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Can cycloalkanes undergo electrophilic addition? If so, what is their mechanism?
Cycloalkanes have the same general formula as alkenes. Doesn't this mean that cycloalkanes can be saturated by turning into alkanes the same way alkenes use? If so, what is the reaction mechanism? In the reaction mechanism of the electrophilic addition of alkenes, a pair of electrons are taken from a C-C double bond and donated to one atom of the attacking binary compound. In the reaction mechanism of the electrophilic addition of cycloalkanes will one entire single bond be broken and the pair of electrons in it donated to one atom of the attacking binary compound? This way it is possible to produce an alkane from a cycloalkane undergoing electrophilic addition, but in terms of energy requirements, I find it a bit unlikely. I think it would be useful if someone could help me clarify this.

You are generalising on a far too high level, and have lost ground. ;) Take cyclohexane and bromine. Do they react?

Pretty much only cyclopropane reacts somewhat similarly to alkenes.

Then why don't cyclohexane and bromine react? Because of energy requirements?


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Standard practice for keeping salts dry
I should point out first that I'm not a chemist, but I'm currently doing a little bit of chemistry; this should clarify that I don't have formal training in a chemistry lab (other than some basic chem classes in my BS).
So the question is this: what is standard practice for keeping salts moisture free if you can't afford a nitrogen generator system (or afford to swap nitrogen tanks out continuously with a desiccator chamber)?
I currently use iridium tetrachloride and I find that a few months after receipt, I have sludge (it's highly hygroscopic). I've tried storing in a tightly sealed bag with desiccant packs (the sealed kind) to no avail. I tried storing in a vacuum chamber but I can't hold vacuum for more than a few hours unless I have it on vacuum continuously (or hook it back up regularly).
I've been told by a local chemist that a simple vacuum desiccator chamber with a bunch of consumable silica gel at the bottom is about the best I can do, I just have to regularly pull vacuum again and swap the silica gel as soon as I see color change.
Thoughts?

You didn't say what your "tightly-sealed bag is made of, but if it's plastic, those are generally permeable to small gas molecules like water vapor.  What you need is a container made from metal or glass that can be hermetically sealed.  I wonder if a mason jar would work.  You also want a dessicant that attracts water more strongly than your hydgroscopic salt.   It would also be helpful to drive off the moisture by heating prior to sealing—IrCl4 decomposes at 763 C, giving off toxic HCl(g), but I assume heating to ~100 C wouldn't be a problem.  But check this first w/ the manufacturer.

Yup, it seems to be some weight of polyethylene, good call.
Do you mean heat the whole chamber or just the salt container?

My suggestion was about the container holding the salt.  Put the salt and the dessicants in something like a mason jar, heat it to drive off the moisture from the salt and the dessicants (check to make sure this works for the dessicants you are using), and seal it.  But before doing any of the above, check with the chemical company that supplies the salt for their guidance.  They may even be able to supply appropriate dessicants and/or a container.

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Are there terms for ions of chlorine other than chloride?
An ion of chlorine with a charge of −1 is called a chloride ion/anion.
Are there terms for other ions of chlorine, for example, $\ce{Cl+}$ or $\ce{Cl^2-}?$

Chloride ion is ok. Everyone will understand chloride anion but that phrase is redundant. Chloride already specifically means the ion $\ce{Cl-}$. // For $\ce{Cl+}$ is use chlorine(I) cation or chlorine(1+) ion. That is such an odd species that I'd explicitly specify the charge and not just say chlorine cation.

Questions regarding halogen forming cations: https://chemistry.stackexchange.com/questions/40171/why-dont-halogens-form-cations-like-h ... https://chemistry.stackexchange.com/questions/40643/why-does-iodine-form-cations-sometimes

Assuming that "term" in your question refers to chemical nomenclature, the authoritative source of information would be the current edition of IUPAC Red Book [1]. From [1, p. 70]:

IR-5.3.2.2 Monoatomic cations
The name of a monoatomic cation is that of the element with an appropriate charge number
appended in parentheses.
[…]

$\ce{I+}$ iodine(1+)

Analogously, the name of $\ce{Cl+}$ is chlorine(1+).
Further from [1, pp. 72–73]:

IR-5.3.3.2 Monoatomic anions
The name of a monoatomic anion is the element name (Table I) modiﬁed so as to carry the
anion designator ‘ide’, either formed by replacing the ending of the element name (‘en’,
‘ese’, ‘ic’, ‘ine’, ‘ium’, ‘ogen’, ‘on’, ‘orus’, ‘um’, ‘ur’, ‘y’ or ‘ygen’) by ‘ide’ or by directly adding ‘ide’ as an ending to the element name.
Examples:

chlorine, chloride

[…]
Charge numbers and radical dots may be added as appropriate to specify anions fully.
Examples:
14. $\ce{O^2-}$ oxide(2−), or oxide

So, the name for $\ce{Cl^2-}$ is chloride(2−).
Reference

IUPAC. Nomenclature of Inorganic Chemistry, IUPAC Recommendations 2005 (the “Red Book”), 1st ed.; Connelly, N. G., Damhus, T., Hartshorn, R. M., Hutton, A. T., Eds.; RSC Publishing: Cambridge, UK, 2005. ISBN 978-0-85404-438-2.

@NilayGhosh I was wrong in that discussion, Mithoron was correct. Chloronium is not $\ce{Cl+},$ it is $\ce{R2Cl+}$ (where $\ce{R}$ might as well be $\ce{H}).$ The original paper explains it more clearly IMO—by the way, you could answer that question of yours by simply quoting the onium compounds section:)

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Pyranose v/s Furanose form of Glucose and Fructose
According to my textbook (Class 12 NCERT Chemistry):

Fructose also exists in two cyclic forms which are obtained
by the addition of —OH at C5 to the C = O group. The ring, thus formed
is a five membered ring and is named as furanose with analogy to the
compound furan.

I understood why the pyranose structure for glucose is more stable than the furanose structure ( The pyranose form has the perfect 60° dihedral angle between two non-ring atoms which minimises this type of strain Source: Stability of furanose vs. pyranose form of glucose?). But I do not understand why the same cannot be applied to the fructose structure.
So, to summarize : Why is fructose present in furanose form and not in pyranose form?

Actually the pyranose form of fructose is favored at equilibrium in water. . .

https://www.ncbi.nlm.nih.gov/books/NBK22547/

@Andrew is correct about the pyranose form of fructose. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/carbohydrates44.pdf That fructose in sucrose is a furanoside is irrelevant.

You are correct that the same reasoning can be applied to fructose, and, as with glucose, the beta-pyranose form is most stable.
It is common in biochemistry textbooks, however, to draw fructose as a furanose. The reason for this is that fructose 6-phosphate is a common metabolite, and the phosphate group prevents the 6'-hydroxyl group from participating in ring closure. As a result, the free 5'-hydroxyl attacks the carbonyl, and a furanose is formed. Similarly, fructan biopolymers often contain 1,6 linkages that also prevent the pyranose formation.


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Phosphoric acid spill cleanup
While working with some rust converter (MSDS, ~20% phosphoric acid by weight), a sequence of events that would make Rube Goldberg proud led to me spilling about 500mL of it all over the place. Yes I was wearing gloves and an acid gas respirator.
I'm not really well versed in chemical cleanup procedures and I needed to act fast so I kind of winged it by:

Wiping the majority with a bath towel.
There happened to be a box of baking soda right there so I dumped a bunch of it on the remaining spill.
It fizzed for a bit then I wiped that up.
I shop-vac'd the rest of the baking soda.

My question is just: Did I handle that OK? And, if not, how should I be prepared if that happens again? (I've been working with the chemical a lot lately.)
The reason I ask is, when I took off my respirator, it still smells like rust converter.
Also, it just happened so I won't be able to see any effects for a few hours but I'm hoping I don't start seeing rust spots on the metals it contacted, or dried crystals, or tear up any of the finish on the wood it contacted. But I'm hoping I got it.
Also, there are cats here and I don't really want residue to be left laying around.
So I'm wondering if that was a decent reaction, or if it was dangerous (I dunno what phosphoric acid + baking soda produces, although the space was well ventilated); and if it was effective, or if there is a better way.
Context is home shop, primarily woodworking; not a lab. So I don't have those nice metal lab tables or anything; lots of cracks, crevices, porous surfaces (concrete and wood), and stuff. And while I normally work with this stuff on a tray for small splash containment, this was a rather... violent spill (heh).

Don't panic. Phosphoric acid is one of the components of Coca-Cola, so it is not poisonous by itself. Avoiding contact to high concentrations should keep you save.

You did well. I'd add a solution of baking soda, just to chase the last traces of acid in all those cracks and crevices. On a side note, phosphoric acid by itself does not smell.

@ivan Ah must be something else. I think it's the smell after it reacts with iron because it's definitely common to rust remover and converter products. It's sort of like metallic fruit. I find it nauseating.

The product you were using is very likely a "rust converter". Used to convert rust to a stable products before painting . I have used it several times , usually I just rinse the  surface , let it dry and paint . It is also used for industrial maintenance coating of poorly prepared surfaces ( those not sand blasted). That is ;it is commonly used and when I worked with coatings I never heard of a problem. Baking soda will neutralize it . I once inspected super phos storage tanks : it starts as 107 % phosphoric acid used to blend liquid fertilizers. When I stepped out of the tank and walked across the limestone gravel ,each step of the rubber boots left some jelly-like acid which foamed on the gravel. The people that worked at the facility thought nothing of it; from that I conclude neutralized phos acid it not a problem . Sorry this is all anecdotal ,but I have never needed to research the situation.

"It is also used for industrial maintenance coating of poorly prepared surfaces ( those not sand blasted). " ... my dentist uses it, too, thankfully not sandblasting my teeth.


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Surfactant micelles from glassware
FYI to start- I'm not a chemist by training, so assume I'm missing some basic knowledge of the field.
I've been trying to do some electrodeposition with an iridium tetrachloride-based solution, and my success at doing so is random at best. In talking to a chemist, they suggested that I buy new glassware and perform nitric acid or sulfuric acid cleaning on all of my glassware; according to them, my cleaning the glassware with regular dish soap (I've been using Dawn) has most likely coated everything in surfactants and those are releasing micelles of surfactant into all of my solutions.
I believe what this person is telling me, as I can read about surfactants forming micelles. However, in a search of both academic (e.g. Scopus) and non-academic sources (read: Google), I can't find any  detailed information on this issue, or its prevention, in lab settings; I also can't find details about how these surfactant micelles actually affect a process like electrodeposition. I seem to only be able to find random informal glassware cleaning SOPs from different labs, but no good reading on surfactant contamination in electrochemistry experiments.
Am I searching the wrong keywords? Is this too broad a question? Is this just common knowledge among chemists because of their knowledge base?
Thanks!

Well, yes, consider this a common knowledge. Surfactant contamination is a thing.

@PoGami, Is this a part of a formal academic research project or just a home experiment? The actual glassware cleaning systems are too dangerous, they are used where really clean glassware is needed. On the other hand, micelle can only form under a certain concentration and they cannot permanently stick to glass. You might have surfactant contamination due to improper rinsing but not as micelles.

@M.Farooq - I view it a bit differently. I'd absolutely believe that the surface of the glassware could be contaminated, that the contaminating molecules could go into solution and then form micelles in the solution.   I totally agree that micelles themselves are not going to be sticking to the glassware and then dissolve into the solution.

@MaxW, But remember that micelles can only be formed when the surfactant conc. exceeds the critical micelle concentration in solution. Suppose a clean glassware has been washed several times, by no means, it can exceed critical micelle concentration in a solution. Yes there will be surfactant in solution but in ppm levels or even less.

@M.Farooq: yes, this is formal research being carried out in an academic lab. I've viewed SDS' and have done the cleaning in a fume hood with proper secondary storage.

@PoGaMi, In order to get rid of surfactants from glass surfaces one can wash or rather soak glassware in 1 M sodium hydroxide for 15-20 min (Caution with eyes and skin) followed by copious amounts of deionized water. I am assuming no other electrode or metallic part will come in contact with base and the glassware is standard borosilicate glass. Remember, dishwashing detergent is the worst thing to use for electrochemical work. There are specialized detergents such as Alconox. Even they should not be used for electrochemistry but still better than liquid soap which sticks forever.

@M.Farooq - got it, thank you, I'll try it if the nitric acid wash I did already doesn't end up working. I guess my original question still stands, though- other than just the understanding that "surfactants eventually form micelles and bind to things", is there any more formal literature where I can read about why this affects processes like electrodeposition?

I am not familiar with electrodeposition but dealt a lot with surfactants affecting silica capillaries. Our standard protocol was to wash the silica capillary with strong base rather than acid to get rid of micelles. Keep in mind that surfactants affect the resistance or better conductivity of the solution. Search conductivity of micellar media.


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In aldehydes with an alkene substituent, what conditions favor Wolff–Kishner reduction over Clemmensen reduction?
This module I am answering asked me to provide a reason why Wolff–Kishner reduction is favored over Clemmensen reduction in reducing (3Z)-pent-3-enal to (2Z)-pent-2-ene but either Wolff–Kishner or Clemmensen can be used to reduce pentanal to pentene.
I am currently running with the notion that the alkene substituent double bond's reactivity is the reason behind this since acid is used in Clemmensen reduction compared to Wolff–Kishner's base, and it might interfere with this alkene substituent instead of the carbonyl group in aldehyde. Is this notion correct?

Correct, the cHCl conditions can add the elements of water or HCl across the double bond or just isomerise it

Why is the Wolff-Kishner preferred? Isomerization of the aldehyde or hydrazone is likely in basic medium.

Doesn't it depend if you have 1,2 vs 1,4 addition? Hydrazine is a strong nucleophile. Also enolizable carbonyls can create an unholy mess in these reactions


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Photodissociated iodine laser and population inversion
Iodine molecules ($\ce{I2}$) can absorb in the visible region and dissociate into $\ce{2 I^.}$ radicals. One of the I atoms is in ground electronic state $\mathrm{^2P_{3/2}}$ and the other I atom is in the excited $\mathrm{^2P_{1/2}}$. So, photodissociation of $\ce{I2}$ produces 50% of the species in the excited state and 50% in ground state.
The $\mathrm{^2P_{1/2}\rightarrow^2P_{3/2}}$ of iodine has been extensively used in laser production, in chemical iodine laser or methyl iodide photodissociation laser for example. But in those cases, most of the $\ce{I}$ atoms are produced in the excited state. The criteria for laser production is population inversion, where >50% of the species are in the excited state.
Here I am not sure if the photodissociation of iodine molecule satisfies the criteria of population inversion, as exactly 50% of the species is in excited state. So my question is—can we consider 50% excited species as population inversion and if so can photodissociated $\ce{I_2}$ generate laser?

You are making a chemical laser. A laser depend on keeping a difference in population between the two levels involved, and of course feedback for a laser oscillator. It does not matter how many I* you make as long as the lasing transition does not go the the lowest ground state level. You should make the transition end at a higher level in the ground state I atoms that has no thermal population, (i.e. a make a 3 level laser) so that the laser radiation cannot pump  this level to the higher level from which lasing occurs.  You must of course keep producing I* atoms.

You are making a chemical laser. A laser depend on keeping a positive difference in population between the two levels involved (upper >lower), and of course feedback for a laser oscillator. It does not matter how many I* you make as long as the lasing transition does not go the the lowest ground state level. You must make the transition end at a higher level in the ground state I atoms that has no (i.e. insignificant) thermal population, thus a make a 3 level laser meaning that the laser radiation cannot pump  this lower level to the higher level from which lasing occurs.  You must of course keep producing I* atoms.

So, if there are two levels (as in the question), there would be no laser, as the emission from excited states would be reabsorbed by the ground state atoms. Three level laser would solve the problem. Is that correct?

Not quite, for two levels the stimulated emission and absorption are at the same wavelength so the population of the two levels equalise. Look up 3 and 4 level lasers.


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Preparation of ammonium acetate buffer for HPLC
Buffer solutions are typically prepared by dissolving an appropriate salt in water and then adjusting the $\mathrm{pH}$ to the desired position by addition of the conjugate acid or base.
For example, to prepare a $\pu{100 mM}$ ammonium acetate solution at $\mathrm{pH}~9,$ you’d first prepare a $\pu{100 mM}$ ammonium acetate, and then add ammonium hydroxide dropwise until the desired $\mathrm{pH}$ was achieved (a titration, using a pH meter to determine endpoint).
I’d like to work out (roughly—I understand that what I’m about to describe is nothing more than an approximation, but it’s a reproducible approximation at least which is good enough for our needs) directly how much ammonium hydroxide and acetic acid I would need to make up the same solution.
I understand the fundamental Henderson–Hasselbalch equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A}]}{[\ce{HA}]}$$
But as I see $\ce{A}$ is ammonium and $\ce{HA}$ is ammonium acetate. Thus, in my example I don’t really simplify things that much.
Additionally the majority of examples I’ve found look at acidic buffers rather than alkali ones. Does my equation therefore need to be rearranged further to take account of this?

HA is ammonium NH4+ and A is ammonia NH3, with pKa approx. 9.25.

I know the pKa of ammonium, my issue is piecing everything together to work out the acetic acid amount. All of the examples on the internet seem to be for acid buffers

Eg. Many examples of preparing acidic ammonium acetate solutions calculated from a HH approximation

pKa of ammonia is very high, you mean pKa of ammonium.

to prepare a 100 mM ammonium acetate solution at pH 9, you’d first prepare a 100 mM ammonium acetate, and then add ammonium hydroxide dropwise until the desired pH was achieved.

Well, this procedure is good enough for a rough screening solution- call it a quick and dirty method. The main issue is that one would not prepare a solution of 100 mM ammonium acetate beforehand and add drops of ammonia to that. Whenever you need to prepare a buffer solution, you should know what volume is desired and what concentration is needed.
For HPLC, let us say you need 1000 mL of buffer containing 100 mM ammonium acetate. One would dissolve the required amount of ammonium acetate in about 900 mL water in a 1 L beaker, add ammonium hydroxide dropwise till you reach the desired pH, add more water till you are close to 950 mL. Check the pH again, and adjust as desired, and then quantitatively transfer the solution to a 1 L volumetric flask and then make up the volume to 1 L. This will be your analytically correct 100 mM ammonium acetate buffer.
Now, coming to the buffer equation. Again, for that you need to define the concentration and volume. What would ask yourself what should be concentration of ammonia should be present in 1 L of 0.1 M ammonium acetate solution such that the final pH is 9 solution.
You would use the pK$_a$ of ammonium ion and from the equation you would figure out the concentration of ammonia.
See example no. 2 here: Buffer calculation
And also study example 3. Ammonia buffers
Now the problem is that conc. ammonia is sold as w/w % solution. Using the density and percentage, you need to determine the molarity of ammonia solution as well to actually prepare a buffer.

Thanks for your answer. The main purpose of my problem is that we don’t have a reliable way of measuring pH and such so I was trying to estimate how much of A and B I’d need to mix to theoretical have the solution at the right pH. I’ll work it through and see where I get. Thank you

I’m still lost about the ammonium acetate since I don’t want to use the salt but rather just mix HPLC grade ammonium hydroxide and acetic acid - would they just be mixed 1:1 to make the 100mM solution and then work out how much overall ammonia is needed based on your links ?

@user103231, You have an HPLC but not a pH meter?

@user103231 For given pH, you need to fit pH=9.25 + log (( c1.V1 - c2.V2)/(c2.V2)), where c1,V1 are molar concentration and volume of ammonia and c2, V2 the same for acetic acid. Nothing beats pH meter, but if this is good enough for you... At this pH, acetic acid can be considered like if it were a strong acid.

This is not correct for a $\ce{100 mM}$ ammonia buffer.  When you add more ammonia to adjust the pH to the desire value, you increase the molarity of the buffer so that the total amount of $\ce{NH4+/NH3}$ is more than $\pu{100 mol}$. Here the critical buffer ion is the ammonium/ammonia couple, not the acetate which has no role as a buffer in this case.
To prepare an exact ammonia buffer at $\ce{100 mM}$:

pour the quantity (volume) of ammonium necessary for $\pu{100 mM}$ in water (less than the final volume)
adjust to the desired pH with acetic acid
complete with water to full volume.

NB: regardless of the pH, this will be a $\pu{100 mM}$ buffer


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Electrolysis of copper chloride in solution of Vinegar + H2O2 + NaCl after copper etching
I did copper etching (to make a PCB) using a solution of vinegar + H202 + NaCl and now I need to dispose the solution. There is no waste collection for this kind of stuff where I live, so after some research I saw that the cleanest dispose process seams to be to make electrolysis.
I used a copper electrode at the negative pole (cathode), a nail on the positive (anode) and plugged a 9V power supply. I was expecting some bubbling on the nail (Cl2 gas) and the Cu2+ to accumulate on the copper electrode, but the bubbling happened only around the copper wire and after  15 min the solution started to get really dark brown.
My goal is to get the copper out of the solution, so that I can throw the solution away in the drain and the copper in trash.

Is this what is supposed to happen, or did I do something wrong? How do I know when it's done taking the copper from the solution? when done, can I just filtrate and then pour the solution in the drain?

Why the bubbling happened at the cathode and not the anode and why the solution turned brown?

Update
the result after some 2-3 hours. how to know if it is done?

My goal is to get the copper out of the solution || Make a solution of baking soda, mix it with your remains, leave it for a few days, decant clean water, filter off the precipitate, dry it. The result is basic copper carbonate which is mostly harmless, and the solution would be sodium chloride with minor impurities, reasonably safe to go down the drain || You CAN try to reduce it to get metallic copper out, but it is quite slow and isn't easy to do right (I tried).

thanks for reply. but what is mostly harmless? can I just throw it in the trash or is there a "cleaner" option? and how to know how much baking soda is necessary? only by the color?

You should not eat it. Otherwise, it's safe. In my country I would wash it down the drain. If I wanted to be particularly responsible, I would dry and heat the solids until it becomes completely black before washing it down the drain. As for amount of backing soda, you need need one molecule of soda per one $HCl$ molecule in the the etching solution and then a slight excess. Assuming ~35% $HCl$, this should be around 90-100 g of solid baking soda per 100 ml of $HCl$ solution used for the etching solution.

looking at density table Actually, to be safe, I would go for about 110 g. I didn't check the solution density.

A simple test to know whether a solution does contain copper ions is to filtrate it and then to add some milliliters of ammonia solution. If the solution contains copper ions, it will take a deep blue color, and no precipitate. If not it will make a brown precipitate, the solution contains Fe ions.

Mix with sodium hydroxide drain opener (outside!) The metal ions (copper and iron now) will combine with the OH- ions and drop out of solution as harmless iron and copper hydroxide and can be thrown out. The Na+ ions will combine with the Cl- ions and form aqueous NaCl and can safely be put down the drain.
Even if you put too much sodium hydroxide in your solution, it’s still safe to put down the drain because it’s less caustic than it started as when it was drain opener.
If you’re still uneasy about disposing of the copper and iron hydroxide, burn it and the hydroxides will turn into metal oxides.

interesting. should the copper and iron hydroxides formed be filtered out of the solution? or can I throw them together down the drain?

I wouldn't put the hydroxides down the drain - there's no need to add additional metals to your wastewater for your local water treatment facility to deal with. Also, the inside of most people's plumbing isn't as smooth as people expect it to be. Put solids in the drain and it's going to cause you problems when it will be most inconvenient (i.e. when you cannot get in touch with a plumber!)

The best thing to do is to put some nails in the copper solution, or anything made of iron. You will soon see that the nails ge. covered by metallic copper, which is red-brown, After maybe one hour, the solution is not blue any more. So it does not contain copper ions anymore. You can throw this solution down the drain : it will contain iron instead of copper, and it looks like rust. And you can throw away the nails covered with copper. The equation of the chemical reaction is :  $$\ce{Fe  +  Cu^{2+}->  Fe^{2+} + Cu(s)}$$ Quite often, these $\ce{Fe^{2+}}$ions are then oxidized in the air to produce a brown deposit of $\ce{Fe(OH)3}$ according to $$\ce{4 Fe^{2+} + 3 O2 + 6 H2O ->  4 Fe(OH)3}$$ It is a typical chemistry demonstration that probably all chemistry teachers in the world present in their high school classes. It never fails.

Even better could be a galvanic cell , formed by connected iron and copper plates. This would have advantage iron would not get covered by copper.

@Poutnik. You are right. But my method is quickly done, and does not require any equipment.

Iron +3 is very good in masking copper coloration. BTW, copper-chloride solutions are often green.


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How do I understand exactly how hazardous a material is based on SDS?
I work with a lab occasionally and get asked often if a chemical is safe to breathe, how hazardous is it if it lands on skin, etc. For the most part, I'll try to go to an SDS and look it up. The pictograms and diamond are what I search for, so I try to give a general answer based on that. I'm never entirely confident in what I suggest, though.
We had a lab supervisor who could tell us something like, "You don't want that on your skin for a long time, but if it falls on you, don't worry about it too much, just wash your hands thoroughly when you can." I'd like to be able to give answers like that without going over the top.
Do you have any suggestions, or could you point me to some other sources that go more in depth?

If you find SDS too excessive or vague, refer to R/S statements or H/P statements. It doesn't get any more general or strict than that. By the way, you can find more or less complete lists for different languages in the mhchem manual.

Your instructor is right, for the huge majority of the chemicals. But not for all. There exist a couple of chemicals where you must react immediately if a drop touches your skin. I am thinking of hydrogen peroxide for example :  a tiny droplet may slowly but deeply destroy your skin, up to the blood.

@andselisk +1 You should edit/upgrade that comment to an answer. Without getting a chemistry degree, having years of lab experience, and reading constantly to keep up to date, using something like the GHS to guide you is about the best most people can do.

This is what Industrial Hygiene folks are used for. Use them, and if the chemistry lab doesn't have access to one I would not work there.


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Why is the pairing energy not considered when calculating the crystal field splitting energy?
Consider the low-spin octahedral crystal field splitting diagram for a d4 system.

When we calculate the crystal field splitting energy we simply observe the absorbed color of light and calculate the frequency of a photon required to promote an electron from the t2g to the eg orbitals. Why is the spin pairing energy not considered (PE) in these calculations? If we had a simple d3 system, I would understand why the energy of the photon absorbed is equal to the crystal field splitting energy since we are not dealing with any paired electrons; however, in the case of the d4 system, doesn't "removing" one of the paired electrons actually release energy? Why do we say the energy of the photon absorbed is equal to Δo and not Δo - PE?

It is actually the two single-paired electrons are more likely to be promoted than the two paired electrons.
Promoting the electrons always cost energy whether the electrons in question are paired or not. In the classical examples of hybridization (where they might actually promote a paired electron), this energy cost is compensated by the formation of stable covalent bonds, but in this case, a complex, no bonds formed here.
So why the single electrons are promoted before the paired ones?
The spin rule in the coordination complexes tell that the allowed d-d transitions must not involve the change in the total spin of the electrons.
In your case of a low-spin d4, promoting the the paired electrons will leave the d-orbitals with single electrons with opposite spin, which is highly unfavorable. The changing of the spin of the opposite-spin electron is forbidden per the spin rule since it will change the total spin of the system.


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Does water have properties suggesting it has a phase transition at 50-63 Celsius so it has two phases?
https://www.youtube.com/watch?v=YsQXN-Kh2-w&t=497s
In this video, he says that water has four properties (surface tension, refractive index... ) that transition from 50 - 63 Celsius between totally separate values? Eminently testable but I don't have the facilities. I tried searching but there some statements of liquid water having two phases but related to supercooled water. What experimental evidence is there that  water has 2 phases and a transition occurs at 50C?

Sounds like nonsense.

@Ivan Neretin, The video is non-sense, but the report in Science on the two states of supercooled water is not. The video is apparently made on this paper. The title of the paper "Experimental observation of the liquid-liquid transition in bulk supercooled water under pressure." There are no two phases like hexane and water mixtures, but some transitions were observed at nano to microsecond timescales.


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Volume of sample for quantitative NMR
I’m interested in quantitating lithium via NMR using a Magritek Spinsolve 60MHz bench top NMR spectrometer. Is there any evidence that suggests the volume of the liquid sample used in the NMR tube has any effect on the area of the spectra peaks? I’m interested in hearing opinions on the matter before running tests verifying if such correlations in-fact do exist.
I’ve seen in relevant literature that 0.6-1mL is generally a sufficient amount of solution for the 5mm tubes. However it’s not clear to me yet if this volume should be consistent between the calibrants and experimental solutions for analytical NMR. Thanks!

Is this spectrometer designed to detect lithium? As per their brochure, the standard configuration is for proton and fluorine nuclei. If this is not an pedantic academic exercise or an educational experiment, I would suggest to use a simple flame photometer to quantitate lithium for practical purposes.

Yes it is designed to detect lithium @M.Farooq

If you have too little volume there will be shimming problems. The actual volume is different for simple or Shigemi tubes, see e.g. http://sopnmr.ucsd.edu/sample-preparation.htm

No, sample volume will not effect quantitation if the following conditions are met: 1) you are comparing samples in tubes with the same inside diameter, 2) you have enough sample volume so that the sample liquid extends at least a few diameters past the ends of the RF coil, 3) your shim is good enough so that the peaks you want to measure do not significantly overlap other peaks. Field distortion due to susceptibility discontinuity at the ends of the sample liquid can degrade shim, and people often use samples that extend far beyond the RF coil for that reason.  However, provided the above conditions are met, shim does not change peak integrals. This is because the integral of the line shape is exactly equal to the initial amplitude that the line contributes to the FID, and is independent of how it decays in time.


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What is the chemical reaction behind ginger milk curd?
Pouring hot milk into ginger juice solidifies the milk. All I know is that ginger juice contains Zingibain that could break down milk proteins into smaller polypeptides. I'm confused as though why is this sufficient to solidify the milk?

Here's a whole pdf on the exact same thing

also here's more information from cooking stack exchange

@Desai Sorry the pdf link isn't working for me. The cooking stackexchange only says that it will curdle, but I'd like to know what's happening at the molecular level?

Ginger Milk Google Drive Link


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In the Gold Foil experiment, did Ernest Rutherford's alpha particles hit the same spot every time or did he direct the emitter to different spots?
When reading and learning about the gold foil experiment, the alpha particles mostly go straight through the gold foil and some get deflected or bounce back but it doesn't say anything about the direction of the alpha particle emitter. Did the alpha particle emitter stay in the same spot or get directed to different areas of the foil? How could they all hit the same spot and get deflected in so many different directions?

The source had area, and the foil had area. The alpha particles were collimated. This is just how scattering (and reaction cross sections) work.

The original set-up had the alpha source some distance away so the particles must be were randomly hitting the gold foil not exactly at the same atom, if that is in the mind of OP. The alpha particles was "collimated" by hole in a lead chamber. Jon Custer can better explain about the divergence of the alpha particles. Do they diverge from the hole or just travel in straight lines?

In this article on microbeams  (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6092239/)  it clearly states that collimaters are used to focus not disperse as quoted " The most important aspect for a microbeam facility is the capability to confine the beam to a micrometer or even sub-micrometer diameter size. The two most popular approaches are collimation and magnetic focusing.". and "Collimators of various materials and thicknesses were tested to minimize the scattering and to achieve a better accuracy (Folkard et al. 1997)"   Do alpha particles then differ from photons in this way?

Rutherford performed his experiments 100 years ago, at that time these facilities did not exist. You should see his original paper. There is nothing in common in photons and alpha particles. Collimations means to make beams parallel not focus anything.

Even a small beam is very big compared to a gold atom
It seems you are making an assumption that, if the beam were perfectly focussed it would always show the same deflection (because geometry).
But whatever level of focus Rutherford could achieve with alpha particles, the scale is all wrong for this to be true. Even if the beam was only micrometers wide (very tight for the early technology of the day) it would be more than 10,000 times wider than a single gold atom. And, since the nucleus is what actually scattered the alpha particles, there is another factor of 10,000 or so in scale to consider since the nucleus is very much smaller than the atom.
So the question makes a false assumption. The focus of the macroscopic beam is irrelevant and can never, given reasonable technology, come close to focussing on  single nucleus. The beam area in Rutherford's experiment will cover vast numbers of atoms and will hit them at random.

They were hitting the same spot in the sense they landed on an area that was predicted by the spread of the beam. They weren't all traveling exactly the same path. This randomness allowed some of them to bounce from atom nuclei.


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Is this Ferrous Chloride powder?
I recently ordered a 500g sample of Ferrous Chloride powder. However the powder is in light brown color, and on dissolving it in water, it gives a similar brown hue and it is not a clear solution. From what I saw on various preparation methods on YouTube, all the solutions are green in color. Is there something wrong with my sample? Could this be due to Iron Oxide formation? Kindly help.

Does it do the same if dissolved in water acidified by HCl ? Is possible it is ferric chloride(III) by packing mistake, or got oxidized ? Interesting result could be provided by analysis of Fe(III) content.

That powder could be Fe II but the solution not. The latter seems to scatter a lot....


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Ambiguity in naming haloalkanes
My question in quite simple why a compound such as chloro-ethane is called as such why is if not chloro-ethyl??
Since it is a hydrogen short.

Holoalkanes...?

Chloroethane is ethane with 1 hydrogen replaced by chlorine. So it is substituted ethane.
$$\ce{CH3-CH2Cl}$$
1-Chloroethyl or 2-Chloroethyl  is ethyl with 1 hydrogen replaced by chlorine. So it is substituted ethyl radical, standalone or bound to a larger molecular structure.
$$\ce{CH3-CHCl.}$$
or
$$\ce{CH2Cl-CH2.}$$


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Reaction/complexation of aluminium chloride and acetic acid
Thinking about Friedel-Crafts acylation using carboxylic acid anhydride got me wondering about how exactly aluminium chloride ends up in that reaction.
Generally, despite the fact that Friedel-Crafts alkylation can be carried out using catalytic amounts of aluminium chloride - when chloride of a carboxylic acid is used as an acylating agent in Friedel-Crafts acylation at least 1 equivalent of aluminium chloride has to be used. This is due to the fact that aluminium chloride is complexed by resulting ketone (please see figure below).

Formation of such complex takes aluminium chloride out of action: 1 mole of aluminium chloride gets deactivated for every 1 mole of ketone produced, so it can not activate another molecule of acid chloride. And this is the reason why Friedel-Crafts acylation needs stochiometric amounts of aluminium chloride. (This is often omitted when Friedel-Crafts acylation is being introduced in organic chemistry textbooks, but it is sometimes mentioned later on - especially when it comes to actual lab work and real life conducting Friedel-Crafts acylation.) In the end formation of a ketone-aluminium chloride complex is not a big problem because when chemist who was running the reaction decides to isolate the product, the ketone can be liberated by hydrolysis of that complex using aqueous acid (note that water is unwanted in Friedel-Crafts reaction).

In instance of using carboxylic acid anhydride at least 2 equivalents of aluminium chloride are needed. One surely ends up bounded to the ketone - similarly as described above, but I am not sure what exactly happens with the second equivalent. My guess is that it reacts somehow with carboxylic acid that was produced from anhydride while the other part of anhydride got bounded to aromatic unit.
Hence my questions:
(1) What happens with that second equivalent of aluminium chloride?
(2) How does aluminium chloride react with acetic acid in nonaqueous medium? (My guess would be that this is something like in the figure below. However, I don't know if reaction goes all the way to chelate or not.)

Does this answer your question? Mechanism of Friedel-Crafts acylation with succinic anhydride

@cngzz1 - thank you for the comment but it doesn't answer my question. What's under link provided makes it look like if AlCl3 was restored - at the end of therein presented mechanism H+ protonates carboxylate bonded to Al and AlCl3 is released. If that was the end then only catalytic amount of AlCl3 would be needed (as in F-C alkylation). Synthetic procedures for F-C acylation require stoichiometric amounts of AlCl3. At least 1 eq. for acid chlorides and 2 eq. for anhydrides. What exactly happens with 2nd eq. of AlCl3 is still unclear. Hence, I think this question needs to be reopened.

For your first question: After aqueous work up, AlCl3 is transformed into aluminium hydroxide, which is insoluble, so can be filtered. This might help: https://www.organic-chemistry.org/namedreactions/friedel-crafts-acylation.shtm

@cngzz1 - Actually, from the very beginning my question is about what happens with 2nd eq. of AlCl3 in F-C acylation using carboxylic acid anhydride in reaction mixture (before work up). In other words: how exactly this 2nd eq. of AlCl3 is blocked from activating acylating agent (anhydride) while knowing that 1st eq. of AlCl3 is probably bounded by ketone carbonyl group - as in case of using acid chlorides. (This is why I put this suggestion at the end of my question that maybe AlCl3 reacts with carboxylic acid produced from anhydride in acylation process, but I am not sure if thats the case.)


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What is the best way to clean Lovibond tintometer quartz cuvette used for oil?
I work in a laboratory where we regularly check edible oil color by using a Lovibond colorimeter. Some people suggested me to wash the quartz cuvette using soap solution.
But I am thinking about washing the quartz cuvette by using acetone solution and then leave it to dry in air. Is this method correct?

The best way is to refer to the manual, or ask a service engineer.

Any solvent that is both reasonably volatile and in which your oils are highly soluble will work. If your oils dissolve readily in acetone, go ahead and use it. But if they don't dissolve easily in it, the residue will just stay on the cuvette during the acetone wash


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Is azonine planar?
I came across aza[9]-annulene, which is given to be aromatic. I can clearly see it follows Hückel's rule with 8 π-electrons in the ring and the lone pair of nitrogen.

But the thing that concerns me is the planarity of structure. I'm not convinced that this structure is planar with so much angle strain (and thus aromatic).
Can anyone please explain this to me?

There is a relevant article on Wikipedia regarding the molecule, azonine.

@NicolauSakerNeto Thanks for the link. That surely answers my question to some extent. :)

related: https://chemistry.stackexchange.com/q/51484/102629

Please check the following link    https://en.m.wikipedia.org/wiki/Azonine .  It clearly states , this molecule is planar.

There is no much of strain, even. It could be higher for closing a non-plantar ring!

@Alchimista Sir, So why do we consider angle strain (~24°) in [10]-annulene, which is merely 4° greater than in the given compound?

@Nex likely the non planar, 10 annulene counterpart is less strained. As far as a molecule do exist, there is not a clear border between stability or not. Also because strain is not the only parameter that affects the overall energy. Rather than compare different molecules you should compare the possible structures of the same molecule. If you would have molecular models at hand, you will certainly have to apply a force to close the azonine cycle while using localized double bonds, too.

@Nex above, where I say "there is no clear border between...." I mean you cannot set a fixed value for angle strain as a discriminant, and the same is true for every strain or factors contributing to the overall energy. When everything is balanced and results are compared, then you can conclude what is stable and what not. Still, nothing is better than experiments: most arguing is done at posteriori.

@Alchimista Got it. And yes, I completely agree with you on the posteriori part :P. Thanks for the input.

Comments cite Wikipedia as stating the molecule is planar. Here that is backed up with a technical reference from Somers et al 1.
Ref. 1 gives a drawing of the molecular structure, reproduced below from that source, showing bond angles in the ring whose average is $140°$ with good accuracy; this value is consistent with a planar enneagon.

The authors point out that this planar structure is easily distorted, but without necessarily losing aromaticity:

[I]nteraction with surrounding H2O molecules and alkali ions and substitution of the N−H hydrogen distorts the planarity of the ring. This distortion is such that the aromaticity remains. N-Methylazonine is characterized as nonplanar and the global minimum structures of the alkali salts have the metal residing on top of the distorted ring (cation−π interaction).

One reason for the methylated compound going away from planarity has to do with the $140°$ ring angles in the planar structure, which bring neighboring substituents close to each other. Substituents larger than hydrogen atoms are more prone to distort in such a situation. Hydrogen bonding of the protic hydrogen with water may also be regarded as a larger substituent.
Reference
K. R. F. Somers, E. S. Kryachko, and A. Ceulemans (2004). "Azonine, a 'Nearly' Forgotten Aromatic Molecule". J. Phys. Chem. A 108, 18, 4059–4068.
https://doi.org/10.1021/jp037046+


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Linking an electrochemical cell to an electrolytic cell
Electrolysis occurs when a non-spontaneous redox reaction occurs when electrical energy is put into it. An electrochemical cell functions by using two redox reactions to facilitate movement of electrons. So for an alkaline battery, the zinc anode will oxidize and release electrons, which will flow through the circuit to reach the manganese ($\ce{MnO2}$) cathode, which will be reduced, allowing more electrons to flow through the circuit without buildup of electric charge. I assume here that both reactions are spontaneous, and the electron buildup in the anode prevents further zinc oxidation when the circuit is incomplete.
$$\ce{Zn(s) + 2OH- (aq) \rightarrow ZnO(s) + H2O(l) +2e-} [E^\theta= 1.28V]$$
$$\ce{2MnO2(s) + H2O(l) + 2e- \rightarrow Mn2O3(s) +2OH- (aq)}\ [E^\theta= 0.15V]$$
If the alkaline battery is linked to an electrolytic cell, won't it just be like two electrochemical cells working in conjunction to each other? The zinc anode will be linked to the cathode of the electrolytic cell, and then the ensuing redox reaction will (probably) be spontaneous. So the "electrical energy" put into the electrolytic cell will be instead chemical energy?
Also, seeing that the $\ce{MnO2}$ cathode has a relatively small $E^\theta$ value, won't it be hard to trigger oxidation of the electrolytic anode? When connected to the electrolytic cell, do the cathode and anode of the electrochemical cell now work totally independently? If that is the case, the voltage of the battery should not be considered, but instead the electrode potentials of the anode and cathode, but that is usually not the case. What am I misunderstanding about this?
edit: I'm asking about the mechanism when an electrochemical cell is linked to an electrolytic cell, not about the reactions taking place in the electrochemical cell itself, so the question suggested doesn't quite fit the bill.
edit 2: Added a picture to better show my doubts. If the galvanic cell is linked to an electrolytic one as shown here, the galvanic anode and the electrolytic cathode are linked and seemingly acts as a separate cell. The same thing can be said about the other two electrodes. If this is the case, doesn't the galvanic anode and cathode each act independently of one another, and therefore the electrode potential of the individual half cell must be considered? If they are actually linked, how?

related: Reaction in Zinc-Carbon-Ammonium Chloride battery

It is not clear to me just what you mean by linking an electrochemical cell to an electrolytic cell. For example, you can use a 9 V battery to electrolyze a salt water solution. In fact, aspects of this are a common query here. Perhaps you could edit down to what you really want to ask about and a picture is worth a thousand words, as they say.

You seem to be mixing a couple of things. One important clarification regarding

Also, seeing that the MnO2 cathode has a relatively small E$^o$ value, won't it be hard to trigger oxidation of the electrolytic anode? When connected to the electrolytic cell, do the cathode and anode of the electrochemical cell now work totally independently?

The listed electrode potentials are valid only in relatively dilute solutions under very specific conditions. Don't apply the half cells on a dry cell or an alkaline cell. In short, those numbers are not meaningful for an alkaline battery or a carbon-zinc dry cell. Those commercial cell compositions is just a magical mixture, which is more art than science. Battery technology is not that trivial. Think why there are not many types of batteries despite 300 years of research?
If you consider a galvanic cell such as your alkaline battery as a simple water pump, then a lot of your confusions will go away. Its only job is to "circulate" the electrons. When you connect a galvanic cell to an electrolytic cell, the anode and cathodes can no longer be considered independent. If the galvanic cell pumps out one electron, the electron has to travel the cathode, anode and the external circuit of the electrolytic cell. In that process, something in the electrolytic cell has to chemically decompose!

So what I understand about this is that alkaline cells cannot be thought of as just a simple anode and cathode, and are more than the sum of its parts. Also, kinda dumb question; for linking a galvanic cell to a electrolytic one, you mentioned that the electron travels from the cathode to the anode. If you can detail this process a bit more, I would be very grateful.

Check this answer https://chemistry.stackexchange.com/questions/144334/how-do-you-connect-a-battery-to-a-galvanic-cell-in-order-to-reverse-the-flow-of/144341#144341

Electron current always travels from anode to cathode whether it is a galvanic cell or electrolytic cell. There is no dumb question.

I have another doubt, sorry to bother. I have pinned a picture onto the post, which supposedly shows a galvanic cell linked to an electrolytic cell. The way this is set up is as if the anode of the galvanic cell and the cathode of the electrolytic cell act as a separate cell altogether, same goes with the other two electrodes. If what you said is true and the electrons travel a full circuit (galvanic anode to galvanic cathode), how are the cells linked?

For an electrolytic cell, a galvanic cell is nothing else but a power source.
For a galvanic cell, an electrolytic cell is nothing else but a powered electronic circuit.
The galvanic cell anode/cathode does not form with the electrolytic cell cathode/anode a respective separate cell, as there is no ionic flow between them. Similarly for a simpler case, 2 half cells do not form a cell if there is no ionic connection between them.
The reason for it is simple, based on the electrostatic neutrality requirement and enormous forces between unbalanced charges. Imagine you managed to push the current 1 A for 20 s through the wire, without being balanced by ionic motion. The comparable unbalanced charge causes in atmosphere a lightning several kilometres long.

I think I'm catching your drift now! So for the image above, since there isn't a salt bridge or something of the sort, for each electron used to reduce a cation in the electrolytic cell, an electron is produced by oxidation of an anion, as to maintain neutrality of charge. Am I getting this right?

Therefore, the galvanic cell as a whole is used to drive electrons into the electrolytic anode and out the electrolytic cathode. Makes much more sense if that's the case!

Like that  You can consider the galvanic cell as a water pump and the electrolytic cell as the piping system. But there is circulation of charge,instead of water.

If there were charge disbalance, it would very quickly build an electrostatic potential acting against the charge flow leading to such a disbalance, and the current would cease.


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Why is the atomic mass unit less than the mass of both a neutron and a proton?
The atomic mass unit is $1.6605 \times 10^{-27}$ kg.
This is less than the mean of the masses of 6 protons and 6 neutrons.
How do we account for the lower mass ?
My understanding is that some of the mass is in the form of energy somewhere, in the bonding maybe or in the kinetic energy.
Can someone clarify where this mass is please ?

Binding a proton to a neutron releases energy.  I've read that some gentleman named Albert Einstein figured out that this energy carries off the missing mass, $m=E/c^2$ is how he originally rendered it I think.

Does this answer your question? Units of mass on the atomic scale

related: https://chemistry.stackexchange.com/q/70811/102629

The source of the discrepancy is the mass defect, which is described in the article below as :
The difference between the sum of the masses of the components and the measured atomic mass is called the mass defect of the nucleus. Just as a molecule is more stable than its isolated atoms, a nucleus is more stable (lower in energy) than its isolated components. Consequently, when isolated nucleons assemble into a stable nucleus, energy is released. According to Equation 4, this release of energy must be accompanied by a decrease in the mass of the nucleus.
The equation four referenced is just E=mc^2.  When these nucleons form carbon-12, which is the reference isotope for the definition of the amu, the mass defect appears, as you mentioned, by conversion to binding energy.

https://chem.libretexts.org/Courses/Grand_Rapids_Community_College/CHM_120_-_Survey_of_General_Chemistry/2%3A_Atomic_Structure/2.07_Mass_Defect_-_The_Source_of_Nuclear_Energy


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What is the composition, function, and classification of “6-Thio-2-Deoxyguanosine” and “13-mer thio-phosphoramidate”? How do I learn more about them?
I am assigned the task of explaining a biomedical research paper, it is about telomere and telomerase , and it talks a lot about the molecule Thio-2-Deoxyguanosine , 13-mer thio-phosphoramidate and " 6-thio-2-deoxyguanosine 5′-triphosphate".
I have a good understanding of what the paper is saying , but I would really like to know as much as I can about those molecules, their chemical classification and biological function and chemical composition, any ideas?
Thanks in advance
(Please do not mark this as off-topic , as I asked this in biochemistry stack exchange, and they wanted me to put it here )

Google is your friend! 6-thio-2'-deoxyguanosine for example is commercially available  https://www.sigmaaldrich.com/catalog/product/sigma/sml1296?lang=en&region=GB

"How do I learn more about them ?". Go to Chemspider, Pubchem, ChEMBL, and literally all the databases which can be found under the "Identifiers" section of Wikipedia for that specific molecule. The 3 databases mentioned above give a summary of molecules, their physical and chemical attributes as well as a link to further readings/papers. You can also find more information about your molecules in the references of your first paper. http://www.chemspider.com/     https://pubchem.ncbi.nlm.nih.gov/     https://www.ebi.ac.uk/chembl/    good luck!

we all know those sites, zero results in all databases


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Intermolecular forces of attraction between positional isomers of alcohols
This seemingly trivial question is as follows: Why is the boiling point of 1-butanol ($\ce {117.7 ^\circ C}$) higher than that of 2-butanol ($\ce {99 ^\circ C}$)?
The only reason I can think of is that the former molecule has a "straighter" chain than the other and a "straighter" chain has a greater surface area allowing for stronger van der Waals forces to exist between the molecules. As molecules in the liquid state do not pack themselves close to each other, ideas of "packing efficiency" would not be useful in explaining these boiling point differences. Perhaps, it is instructive for us to first observe if the trend is persistent for longer chains?
\begin{array}{|c|c|c|c|} \hline
\text {No. of C} & \text{1-alcohol} & \text{2-alcohol} & \text{3-alcohol}\\ \hline
\ce{5} &  138 & 119.3 & 116\\ \hline
\ce{6} & 157 & 140 & 135\\ \hline
\ce{7} & 175 & 159 & 156\\ \hline
\ce{8} & 195 & 178.5 & 173\\ \hline
\ce{9} & 214 & 194 & -\\ \hline
\end{array}
As can be seen from the above table, where the boiling points are all sourced from Wikipedia, with units of $\ce { _^\circ C}$, the trend is indeed persistent. The alcohol isomer with the $\ce {-OH}$ group attached to the 2-position consistently exhibits lower boiling points of around $\ce {16}$ to $\ce {20}$ $\ce {^\circ C}$. Shifting the hydroxyl group further along the chain by one more position also lowers the boiling point of the molecule but to a lesser extent.
The explanation I can come up with for this is the same as aforementioned: By placing the hydroxyl group at the end, there is the maximum surface area for interaction of the alcohol molecules with each other. Furthermore, the hydroxyl groups, being placed at the ends, would cause the least amount of disruption to the hydrophobic interactions between the alkyl chains. Does anyone have better explanations for the above trends?

A detailed answer might even turn out somehow opposite as far molecular region are considered, but ending at the same and actual fact... You neglect that a source of interaction is H bond. Consider that and go through it again. Using the same concepts I am pretty sure you can find a rationale. Don't worry that liquid do not pack. They certainly do it with the nearest neighbours.

@Alchimista I am well aware that alcohols can undergo hydrogen bonding. Are you suggesting that by having the hydroxyl group at the terminal carbon, hydrogen bonding between the molecules is facilitated?

basically you have to look as you did in the question but bonding molecule first. This would make the molecules differently arranged. Then you look if the hydrocarbon chains can intercalate and how, etc. You shift the analysis done before for individual molecules to an H bound ensemble.

If liquids didn't pack, there would be a tremendous jump in density at the mp. Some substances even expand on crystallisation!

@Karl Hmmm... So what are you trying to imply by that comment? You mean to say that packing of the different alcohol isomers can possibly account for differences in the boiling points?

Recommend you also compare the density of those alcohols.

@Karl The difference in density appears be rather marginal, strangely. 1-butanol has density of $\ce {810 kg m^{-3}}$ while 2-butanol has a density of $\ce {806 kg m^{-3}}$

It does seem that placing the hydroxyl group at the terminal carbon maximises the hydrogen bonding interactions between alcohol molecules and also the van der Waal forces between them. This is best illustrated with a crude diagram I have made:

Finer differences due to the position of the hydroxyl group along the chain, such as the differences between the 2- isomer and the 3- isomer, are not explained here. My explanation does explain the markedly higher boiling point of the 1- isomer.

Upvoted because it is most likely a very pictorial and rough description but it gives the sense of the situation and of the type of reasoning involved. It is what I have suggested to OP - who, by the way, was already on the right track.


text:
Columnar crystallization of ice under small pieces of eroded granite
Walking through mountains in northern Portugal I noticed a curious formation of ice cristals whose explanation I was unable to give to my grand children (Photo 1). I searched in the internet but I couldn't find it.
On the soil, in the areas covered with raw sand, resultant of erosion of granit, the grains of sand were raised up by long ice cristals (photo 2). With the weihgt of the sand grains the "prismatic" cristals of ice curved (photo 3)
Why cristals elongate this way, raising the grains of sand? Why they do it as separate columns of ice?
I would be very glad to obtain an explanation
Thank you very much
PS I couldn´t send the images

Too bad there are no photos.

The photos are missing but I suspect it is related to this https://physics.stackexchange.com/questions/70974/how-do-ice-spikes-form Look also at the comment of mine at the very bottom of the thread. It should be the same with just sand on top.


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Determination of Cell constant
I read that cell constant is measured indirectly using a solution with known conductivity(like KCl).
But then how was the conductivity of that solution measured? (Is it by some theoretical consideration or another accurate experiment?)

The gist is that  using a detailed procedure a chemist should be able to make the standard solution very reproducibly.

That is not is question. His is more like a chicken and egg query. Which came first?

The conductivity or the cell constant?

Conductivity measurements were done in late 1800s by Friedrich Kohlrausch, who painstakingly measured the conductivities of solutions so accurately that they are still valid today. Most of his work is in German was published in relatively unknown journals. In his time, conductivity theory was not advanced enough to predict the conductivity of simple salt solutions in water, so there was no a priori theoretical prediction of conductivity, from the first principles. I don't think it is doable today after 100 years (may well be wrong(.
However, the accurate measurement of resistance was well established in physics. See below.
A paper in the J. Am. Chem. Soc. 1933, 55, 5, 1780 describes what Kohlrausch and his meticulous colleagues they did. Note the level of details which is remarkably missing from papers today.

After Kohlrausch had developed his method of measuring the electrical
conductance of solutions by means of an alternating current bridge, he
undertook with the aid of Holborn and Diesselhorst’ the determination
of the specific conductance of seven standard reference solutions in
absolute units in order that the results might be available to himself
and to other experimenters for the calibration of conductance cells.
For this purpose they used two cells whose cross sectional areas, a,
and length, l, were determined mechanically. The cell constant,
l/a, could then be computed in absolute units (cm$^{-1}$) from the measured dimensions.
The cell constants of a considerable number of other cells with
sealed-in electrodes of the usual type were determined electrically by
comparison with the primary cells by measurements of resistance when
filled with a common solution. These cells, whose cell constants were
thus determined indirectly, were then used for the determination of
the specific conductance of the seven standard reference solutions at
approximately 0, 9, 18, 27 and 36 C. The specific conductance of each
of these solutions for every degree between 0 and 36 C was obtained by
interpolation. No actual measurements were made at either 20 or at
25’. The water-bath used to fix the temperature of the cells was not
provided with an automatic thermostatic control. The majority of the
data on conductance in the literature are based on the values ascribed
to these standard reference solutions by Kohlrausch, especially the 1
N and 0.1 N potassium chloride solutions.
In the original paper by Kohlrausch, Holborn and Diesselhorst it is
stated that the normal potassium chloride solution was prepared by
dissolving 74.555 g. of potassium chloride weighed in air and making
up to 1 liter at 18 C. The 0.1, 0.02, and 0.01 normal solutions were
prepared by dilution by volume. The book “Das Leitvermogen der
Elektrolyte” by Kohlrausch and Holborn, which was published in the
same year as the paper referred to, contains alternative directions
for the preparation of these solutions by weight. This book has
probably been used by other experimenters more than the original paper
because it is more generally available.


text:
What is change in internal energy of a system in which combustion occurs at constant temperature?
We got a question in a test, in which we were asked which system has zero change in internal energy and it had an option which was combustion of methane at constant temperature. I imagined this to be a situation in which the combustion is carried out in a system, where the change in internal energy of system due to heat released, is cancelled by the work done by the system. The reason why I thought of this is because of the formula,
∆U=nCv∆T,
But this was wrong. When I asked my sir how this was wrong, he was not able to say why the ans using above stated formula is wrong. He tried giving me a logical explanation in which he said that a small part of the system releases heat which is used by the remaining part for it's combustion, thus the total temperature remains constant. but I am unable to understand why we choose to only study a part of the system while we have the entire system. It makes no sense to me. Could someone please explain how my sir's explanation is valid here and why the formula is wrong to use here?

The change in internal energy of a reaction mixture is a function not only of the temperature change but also of the changes in the amounts of reactants and products in the mixture.  This is because energy is consumed and released when we break- and make chemical bonds.  The equation you wrote for the internal energy change applies only to a pure species or to a mixture of constant chemical composition.  The occurrence of chemical reactions causes a change in the internal energy even at constant temperature.

So you mean, the equation I wrote isn't suitable to be used for chemical reactions? One more thing, so can I say that when a chemical reaction occurs, even at constant temperature, it always has some enthalpy of reaction associated to it, so whenever there is change in enthalpy, there is change in internal energy using the relation, ∆H=∆U+P∆V?

Yes and yes to your questions.


text:
Write the detailed, stepwise mechanism of this reaction:

I couldn’t figure out how to turn the five membered ring into 6 membered ring. Pls help

Where is your attempt?

Protonate -OH, after which water leaves, generating a carbocation, which rearranges, and then an elimination occurs. You should have a look at E1 mechanism.

I would like to see a solution to this, it looks less than straightforward.

Looks like an acid treatment of tetrahydrofurfuryl alcohol leads to 2,3-dihydropyran. That's all good. But I'm not sure how the inclusion of a methyl in the reactant molecule as above leads to the resulting molecule above instead of 6-Methyl-3,4-dihydro-2H-pyran. I'd like to see the solution as well indeed.

Despite the OP having made no effort to answer this, I was intrigued and wanted to see a solution.
The secondary alcohol is protonated by the strong acid, water leaves to generate a secondary cation. With assistance from the ring oxygen, a methylene shift occurs to generate a more stable cation centred on the oxygen. Solvent removes the methine proton to generate the observed dihydropyran product. This only works because of the presence of the oxygen.

Please explain how the 'oxygen-centred' cation is more stable?
I have very basic knowlege regarding this and I've learnt that positive charge on less electronegative atom would be much more stable than on a more electronegative one.
Here, oxygen should be the least stable place the positive charge could be put on, right?

Oxygen has electron pairs that it can donate, enough to make this process viable


text:
Is the average kinetic energy of evaporating water molecules (at room temperature) equivalent to the average kinetic energy of boiling water?
Purpose:
On new year's eve, after a splendid red and an assortment of sumptuous repasts, I made a bold remark which, on further consideration, may turn out to be incorrect.  Unless!  Unless I can concoct an impressive scientific explanation - for which I will need your input
Context:

Temperature is the average kinetic energy of a "bulk" liquid
Particles within the liquid will have a range of kinetic energies
according to kinetic molecular theory (Maxwell-Boltzmann curve etc.)
Water boils at 100 deg Celsius (101.3kpa etc)
Evaporation can happen at any temperature (>273K)
Individual particles don't have temperature but rather kinetic energy

Question:
Assuming we could measure the kinetic energy of multiple evaporating water molecules just as they left the surface of liquid water (at standard temp and pressure) over time, would the average kinetic energy of the sample of evaporating water molecules equal the average kinetic energy of boiling water (at 101.3kpa)?   That is, if the evaporating water molecules had a temperature - would it be approximately one hundred degrees.  Or, as I may have put it at the time, at a molecular level, do my drying undies effectively boil?  (Note: I understand that the bulk water is at room temperature)
Thoughts:
If I knew the speed of evaporating water molecules I could calculate their energy and compare this to the average energy of boiling water and see if they were similar - but I can't see how to estimate the speed of evaporating water molecules?
Also:
...I have read "Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?" on this site.  I don't think it answers this question.
Diagram for discussion

If I assume that Boltzmann's distribution works for liquids (I get that it's meant for ideal gases) and assume 9 degrees of freedom for water molecules then:

This suggests (if it's even remotely correct) that a very small number of molecules in room temperature water are moving very fast and therefore are at very high temperatures?  It's not a relationship per se but, if correct, it does affirm the initial idea that evaporating water molecules are "hot"...even boiling?
Final Comments?
Thanks for the input and careful consideration.  The chart below is my attempt to summarise my thoughts inspired by your comments.  It's rendered in excel from the equation shown and accords well with the chart for water included in Boltzmann distribution for water which didn't extend far enough on the x axis for what we're trying to show here.  Thanks for the heads up re energy distribution rather than speed - much easier to understand.

The equation comes from BC Campus Molecular Speeds I have ignored degrees of freedom effects in this and the other equations on the chart.
Clearly the energy of evaporating water molecules (let's say liquid immediately before take off), at room temperature (25 deg) is significantly higher than the average energy of boiling water.
What can be said about the little molecules about to liberate themselves from my drying undies then?  The original question precipitated from the idea that molecules evaporating from washing 'boiled'.

They are at similar energy, a little greater in fact, than the energy of molecules about to jump from a pot of boiling water (44kJ vs 41kJ, 7% difference), and massively more energetic than the average energy of boiling water (4.7kJ)
If we could measure the temperature of a bunch of them they would be at an energy equivalent temperature some ten fold greater (3527K/373K) than the average temperature of boiling water (T = 2E/3Nk, bc campus university physics internal energy), but close to the equivalent (theoretical) temperature of the 'boiling' molecules.  I know we can't measure their temperature as it is related to the average energy of the bulk liquid - but theoretically... (there's a case of beer in this)
The proportion of molecules ready to let loose from my y-fronts is much smaller per unit of room temperature water than it would be for boiling water (see the area under the curve right of the 41kJ and 44kJ points)

So...to all intents and purposes, the undie vaporizing molecules are doing what the boiling water vaporizing molecules are doing ....boiling (but they're not at 100 degrees, and we can't really measure their temperature - maybe half a case of beer then?).  We don't feel the heat of the highly excited little blighters because it's a (relatively) small number of individual molecules within a bulk liquid at an average temperature of 25 degrees.
Other Stuff
In coming to the answer from a kinetic energy perspective, I found this paper Velocity of a droplet evaporated from waterwhich measures and models the speed of protonated water molecules evaporated from nano droplets.  Speeds of non-protonated molecules in bulk liquid water will vary but it was encouraging to see that the speeds (say 2000m/s) were of an order of magnitude commensurate with with Boltzmann's distribution for water in Boltzmann distribution for water.  On this curve, the experimental speeds would also be out in the flat area of the curve equivalent to my energy curve above.

https://chemistry.stackexchange.com/questions/88803/kinetic-energy-of-molecules-in-liquid-state

Does this answer your question? Kinetic energy of molecules in liquid state?

You do know the average kinetic energy of a gaseous water molecule if you know the temperature of the gas ...

This Temperature is the average kinetic energy of a "bulk" liquid is false.

Thanks @Alchimista.  Your first reference refers to the average kinetic energy (KE)  I think.  I've added a diagram to the original question.  I think 3/2kT will give me the average KE of the distribution.  My hypothesis is that within the distribution at room temperature the small volume of molecules with sufficient energy to evaporate probably have similar energy to the average kinetic energy of all the molecules in the 100 deg distribution.  So I can use 3/2kT to find the KE of the 100 deg distribution but how do I find the KE of the evaporating particles on the 25 deg curve (coloured)

Ditto reference number two I think

Thanks @Karl.  That's true but it's not the average kinetic energy that i'm interested in here as that would include all the gaseous particles even those that have lost their initial escape velocity and at point of condensation.  I'm interested here in the energy / temperature of the escaping molecules only - just as they leave the surface of room temperature water.

Fair point @Poutnik. As KE = 3/2kT I could more accurately say that temperature is proportional to KE and that the liquid's average temperature is equal to its average KE times some constant.

In the added diagram, if "A" is the KE above which water molecules at the liquid's surface have sufficient energy to evaporate (in a room temperature liquid), how do I find "A"?

@jeff  KE=3/2kT is valid only for  monoatomic gases, and exactly just for ideal ones.

@Jeff I must admit that I was thinking something else, like you asked if the av K in the vapour phase is equal to av K in the liquid. I see you are comparing vapour in two different situation (if they are). I must read it again as well as I must read all the thread.

@Poutnik.  Yes, it's ideal, and perhaps I'm stretching the boundaries of it's usefulness too much here.  Just not sure what else to use.

What do you mean by the average kinetic energy of evaporating water molecules at room T? Those molecules that actually escape the liquid phase(K > A) ? I do not think there is a base for the comparison you want. Even in boiling water there will be a distribution of K, with not yet boiling molecules. But the distribution has changed. If there is a quantitative meaning in your graph, then the av K of the actually boiling molecules is higher. Of all water molecules it is hard to say by eyeballing. But the differences are determined by the different distribution.

At equilibrium, average kinetic energy of evaporating molecules is equal to average kinetic energy of condensing molecules. If not, it would not be possible for vapour and liquid to be in thermal equilibrium. This energy is lower for room temperature, compared e.g. with boiling water temperature.

@Poutnik you should make it an answer

I did upvoted @Poutnik below and even told him to publish. But I got to think that vapour pressure / T dependance prevents that molecules evaporating at T1 have average K equal to that of molecules from a boiling bulk at higher T2. Right because of the equilibrium logic given by Poutnik. What you both, and readers, think?

Molecules need their energy above the threshold limit, related to evaporation enthalpy. Warmer water means greater fraction of these molecules, and also the mean energy of this fraction is higher, leading to higher residual energy after evaporating. So no, water molecules that left 100degC boiling water have higher mean energy than those just evaporated from 20 degC water.

[OP] the average kinetic energy of evaporating water molecules

You have to specify whether you are talking about the kinetic energy just before the water molecule breaks the hydrogen bonds to its neighbors or just afterwards. A millisecond before or after the event, of course, the average kinetic energy will be determined by the bulk temperature.
One way to picture this is two water molecules "colliding" at the surface of the liquid. In order for one water molecule to leave the liquid phase, the collision needs to have at least sufficient energy to break the hydrogen bonds (and perhaps some more to overcome an activation energy). When that happens, the total kinetic energy of the molecules in the collision will be lowered (because energy is conserved). What exactly we mean by "collision" in the liquid phase is not so important because the temperature-dependence of the kinetics of the process is still governed by the Arrhenius equation (and the Boltzmann distribution of collision energies) even when we go from simple mono-atomic gases to wicked-complicated liquids.
The energy required to break the water loose (~40 kJ/mol, according to Poutnik's answer) is much higher than the median collision energy at room temperature (~2 kJ/mol). Whether the water molecule has more or less kinetic energy than the average after it enters the gas phase is not so important. The important part is that it had an unusually high energy, and most of that excess energy went into breaking the hydrogen bonds.

[OP] Assuming we could measure the kinetic energy of multiple evaporating water molecules just as they left the surface of liquid water (at standard temp and pressure) over time, would the average kinetic energy of the sample of evaporating water molecules equal the average kinetic energy of boiling water (at 101.3kpa)?
https://chemistry.stackexchange.com/a/121656/

The kinetic energy of the water molecules that just evaporated is not exceptionally high or low. As the molecule leaves the liquid phase, the available energy is somehow partitioned between that water molecules and the ones it leaves behind. Your question boils down to whether the evaporative cooling of the liquid is temperature-dependent. I'm not sure if there is an experiment that would address this, and whether it has been done already.

[OP] Diagram for discussion

The diagram has a lot of problems, so I would not use it for discussion. The curve does not change that much with temperature, the shaded area is too big by far (according to Poutnik's answer, a molecule on the surface of the liquid has about a one in a million chance per collision to get into the gas phase at 373 K). Here is a diagram showing how the energy available from collisions changes when the temperature is raised from room temperature to body temperature. You can see that the changes for the bulk of energies are subtle. If you zoom in to the regions of exceptionally high collision energies, however, you can see that the changes are substantial.

Thanks.  Good call.  Found and then created similar chart from Boltzmann as shown in edit.  Much appreciated.  Will take the liquid - just before evaporation option.

At gas/liquid phase  equilibrium, average kinetic energy of evaporating molecules, i.e. those just passed to a gas phase, is equal to average kinetic energy of condensing molecules.
The latter is then approximately proportional to $T$.
If these average values were not equal, the system would not be in thermal equilibrium.

Feedback to comments:
@theorist In case the gas phase consists exclusively ( or almost ) from water molecules, then majority of gas molecules collisions with liquid water leads for them to becoming a part of the liquid.
But still, if the mean kinetic energy of condensing molecules differs from the mean kinetic energy of molecules that during the same time interval evaporated liquid, Houston has a problem.
Those collisions that are not part of mass exchange would have to cause nonzero thermal transfer, what is not possible when temperatures of phases is equal.
BTW, I do not think, imply nor say that liquid leaving probability is KE independent. I suppose there is a KE threshold to leave liquid, to overcame the mean bonding energy. OR, perhaps, rather a threshold of the normal component of the velocity vector ( or equivalent energy for 1 degree of freedom ), as the molecule motion has to be properly oriented.

I have created a quick and dirty model to estimate the percentage of vapour molecules that condense after hitting the liquid phase.
I assumed the Boltzman energy distribution for 1 degree of freedom - the normal direction wrt the liquid surface. Then I recalculated the molar water enthalpy of evaporation to 1 water molecule as the energy to overcome when evaporating and calculated the fraction of molecules with enough energy in this degree of freedom.
The water molar enthalpy of evaporation: $\Delta H_{w,mol,evap}=\text{40.7 kJ/mol}$
The Boltzmann statistic:  $\exp{\left(-\frac{\Delta H_{w,mol,evap}}{N_\mathrm{A} \cdot kT}\right)} \approx \exp{\left(-\frac{4895}{T}\right)}$
Than I have calculated saturated vapour pressure and related vapour density from the Clausius-Clapeyron equation.
Then I estimated the condensing fraction as the Boltzmann fraction multiplied by water/vapour density ratio.
But still, it seems to me the system would be just much closer to equilibrium than the prior assuming of all condensing vapour molecules, that is not in equilibrium.

t$\pu{[^{\circ}C]}$
T$\pu{[K]}$
x(Boltzmann)
p$\pu{[Pa]}$
Vap. Density$\pu{[kg/m^3]}$
Rel. density
x condensing

0
273.15
1.65E-08
831
0.0066
6.60E-06
0.0025

10
283.15
3.11E-08
1566
0.0120
1.20E-05
0.0026

20
293.15
5.60E-08
2824
0.0209
2.09E-05
0.0027

30
303.15
9.71E-08
4899
0.0350
3.50E-05
0.0028

40
313.15
1.63E-07
8205
0.0568
5.68E-05
0.0029

50
323.15
2.64E-07
13309
0.0892
8.92E-05
0.0030

60
333.15
4.16E-07
20972
0.1364
1.36E-04
0.0030

70
343.15
6.38E-07
32182
0.2032
2.03E-04
0.0031

80
353.15
9.56E-07
48200
0.2957
2.96E-04
0.0032

90
363.15
1.40E-06
70603
0.4213
4.21E-04
0.0033

100
373.15
2.01E-06
101325
0.5884
5.88E-04
0.0034

Interesting would be the result, if

The Maxwell-Boltzmann statistics would be taken as the energy distribution model for liquid phase molecules
It would be taken its subset, fitting the Boltzmann 1-degree-of-freedom energy threshold criterium
It would be calculated the mean kinetic energy of this subset, then subtracted the heat of evaporation.
Finally, it would be calculated the equivalent temperature.

I don't believe that is the case. Thermal equilibrium can be maintained by the enormous number of ongoing collisions between non-condensing gas molecules (both air and water vapor water) and the surface of the liquid water.  Besides, it seems non-physical that the likelihood of a molecule evaporating is independent of its KE. We know the likelihood of molecules reacting (getting over an activation barrier) depends on KE, because their KE needs to be converted to PE to break bonds.  So why wouldn't this also apply to an evaporating molecule, which needs to break the water-water bonds to escape?

"In case the gas phase consists exclusively ( or almost ) from water molecules, then majority of gas molecules collisions with liquid water leads for them to becoming a part of the liquid." What is your basis for that assertion? Yes, at equil., the rate of evaporation and the rate of condensation are equal. And if the vapor pressure is, say, 0.1 atm, and there are no other gases, the concentration of gas molecules at the liquid-vapor boundary will only be ~1/10,000 that of the liquid molecules. But that still doesn't mean that most of the gas-liquid collisions have to lead to condensation.....

....For instance, even with the enormous difference in density, it could still be that the rate of evaporation from the liquid phase is only 1/100 the rate at which gas molecules strike the liquid phase, in which case only 1% of the collisions of gas molecules with the liquid phase would lead to condensation.  The remaining 99% of collisions could maintain the thermal equilibrium.

@theorist About this 1%, you seem to be actually very close to real thing, see the incoming answer update after a while. But it would IMHO just mean the state would be 100 times closer to equilibrium, not that it would be in equilibrium. (BTW, I hope you have not taken the prior Vdp + pdV thing personally.)

Your "interesting results" comments @Poutnik, was where I was headed too....see edit

@Jeff Are you sure you do not confuse Boltzman and Maxwell-Boltzmann distributions ?. The former is general one, not limited to gases nor motion degrees of freedom. It applies also e.g. to   bound electron energy levels.

Maybe.  Thanks @Poutnik.  I wrongly abbreviate M-B as Boltzmann.

@Poutnik  Actually, I didn't couldn't find your responses to my last comment on our Vpd+pdV conversation, since I couldn't figure out a way to look up that Discussion (I haven't figured out how to search this site specifically for Discussions).  So, anyways, no, we're fine.

@Poutnik   As to the current issue:  There's actually a term for what we're discussing here (the % of vapor molecules striking the liquid surface that condense).  It's called the "condensation ratio".  Langmuir thought its value was $1$.  However, subsequent experiments and computer simulations put the condensation ratio for water between $\approx 1 \text{ and} \approx 10^{-3}$.   And it remains unresolved.  Here's a review article about it, which notes how surprising it is that we still don't know the condensation ratio for water:  https://link.springer.com/chapter/10.1007/978-3-319-96304-4_7

This is a very interesting question!
Also, I don't believe it is a duplicate of the other cited questions, since those seem to concern the average kinetic energy of liquid vs. gaseous water, while the OP is asking whether those that evaporate from the liquid are at the tail of the distribution, and whether there's a relationship between the KE of this subpopulation and the average KE of those in boiling water.
I'll give a somewhat speculative answer, which might serve as a springboard for others to contribute more.
We know that, for chemical reactions, the subpopulation that reacts comes from those with sufficiently high energy (which can be both kinetic and potential) to get over the activation barrier.
Intuitively, one would expect that to be the same for evaporation—it's those molecules that are at the surface, and have sufficiently high KE, that evaporate.  And the reason the evaporation rate increases with T is that a greater proportion of molecules meet that criterion.
So what you're asking is whether there is some relationship between the cutoff for sufficient KE to evaporate, and the KE needed for boiling.
I suspect the answer is no, since the boiling temp changes with ambient pressure, yet I don't believe the ambient pressure significantly affects the KE needed to evaporate (it has little effect on the chemical potential of the liquid water).
For instance, the boiling point of water at 1 atm is 100 C, while at 0.1 atm it is only 41 C.  If the boiling point determined the KE needed to escape from the liquid to to the gas, then the KE needed at 100 C would be much greater than that needed at 41 C.
However, I don't believe the KE needed to escape the liquid would be significantly affected by ambient pressure because, as I mentioned above, there is little difference between the chemical potential of liquid water under 1 atm pressure and 0.1 atm pressure.
Rather, it appears that, to the extent that the ambient pressure does affect the evaporation rate, it's because reduced ambient pressure reduces the density of the ambient gas and thus increases the diffusion rate, and hence the evaporation rate.  See:
Sefiane, K., et al. "On the effect of the atmosphere on the evaporation of sessile droplets of water." Physics of Fluids 21.6 (2009): 062101.
https://aip.scitation.org/doi/10.1063/1.3131062

Thanks @theorist. Precisely; is a "relationship between the cutoff for sufficient KE to evaporate, and the KE needed for boiling."  I will need to consider your in depth response further.  Off the cuff I would have thought a lower air pressure would reduce the energy required for particles to evaporate at room temperature.  (KE = 3/2kT, PV = nRT, so less P = less T = reduced kinetic energy).  Or, less weight of air pushing on the liquid surface means less energy is required for particles to escape.  Not sure, will need to cogitate on your thoughts further.

@Jeff  Less p does not mean less T here.  T can be whatever you designate it to be.  There are other variables in pV = nRT.    Think about what you're comparing when you want to understand the effect of changing ambient pressure:  You're comparing the rate of water evaportating at some T (say 25 C) under 1 atm pressure, with the rate of water evaporating at the same T at a lower p.   So you already know that, in this comparison, when you are changing p, you are not changing T.

Pressure
Evaporation happens faster if there is less exertion on the surface keeping the molecules from launching themselves.  (Wiki).  I was referring to your earlier comment that pressure may not affect the KE required to evaporate.  I'm assuming that this Wiki comment infers that less pressure on the surface means less energy is required to escape.

@Jeff (a) I don't know what "pressure evaporation" is supposed to mean.  (b)  "Wiki" is not a reference.  You need to give the actual link.   Wikipedia could be talkng about pressure and evaporation in any of numerous diff. articles, so there's no way to tell which article you saw it in without a specific link.   In the article I reference, the authors explain that ambient pressure from inert gases reduces the evaporation rate by reducing the diffusion rate, not by changing the KE needed to evaporate.   It is true that high p can reduce the energy needed to evap., but it's a tiny effect.

Thanks @Theorist.  It was Pressure (heading).  Evaporation happens......Link :  https://en.wikipedia.org/wiki/Evaporation.  Apologies, just getting used to this site.

I found it:  "Evaporation happens faster if there is less exertion on the surface keeping the molecules from launching themselves."  I could try to parse what that means, but it's so badly-writtten I don't want to try.   That sentence needs to be fixed by the Wikpedia editors. Indeed, that entire article reads like it was written by a student who somewhat, but doesn't completely, understand the subject. Also, If you quote something verbatim, you should put it in quotes.   [That's not about this site, it's the general proper practice.]

How about looking at the partial pressures of two gases in a perfectly insulated pair of insulated bottles connected by a Maxwell's demon opening that can tell us which kind of gas molecule went through the opening between the bottles and which way the molecule was moving. Let us say the bottles are exactly the same size and there are in total 10 molecules of gas C and 10 molecules of gas H. In bottle A let us say there are x molecules of gas C and y  molecules of gas H. In bottle B let us say there are 10-x molecules of gas C and 10-y molecules of gas H.
Let us start with x=6, bottle A has 6x bottle B has 4x. For the flow through the Maxwell gate to be 0, the count of molecules on each side must be equal. Bottle A must have x=6 and y=4 while bottle B must have x=4 and y=6. Bottle A has 10 and Bottle B has 10 molecules each. In this very simple system, the temperature is the same in each bottle. If you increase the temperature of one bottle(a very small amount), warmer gas molecule will flow to the other bottle until the temperature on each side is again the same.
Referring back to " do my drying undies effectively boil?" I think this model indicates whatever temperature they started at, they will converge on the same partial pressures as the surrounding environment.

Thanks Lee, a useful conceptualisation.  I think they will too - as you've illustrated.  Before that time, just prior to launching from the undies, what are the molecules doing?  Have made some summaries in the edited question.  Thanks for your help.

As the water body is colder, evaporating water obviously has a lower speed of gas particles as they leave the surface. If that were otherwise, evaporating water would cool tremendously, and at the same time heat the air above.
While it certainly does the former (minus the "tremendously"), the latter for sure does not happen.
The quantitative answer is that any water molecule that leaves the surface has just lost the evaporation enthalpy (strictly speaking its binding energy to the molecules around and below it). If it has any kinetic energy left, it can fly off.

Thanks @Karl.  Help me understand how it's obvious that the molecules have lower speeds.  According to Maxwell-Boltzmann, water, even at room temperature can have molecules travelling at very high speeds - with very high kinetic energies, even though the average (bulk liquid) kinetic energy is relatively low.  You've nailed exactly the problem at hand I think - how to determine the speed and therefore kinetic energy of evaporating particles so it can be compared to the average kinetic energy of boiling water. The quantitative answer requires maths and I don't know what math to use.

@Jeff Note that the M-B distribution is applicable to gases only. For liquids, it is just a very rough approximation.