poutnik's answer).
If you have an abbreviated set of solubility rules, you might think that magnesium chromate is insoluble:

The carbonates, phosphates, borates, sulfites,
chromates and arsenates of all metals except sodium,
potassium and ammonium are insoluble in water but
soluble in dilute acids. $\ce{MgCrO4}$ is soluble in water;
$\ce{MgSO3}$ is slightly soluble in water.
Source: https://www.chem.tamu.edu/rgroup/hughbanks/courses/102/slides/slides13_2.pdf

So if you miss the statement about $\ce{MgCrO4}$ and $\ce{MgSO3}$ being an exception, you might think otherwise.


text:
What is the exact Hamiltonian core of the Fragment Molecular Orbital Method?
My major is not chemistry and I am trying to understand the Hamiltonian core of the Fragment Molecular Orbital (FMO) method.
I think it should be the normal Hamiltonian core of the restricted Hartree-Fock plus two other terms: nuclei attraction between other fragments and the current fragment in the calculation and also the potential energy between them.
But I was wondering If I could get the formula, as I saw in different papers that there are different versions and different formulae associated with that.

I must wonder whether this would be better suited at [mattermodeling.se]. Please refrain from asking it there again. If you would like it migrated there, please raise a custom flag and a moderator will take care of this. To make this clear though: I do not think it is off topic here, so there is no urgent need for migation.


text:
Why is Nylon-2,6 biodegradable but Nylon-6,6 isn't?
Both have similar functional groups so what causes Nylon 2,6 to be biodegradable while Nylon 6,6 is not?

Perhaps because used aminoacids ( 1 of 2 is glycine ) in Nylon-2,6 remind peptidic structure of proteins. $\ce{[-NH-CO-CH2-NH-CO-(CH2)5]n}$, in contrary to Nylon-6,6 $\ce{[-NH-CO-(CH2)4-CO-NH-(CH2)6]n}$ where amidic bonds alternate the C-N order.

Biodegradation of polymers generally requires three conditions:

There must be a way to break the high MW polymer molecules down into smaller MW pieces. With condensation polymers like polyesters and polyamides, the logical break points are the ester or amide bonds, which are broken by hydrolysis. Breaking a polyolefin such as polyethylene is much harder and usually is done by oxidation.

The break point must be accessible by an enzyme and any required co-substrate in order for the reaction to happen. In the case of polyamides, this means some sort of amide hydrolase enzyme and water.

The resulting low MW product must be able to be metabolized by microorganisms. With most non-aromatic small organic monomers, this is the easiest step, especially if the monomers are aliphatic acids and amines, as they are for simple polyamides. An example of monomers that cannot be metabolized is halogenated organics such as perfluoroalkanoic acids, where the fluorine atoms interfere with the reactions necessary for breakdown.

For most organic plastics, the first condition is the hardest to satisfy. In particular, polymers with high crystallinity are resistant to degradation because enzymes cannot access the bonds. Similarly, highly hydrophobic polymers are more resistant to hydrolytic degradation because water is excluded from the structure.
As an example, cellulose, which is highly crystalline, is not biodegraded as easily as starch, which is non-crystalline. Among plastics, nylon 6,6 (Tm ~ 250 C, Tg ~ 55 C) is much harder to degrade biologically than poly-$\epsilon$-caprolactone (Tm ~ 60 C, Tg ~ -60 C).
I was unfortunately unable to find a value for the melting temp or glass transition temp of nylon 2,6, but I am confident that those values are much lower than for nylon 6,6, indicating a more amorphous structure which is more amenable to enzymatic hydrolysis to soluble oligomers and monomers.


text:
What is the reason behind the stereoselectivity of the conversion of vinyl silanes to alkenes?
I have included some snapshots from clayden organic chemistry:

The photos show a discussion of the addition of an electrophile to the bottom of the vinyl silane. However, what if the electrophile attacks from the top? Why does it still lead to the same product as seen in the photo? I seem to get the opposite product (e.g. E isomer becomes Z isomer instead of remaining as E)

You are starting with an achiral vinylsilane and forming an achiral product with retention of stereochemistry (Si --> E). Addition of the electrophile to either face of the double bond is equally likely. It forms a pair of mirror image intermediates. The collapse of each one leads to the same product. Make sure your intermediates are enantiomeric. If both mirror image transition states had to be drawn, Clayden would cost a lot more!!! Only one transition state is normally shown.

I think the stereochem of the final compound depends on which way the Si group is--up or down. Is there a way to predict this?

BTW, this reaction is stereospecific, not stereoselective. See: https://chemistry.stackexchange.com/questions/77061/what-are-stereospecific-and-stereoselective-reaction

Clayden mentioned that the rxn is stereoselective not specific though... I suspect they are just trying to explain why a particular path is favoured

The terms  have specific meanings. You can google them. If you  have 2-phenylpropanal and add CH3MgBr and get a lot more of one diastereomer than the other, then the reaction is stereoselective. Sometimes synthetic organic chemists, who complete a total synthesis, use the term stereospecific to describe the synthesis when they should say stereoselective. Stereospecific reactions are invariably mechanism controlled. The bromination of 2-butene is another example.

I have used the achiral (Z)-vinylsilane 1 as an example with one of the p-orbitals of the double bond colored red. When the electrophile reacts on the "top" of the black p-orbital the carbon rehybridizes from sp2 to sp3 producing the chiral carbocation conformation 2. The same operation on the "bottom" lobe of the black p-orbital affords carbocation conformation 3. These two carbocations are mirror images of one another. In these Newman projections the polar silicon-carbon bond (Si$\rightarrow$C) makes a 60o angle with the vacant p-orbital. There is already partial overlap between the Si-C σ-bond and the vacant orbital. Rotation about the central C-C σ-bond of the sp3 carbon by 60o in a clockwise direction in 2 brings the E-substituent proximate to the phenyl group. Cleavage of the silyl group leads to (Z)-4. The same operation with 3, but in a counterclockwise direction, also leads to (Z)-4. There is no overlap between the Si-C bond and the other lobe of the vacant p-orbital.
Addendum: There can be no rapid rotation about the central C-C bond in 2 or 3 because that would lead to the same distribution of products as afforded by both (Z)- and (E)-1. If so, the reaction would no longer be stereospecific.


text:
Simmons-Smith reaction, example reaction
Does this reaction work in theory? It's based on the Simmons–Smith reaction but with ZnCCl$_4$ instead of ZnCH$_2$Cl$_2$.

I doubt it. S-S is overwhelming done with di-iodo or dibromomethane. Dichloromethane is often used as solvent but does not participate.

ZnCCl4 is not possible. At least, I don't know. Can you link me where did you find this compound? $\ce{ZnCH2Cl2}$ also does not exist as a sole compound as it is a carbenoid intermediate made from the reaction of DCM and zinc. $$\ce{2CH2Cl2 + 2Zn -> [2ClCH2ZnCl] <=> (ClCH2)2Zn + ZnCl2}$$

@NilayGhosh I didn't see it anywhere. My thought was that if $\ce{ZnCH2Cl2}$ makes a compound with 2 hydrogens, $\ce{ZnCCl4}$ would make a compound with 2 chloride atoms attached.

But that compound isn't possible.


text:
Novalucol back titration lab calculations
I had a lab today where we had the goal of figuring out how many moles of HCl were neutralized by one tablet of Novalucol. Novalucol is an antacid tablet containing magnesium hydroxide and calcium carbonate, 180.4 mg and 55 mg respectively. We had 0.984 grams of the tablet, we took one sixth of that meaning 0.164 grams and dissolved it in 100 ml 0.1 M HCl. We then divided these into four 20 ml beakers and titrated them with NaOH, our average value being 16.1 ml NaOH needed. My issue is these calculations and figuring out how I adjust for us having taken 1/6th of the tablet and 20 ml of the 100 ml solution to find the whole thing of how many mols 1 tablet could neutralize. I gave these calculations an attempt without adjusting for those factors if someone could let me know if I'm on the right track?

Please typeset this per style guide for the site. It will also be helpful to link to the chemical composition of Novalucol for our non-Swedish audience.

I apologise! Where can I find information about a style guide for questions? I just figured being able to see my calculations on paper would make it easier to understand what I was doing. Will add info about Novalucol as well, was honestly not aware this was a swedish thing only.

https://chemistry.stackexchange.com/help/notation


text:
Does nascent Oxygen/Hydrogen really exist or just or is it just electron from ionic equation?
Recently I was taught by my teacher that there is no such thing as nascent oxygen neither hydrogen. It just represents Electrons from ionic equations. Then he proceeds to give example of following partial equations.
$$\ce{SO_2 + 2H_2O -> H_2SO_4 + 2H}$$
$$\ce{Cl2 + 2H -> 2HCl}$$
This equation in ionic form can be written as:
$$\ce{SO2 + 2H2O ->SO4^{-2} + 4H^+ + 2e^-}$$
$$\ce{Cl2 + 2e^{-} -> 2Cl-}$$
I want some more insight on this matter and some conceptual understanding.

Search the site, it has been discussed at length. Nascent H2 or O2 is not an ionic version.

https://chemistry.stackexchange.com/questions/131162/is-there-any-evidence-any-evidence-at-all-that-nascent-hydrogen-actually-exi

$\ce{SO2}$ does react with $\ce{H2O}$ according to your second reaction (which is a half-equation) : $$\ce{SO2 + 2 H2O -> SO4^{2-} + 4 H+ + 2 e^- }$$ But $\ce{SO2}$  cannot react according to the first equation, which produces two non-combined (nascent) $\ce{H}$ atoms. Usually the electrons produced in this half-equation do not react with $\ce{H+}$ to produce free $\ce{H}$ atoms as you state. These electrons react with some stronger oxydant like $\ce{Cl2}$ or $\ce{O2}$ in equations like $$\ce{Cl2 + 2 e^- -> 2 Cl-}$$ or $$\ce{O2 + 4 e- + 4 H+ -> 2 H2O}$$ So "nascent" $\ce{H}$ atoms are not produced by $\ce{SO2}$ in water.


text:
By what chemical process does shellac in alcohol degrade over time?
It is well established that shellac dissolved in alcohol ('liquid shellac') has a limited shelf life. The actual shelf life is highly variable, depending on the source, manufacture, storage. But after sufficient time, a batch of liquid shellac will no longer dry hard. A good test is putting a little shellac on a piece of glass. Fresh liquid shellac will dry hard in about 15 minutes. Sufficiently old shellac will stay tacky/gummy indefinitely.
What is the mechanism behind this change in properties?

I think you should give us a sense of what level answer you expect. Shellac is complex. I suspect that many degradation routes are possible. For instance esterification can occurs. Or the water content might get, although slowly, over the threshold for hardening (alcohol absorbs water..). And so on.

I guess my real question is, "Why is shellac stable by itself, but degrades when dissolved in alcohol?" And the level of detail I'm looking for is something like, " classes of processes result in  classes of changes. Those changes prevent hardening because ". I'm used to thinking of solutions as stable, and reactions (from college chemistry) as things that happen within minutes or hours and result in significantly different chemicals.

that is correct. But as I said, and having no specific idea, I believe that even specialists of that material are still investigating. At least not because interested on shellac itself, but as working horse for techniques and methods. For instance, if the water content rises, it can reach a point where hardening does not take place. Esterification reaction can occur internally, or the material can degrade (kind of molecular weight reduction) so that smaller fragment do not hard. Etc. I am convinced that multiple path are involved depending on the batch origin and treatment.

Reaction can be also very slow. Among the immense number of synthetic and natural molecules know, relatively few are indeed both energetically and kinetically stable to avoid alterations along days, years or decades. Shellac might be one of that, but apparently not in hydroalcoholic solutions.

Since shellac arguably can be described as a bioplastic, now I'm thinking along these lines: Plastics consist of long polymer chains. Hard plastics are hard because the chains are long and cross linked. If the chains in shellac are 'cut' shorter, that reduces its ability to harden because shorter chains result in less bulk rigidity. Is that on the right track? Does esterification reduce cross linking?

Would water permanently prevent hardening? It makes sense that water present in the bulk material would result in less hardness, but wouldn't the water eventually evaporate? Or does the water end up (hydrogen) bonded to the chains, as in a hydrated mineral crystal? Or does the water chemically bond to the polymers, as in hydration reaction like the formation of sucrose?

First. I am just saying what could make sense in a broad context. And what you said above is about the same line of thinking. Another possibility is partial cross linking. Even obtaining bigger chains can interfere whit proper hardening at the right moment. Water can be trapped inside kind of cage. And so on. For details, I suspect you have to refer to a real search, Google might be not enough. With luck, someone who indeed worked on shellac chemistry would perhaps respond. But I doubt that we have such an user here.


text:
What is the pH of the universal indicator?
I know that the universal indicator is a solution of various pH indicators used to find out the pH of a substance by displaying different colors.
But, what is the pH of the universal indicator itself?
I found this line on the Wikipedia page for pH indicator:

In and of themselves, pH indicators are frequently weak acids or weak bases.

A quick google search indicates that a universal indicator is usually composed of thymol blue, methyl red, bromothymol blue and phenolphthalein.
How can this information be used to infer the actual pH of the universal indicator itself?

I think you should read about pKa...

Interesting question. You successfully found the components but forgot the solvent(s)? Do you remember that pH is defined for aqueous solutions only. All these indicator molecules are large relatively hydrophobic organic molecules so they are not water soluble. Thymol blue and phenolphthalein are such examples. One needs to add alcohol to dissolve these indicator dyes. Now, that you have alcohol + water mixture and dyes, the pH concept becomes murky. In short, for the universal pH indicator solution, we cannot state its pH.
Yes, pH concept can be extended to organic solvents and their aqueous mixtures but the question of measuring it is a headache!


text:
Synthesis of allylic ether, starting with phenalene
I stumbled across a retro-synthesis question, for which I don't know the exact answer; I prepared two synthesis ways, but I don't know if they could be realized in a lab. The question is the following:

Here are my two ways:

Are those ways feasible or should I take another approach?

The method of installing the allyl ether in your top method will not work. NaH + allyl bromide would be better.

Ah, so basically, the alcohol gets deprotonated, forms an alkoxide ion, and attacks the allyl bromide in an Sn2-reaction to form the ether?

Exactly so, NaH is easier to handle than sodium metal for creating the alkoxide

You could also epoxidise the double bond and do a reductive opening to give the same alcohol as in method 1. There are a number of methods for this, catalytic transfer hydrogenation with ammonium formate and Pd/C should work as described here:  https://www.researchgate.net/publication/232995596_Pd-Catalysed_Regiospecific_Reductive_Ring_Opening_of_Epoxides_and_Glycidic_Esters

In your second solution, the allylic alkoxide is likely to effect E2 elimination on the bromide. Go with @Waylander.

@Waylander is right. For the top method you need a stronger base and one hydroxyl group to deprotonate, plus another halide where the deprotonated alcohol can attack (hence his suggestion of allyl bromide). NaH should be good enough for deprotonation. In the second method (Williamson synthesis), alkoxide can act as a base and will give significant amounts of elimination product, so though the reaction works, the yield will be low.


text:
Comparing average radius of subshells for multielectron atoms
I recently came across a question comparing the average radius of subshells.
A search on the internet gave the following result for single electron atoms:
$$\langle r\rangle_{n,\,l}=\frac{a_0 n^2 \left(\frac{1}{2} \left(1-\frac{l (l+1)}{n^2}\right)+1\right)}{Z}$$
Thus, the average radius should decrease in the order s>p>d for same n
But, I remember the usual saying that goes like s orbitals are more close to the nucleus and d orbitals are more spread out. Or s orbitals have more penetrative power than d orbitals. This same argument is given when computing Zeff, that s shield better than d.
So, are these arguments correct? Also, these arguments come into play in multielectron species. So, does the average radius of subshells follow a different order in multi-electron species? I have tried searching the internet but couldn't find any such data for multi-electron species.
I have already seen this answer and it explains well but it doesn't answer why the above arguments give the wrong result. So, I posted a new question.

Comparing avg r may be very misleading, as even for about the same avg r, different radial and angular distribution plays big role.

Ask yourself, for the given avg radius, what happens to the orbital energy, screening and being screened factors, if the electron distribution gets more or less spreaded radially or angularly. The classical electrostatic mental analysis is sufficient for it. Include this analysis to the question. Readers can then guide you by their confirmation or refutation.

Actually, I want to ask that whether the same order follows for multi electron atoms? If yes, then what about the shielding arguments that I gave; if no, then why is it different from single electron atoms.


text:
LEED - determining surface structures
How can one describe surface structures (before and after adsorption)? I understand the diffraction theory but I have never really tried to determine the structures of samples by myself.
LEED (low electron energy diffraction images of Pd(1,1,0) surface (clean and with adsorbed CO):

All I can see is a rectangular diffraction pattern (which is regular so we are dealing here with a single crystal) and that the pattern has changed a bit after the adsorption. Is there any other information I can get from the images? I searched online for some tutorials but all I could find were scientific articles and I really could not grasp it fully. I need to work out some examples.

EDITED:

An incomplete list of suggestions:

From the position of the spots, did you determine their mutual distances (e.g., to discern a square from a rectangle), and angles if you connect «the dots»?  Beside symmetry, this eases to relate between direct and reciprocal space.

What about the intensity of the spots (in either one of the frames) relative to each other, is there a regular (symmetry related) pattern?  Depending on the space group (in 3D crystallography), there are systematic extinctions, so I think 2D and 2.5D crystallography may have an equivalent to this, too.  For the clean substrate, match the spots the relative anticipated position (e.g., $d$-spacing) and intensity pattern?

Do you recognize a second, weaker pattern if you extend the time of exposure for the sample (substrate + deposit)?  (Again in analogy from 3D crystallography, where an additional set of diffraction spots may be observed in presence of a sample's superstructure).  If the experiment anyway is recorded with a CCD detector, this may be easier to recognize visually if intensity $I$ is not scaled linearly, but follows e.g., $\sqrt(I)$ (this pushes a bit the weaker signals relative to the stronger ones) and good color map (e.g., viridis, cross-reference) for visualization.  For this, a good correction vs. background and empty substrate is helpful.

I think it is a FCC structure, I finally found a nice database.

Nice answer @Buttonwood! We've had a bunch of LEED or diffraction-related questions unanswered on MMSE for quite a while, do you happen to have any insight on any of them? For example: https://mattermodeling.stackexchange.com/q/2380/5, https://mattermodeling.stackexchange.com/q/4459/5, https://mattermodeling.stackexchange.com/q/6501/5,
https://mattermodeling.stackexchange.com/q/6171/5 https://mattermodeling.stackexchange.com/q/3865/5, https://mattermodeling.stackexchange.com/q/6095/5, https://mattermodeling.stackexchange.com/q/6269/5,
https://mattermodeling.stackexchange.com/q/6574/5

@user1271772 While I recall passing answered questions which mention Vesta, CCDC Mercury, .cif files, etc., I wasn't aware MMSE would touch this much diffraction experiments.  Your comment expands nicely my perception of MMSE, that not only holds a tag x-ray-diffraction (like the one I knew from chemistry.se), but equally about (what small-molecule crystallography may consider as «on occasion, work on much smaller crystals») electron-diffraction.  I'm going to read these other questions, too.

Palladium is fcc so you are looking at the fcc(110) surface.
If the 3D lattice constant is $a$ then the (110) surface has a rectangular 2D lattice with lattice constants $a$ and $a/\sqrt{2}$.
From https://www.ibiblio.org/e-notes/Cryst/FCC.html
 click for larger
The longer direction in space is the shorter direction in reciprocal space as shown by LEED.
In the drawings below the added red spots could represent added features to the real and reciprocal lattices, perhaps CO adsorption on the Pd(110) surface. We don't know if they are on top of Pd atoms or in the bridge or hollow points and can't tell from normal LEED images.
A: fcc(111) surface
B: one explanation for the LEED image seen
C: another explanation for the LEED image seen
But I don't have those extra spots
With only one image at one beam energy, it's impossible to tell if those spots are missing due to interference at 76 eV only, or if they never appear at any energy.
If a spot appears, you know you have some structure with the periodicity it suggests, but the mere absence of a spot in one image does not by itself guarantee that there is no corresponding periodicity present.
The electrons pass through several atomic layers before attenuating and so the returning reflections can interfere constructively and destructively in complicated ways.
If you have many images at many energies and you never see the spots appearing in C but not B, then you can *tentatively conclude your structure is most likely B and probably not C, but without more advanced studies like IV-LEED (plots of spot intensity versus beam energy for a long series of closely spaced energies, fit by a theoretical calculation or surface XRD or electron holography, it's sometimes hard to differentiate definitively between possible solutions.
From Matter Modeling SE:

How do I know if the reconstruction in the picture is a 2x2 reconstruction? Highly Relevant!
Overview of how self-consistent dynamical low-energy electron diffraction simulations are performed
Simulated low energy electron diffraction (LEED) patterns
Have FDTD methods made inroads into dynamical simulation of electron and/or X-ray scattering by crystals?
Straightforward formalism to get four sets of 2D hexagonal lattice vectors of fcc(111) planes that I can also cite?

@uhoh  And I actually do have other images, less spots there, i Edited my question.

@Leif I see you r edit. Those are different beam energies for the sample before adding CO, right? The 76 eV image in the new group of images doesn't show the extra spots. But it's a beautiful demonstration of how each spot changes intensity differently as a function of energy. Very nice!

+1 I found this because you posted it in the "What's going on elsewhere" chat room. Since I advertised one of your questions on the other answer to this question, perhaps I can also ask if you could help on some of our related questions at MMSE? https://mattermodeling.stackexchange.com/q/4459/5  https://mattermodeling.stackexchange.com/q/6501/5
https://mattermodeling.stackexchange.com/q/6171/5 https://mattermodeling.stackexchange.com/q/3865/5 https://mattermodeling.stackexchange.com/q/6095/5 https://mattermodeling.stackexchange.com/q/6269/5
https://mattermodeling.stackexchange.com/q/6574/5

@user1271772 It seems that I need to wean myself off of Space SE and move over there, my "day job" MM-light so it would behoove me to dig in and start learning from that trove of treasure. I'll take a look at these. Thanks!


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What is the colour of cesium?
I am confused with the color of cesium. Some websites and books say that it is a silvery white metal, and some say it has a golden colour, and some say it has a silvery golden colour.
Among these colours, which is the true colour of the cesium? If it has a golden or silvery golden colour, why does its colour deviate from colours of the remaining alkali metals (i.e., silvery white colour)?

https://en.wikipedia.org/wiki/Caesium is pretty clear about it.

Yeah I saw this website. But in my text book they gave that cesium is in silvery white colour and   I also saw some pics in internet in which they showed silvery white colour of cesium.

This answer says caesium is silvery-golden.

A picture, specifically from  WP user Dnn87, beats a thousand words. The gray background is ideal for bringing out the true golden color of caesium metal.

The color comes from plasma oscillations, in which  electrons in a conducting medium spontaneously oscillate due to inherent instabilities. These may be traced to the electromagnetic interactions as described by Maxwell's Equations. In most metals the plasma oscillations are at too high a frequency to interact with visible light, but in caesium the plasmonic frequency is lower (due in part to the large atomic size and only one conductive electron per atom, which lowers the electron density); thus the absorption range enters the blue/violet region of visible light.

I do like this video showing caesium metal being distilled - it is both extraordinarily beautiful and extraordinarily dangerous! Interestingly, some portions appear golden, and others appear silvery. It's possible that sufficiently thin layers of caesium (maybe a few dozen atoms thick) can reflect enough light to look metallic, but not absorb enough blue light to appear golden.

Also, I believe it's useful to mention the importance of relativity in describing why caesium is golden, while rubidium has no such tint.

@nicolau the article you reference de-emphasizes the relativistic effect on electron orbitals for caesium, in favor of the plasmonic effect.

Hmm,  caesium vapor absorption spectrum is not so much decisive to support the plasmonic effect.

You don't have plasmons in the gas phase. They are a property of the condensed metal.

I know, that is why I have posted that. :-)

What is the argument here? anyone who has ever seen clean cesium in an ampoule knows it has a notable gold tinge (it isn't as strong a colour as gold, but it is clearly not a "silvery" metal.)

@matt_black It seems to be rather about the reason than the fact itself.


text:
What is the mass of a proton?
My textbook explains that the deviation from integer atomic mass is caused by mass difference between proton and neutron, which are $1.67262·10^{-27}$ kg and $1.67493 ·10^{-27}$ kg, respectively.
If SI system currently define $1$ mole as the number of entities in exactly $12$ g of $C^{12}$, which is Avogadro's number $N_A = 6.02214\times 10^{23}$, then $$6.02214\times10^{23} ( C^{12}) = 12g$$ Likewise atomic mass unit (u) for relative mass is also defined by $C^{12}$ isotope as $12 u = C^{12}$ atom. So $$1 g = 6.02214\times 10^{23}u$$
Then if $12 u = 6(m_{proton}+m_{neutron}+m_{electron})=1C^{12}$,
and $m_{proton}<m_{neutron}$ (as stated by the text),
$$6\times\frac{m_{proton}}{m_{total}}<\frac1{2}$$proton mass ratio in carbon-$12$ is less than half, and one proton mass will be $m_{proton}$ $1 u$ = $1.66054 ·10^{-27}$ kg
This proton mass is less than the value in the book $1.67262·10^{-27}$ kg and would produce a much lower atomic mass for $H^1$ than $1.00783$ u. Catastrophic! Why is this and what am I missing here?

mass defect, see e.g. https://www.chem.purdue.edu/gchelp/howtosolveit/Nuclear/nuclear_binding_energy.htm

The 1.00783 u is the average atomic mass of hydrogen and also includes a small fraction of deuterons. Note also that the definition of the Avogadro constant has been changed and is an exact number since 2019. Therefore, the mass of a carbon atom now carries uncertainty.

The key question is why the atomic mass be integers and integers in which units. I don't think there is such a fundamental requirement.

I apologise for the wording of my original post which may have caused some confusion. All the atomic mass values I quote (from Clayden's Organic Chemistry, mass spectrometry, p.51) are not average isotope abundance but the exact mass of a single isotope. @KarstenTheis and matt_black pointed out that this difference between calculated value and actual mass of atoms is contributed by mass defect and formation energy, therefore it's observed that calculating atomic mass from the sum of its parts would result in a value higher than the actual exact mass.

The definition of Mole has been changed (tough this does not change the issue that you have). Just for sake of precision as this is a chemistry site. Similarly a proton does not have atomic mass.

A free proton is heavier than a proton included in a nucleus. When coming together to create a nucleus, protons and neutrons are loosing  a fraction of their masses, which is emitted as energy. This process is called "fusion".

Your textbook is wrong and there are several confusions in your argument
The first issue is the textbook claim. Which is wrong for two possible reasons. One is that the masses quotes for atoms ni most periodic tables are for the natural abundance of the element given multiple possible isotopes (hydrogen has three: one with a single proton, one with an extra neutron and one with two extra neutrons but both are very rare). Chlorine has two common isotopes (one with 18 neutrons and one with 20; about 3/4 of chlorine atoms have 18 neutrons giving an average mass of ~35.5). Are you sure you consistently quote the atomic mass of specific isotopes?
The other is that the atomic nucleus has a binding energy (which from E=MC2 makes a difference to the net mass which isn't just the combined mass of naked protons plus naked neutrons). In other words the energy it takes to hold a nucleus together actually makes a notable difference to the net mass of the nucleus.
The second effect is at least as important as the difference in mass of naked nuclei. So any claim that the "deviation" is caused by that is just wrong.
By the way, these small details are rarely important in normal chemistry and only matter to nuclear chemists or physicists.


text:
How can we synthesize this cycloester, starting with methyl 4-phenylbutanoate?
I was wondering how to realize following synthesis:
I am struggling with this one; for sure, this has something to do with the benzyl position which is particularly reactive, and something nucleophilic which can close the ring and form the cycloester. But how to do this? Thanks in advance.

Step 1 - benzylic bromination with NBS, Benzoyl peroxide
Step 2 - displace bromine with cyanide using KCN/DMSO
Step 3 - ester hydrolysis to acid with LiOH/THF/water
Step 4 - acid reduction to alcohol with BH3
Step 5 - nitrile hydrolysis with KOH
Step 6 - lactone formation with cat. PTSA/toluene reflux.

An alternative approach to the proposal of @Waylander for the synthesis of lactone 7 from ester 1 depends on the efficacy of the Barton nitrite photolysis, 2$\rightarrow$3. Examples of this reaction are typically conducted with secondary nitrite esters in rigid, steroidal frameworks with few degrees of freedom, an advantage that the transformation 2$\rightarrow$3 lacks. The generation of an alkoxy radical (8$\rightarrow$9) allows for 1,6-H transfer. In this case, the C-H bond has the advantage of being benzylic. Hydrolysis of the oxime 3 affords the hydroxyketone 4 that may be in inconsequential equilibrium with its hemiketal 5. A Wittig reaction with (methoxymethylene)triphenylphosphane produces an enol ether that is a masked aldehyde group. Acid hydrolysis of the enol ether affords the hemiacetal 6, which can be readily oxidized by Jones reagent. In the event that a keto acid is produced upon oxidation, $\ce{NaBH4}$ reduction of the ketone and acid-catalyzed cyclization may be effected to form lactone 7.

Neat! Silver Carbonate/Celite will do the lactol to lactone selectively if there is any problem with Jones reagent

@Waylander: I couldn't find 2-->3 in Chem. Absts. Who knows? Another Tet. Lett. for that ambitious young chemist? ;)

if only I still had a lab

@Waylander: Unfortunately, what we are doing currently is our default position.


text:
General cleaning solution for clinical chemistry applications
I want to reuse some plastic and glass items that I am using to do colorimetric concentration analysis of serum and urine. There are some enzymatic assays, mineral assays and so on. I have been trying to soak in hydrochloric acid (pH 1.5) for 30 minutes and rinse in distilled water. But it seems the items are not satisfactory clean because when I use them, it seems drops cling to the surface as if there are microscopic impurities on the surface.
What is recommended for cleaning in this situation?

I used to use chromic acid for cleaning glassware for analytes in HPLC applications, and it worked very well. This guide might be of use.

Please don't suggest chromic acid. It is too dangerous and carcinogenic. We have no idea of the student's handling expertise. Plastic cuvets have to land in garbage.

Likely, there won't be one solution to everything for the history (what kind of sample/sample digest/derivatization) took place in the container made of glass or plastic.  «Plastic» itself comprises a wide range of polymers with different resistance toward chemicals, too.  I speculate biology.se is better suited to answer this kind of question with currently 500 entries found for keyword «cleaning», including this one.


text:
Do solid phase pH buffers exist?
I want to buffer the pH of my solution but I don’t want to add any mobile solutes. I imagine that large molecular mass buffers could be used such as phosphorylated polysaccharides or proteins. The limit of such buffers would be solid phase materials that can protonate or deprotonate at their surfaces, but aren’t mobile.

Solid acids / bases don't affect the pH of a solution, because hydronium / hydroxide ions stay close to their surface.

Just like wave-particle duality eventually ends up in meta-physics, you are extending the concept of pH too far where the concept of pH becomes murky. The simple answer to your query is ion-exchange resins, which readily protonate and deprotonate at the surface. You have dozens of strong acid, weak acid, strong base, weak base resins. They work perfectly!
Let us consider a very simple case using a sulfonic acid resin, which consists of a polymer with sulfonic acid groups ($\ce{-SO3H}$. The sulfonate $\ce{-SO3-}$ group is fixed and the associated proton is exchangeable with another cation in a quantitative way. It looks something like this...

Suppose this resin is in contact with 0.01 M HCl. The pH of the solution is 2, but what is the local pH of the resin surface? The number of protons at the resin surface are pretty high hence the pH is different. Such difficulties render the extension of pH to solid surface quite difficult. This type of work of done in the 1940s with indicators adsorbed at the resin.
In short, such materials exist but we cannot call them solid phase buffers, because just like pH buffering concept cannot be applied to surfaces. One can ask the question of a pH at a solid surface in contact with an aqueous solution but we cannot ask what is the pH or the buffer capacity of solid surfaces.
EDIT The comment asked about the titration of a resin. Yes, direct titration of ion-exchange resins is doable and we do get typical titration curves. As stated earlier, this work was extensively done in the 1940s-50s. This figure is an example of a sulfonic acid resin (phenol+aldehyde type) with NaOH directly.

Ref: Gregor, H. P., & Bregman, J. I. (1948). Characterization of Ion Exchange Resins. I. Acidity and Number of Constituent Cation Exchange Groups1, 2. Journal of the American Chemical Society, 70(7), 2370-2373.

The pH/buffering of the resin itself or near the surface may be ill defined, but can't you reasonably talk about the effect of the resin on the aqueous bulk, at least on a phenomenological level? (That is, not the pH of/near the solid surface, but of the separate aqueous phase.) E.g. take your aqueous solution in a flask, dump in the ion exchange resin. Shake and let settle. Measure the pH of the aqueous supernatant. Add a quantity of acid/base. Repeat. Does the resulting curve look anything like a titration of a soluble buffer, or is it like the resin isn't there, or something else?


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How to "draw" a coordination bond (dashed bond) in MOLFILE?
How can we "draw" a coordination bond (dashed bond) in MOLFILE?
I cannot find a bond property or option that would allow me to choose such a thing.
So many drawings of organo-metallic compounds (e.g., metal ion and chelator) contain that.
How come the MOLFILE format doesn't support such a thing?

Generally, Lewis structures do not support dative bonds (dashed or arrows and the like); V2000 simply might be too old to have this implemented. IUPAC recommends something different for schemes for organometallic compounds, where the lines lose their strict character (and are definitively not Lewis(-like) structures).

The BIOVIA documentation about CTfiles, August 2020 edition, pdf file documents both v2000 and v3000 format, with preference for the v3000 format (p. 1). Chapter 2, p. 11 includes a table «Meaning of the values in the bond block [of v3000 format]» which just follows the atoms and their coordinates, this mentions type 9 with two sub-types about coordination, either for metal complexes, or for Lewis acids/bases.

While currently not functional for me, RDKit's cookbook includes a relevant entry to this, though.  Instead of dashes, the dative bond would be represented by an arrow pointing toward the transition metal, e.g.

(from RDKit's Cookbook)
in a section called Organometallics with Dative Bonds.
If you use .sdf in the v3000 format,* RDKit may be adjusted to visualize dative bonds as dotted lines, too:

(Esben Jannik Bjerrum, cheminformania.com)
*) The current definition of .sdf files by Biovia (August 2020, link to .pdf) describes in detail the types of bonds available in the v3000 format to the bond block (cf. p. 11).  Type 9 is set for coordination with dipolar bonds of either sub-type COORD for coordination bonds in metal complexes, or DATIVE for bond between Lewis acids and bases. (There might be a concptual overlap here.)  Bond type 10, by the way, offers two sub-types to handle hydrogen bonds, too.

So that's not really anything explicit in the MOLFILE, but I suppose we could use the same logic they used: "Process organometallic SMILES by detecting single bonds between metals and replacing with dative bonds." So whenever in the bond table in a MOLFILE I find one of the atoms a metal, I would draw a dashed line instead of a solid line. I have my own MOLFILE renderer so I can do that.

I noticed that PubChem also uses that convention, and it was easy enough for me to implement. Single bond with any ligand being a metal --> dashed line.

Alex Clark published a proposal for zero order bonds as an extension to SD files. It’s supported by some software, but I don’t think there’s a list anywhere.

@GuntherSchadow Just out of curiosity about «I have my own MOLFILE renderer so I can do that», is this program somewhere presented in a publication, mirrored on a repository like GitHub/commercialized? An answer in lines of «No, it is an application working fine for me, not intended for publication» were acceptable, too.

@GeoffHutchison I just passed a fairly recent documentation of the ctfiles (BIOVIA, August 2020) here.  Page 11 (or, counting the pages in the pdf, p. 17) describes 10 different bond types (single, double, triple, etc.) without even touching the bond's configuration as either «up», «down», «either», or «none».  Possibly «no bond at all» does not fit level 8, any bond order.

Just a technicality nitpick: instead of writing "credit" and hiding the link, I find it preferable to actually name the source. That makes it easier to recognise the credit once the link rots away and is easier recognised by most readers. Additionally, it might also be a criterion that allows use of an image; in any case, I find it better to be on the safe side with this.

@Martin-マーチン With some thought, this is a convincing argument. It is going to be implemented for illustrated answers in future and, when encountered, in those already given (about 0.5k).

@Buttonwood - The zero bond order proposal is an extension https://doi.org/10.1021/ci200488k and as I said, I don't think there's a list of programs that support ZBO.


text:
Why is the enthalpy of solution of hydrated calcium chloride positive?
My book says that the enthalpy of solution of anhydrous calcium chloride is negative because it produces hydrated calcium chloride. But when hydrated calcium chloride is dissolved in water, the enthalpy turns out to be positive because it is already hydrated. What I don't understand here is: if it is already hydrated and doesn’t want to interact with water anymore, then why can't the enthalpy be zero? Why does it have to be positive? Can it be logically deduced somehow?

Dissolution enthalpy is kind of scale balancing as being combination of lattice energy ( positive solid matrix breaking enthalpy) and hydration enthalpy(usually negative), following the formal process
$$\ce{CaCl2(s) -> Ca^2+(g) + 2 Cl-(g) ->[H2O]  Ca^2+(aq) + 2 Cl-(aq)}$$
The hydration enthalpy is very negative for anhydrous $\ce{CaCl2}$, but just little negative for $\ce{CaCl2.6H2O}$. It leads to negative dissolution enthalpy for the former, but positive one for the latter.
There is more of known $\ce{CaCl2}$ hydrates. Aside of the hexahydrate, the most common $\ce{CaCl2.2 H2O}$ has the dissolution enthalpy much closer to zero than for the above two cases.
For more, see Wikipedia.org: Enthalpy change of solution

Which are the "above two cases" ?

There was confusing typo -should be 2 H2O instead of 6


text:
Isn't hydrogen chloride a salt, because it is an ionic compound?
So I just learned in chemistry that salts are ionic compounds.
Shouldn't,
$\ce{H+Cl-}$ be a salt since hydrogen ion has positive charge and chloride ion has negative charge? If $\ce{HCl}$ is a salt then why do we call it an acid?
Sorry if my question is stupid.

A is B does not generally mean B is A.  Salts have ionic structure when in pure form ( or would have if not stable enough to isolate them ). It is not the case of HCl // BTW, a compound can be a salt, an acid and a base at the same time, like solution of KH2PO4. As being a salt, aan acid or a base are different, not excluding behaviour aspects of compounds.

Pure $\ce{HCl}$ is not a salt. Pure $\ce{HCl}$ is a gas, and a molecular gas. It is not made of ions. A salt is made of ions. There are no ions in pure $\ce{HCl}$. OTOH, $\ce{HCl}$ does not react with $\ce{CO3}$

HCl is not ionic. True, it becomes ionic when dissolved in water, but that's another story. Salts are compounds of metals, and there are no metals in HCl.

"Salts are compounds of metals". Like ammonium chloride. OK.

Hydrogen and chlorine atoms of $\ce{HCl}$ in gaseous state are covalently bound and is termed hydrogen chloride. When this gas is bubbled into water, it ionizes completely to give $\ce{H3O+}$ (free proton + water molecule) and $\ce{Cl-}$ ions and becomes an acid solution which is termed hydrochloric acid. Even in gaseous $\ce{HCl}$, the charge is not distributed evenly. The chlorine is partially negative and the hydrogen is partially positive.

(image source)

The free ion $\ce{H+}$ does not exist in water, because it is not an atom : it is a proton, and it is much smaller than an atom. So it is attached to the Oxygen atom of the next water molecule, producing the ion $\ce{H3O+}$

@Maurice You are correct. Edited.

No bond is completely ionic or completely covalent, Fajan's rule. Also $HCl$ is predominantly a covalent compound rather than ionic.

Shouldn't, H+Cl- be a salt since hydrogen is positive and Cl is negative?

Sure, $\ce{HCl}$ does dissociate into $\ce{H+}$ and $\ce{Cl-}$ in a polar solvent, but that doesn't mean $\ce{HCl}$ is simply $\ce{H+}$ and $\ce{Cl-}$. It is largely covalent. Now you can answer your question. Hope that helps.

You can find useful  Notation basics and How can I format math/chemistry expressions on Chemistry SE. Like $\ce{H2SO4}$ leads to $\ce{H2SO4}$


text:
Why exactly is the conduction band in semiconductors conducting?
Is it because when the electrons are excited to a higher energy level, they can lose energy and fall back down to lower energy levels? Due to the sheer number of antibonding MOs in the conduction band, this could mean that there are more levels for the electron to fall down to rather than instantly go back to the bonding MO, hence leading to more movement of electrons i.e. conduction of electricity?

At least in principle the band does extend physically through the material

Look into solid state physics and, in particular, semiconductor physics.

Please don't edit your questions adding a paragraph asking to like and upvote your posts. This adds zero value to the actual question, bumps it to the main page and your posts are likely going to receive even more downvotes. Instead, look for the ways to improve the content. There were plenty of suggestions in the comments as to how to do so by adding your research or improving writing or formatting, but you seem to ignore most of it. See What can I do when getting "We are no longer accepting questions/answers from this account"?

The question has already been answered, what is the point of me editing it? I believe a better way to go about this would be to let me create a new question and apply the lessons I have learnt from the comments given in my other questions

Semiconductors conduct because some electrons are easily bumped into the conduction band
The basic definition of a semiconductor is a substance where there is an empty bundle of orbitals at an energy just a little higher than the valence band where the bonding electrons reside.
This gap is close enough to the valence band in energy that even thermal energy is enough to boost some electrons into the conduction band. This also explains why the properties of semiconductors are dependent on temperature.
But, when this happens, there are "holes" in the valence band as well as electrons in the conduction band. Both the holes and the electrons can carry current. Moreover, semiconductors can be doped by small amounts of other elements to either add electrons (n-type) or subtract electrons (p-type) from the material (this is vital in semiconductor manufacture for electronics). When doped, the balance of conductors between electrons and holes changes (n-type dominated by electrons, p-type by holes).
So the conductivity of semiconductors depends on a small number of electrons being in the conduction band. But the actual conduction happens by movement of both electrons in the conduction band and holes (the absence of electrons) in the valence band.